eee3055f_notes.2012_V36

UNIVERSITY OF CAPE TOWN

DEPARTMENT OF ELECTRICAL ENGINEERING

EEE3055F

2012

Module A: Electromagnetic Field Theory

Lecture Notes: Version 2.33 2012

Lecturer: Assoc. Prof. A.J.Wilkinson

Room: 6.09.1

Email: [email protected]

30th May 2012

• Feel free to send me an email if you have questions regarding the course,or would like to arrange a time to see me. This course module has a

web page where marks and notices will be posted (see courses link onwww.ee.uct.ac.za)

• This part of EEE3055F consists of 24 lectures (two per week). Check

web site and notice board on 4th floor Menzies for venues.

• Final mark for module A = exam (76%) + test1 (10%) + test2(10%)

+ lab (4%) + assignments (6%).

• Two class tests, provisional dates to be announced.

• There will be one laboratory practical on microwave antenna measure-ments. You will be required to submit a short report that will count

4% of your total mark.

• Computer simulation assignments will be set that will contribute up to

6% of the final mark.

• The examination counts 70% of the mark awarded for this module.

• There is a sub-minimum of 35% required for each module to pass the

course.

• To qualify for a Duly Performed certificate, all assignments and lab

reports must be submitted by the due date. All submissions mustinclude a signed declaration, declaring that the work is one’s own, and

has not been plagiarised.

Course notes will be handed out in class, supplemented by material on theblackboard. There is no prescribed text, but the following books may be

consulted for supplementary reading:

• Fundamentals of Physics, D. Halliday & R. Resnick

• Introduction to Electrodynamics, D.J.Griffiths

• Fields and Waves in Communication Electronics,

S. Ramo, J.R. Whinnery and T. Van Duzer.

2 AJW, EEE3055F, UCT 2012

• Electromagnetism, I.S. Grant & W.R.Phillips

• Electromagnetic Wave Propagation & Antennas, S. Cloude

• Antenna Theory, C.A. Balanis

Lecture Topics

Lectures will be given on aspects of the following topics:

1. Review of basic static field theory: electrostatics; Gauss’ law; magne-

tostatics: Gauss’ law for magnetic fields; Ampere’s (Oersted’s) law

2. Dynamic laws: Faraday’s law of electromagnetic induction

3. Maxwell’s equations & displacement current

4. Electromagnetic boundary conditions

5. Relationship to circuit theory an Kirchhoff’s laws

6. Radiation and electromagnetic Waves; the radiation mechanism; thewave equation and solutions

7. Sinusoidal EM waves

8. Plane waves in (i) free space (ii) non conducting dielectrics; polarization

9. Simulation of propagating waves using FDTD method

10. Power flow in electromagnetic waves; electromagnetic safety consider-

ations

11. Reflection and refraction at boundaries

12. Propagation in conducting media and the skin effect

13. Radiation from antennas: Hertzian dipole, wire antennas

14. Thermal Radiation from warm objects


Laboratory session

The laboratory session is on radiation from microwave horn antennas. It

covers:

• experimentation with 10 GHz horn antennas.

• exposure to some microwave components and instrumentation.

• measurement of received power as a function of distance.

• linear polarization and its effect on the orientation of the antenna.

• measurement of the antenna beam patterns using a rotating turntable.

A formal report must be submitted, documenting the experiments and re-

sults obtained.The laboratory venue is the Microwave Laboratory on the 7th floor of the

Menzies Building.


Contents

1 Introduction and course objectives 1-11.1 Course aims . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-1

1.2 Course outcomes . . . . . . . . . . . . . . . . . . . . . . . . 1-11.3 Examples of the applications of EM field and wave theory: . 1-11.4 Prerequisite Knowledge . . . . . . . . . . . . . . . . . . . . . 1-2

1.5 List of Symbols and Units . . . . . . . . . . . . . . . . . . . 1-2

2 The laws of Electromagnetics 2-1

2.1 Forces between charges (Coulomb’s Law) . . . . . . . . . . . 2-12.2 The Electric Field . . . . . . . . . . . . . . . . . . . . . . . . 2-2

2.3 Forces on charges from Magnetic Fields . . . . . . . . . . . . 2-32.4 Lorentz Force Law . . . . . . . . . . . . . . . . . . . . . . . 2-5

2.5 Biot-Savart Law . . . . . . . . . . . . . . . . . . . . . . . . . 2-52.6 Work and Potential . . . . . . . . . . . . . . . . . . . . . . . 2-72.7 Magnetic Vector Potential . . . . . . . . . . . . . . . . . . . 2-9

2.8 Energy stored in an Electrostatic Field . . . . . . . . . . . . 2-102.9 Energy stored in a Magnetostatic Field . . . . . . . . . . . . 2-11

2.10 Other useful Vector Quantities . . . . . . . . . . . . . . . . . 2-122.11 Fields in Linear Isotropic Media . . . . . . . . . . . . . . . . 2-12

2.12 Electric Fields in Dielectric Media . . . . . . . . . . . . . . . 2-122.13 Magnetic fields in Materials . . . . . . . . . . . . . . . . . . 2-14

2.14 Maxwell’s Equations governing Electromagnetic Behaviour . 2-182.15 Gauss’ Law for Electric Fields . . . . . . . . . . . . . . . . . 2-182.16 Gauss’ Law for Magnetic Fields . . . . . . . . . . . . . . . . 2-23

2.17 Faraday’s Law of Electromagnetic Induction . . . . . . . . . 2-242.18 Ampere’s Circuital Law . . . . . . . . . . . . . . . . . . . . 2-29

2.19 Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . 2-32

5

3 The Differential Forms of Maxwell’s Equations 3-13.1 The Static EM Equations in Integral and Differential Form . 3-3

3.2 Electromagnetic Equations before Maxwell . . . . . . . . . . 3-103.3 Continuity of Charge . . . . . . . . . . . . . . . . . . . . . . 3-113.4 Fixing the Problem with Ampere’s Law . . . . . . . . . . . . 3-13

3.5 Maxwell’s Differential Equations . . . . . . . . . . . . . . . . 3-143.6 Summary Table of Maxwell’s Equations . . . . . . . . . . . . 3-15

4 Electromagnetic Boundary Conditions 4-14.1 Boundary Conditions for the Electric Field . . . . . . . . . . 4-1

4.2 Boundary Conditions for the Magnetic Field . . . . . . . . . 4-54.3 Examples of Boundary Conditions . . . . . . . . . . . . . . . 4-9

5 Relationship between Field Theory and Circuit Theory 5-15.1 Resistors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-25.2 Capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-5

5.3 Inductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-65.4 Formulas for Practical Coils . . . . . . . . . . . . . . . . . . 5-11

5.5 Mutual Inductance . . . . . . . . . . . . . . . . . . . . . . . 5-155.6 Kirchhoff’s Voltage Law . . . . . . . . . . . . . . . . . . . . 5-18

5.7 Kirchhoff’s Current Law at a Node . . . . . . . . . . . . . . 5-225.8 The Relaxation Time of Conducting Materials . . . . . . . . 5-26

5.9 Shielding and The Faraday Cage . . . . . . . . . . . . . . . 5-295.10 Twisted Pair Cables . . . . . . . . . . . . . . . . . . . . . . 5-29

6 Electromagnetic Waves 6-1

6.1 Mathematical Description of Travelling Waves . . . . . . . . 6-26.2 Wave Equation derived from Maxwell’s Equations . . . . . . 6-6

6.3 Physical Interpretation of the Radiation Generation and Prop-agation Mechanism . . . . . . . . . . . . . . . . . . . . . . . 6-9

6.4 Some Additional Notes on Wave Equations . . . . . . . . . . 6-126.5 Travelling Waves in a Plucked Guitar String . . . . . . . . . 6-14

7 Sinusoidal Waves 7-17.1 Signals as Sums of Sinusoidal Functions . . . . . . . . . . . . 7-1


7.2 Real Sinusoidal Travelling Waves . . . . . . . . . . . . . . . 7-27.3 Complex Phasor Representation . . . . . . . . . . . . . . . . 7-4

8 Plane Waves 8-18.1 Plane wave propagating in z direction . . . . . . . . . . . . . 8-28.2 Characteristic Impedance . . . . . . . . . . . . . . . . . . . . 8-4

8.3 Sinusoidal Representations . . . . . . . . . . . . . . . . . . . 8-58.4 Plane Wave Propagating in an Arbitrary Direction . . . . . 8-7

8.5 Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . 8-9

9 Simulating Electromagnetic Waves using the Finite Difference

Time Domain (FDTD) Method [not covered in 2009] 9-19.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 9-1

9.2 FDTD Solution to Maxwell’s equations in 1-D . . . . . . . . 9-39.3 FDTD Solution to Maxwell’s equations in 2-D Space . . . . 9-159.4 Solving Maxwell’s Equations in 3D . . . . . . . . . . . . . . 9-22

9.5 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9-229.6 MATLAB CODE . . . . . . . . . . . . . . . . . . . . . . . . 9-23

10 Power Considerations and the Poynting Vector 10-110.1 Power in a Sinusoidal Plane Wave . . . . . . . . . . . . . . . 10-3

10.2 Power density from Radiating Antennas . . . . . . . . . . . 10-510.3 Power Flow in a Simple Circuit . . . . . . . . . . . . . . . . 10-9

10.4 Electromagnetic safety considerations . . . . . . . . . . . . . 10-14

11 Reflection and Refraction at Boundaries 11-1

11.1 Reflection of Normally Incident Plane Waves from a PerfectConductor . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11-1

11.2 Transmission Line Analogy . . . . . . . . . . . . . . . . . . . 11-4

11.3 Normally Incident Plane Wave on a Dielectric Interface . . . 11-511.4 Reflection and Transmission (refraction) at a dielectric inter-

face - arbitrary incident angle . . . . . . . . . . . . . . . . . 11-6

12 Propagation in Conducting Media and the Skin Depth 12-1

12.1 Wave propagation in a conducting medium . . . . . . . . . . 12-1


12.2 Skin Depth . . . . . . . . . . . . . . . . . . . . . . . . . . . 12-512.3 Skin Effect in Conducting Wires . . . . . . . . . . . . . . . . 12-7

13 Radiation from Antennas 13-113.1 Hertzian Dipole . . . . . . . . . . . . . . . . . . . . . . . . 13-113.2 Dipole Antennas . . . . . . . . . . . . . . . . . . . . . . . . 13-1

13.3 Phase array pattern . . . . . . . . . . . . . . . . . . . . . . 13-113.4 Aperture Antennas [simplified treatment; lab session] . . . . 13-1

13.5 Discussion on why circuits that are small compared to thewavelength do not radiate . . . . . . . . . . . . . . . . . . . 13-1

14 Thermal Radiation from Warm Objects 14-1


1 Introduction and course objectives

1.1 Course aims

This course aims to provide students with an understanding of electromag-netic field theory and wave propagation in the context of applications in

electrical engineering.

1.2 Course outcomes

• Knowledge and understanding of Maxwell’s equations in integral and

differential form.

• Understanding of the relationship between circuit theory and field the-

ory.

• Understanding of electromagnetic wave phenomena: radiation, plane

wave propagation, reflection and refraction.

• Exposure to tools and methods for simulating wave propagation

• Basic antenna concepts and understanding radiation from wire anten-

nas.

1.3 Examples of the applications of EM field and wave

theory:

• Understanding propagation of electromagnetic waves, and their inter-

action with matter.

• Design of structures for efficient radiation of EM power in the form of

antennas for communication links and radar applications.

1-1

• Predicting the behaviour of electronic circuits, particularly at higherfrequencies, where the dimensions of components approach the wave-

length.

• Solving field problems, e.g. magnetic fields for the analysis of motorsand generators

• In electrical impedance tomography, an image of object is reconstructedfrom impedance measurements. These are obtained by injecting elec-

trical currents and measuring voltages around the boundary; the re-construction algorithms require modelling the electric fields within the

medium under investigation.

1.4 Prerequisite Knowledge

• Previous course covering electrostatics, magnetostatics.

• Vector calculus to describe such fields, e.g. concepts such as div, curl,Gauss’ theorem, Stokes’ theorem etc.

1.5 List of Symbols and Units

A - vector potential

B - the magnetic flux density vector in Wb m−2 or Tesla T

Bt, Bn - tangential and normal B-field components at a surface

c - speed of propagation in ms−1

C - used to label an integration contour

dl - vector used to define elemental section of a contour for contour integra-

tion

dS - vector dS ≡ ndS used to define a surface element for surface integra-tion

dV - volume element for volume integration

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D - the electric flux density vector in Cm−2

E - electric field strength in in NC−1 or Vm−1

Et, En - tangential and normal E-field components at a surface

E - the electric field intensity vector in NC−1 or Vm−1

F - force vector in N

H - the magnetic field intensity vector in Am−1

J - current density vector in Am−2

k - electric field constant k = 1/(4π/ε0)

k - wavenumber in rad m−1

n - unit vector normal to a surface

P - Poynting vector in Wm−2

S - used to label a surface for integration

U - energy in joules

UE energy in electric field in J

UM energy in magnetic field in J

I - current in A or Cs−1

I - used occasionally to denote power density in Wm−2

L - inductance in H.

M - mutual inductance in H.

t - time in s

f - frequency in Hz

G - antenna gain

r - vector denoting position in space

1-3 AJW, EEE3055F, UCT 2012

q, Q - electric charge in Coulombs C

T - period of a wave in s

V - voltage V or JC−1

V - used to label a volume for volume integration

v - velocity vector of a particle in ms−1

W - work in J

ρ - charge density in Cm−3

ρs - surface charge density in Cm−2

ρ - reflection coefficient

σ - conductivity in Sm−1

ε - permittivity constant. In free space ε0 = 8.854 × 10−12 Fm−1 (or

C2N−1m−2)

µ - permeability; in free space µ0 = 4π × 10−7 in NA−2

η - impedance of a medium of propagation in Ω

λ - wavelength in m

Φ - potential difference or potential relative to zero reference in V or JC−1

ΨM - magnetic flux in webers Wb

ΨE - electric flux in coulombs C

ω - frequency in rad s−1

1-4 AJW, EEE3055F, UCT 2012

2 The laws of Electromagnetics

This section contains a review of the relationships covered in the static fields

course PHY210S. This leads on to the incorporation of the dynamic fieldlaws, including Faraday’s law and the modified form of Ampere’s law. These

laws are summarized in an elegant set of equations known as Maxwell’sequations.

The fundamental electromagnetic (EM) field quantities are

E the electric field intensity vector in NC−1 or Vm−1

B the magnetic flux density vector in Wb m−2 or Tesla T

• These vector field quantities have been introduced to explain and pre-

dict the observed effects that charged particles have on one another instatic and dynamic situations.

• EM fields are not directly visible to the eye; but their effects are.

• Our eyes are of course sensitive to EM radiation in the optical band,although we are not observing the field quantities E and B themselves.

• Although field quantities are, in general, functions of position (x, y, z)

and time t, e.g. E ≡ ~E(x, y, z, t), for compactness, the arguments willusually be omitted in the written notation. Bold symbols will be used

for vector quantities.

2.1 Forces between charges (Coulomb’s Law)

Two charged particles q1 and q2, separated by a distance r in free space(vacuum), experience a force of either attraction or repulsion modelled by

F = kq1q2R2

2-1

where k = 14πε0

is a constant. The permittivity constant ε0 = 8.854× 10−12

Fm−1

Force onq2

q2

q1

F = q1q24πε0r2

R

F

R

R

Figure 2.1: Force on charge q2 owing to q1.

• Note that the force is proportional to the charge values and inversely

proportional to r2.

• The effect of such forces can be observed and measured.

• We consider the force on q2 to be a result of placing q2 in the electric

field surrounding q1.

If we place the points within a coordinate system at locations ~P1 and ~P2,the force on charge q2 can be expressed in vector form as

F =q1q2

4πε0R2R

where R = R|R| is a unit vector in the direction of the vector R = −~P1 + ~P2,

pointing from q1 to q2.

2.2 The Electric Field

The electric field arising from a charge q is defined in terms of the force itexerts on a test charge ∆q

E =F

∆q=

q

4πε0r2R

in units of NC−1.

• The test charge ∆q would generate its own field, which acts on q.

2-2 AJW, EEE3055F, UCT 2012

• Superposition holds and so the total field at any point is the sum ofthe components from all charges, i.e.

E =N∑

i=1

Ei

where Ei results from each of the N charges as illustrated in the fol-

lowing diagram.

Total Electric Field

q3

P

E3

E2

E1

q1

q2

E =∑N

i=1Ei

Figure 2.2: Electric field at point P is the vector sum of components all sources.

2.3 Forces on charges from Magnetic Fields

Magnetic fields were originally introduced to explain the effects of forces

between magnetized materials, and also current carrying conductors. Mag-netic fields are created by moving charges. The magnetic fields produced by

permanent magnets can be explained in terms of the sum of fields arisingfrom the electrons orbiting atoms.Recall, for a long straight wire conductor carrying current I, the sur-

rounding field at a radial distance r from the wire is given by B = µ0I2πr . To

determine the direction of the magnetic field:

2-3 AJW, EEE3055F, UCT 2012

into page

with thumb pointing in direction of current.

Fingers point in direction of magnetic field

r

Grasp the conductor with right hand

I

BB

I

Figure 2.3: Magnetic field surrounding a long straight current-carrying wire.

A charge moving through a magnetic field experiences a force, describedby the vector relationship

F = qv ×B

where v is the velocity vector of the particle. The directions of the field

quantities are depicted below (for a positive charge):

(into page)

(down)

on moving chargeMagnetic force

F = qV ×B

B

V

Figure 2.4: Force on a charged particle moving through a magnetic field.

The units of B are deduced from the force per coulomb per unit of speed,i.e. (NC−1)/(ms−1) = NC−1m−1s. For convenience the equivalent unit of

“tesla” T is commonly used (1 T = 1 NC−1m−1s).

2-4 AJW, EEE3055F, UCT 2012

2.4 Lorentz Force Law

The total electromagnetic force on a charged particle is the sum of forces

resulting from the electric and magnetic fields and is known as the Lorentzlaw

F = qE+ qv ×B

A conductor carrying a current experiences both electric and magnetic

forces from the EM fields in its environment. A stationary charged particlecan experience only an electric force, since the term qv ×B = 0 if v = 0.

2.5 Biot-Savart Law

The Biot-Savart law (1820) relates the current flowing in a wire to themagnetic field resulting from it. Consider a linear homogeneous medium inwhich a wire carries current.

R

B

φ

P

(points out of page)

Ir

r′

dl′

Magnetic Field

Figure 2.5: Magnetic field at point P owing to current I.

The contribution to the magnetic field at any point P resulting from an

elemental section of the wire dl′ carrying current I in the direction described

2-5 AJW, EEE3055F, UCT 2012

by the vector quantity dl′ is found experimentally to be

dB(r) =µIdl′ sinΦ

4πR2

where Φ is the angle between vector R and dl′, and µ is the permeability ofthe medium. The magnetic field at P points out of the page. The field can

be expressed in vector form as

dB(r) =µIdl′ × R

4πR2

where R = (r−r′)|r−r′| =

R|R| is a unit vector, given in terms of r being the position

of point P , and r′ being the position of the current element. The expressionis sometimes written as

dB(r) =µIdl′ × (r− r′)

4π |r− r′|3

To calculate the total magnetic field resulting from a section (or entire

circuit loop), we must add contributions from all elemental sections by in-tegrating along the length of the wire, usually a closed loop

B =

∮dB =

µI

4π

∮dl′ × R

R2

Homework

Use the Biot-Savart law to show by integration, that the field arising from aninfinitely long straight conductor carrying current I is BΦ = µI

2πR where R is

the perpendicular distance from the point of consideration to the conductor.

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2.6 Work and Potential

The energy required to move a test chargeq from point A to point B along

a contour C is given by the contour integral

W = − lim∆l→0

∑

i

(Fi cos θi)∆l

= −∫ B

A

F · dl = −∫ B

A

(qE+ qv ×B) · dl = −∫ B

A

qE · dl

A

Bcontour C

dlθF

Figure 2.6: Contour C joining points A and B.

• The dot product F · dl = F∆l cos θ extracts the component of F tan-gential to the contour C (in the direction of dl) and multiplies it by

the incremental distance ∆l, to yield the incremental amount of workdW = −F · dl.

• It is interesting to see that no work is done against the magnetic field

because the magnetic force component of F is always perpendicular todl, i.e. qv×B ⊥ dl, and hence the dot product is zero (note that F⊥v

and v‖dl hence F⊥dl). Work is only done against the electric field.

In electrostatics, it is often convenient to introduce the concept of ‘poten-

tial’. The potential difference between two points A and B is defined as thework per Coulomb involved in moving a test charge q from A to B

ΦB−A =W

q= −

∫ B

A

E · dl

• The units of potential difference are JC−1 or volts V.

• For static fields, this integral is path independent, i.e. electrostatic

fields are conservative.

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• It is often convenient to define a reference point (e.g. in circuit analysis)as the ‘zero potential’, and specify potentials relative to that point.

• For analysing problems involving point charges, the reference point issometimes set “at infinity” and thus the potential at any point P isgiven by

ΦP = −∫ P

∞E · dl

It can be shown (in many standard textbooks) that the potential at a

point P resulting from N point charges is given by

ΦP = −∫ P

∞E · dl =

N∑

i=1

qi4πε0ri

where ri is the distance from charge qi to point P.

• The electric field can be obtained from the gradient of an electrostaticfield,

E = −

∂Φ∂x∂Φ∂y∂Φ∂z

= −gradΦ = −∇Φ

This is a useful relationship, as the approach to finding the electric fieldis often approached by first finding the (scalar) potential distribution1,

followed by calculation of the electric field from the gradient of thepotential function.

1

– More generally, for solving dynamic fields problems, potential functions may also be defined. Theseare known as “retarded potentials” and incorporate the wave propagating nature of electromagneticfields. Retarded potentials are functions of both position and time. It is often easier to solve aproblem (either analytically or numerically) by first solving for the potential functions, and thensubsequently calculating E and B field quantities. For example, the electric field is found from

E = −∇Φ− ∂A

∂t

where Φ(r, t) is the retarded electric potential function and A(r, t) is the retarded magnetic vectorpotential.

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2.7 Magnetic Vector Potential

In the same way that a scalar potential function is useful in solving electric

field problems, it is possible to define a special potential for solving magneticfield problems.

Since∇·B = 0, and∇·(∇×F) = 0 ∀ F, it follows thatB can be expressedas the curl of another vector, A, i.e.

B = ∇×A

As this is analogous to E = −gradΦ, the vectorA is known as the“magnetic

vector potential”.It is interesting to note that the flux through an arbitrary surface S is

related to A by a line integral around the perimeter:

ΨM =

∫

S

B · ds =∫

S

∇×A · ds =∮

A · dl (by Stokes’ theorem)

Although we shall not analyse this here, the magnetic vector potentialA can be determined at any point in space r by integrating contributions

from all currents or current densities that exist, i.e. for the case of a currentcarrying conductor (see Figure 2.5),

A(r) =

∫µI(r′)dl′

4πR

where dl′ is the elemental section of wire at position r′ on the contour, andR = |r− r′|.For a distributed current density within volume V ,

A(r) =

∫

V

µJ(r′)dV ′

4πR

Once A(r) has been obtained, the B field is calculated from:

B = ∇×A

The introduction of the magnetic vector potential is useful for solving bothstatic B field problems as well as for finding the fields surrounding an an-

tenna.

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2.8 Energy stored in an Electrostatic Field

q3

q2

P2

P3

P1q1

Figure 2.7: Collection of charges.

It can be shown [Griffiths and other texts] that the work required to assemble

a configuration of point charges by moving charges from infinitely far apartinto fixed position, is given by

W =1

2

N∑

i=1

Φiqi

where Φi is the potential at point Pi resulting from all other charges, exclud-ing the i′th charge itself2. More generally, to assemble a continuous chargedistribution ρ(x, y, z) with corresponding potential field Φ(x, y, z), is found

by integrating elemental volume contributions

W =1

2

∫

all space

ΦρdV

An important result is that this expression can be re-written purely in termsof the electric field [Griffiths] to yield

W =1

2ε0

∫

all space

E2dV

2Note: Φiqi is the work done to bring qi from far away into position Pi against the force from all othercharges, i.e. Φiqi counts work against qj (j 6= i), but Φjqj counts work against qi (j 6= i). Thus the

quantity∑N

i=1 Φiqi counts the work in positioning qi against qj twice (for each pair). Hence we must

divide by two, giving W = 12

∑Ni=1 Φiqi.

2-10 AJW, EEE3055F, UCT 2012

The term 12ε0E

2 has units of Jm−3 and can be thought of as the energydensity in the electrostatic field. This will have some significance in our

later discussion on radiating electromagnetic fields.

2.9 Energy stored in a Magnetostatic Field

To set up a current in a conducting wire coil requires work; the energyinvolved can be recovered. The figure below shows an inductor carrying a

steady state current of I amperes. If the switch is opened, the inductor willforce the current through the resistor markedR, with a 1st order exponential

decay. The energy stored in the inductor is dissipated in the resistor.

Open

RLI

Figure 2.8: Energy is stored in the magnetic field of the inductor.

It can be shown [Griffiths and other texts], that the work required to

establish the current is W = 12LI

2, which in turn, can be expressed purelyin terms of the surrounding magnetic field arising from the current as

W =1

2µ0

∫

all space

B2dV

and thus we can consider the term 12µ0B2 as the energy density in Jm−3 and

can be thought of as the energy density in the magnetic field. This will have

some significance in our later discussion on radiating electromagnetic fields.

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2.10 Other useful Vector Quantities

For convenience, two additional EM field quantities are introduced:

D = εE is called the electric flux density in Cm−2. It is also known as thedisplacement vector.

H = Bµis called the magnetic field intensity in Am−1

These quantities are useful for

• writing compact expressions of the laws of electromagnetics

• handling the case of fields in materials

• aiding in conceptualising certain physical relationships e.g. Gauss’ law

in for electric fields; Ampere’s circuital law.

The D vector has a close analogy to the concept of fluid flow (hence thename“flux density”. The word“flux”literally means the“rate of flow”(whichwould be in units of litres per second in the case of fluid flow).

2.11 Fields in Linear Isotropic Media

In the aforementioned equations, the effects of the medium in which a field

exists can be conveniently handled by introducing the concepts of permit-tivity ε and permeability µ of a medium.For the case of a linear isotropic medium, the parameters are related to

the free space values by constants, i.e.ε = εrε0 where εr ≥ 1 is the relative permittivity

and similarly µ = µrµ0 where µr is the relative permeability.

2.12 Electric Fields in Dielectric Media

When a dielectric (non-conductor) is subjected to an electric field, the

molecules deform and ‘polarize’ forming tiny dipoles. The electric fieldfrom the induced molecular dipoles is in the opposite direction to the ap-

plied electric field. The net effect of the polarization of the material is to

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reduce the net electric field within the material. The polarization effect alsochanges the direction of an electric field vector at the interface between two

different dielectric media (e.g. air and glass). This is studied later in thechapter on boundary conditions.

air

glass

air

z

E(z)

z

D(z)

Eapplied

zEinduced

Figure 2.9: Polarization of molecules caused by an applied electric field.

The E field (normal component) discontinuity arises from the polarizationeffect in the dielectric which results in an induced layer of bound charge at

the interface between dielectrics. In the illustration, it can be seen that theslight polarization of the molecules causes a thin positive layer on the upper

surface of the glass, and a thin negative layer on the bottom surface. Anelectric field is induced within the material that is opposite to the applied

field. The total field within the dielectric is the vector sum of the two,and is hence reduced. A plot of Ez(z) will show a step change at theinterfaces between glass and air. In a linear medium, Einduced ∝ Eapplied.

i.e. E = Eapplied − kEapplied = (1 − k)Eapplied =1εEapplied. In this situation,

the quantity D = εE with be the same inside and outside of the medium -

hence the D vector is continuous (at least the normal component is) acrossa boundary, analogous to fluid flowing.

The table below lists values of εr for several materials.

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Medium Dielectric constant εrVacuum 1

Gases He,Ne,H,Ar,Ni <1.0007Air (sea level) 1.00059

Air (pressurized 100 atm) 1.05Water vapour 1.01Polystyrene 2.5

Paper 3.5Glass 4 - 7

Methyl Alcohol 35Water 80.4

Table 2.1: Some permittivity values

2.13 Magnetic fields in Materials

For nearly all materials the permeability is very close to that of free space,i.e. for practical problems, µ ≈ µ0, i.e. µr ≈ 1.00. Iron (Fe) is one notable

exception, for which µr is in the thousands or tens of thousands, dependingon the purity of the iron. Coils are wound on iron cores to increase the

magnetic flux density. It should also be noted that behaviour of iron isnonlinear, and described by the “BH” curve.

Magnetic Domains

Insertion of a material into a magnetic field will modify the field. For

example, an electromagnet can be made by winding a coil around an ironbar; the net effect of the presence of the iron bar is to boost the magnetic B

field strength, by a factor as much as 1000. To understand this phenomenon,one must examine the microscopic structure of the material. The image in

Figure 2.10 [Ramo et al. 3rd ed. p 692] is a highly magnified picture of anickel crystal. The boundaries (revealed by an etching technique) shown arelocal magnetic domains (typically 10−10 to 10−12 m3 in volume; or about 0.1-

0.5 mm in diameter) that have a net magnetic field. Under unmagnetizedconditions, the domains point in random directions. If an external magnetic

field is applied and increased with time, initially, the domains with directionsagreeing with the applied field enlarge, shifting the boundary walls of each

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domain. Increasing the applied field causes all domains to rotate to alignwith the applied field.

Figure 2.10: Ferromagnetic domains in nickel. [Ramo et al. 3rd ed. p 692]

Permanent Magnets

Subjecting an iron rod to a strong DC magnetic field causes the magneticdomains to align with the applied field. If the applied field is reduced tozero, the domains will tend to rotate back to their original positions, but

not entirely. The remaining or residual effect is a net magnetic polarization,with a resulting magnetic field.

It should be noted that energy is required to rotate the domains. Someof this energy is lost as heat, and some is stored in the magnetic field.

In a permanent magnet, the total magnetic field is the sum of contribu-tions from magnetic fields generated by the electron current surrounding

individual atoms. The illustration below show two atoms within the mag-netized material. The sum of many such current loops can be modelled asa single current sheet circulating around the boundary of the material as

illustrated on the right (interior current contributions tend to cancel).

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Figure 2.11: Permanent magnet modelled by an equivalent field-generating current sheet.

The B-H Curve

The figure below shows the B-H curve traced out for commercial iron [Grant& Phillips]. On the x-axis is the applied H field to which the material is

subjected3; on the y-axis is the resulting B field. The H field arises fromcurrent I in a coil of wire, i.e. H ∝ I and the resulting field is a non-linear

function of I (the B-H curve exhibits a hysteresis cycle). The permeabilityin the linear region (H < 200A/m) is µ = dB

dH ≈ 1000µ0. At H = 250 A/m,the B field is a little over 1.5 Tesla. As H is increased beyond 250 A/m, the

curve flattens and converges on a line of slope dBdH = µ0. This is explained

by the fact that once all domains are aligned with the applied H field,

increasing H has no further effect on material, and B = µ0H + constant.Reducing H to zero leaves the so-called remanence magnetization Br. The

H required to reduce B back to zero is called the coercive force Hc.

3The applied H field can also be seen as the applied B field divided by µ0, i.e. H = Bapplied/µ0. Thusthe total B is Binduced by rotation of domains + µ0H = µH where µ = µrµ0.

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µ0

µ0µ

r

linear regionslope

0.5

500250

1.5

1.0

Tesla

A/m

B−H curve for commercial iron used to make mains transformers [Grant & Phillips]

saturation

saturationslope

magnetisationremanence

cH

Br

H

B

Figure 2.12: B-H curve for iron.

In a 50 Hz AC machine or transformer, the iron is driven repeatedlyaround the B-H curve 50 times per second. Each cycle results in an energy

loss in the core (and hence heating), proportional to the area within theS-curve. The hysteresis power loss increases with frequency. If the core isnot driven close to saturation, the area is less, and so are the losses.

Example

Calculate the power loss in a 50 Hz transformer containing 1000 cm3 of iron,

operating along the hysteresis curve in the illustration above.

Solution

By integrating to obtain the area enclosed by the S-curve, the energy loss percubic metre is found to be approximately 350 Jm−3 in one cycle. The time-

averaged power loss in 1000 cm3 is therefore 50×350×1000×(0.01)3 = 17.5watts (which is quite high). In practice, the iron core is not driven so far

into saturation, and the area (and power loss) is about 1/6 of the B-H curveshown above.

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Magnetic Materials for Machines

Magnetic materials are divided according to their properties and applica-

tion.

• “Hard” Magnetic Materials - have a wide B-H curve and can retain

a high degree of magnetization; used for making permanent magnets,coating the surface of magnetic storage media like stiffy disks and audiotapes.

e.g. Iron (bonded powder) Br = 0.6T and Hc = 0.0765Am−1.

• “Soft” Magnetic Materials - have a narrow B-H curve, low hysteresis

losses, high initial permeability (µr)0 ≡ dBdH at H = 0, high maximum

permeability (µr)max =BH , and high saturation Bsat.

e.g. Commercial Iron (99 Fe), (µr)0 = 200, (µr)max = 6000, Hc = 9000Am−1 and Bsat = 2.16 T.

Pure Iron (99.9 Fe),(µr)0 = 25000, (µr)max = 350 000, Hc = 100 Am−1

and Bsat = 2.16 T.

• Note that the above mentioned permeabilities (µr) are relative perme-

abilities i.e. µ = µrµ0.

2.14 Maxwell’s Equations governing Electromagnetic

Behaviour

Maxwell’s equations are a set of equations that govern the behaviour ofelectromagnetic fields for both static and dynamic situations. Some of these

equations you will have already met in PHY2010S covering electrostaticsand magnetostatics. In this section we shall describe the basic laws and

their physical interpretation.

2.15 Gauss’ Law for Electric Fields

Electric fields arise from charges. In free space, Gauss’ law relates a surfaceintegral of the quantity ε0E · dS over a closed surface to the total charge

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therein: ∮

S

ε0E · dS =

∫

V

ρdV

In a mixed medium, the electric field is affected by the presence of dielectricmaterial, and Gauss’ law is written as∮

S

ε(r)E · dS =

∫

V

ρdV

where ε(r) is the position dependent permittivity. A useful analogy to fluidflow can be created by defining the quantity D = εE , which we call the

“flux density” arising from the free charge enclosed within S. Bound charge(which may be found on the surface of a dielectric inteface) is not considered

- it is taken into account in the variation of ε(r).Gauss’ law states that the total electric flux emanating from a closed

surface is equal to the (unbound) charge enclosed by that surface. It iswritten as ∮

S

D · dS =

∫

V

ρdV

q+

closed surface S surrouding charge q.

dS

D

n D · n

Figure 2.13: Illustration of Gauss’ law applied to a single charge.

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• The quantity D · dS ≡ D · ndS is the amount of flux emanating froma small elemental surface area dS on S.

• D · n is simply the normal component of D at the surface.

• ρ is the charge density4 in Cm−3 that is integrated over the enclosedvolume to give the total enclosed charge Q.

• The concept of flux from a charge is a bit like thinking of fluid from asource. The total ‘flow’ from a closed surface is numerically equal to

the charge contained within.

• The law can also be expressed in terms of E by substituting D = εE

where ε = εrε0.

• The advantage in working with D, apart from the slightly more com-

pact notation in some equations, is that the flux density/flow conceptholds in all types of media as well as free space. Boundary conditions

are specified in terms of the continuity of flux i.e. D · dS is the sameon either side of a boundary. In keeping with the fluid flow analogy,‘what flows in on one side of an interface between two materials must

flow out on the other’. This does not hold for E · dS on the interfacebetween two different dielectrics i.e. E · dS is discontinuous on either

side of a dielectric interface. Shrinking dS to an infinitesimal area,one concludes that the normal component of D is continuous across

a dielectric interface (i.e. at some point on the interface, the normalcomponent of D does not change) but the normal component of theE vector does change. The E field (normal component) discontinuity

arises because of the polarization effect in a dielectric which resultsin an induced layer of bound charge ob the boundary surface of each

dielectric.

• Gauss’ law can be used to simplify calculations for fields in certain sym-

metrical situations e.g. determining the field surrounding a charged

4

– Note: There is one subtlety here: in matter, ρ refers only to “free” charge i.e. charge that is free tomove or form a current. It excludes the surface charge on dielectrics induced by an electric fields.

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sphere, determining the field surrounding a long line of charge; deter-mining the field from a flat (capacitor) plate of uniform charge.

• Gauss’ law can in fact be derived from Coulombs force law and visaversa. As an exercise, derive Coulombs law by applying Gauss’ law toa sphere with a point charge located at its centre.

Example: Line of charge

The field surrounding a long line of charge of density ρl Cm−1 can be esti-

mated by enclosing the line in an imaginary Gaussian cylinder, or radius r,and length l. From symmetry, the electric field must point radially outwards.

Gauss’ law states:∮S D · dS = Qenclosed. Ignoring the small contributions

of flux leaving the areas at the ends of the cylinder, Gauss’ law becomes

2πrl Dr = ρll from which we obtain Dr =ρl2πr , or Er =

1ε0Dr =

ρlε02πr

. BothDr and Er decay smoothly as a function of r.

r

r

Dr(r)

Er(r)

Figure 2.14: Application of Gauss’ law to a long line of charge.

Note: If we placed the line of charge inside a long hollow cylinder of glass

(along its axis), keeping cylindrical symmetry in the problem, then we canstill apply Gauss’s law to establish the field quantities. In this new situation,

Dr remains unchanged (compared to the previous case) - the D flux passesthrough the glass unaffected (it is normal to the glass surface). The Er fieldhowever abruptly decreases on entering the glass (i.e. is discontinuous), i.e.

Er = Dr/ε(r) has a step change as it enters and then leaves the glass owingto the step change in ε(r). This modified example emphasises why is is

better to view the D field as a “flowing fluid” and not the E field.

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ǫ0

ǫ0

r

Er(r)

r

Dr(r)

ǫglass

Figure 2.15: Illustration of Gauss’ law applied to a line charge within a cycilinder.

Example: Parallel Plate Capacitor

The capacitance of a capacitor is defined as C = Q/V where Q is thecharge on either plate (+Q on one plate, and -Q on the other), and V is

the potential difference between plates. To estimate the capacitance of aparallel plate structure, with surface area A and small gap d, we must relateQ to V. Gauss’ law allows us to relate the electric field strength between the

plates, to the charge on the plates.Consider first, a single flat plate, area A and containing surface charge Q.

Close to the plate, we can apply Gauss’ law to a small pillbox with upperand lower surface of area a (though which the field lines penetrate), and

with sidewalls parallel to the field, as shown. Let Dn be the magnitude ofthe D field normal to the plate.∮

S D · dS = Qenclosed => aDn + aDn ≈ ρsa => Dn = ρs/2 = (Q/A)/2 =

Q/(2A). If two plates are placed a small distance d apart, one charged to Qand the other to−Q, the total field between the plates isD = 2×Dz = Q/A.

The field vectors in the gap add constructively, doubling the magnitude, i.e.D = 2×Dn = Q/A. Above the upper plate (and below the lower plate), the

field vectors add destructively (being of equal magnitude Dn but opposit indirection), and hence will be zero. The result is illustrated in Figure 2.16.

Thus the capacitance is

C =Q

V=AD

Ed=AεE

Ed=εA

d

which is a (should be) familiar result. Two plates, size 5cm × 5cm, sepa-

rated by 5mm in air, have a capacitance of about 8.85E−12×0.05×0.050.005 = 4.4pF.

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Think of the physical size of a 5pF capacitor used for electronic circuits -it is physically much smaller. This small size is made possible by reducing

the gap between plates, as well as the insertion of a dielectric material be-tween the plates. Much larger capacitances in the micro-farad range canbe created in a compact package by sandwiching an electrolytic dielectric

between two thin metal sheets that are then rolled up like a swiss roll (so-called ‘electrolytic’ capacitors). More recently, ‘supercapacitors’ have be-

come available, which make use an extremely thin dielectric layer to achievesignificantly higher (>1000 times) capacitances per unit volume than com-

mon electrolytic capacitors.

surface charge densityCharged ParallelPlate Capacitor

Field from aplate of positive charge

Field cancels

Field cancels

Field adds

plates

Result of addition of fields frompositive and negative

Gaussian surface

D

Area A

d

D = Q/A

ρs a

Figure 2.16: Deriving the electric field for a charged parallel plate capacitor.

2.16 Gauss’ Law for Magnetic Fields

Gauss’ law for magnetic fields is∮

S

B · dS = 0

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• The law reflects the fact that no point magnetic charges have beenfound. Magnetic field lines arise from moving electric charges (e.g.

current in a piece of wire)

• Magnetic flux lines form continuous loops; they do not begin and endat fixed points in space (like electric field lines on point charges).

• Any magnetic flux line that enters closed surface S at some point mustexit somewhere, as illustrated below:

North

South

Magnetic field lines

form continuous loops

(closed surface)Gaussian surface

B

Figure 2.17: Illustration to show that magnetic field lines form continuous loops; hence anyline that enters a close surface must exit it.

2.17 Faraday’s Law of Electromagnetic Induction

Faraday discovered that a changing magnetic field can induce an electriceffect. Faraday’s law is stated in integral form in terms of a relationship

between the line integral of the tangential component of the electric field

2-24 AJW, EEE3055F, UCT 2012

around a closed loop and the rate of change of magnetic flux threading theloop. Stated in integral form

∮E · dl = − d

dt

∫

S

B · dS = −dΨM

dt

where ΨM =∫SB ·dS is the magnetic flux flowing through the open surface

S, bounded by the integration contour. The law may also be written as∮

E · dl = −∫

S

∂B

∂t· dS

• In evaluating the line integral, the right hand rule must be observed:if the vector dS is defined to point up as shown above, then

∮E · dl is

made anticlockwise around the loop as illustrated below:

Integration

direction

dS

Figure 2.18: Illustration showing the direction of integration when applying Faraday’s lawto an open surface with surface normal dS. The right hand rule applies.

• This implies that for the case where ∂ΨM

∂t > 0 and uniform, the electricfield vectors point clockwise around the loop as illustrated for the pole

face of an electromagnet shown below.

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TOP VIEW

Increasing

Increasing

Case of uniform (changing)magnetic field piercingthrough flat circular areae.g. on the pole face of an electromagnet.

B

∂ΨM

∂t> 0

E

i(t)

E

E(r)

Emax

rmax

∝ 1/r∝ r

r

E

Figure 2.19: Induced E-field vectors surrounding the pole face of an electromagnet for thecase where the magnetic field is increasing in the direction shown. Note:although the B-lines must eventually curl back and form continuous loops, wecan regard the flux density for r > rmax as negligible.

• In electrostatics,∮E · dl = 0 indicating the conservative nature of

electrostatic fields and the fact that the net energy required to move a

test charge around a loop is zero. In general, electric fields are NOTconservative.

• The effect of the induced E field loop can be measured by inserting awire ring into a changing magnetic field and observing that a currentflows as a result of the induced electric field ‘pushing the electrons

round the circuit’. Note: current density J = σE inside/on wire.

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Increasing

B

I = Vinduced

Rwire

J = σE

Recall

Current flow

in wire ring

∂ΨM

∂t> 0

Figure 2.20: Current flows in a wire ring placed on the pole face.

• If the ring is broken by a small cut, the net effect is that the chargespile up on either side of the cut, resulting in a local concentration ofelectric field as illustrated below.

inside conductor

B

E = 0

∂ΨM

∂t> 0

Figure 2.21: Top view of a wire ring with an air gap. Note: the field inside the metal is zero.Thus he field from the induced dipole must exactly cancel the Faraday-inducedelectric field within the metal - try to draw the two fields.

• If a voltmeter is connected as shown in the figure shown below left, thevoltmeter will read a value Vinduced =

dΨM

dt . The voltmeter wires extend

the definition of the loop i.e. the boundary of a surface through whichthe magnetic flux passes.

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V voltmeter

inside conductor

V

Voltmeter Reading ?

B

E = 0

Vmeasured =∂ΨM

∂t

∂ΨM

∂t> 0

B

∂ΨM

∂t> 0

Figure 2.22: Wire ring with voltmeter placed in different positions. Are the readings thesame?

QUESTION: What happens if the voltmeter is moved to the opposite side

as depicted above on the right? What value will the voltmeter read?ANSWER: virtually zero! Why? Ask yourself how much changing flux willthread the loop.

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2.18 Ampere’s Circuital Law

The dynamic form of Ampere’s law states that the line integral of the tan-

gential component of magnetic field intensity around a closed loop, boundingan open surface S is equal to the sum of (i) the conduction current flowing

through S, and (ii) the rate of change of electric flux cutting through S,∮

H · dl =∫

S

J · dS+

∫

S

∂D

∂t· dS

or ∮H · dl =

∫

S

J · dS+d

dt

∫

S

D · dS

• The term∫S J · dS = Ic is the conduction current and refers to the

physical movement of charge.

• In linear conducting media, the current density is proportional to theelectric field strength J = σE where σ is the conductivity constant inSm−1 (Siemens per metre).

• The term∫S∂D∂t · dS = d

dt

∫S D · dS = dΨE

dt = Id is known as the ‘dis-

placement current’ as it has the same units of current Ic, although itdoes not refer to the physical movement of charge. The displacement

current is the rate of change of electric flux on S.

• The time varying term∫S∂D∂t

· dS was added by Maxwell as a result

of a careful theoretical analysis that revealed that the static form ofAmpere’s law was incomplete.

• The ∂D∂t is the displacement current density in units of Am−2.

In the next section, we shall examine the dynamic form of Ampere’s law in

detail and establish a deeper understanding of displacement current.

2.18.1 Physical interpretation of Displacement Current

A classic example for gaining a physical understanding of the concept ofdisplacement current involves the consideration of Ampere’s circuital law in

a simple circuit involving a capacitor as depicted in Figure 2.23

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I

S1 (shaded)

S2 (balloon shaped surface passing

between plates)

C

++++++

−−−−−−

−+H

Figure 2.23: Illustration showing how Ampere’s static law yields inconsistant results, de-pending on the definition of the surface S.

Consider applying the magnetostatic form of Ampere’s circuital law tothe closed loop C, where the current passes through surface S1 in the plane

of the loop. Loosely speaking, the static form of Amperes law says thatthe line integral of the tangential component of H around some contour C

equals the conduction current (i.e. physical movement of charge) passingthough a surface S bounded by loop C. For a long section of conductor,from considerations of symmetry, one concludes that the H field lines are

circles centred on the wire. The magnitude of the ‘circulating’ field can befound by applying Ampere’s law to a circular contour, of radius r from the

wire. From Ampere’s static law∮

H · dl = I = H2πr

which implies

H =I

2πr.

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If we consider an alternative choice of S = S2, which is carefully chosen topass between the plates of the capacitor and not cut the wire as illustrated

in the sketch. Clearly the conduction current passing S2 is zero, implyingthat the static form of Ampere’s law is either wrong, or is an incompletedescription.

Maxwell accounted for this apparent paradox, by introducing an addi-tional term which he called the “displacement current density”. The line

integral∮H · dl around closed loop C bounding some arbitrary surface S

equals the sum of the conduction current Ic =∫S J·dS plus the displacement

current Id =∫S∂D∂t · dS passing through S.

If the conduction current is Ic = I, then between the plates of the ca-

pacitor, an electric field will build up, resulting from the build up of a netpositive charge on one plate, and a negative charge on the other.We wish to explore the displacement current term

Id =

∫

dS

∂D

∂t· dS =

d

dt

∫

S

D · dS =dΨE

dt

for the case of S = S2. We can apply Gauss’s law to determine dΨE

dt andrelate it to other parameters.

Gauss’s law states that for a closed surface S,∮

S

D · dS =

∫

V

ρdV = Qenclosed

Although S2 is not closed (having an opening at its “mouth”), most of the

flux is highly concentrated between the plates, and hence the amount ofelectric flux passing through surface S2 is approximately ΨE ≈ Q where

Q is the charge on the left plate. Only a tiny amount of flux would passthough the “mouth” at contour C.Thus dΨE

dt≈ dQ

dt= Ic, being the conduction current in the wire. If

we shrink the mouth of S2 to be a tiny hole then ΨE → Q and henceId (through S2) → I. Thus the displacement current Id flowing through S2

is equal to the conduction current Ic flowing through S1.Maxwell’s form of Ampere’s law works for any definition of S, notably

• for S = S1, Ic = I and Id ≈ 0, which implies∮H · dl = Ic + Id ≈ I + 0

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• for S = S2, Ic = 0 and Id ≈ dQdt = I, which implies

∮H · dl = Ic + Id ≈ 0 + I

.

Figure 2.24: The magnetic field ‘ring’ exists both in the region around the conducting wire,and also around the displacement current in the gap between the plates.

2.19 Maxwell’s Equations

The set of four laws together are known as Maxwell’s equations:

∮

S

D · dS =

∫

V

ρdV

∮

S

B · dS = 0

∮E · dl = − d

dt

∫

S

B · dS∮

H · dl =∫

S

J · dS+d

dt

∫

S

D · dS

In the following chapter, the differentials forms will be derived.

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3 The Differential Forms of Maxwell’sEquations

In the previous chapter, we examined the integral forms of Maxwell’s equa-

tions being:

1. Gauss’ law for electric fields∮

S

D · dS =

∫

V

ρdV

2. Gauss’ law for magnetic fields∮

S

B · dS = 0

3. Faraday’s law ∮E · dl = − d

dt

∫

S

B · dS

4. Ampere’s (modified) law∮H · dl =

∫

S

J · dS+d

dt

∫

S

D · dS

In this chapter, we shall show how a set of differential equations can be

derived from Maxwell’s four equations. These are known as the “differentialforms”and serve to describe field properties at a point in space and time; this

contrasts with the integral forms, which concern integrals along contours,over surfaces and over volumes. Both the static laws and the dynamic laws

will be examined. Particulary, we shall also show how the displacementcurrent term introduced by Maxwell can be derived by consideration of thedifferential forms.

The differential forms are found by applying two theorems from vectorcalculus that apply to any vector field F = F(x, y, x, t) at any point in

space and time:

3-1

Gauss’ Divergence Theorem

For any (differentiable) vector field F, and any closed surface S defined in

space enclosing a volume V ,

∮

S

F · dS =

∫

V

∇ · F dV

where ∇ · F = divF is the divergence1 of the vector defined as

divF = lim∆V→0

∮F · dS∆V

where the quantity∮F · dS is the net flux leaving volume ∆V .

The divergence theorem alows one to express a surface integral (over aclosed surface) as a volume integral.

The physical interpretation of the term on the right hand side of thethereom

∫V ∇ ·F dV is that we are summing up, for each elemental volume

dVi within V , the quantity ∇ · F dVi, i.e. we are calculating∑

i∇ · F dVi.The quantity ∇ ·F dVi =

∮SiF · dS is the flux leaving a volume element dVi

through its surrounding closed surface Si. The flux contributions through

the side walls of adjacent dV ’s cancel, leaving only the contributions throughthe side walls touching the outer surface S of the volume V , which we write

as∮S F · dS. Hence

∮S F · dS =

∫V ∇ · F dV .

Stokes’ Curl Theorem

For any (differentiable) vector field F, and contour C bordering an opensurface S,

∮

C

F · dl =∫

S

∇× F · dS

where ∇ × F = curlF is a vector2 that measures ‘rotation’ in the vectorfield. The curlF vector points along the ‘axis or rotation’, and has mag-

1In Cartesian coordinates, ∇ ·F =

∂∂x∂∂y∂∂z

·

Fx

Fy

Fz

= ∂Fx

∂x+

∂Fy

∂y+ ∂Fz

∂z.

2and ∇×F =

∂∂x∂∂y∂∂z

×

Fx

Fy

Fz

=

∣∣∣∣∣∣

x y z∂∂x

∂∂y

∂∂z

Fx Fy Fz

∣∣∣∣∣∣=

(∂Fz

∂y− ∂Fy

∂z

)x+

(∂Fx

∂z− ∂Fz

∂x

)y+

(∂Fy

∂x− ∂Fx

∂y

)z

3-2 AJW, EEE3055F, UCT 2012

nitude equal to lim∆S→0

∮F·dl∆S

where ∆S is an elemental area in the plane

perpendicular to the axis of rotation; the direction of integration along thecontour is defined by the right hand rule applied to vector dF.

If one imagines suspending a ball at some point in a fluid velocity vectorfield, with non-zero curl at the point, then the ball will tend to rotate about

the curlF axis, and in a direction defined by the right hand rule applied tovector curlF.Stokes’ theorem relates a contour integral around the boundary of an open

surface to a surface integral over the surface.The physical interpretation of the right hand side

∫S ∇×F ·dS, is that we

are summing up for each surface element dSi on S, the quantity ∇×F ·dSi,i.e we are calculating

∑i∇× F · dSi. The quantity ∇× F · dSi =

∮CiF · dl

is the line integral on the contour Ci bordering elemental area dSi. Thecontributions of the side edges of adjacent dSi’s cancel, leaving only the

contributions from the outer boundary C bordering the surface S, which wewrite as

∮C F · dl. Hence

∮C F · dl =

∫S ∇× F · dS.

3.1 The Static EM Equations in Integral and Differential

Form

The static forms of Maxwell’s equations in integral form are:∮

S

D · dS =

∫

V

ρdV

∮

S

B · dS = 0

∮E · dl = 0

∮H · dl =

∫

S

J · dS

where D = εE and B = µH.

• The term “static” refers to the fact that the field quantities are nottime varying, i.e. ∂E

∂t = 0, ∂D∂t = 0, ∂H∂t = 0 and ∂B∂t = 0.

3-3 AJW, EEE3055F, UCT 2012

The differential forms are derived by employing the divergence theorem andStokes’ theorem as follows:

Gauss’s law for electric fields states∮S D · dS =

∫V ρdV . From the diver-

gence theorem,∮D · dS =

∫V ∇ · D · dV , hence

∫V ∇ · D · dV =

∫V ρdV .

Since this must be true for any closed surface, independent of its shape orsize, we conclude that ∇ ·D = ρ.

Gauss’s law for magnetic fields states∮S B · dS = 0. From the divergence

theorem,∮S B · dS =

∫V ∇ · BdV . Again, since the closed surface can be

arbitrarily chosen, we conclude that ∇ ·B = 0.

The conservative property of electrostatic fields states∮E · dl = 0. Ap-

plying Stokes’ curl theorem,∮E ·dl =

∫S∇×E ·dS. Since

∫S ∇×E ·dS = 0

for any surface, we conclude that ∇×E = 0.

The static form of Ampere’s law states that∮H · dl =

∫S J · dS. Applying

Stokes’ curl theorem,∮H · dl =

∫S∇×H · dS, we obtain

∫S ∇×H · dS =∫

S J · dS for any surface, from which we conclude that ∇×H = J.

Thus the four static equations in differential form are

∇ ·D = ρ

∇ ·B = 0

∇×E = 0

∇×H = J

These differential forms are used to solve for field distributions in staticsituations. Like all differential equations, the solutions require the specifi-

cation of boundary conditions.

3-4 AJW, EEE3055F, UCT 2012

Examples of Static Field Problems

Example 1

In an electrostatic situation, we may wish to know how to calculate the fieldquantities E(x, y, z) and Φ(x, y, z) in space surrounding two metal plates at

fixed potentials relative to ground. As a more specific example, consider thefields surrounding a charged capacitor. The potentials on the metal plates(measured relative to a chosen reference ground) are the so called“boundary

conditions”.A differential equation describing the surrounding electric field can be

found from Gauss’ law, ∇ ·D = ρ, and observing that the density ρ is zeroin the surrounding space (there may and will however be a thin layer of

surface charge on the plates themselves).Substituting ρ = 0 and D = εE into Gauss’ law we get

∇ · (εE) = ρ = 0

The above relationship describes a relationship between the three compo-

nents of E. Since E = −gradΦ ≡ −∇Φ, we substitute into the aboveequation to obtain a differential equation in terms of the single scalar func-

tion Φ(x, y, z):∇ · (ε∇Φ) = 0

or∂

∂x

(ε∂Φ

∂x

)+

∂

∂y

(ε∂Φ

∂y

)+

∂

∂z

(ε∂Φ

∂z

)= 0

This is a second order partial differential equation that may be solved using

numerical methods like the finite element method, together with the spec-ified boundary conditions. An equation in terms of the potential Φ is also

convenient as the boundary condition(s) are given in terms of the values ofthe potential Φ on the plates of the capacitor.If the surrounding medium homogeneous, ε is a constant, the equation

reduces to a well-known equation called Laplace’s equation

∇ · ∇Φ = 0

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Laplace’s equation is sometimes written in compact form as ∇2Φ = 0. Ex-panding in Cartesian coordinates, the differential equation is

∇ · ∇Φ =

∂∂x∂∂y∂∂z

·

∂∂x∂∂y∂∂z

Φ(x, y, z) =

∂∂x∂∂y∂∂z

·

∂Φ∂x∂Φ∂y∂Φ∂z

or∂2Φ

∂x2+∂2Φ

∂y2+∂2Φ

∂z2= 0

Having obtained Φ(x, y, z) using a numerical method, one can obtain E bydifferentiation since

E = −∇Φ.

Example 2

In electrical resistance tomography3 (ERT), the goal is to reconstruct an im-age of the electrical conductivity σ(x, y, z) within a tank from measurements

taken at electrodes placed around the circumference. ERT is applicable insituation where the medium under investigation can be modelled as a resis-

tive medium. Applications include imaging inside mixing tanks and pipescontaining fluids. Any fluid containing dissolved salts is conductive.

Currents are injected between a pair of electrodes, and voltages are mea-sured at all nodes. The reconstruction algorithm needs to be able to predictthe voltages at all electrodes around the boundary, given either specified po-

tentials or currents at the two driving electrodes. This relationship betweeninjected and measured quantities called the “forward model”. Reconstruc-

tion algorithms typically solve the so-called “inverse problem”by iterativelyfinding a solution (i.e. the conductivity distribution σ(x, y, z)) for which

the voltages predicted by the forward model closely match the measuredvoltages.

3“Tomography”refers to a method by which an image of the interior of an object is obtained via measure-ments taken around the boundary. Tomography can be implemented via the injection and measurementof EM waves such as X-rays or microwaves, or using ultra-sound, or via low frequency impedance meth-ods. The word “tomography” comes from the Greek words “tomos” which means “part” or “section”,and the word“graphein”(“to write”). In some types of tomography (e.g. X-ray tomography), the imageis made up of slices taken at different angles and combined using an algorithm.

3-6 AJW, EEE3055F, UCT 2012

JI

v1

v2

v3

v6

v8

v7I

Electrical Resistance Tomography

Inject current

v5

v4 v4 − v5 Measure voltagesbetween pairs

Figure 3.1: Illustration of current injection in electrical resistance tomography (ERT). Volt-age measurements are made between pairs of electrodes.

Stated mathematically, the goal is to find the distribution σ(x, y, z), de-scribed by a set of parameters σ , which minimizes a cost function C(σ)derived sum of the squares of the difference, i.e. minimise as a function ofσ,

C(σ) =N∑

n=1

(Vpred.(n, σ)− Vmeas.(n))2

where Vmeas.(n) is the n’th voltage measurement, and Vpred.(n, σ) is the

voltage predicted by a simulator. In practice, a smoothness constraint onσ(x, y, z) is usually applied as well. Minimizing C(σ) is a multi-variable

optimization problem for which iterative solutions have been developed.The forward model is a partial differential equation that must be solved

numerically to get Φ. To construct the differential equation to be solved, westart with the continuity equation4, ∇ · J = 0, and substitute the relationsJ = σE and E = −∇Φ. The resulting differential equation is ∇ · σ∇Φ = 0.

For the special case where σ(x, y, z) = constant, the equation reduces toLaplace’s equation ∇ · ∇Φ = ∇2Φ = 0.

The value of the Φ(x, y, z) on the boundary electrodes provides the valuesrequired to predict the measurements Vpred.(n, σ) in the cost function

given above.

4Explained in a later section. ∇ · J = 0 is the differential form of the continuity of charge relationship∮SJ · dS = −

∫V

∂ρ∂tdV , which describes the flow of charge within a conducting medium.

3-7 AJW, EEE3055F, UCT 2012

Figure 3.2: Electrical resistance tomography performed on a tank of water containing twonon-conducting objects.

One can also extract other parameters of interest from the simulated po-tential function, for example the electric field can be found by differentiation,since E = −∇Φ. The current distribution can also be found via J = σE.

This is all implemented using discrete approximations to the derivatives.

Example 3

In electrical capacitance tomography (ECT), the goal is to reconstruct animage of the electrical permittivity ε(x, y, z) within a non-conducting object

(i.e. a dielectric medium) from capacitance measurements between electrodeplates placed around the circumference of the object.

The equation describing the potential distribution within the dielectric

medium is∇ · ε∇Φ = 0

This partial differential equation can be solved for a given permittivity dis-tribution ε(x, y, z) and an applied boundary condition (being the potential

difference applied between two plates), using a numerical method. Tomo-

3-8 AJW, EEE3055F, UCT 2012

Figure 3.3: Electrical capacitance tomography - the image shown is of three objects insertedinto an (air-filled) container.

graphic algorithms are designed to find an estimate of ε(x, y, z) for whichsimulated measurements best match the actual measured data. The im-age shown is of three objects (a blackboard duster, and two plastic rods)

standing vertically inside in a plastic container.A present research project at UCT concerns the condition monitoring of

wooden electricity distribution poles using electrical capacitance tomogra-phy. An instrument has been built that allows one to detect decay and

termite damage within poles. A specially designed jacket of 12 electrodes isattached to the base of a wooden pole, and an image of the interior can beseen on a computer screen.

More generally, the method known as electrical impedance tomography

3-9 AJW, EEE3055F, UCT 2012

(EIT) is used in cases where both the conductivity and permittivity pa-rameters are to imaged. Typically, a sinusoidal signal is injected (kHz or

MHz frequency range), and both the magnitude and phase at the sensingelectrodes are recorded. Pulse-based methods are also sometimes used.Another tomography method involves stimulating the medium under in-

vestigation with a sinusoidal magnetic field (driving and sensing coils areplaced around the boundary). This method is known both as magnetic

induction tomography (MIT) or eddy current tomography. The appliedtime-varying magnetic field creates eddy currents in/on conducting material

within the medium, which, in turn, creates a magnetic field which modifiesthe magnetic field sensed by the coils.

A prototype MIT system is also being developed at UCT.

3.2 Electromagnetic Equations before Maxwell

Before Maxwell introduced the concept of displacement current, the fourequations describing the field relationships stood as

∮

S

D · dS =

∫

V

ρdV

∮

S

B · dS = 0

∮E · dl = −

∫

S

∂B

∂t· dS

∮H · dl =

∫

S

J · dS

and in differential form∇ ·D = ρ

∇ ·B = 0

∇×E = −∂B∂t

∇×H = J

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The third differential equation (Faraday’s law) is obtained by applyingStokes’ equation to E, i.e.

∮E · dl =

∫

S

∇×E · dS

and comparing to∮E ·dl = −

∫S∂B∂t ·dS. Since

∫S ∇×E ·dS = −

∫S∂B∂t ·dS

must hold for all open surfaces, this implies, ∇×E = −∂B∂t .

A modern approach to justifying the addition of the displacement cur-rent term is to show that the static form of Ampere’s law is inconsistentwith an relationship known as the “continuity of charge equation”, and that

the inclusion of the displacement current (density) term fixes the problem(Maxwell may not have done it quite this way). For this approach, we need

the concept of the “continuity of charge”.

3.3 Continuity of Charge

If we consider a closed surface S, surrounding a volume V , the fact thatcharge is neither created or destroyed leads us to the continuity requirement

∮

S

J · dS = − d

dt

∫

V

ρdV = −dQdt

which in words, simply states that the current leaving the volume equals(minus) the rate of change of charge within the volume.

• Note that if the net outflow is positive, the dQdt will be negative, and

hence the minus sign in the above equation.

• Note also, that dQdt = d

dt

∫V ρdV =

∫V∂ρ∂tdV since differentiation and

integration are linear operations, which allows the order can be inter-

changed.

Thus we can write the continuity equation as∮

S

J · dS = −∫

V

∂ρ

∂tdV

3-11 AJW, EEE3055F, UCT 2012

The above integral form can be expressed as a differential form by applyingthe divergence theorem to J, i.e.

∮

S

J · dS =

∫

V

∇ · JdV

Thus ∫

V

∇ · JdV = −∫

V

∂ρ

∂tdV

Since this must hold for any chosen closed surface S containing volume V ,

we conclude that

∇ · J = −∂ρ∂t

This is the differential form of the continuity of charge equation.

SChargeleaving

dQ/dt<0

net currentflowing out

J

J

If dQ/dt < 0then there is a

Figure 3.4: The continuity of charge equation implies that the rate at which charge leavesa closed surface equals (minus) the rate of change of the total charge withinthe enclosed volume.

An alternative, and more fundamental approach to obtain the continuityof charge relationship in differential form is directly from the definition of

divergence:

∇ · J = limV→0

∮S J · dSV

= limV→0

−∫V∂ρ∂tdV

V= lim

dV→0

−∂ρ∂t dV

dV= −∂ρ

∂t

3-12 AJW, EEE3055F, UCT 2012

3.4 Fixing the Problem with Ampere’s Law

A fundamental vector identity states that div(curlF) = ∇ · (∇× F) = 0 for

any differentiable vector field F. If we take the divergence of the left andright hand side of Ampere’s static form

∇×H = J

we get

∇ · (∇×H) = ∇ · Jwhich would imply that ∇ · J = 0. This clearly is not consistent with the

continuity equation ∇ · J = −∂ρ∂t , and so we must conclude that ∇×H 6= J

in general, although we know it holds for certain measurable cases (e.g.

magnetic field surrounding a conductor).To fix the problem, we try adding a new term T, and see what we come

up with - i.e. let

∇×H = J+T

Taking the divergence on both sides, and applying again our vector identity,

we get0 = ∇ · J+∇ ·T

Rearranging and substituting the continuity equation we get

∇ ·T = −∇ · J =∂ρ

∂t

Noticing that we can substitute ∇ ·D = ρ, we see

∇ ·T =∂ρ

∂t=∂∇ ·D∂t

= ∇ · ∂D∂t

Mathematically,

T =∂D

∂t+ k

where k is a constant, satisfies the equality. The “integration constant”does however not model anything physical5 and so, T = ∂D

∂t as introduced

5For example, if k 6= 0, then it would suggest that ∇×H = k 6= 0 even when J = 0 and ∂D∂t

= 0. Untilsomething is observed physically, the constant is zero.

3-13 AJW, EEE3055F, UCT 2012

by Maxwell, is sufficient to resolve the inconsistency in Ampere’s law. Thisaddition to Ampere’s magnetostatic law provides Maxwell’s fourth equation

∇×H = J+∂D

∂t

• This simple fix makes Maxwell’s equations a mathematically consistentset, and predicts the existence of an additional physical quantity called

the “displacement current” (density) ∂D∂t .

• The units of ∂D∂t

must be the same as the conduction current J, being

Am−2, although ∂D∂t does not refer to any physical movement of charge.

• The extended model has been shown to be accurate by experiment,and predicts a variety of phenomena, some of which had not been

previously realised, including the fact that light is an electromagneticphenomenon, i.e. a propagating EM wave (discussed later).

We have derived Maxwell’s modified form of Ampere’s law in differential

form. To obtain the integral form, we again write down Stokes’ theorem forH, and substitute ∇×H = J+ ∂D

∂t,

∮H · dl =

∫

S

∇×H · dS =

∫

S

J+∂D

∂t· dS =

∫

S

J · dS+

∫

S

∂D

∂t· dS

giving ∮H · dl =

∫

S

J · dS+

∫

S

∂D

∂t· dS

3.5 Maxwell’s Differential Equations

The complete set of Maxwell’s Equations in differential form is

∇ ·D = ρ

∇ ·B = 0

∇×E = −∂B∂t

∇×H = J+∂D

∂t

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whereD = εE andB = µH. It is pointed out that this groups for equations,together with the Lorentz force law

F = qE+ qv ×B

summarize all the behaviour of classical electromagnetic field theory in both

matter and free space.

• All other laws e.g. Coulomb’s force law, the Biot-Savart law and thecontinuity equation can be derived from the above equations.

• The only place where these differential forms do not hold are on bound-aries between different materials where the derivatives are not finite.

At boundaries, we can derive a set of “boundary conditions” that relatethe field quantities immediately on either side of the boundary (see

section on boundary conditions).

• The quantities ρ and J refer to the “free charge” in the medium - i.e.

the charge that is free to move as current, as opposed to the chargebound to atoms. (When a dielectric is polarized via exposure to anelectric field, a layer of surface charge forms on the polarized dielectric,

but this kind of bound charge is not included in the “ρ” .)

3.6 Summary Table of Maxwell’s Equations

Gauss’ law∮SD · dS =

∫V ρdV ∇ ·D = ρ

“Gauss’ law”∮SB · dS = 0 ∇ ·B = 0

Faraday∮E · dl = − d

dt

∫S B · dS ∇×E = −∂B

∂t

Ampere/Maxwell∮H · dl =

∫S J · dS+ d

dt

∫SD · dS ∇×H = J+ ∂D

∂t

Auxiliary equations:

F = qE+ qv ×B

D = εE

B = µH

J = σE

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4 Electromagnetic Boundary Conditions

The integral forms of Maxwell’s equations describe the behaviour of elec-tromagnetic field quantities in all geometric configurations. The differential

forms of Maxwell’s equations are only valid in regions where the parametersof the media are constant or vary smoothly i.e. in regions where ε(x, y, z, t),µ(x, y, z, t) and σ(x, y, z, t) do not change abruptly. In order for a differen-

tial form to exist, the partial derivatives must exist, and this requirementbreaks down at the boundaries between different materials.

For the special case of points along boundaries, we must derive the rela-tionship between field quantities immediately on either side of the boundary

from the integral forms (as was done for the differential forms under differ-entiable conditions).

Later, we shall apply these boundary conditions to examine the behaviourof EM waves at interfaces between different materials - from them we canderive the laws of reflection and refraction (Snell’s law).

4.1 Boundary Conditions for the Electric Field

Consider how the electric field E may change on either side of a boundarybetween two different media as illustrated below.

The vector E1 refers to the electric field in medium 1, and E2 in medium 2.One can further decompose vectorsE1 and E2 into normal (perpendicular tointerface) and tangential (in the plane of the interface) components. These

components labelled En1, Et1 and En2, Et2 lie in the plane of vectors E1 andE2.

To derive boundary conditions for E, we must examine two of Maxwell’sequations: ∮

E · dl = −∫

S

∂B

∂t· dS

4-1

Medium 2

Medium 1

Et2

Et1

E1

E2En2

En1

Figure 4.1: Normal and tangential components of the electric field on either side of theinterface between two media.

and ∮D · dS =

∫

V

ρdV

which will allow us to relate the tangential and normal components of E oneither side of the boundary. Note that ρ refers to the free, unbound changewithin V , i.e. it excludes the charge bound to atoms.

Normal Component of D

The boundary condition for the normal component of the electric field can

be obtained by applying Gauss’s flux law∮

D · dS =

∫

V

ρdV

to a small ‘pill-box’, positioned such that the boundary sits between its

‘upper’ and ‘lower’ surfaces as shown in the illustration.

4-2 AJW, EEE3055F, UCT 2012

Surface charge

Medium 2

Medium 1

ρs

ε2, µ2, σ2

ε1, µ1, σl dS = n∆s

dS = −n∆s

∆h

n

∆s

Figure 4.2: Gaussian pill box straddling the interface between two media.

If we shrink the side wall ∆h to zero (keeping the interface sandwichedbetween the upper and lower surface) then all electric flux enters or leavesthe pill-box through the top and bottom surfaces, and

∮D · dS → D1 · n∆s+D2 · (−n)∆s = Dn1∆s−Dn2∆s

where Dn1 and Dn2 are the normal components of the flux density vector

immediately on either side of the boundary in mediums 1 and 2, and ∆s isthe elemental surface area.

The amount of charge enclosed as ∆h → 0 depends on whether thereexists a layer of charge on the surface (i.e. an infinitesimally thin layer of

charge)1. If a surface charge layer exists then∫

V

ρdV = ρs∆s

and thusDn1∆s−Dn2∆s = ρs∆s

1In perfect conductors, any excess free charge always resides on the surface of the conductor and isdenoted by ρs in units of Cm−2. Within the conductor, the charge density very rapidly goes to zero -this is discussed in a later section on “relaxation time”.It should also be noted that in the case of dielectrics subjected to an electric field, the material

polarizes, which does in fact result in a surface charge layer - however this charged layer is bound

charge caused by the polarization effect, and is not part of the quantity ρs.

4-3 AJW, EEE3055F, UCT 2012

from which we concludeDn1 −Dn2 = ρs

For the case where ρs = 0,Dn1 = Dn2

Or in terms of the electric field E,

ε1En1 = ε2En2

or

En1 =ε2ε1En2

Tangential Component of E

We can derive the tangential component of E by applying Faraday’s law toa small rectangular loop positioned across the boundary, and in the plane

of E1 and E2, as illustrated in the diagram below.

a

c d

b

Medium 2

Medium 1

ε2, µ2, σ2

ε1, µ1, σl∆h

∆ln

E1

E2

Figure 4.3: To determine the boundary condition on the tangential component of the Efield, Faraday’s law is applied to rectangular loop straddling the interface be-tween two media.

Consider the limiting case where the sides ∆h perpendicular to boundaryare allowed shrink to zero. In the limit as ∆h → 0 , the magnetic flux

threading the loop shrinks to zero, and thus∮

E · dl →∫ b

a

E1 · dl+∫ d

c

E2 · dl = 0

4-4 AJW, EEE3055F, UCT 2012

⇒ E1 ·∆l+ E2 · (−∆l) = 0

Writing the tangential components of E1 and E2 along the contour as Et1

and Et2, we have⇒ Et1∆l − Et2∆l = 0

from which we conclude that on either side of the boundary,

Et1 − Et2 = 0

orEt1 = Et2

i.e. tangential components immediately on either side of a boundary are

equal.

4.2 Boundary Conditions for the Magnetic Field

The derivation of boundary conditions for the magnetic field, follows similar

arguments to that of the electric field, but using equations∮

B · dS = 0

∮H · dl =

∫

S

J · dS+

∫

S

∂D

∂t· dS

Again we consider the normal and tangential components as illustrated be-low.

Medium 2

Medium 1Bn1B1

Bt1

Bt2

Bn2

B2

Figure 4.4: Normal and tangential components the B field on either side of an interface.

4-5 AJW, EEE3055F, UCT 2012

Normal Component of B

The boundary condition for the normal component of the magnetic field

can be obtained by applying Gauss’s flux law∮B · dS = 0

to a small pill-box.

If we shrink the side wall ∆h to zero, all magnetic flux leaves/enters thepill-box through the top two surfaces,∮

B · dS → B1 · n∆s+B2 · (−n)∆s = Bn1∆s− Bn2∆s

Equating to zero, we findBn1 − Bn2 = 0

and hence the normal component of B is continuous at boundaries.

Tangential Component of H We can derive the tangential component of Hby applying Ampere’s law to a closed loop as illustrated below. Again, the

rectangular loop is in the plane of vectors H1 and H2.

a

c d

b

Medium 2

Medium 1

ε2, µ2, σ2

ε1, µ1, σl∆h

∆ln

H1

H2

Figure 4.5: To determine the boundary condition on the tangential component of the Hfield, Ampere’s law is applied to rectangular loop straddling the interface be-tween two media.

Ampere’s law states∮H · dl =

∫

S

J · dS+

∫

S

∂D

∂t· dS

4-6 AJW, EEE3055F, UCT 2012

Consider the limiting case where the sides ∆h perpendicular to the boundaryare allowed shrink to zero. The left hand side becomes∮

H · dl →∫ b

a

H1 · dl+∫ d

c

H2 · dl

= H1 ·∆l+H2 · (−∆l)

On the right hand side, the displacement current term Id =∫S∂D∂t·dS shrinks

to zero. For physical media, the conductivity σ is finite, and J is also finite.Thus within the loop Ic =

∫S J · dS also shrinks to zero, and so

H1 ·∆l+H2 · (−∆l) = 0

which implies the tangential component of H does not change immediatelyon either side of the boundary, i.e.

Ht1 = Ht2

For the special case of an idealised perfect conductor, σ → ∞, a surface

current may exist (i.e. current flowing within a vanishingly thin layer onthe surface). Some physical situations involving good conductors like metals(e.g. skin effect and reflection of EM waves off metallic objects) may allow us

to treat currents concentrated on the surface as a surface current modelledby a vector Js in units of Amps/m (NB not m2) flowing in an infinitesimally

thin layer. To get the correct orientation of the tangential component of Ht,we can orientate the rectangular loop until the line integral is a maximum.

This is achieved when Js flows perpendicular to our rectangular loop. Thecurrent flowing through S is then

Ic =

∫

S

J · dS → Js∆l

where Js is the magnitude of the surface current density.We conclude that for the case where a surface current exists, the boundary

condition on the tangential component of H is therefore

Ht1 −Ht2 = Js

and the tangential components, Ht1 andHt2, lie perpendicular to the surface

current vector Js. The direction of vectors Ht1 and Ht2 can be found byapplying the right hand rule with one’s thumb pointing in the direction of

Js.

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Summary of Boundary Conditions

Below are depicted the components on either side of the boundary in side

view.

Medium 2

Medium 1

ε2, µ2, σ2

ε1, µ1, σlEt2

En2

Et1

En1

Bt2

Bt1

Bn2

Bn1E1

E2B2

B1

Figure 4.6: Normal and tangential components illustrated for the cases of the E field andthe B field.

The boundary conditions are summarised below.

Dn1 −Dn2 = ρs

Et1 − Et2 = 0

Bn1 − Bn2 = 0

Ht1 −Ht2 = Js

The boundary conditions can be expressed in vector form2 as:

D1 · n−D2 · n = ρs

n× E1 − n×E2 = 0

2These vector forms require careful 3-D visualization. Taking the cross product between the surfacenormal and a vector, say n×E1 extracts the tangential component (as a vector) .

4-8 AJW, EEE3055F, UCT 2012

B1 · n−B2 · n = 0

n× (H1 −H2) = Js

These general conditions can be further refined depending on the specific

media on either side of the interface. Some examples are given below.

4.3 Examples of Boundary Conditions

4.3.1 e.g. Dielectric - Dielectric Interface

Dielectrics are materials for which all electrons are bound to atoms, and arenon-conducting, i.e σ → 0; no currents flow, and no unbound surface chargeexists unless explicitly put there (i.e. ρs = 0). Thus we have

Dn1 = Dn2 or ε1En1 = ε2En2

Et1 = Et2

Bn1 = Bn2

Ht1 = Ht2 or1

µ1Bt1 =

1

µ2Bt2

• An example of a dielectric-dielectric interface is the interface between

air and glass.

• The above boundary conditions are applied when analysing the reflec-

tion and refraction of plane waves (studied in a later section).

For example, consider an air-glass interface where ε1(air) = ε0 and ε2(air) =

5ε0. Given the E1 vector in the air, how can we sketch the E2 vector?

4-9 AJW, EEE3055F, UCT 2012

Medium 2

GLASS

Medium 1

AIR

ε1 = ε0

E1

ε2 = 5ε0, µ2

E2 =?

Figure 4.7: E1 field at the the interface between two dielectrics.

We already know the tangential component must be sketched as Et2 = Et1.The normal component is related by En2 =

ε1ε2En1 =

15En1. The E2 can thus

be sketched as shown below:

Medium 2

GLASS

Medium 1

AIR

ε1 = ε0

E1

E2

En1

ε2 = 5ε0, µ2

En2Et2 = Et1

Figure 4.8: E1 and E2 field at the the interface between two dielectrics.

4.3.2 e.g. Dielectric - Perfect Conductor

If one of the media is a dielectric (say medium 1 is air), and the othermedium (medium 2) is a perfect conductor σ2 → ∞, then En2 = 0 andEt2 = 0 inside the perfect conductor3.

3The conductor is assumed to be stationary within the xyz frame of reference. If a conducting metalobject moves through a magnetic field, a non-zero E field can exist within the conductor. For example,if a long thin metal rod moves through a uniform B field with velocity v, the electrons inside the rod

4-10 AJW, EEE3055F, UCT 2012

• Since Dn1 −Dn2 = ρs, we conclude that Dn1 = ρs

• Since Et1 = Et2 and Et2 = 0, we conclude that Et1 = 0, i.e. there exists

no tangential component on the dielectric side of the interface.

In vector form we state the boundary conditions for the field in the dielectricas

D1 · n = ρs

and

n×D1 = 0

• The E field lines always meet a perfect conductor perpendicular to the

surface, and magnetic field lines parallel to the surface as is illustratedin the figure below:

experience a force qv×B, perpendicular to the direction of motion and perpendicular to B. Electronsdeplete on the one side and increase on the opposite side, causing an induced dipole.

through uniformmagnetic field

Rod moving

Fields insidethe rod

Ev×B

v

B

As the dipole forms, an electric field builds up until the internal forces balance, i.e. qE = −qv ×B,and the electron current no longer flows. The internal field strength is E = −v × B. If the rod isorientated in the direction of v ×B, then the potential difference between the ends of the moving rod

is φ =∫ l

0 E · dl = vBl where l is the length of the rod. In cases where a metal object is stationarywithin the frame of reference, the electrons will rearrange rapidly (if placed in an EM field) such thatthe internal electric field goes to zero; the potential difference between any two points on the conductorwill also then be zero.

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Medium 2

Medium 1

(AC fields)

Perfect conductor

dielectric e.g. air

Bn1 = 0

ε2, µ2, σ2 = inf

E1

B2 = 0

B ε1, µ1, σlEt1 = 0

E2 = 0

En1 = ρ

n×H1 = Js

Figure 4.9: E field and the B field at the interface between air and a perfect conductor.

For AC fields, no time-varying magnetic field exists in a perfect conductor- why? Recall that ∇× E = −∂B

∂t and since E = 0 in a perfect conductor,∇× E = 0 and hence ∂B

∂t= 0. In other words, no changing magnetic field

can exist in a perfect conductor, and hence Bn2 = Bn1 = 0, i.e. the normalcomponent of the magnetic field is zero. A surface current can still exist,

implying a tangential component of B1 can exist. These two conditions canbe expressed in vector form as

B1 · n = 0

n×H1 = Js

• These boundary conditions are useful for establishing, for example, the

charge density or current distribution on the surface of a conductor,when the field quantifies in the dielectric are known or specified.

• These boundary conditions will be applied when analysing the reflec-

tion of an electromagnetic plane wave off the surface of a perfect con-ductor.

4.3.3 e.g. Conductor-Conductor (steady state current)

For the case of two conductors under static field conditions (i.e. ∂E∂t = 0 and

∂B∂t = 0), there can be no charge build up at the interface and hence

Jn1 = Jn2

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Since Jn = σEn, we have an additional constraint on the normal componentof the electric field, i.e.

σ1En1 = σ2En2

For non-steady state conditions a more complicated boundary constraintrelates J1 and J2, which can be derived by application of the continuity of

charge equation∮S J · dS = − ∂

∂t

∮V ρdV at the boundary. The result is

(J1 − J2) · n+∇t · Js = −∂ρs∂t

where ∇t is the two-variable divergence in the tangent plane applied to the

surface current Js, and∂ρs∂t is the rate of change of surface charge density in

Cm−2s−1. See [Griffiths] for details.

Medium 2

Medium 1

Side View

surface chargeε2, µ2, σ2

ε1, µ1, σl

Jn1 = J1 · n

Jn2 = J2 · n

n

ρs

Figure 4.10: Boundary condition for the normal component of J for conductors (for thenon-steady state case for which there may be charge build-up at the interface).

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5 Relationship between Field Theoryand Circuit Theory

(ref: Ramo et al.)

At lower frequencies where physical circuit dimensions are small comparedto the wavelength1 of electromagnetic waves, the behaviour of circuits is ac-curately modelled using“lumped element”component models, together with

Kirchhoff’s laws. At higher frequencies where the distances between compo-nents are a significant fraction of a wavelength and greater, the signals car-

rying information or power from one place in a circuit to another are treatedas waves. Signals must be routed from one point to another using trans-

mission lines, modelled using transmission line theory. If the componentdimensions be comparable to the wavelength then accurate understanding

and prediction of behaviour may require modelling using electromagneticfield and wave theory.In this section we examine the relationship between Maxwell’s equations

and circuit theory. Both Kirchhoff’s voltage law, relating to voltage dropsVi around a loop

N∑

i=1

Vi = 0

and Kirchhoff’s current law, relating relating currents Ii leaving a node

N∑

i=1

Ii = 0

can be explained in terms of field theory.Consider for example, the circuit shown below shown firstly using standard

circuit symbols for R, L and C, and secondly as a physical representation

1 f 50 Hz 100 kHz 1 MHz 10 MHz 100 MHz 1 GHz 10 GHz 100 GHzλ = c/f 6000 km 3 km 300 m 30 m 3 m 30 cm 3 cm 3 mm

5-1

with wires of finite width, a resistor as a rod of carbon, an inductor as acoiled wire, and the capacitor as a pair of metal plates. We shall examine

briefly each lumped component model from a field theory perspective.

V(t)

C

R

L

I(t)

2

0 3

1

0

12

3

Coil

RL

C

V(t)

0 3

3

2

0

21

1

I(t)

Figure 5.1: Circuit diagram showing component symbols (above) and a more physical de-piction of the components (below).

5.1 Resistors

A resistor can be constructed from a resistive material of conductivity σ,length l and cross-sectional area A, as depicted below.

5-2 AJW, EEE3055F, UCT 2012

e1 2

Area A

σ

J = σE

l

E

Figure 5.2: Resistor made from a cylinder of carbon. A current flows as a consequence ofthe (axial component) of the electric field.

If the material is subjected to an electric field orientated along the length

of the cylinder, a current will flow, explained as follows:

• Electrons move under the influence of the electric field to reach anaverage drift velocity.

• A classical model explains this as follows:

Electrons initially accelerate under the influence of the field, but repeat-edly collide with bound atoms, and “bounce off”, resulting in deceler-

ation. The net result is a constant average velocity for the electrons.This has some analogy to the terminal velocity reached by a fallingobject as a results of the resistance from the air molecules.

Area A

eelectron, which collideswith atoms.

Path of an accelerating

Average drive velocityof the electrons

J = σEE

Figure 5.3: The electrons accelerate, but are impeded by the atomic structure, hence reach-ing a finite (average) terminal velocity. An imagined path of a single electronis shown.

Because of the high density of electrons, the average drift speed is sur-prisingly slow. For example, Halliday, Resnick & Walker 6th Ed, do anexample calculation (Problem 27.3) in which the drift velocity with a cop-

per wire or radius 0.9 mm, carrying a current of 17 mA, is calculated to be4.9× 10−7m/s or 1.8 mm/hr.

5-3 AJW, EEE3055F, UCT 2012

• The average current density in Am−2 is proportional to the strength ofthe electric field, i.e.

J = σE

where σ has units [Sm−1] and is a property of the medium.

• It is noted that any particular electron continually accelerates and de-

celerates with each collision as illustrated.

• Energy is dissipated as a result of the collisions (i.e. in the form of

heat).

The voltage developed across the resistor is found by integrating the electric

field through it from node 1 to node 2:

V21 = V2−V1 = −∫ 2

1

E·dl = −∫ 2

1

Edl = −∫ 2

1

J

σdl = −

∫ 2

1

I/A

σdl = −I l

σA

The constant lσA

is identified as the resistance of the rod, i.e.

R =l

σA

In the labelled circuit loop,

V21 = V2 − V1 = −IR

5.1.1 Calculation of average drift velocity

To calculate the average drift velocity of the electrons in a cylindrical wireconductor, one needs to know the current I, the charge on one electron

(qe = −1.6 × 10−19 coulombs), the thickness of the wire, and the electrondensity ρe [m

−3] (not to be confused with charge density). Copper containsρe = 8.5× 1028 electrons per m3 - a very large number!

Imagine a slug of electrons moving along a wire rod in the x direction.The electron drift velocity in metres per second (in the x direction) can be

expressed in terms of the current as

vdrift =dx

dt=dx

dQ

dQ

dt=

1dQdx

(−I)

where

5-4 AJW, EEE3055F, UCT 2012

• dxdt is the velocity of the leading edge of the slug as it passes some pointx0,

• dQdx is the ratio of charge passing the point x0 per distance dx movedin the x direction. This is identical the charge density in coulombs per

metre of wire.

• dQdt is the charge per second passing point x0 per second. If the conven-tional current is I amperes moving in the negative x direction, thendQdt = −I.

To determine dQdx, consider a wire segment of length dx, and of thickness 2r.

The volume of the segment is πr2dx. The number of coulombs per metre is

dQ

dx=

(charge/vol)× vol

length=

(ρeqe)(πr2dx)

dx= ρeqeπr

2

and hence

vdrift =−I

ρeqeπr2

For copper wire of 1 mm thickness,

dQ

dx= ρeqeπr

2 = 8.5× 1028 ∗ (−1.6× 10−19) ∗ 3.14 ∗ (0.5× 10−3)2 = −10676Cm−1

If a current of I = 1 ampere flows in the wire, the electron drift velocity is

vdrift =−I

ρeqeπr2=

−1 [Cs−1]

−10676 [Cm−1]= 9.37× 10−5ms−1 = 34 cm/hr.

The number of electrons making up one coulomb is 1/ |qe| = 6.25× 1018.

Thus for a 1 ampere current, 6.25× 1018 electrons pass per second.

5.2 Capacitors

Consider a parallel plate capacitor. As the current flows through the wires,

a surface charge builds up on the inner sides of the capacitor’s plates.

5-5 AJW, EEE3055F, UCT 2012

• Note that all excess charge will sit on the surface of the capacitorplates, in a thin layer (not inside the metal). Recall that E = 0 inside

a perfect conductor, and since div D = ρ or div εE = ρ this implies,ρ = 0 inside the conductor. All excess charge must therefore lie on thesurface, described by a surface charge density ρs in Cm−2.

The charge Q on the plate onto which the conventional current flows is

found by integrating the current flowing onto the plate, i.e.

Q(t) =

∫ t

t0

I(t)dt+Q0

where Q0 is the initial charge at some starting time t0. The other plate willhave a charge of −Q(t).A potential difference builds up between the places. The potential differ-

ence can be shown by careful argument [Griffiths] to be proportional to the

charge Q on the plates, i.e.

Vc(t) =1

CQ(t)

where C is the constant known as the capacitance.

Substituting for Q(t), we get

Vc(t) =1

C

∫ t

t0

I(t)dt+Q0

C

In the circuit loop, V03 = V0 − V3 = −Vc(t).

5.3 Inductors

Inductors are made by winding several turns of wire either in air, or aroundsome high permeability material (which boosts the inductance, requiring

fewer turns).We shall explain the operation of an inductor by considering first a single

turn, and then a coil of several turns, in the context of the series circuit

under analysis.As already discussed, we are interested in applying Faraday’s law around

the dashed loop shown in the physical circuit. For the inductor, we are

5-6 AJW, EEE3055F, UCT 2012

interested in the integral of the electric field through the air gap betweenthe terminals as indicated by the dotted line between nodes 2 and 3 in the

circuit.

5.3.1 Single-turn Inductor

Consider a single turn inductor made from a thin piece of wire and illus-trated in Figure 5.4.

+

−

contour CIntegration

B

V

I

B

Figure 5.4: Single turn inductor.

We can apply Faraday’s law locally to a closed contour C that goes clockwise

around the inside of the wire and then across the air gap (in the shape ofthe dotted path),

∮

C

E · dl =∫

(air)

E · dl+∫

(wire)

E · dl = − d

dt

∫

S

B · dS =−dΨM

dt

where and dS points into the page, and ΨM is the flux threading the inte-gration loop (and cutting a chosen surface S, bounded by C).Since E = 0 in the wire, the potential difference is then

V = −∫

(air)

E · dl = d

dt

∫

S

B · dS

The magnetic field B is linearly proportional to the current I flowing in thewire2, i.e. B ∝ I, and so is

∫S B · dS ∝ I. The constant of proportionality

2The magnetic field vector owing to a short current segment can be computed using the Biot-Savartlaw (reviewed in Section 2.5). The total field is found by integration of all contributions from currentelements in the wire.

5-7 AJW, EEE3055F, UCT 2012

is known as the inductance, i.e.∫

S

B · dS = LI

or

L =

∫SB(t) · dSI(t)

=ΨM

I

The units of inductance3 are henrys [H], The voltage across the inductor is

V =d

dt

∫

S

B · dS =d[LI]

dt= L

dI(t)

dt

In the labelled series circuit,

V32 = V3 − V2 = −LdI(t)dt

5.3.2 Multi-turn Inductor

A multi-turn inductor is constructed by winding a coil of wire as depicted

in the figure below.

+

−

V

V3

V2

V1

I

Figure 5.5: Multi-turn inductor (N = 3 here).

The voltage drop across the terminals is the line integral along the dashedline:

V = −∫ +

−E · dl

= −[

∫

gap1

E · dl+∫

gap2

E · dl+ ...

∫

gapN

E · dl]

= V1 + V2 + ...VN3The units of inductance are equivalently [H] = [Wb A−1] = [VA−1s−1] = [VCs−2] = [NC−1mCs−2] =[Nms−2].

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If we further assume that the flux Ψ linking each turn is the same, then

V1 =dΨ

dt= V2 = ... = VN

and

V = NV1 = NdΨ

dtBecause there are N turns, the flux Ψ threading the coil will be N times

stronger than the contribution from a single turn, i.e.

Ψ = N ×Ψ1turn

where Ψ1turn is the flux contribution from a single turn. Substituting, weobtain

V = NV1 = NdΨ

dt= N2dΨ1turn

dt= N2d(L1turnI)

dt= N2L1turn

dI

dt

Thus the inductance for a tightly wound N -turn coil is

L = N2L1turn

where L1turn is the inductance of a single turn.

It is worth remembering that the inductance increases as a function of N2.

If one doubles the number of turns, the inductances increases by a factor offour.

Alternative Explanation

Analysis of a multi-turn coil is similar to the case of a single turn coil, withthe added complication that the integration contour C is not a circle, but

rather made up of a spiral that follows the wire (and a short section in theair gap between its terminals). As argued for a single turn case, the voltageacross the terminals is

V = −∫

(air)

E · dl = d

dt

∫

S

B · dS

where surface S is now a complicated-to-visualise surface that is bounded

by the contour C. It helps to imagine the surface within the coil as a smooth

5-9 AJW, EEE3055F, UCT 2012

“spiral staircase” winding around an imaginary centre line. The total fluxΨM passing through S is proportional to the current I in the wire, and is

given by

ΨM =

∫

S

B · dS = LI

where L is the inductance of the multi-turn coil. Thus

L =ΨM

I=

∫S B(t) · dSI(t)

where ΨM must be carefully evaluated for the particular coil structure. For

a short, compact coil of N turns, it is not difficult to show that

L = N2L1turn

where L1turn is the inductance of a single turn coil of the same radius. Tosee this, one must grasp two points:

• the spiral surface S through with the flux lines pass consists of a stackof N identical contributing “flattish” discs (the total surface area isapproximately N times larger than for a single turn)

• the flux density on each component“disc” is N times stronger than theflux density generated by a single turn (superposition of contributions

from N turns, each carrying current I)

Thus the total flux threading the surface of the multi-turn contour is

ΨM =

∫

S

B · dS ≈ N ×(N ×Ψ1turn

)

where Ψ1turn is equal to the flux generated by a single turn coil carrying

current I.The inductance of the multi-turn coil is then

L =ΨM

I=N2Ψ1turn

I= N2L1turn

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5.4 Formulas for Practical Coils

In order to calculate the inductance accurately, we need to consider both the

field inside, and outside of the wire making up the coil. The total inductanceis usually computed in two parts:

L = Linternal + Lexternal

where Linternal is the contribution arising from the magnetic field within thewire, and Lexternal is the contribution from the field outside the wire.

5.4.1 Inductance of a Circular Wire Ring

Internal field

in the surroundingair.

wire 2a

inside the metal

External field

2ar

wire

diameter

radiusof circle

Figure 5.6: Circular wire ring, and its cross section.

The inductance of a wire ring can be found from

L =

∫S B · dSI

=ΨM

I

where B can be found by integrating the field contributions (using the Biot-

Savart law, reviewed in Section 2.5) from each elemental current segmentaround the ring.

Considering only the contribution from the flux outside the wire, it canbe shown, by integration, that ΨM ≈ Iµr[ln8r

a − 2] and hence the “external”

inductance (Ramo et al.) is

Lext =ΨM

I≈ µ0r[ln

(8r

a

)− 2]

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for the case where a << r.The “internal” inductance of a long straight wire (in henrys per metre of

wire), assuming uniform current density in the wire4, can be shown to beindependent of wire radius a, and for non-magnetic metal wire is given by(ref Ramo et al.)

L′int =

µ08π

= 0.5× 10−7 Hm−1

Example Calculation

Calculate the inductance of a circular copper wire ring, radius r = 10 cm,wire radius a = 0.5 mm. NOTE µ ≈ µ0 in copper.

The external component is

Lext ≈ µ0r[ln

(8r

a

)− 2] = 4π× 10−7× 0.1[ln

(8× 0.1

0.0005

)− 2] = 0.7× 10−6H

The internal inductance in henrys per metre is

L′int =

µ08π

=4π × 10−7

8π= 0.05× 10−6Hm−1

The total internal inductance

Lint ≈ L′int2πr = 0.050× 10−6 ∗ 0.63 = 0.03× 10−6H

which is relatively small compared to Lext. The total inductance is

L = Lint + Lext = 0.73× 10−6H

4At sufficiently low frequencies (kHz down to DC), the current density is fairly uniform across the crosssection of the wire. As the frequency increases from DC, the current tends to concentrate increasinglytowards the outside of the wire. This effect is known as the ’skin effect’, and is discussed later in thecourse.

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5.4.2 Inductance of a Short Coil (short length to radius ratio)

r

Radius

Figure 5.7: Short coil.

For a short length-to-radius ratio coil such as that depicted in Figure 5.7,

the external inductance of an N turn coil is N2 times that of a single turn,i.e.

L ≈ N2µ0r[ln

(8r

a

)− 2]

When winding a coil, it is useful to remember that the inductance isproportional to the square of the number of turns. The inductance may beincreased by winding the coil on an iron or ferrite rod or on a toroid, which

has a relative permeability of hundreds or thousands that of air.

5.4.3 Inductance of a Long Multi-turn Coil (long length to radius

ratio)

Figure 5.8 depicts a long coil of length l and coil-radius r ≪ l, containing N

closely-wound turns carrying current I amperes. Such a coil can be treatedas a wrapped “sheet of current” (the row of dots in the illustration), from

which H can easily be obtained by application of Ampere’s law∮

H · dl =∫

S

J · dS

. Consider the “dashed” integration contour shown in the sketch, which

contains a horizontal length along the inside of the coil, and a length on theoutside. The vertical side contributions to the integral are negligibly small

because the flux lines are run perpendicular to the contour at the sides.

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length l

Integration contour

B ~ 0 outsideI

I

rRadius

B field

Figure 5.8: Long coil.

Outside the coil, the flux lines spread out, and H becomes negligibly small

compared to inside the coil. Consequently, we can also ignore the horizontalsegment of the integral on the outside of the coil. Thus

∮H · dl ≈ H l

where H is the magnetic field inside the coil. The total current passing

through the integration contour is∫S J · dS = N I.

Thus Ampere’s law impliesH l ≈ NI

from which

H ≈ NI/l

and

B ≈ µ0NI/l

An additional point to note is that this result is independent of the exact

position of the lower horizontal segment of the contour within the coil. Thisimplies that the field intensity is uniform on a cross section of a long coil, andhence the flux through the cross section of such a coil is Ψ = Bπr2 = µ0Hπr

2

5-14 AJW, EEE3055F, UCT 2012

(i.e. linking one turn of the coil)

The inductance is the ratio of the total flux linking the coil to the current,being

L =

∫S B · dSI

=NΨ

I=N Bπr2

I≈ N (µ0NI/l)πr

2

I=πµ0r

2N2

l

5.4.4 Inductance of an Intermediate Length Multi-turn Coil

In cases where the coil can neither be considered very long or very short,

the following approximate formula is commonly used:

L ≈ πµ0r2N2

l + 0.9r

The formula incorporates an empirical correction factor (+0.9r) in the de-nominator.

5.5 Mutual Inductance

If two wire coils are close to one another, flux resulting from current flowing

in one coil will thread the coil of the other. Thus a changing current in onecoil, will result in a changing flux in the other and hence induce a voltageacross its terminals. This is the basis of a transformer.

I2(t)I1(t)

V1(t) V2(t)

ψ1 + ψ1from2ψ2 + ψ2from1

Figure 5.9: Coupled coils.

Consider two coils in close proximity, one containing N1 turns, and theother containing N2 turns. Although not shown in the sketch, these coils

are parts of circuits and carry currents.

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Let Ψ1(t) be the component of the flux threading coil 1, resulting purelyfrom the current I1(t) flowing in coil 1. Let the Ψ1 from 2 be the flux threading

coil 1 arising from the current I2(t) in coil 2. From Faraday’s law, the voltageacross the terminals of coil 1 is

V1(t) =d (N1Ψ1 +N1Ψ1 from 2)

dt= N1

dΨ1

dt+N1

dΨ1 from 2

dt

where N1Ψ1 is the total flux threading the multi-turn coiled surface. Since

Ψ1 is proportional to I1 and Ψ1 from 2 is proportional to I2, we have,

V1(t) = L1dI1dt

+M12dI2dt

where L1 is the self inductance constant of coil 1, and M12 is another con-

stant. Since N1dΨ1 from 2

dt =M12dI2dt ,

M12 =N1Ψ1 from 2

I2

Similarly, the voltage across the terminals of coil 2 is

V2(t) =d (N2Ψ2 +N2Ψ2 from 1)

dt= N2

dΨ2

dt+N2

dΨ2 from 1

dt

and

V2(t) = L2dI2dt

+M21dI1dt

where L2 is the self inductance constant of coil 2, and M21 is a constant:

M21 =N2Ψ2 from 1

I1

It can be shown (consult more detailed texts), that regardless of the ge-ometry, M12 = M21. The constant M = M12 = M21 must have the same

units of L1 and L2 being henrys, and is known as the “mutual inductance”between the coils5.5One could in principle, calculate M12 =

N1Ψ1 from 2

I2for a given coil geometry by doing a surface inte-

gration of B(i2) · dS to obtain Ψ1 from 2(I2) =∫S1

B(i2) · dS. B(i2) can be obtained directly from theBiot-Savart law (which requires a contour integration along coil 2). There is a better way to do it(consult other texts for details).

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Mutual Inductance for Tightly Coupled Coils

A special case is when the coils are tightly coupled, e.g. stacked on top

of one another such that Ψ1 from 2 = Ψ2 and Ψ2 from 1 = Ψ1 (or coils wouldaround a common toroid). For this case,

M12dI2dt

= N1dΨ1 from 2

dt= N1

dΨ2

dt= N1

[L2

N2

dI2dt

]

and hence M12 =N1

N2L2. Similarly, M21 =

N2

N1L1.

Since M =M12 =M21, the product M12M21 yields

M2 =N1

N2L2N2

N1L1 = L1L2

orM =

√L1L2

for tightly coupled coils.The voltage ratio is

V1V2

=L1

dI1dt +M dI2

dt

L2dI2dt +M dI1

dt

=N2

1L1turndI1dt +

√N2

1L1turnN22L1turn

dI2dt

N22L1turn

dI2dt

+√N2

1L1turnN22L1turn

dI1dt

=N1

N2

which is a well known result for tightly coupled coils.

5-17 AJW, EEE3055F, UCT 2012

5.6 Kirchhoff’s Voltage Law

The relationship between Kirchhoff’s law for a lumped element circuit model

and the physical component layout, is established by application of Fara-day’s law.

Consider applying Faraday’s law to the closed contour indicated by thedotted line in the following physical circuit representation. Define dS point-

ing into the page, which implies the integration direction is clockwise.

Coil

L

C

V(t)

R

0 3

3

2

0

1

1

I(t)

V10

2

V21

V03

V32

Figure 5.10: Series circuit loop - Faraday’s law is applied along the dotted line to deriveKirchhoff’s voltage law.

Since E → 0 in the wires, the voltage drops around the circuit occur

across the components. Thus, we can write

−∮

E·dl = −∫ 1

0

E·dl−∫ 2

1

E·dl−∫ 3

2

E·dl−∫ 0

3

E·dl =∫

S

∂B

∂t·dS =

dΨM

dt

or

V10 + V21 + V32 + V03 =dΨM

dtThe flux threading the loop can be split into three contributions:

ΨM = Ψapplied +Ψself +Ψmutual

where

5-18 AJW, EEE3055F, UCT 2012

• Ψapplied refers to any flux imposed on the circuit e.g. wave a bar magnetpast the circuit.

• Ψself refers to the flux generated by the current flowing in the circuitloop itself (the circuit can be thought of as a single turn inductor).Ψself = LselfI where Lself is the self inductance of the loop, which

carries current I.

• Ψmutual refers any leakage flux from other parts of the circuit (notably

the inductive element) that threads the loop.

Substituting the lumped element relationships derived above,

V (t)− IR− 1

C

∫ t

t0

I(t)dt− LdI

dt=dΨM

dt

• Kirchhoff’s law∑N

i=1 Vi = 0 describes the circuit model, and hence we

must introduce additional model component(s) into the circuit modelto account for the term dΨM

dt .

• It is noted that the term dΨM

dt will modify the current flowing in thecircuit, and should be included for accurate prediction of the behaviourof the circuit.

• In practice however, this term is usually small compared to the otherterms, and is often neglected in practical circuit design.

For the complete model shown below, the equation in the form of Kirchhoff’slaw is written as

V (t)− IR − 1

C

∫ t

t0

I(t)dt− LdI

dt− dΨapplied

dt− dΨself

dt− dΨmutual

dt= 0

orN∑

i=1

Vi = 0

5-19 AJW, EEE3055F, UCT 2012

(the direction of positive flux)

R

L

I

C

V (t)

dS points into the page

Lself

−dψapplied

dt

Figure 5.11: Circuit modified to incorporate an additional series inductor Lself which mod-els the series inductance of the loop, and an additional voltage source whichmodels unwanted external signals.

The termdΨself

dt = LselfdIdt resulting from the current in the loop, is mod-

elled by a (small) series inductance Lself . A feeling for the magnitude ofthis self inductance can be gained by considering a circuit arranged in a

circular loop of radius 10 cm. We previously calculated the self inductanceof a wire ring of radius 10 cm and wire radius of 0.5 mm to be 0.7 µH.

At an operating frequency of say 1 kHz, the AC reactance of this termis XL = 2πfLself ≈ 4 × 10−3 ohms, which is usually small enough to be

neglected from calculations. At higher frequencies, this term may becomesignificant.The term

dΨapplied

dtarises if the circuit is exposed to some externally gen-

erated AC field, e.g. a nearby transmitter like a cell phone, or perhaps amotor, or close-by transformer. Usually this term can be neglected. Of

course radio waves are ever present, but their contribution is usually in-

5-20 AJW, EEE3055F, UCT 2012

significant compared to the voltage signals of interest in the circuit. Higherfrequency fields are naturally suppressed by the series inductance of the cir-

cuit or intentional bandlimiting in the design of circuits with RF immunity.Circuits can also be shielded from external sources by placing them in ametal enclosure such as a Faraday cage6.

The term dΨmutual

dt in this context arises from the leakage flux from theinductor L, and is in practice usually small compared to the voltage drop

across the (multi-turn) inductor. The net effect may either be to increaseor decrease the current in the circuit, depending on the physical orientation

of the inductor.NOTE: The flux

∫S B ·dS requires the direction of dS to be defined. Since

the integral of E was taken clockwise around the loop, the right hand ruletells us that dS points into the page. The flux will be a positive quantity ifB (threading the loop) points into the page.

6A Faraday cage will provide good shielding from electric fields. DC or slowly varying magnetic fieldshowever do penetrate metal enclosures. e.g. the earth’s magnetic field is still detected by a magneticcompass within a Faraday cage. The degree of penetration of time-varying AC electromagnetic fieldsis a function of a frequency dependent parameter of the metal known as the “skin depth”, which willbe studied later in this course. For good shielding at a particular frequency, the enclosure wall shouldbe considerably thicker that the skin depth (at that frequency).

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5.7 Kirchhoff’s Current Law at a Node

Consider the illustration in Figure 5.12 showing four wires connecting to a

node carrying currents I1, I2, I3 and I4. Kirchhoff’s node current law statesthat the sum of all currents leaving the node equals zero, i.e.

N∑

i=1

Ii = I1 + I2 + I3 + I4 = 0

closed surface S2

Displacement current "flows" between plates

closed surface S1

I1

I2

I3

I4

n

dS

Figure 5.12: The relationship between Kirchhoffs cuurent law at a node the continuityequation.

Consider now the continuity of charge relationship∮

S

J · dS = − d

dt

∫

V

ρdV = −dQdt

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which states that the total conduction current leaving an arbitrary closedsurface S is equal to (minus) the rate of change of charge within the volume

V enclosed by S. One can re-express the continuity relationship in a formthat looks similar to Kirchhoff’s law by moving the charge term to the lefthand side: ∮

S

J · dS+d

dt

∫

V

ρdV = 0

The 2nd term can be expressed as a surface integral over S by substituting

Gauss’ law∫ρdV =

∮S D · dS,∮

S

J · dS+d

dt

∮

S

D · dS = 0

Moving the time derivative within the integral, the continuity equation be-comes ∮

S

J · dS+

∮

S

∂D

∂t· dS = 0

which says that the sum of the conduction current Ic and the displacementcurrent Id leaving an arbitrary closed surface7 is zero. I.e. for any closed

surface S,Ic + Id = 0

where

Ic =

∮

S

J · dS

and

Id =

∮

S

∂D

∂t· dS

Thus we have derived a generalised form of Kirchhoff’s current law, which

can be applied to an arbitrary closed surface.

For example, consider the closed surface S = S1 surrounding the node inFigure 5.12. There are N = 4 wires piercing the surface and joining at thenode.

7It is worth noting that the total displacement current flowing out of a closed surface is equal to the timerate of change of charge enclosed by the surface, i.e. Id =

∮S

∂D∂t

· dS = dQ/dt.

5-23 AJW, EEE3055F, UCT 2012

For S1, the total conduction current leaving the surface is

Ic =

∮

S1

J · dS = I1 + I2 + I3 + I4

The displacement current is typically insignificant (there is no significant

charge build up within S1), i.e.

Id =

∮

S1

∂D

∂t· dS =

dQ

dt≈ 0

Thus we haveI1 + I2 + I3 + I4 ≈ 0

If one shrinks surface S1 to a tiny surface surrounding the node, the dis-placement current shrinks to zero and the relationship converges exactly to

Kirchhoff’s law.

If however, we choose a surface S = S2 in such a way as to pass between the

plates of the capacitor as illustrated in Figure 5.12, then we have a slightlymore subtle situation.

As there is one less wire cutting the surface, the total conduction current is

Ic =

∮

S2

J · dS = I1 + I2 + I3

There is however a significant charge build-up on the plate(s) of the capac-

itor as a result of current I4. The charge Qplate on the plate (within S2)

builds up at a rate ofdQplate

dt = I4.

Thus the continuity relationship∮

S

J · dS = −dQdt

applied to surface S2 becomes

I1 + I2 + I3 ≈ −dQplate

dt= −I4

Rearranging, we get

I1 + I2 + I3 + I4 ≈ 0

5-24 AJW, EEE3055F, UCT 2012

which is consistent with the case where S = S1 and Kirchhoff’s law.The approximation (” ≈ ”) is present in the above expression because a

small (and negligible) charge will exist on the surface8 of the conductors 1to 4

Another way to look at the situation is to observe that the sum of allcurrents, both conduction and displacement current, flowing out of a closed

surface is zero, i.e. for surface S2,

I1 + I2 + I3 + Id = 0

where Id =∮S∂D∂t · dS is the displacement current leaving the surface. Id

is concentrated primarily between the plates of the capacitor (where the

electric field is strongest).

Also, since we have shown that dQdt =

∮S∂D∂t ·dS = Id (for any closed surface),

and if the only significant dQdt within S2 is the charge build up on the inner

plate of the capacitor due to I4, then

Id =dQ

dt≈ dQplate

dt= I4

which again for S2 implies

I1 + I2 + I3 + I4 ≈ 0

8Any excess charge must be the surface, because ρ→ 0 very rapidly inside a metal conductor - see sectionon relaxation time.

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5.8 The Relaxation Time of Conducting Materials

The term ‘conductor’ refers to a material that will carry current when sub-

jected to an electric field. In solid materials, like metals, electrons are freeto move, and the net movement of electrons constitutes a current. In liquids

(e.g. a salt solution), charged ions in solution are free to move allowing acurrent to exist. Insulating materials, in contrast, are materials for which

the electrons are tightly bound to particular atoms, and hence no currentcan flow.A “perfect conductor” is one for which there is an unlimited abundance

of free electrons. The conductivity of a perfect conductor is infinite - aninfinitesimal electric field will create a large current. Metals can often be

approximated as perfect conductors in the analysis of their behaviour undercertain conditions.

If a conducting object is placed in a stationary position within an electricfield, the electrons will, given time, rearrange themselves such that:

• E goes to zero inside the conductor (electrons quickly re-arrange them-selves until the total E = 0 inside conductor). Note: in the steady statesituation, the net force on the electrons must go to zero - the electrons

will rearrange themselves to achieve this. Since the conductor is notmoving (stationary), in the steady state situation, the magnetic force

on the electrons will be zero, and hence the electric force must also bezero9.

9If a conductor is moving through a static magnetic field - then the E field inside the metal can benon-zero - electrons will always rearrange themselves such that the sum of the magnetic and electricforces equals zero. For example, a rod of length l moving at velocity v through a static magnetic fieldB will experience a magnetic force on the electrons F = qv ×B. Electrons will re-arrange themselvessuch that the total force on an electron of chage q is qv×B+qE = 0, i.e. inside the metal, E = −v×B

once the electrons have rearranged themselves.

through uniformmagnetic field

Rod moving

Fields insidethe rod

Ev×B

v

B

There will also exist a potential difference between the end points of the rod, i.e. Φb − Φa =

−∫ b

aE · dl =

∫ b

a(v×B) · dl. If v is perpendicular to B and the rod is orientated such that its length is

perpendicular to v and B, then the potential difference between the ends will be vBl.

5-26 AJW, EEE3055F, UCT 2012

• The charge density ρ = 0 inside the conductor (since divD = ρ andE = 0, it means ρ = 0) .

• Any net charge (excess charge) resides on the surface in an infinitesi-mally thin layer (we refer to this charge as a ‘surface charge’).

• The potential Φ(x, y, z) is constant throughout the conductor (since

the electric field is zero inside).

• E is perpendicular at the boundary (i.e. no tangential component).

In practice, one might wonder just how long it would take for the electronsto re-arrange themselves. Imagine setting up an arrangement of charge

ρ(x, y, z) within a homogeneous conducting material and then releasing thecharge at some instant. The charge will redistribute itself such that the

electric field goes to zero at every point within the conductor10. This hap-pens very rapidly in metals, so fast that it can be considered instantaneousin many practical situations.

charges moveJ J = σE

J

J

ρ(x, y, z, t)

t

ρ0

ρ(t)

ρ(t) = ρ0e−

σ

εt

τ = εσ

0.37ρ0

Figure 5.13: The illustration shows an initial charge distribution within a homogeneousconducting medium. The charge will rearrange as time progresses, with thecharge density at any monitored point decaying over time.

Consider the charge within the conducting body. The movement of chargewill be governed by the continuity equation, for which the differential form

∇ · J = −∂ρ∂t

10Here we are ignoring the granularity of electrons and treating the charge as a kind of fluid.

5-27 AJW, EEE3055F, UCT 2012

describes the relationship between current leaving a small volume element,and the rate of change of charge within the element. If we substitute

J = σE, we get

∇ · (σE) = −∂ρ∂t

and then eliminate E via Gauss’ law (∇ · E = ρ/ε) we obtain a first orderdifferential equation

σ

ερ+

∂ρ

∂t= 0

which has solution

ρ(t) = ρ0e−σ

εt

where ρ0 is the initial charge density at time t = 0.

Thus the charge density at any point within the material will dissipateto zero with an exponential decay. The decay curve is characterised by thetime constant τ = ε

σ, also known as the relaxation time, which is the time

at which the charge density has reduced to e−1 ≡ 36.8% of its initial value.After 5τ the charge density will have decayed to less than 1% of the initial

value.To see how quickly this happens in practice, the time constant may be

calculated for various materials. For example for a metal conductor likecopper (σ = 5.8 × 107 Sm−1, ε ≈ ε0 = 8.85 × 10−12 Fm−1), the time

constant is τ = εσ = 1.5 × 10−19 s, which is extremely short compared to

say the period of a 100 GHz microwave sinusoid, being 10−11 seconds. Inelectronic circuits, the charge in the wires rearranges itself very quickly in

response to the dynamics of the circuits (i.e. to a very good approximation,we consider E ≈ 0 and ρ ≈ 0 inside the connecting copper wires - a small

component of E must however exist to drive the current).For a weakly conducting liquid like tap water (σ ≈ 10−2 Sm−1, ε ≈ 81ε0

Fm−1), the relaxation time is about 70 × 10−9 s. For a good insulator likeglass (e.g. σ = 10−14 Sm−1, ε = 5ε0), the relaxation time is calculated tobe about 4000 seconds (67 minutes).

Exercise: Calculate the relaxation time of iron (σ = 0.9× 107 Sm−1).

5-28 AJW, EEE3055F, UCT 2012

5.9 Shielding and The Faraday Cage

Circuitry may be shielded from external electric fields by enclosing the cir-

cuit inside a metal box known as a Faraday cage. External electric fieldshave no influence on the circuitry within a box made from a perfect con-

ductor - an electrically quiet zone exists within the box. On a larger scale,Faraday cages can provide protection against lightning strikes. “Low fre-

quency” magnetic fields can however penetrate a real metal enclosure andinfluence the circuitry inside it. Try for yourself to see if a permanent mag-net is able to attract iron pieces through the walls of a metal box. DC

magnetic fields penetrate through metal enclosures. For example, a mag-netic compass will still detect the earth’s magnetic field inside a Faraday

cage. Effective shielding from a magnetic field at 50 Hz requires a thickwall (several mm), preferably made from a high permeability material. As

the frequency increases, a metal wall becomes more effective in attenuatingmagnetic fields. In the MHz range and higher, metal enclosures are very

effective (if well sealed) for shielding circuits from external electromagneticfields, and also for preventing radiation leakage from the circuitry withinthe enclosure. Electromagnetic waves “reflect off” the enclosure, and what

does enter the metal, decays exponentially with a decay constant called the“skin depth” (covered later in the course).

5.10 Twisted Pair Cables

An interesting application of field theory concerns the understanding of howtwin-wire transmission lines are influenced by electric and magnetic fields.Figure 5.14 depicts a parallel wire (non twisted) transmission line, which

could be used to carry a signal from one location to another. Such parallelwire transmission lines are particularly susceptible to inductive coupling

of magnetic fields, especially when several signals need to be carried in thesame bundle. Changing magnetic flux dΨ/dt within the circuit loop induces

an additional voltage which adds to the signal voltage in accordance withFaraday’s law. A clever solution to minimising inductive coupling is toreduce the net flux, by twisting the pair of wires as illustrated. The flux

5-29 AJW, EEE3055F, UCT 2012

contributions B · dS in adjacent twists are opposite in polarity and willtend to cancel, resulting in reduced dΨ/dt and hence reduced magnetic

field interference. This type of cable is known as ‘twisted pair’ and is verycommonly used for data networks.

V (t)

V (t)

dψm/dt

net dψm/dt ≈ 0

Figure 5.14: Illustration comparing a straight wire transmission line with a twisted pairtransmission line.

In addition to the minimization of magnetic coupling, the twisting alsoimproves the immunity to capacitive coupling (an electric field effect). If,

for example, the cable lies close to a conductor that is varying in poten-tial relative to ground, like the ‘live’ wire wire 50 Hz mains supply, this 50Hz signal will capacitively couple to the conductors (imagine small valued

capacitors (C1 and C2 in Figure 5.15 and Figure 5.16) between the 50 Hzconductor and cable’s two wires). Twisting the cable, creates a more sym-

metrical coupling arrangement, independent of the of the orientation of thepair, hence causing the effect to be common to both wires. A differential

amplifier at the receiver with a high common-mode rejection ratio extractsthe desired differential signal, and removes the common capacitively cou-

pled interference. It is also evident in Figure 5.15 that a balanced drivecreates a symmetrical arrangement in which the interfering signal will bet-ter cancel at the receiver (if C1 ≈ C2 and the source resistors labeled R are

identical). Twisting the wires as in Figure 5.16, better matches the coupling

5-30 AJW, EEE3055F, UCT 2012

capacitance to each wire i.e. makes C1 ≈ C2.

Figure 5.15: Illustrations of capacitive coupling onto parallel wire transmission lines for thecase of balanced versus unbalanced driving circuitry.

Sometimes twisted pairs are also shielded (i.e. wrapped with an outerbraiding or foil sheath), offering increased immunity to electromagnetic in-terference and noise. The shielding also further reduces radiation from the

cable itself. Several twisted pairs are sometimes bundled within the samecable. The use of twisted pairs offers significantly lower cross talk between

data channels compared to non-twisted “side-by-side” wires within a cable.Unshielded twisted pair (UTP) cable is now used for connecting standard

PCs in in-door local area networks (LANs). UTP network cables replacedpreviously used 50 ohm coaxial cables for LANs because UTP cables are

cheaper to manufacture than coaxial cable, and offer adequate immunityto electromagnetic interference. UTP network cable is typically used for

5-31 AJW, EEE3055F, UCT 2012

Figure 5.16: Illustrations of capacitive coupling onto wire transmission lines for the caseof straight-wire versus twisted pair cables. The twisting of the wires makesthe capacitive coupling between each wire and the interfering source moreequal. The interference is common to both wires and will be canceled at thedifferential input of the receiver.

5-32 AJW, EEE3055F, UCT 2012

distances up 100m.The Cat-5e series cable is the cable commonly used for PC LANs (for both

100 Mbit/s and gigabit ethernet networks), and is designed to carry frequen-cies up to 100 MHz. PC LAN network cables contain four unshielded twistedpairs, with RJ-45 connectors on each end. The characteristic impedance of

Cat-5e is 100 ohms.Mechanical arrangement within the cable can further reduce coupling be-

tween pairs. For example the “Power Cat-6 four pair cable” sold by RSElectronics contains four unshielded twisted pair (UTP) cables, with a cen-

tral separator, and is designed to support high speed data transmissionsystems (frequencies up to 250 MHz).

Cat-5e Cat-6

RJ-45 connector

Figure 5.17: Photos from from RS website; connector from Intel website

5-33 AJW, EEE3055F, UCT 2012

6 Electromagnetic Waves

In this section we shall examine the generation and propagation charac-teristics of electromagnetic radiation as solutions of Maxwell’s differential

equations.One can divide the subject into six categories:

• Generation of radiation (for which we design antennas)

• Propagation through various media (e.g. free space and lossless media,

lossy conducting media)

• Reflection at interfaces between different media

• Refraction i.e. change of direction as a ray of light passes from onemedium to into another (consider a laser beam passing from air into a

prism)

• Diffraction (e.g. bending around corners, and slit diffraction effects)

• Scattering from objects (e.g. radars detect the energy scattered fromtargets)

Maxwell’s Differential Equations

The complete set of Maxwell’s Equations in differential form is

∇ ·D = ρ

∇ ·B = 0

∇×E = −∂B∂t

∇×H = J+∂D

∂t

6-1

where D = εE and B = µH.It is pointed out that this set, together with the Lorentz force law

F = qE+ qV ×B

summarize all the behaviour of classical electromagnetic field theory in both

matter and free space.

• All other laws e.g. Biot-Savart law and the continuity equation can be

derived from this set.

• The only place where these differential forms do not hold are on bound-

aries between different materials where the derivatives are not finite.At boundaries, we can derive a set of “boundary conditions”which re-

late the field quantities immediately on either side of the boundary (seesection on boundary conditions).

6.1 Mathematical Description of Travelling Waves

(ref S.Cloude 1995)

Consider the signal received at a distance z metres from a radiating source.If the signal progates at speed of v metres per second then the time delay

will be z/v seconds.In one dimension, a waveform propagating at a fixed speed v [ms−1] in

the positive z-direction, can be described by a function of time and positionf(z, t) of the form

f1(t−z

v)

where zv is the delay (time shift). A waveform travelling in the negative

z-direction can be expressed in the form

f2(t+z

v)

See hand drawn illustrations of plots of f1(t− zv), plotted

1. as a function of time for different, fixed, values of z.

2. as a function of z at different time instants.

6-2 AJW, EEE3055F, UCT 2012

Figure 6.1: Illustration of wave propagation.

6-3 AJW, EEE3055F, UCT 2012

It is perhaps easiest to see why f1(t− zv) represents a forward propagating

wave by re-expressing the argument as

f1(−1

v(z − vt))

and defining a new function w1(.) such that

w1(z − vt) = f1(−1

v(z − vt))

The function w1(.) is a ‘flipped’ and stretched version of f(.), i.e. w1(u) =

f(−uv ).

You should recognise that in w1(z − vt), the quantity “vt” represents a(time varying) shift along the z-axis. As time increases, the waveform w(z−vt) will therefore slide along the z − axis in the positive z direction, at arate of v metres per second.

Similarly, the function f2(t+zv) can be re-expressed as

f2(1

v(z + vt)) = w2(z + vt)

which reveals that waveform w2(.) will move in the negative z direction as

vt increases.

6.1.1 The Wave Equation

Waves travelling in the ±z directions are solutions to a 2nd order partialdifferential equation known as the “wave equation” of the form

∂2f

∂z2=

1

v2∂2f

∂t2

or∂2f

∂z2− 1

v2∂2f

∂t2= 0

A more compact notation uses subscripts to indicate the derivatives, i.e.

fzz =1

v2ftt

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This equation arises in many physical contexts e.g. propagation of soundwaves in air and fluids, electromagnetic waves in space, transmission line

theory.The wave equation can be generalized to model propagation in 3-D space

and also to include a driving term that represents the source of the radiation.

The generalized form in Cartesian coordinates is(∂2f

∂x2+∂2f

∂y2+∂2f

∂z2

)− 1

v2∂2f

∂t2= g(r, t)

where r =< x, y, z > represents position in space, and g(r, t) is the generalsource term creating the waves. Away from the souce, g(r, t) = 0.The source term g(r, t) models the underlying radiation generation mech-

anism e.g. a loudspeaker in acoustics, or an antenna in electromagneticradiation.

The 3-d wave equation may be more compactly written as

∇2f − 1

v2∂2f

∂t2= g(r, t)

where ∇2 ≡ ∂2

∂x2 +∂2

∂y2 +∂2

∂z2 is the Laplacian operator.

d’Alebert’s solution to the 1-D wave equation

Solutions to the wave equation are propagating waves. It is easy to verify

that f1(t− zv) and f2(t+

zv) (or equivalently w1(z − vt) and w2(z + vt)) are

solutions of the 1-D wave equation:

∂2f

∂z2− 1

v2∂2f

∂t2= 0

The solution known as d’Alembert’s solution is

f(z, t) = f1(t−z

v) + f2(t+

z

v)

being the sum of forward and reverse propagating waves.

As an exercise, it is worth verifying the solution by expanding out the2nd partial derivatives with respect to z and t and substiting into the wave

equation (do this yourself):

6-5 AJW, EEE3055F, UCT 2012

∂f

∂z=

−1

vf

′

1(t−z

v) +

1

vf

′

2(t+z

v)

∂2f

∂z2=

1

v2f

′′

1 (t−z

v) +

1

v2f

′′

2 (t+z

v)

and

∂f

∂t= f

′

1(t−z

v) + f

′

2(t+z

v)

∂2f

∂t2= f

′′

1 (t−z

v) + f

′′

2 (t+z

v)

Clearly ∂2f∂z2 = 1

v2∂2f∂2t and f(z, t) is a solution to the wave equation.

In any physical problem, the solutions are constrained by the so called“boundary conditions”. In the case of electromagnetic radiation, these arethe constraints on E,D,B and H at interfaces between different media, and

the constraint on the field quantities at the source of the radiation.

6.2 Wave Equation derived from Maxwell’s Equations

In this section, we shall see how the wave equation may be derived directly

from the differential forms of Maxwell’s equations. We shall also see thatthe source of electromagnetic radiation is the acceleration of electric charge.We must show that the field quantities described by Maxwell’s equations

can be manipulated into a standard form of the wave equation

∇2f − 1

v2∂2f

∂t2= g

where g ≡ g(r, t), and f ≡ f(r, t).

Following the approach described in (S.Cloude 1995), we can find an ex-pression for the electric field as follows:

• Take the curl of Faraday’s law on both sides, i.e.

∇× E = −∂B∂t

6-6 AJW, EEE3055F, UCT 2012

becomes

∇× (∇× E) = ∇× (−∂B∂t

)

Making use of the interchangeability of the partial derivatives

∇×∇×E = − ∂

∂t(∇×B)

• To obtain an equation involving only E and not B, we can substitutefor ∇×B which we obtain from Ampere’s law,

∇×H = J+∂D

∂t

with D = εE and B = µH, i.e.

∇×B = µJ+ εµ∂E

∂t

Substituting for ∇×B we get

∇×∇×E = − ∂

∂t(µJ+ εµ

∂E

∂t) = −εµ∂

2E

∂t2− µ

∂J

∂t

• Note that the right hand side looks a bit like the wave equation.

• To make further progress, we make use of a vector identity (proofs maybe found in texts on vector analysis),

∇×∇× E = ∇(∇ · E)−∇2E

where ∇2E ≡ ∇ · ∇E =

∇2Ex

∇2Ey

∇2Ez

, and ∇2Ex =

∂2Ex

∂x2 + ∂2Ex

∂y2 + ∂2Ex

∂z2

• Lastly, we use Gauss’ law

∇ ·E =ρ

ε

to write the 1st term on the right hand side as

∇(∇ · E) = ∇ρ

ε= grad(

ρ

ε)

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In a charge free region ρ = 0 (e.g. a vacuum, or air, or a dielectricmaterial), grad(ρε) = 0, and the equation simplifies further to

−∇2E = −εµ∂2E

∂t2− µ

∂J

∂t

or

∇2E− εµ∂2E

∂t2= µ

∂J

∂t

This is a vector form of the wave equation. For each Cartesian coordinate

∇2Ex − εµ∂2Ex

∂t2= µ

∂Jx∂t

∇2Ey − εµ∂2Ey

∂t2= µ

∂Jy∂t

∇2Ez − εµ∂2Ez

∂t2= µ

∂Jz∂t

NOTE:

• In free space, J = 0, and so all three components of E are of theform

∇2Ex − εµ∂2Ex

∂t2= 0

• If we compare this equation to the standard form wave equation

∇2f − 1

v2∂2f

∂t2= 0

we identify 1v2 ≡ εµ, and hence we conclude that electromagnetic waves

must propagate with velocity

v =1√ε0µ0

=1√

8.85× 10−12 4π × 10−7= 2.998× 108 ms−1

This is the same as the measured value for light, and hence sug-gests that light is an electromagnetic wave - Maxwell’s great revelation

[around 1865]. At the time of Maxwell, this was an amazing result.The constants ε0 and µ0 were purely the results of electrostatic and

6-8 AJW, EEE3055F, UCT 2012

magnetostatic theory and experiments, and had not been associatedwith light. Maxwell also predicted the existance of EM waves at all

frequencies including radio waves, first demonstrated experimentallyby Hertz in 1888.

• In other non-conducting media (e.g. dielectrics), the speed of

propagation is

v =1√εµ

=1√

εrε0µrµ0=

1√εrµr

1√ε0µ0

=c√εrµr

Since, for most materials µ ≈ µ0, the speed of propagation is v ≈ c√εr.

The term n =√εrµr is called the “refractive index” of the material.

• Although we have derived the wave involving E, in exactly the same

way, we can derive a differential equation for B. In a non conductingmedium, the equation is

∇2B− εµ∂2B

∂t2= 0

identical in form to that of the electric field.

6.3 Physical Interpretation of the Radiation Generation

and Propagation Mechanism

By comparing the derived differential equation involving E (or its compo-nents Ex, Ey or Ez) with the standard form(s) of the wave equation, the

source of radiation is clearly identified as the term g(r, t) = µ∂J∂t . Thus wecan conclude that the source of electromagnetic radiation is a time-varying

current (density). A constant current implies charge moving at a constantspeed. The time-derivative of a current corresponds to the acceleration of

charge1.

Thus radiation arises from accelerating charges.

1The current density J resulting from current I passing through a small area ∆S (∆S in a plane perpen-

dicular to the direction of movement), is J = I∆S

∝ ∂x∂t. Thus ∂J

∂t∝ ∂2x

∂t2

6-9 AJW, EEE3055F, UCT 2012

6.3.1 Propagation Mechanism

To understand how the radiation propagates, examine the illustrations of

the electromagnetic wave propagating aways from a dipole radiating source.

SEE:

• (1) HANDOUT OF ILLUSTRATIONS of EM Propagation [from bookon Antennas by J.D. Kraus]

• (2) Simulations of EM propagating waves via links on the EEE3055F

course website.

Figure 6.2: a) Side view illustration of the E field propagating away from a dipole antennadriven by a sinusoidal source. b) Top view illustration of the correspondingB field propagating away from the dipole antanna. [images obtained fromsimulation made available by Hsiu C Han of Iowa State University]

A short dipole anntena radiator can be constructed from two bits of wire.

Driving the dipole with a sinusoidal voltage source sets up time varyingcurrents in the two arms. The time varying current sets up a time-varying

electromagnetic field that propagates away from the source. A cross sectionshowing the E field at a particular time instant can be seen in Figure 6.2.

6-10 AJW, EEE3055F, UCT 2012

This disturbance propagates away from the accelerating charge governedby two equations:

1. Faraday’s law:

∇× Et = −∂B∂t

or (by Stokes’s theorem)∮

Et · dl = −∫∂B

∂t· dS

2. Maxwell’s equation (from Ampere’s law):

∇×Bt = εµ∂E

∂t

or (by Stokes’s theorem)∮

Bt · dl = εµ

∫∂E

∂t· dS

The radiation pattern can be used to depict the power density (W/m2) ina polar format in a particular direction. For an accelerating charge, a crosssection of the pattern is shown below - the distance from the origin at a

particular angle represents the power density radiated in that direction.

Radiation pattern from accelerating charge∂J∂t

PowerDensity(θ)θ

Figure 6.3: Radiation pattern from an accelerating charge.

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6.4 Some Additional Notes on Wave Equations

There are many forms of the wave equation, describing wave propagation

in 1, 2 and 3 dimensions, in Cartesian, cylindrical and spherical coordi-nates, and in scalar as well as compact vector notations. The general form

describing some scalar field quantity is

∇2f − 1

v2∂2f

∂t2= g(r, t)

where f = f((r, t)) is a function of position in space and time. The function

g(r, t) is used to model a radiation source. The constant v is the velocity ofpropagation.

Away from the source, g(r, t) = 0 and the wave equation is written as

∇2f − 1

v2∂2f

∂t2= 0

The term ∇2f is called the Laplacian of f.

In 1-D spatial dimension denoted by z, ∇2f = ∂2f∂z2 and the equation is

written as∂2f

∂z2− 1

v2∂2f

∂t2= g(z, t)

It is easy to show (by differentiation) that a solution to the 1-D wave equa-tion is of the form

f(z, t) = f1(t−z

v) + f2(t+

z

v)

or alternatively written as

f(z, t) = f1(−1

v(z − vt)) + f2(

1

v(z + vt))

= w1(z − vt) + w2(z + vt)

where w1(u) = f1(−vu) and w2(u) = f2(vu).The functions f1 and f2 (or alternatively w1 and w2) represent waves

travelling in the positive and negative z directions. This solution is knownas d’Alembert’s solution. It can be proved that this f(z, t) is a complete

solution to the 1-D wave equation.

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The exact forms of the functions f1 and f2 are only known once a problemhas been more fully specified i.e. specification of (1) the initial conditions

and (2) the boundary conditions (an example of a 1-D problem involving aplucked string follows).In 3-D, the Laplacian can be expanded in Cartesian coordinates to

yield (∂2f

∂x2+∂2f

∂y2+∂2f

∂z2

)− 1

v2∂2f

∂t2= g(r, t)

which has solutions of the form

f(x, y, z, t) = f1(ax+ by + cz − vt) + f2(ax+ by + cz + vt)

which can be verified by differentiation. This form is used for modelling

problems with rectangular symmetry e.g. a plane wave propagating in space.The values of the constants define the direction of propagation of the wave.

Waves can propagate simultaneously in several directions, i.e. f(x, y, z, t)can be the superposition of one or more waves propagating in different

directions.The wave equation can also be expressed in spherical coordinates,

which is convenient form modelling waves with spherical symmetry e.g.

spherical waves radiating away from a source. The Laplacian in sphericalcoordinates is

∇2f =1

r2∂

∂r

(r2∂f

∂r

)+

1

r2sinθ

∂

∂θ

(sinθ

∂f

∂θ

)+

1

r2sin2θ

∂2f

∂Φ2

The wave wave equation becomes unwieldy to write, and so the compactnotation

∇2f − 1

v2∂2f

∂t2= g(r, t)

is very convenient. If the wave amplitude depends only on the distance fromthe source (e.g. radiation from a point source), ∂

∂θ≡ 0 and ∂

∂Φ≡ 0 and the

Laplacian simplifies to

∇2f =1

r2∂

∂r

(r2∂f

∂r

)+ 0 + 0 =

∂2f

∂r2+

2

r

∂f

∂r

6-13 AJW, EEE3055F, UCT 2012

and the wave equation in spherical coordinates simplifies to(∂2f

∂r2+

2

r

∂f

∂r

)− 1

v2∂2f

∂t2= 0

which has a general solution the form

f(r, t) =1

rf1(r − vt) +

1

rf2(r + vt)

where the first term represents a spherical wave travelling radially outwardfrom the source, with a decaying amplitude proportional to 1

r ; the 2nd termrepresents a wave converging on the origin. This solution can be verified by

substitution.A 2-D example of a ‘spherical’ wave, is the wave front propagating out on

the surface of water when a stone is dropped into onto the surface.Sufficiently far from a radiating antenna, the amplitude of the electric and

magnetic fields decay as a function of 1r , like a spherical wave. The radiated

power density 2 decays proportional to 1r2 . Unlike a spherical wave, practical

antennas focus the radiated energy in particular directions, like the beam

of a torch.

6.5 Travelling Waves in a Plucked Guitar String

(references: Wylie & Barrett, and also Griffiths)A simple example of the wave propagation is found in the analysis of a

plucked guitar string, clamped at x = 0 and x = L as shown in Figure 6.4.

(transverse displacement)

xx = Lx = 0

f(x|t)

Figure 6.4: A plot of the displacement of the string as a function of position x (at somegiven instant t in time).

2As we shall see in a later section, the power density vector P = E×H points in the direction of powerflow, and far from a source, the power density |P| ∝ 1

r2(in the so-called “far field”).

6-14 AJW, EEE3055F, UCT 2012

The function f(x, t) models the transverse displacement of the string asa function of position x and time t, and turns out to be the solution of a

1-D wave equation. The wave equation is derived by considering the forcesacting on a small elemental section of the string shown in Figure 6.5.

Horizontal and transversecomponents of

Displacement

∆x

xT1

θ2T2

force

Restoring

x1

f(x|t)

Elemental section of string

T2

θ2θ1

T2sinθ2T2

T2cosθ2

Figure 6.5: Forces on an element of the string; the magnitudes of the angles are exaggerated.

We shall make the following assumptions:

• the transverse displacement of the string is small (typically mm) com-pared to its length (normally 65 cm).

• the resulting small increase in string length has negligible increase intension, i.e. tension T , in Newtons, is approximately constant i.e. T1 =

T2 = T . The tension in a real guitar string is in the range 70 N to 180 N.

• the horizontal displacement of the elemental section is negligibly small.

• the tension is high enough such that gravity can be neglected i.e. string

does not sag under its own weight.

• air resistance is also negligible - although it would in practice contributeto bringing a sting in motion to rest over time. Energy is lost as it is

converted into an audible sound wave.

• the string has a uniformly distributed mass of w kg m−1 (typically

0.4× 10−3 to 8× 10−3 kg m−1).

• for any segment considered, the angles θ1 and θ2 are small enough suchthat we can make the small angle approximations sinθ1 ≈ tanθ1 ≈ θ1and sinθ2 ≈ tanθ2 ≈ θ2.

6-15 AJW, EEE3055F, UCT 2012

The elemental section experiences a force T1 = T to the“left”directed alongthe string, and a force T2 = T directed to the “right” along the string as

illustrated. The horizontal force components are approximately equal, i.e.T2cosθ2 ≈ T1cosθ1. The net transverse force on the elemental section is thedifference in the transverse components, i.e.

F = T 2sinθ2 − T1sinθ1 = T (sinθ2 − T1sinθ1)

≈ T (tanθ2 − tanθ1) = T (∂f

∂x

∣∣∣∣x1+∆x

− ∂f

∂x

∣∣∣∣x1

)

For a positive lateral displacement (i.e. f(x1|t) > 0) the value of F will benegative, indicating a restoring force in the direction of string’s rest position

being the x-axis.According to Newton’s 2nd law, the force equals mass times acceleration.

For the small section of length ∆l, its mass m = w∆l ≈ w∆x since θ1 and

θ2 are small angles. Thus Newton’s 2nd law implies

F = T (∂f

∂x

∣∣∣∣x1+∆x

− ∂f

∂x

∣∣∣∣x1

) = w∆x∂2f

∂t2

or( ∂f∂x

∣∣∣x1+∆x

− ∂f∂x

∣∣∣x1)

∆x=w

T

∂2f

∂t2

Taking the limit ∆x → 0, the left hand side becomes ∂∂x(

∂f∂x) =

∂2f∂x2 , being

the second partial derivative evaluated at position x = x1. Thus we have

derived the standard form of the 1-D wave equation

∂2f

∂x2=w

T

∂2f

∂t2

which describes the lateral displacement of the string at any position x andtime t between its clamped endpoints. The propagation speed of the wave

(in the +x or −x direction) is identified as v =√

Tw .

How do we obtain a solution? d’Alembert’s solution, we know is a com-

plete solution of the form

f(x, t) = f1(x− vt) + f2(x+ vt)

6-16 AJW, EEE3055F, UCT 2012

where f1() and f2() to be determined from the initial and boundary condi-tions. In this problem, the initial conditions are (i) the shape of the string

at time t = 0 when it is released, i.e. f(x, 0) = f(x, t)|t=0 = U0(x), and (ii)

the initial transverse velocity ∂f(x,t)∂t

∣∣∣t=0

= V0(x) over the length.

From these initial conditions, we have two equations

U0(x) = f1(x) + f2(x)

and

V0(x) =∂f(x, t)

∂t

∣∣∣∣t=0

= −vf ′1(x) + vf ′

2(t)

Differentiating the first equation, U ′0(x) = f ′

1(x) + f ′2(x) and substituting

f ′2(x) = U ′

0(x)− f ′1(x) into the second, we get

V0(x) = −vf ′1(x) + v(U ′

0(x)− f ′1(x)) = −2vf ′

1(x) + vU ′0(x)

Rearranging

f ′1(x) =

1

2U ′0(x)−

1

2vV0(x)

and by integration we get

f1(x) =1

2U0(x)−

1

2v

∫ x

0

V0(u)du− k

where k is a constant, and

f2(x) = U0(x)− f1(x) =1

2U0(x) +

1

2v

∫ x

0

V0(u)du+ k

If we assume that the string is stationary at the time of release, V0(x) = 0;the integral terms are zero, and f(x, t) simplifies to

f(x, t) = f1(x− vt) + f2(x+ vt)

=1

2U0(x− vt) +

1

2U0(x+ vt)

The plucked string is the shown below as a ‘triangular’ initial displacement3

at t = 0. The time-varying solution shows that f(x, t) is the sum of two

6-17 AJW, EEE3055F, UCT 2012

Initial displacement

Wave solution to a plucked string

No longer overlapping

Waves moving apart

at t=0

Waves invert on reflection at clamped end points.

U0(x+ vt)/2 U0(x− vt)/2

U0(x)/2

U0(x)

Figure 6.6: The illustration shows an initial condition U0(x). Releasing the string at t = 0sets up two propagating waves that reflect off the clamped ends of the string.There is an inversion on reflection (similar to an voltage wave travelling downa transmission line and reflecting off a short circuit termination).

6-18 AJW, EEE3055F, UCT 2012

waves, each being 12U0( ), one propagating to the left and the other to the

right.

If a finite length string is considered and the boundary conditions areapplied (i.e. f(0, t) = 0 and f(L, t) = 0), then the full solution which canbe derived which shows that the waves will bounce back and forth off the

end points, reversing polarity at each bounce.The time taken for the string to return to its initial state is the period

2L/v seconds - each pulse must reflect twice to return to its original state,travelling a distance of 2L. The frequency of the lateral oscillation is there-

fore

f =1

period=

v

2L=

1

2L

√T

w

which is the fundamental frequency of sound that one hears. The vibration

of the string creates a compression wave in the air that propagates to one’sears. Higher frequencies are also heard at integer multiples of the funda-

mental. If one plots the lateral displacement of a point on the string as afunction of time, it will be a periodic function. Fourier analysis will reveal

the amplitude of the harmonic components.

Example

A guitar string has the following parameters: T = 100N, w = 2×10−3kg m−1

and L = 0.65m. The speed of wave propagation is v =√

Tw = 224ms−1.

The fundamental frequency is

f =v

2L=

224

2 ∗ 0.65 = 172Hz

Tightning the string will increase speed of propagation and hence the fre-quency.

3One could imagine creating such an initial displacement using three drawing pins that are simultaneouslyremoved at time t = 0. A more realistic initial condition for a guitar string is a “pluck”made by a singlefinger.

6-19 AJW, EEE3055F, UCT 2012

7 Sinusoidal Waves

7.1 Signals as Sums of Sinusoidal Functions

Analysis of a waveform can often be greatly simplified by representing thewaveform as the sum of functions that are more easily analysed that the

waveform itself. Examples of such representations for general signals arethe Fourier series (for periodic waveforms)

f(t) =

∞∑

n=1

Cncos(nω0t+ θn)

the complex Fourier series (for periodic waveforms)

f(t) =∞∑

n=−∞Fnexp(jnω0t)

and the (inverse) Fourier transform for non-periodic waveforms, i.e.

f(t) =1

2π

∫ ∞

−∞F (ω)exp(jωt)dω

Well-known formulas exist for calculating the weightings of the sinusoids inthe above representations, i.e. Cn (and phase θn), Fn and F (ω).

If, for example, one wishes to predict how a transmitted pulse propagatesto a receiver through a complex medium (the response of which may befrequency dependent), one can carry out the analysis for a single frequency

sinusoid, and having done so, build the final output as the superposition ofall the sinusoidal components. One could in fact represent the relationship

between transmitter and receiver in terms of a frequency dependent transferfunction H(ω), for which

Vrec(ω) = H(ω)Vtran(ω)

7-1

where Vrec(ω) andVtran(ω) are the Fourier transforms of the transmittedsignal Vtran(t) and received signal Vrec(t). The time domain waveform Vrec(t)

found by inverse transforming Vrec(ω).This motivates the detailed study of sinusoidal solutions to Maxwell’s

equations.

7.2 Real Sinusoidal Travelling Waves

Any function f(t) may be transformed into a travelling wave moving inthe x direction at speed v by replacing the argument t by (t − x

v ), i.e.

f(t − xv ). A forward travelling sinusoidal wave f(x, t) can be constructed

from cos(ωt+Ψ) simply by replacing the variable t by (t− xv ), i.e.

f(x, t) = A cos(ω(t− x

v) + Ψ)

= A cos(−ωv(x− vt) + Ψ)

The function is also sometimes written as

f(x, t) = A cos(ωt− ωx

v+Ψ) = A cos(ωt− kx+Ψ)

where ω is the angular frequency in radians per second, and the constantk = ω/v is called the wave number and is the spatial frequency in radians

per metre, and Ψ is a constant phase shift.The waveform may be sketched as a function of x at a fixed time, or as a

function of t at a fixed position x.

Below are shown sketches of the function cos(ωt) and the travelling wave

A cos(ωt− kx+Ψ) = A cos(ω[t− kx−Ψ

ω])

as a function of time t at a fixed position x.

7-2 AJW, EEE3055F, UCT 2012

cos(ωt)

as a function of t

cos(ωt− kx+ ψ)

as a function of t kx−ψω

shift

period T = 2πω

t

t

Figure 7.1:

It is pointed out that A cos(ωt) is shifted to the right by a time kx−Ψω

=

x/v −Ψ/ω.

• The period T of the waveform is the time over which the phase changes

by 2π, i.e. solving ω(t+ T )− ωt = 2π yields

T =2π

ω

A sketch of cos(kx) and the wave as a function of x at a fixed time t areshown below:

7-3 AJW, EEE3055F, UCT 2012

x

cos(kx)

cos(ωt− kx+ ψ)

x

wavelength λ = 2πk

as a function of x

as a function of x

shiftωt+ψk

Figure 7.2:

The wave A cos(ωt− kx+Ψ) = A cos(k[−x+ ωt+Ψk ]) which is A cos(−kx)

shifted to the right by a distance ωt+Ψk = vt+Ψ/k

• The wavelength λ is defined as the distance between consecutive points

of identical phase, i.e. the distance over which the phase changes by2π, i.e. solving k(x+ λ)− kx = 2π yields

λ =2π

k=

2π

ω/v=

v

ω/2π=

v [ms−1]

frequency [Hz]

The complete sinusoidal solution to the 1-D wave equation is the sum of

forward and reverse travelling waves of the form:

f(x, t) = A1 cos(ωt− kx+Ψ1) + A2 cos(ωt+ kx+Ψ2)

where the negative travelling wave is created by replacing x with −x.

7.3 Complex Phasor Representation

Complex exponentials are often used in the analysis of linear system for anumber of reasons:

• Complex exponential function are the basis functions used in the com-

plex Fourier representations of signals.

7-4 AJW, EEE3055F, UCT 2012

• Mathematical notation and analysis can be simpler (e.g. in circuitanalysis)

• The magnitude and phase shift changes that occur to a real sinusoid

when passed through a linear system, can be obtained directly fromanalysis of the response to a complex sinusoid.

The real function

f(x, t) = A1 cos(ωt− kx+Ψ1) + A2 cos(ωt+ kx+Ψ2)

is represented by the complex exponential form

f(x, t) = A1exp [j(ωt− kx+Ψ1)] + A2exp [j(ωt+ kx+Ψ2)]

It is noted that

f(x, t) = Ref(x, t)If we pass a real signal through a physical linear system, the output is aconvolution, i.e.

f(t)⊗ h(t)

where h(t) is the (real) impulse response of the system.Passing the complex signal f(t) through the same linear system we get

f(t)⊗ h(t) =[Ref(t)+ jImf(t)

]⊗ h(t)

= Ref(t) ⊗ h(t) + jImf(t) ⊗ h(t)

= f(t)⊗ h(t) + jImf(t) ⊗ h(t)

We see thatRef(t)⊗ h(t) = f(t)⊗ h(t)

from which it is clear that we can perform the analysis using the complex

function f(t) and simply take the real part afterwards to obtain f(t)⊗h(t).The complex function f(x, t) = A1exp [j(ωt− kx+Ψ1)] is sometimes

written more compactly as

f(x, t) = A1ejΨ1e−jkxejωt = Ae−jkxejωt

where the factor A1ejΨ1 has been replaced by a single complex constant

A = A1ejΨ1 . It is also common in wave analysis to drop the sinusoidal

7-5 AJW, EEE3055F, UCT 2012

component ejωt in the written notation, and write the forward and reversetravelling waves as

A1e−jkx + A2e

jkx

The term A1e−kx is called a complex phasor representation of the (forward)

wave.

The derivative of a complex sinusoid is

d

dtA1exp [j(ωt− kx+Ψ1)] = A1exp [j(ωt− kx+Ψ1)] jω

Thus time derivatives of complex phasors are simply obtained by multi-

plying by jω. Integration with respect to time is achieved by dividing byjω.

7.3.1 Maxwell’s Equations in Complex Phasor Notation

If all time varying quantities are sinusoidal, the set of equations

∇ ·D = ρ

∇ ·B = 0

∇×E = −∂B∂t

∇×H = J+∂D

∂tcan be written in phasor notation as

∇ · D = ρ

∇ · B = 0

∇× E = −jωB∇× H = J+ jωD

where ρ(~r, t) = ρ(~r)ejωt, E(~r, t) = E(~r)ejωt, B(~r, t) = B(~r)ejωt, etc. Sincethe ejωt factors cancel on left and right hand sides, the phasor form of

Maxwell’s equations relate the non-time dependent portions of ρ, E and B.

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7.3.2 The Wave Equations in Complex Phasor Notation

The wave equations (derived for a non-conducting dielectric medium)

∇2E− εµ∂2E

∂t2= 0

∇2B− εµ∂2B

∂t2= 0

are written in phasor notation as

∇2E+ ω2εµE = 0

∇2B + ω2εµB = 0

since ∂2

∂t2≡ (jω)(jω) = −ω2. It is noted that in phasor form, the wave

equations are purely a function of spatial coordinates and not time (theejωt terms cancel). We shall use the phasor notation in studying wave

propagation in lossy conducting media.

7-7 AJW, EEE3055F, UCT 2012

8 Plane Waves

The 3-D wave equations for E and B in free space in vector notation are

∇2E− εµ∂2E

∂t2= 0

∇2B− εµ∂2B

∂t2= 0

Expanding the components of E in Cartesian coordinates,

∂2Ex

∂x2+∂2Ex

∂y2+∂2Ex

∂z2− εµ

∂2Ex

∂t2= 0

∂2Ey

∂x2+∂2Ey

∂y2+∂2Ey

∂z2− εµ

∂2Ey

∂t2= 0

∂2Ez

∂x2+∂2Ez

∂y2+∂2Ez

∂z2− εµ

∂2Ez

∂t2= 0

we see that each (scalar) component of the E vector obeys a 3-D wave

equation; and similarly for the components of B. Functions of the formf(x, y, z, t) = fn(ax+ by+ cz− vt) are solutions, and represent waves trav-

elling in a particular direction.A plane wave is a propagating wave for which the field is uniform in any

plane perpendicular to the direction of propagation. This kind of field is anidealization, as the fields radiated from sources spread out as a function of

distance with a curved wavefront (see handout of E field lines propagatingaway from a dipole antenna). Far from the source, the wavefront may beanalysed locally as a plane wave (by locally, is meant a region over which

the curvature of the field lines is negligible).

8-1

8.1 Plane wave propagating in z direction

To simplify analysis and understanding, let us consider a wave front prop-

agating in the positive z-direction. Since the components of a plane waveare by definition uniform within any plane perpendicular to the direction

of propagation, we conclude that E is only a function of z, independent ofx and y. Thus ∂

∂x = 0 and ∂∂y = 0 and the 3-D wave equations in Cartesian

coordinates simplify to 1-D wave equations

∂2Ex

∂z2− εµ

∂2Ex

∂t2= 0

∂2Ey

∂z2− εµ

∂2Ey

∂t2= 0

∂2Ez

∂z2− εµ

∂2Ez

∂t2= 0

Additionally, we shall assume that only an Ex component exists, and thatthe Ey and Ez components of the E vector are zero.1i.e. E = 〈Ex(z, t), 0, 0〉.Now we need only consider a single equation

∂2Ex

∂z2− εµ

∂2Ex

∂t2= 0

d’Alembert’s complete solution is of the form

Ex(z, t) = f1(z − vt) + f2(z + vt),

with v = 1√εµ.

1The orientation of E field depends on the radiating source. If a wire dipole antenna is placed at theorigin and aligned with the x axis, then in the y − z plane, the electric field will have only an Ex

component. Away from the y − z plane, other components are non-zero, because of the curvature ofthe E-field lines being closed loops.

8-2 AJW, EEE3055F, UCT 2012

Electromagnetic propagation of a short pulse

E(z, t) = 〈Ex(z, t), 0, 0〉

y

z

c

c

H(z, t)

x

Figure 8.1: Propagation of a short EM pulse.

What about the magnetic field?

The magnetic field is coupled to the electric field in Maxwell’s equations,

particularly,

∇×E = −∂B∂t

Expanding the curl of E in Cartesian coordinates,

∇×E =

∣∣∣∣∣∣

x y z∂∂x

∂∂y

∂∂z

Ex Ey Ez

∣∣∣∣∣∣=

∣∣∣∣∣∣

x y z0 0 ∂

∂z

Ex 0 0

∣∣∣∣∣∣=∂Ex

∂zy

Thus∂B

∂t= −∂Ex

∂zy

from which we conclude that the time-varying magnetic field has only a ycomponent! i.e.

B =

(−∫∂Ex

∂z∂t

)y

It is pointed out that, mathematically, time-independent x, y and z com-ponents could exist, however these “DC” components are not generated by

any source, and are therefore zero. Ex is a travelling wave of the general

8-3 AJW, EEE3055F, UCT 2012

form Ex(z, t) = f(z − vt). Substituting, we get

By(z, t) = −∫ (

∂Ex

∂z

)∂t

= −∫

(f ′(z − vt)) ∂t

=1

vf(z − vt) + h(z) + C

where h(z) is some function of z and C is a constant. We have the additional

condition that By(z, t) is a travelling wave of the form By(z, t) = fn(z− vt)(the solution to the wave equation). The first term is of this form, thus

By(z, t) =1

vEx(z, t)

8.2 Characteristic Impedance

Often, in engineering, the H field is described; the Hy component is related

to the Ex component by

Hy(z, t) =1

µBy =

1

µvEx(z, t) =

1

ηEx(z, t)

where the constant η = µv = µ 1√εµ =

√µε is called the characteristic

impedance of the medium and is the ratio of the electric to magnetic fieldsi.e. η = Ex

Hy. The units of η are V m−1

Am−1 = V A−1 = ohms, hence of the use of

the term characteristic impedance.For a plane wave propagating in free space, the characteristic impedance

is

η0 =

√µ0ε0

=

√4π × 10−7

8.85× 10−12≈ 377 ohms (in free space)

In air, µ ≈ µ0 and ε ≈ ε0, and so the characteristic impedance of free space

is usually used for propagation in air.In other media, for example a dielectric material like glass with µ ≈ µ0

and ε = εrε0 where εr ≈ 5 say, the characteristic impedance is reduced to

η =

√µrµ0εrε0

≈ η0√εr

=377√5≈ 170 ohms (in glass)

8-4 AJW, EEE3055F, UCT 2012

8.3 Sinusoidal Representations

The sinusoidal representation of the forward travelling wave is then the EM

pairEx(z, t) = E1 cos(ωt− kz +Ψ1)

Hy(z, t) =1

ηEx(z, t) = H1 cos(ωt− kz +Ψ1)

where k = ω/v = ω√εµ, and H1 =

1ηE1.

The following illustration depicts the sinusoidal electric and magnetic fieldsat a fixed instant in time:

x

E(z, t) = 〈Ex(z, t), 0, 0〉

y

z

E = Exx

n direction of propagation

H = 1ηExy

E

H

Figure 8.2: Sinusoidal EM wave propagating in the z direction.

NOTES:

• The electric and magnetic components are perpendicular and are bothnormal to the direction of propagation.

• For the more general case of a plane wave propagating in an arbitrary

direction, involving Ex, Ey and Ez components , the associated Hvector is perpendicular, and concisely related by

H =1

ηn× E

where n is a unit vector pointing in the direction of propagation.

• The cross product E×H points in the direction of propagation.

8-5 AJW, EEE3055F, UCT 2012

Complex Phasor Representation

By assuming steady state complex exponential solutions, the standard wave

equation can be simplified and the solutions obtained from the simplifiedequations. The 1-D wave equation representing the electric field component

of a plane wave propagating in the z direction is

∂2Ex

∂z2− εµ

∂2Ex

∂t2= 0

We assume a solution of the form Ex(z, t) = Ex(z)ejωt, where Ex(z) is a

(complex) function to be determined. Substituting into the above wave

equation, we get∂2Ex

∂z2ejωt + ω2εµExe

jωt = 0

The term ejωt term cancels yielding

∂2Ex

∂z2+ ω2εµEx = 0

which is now a simpler differential equation as it involves only functionsof z and not of time. The complete solution to this 2nd order ordinary

differential equation is known to be

Ex(z) = E1e−jkz + E2e

jkz

where E1 = E1ejΨ1 and E2 = E2e

jΨ2 are constants (in general complex).The real signal it represents is found by multiplying by ejωt and taking the

real part, i.e.

Ex(z, t) = E1 cos(ωt− kz +Ψ1) + E2 cos(ωt+ kz +Ψ2)

8-6 AJW, EEE3055F, UCT 2012

8.4 Plane Wave Propagating in an Arbitrary Direction

A plane wave propagating in an arbitrary direction in Cartesian coordinates

is related to the plane wave propagating in the z direction by a rotationof the axes. The diagram illustrates a plane wave propagating in the k

direction.

phase

planes of constant

z

y

l-axis

r

x

k direction of propagationθ

θ

l-axisl = 0

k · r= r co

s θ

Figure 8.3: Plane wave propagating in an arbitrary direction.

The phasor representation of an electromagnetic wave propagating in anarbitrary direction can be compactly written as the pair

E(r, t) = E1ej(ωt−k·r+Ψ) = E1e

−jk·rejωt

H =1

ηk× E

where

• r =

x

yz

specifies position in space.

8-7 AJW, EEE3055F, UCT 2012

• k =

kxkykz

= kk is a vector that points in the direction of prop-

agation. The wave number is given by k =√k2x + k2y + k2z , and the

wavelength λ = 2πk .

• The term k · r accounts for the phase change in the direction of prop-agation. To visualise this refer to the accompanying sketch. The term

k · r = kk · r is just k times the component of r in the k direction, i.e.k · r = k r cos θ.

One can define an axis (labelled “l-axis”) pointing in the k direction,as shown in the sketch. The quantity l = k · r = r cos θ, is the distance

measured along the l-axis from its origin. Then factor

e−jk·r = e−jkl

is easily recognised as spatially varying phase factor, along the axis ofpropagation.

• E1= E1ejΨ is a (complex) vector that is perpendicular to the direction

of propagation (i.e. E1 · k = 0), and lies in the “plane” of the plane

wave.E1 determines the orientation of the E field in the plane, i.e. in the ori-entation of the linear polarization. E1 determines both the amplitude,

phase shift, and polarization orientation of the plane wave. It can bemade complex to allow for an arbitrary phase shift Ψ.

8-8 AJW, EEE3055F, UCT 2012

8.5 Polarization

The polarization of a plane wave refers to the orientation of the E-field

vectors in the plane perpendicular to the direction of propagation. Up tillnow, we have considered only the simplest case known as linear polarization,

in which the E field lines are orientated at a fixed angle in the plane.More generally, a forward travelling sinusoidal EM wave travelling in the

+z direction can be represented as the sum of two orthogonal components,

E(z) = xE1e−jkz + yE2e

jΨe−jkz

where E1 and E2 are the amplitudes of the x and y components, and Ψrepresents a possible relative phase shift between them. The associated H

field is given by

H(z) = −x1ηE2e

jΨe−jkz + y1

ηE1e

−jkz

The physical sinusoidal electric and magnetic fields are modelled by the pair

E(z, t) = xE1 cos(ωt− kz) + yE2 cos(ωt− kz +Ψ)

H(z, t) = −x1ηE2 cos(ωt− kz +Ψ) + y

1

ηE1 cos(ωt− kz)

From the real forms can be derived the magnitude and orientation of theE-field vector as a function of time. There are three classes of polarization,

which depend on the relative amplitudes of E1 and E2, and the phase shiftΨ. These are described in detail in the following subsections.

8.5.1 Linear Polarization: Ex and Ey are in phase i.e. Ψ = 0

The electric vector is orientated at an angle determined by the relative values

of of amplitudes E1 and E2. The amplitude of the vector varies sinusoidally,illustrated by a sequence of snapshots in a fixed plane (e.g. at z = 0), atdifferent fractions of the period T of the sinusoidal wave.

8-9 AJW, EEE3055F, UCT 2012

t = 0 t = T/8 t = T/4 t = 3/8 t = T/2

Linear Polarization - snapshots in time.

Ex

Ey

E

α

Figure 8.4: Linear polarization.

The time varying magnitude of the resultant E vector is given by

√E2x + E2

y =√E2

1 cos2(ωt− kz) + E2

2 cos2(ωt− kz) =

√E2

1 + E22 |cos(ωt− kz)|

The orientation angle α may be calculated from

tanα =Ey

Ex=E2

E1

and jumps 180 degrees, when the resultant passes through zero. It is notedthat the magnitude of the resultant H vector is 1

η times the magnitude of

the resultant E vector, as can be seen from

√H2x +H2

y =

√(−Ey

η

)2

+

(Ex

η

)2

=1

ηE

It is common in telecommunications and radar engineering to describe the

polarization by the orientation of the electric field vector: “vertical polariza-tion”if the electric field is orientated vertically, and“horizontal polarization”

if the electric field is horizontal. The orientation of the polarization is de-pendent on the orientation of the radiating antenna.

8-10 AJW, EEE3055F, UCT 2012

8.5.2 Circular Polarization: E1 = E2 and Ψ = ±π2

If the two orthogonal components are equal in amplitude, but with a rel-

ative phase shift of 90 degrees, the total electric field vector will “rotate”as a function of time if viewed in a plane perpendicular to the direction of

propagation at a fixed location in space.Imagine holding up a transparent sheet of paper through which the EM

wave passes. The electric field lines in the plane of the paper will rotate as

a function of time at a rate of ω rad/s, but the magnitude, indicated by theline spacing, does not change as illustrated below.

t = 0 t = T/8 t = T/4 t = 3/8 t = T/2

E

Circular Polarization - snapshots in time.

Figure 8.5: Circular polarization - snapshots at increasing times, in the same plane.

The E-vector at a point on the plane rotates as illustrated below, wherein this alternative representation, the magnitude is indicated via the length

of the vector.

t = 0 t = T/8 t = T/4 t = 3/8 t = T/2

E

x

y

α(t)

Circular Polarization - snapshots in time (at a fixed position, z=0)

Wave propagates in +z direction out of page.

Figure 8.6: Circular polarization - snapshots at increasing times, in the same plane.

The diagram below illustrates a side view snapshot along the axis of propa-

gation at a fixed instant in time - the vectors wind around like a“corkscrew”.

8-11 AJW, EEE3055F, UCT 2012

The corkscrew moves forwards in the z direction as time increases.

Figure 8.7: Circular polarization - E-vectors shown at a fixed instant in time. Note: thisis “right-hand” circular polarization.

To analyse the case of circular polarization, we set E1 = E2 = A and

Ψ = ±π2 . The magnitude of the resultant E-vector is given by

√E2x + E2

y =√A2 cos2(ωt− kz) +A2 sin2(ωt− kz) = A

which is independent of z and t. The orientation angle is however time-varying, and may be calculated from

tanα =Ey

Ex=A cos(ωt− kz ± π

2 )

A cos(ωt− kz)=

∓A sin(ωt− kz)

A cos(ωt− kz)= ∓ tan(ωt− kz)

Thus the angle is

α = (ωt− kz)mod 2π if Ψ = −π2

or

α = −(ωt− kz)mod 2π if Ψ =π

2NOTE: At a fixed position z, the vector rotates at a rate of ω radians per

second in a circle in the plane. At a fixed time t, the vector rotates at kradians per metre (with a cork-screw locus).

8-12 AJW, EEE3055F, UCT 2012

The IEEE defines two types of circular polarization, according to thedirection of rotation:

Left-hand circular: If the E vector of a circularly polarized plane wavepropagating out of the page rotates in the clockwise direction in the planeof the page. You can imagine drawing the E vector’s position on the page,

and watching it rotate clockwise. Another way to remember the definitionis to point the thumb of your left hand in the direction of propagation (i.e.

z-direction) with your fingers bent - your curled fingers will point in thedirection of rotation of the E vector in the x-y plane (i.e. the x-y plane is at

a fixed position, say z=0). It is noted that in our particular example, thiscorresponds to the case where Ψ = π

2 .

Right-hand circular – if you use your right hand, with thumb pointing inthe direction of propagation (z direction), your curled fingers will point in

the direction of rotation of the E vector in the x-y plane (i.e. the x-y planeis at a fixed position, say z=0). In our particular example, this corresponds

to the case where Ψ = −π2 .

The illustrations below show the directions of rotation in an x-y plane forthe case of a plane wave travelling in the z-direction, being out of the page.

E

x

Right-Hand Circular Polarization

y

z

E

x

Left-Hand Circular Polarization

y

z

(Propagation out of page in z-direction)

Figure 8.8: Left-hand versus right-hand polarization.

One can easily plot the E vector in the x−y plane for t = 0, T/8, 2T/8, 3T/8, 4T/8

by considering the Ex and Ey components in our polarization model

E(z, t) = xE1 cos(ωt− kz) + yE2 cos(ωt− kz +Ψ)

8-13 AJW, EEE3055F, UCT 2012

For example, for the case of Ψ = −π2 , the E field components at z = 0 are

Ex = A cos(ωt)

Ey = A cos(ωt− π

2)

The Ex and Ey values are plotted as functions of time below:

Ex = cos(ωt)

Ey = cos(ωt− π/2)

T/4 T/2

T/2

T/4

T/4

t = 0

T/83T/8

T/2

y

x

t

t

E vector rotates in z = 0 plane

Wave propagates in +z direction out of page; hence this is right hand circular polarization

Circular Polarization - snapshots in time (at a fixed position, z=0)

Figure 8.9: Circular polarization - Ex and Ey components plotted against time.

8.5.3 Elliptical Polarization: Either E1 6= E2 or Ψ 6=0, π2 ,−π

2

Elliptical polarization is the general case, in which the vector not only ro-

tates, but also varies in length, tracing out an ellipse in a plane at a fixedposition, as illustrated in the sequence below (which is left-hand ellipticalpolarization):

t = 0 t = T/8 t = T/4 t = 3/8 t = T/2

E

Elliptical Polarization - snapshots in time.

y

x

Figure 8.10: Elliptical polarization - snapshots in time.

8-14 AJW, EEE3055F, UCT 2012

In all three classes of polarization, the H-field is always perpendicular tothe E-field, and in phase with it, as illustrated below:

Circular Polarization

- rotating vector traces a circle

Linear Polarization

- fixed orientation of resultant

- amplitude varies sinusoidally.

- rotating vector traces an ellipse

Elliptical Polarization

Ex(t)

EEEy(t)

HH

E

H

Figure 8.11: All three classes of polarization, showing both E and H vectors, and the lociof the tips of the vectors.

8.5.4 Applications of polarization

Communication Links

In broadcast transmitters e.g. radio, TV signal, linear polarization is most

commonly used. Wire antennas like dipoles, monopole and Yagi antennasradiate linearly polarized radiation. Both horizontal and vertical polariza-

tion is used. The orientation of the transmitting antenna determines theorientation of the polarization. The receiving antenna must be orientatedcorrectly to receive the maximum signal strength.

Consider for example, a dipole antenna receiving a vertically polarized EMwave. The dipole must be orientated vertically to receive the signal - the

8-15 AJW, EEE3055F, UCT 2012

electrons rush vertically up and down the wires in response to the incomingE-field, resulting in a “voltage response” across the centre terminals. If the

antenna is rotated 90 degrees to the electric field, it should be clear that theelectrons do not move up and down the rod as before, and so a potentialdifference does not appear across the terminals. At an arbitrary angle, the

signal drops off as the cosine of the angle from the vertical (i.e. with thecomponent of the electric field in the direction of the dipole rod).

Figure 8.12: Vertical versus horizontal polarization from a radiating dipole.

Radar - “Radio detection and ranging”

Radar is a technique used to detect far away targets like aeroplanes ap-

proaching an airport, or ships out at sea. A pulse is transmitted in thedirection of interest, and the return echo is received and used to detect the

presence of targets. The range to the target is inferred from the time delayof the echo, and the direction obtained from the pointing direction of the

antenna.Circular polarization is sometimes used in radar applications as some scat-

8-16 AJW, EEE3055F, UCT 2012

tering structures only reflect electromagnetic energy of a particular polar-ization. In circular polarization, the E-field vector rotates in a plane at the

reflector (being a fixed distance from the source), which ensures that therewill aways be some reflection from such structures.For example, a vertically orientated thin wire rod will only reflect vertical

polarization, where as a horizontally orientated wire rod will only reflecthorizontal polarization. Circular polarization can be thought of as the sum

of two orthogonal, linearly polarized components. On reflection, only onecomponent, in line with the orientation of the reflecting rod will be re-

flected. An incident circularly polarized wave will be linearly polarized onreflection. The received signal will be 3dB lower than one could obtain from

transmitting linear polarization, as only half the power is reflected.Circular polarization can be generated by using two perpendicular dipoles

(crossed-dipoles), driven 90 degrees out of phase. Alternatively, the perpen-

dicular dipoles may be driven in phase, but physically separated by a quarterwavelength in the propagation direction to achieve the required 90 degree

phase shift.

8-17 AJW, EEE3055F, UCT 2012

9 Simulating Electromagnetic Wavesusing the Finite Difference TimeDomain (FDTD) Method [notcovered in 2009]

With the availability of fast computers with ever increasing memory andcomputational performance, the simulation of wave propagation has been

made possible. In this section, a powerful method known as the finite dif-ference time domain (FDTD) method for simulating the propagation ofelectromagnetic waves will be introduced. To simplify the discussion, the

method will be described initially for wave propagation in 1-D space, fol-lowed by the extension to 2-D, and 3-D1. This chapter is meant to provide

only a brief overview.

9.1 Introduction

The finite difference time domain (FDTD) method is a full-wave, dynamic,

and powerful solution tool for solving Maxwell’s equations, introduced byK.S. Yee in 1966 [Yee, 1966]. The algorithm involves direct discretizations of

Maxwell’s equations by writing the spatial and time derivatives in a centralfinite difference form2.

The time-dependent Maxwell’s curl equations in a homogeous dielectricmedium (ǫ = ǫ0ǫr, µ = µ0, µr = 1) are

1The course coordinator wishes to thank doctoral student Pradip Mukhopadhyay for his assistance inpreparing the material in this section on FDTD simulation.

2The central finite difference approximation for the derivative of the function f(x) at point P (x0) can bewritten as

df(x0)

dx= f ′(x0) ∼=

f(x0 +x/2)− f(x0 −x/2)x

9-1

∂E

∂t=

1

ǫ0ǫr∇×H (9.1)

∂H

∂t= − 1

µ0∇× E (9.2)

E andH are vectors in three dimensions. The constants ǫ0 and µ0 are knownas the permittivity and permeability of free space and ǫr is the relative

permittivity of the material.

9.1.1 Curl Equations in Cartesian Coordinates

Expanding∂E

∂t=

1

ǫ0ǫr∇×H

in Cartesian coordinates3,

x∂Ex

∂t +y∂Ey

∂t +z∂Ez

∂t = 1ǫ0ǫr

[x(∂Hz

∂y − ∂Hy

∂z

)+ y

(∂Hx

∂z − ∂Hz

∂x

)+ z

(∂Hy

∂x − ∂Hx

∂y

)]

Equating the vector components, we obtain three equations, one for eachvector component

∂Ex

∂t=

1

ǫ0ǫr

(∂Hz

∂y− ∂Hy

∂z

)

∂Ey

∂t=

1

ǫ0ǫr

(∂Hx

∂z− ∂Hz

∂x

)

∂Ez

∂t=

1

ǫ0ǫr

(∂Hy

∂x− ∂Hx

∂y

)

Similarly expanding∂H

∂t= − 1

µ0∇×E

as

3∇×H =

x y z∂∂x

∂∂y

∂∂z

Hx Hy Hz

= x

(∂Hz

∂y− ∂Hy

∂z

)+ y

(∂Hx

∂z− ∂Hz

∂x

)+ z

(∂Hy

∂x− ∂Hx

∂y

)

9-2 AJW, EEE3055F, UCT 2012

x∂Hx

∂t +y∂Hy

∂t +z∂Hz

∂t = − 1µ0

[x(∂Ez

∂y − ∂Ey

∂z

)+ y

(∂Ex

∂z − ∂Ez

∂x

)+ z

(∂Ey

∂x − ∂Ex

∂y

)]

we obtain three more equations,

∂Hx

∂t=

1

µ0

(∂Ey

∂z− ∂Ez

∂y

)

∂Hy

∂t=

1

µ0

(∂Ez

∂x− ∂Ex

∂z

)

∂Hz

∂t=

1

µ0

(∂Ex

∂y− ∂Ey

∂x

)

In 1-D, we consider (i) exciting an Ex component, and assume Ey = 0and Ez = 0 and (ii) no variation in the x-y plane, i.e. ∂

∂x = 0 and ∂∂y = 0.

The equations reduce to

∂Hx

∂t=

1

µ0

(∂Ey

∂z− ∂Ez

∂y

)= 0

∂Hy

∂t=

1

µ0

(∂Ez

∂x− ∂Ex

∂z

)= − 1

µ0

∂Ex

∂z

∂Hz

∂t=

1

µ0

(∂Ex

∂y− ∂Ey

∂x

)= 0

∂Ex

∂t=

1

ǫ0ǫr

(∂Hz

∂y− ∂Hy

∂z

)= − 1

ǫ0ǫr

∂Hy

∂z

Furthermore, if the source has no DC component, then Hx = 0 and Hz =

0, leaving only the Ex and Hy components.

9.2 FDTD Solution to Maxwell’s equations in 1-D

In a simple one-dimensional case, we will consider the case where only theEx and Hy components exist - consistent with modelling plane wave prop-

agation far away from an antenna. Equations 9.1 and 9.2 become

∂Ex

∂t= − 1

ǫ0ǫr

∂Hy

∂z(9.3)

∂Hy

∂t= − 1

µ0

∂Ex

∂z(9.4)

9-3 AJW, EEE3055F, UCT 2012

These equations represent a plane wave with the electric field orientatedin the x-direction and magnetic field oriented in the y-direction and trav-

elling in the z-direction. In the FDTD formulation, the central differenceapproximations for both the temporal and spatial derivatives are obtainedat (z = k∆z, t = n∆t) for the first equation:

En+1/2x (k)− E

n−1/2x (k)

t = − 1

ǫ0ǫr

Hny (k + 1/2)−Hn

y (k − 1/2)

z (9.5)

and at (z +∆z/2, t+∆t/2) for the second equation:

Hn+1y (k + 1/2)−Hn

y (k + 1/2)

t = − 1

µ0

En+1/2x (k + 1)− E

n+1/2x (k)

z (9.6)

In the equations above, n is the time index and k is the spatial index, whichindexes times t = nt and positions z = kz, or positions t = (n±1/2)tand positions z = (k± 1/2)z. The time index is written as a superscript,and the spatial index is within brackets.

Equations 9.5 and 9.6 can be rearranged as a pair of ‘computer updateequations’, which can be repeatedly updated in loop, to obtain the next timevalues of E

n+1/2x (k) and Hn+1

y (k + 1/2), corresponding the Ex(t+∆t/2, z)

and Hy(t+∆t, z +∆z/2).Figure 9.1 illustrates the interleaving of the E and H fields in space and

time in the FDTD formulation.

9-4 AJW, EEE3055F, UCT 2012

k−2 k k+1 k+2k−1

xEn−1/2

k−2 k k+1 k+2k−1

k−1 1/2 k−1/2 k+1/2 k+1 1/2 k+2 1/2

xE

Hyn

n+1/2

Figure 9.1: Interleaving of the E and H fields in space and time in the FDTD formulation.To calculate Hy(k + 1/2), for instance, the neighbouring values of Ex at k andk + 1 are needed. Similarly, to calculate Ex(k + 1), the value of Hy at k + 1/2and k + 11

2are needed (Sullivan 2000).

In equations 9.5 and 9.6 ǫ0 and µ0 differ by several orders of magnitude,

Ex and Hy will differ by several orders of magnitude. Numerical error isminimised by making the following change of variables as

Ex =

√ǫ0µ0

· Ex (9.7)

which bring the field quantities to similar levels. Implementing the changing

of variables, equations 9.5 and 9.6 become

En+1/2x (k) = En−1/2

x (k)− 1

ǫr√ǫ0µ0

tz

[Hny (k + 1/2)−Hn

y (k − 1/2)]

(9.8)

Hn+1y (k + 1/2) = Hn

y (k + 1/2)− 1√ǫ0µ0

tz

[En+1/2x (k + 1)− En+1/2

x (k)]

(9.9)

Stability and the FDTD method: For stability purposes, we needto choose the cell size z to allow 10 to 15 points per wave length. Infree space, an electromagnetic wave travels a distance of one cell in time

t = zc0, where c0 is the speed of light in free space. This limits the

9-5 AJW, EEE3055F, UCT 2012

maximum time step. In the case of a 2-D simulation, we have to allow forthe propagation in the diagonal direction, which brings the time requirement

to t = z√2c0

. Obviously, three-dimensional simulation requires t = z√3c0

.We will use in all our simulations a time step t of

t = z2 · c0

, (9.10)

where c0 is the velocity of light in free space, which satisfies the requirements

in 1-D, 2-D and 3-D for all media. The factor

1√ǫ0µ0

tz = c0 ·

z/(2 · c0)z/2 = 0.5 (9.11)

Making use of equation 9.11 in equation 9.8 and 9.9, we obtain the followingcomputer update equations:

ex(k) = ex(k) +0.5

ǫr(k)[hy(k − 1)− hy(k)] (9.12)

hy(k) = hy(k) + 0.5 [ex(k)− ex(k + 1)] (9.13)

which are used repeatedly in a loop to update the field quantities at every

position at all positions in space, as time progresses. Note that the n orn+ 1/2 or n− 1/2 in the superscripts do not appear. In equation 9.12, theex on the right side of the equal sign is the previous value at n − 1/2, and

the ex on the left side is the new value, n+ 1/2, which is being calculated.In case of the spatial index, k+1/2 and k−1/2 are replaced by k and k−1

in order to specify an integer position in an array. It is understood fromthe derivation, however, that the value stored in hy(k) is the H value at

position k + 1/2.

The value of ex(k) on the n’th iteration, represents En+1/2x (k), which is

related to the electric field by Ex = Ex/√

ǫ0µ0.

9.2.1 1-D FDTD code in Matlab

This section contains a code segment written in Matlab, in which the 1-D

FDTD algorithm is implemented. An electromagnetic pulse is radiated from

9-6 AJW, EEE3055F, UCT 2012

a source located in free space. The source waveform is a Gaussian shapedpulse, injected at the centre of the array used to store the pulse in space.

This is achieved by repeatedly updating the E-field pulse value at the sourcelocation; the values at the other locations are computed using the updateequations. Figure 9.2 shows snapshots of the simulation as time progresses.

The E-field pulse is seen to propagate away from the source to the left andto the right. The corresponding H field is also calculated, but not plotted

here.Note: In free space ǫr = 1

1) Solution: The MATLAB file : fdtd 1d 1.m

KE = 201; % number of z samples.

kc = fix(KE/2) ; % centre of the grid

ex = zeros(1,KE); % initialize Ex field

hy = zeros(1,KE); % initialize Hy field

cb = zeros(1,KE);

cb(1,:) = 0.5;

% A Gaussian pulse parameters

t0 = 36.0 ; % location in time of the pulse

spread = 12 ; % Width of the pulse

NSTEPS = 190 ; % number of time step

for n=1:NSTEPS

for k=2:KE

ex(1,k) = ex(1,k) + cb(1,k).*(hy(1,k-1) -hy(1,k));

end

pulse = exp(-0.5*((t0-n)/spread)^2.0); %The Gaussian pulse

% adding the pulse at the centre of the grid

ex(1,kc) = pulse;

for k=1:KE-1

hy(1,k) = hy(1,k) + 0.5*(ex(1,k) -ex(1,k+1)) ;

end

plot(ex);

pause(0.05);

end

9-7 AJW, EEE3055F, UCT 2012

20 40 60 80 100 120 140 160 180 200−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

11D EM Propagation

Time step=35

20 40 60 80 100 120 140 160 180 200−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

11D EM Propagation

Time step=60

20 40 60 80 100 120 140 160 180 200−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

11D EM Propagation

Time step=120

Source location

Figure 9.2: 1-D Ez propagation. Propagation of a Gaussian pulse away from a sourcelocated at the centre.

9-8 AJW, EEE3055F, UCT 2012

9.2.2 Wave hitting a dielectric medium (Matlab code)

We now consider the case where a plane wave travelling in free space

(medium 1) strikes a dielectric medium (medium 2), as is illustrated in Fig-ure 9.3, which shows a source in free space on the left side, and a dielectric

slab on the right.

z

kstart

Source

0

ε=ε0 εr

(or k)

ε=ε

Figure 9.3: EM wave hitting a dielectric surface

When the wave strikes the interface, a fraction of the incoming wave is

reflected, and a fraction is transmitted into the medium. The amplitudeof the reflected and transmitted waves, relative to the incident wave, are

described by the reflection coefficient ρ and the transmission coefficient τ ,which relate the amplitudes of the E field waves. From theoretical analysis,these can be determined in terms of the characteristic impedances of the

media, as

ρ =Eref

Einc=η2 − η1η2 + η1

(9.14)

τ =Etrans

Einc=

2η2η2 + η1

(9.15)

where η1 and η2 are the impedances of the respective media given by

9-9 AJW, EEE3055F, UCT 2012

η =

√µ0µrǫ0ǫr

(9.16)

In case of a non-magnetic medium (µr = 1), equations 9.14 and 9.15 become,

ρ =

√ǫ1 −

√ǫ2√

ǫ1 +√ǫ2

(9.17)

τ =2 · √ǫ1√ǫ1 +

√ǫ2

(9.18)

where ǫ1 and ǫ2 are the relative permittivities of medium 1 and 2 respec-tively.The 1-D FDTD code can be easily adapted to model propagation against

a dielectric interface, as shown below. Running the simulation should showreflected and refracted waves as in the snapshots in Figure 9.4. Note here

that the reflected pulse inverts in sign, as indicated by a negative reflectioncoefficient (check this by calculating it).

Modify the Matlab program as follows:

* Choose the dielectric constant: eps_r=4.0 (for example)

* Define were to start the dielec-

tric slab: k_start=kc+kc/2 (for example)

* Put the dielectric material in one side as

cb(1,k_start:KE)=0.5/eps_r

Modify the above program as follows:

.

.

eps_r = 4.0;

k_start = kc+kc/2;

cb = zeros(1,KE);

cb(1,:) = 0.5;

cb(1,k_start:KE) = 0.5/eps_r;

.

9-10 AJW, EEE3055F, UCT 2012

.

for n=1:NSTEPS

.

.

plot(ex);

pause(0.05);

end

%----------------------------------------------------

9-11 AJW, EEE3055F, UCT 2012

20 40 60 80 100 120 140 160 180 200−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

11D EM Propagation − striking a dielectric slab on the right side

Time step=100

20 40 60 80 100 120 140 160 180 200−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

11D EM Propagation − striking a dielectric slab on the right side

Time step=190

Transmitted

Reflected

Figure 9.4: Reflection and transmission at a dielectric (1-D simulation)

9-12 AJW, EEE3055F, UCT 2012

9.2.3 Removing the unwanted reflection at the boundary - “TheAbsorbing Boundary Condition (ABC)”

In calculating the E field, we need to know the surrounding H values; thisis the fundamental assumption in FDTD. At the edge of the problem space

we will not have the value to one side, but we know there are no sourcesoutside the problem space. The wave travels z

2 (=c0 · t) distance in onetime step, so it takes two time steps for a wave front to cross one cell.

Suppose we are looking for a boundary condition at the end where k=1.Now if we write the E field at k=1 as

Enx(1) = En−2

x (2)

then the fields at the edge will not reflect. This condition must be applied

at both ends.

Implementation

Modify the afore program as follows:

This is easy to implement, store the value of Ex(2) for two time steps andthen put it in Ex(1).Modify the above program as follows:

.

.

ex_left_m1 = 0.0;

ex_left_m2 = 0.0;

ex_right_m1 = 0.0;

ex_right_m2 = 0.0;

cb = zeros(1,KE);

cb(1,:) = 0.5;

% You may choose cb(1,k_start:KE)=0.5/eps_r

.

.

for n=1:NSTEPS

.

.

% add the pulse at the centre of the grid

ex(1,kc) = pulse;

9-13 AJW, EEE3055F, UCT 2012

% add these additional lines after ex(1,kc)=pulse

ex(1,1) = ex_left_m2; % left boundary

ex_left_m2= ex_left_m1;

ex_left_m1 = ex(1,2);

ex(1,KE) = ex_right_m2; % right boundary

ex_right_m2= ex_right_m1;

ex_right_m1 = ex(1,KE-1);

.

.

plot(ex);

pause(0.05);

end

%-----------------------------------------------

Solution: The MATLAB file : fdtd_1d_3.m

% modified programme (use ABC)

KE = 201; % number of sampled in z direction


ex = zeros(1,KE); % initialize the Ex field

hy = zeros(1,KE); % initialize the Hy field

cb = zeros(1,KE);

cb(1,:) = 0.5;

ex_left_m1 = 0.0;

ex_left_m2 = 0.0;

ex_right_m1 = 0.0;

ex_right_m2 = 0.0;

% A Gaussian pulse

t0 = 36.0 ;

spread = 12 ;


for n=1:NSTEPS

for k=2:KE


end

%The Gaussian pulse

pulse = exp(-0.5*((t0-n)/spread)^2.0);


ex(1,kc) = pulse;







9-14 AJW, EEE3055F, UCT 2012

for k=1:KE-1

hy(1,k) = hy(1,k) + 0.5*(ex(1,k) -ex(1,k+1)) ;

end

plot(ex);

pause(0.05);

end

%---------------------------------------------------

9.2.4 Some Exercises

1. In free space, put the source at the middle and see how the wave travels.

See what happens when the pulse hits the boundary ?

2. Modify the program so that it has two sources 20 cells away from themiddle. What happens when the pulse meet each other.

3. Divide the problem space into two, one side being free space and theother a dielectric material. What happens after the pulse hits the

boundary? Look at the relative amplitudes of the reflected and trans-mitted pulses. Check the values with equation 9.17 and 9.18.

9.3 FDTD Solution to Maxwell’s equations in 2-D Space

In deriving 2-D FDTD formulation, we choose between one of two groupsof three vectors each:

1. The transverse magnetic (TM) mode, which is composed of Ez, Hx,

and Hy

or

2. The transverse electric (TE) mode, which is composed of Ex, Ey, andHz

We will work with the TM mode wave propagation. Expanding the curlequations4 9.1 and 9.2 with ∂

∂z = 0, Ex = 0, Ey = 0 and Hz = 0, we obtain

4∇×E =

x y z∂∂x

∂∂y

0

0 0 Ez

= x

(∂Ez

∂y

)+ y

(−∂Ez

∂x

)for the TM mode.

9-15 AJW, EEE3055F, UCT 2012

for the non-zero components of ∂E∂t and ∂H

∂t :

∂Ez

∂t=

1

ǫ0ǫr(∂Hy

∂x− ∂Hx

∂y) (9.19)

∂Hx

∂t= − 1

µ0

∂Ez

∂y(9.20)

∂Hy

∂t=

1

µ0

∂Ez

∂x(9.21)

Now we can write the above three equations in the finite difference schemeas

En+1/2z (i, j)− E

n−1/2z (i, j)

t =1

ǫ0ǫr

(Hny (i+ 1/2, j)−Hn

y (i− 1/2, j)

x −

Hnx (i, j + 1/2, )−Hn

x (i, j − 1/2)

y

)(9.22)

Hn+1x (i, j + 1/2)−Hn

x (i, j + 1/2)

t = − 1

µ0

En+1/2z (i, j + 1)− E

n+1/2z (i, j)

y(9.23)

Hn+1y (i+ 1/2, j)−Hn

y (i+ 1/2, j)

t =1

µ0

En+1/2z (i+ 1, j)− E

n+1/2z (i, j)

x(9.24)

Rearranging the above equations, we can write the final expression as

En+1/2z (i, j) = En−1/2

z (i, j) +tǫ0ǫr

(Hny (i, j + 1/2, )−Hn

y (i, j − 1/2)

x −

Hnx (i, j + 1/2, )−Hn

x (i, j − 1/2)

y

)(9.25)

∇×H =

x y z∂∂x

∂∂y

0

Hx Hy 0

= z

(∂Hy

∂x− ∂Hx

∂y

)

9-16 AJW, EEE3055F, UCT 2012

Hn+1x (i, j+1/2) = Hn

x (i, j+1/2)− 1

µ0· ty

(En+1/2z (i, j + 1)− En+1/2

z (i, j))

(9.26)

Hn+1y (i+1/2, j) = Hn

y (i+1/2, j)+1

µ0· tx

(En+1/2z (i+ 1, j)− En+1/2

z (i, j))

(9.27)For computer implementation we can write the above equations update

equations as

ez(i,j)=ez(i,j)+dt/(eps_0*eps_r)

*[Hy(i,j)-Hy(i-1,j)/dx

-Hx(i,j)-Hx(i,j-1)/dy]

Hx(i,j) = Hx(i,j)-dt/(mu_0*dy)*Ez(i,j+1)-Ez(i,j)

Hy(i,j) = Hy(i,j)+dt/(mu_0*dx)*Ez(i+1,j)-Ez(i,j)

9.3.1 2-D FDTD code in Matlab

The code listing in this section simulated the case of 2-D EM wave propa-gation in free space, source at the centre (ǫr = 1).

Solution: The MATLAB file : fdtd_2d_TM_1.m

clear all;

cc=2.99792458e8; % Speed of light in free space

mu_0=4.0*pi*1.0e-7; % Permeability in free space

eps_0=1.0/(cc*cc*mu_0); % Permittivity in free space

eps_r = 1.0; % Dielectric constant in free space

ie = 201; % Grid pixels in Y direction

je = 201; % Grid pixels in X direction

9-17 AJW, EEE3055F, UCT 2012

is = floor(ie/2) ;

js = floor(je/2) ;

% Electric and Magnetic field

Ez = zeros(ie,je);

Hx = zeros(ie,je);

Hy = zeros(ie,je);

nmax = 150; % Number of time steps

ddx = 1.0e-3; % X grid size

ddy = ddx; % Y grid size

dt = 0.98/(cc*sqrt( (1/ddx)^2 + (1/ddy)^2 ));

%*********************************************************

tw = 26.53e-12;

t0 = 4.0*tw;

T = (0:1:nmax-1)’.*dt;

source = -2.0*((T-t0)./tw).*exp(-1.0*((T-t0)./tw).^2.0);

% Plot injected pulse figure

plot(source) title(’Source pulse’)

pause(1);

Emax = max(source);

Emin = min(source);

C1 = dt/(eps_0*eps_r);

C2 = dt/mu_0;

figure

for n = 1:nmax

%------------------- Ez -----------------

for jj = 2:je

for ii = 2:ie

Ez(ii,jj) = Ez(ii,jj) + C1*((Hy(ii,jj) - Hy(ii-

1,jj))./ddx -(Hx(ii,jj) - Hx(ii,jj-1))./ddy);

end

end

%----------------- Inject source --------

Ez(is,js) = source(n);

%------------------- Hx -----------------

for jj = 1:je-1

for ii = 1:ie

Hx(ii,jj) = Hx(ii,jj) - C2*((Ez(ii,jj+1) -

Ez(ii,jj))./ddy);

end

end

%------------------- Hy -----------------

for jj = 1:je

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for ii = 1:ie-1

Hy(ii,jj) = Hy(ii,jj) + C2*((Ez(ii+1,jj) -

Ez(ii,jj))./ddx);

end

end

%plot the Ez component

imagesc(Ez, 0.4*[Emin Emax]);

title(’Ez component propagating in free space’)

colorbar;

colormap(hot);

pause(0.0002);

end

%--------------------------------------------

0 50 100 150−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1Source pulse

Figure 9.5: Source pulse [differential Gaussian]

9-19 AJW, EEE3055F, UCT 2012

Ez component propagating in free space

20 40 60 80 100 120 140 160 180 200

20

40

60

80

100

120

140

160

180

200

Figure 9.6: Propagation of a pulse away from a source. The pulse moves outwards as anexpanding “circle”, centred on the source.

9.3.2 Wave hitting a dielectric surface (right side), source at thecentre (2-D)

We will examine two examples:

1. Wave hitting a dielectric surface (right side), with the source at the

centre (code: fdtd 2d 2.m)

2. Wave hitting a dielectric surface (top right corner) (code: fdtd 2d 3.m)

To simulate 1 or 2 above, modify the previous programme (fdtd 2d 1.m) asfollows:

9-20 AJW, EEE3055F, UCT 2012

• Specify the position of the dielectric surface (id=....., jd=.....)

• Choose a value of eps r = ......

• Modify the constant value C1 according to that.....

• Run the program, you will see that on one side (free space) the wave

is propagating freely and on the other you will notice reflection andtransmission of the incident wave at the interface.

Figure 9.7 shows the result of case 1, where the pulse strikes the dielectric

interface. Notice that the wave propagates more slowly in the medium onthe right, and the pulse length is shorter.

2D EM propagation −− striking a dielectric slab on the right side

20 40 60 80 100 120 140 160 180 200

20

40

60

80

100

120

140

160

180

200

Figure 9.7: 2-D. Time snapshot of the radiating pulse interacting with a dielectric interface.The reflected and transmitted waves are seen on either side of the interface.

9-21 AJW, EEE3055F, UCT 2012

9.3.3 Multiple source 2-D wave propagation

A last exercise it to add a second source in the 2D simulation. Run the code

(code: fdtd 2d TM multi source.m).Radiation from an aperture antenna (e.g. a microwave horn antenna)

can be simulated as a line of (closely spaced) radiating sources (“Huyguns”wavelets).

9.4 Solving Maxwell’s Equations in 3D

The FDTD concepts described for 1-D and 2-D space can be easily extended

to 3-D. The main difference is that all three components of the E and H fieldsmust be updated within the time loop. The equations are more complicated,

and the simulation is more computationally intensive.Details of 3-D FDTD implementation can be found in the references.

9.5 References

• Yee, K.S., Numerical solution of initial boundary value problems in-

volving Maxwell’s equations in isotropic media, 1966.

• Taflove, A., Computational electrodynamics : the finite-difference time-

domain method, Boston, Mass. : Artech House,1995

• Sullivan, D. M., Electromagnetic simulation using the FDTD method,New York : IEEE Press, 2000.

• Kunz, K.S. and Luebbers, R.J., The Finite Difference Time DomainMethod for Electromagnetics, Boca Raton, FL; CRC Press, 1993.

9-22 AJW, EEE3055F, UCT 2012

9.6 MATLAB CODE

These code segments will be placed on the course web page for download.

1-D code wave hitting a dielectric slab

MATLAB file : fdtd_1d_2.m

KE = 201; % number sample in z dirn




k_start=kc+kc/2;

eps_r=4.0;

cb = zeros(1,KE);

cb(1,:) = 0.5;

cb(1,k_start:KE)=0.5/eps_r;

% A Gaussian pulse

t0 = 36.0 ;

spread = 12 ;

NSTEPS = 190; % number of time step

for n=1:NSTEPS

for k=2:KE


end

%The Gaussian pulse



ex(1,kc) = pulse;

for k=1:KE-1

hy(1,k) = hy(1,k) + 0.5*(ex(1,k) -ex(1,k+1)) ;

end

plot(ex)

pause(0.05);

end

%---------------------------------------------

9-23 AJW, EEE3055F, UCT 2012

1-D code : Reflection at the boundary (Absorbing BoundaryCondition(ABC))

% modified programme (use ABC)

Solution: The MATLAB file : fdtd 1d 3.m

KE = 201; %number of grid in z direction




cb = zeros(1,KE);

cb(1,:) = 0.5;

ex_left_m1 = 0.0;

ex_left_m2 = 0.0;

ex_right_m1 = 0.0;

ex_right_m2 = 0.0;

% A Gaussian pulse

t0 = 36.0 ;

spread = 12 ;


for n=1:NSTEPS

for k=2:KE


end

%The Gaussian pulse



ex(1,kc) = pulse;







for k=1:KE-1

hy(1,k) = hy(1,k) + 0.5*(ex(1,k) -ex(1,k+1)) ;

end

plot(ex);

pause(0.05);

end

%------------------------------------------------

9-24 AJW, EEE3055F, UCT 2012

Wave hitting a dielectric surface (right side), source at the centre


% fdtd_2d_TM_2.m

% Excercise ....

% 2D FDTD free space propagation striking dielec-

tric slab on right side

% For EEE355F AJW

clear all;

cc=2.99792458e8; %speed of light in free space

mu_0=4.0*pi*1.0e-7; %permeability of free space

eps_0=1.0/(cc*cc*mu_0); %permittivity of free space

ie = 200; %Grid pixels in Y direction


is = floor(ie/2) ;

js = floor(je/2) ;


Ez = zeros(ie,je);

Hx = zeros(ie,je);

Hy = zeros(ie,je);

eps_r = ones(ie,je);

jd = js+15; % Position of interface in X direction

eps_r(:,jd:je) = 10.0; % dielectric constant

nmax = 150; % number of time steps



dt = 0.98/(cc*sqrt( (1/ddx)^2 + (1/ddy)^2 ));

%*********************************************

tw = 26.53e-12;

t0 = 4.0*tw;

T = (0:1:nmax-1)’.*dt;


[Emax] = max(source);

[Emin] = min(source);

C1 = dt./(eps_r.*eps_0);

C2 = dt/mu_0;

for n = 1:nmax

%---------------------update Ez--------------------

for jj = 2:je

for ii = 2:ie

Ez(ii,jj) = Ez(ii,jj) + C1(ii,jj).*((Hy(ii,jj) - Hy(ii-


9-25 AJW, EEE3055F, UCT 2012

end

end

%-----------------------source-----------------


%--------------------update Hx---------------------

for jj = 1:je-1

for ii = 1:ie

Hx(ii,jj) = Hx(ii,jj) - C2*((Ez(ii,jj+1) - Ez(ii,jj))./ddy);

end

end

%--------------------update Hy--------------------

for jj = 1:je

for ii = 1:ie-1

Hy(ii,jj) = Hy(ii,jj) + C2*((Ez(ii+1,jj) - Ez(ii,jj))./ddx);

end

end


patch([jd jd],[1 ie],’w’);

title(’2D EM propagation --

striking a dielectric slab on the right side’)

colormap(hot);

pause(0.0002);

end

9-26 AJW, EEE3055F, UCT 2012

Wave hitting a dielectric surface (top right corner)


% fdtd_2d_TM_3.m

% Excercise ....

% 2D FDTD free space propagation striking dielec-

tric slab on top right corner

% For EEE355F AJW

clear all;

cc=2.99792458e8; % speed of light in free space

mu_0=4.0*pi*1.0e-7; % permeability of free space

eps_0=1.0/(cc*cc*mu_0); % permittivity of free space



is = floor(ie/2);

js = floor(je/2);


Ez = zeros(ie,je);

Hx = zeros(ie,je);

Hy = zeros(ie,je);

eps_r = ones(ie,je);

id = is-15; % dielectric interface

jd = js+15;

eps_r(1:id,jd:je) = 40.0; % dielectric constant

nmax = 120; % number of time steps



dt = 0.98/(cc*sqrt( (1/ddx)^2 + (1/ddy)^2 ));

%*****************************************************

tw = 26.53e-12;

t0 = 4.0*tw;

T = (0:1:nmax-1)’.*dt;


[Emax] = max(source); [Emin] = min(source);

C1 = dt./(eps_r.*eps_0);

C2 = dt/mu_0;

for n = 1:nmax

%-------------------update Ez-----------------

for jj = 2:je

for ii = 2:ie

Ez(ii,jj) = Ez(ii,jj) + C1(ii,jj).*((Hy(ii,jj) - Hy(ii-


end

9-27 AJW, EEE3055F, UCT 2012

end

%--------------------source---------------------


%--------------------update Hx------------------

for jj = 1:je-1

for ii = 1:ie


end

end

%---------------------update Hy-----------------

for jj = 1:je

for ii = 1:ie-1

Hy(ii,jj) = Hy(ii,jj) + C2*((Ez(ii+1,jj) - Ez(ii,jj))./ddx);

end

end


patch([jd jd],[1 id],’w’);

patch([jd je],[id id],’w’);

title(’2D EM propagation --

striking a dielectric slab on the right corner’);

colormap(hot);

pause(0.0002);

end

9-28 AJW, EEE3055F, UCT 2012

Multiple source 2D wave propagation

Solution: fdtd 2d TM multi source.m

clear all;

cc=2.99792458e8; % Speed of light in free space

mu_0=4.0*pi*1.0e-7; % Permeability of free space

eps_0=1.0/(cc*cc*mu_0); % Permittivity of free space

eps_r = 1.0; % Dielectric constant in air



is = floor(ie/2) ;

js = floor(je/2) ;


Ez = zeros(ie,je);

Hx = zeros(ie,je);

Hy = zeros(ie,je);

nmax = 400; % Number of time steps



dt = 0.98/(cc*sqrt( (1/ddx)^2 + (1/ddy)^2 ));

%***********************************************

T = (0:1:nmax-1)’.*dt;

f_int = 20.0e+9; %Frequency (20 GHz) of the incident pulse

source = sin(2.0*pi*T.*f_int); %Sine wave

% Plot injected pulse

figure

plot(source)

title(’Source pulse: 20 GHz Sine wave’)

pause(1);

[Emax] = max(source);

[Emin] = min(source);

C1 = dt/(eps_0*eps_r);

C2 = dt/mu_0;

figure

for n = 1:nmax

%------------------------------ Ez -----------------

for jj = 2:je

for ii = 2:ie

Ez(ii,jj) = Ez(ii,jj) + C1*((Hy(ii,jj) - Hy(ii-1,jj))./ddx -

(Hx(ii,jj) - Hx(ii,jj-1))./ddy);

end

end

9-29 AJW, EEE3055F, UCT 2012

%------------------ Inject source -----------------

for ii = is-20:7:is+30

Ez(ii,js) = source(n);

end

%--------------------- Hx -------------------------

for jj = 1:je-1

for ii = 1:ie


end

end

%--------------------- Hy ---------------------

for jj = 1:je

for ii = 1:ie-1

Hy(ii,jj) = Hy(ii,jj) + C2*((Ez(ii+1,jj) -

Ez(ii,jj))./ddx);

end

end


title(’2D EM wave propagation: Multiple source’)

colorbar;

colormap(gray);

pause(0.0002);

end

9-30 AJW, EEE3055F, UCT 2012

10 Power Considerations and thePoynting Vector

Electromagnetic waves carry energy (the capacity to do work) throughspace. At any point in space, the flow of energy can be described by a

power density vector P, which specifies both the power density in watts persquare metre, and the direction of flow. The vector P is called the Poynting

vector, and is a simple cross product of electric and magnetic field vectors,

P = E×H

The units of E are Vm−1 and the units of H are Am−1 and thus the unitsof P are VAm−2 or Wm−2 or Js−1m−2. The total power passing through a

surface S1 is obtained by integration over S1, i.e.

W =

∫

S1

P · dS watts

The power dP passing through an elemental surface dS with normal n is

dP = |P| cosθ dS = P · n dS = P · dS

To understand why P = E × H represents power flow, we consider therate of energy loss though from a volume V enclosed by closed surface S,

and relate this to the electric and magnetic fields. If U is the total energyin V , the rate of energy loss is

−dUdt

=

∮

S

P · dS =

∫

V

∇ ·PdV (by Divergence theorem)

We re-express the term ∇ ·P by substituting P = E×H, and making use

10-1

of a vector identities1 and Maxwell’s curl equations:

∇ ·P = ∇ · (E×H)

= (∇× E) ·H− (∇×H) · E

= −µ∂H∂t

·H− (J+ ε∂E

∂t) ·E

= −µ12

∂(H ·H)

∂t− ε

1

2

∂E · E∂t

− J · E

= − ∂

∂t(1

2µH2)− ∂

∂t(1

2εE2)− J · E

Thus∮

S

P · dS =

∫

V

∇ ·PdV

= −∫

V

∂

∂t(1

2µH2)dV −

∫

V

∂

∂t(1

2εE2)dV −

∫

V

J · E dV

= − d

dt

∫

V

(1

2µH2)dV − d

dt

∫

V

(1

2εE2)dV −

∫

V

J ·E dV

= − d

dtUM − d

dtUE −

∫

V

J · E dV

where

• UM =∫V (

12µH2)dV is the energy in the magnetic field within the vol-

ume V . The quantity 12µH

2 is the magnetic energy density in Jm−3.

• UE =∫V (

12εE

2)dV is the energy stored in the electric field within the

volume V . The quantity 12εE

2 is the electric energy density in Jm−3.

•∫V J · E dV is a term that represents either power dissipated through

ohmic losses or power generated by a source inside V.If representing power dissipated, the term J ·E can be re-expressed asJ · E = σE · E = σE2 in Wm−3.

1Identity 1: ∇ · (E×H) = (∇×E) ·H− (∇×H) ·EIdentity 2: ∂H

∂t·H = 1

2∂(H·H)

∂t

∂H∂t

·H = Hx∂Hx

∂t+Hy

∂Hy

∂t+Hz

∂Hz

∂t= 1

2∂H2

x

∂t+ 1

2

∂H2

y

∂t+ 1

2∂H2

z

∂t

10-2 AJW, EEE3055F, UCT 2012

Thus∮

S

P · dS = magnetic energy loss +electric energy loss +heat loss (in Js−1)

We therefore conclude that∮S P · dS =

∮E×H · dS represents the outflow

of power from a volume, and P = E × H is the power density vector, in

Wm−2.

Example of Electric Energy stored in a Capacitor

As an example of energy stored in a field, let us consider the energy storedin the electric field of a parallel plate capacitor, of surface area A, gap d,and charged to a voltage V. The capacitance is approximately C ≈ εA

dif

the gap is small compared to the plate dimensions. The electric field isconcentrated in the gap between the plates, and has strength of E ≈ V/d.

The energy stored in the electric field is

UE =

∫

V

(1

2εE2)dV

≈ 1

2εV 2

d2Ad

≈ 1

2

εA

dV 2

≈ 1

2C V 2

a familiar result from circuit theory. The relationship is exact (i.e. UE =12C V

2), and can be shown from first principles for a capacitor of any shape.

10.1 Power in a Sinusoidal Plane Wave

A sinusoidal electromagnetic plane wave propagating in the z direction isdescribed by the EM pair

E(z, t) = Ex(z, t)x

H(z, t) = Hy(z, t)y

10-3 AJW, EEE3055F, UCT 2012

whereEx(z, t) = E1 cos(ωt− kz +Ψ1)

Hy(z, t) =Ex(z, t)

η=E1

ηcos(ωt− kz +Ψ1)

The power density is

P = E×H = ExHy sinπ

2z =

E21

ηcos2(ωt− kz +Ψ1)z

At observer or antenna at a stationary point would feel the power pulsatingat twice the frequency of the source. The time averaged power density is

P = |E×H| = E21

ηcos2( ) =

E21

η(1

2+

1

2cos(2ωt− 2kz + 2Ψ1) =

1

2

E21

ηWm−2

10.1.1 Poynting Vector for Complex Notation

Phasor notation is commonly used for treatment of steady state sinusoidalsignals. The phasor representation of the sinusoidal wave EM plane wave is

E = Exx = E1ejΨ1e−jkzx

H = Hyy =E1

ηejΨ1e−jkzy

The Poynting vector for complex phasor representation is defined as

P =1

2E× H∗

such that the real part of P is the time averaged power density, i.e. ReP =P = E×H. To see why the factor of 1/2 is required, we proceeding asbefore, but substituting the phasor forms of Maxwell’s curl equations:

∮(E× H∗) · dS =

∫

V

∇ · (E× H∗) dV

=

∫

V

[(∇× E) · H∗ − (∇× H∗) · E

]dV

=

∫

V

[(−jωµH) · H∗ − (J∗ − jωεE∗) · E

]dV

=

∫

V

[−jωµH · H∗ + jωεE · E∗ − E · J∗

]dV

10-4 AJW, EEE3055F, UCT 2012

In the non-phasor case, J · E represents the instantaneous dissipated (orgenerated) power density in Wm−3 - in other words J · E is a function of

time. Here E · J∗ = σE · E∗ equals twice the time averaged power density(= 1

2σE2). Hence P is defined as P = 1

2E× H∗ in order that the real partequals time averaged power flow in Wm−2.

If we substitute the phasor representations for a plane wave,

P =1

2E× H∗

=1

2E1e

jΨ1e−jkz1

ηE1e

−jΨ1ejkz sin(π

2)z

=1

2

E21

ηz Wm−2

which agrees with the result previously obtained with the real signal repre-sentation.

10.2 Power density from Radiating Antennas

An isotropic antenna is an idealised radiating source that radiates power

uniformly in all directions. The power density at a distance r from thesource is

Pd =Pt

4πr2Wm−2

where Pt is the power in watts flowing from the source and 4πr2 is the

surface area of a spherical shell of radius r.

Isotropicradiator

Figure 10.1: An isotropic radiator radiating in all directions.

10-5 AJW, EEE3055F, UCT 2012

A physical antenna is usually designed to focus the energy in a particulardirection, and hence the power density increases compared to an isotropic

radiator. The increase in power density is specified by a parameter calledthe gain of the antenna, which is defined as the factor by which the powerdensity is increased above that of an isotropic radiator. The power density

in position (r, θ,Φ) is given by

Pd = G(θ,Φ)Pt4πr2

Wm−2

where G(θ,Φ) is the gain pattern of the antenna, which is a function ofangle. A typical gain pattern for a directional antenna is shown in Figure

10.2. The radiated power is concentrated within the main lobe. Additionalside lobes are visible which are usually undesirable, and antenna designers

will often seek to minimise the side lobe levels.The beam pattern can be understood as an interference pattern. The

signal radiated in a particular direction can be modelled as the sum of con-

tributions from elemental radiation sources covering the aperture (imaginea regular grid of tiny sources). The radiation contributions sum coherently

(in phase) in the main-lobe direction (constructive interference), whereas asthe angle increases of the “boresight”, the relative differences in path length

results in destructive interferences, which reduces the total signal strengthin the side lobes.

10-6 AJW, EEE3055F, UCT 2012

−180 deg 180 deg

Typical antenna gain versus angle

Main lobe

Sidelobe

3dB Beamwidth (G drops to 1/2 of max)

Directional Beam

radiator

θ

G(θ)

Ptθ3dB

θ3dB

Figure 10.2: A directional antenna and a typical antenna gain pattern.

The width of a beam is usually measured between the “half power points”on the main lobe, i.e. where G drops to half its peak value, or equivalently

“3 dB” below the peak.A well known result from antenna theory is that the beamwidth from a

radiating aperture (measured in a chosen plane) is approximately

θ3dB ≈ arcsinλ

D≈ λ

D

where D is the diameter of the aperture in the plane, and λ is the wave-

length. The approximation is valid for narrow beam antennas, for whichthe diameter of the aperture (measured in the plane) of the is several wave-

lengths wide.

10-7 AJW, EEE3055F, UCT 2012

Example

An antenna radiates 100 watts. Calculate the electric field intensity and

the magnetic field intensity at a distance r = 1000 m from the antennaassuming the antenna is:

(i) an isotropic radiator(ii) a horn antenna with a peak gain of 10.Assume that the polarization is linear and the field is locally approximated

as a plane wave.

Solution:

(i) For an isotropic radiator, the power density

Pd =100

4π10002= 8.0× 10−6 Wm−2

The power density for a plane wave is P = 12E2

η which implies

E =√2ηP =

√2 377 8.0× 10−6 = 0.08 Vm−1

H =E

η=

0.08

377= 2.1× 10−4 Am−1

(ii) For the horn antenna with a gain of 10,

Pd = 10100

4π10002= 8.0× 10−5 Wm−2

E =√2ηP =

√2 377 8.0× 10−5 = 0.25 Vm−1

H =E

η=

0.25

377= 6.5× 10−4 Am−1

10-8 AJW, EEE3055F, UCT 2012

10.3 Power Flow in a Simple Circuit

Consider the circuit shown below consisting of a resistor connected to a

battery via a pair of wires. The resistor is made of a cylinder of resistivematerial, placed between two metal plates as illustrated.

Metal plate

Battery Resistor

I

Length d

I

V

H

H

Ez

E E

PP

H into page

P

Hφ

J

Figure 10.3: Power flow in a DC circuit.

The magnetic field lines circulate around the wires and resistor as illus-trated (apply right hand rule for direction). The electric field lines extend

from the wire at the positive voltage down to negative voltage wire. Thepower flow at any point in space is described by P = E×H. Below is shown

a top view of the circuit. The Poynting vector points from left to right.

10-9 AJW, EEE3055F, UCT 2012

total power entering load.for calculatingIntegration surface

and resistor)(plane of battery

TOP VIEW

H

E into page

P = E×Hr

P

P

P

P

P

P

Figure 10.4: Top view showing power flow in a DC circuit,

Consider the region local to the resistor. The electric field between theplates is vertically directed (z-direction), and is given by

Ez =V

d=IR

d

and the magnetic field surrounding the resistor is

HΦ =I

2πr

The power flow at any point in space is described by the cross product

P = E×H

= −EzHΦ sin 90 r

= −IRd

I

2πrr

= − I2R

2πrdr

which is a vector that points radially inwards as illustrated2. From this weobserve that

• the power flows from the outside inwards towards the centre axis of thecolumn.

2You should verify the direction E×H by considering the cross product at several points in space.Note that r points radially outwards from the centre axis.

10-10 AJW, EEE3055F, UCT 2012

• the conductor will get warm; i.e. molecules vibrate more vigorously aselectromagnetic energy is converted into kinetic energy.

• energy is re-radiated by the mechanisms of

– thermal radiation, also known as black body radiation (so called

because objects that are dark in colour radiate energy effectivelyin the optical frequency range; dark objects are also good absorbers

in the optical band).

– convection, i.e. kinetic energy transfer to the molecules of thesurrounding air.

– conduction - i.e. through objects that are physically in contact

with the resistor, e.g. metal plates and wires.

The total power entering the resistor can be calculated by integrating thePoynting vector over a closed surface enclosing the resistor. Since E ≈ 0

in the top metal plates, we need only consider the contribution through theside walls of a cylinder of radius r (see Top View figure), which has surface

area = circumference× length = 2πr × d:

P =

∮P · dS ≈ − I2R

2πrd2πrd = I2R Watts

which is in agreement with circuit theory.

Question: How much power flows through the wires to the load?

Answer: For perfect wires, σ → ∞, and E → 0 within the wires. ThusP = E ×H → 0, within the wires, and we conclude that NO power flows

through the wires. From a field theory perspective, the power flows throughthe air. The power (i.e. the capacity to do work) is carried in the EM field,which travels from source to load. Why then do we need wires? The pair of

wires serve to guide the wave from source to load. Transmission line theoryis used to predict the behaviour of such guided waves when the length of

the transmission line is comparable to the wavelength.

10-11 AJW, EEE3055F, UCT 2012

10.3.1 Transmission Lines for 50 Hz AC Mains Electricity Supply

Wires pairs are used for getting EM power from a power station to a home,

which may be many kilometres away. At 50 Hz, however, the wavelengthλ = c/f = 3× 108/50 = 6× 106m = 6000km, which is considerably greater

than the transmission line lengths, and so circuit theory with appropriatelumped element models is more applicable for describing power transfer -even over considerable distances like across a country.

Below is depicted a cross section through a pair of wires: the conductor onthe left is carrying current I into the page (in the direction of the load) and

the conductor on the right is carrying the return current out of the page.Note the directions of the electric and magnetic field lines at this instant in

time. The Poynting vector P = E×H points into the page, in the directionof the load. Since the current is alternating “AC”, what happens to the

Poynting vector when the current reverses direction? (Answer: E×H stillpoints into the page - you should check this yourself).

points into the page

H

out of pageI

E

E

E

H

P = E×H

H

into pageI

Figure 10.5: Electromagnetic field surrounding a parallel wire transmission line. If thepolarity of the current changes, the E and H field lines reverse direction(although P = E × H points always points in the same direction - into thepage).

10-12 AJW, EEE3055F, UCT 2012

10.3.2 Transmission lines for Microwave Circuits

In microwave circuits, the wavelength (e.g. at 10 GHz, λ = 3 cm) is often

comparable to circuit board dimensions, and so wires carrying signals mustbe treated as transmission lines. A common method of circuit construc-

tion is to use a ground plane on the underside of the printed circuit board(PCB), and the signals are routed via tracks on the top side (see figure).The wavelength of the guided wave and the characteristic impedance of the

guiding structure are functions of the dielectric constant of the material, thewidth of the track, and the thickness of the PCB. For standard fibreglass

PCB, a transmission line with a 50 Ohm characteristic impedance can bemade by etching a track of about 2mm wide on the top side; the underside

is a ground plane. Increasing the width of the line reduces the characteristicimpedance; reducing the track width increases the characteristic impedance

of the line.

(copper)

Propagationdirection

fibreglassor speciallow lossmicrowave boarde.g. RT Duroid.

Substrate material

H

Ground plane

Copper track

E

Figure 10.6: Micro-strip microwave transmission line. The EM wave propagates as a guidedwave towards a (matched) load.

10-13 AJW, EEE3055F, UCT 2012

10.4 Electromagnetic safety considerations

The safe exposure level of EM radiation to humans is a function of radiation

frequency, intensity and the particular organs exposed.Detailed guidelines have been drawn up by panels of experts, which specify

maximum safe exposure levels under a wide range of conditions [1,2]. In thissection, we shall briefly summarise some of the key issues. The appropriate

regulations should be consulted for details.

• EM Radiation is classified into two categories:1) ionizing radiation (above visible frequencies f > 7.5 × 1014 Hz or

λ < 0.0004 mm, including ultraviolet 1016 Hz, X-rays 1018 Hz, gammarays 1023 Hz and higher).

2) non-ionising radiation (f < 7.5 × 1014 Hz or λ > 0.0004 mmfromi.e. from the visible frequencies range right down to DC, which includes

the infrared, microwave and RF ranges).

• Ionizing radiation can break chemical bonds (creating “ions” - hence

the name), resulting in tissue damage and damage to DNA. Even lowlevel exposure can lead to severe health problems.

• Non-ionizing radiation primarily causes heating in tissue, and is con-

siderably less harmful than ionizing radiation.

• The safety to humans is a function of the B field and E field strengths

to which one is exposed.

Figure 10.7 contains an illustration of the frequency spectrum showing the

effects on human tissue.In these lecture notes, we shall only discuss safety levels for non-ionizing

radiation in the frequency range from DC to 300 GHz (i.e. wavelengthsgreater than 1 mm).

10-14 AJW, EEE3055F, UCT 2012

Figure 10.7: [http://www.epa.gov/radiation/understand/ionize nonionize.htm, from theU.S. Environmental Protection Agency]. The border between ionizing andnon ionizing radiation is at f ≈ 7.5× 1014 Hz or λ = 0.0004 mm.

10.4.1 Non-Ionizing Radiation

The International Commission for Non-Ionizing Radiation Protection (IC-NIRP) [http://www.icnirp.de/] is an organisation that provides guidelines

[1] relating to exposure to non-ionising radiation.

• In the far field from a radiating source, the ratio of E to B is a constant

(E/H = η). Hence it is sufficient to specify either the E field strength,or the B field strength, or conveniently, the power density P = EH, as

a safety level. Usually the power density in Wm−2 is specified.

• Closer to a radiating source, in the so called near field, the ratio ofE/H is not constant, varying considerably as a function of position

in relation to the source, and the type of source itself (e.g. close to adipole antenna, the E field is comparatively stronger, whereas close to

a loop antenna, the B field is dominant. Thus specifying safety levelsis more complicated in the near field.

• Electric fields inside a human body cause currents to flow: J = σE

10-15 AJW, EEE3055F, UCT 2012

Parameter Units Frequency Range

Current density J Am−2 DC to 10 MHz

Current I A up to 110 MHz

Specific Energy Absorption Rate (SAR) W or Wkg−1 100 kHz to 10 GHz

Specific Energy Absorption (SA) Jkg−1 for pulsed fields 300 MHz to 10 GHz

Power Density P (or symbol S) Wm−2 10 GHz to 300 GHz

Table 10.1: Parameters used for specifying maximum levels for exposure to ionizingradiation.

where σ is the conductivity in Sm−1 in tissue and body fluids. Below100 Hz, typical values of σ are 0.4 Sm−1 in muscle, 0.7 Sm−1 in blood,

and 0.04 Sm−1 in fat. At higher frequencies, muscle in the tongue variesfrom 0.4 Sm−1 at 10 MHz, to 1.0 Sm−1 at 1 GHz, and to 10 Sm−1 at

10 GHz.

• The degree of penetration of the field is also a function of frequency

(the “skin depth” phenomenon).

The ICNIRP guidelines [1] are specified in terms of maximum levels of one

or more of the parameters listed in Table 10.1. The guidelines cover thefrequency range from 300 GHz to DC.

10.4.2 Coupling mechanisms between EM Fields and the Human Body

There are three identified mechanisms:

1. Coupling of low frequency E fields:

E fields cause currents to flow in the body, as well as polarization ofbound charge, and reorientation of electric dipoles in the tissue.

2. Coupling of low frequency E fields:

Changing magnetic fields induce E fields and hence current flow.

3. Absorption of Energy from EM fields:

• for frequencies < 100 kHz, the energy absorption is minimal

• > 100 kHz can lead to significant energy absorption and temperatures

rises in the body.

10-16 AJW, EEE3055F, UCT 2012

Four frequency ranges are described (Durney et al. 1985, in [1]):

1. From 100 kHz to 20 MHz

2. From 20 MHz to 300 MHz: high absorption can occur in the wholebody. The adult human body resonance is between 70 MHz and 100MHz depending on the height of a standing person, for vertical polar-

ization.

3. From 300 MHz to 10 GHz: significant local non-uniform absorption.

4. > 10 GHz: energy absorption is primarily at the body surface. (In thisrange, it is better to specify the incident power density and not the

SAR.)

10.4.3 Guidelines for Limiting EM Exposure

Consult the ICNIRP guidelines [1] (extract distributed as a class handout).In it you will find the following tables listing reference levels for limiting

EM exposure:

1. Table 4: Basic restrictions up to 10 GHz

2. Table 5: Basic restrictions for power density between 10GHz and 300GHz.

3. Table 6: Reference levels for occupational exposure (i.e. people in a

working environment).

4. Table 7: Reference levels for the general public.

Plots of the E field and B field levels are shown in Figure 2 for the casesof “general public exposure”, “peak general public”, “occupational exposure”

and “peak occupational exposure”.A rule of thumb for microwave engineers working above 10GHz, is to avoid

exposure to radiation levels above about 10 W/m2. The reference level forthe range 10 GHz to 300 GHz for occupational exposure is 50 W/m2 andfor general public exposure is 10 W/m2.

10-17 AJW, EEE3055F, UCT 2012

To gain some“feeling”for the numbers, we can calculate the power densityat a distance of one metre from a 50 Watt light bulb, assuming isotropic

radiation. The radiation power density at a distance of 1m would bePtG/(4πr

2) = 50 ∗ 1/(4π12) = 3.9 W/m2. This of the same order as the 10W/m2 specified for microwave radiation (although it is noted that most of

the power emitted by light bulbs is in the optical to infrared region .

References on EM radiation Safety:

1. “Guidelines for limiting exposure to time-varying electric, magnetic, andelectromagnetic fields (up to 300 GHz)”, published by the International

Commission for Non-Ionizing Radiation Protection (ICNIRP), reprinted inHealth Physics, April 1998, vol. 74, no. 4, pp. 494-522.

Downloadable from http://www.icnirp.de2. “A Local Government Official’s Guide to Transmitting Antenna RF

Emission Safety: Rules, Procedures, and Practical Guidance”, Federal Com-munications Commission (FCC), June 2, 2000.3. “Compilation of the dielectric properties of body tissues at RF and

microwave frequencies”, C. Gabriel and S.Gabriel, Technical Report 1996,http://www.brooks.af.mil/AFRL/HED/hedr/reports/dielectric/Report/Report.html

4. “Cellular Phone Antennas (Mobile Phone Base Stations) and HumanHealth”, J. Moulder, Professor of Radiation Oncology,

http://www.mcw.edu/gcrc/cop/cell-phone-health-FAQ/5. U.S. Environmental Protection Agency Website:

[http://www.epa.gov/radiation/understand/ionize nonionize.htm].

10-18 AJW, EEE3055F, UCT 2012

11 Reflection and Refraction atBoundaries

We shall consider the following cases:

1. Reflection from a perfect conductor: wave normally incident.

2. Reflection and transmission at a dielectric interface: wave normallyincident.

3. Reflection from a perfect conductor: wave incident at an arbitrary

angle.

4. Reflection and transmission (refraction) at a dielectric interface: waveincident at an arbitrary incident angle.

11.1 Reflection of Normally Incident Plane Waves from a

Perfect Conductor

Consider a plane wave normally incident on a flat perfectly conducting plate(σ = ∞). The pointing vector tells us that the energy does not pass through

the conductor - why? P = E × H and E = 0 so P = 0. All energy musttherefore be “reflected”.

A good question is: Why should an incoming wave“bounce”off a conduct-ing surface? What is physically going on on the surface? My explanation

is that the incoming wave does not bounce at all - rather the electrons onthe surface react to form an induced electric field of the opposite polarityto that of the incident plane wave. Thus the reacting current sheet on the

surface re-radiates the so-called “reflected wave” (but with an inversion inpolarity).

The next question that arises is: Why does the incident wave not pass

11-1

straight through the metal? In a strict sense, “yes” it does on propagatingforward, but it is also accompanied by a forward propagating wave arising

from the surface currents. These two forward propagating waves are in-verted, with respect to one another, and sum to zero. Thus in speech wesay that the incident wave “bounces off the surface” - but more correctly,

the phenomenon of reflection is a re-radiation effect.

z

xIncident wave

Reflected

Perfectconductor

Figure 11.1: Reflection from a perfect conductor.

We now analyse the steady state plane wave solution to the wave equationsubject to the appropriate boundary condition: tangential components of

E are zero at the boundary, i.e. Ex(0) = and Ey(0) = 0.The phasor forms of the incident and reflected waves are

Ex = E+e−jkz + E−e

+jkz

Hy =1

η

[E+e

−jkz − E−e+jkz

]

Applying the boundary condition

0 = E+e−jk0 + E−e

+jk0

impliesE− = −E+

Thus the total field in terms of E+ is

Ex = E+e−jkz − E+e

+jkz = E+

[e−jkz − e+jkz

]= −E+2j sin kz

11-2 AJW, EEE3055F, UCT 2012

The instantaneous form is thus

Ex(z, t) =ReExe

jωt

=Re −E+2j sin kz [cosωt+ j sinωt]=2E+ sin kz sinωt

This sum of the forward and reflected wave does not contain a “z − vt”term and hence does not appear to move left or right - it is what is known

as a “standing wave”. The peak amplitude |Ex(z, t)|peak = |2E+ sin kz| (or“envelope”) varies with distance from the interface, as plotted below.

zz = 0

Envelope ofelectric field

magnetic fieldEnvelope of

−λ4−3λ

4

zz = 0

2E+

−λ −λ2

2E+

η0

Figure 11.2: Standing wave resulting from reflection of a sinusoidal wave by a perfect con-ducting surface.

The expression for the magnetic field is similarly obtained by substituting

E− = −E+ in the expression for Hy yielding

Hy =1

η

[E+e

−jkz − E−e+jkz

]=

1

ηE+

[e−jkz + e+jkz

]=

1

η2E+ cos kz

with instantaneous form

Hy(z, t) = ReHye

jωt= Re

1

η2E+ cos kz [cosωt+ j sinωt]

=1

η2E+ cos kz cosωt

Notice, that at the metal interface, the peak value Hy(z = 0, t) is in fact a

maximum, as is shown in the plot of the resulting standing wave.

11-3 AJW, EEE3055F, UCT 2012

11.2 Transmission Line Analogy

The equations governing reflection of plane waves are identical in form to

those derived for transmission lines. The following table compares equiva-lent relationships.

Fields Transmission Lines

Ex = E+e−jkz + E−e+jkz Vx = V+e

−jβz + V−e+jβz

Hy =1η

[E+e

−jkz − E−e+jkz]Iy =

1Z0

[I+e

−jβz − I−e+jβz]

k = ω√µε β = ω

√LC

η =√

µε Z0 =

√LC

All transmission line concepts and formulas are applicable to the reflection

of plane waves, e.g. SWR, reflection coefficient, Smith chart calculations etc.The units of E are V/m, and the units of H are A/m, analogous to voltage

and current in the transmission line.The wave impedance at a position z is defined as the ratio of total electric

to magnetic field in the plane

η(z) =Ex(z)

Hy(z)

Directly from the transmission line analogy, the impedance at a distance lfrom a “load plane” of impedance ηL is

η(l) = ηηL cos kl + jη sin kl

η cos kl + jη sin kl

and the reflection and transmission coefficients are

ρ =E−E+

=ηL − η

ηL + η

and

τ =E2

E+=

2ηLηL + η

where E2 is the amplitude of the wave in medium two.

11-4 AJW, EEE3055F, UCT 2012

Example: Application of Transmission Line Analogy to Incidence on aPerfect Conducting Plane

At the interface with a perfect conductor, the electrons on the surface movein such a way as to force the electric field to zero. The load impedance as

seen by the plane wave is

ηL =Ex(0)

Hy(0)=

0

Hy(0)= 0

Substituting, we getρ = −1

and

τ = 0

and

η(l) = jη tan kl

which should be familiar results from transmission line theory.

11.3 Normally Incident Plane Wave on a Dielectric

Interface

Again we apply the results of transmission line theory with η1 =√

µ1

ε1and

η2 =√

µ2

ε2to obtain

ρ =E−E+

=η2 − η1η2 + η1

and

τ =E2

E+=

2η2η2 + η1

The fraction of power that is reflected is |E−|2

|E+|2= |ρ|2 and the fraction of

power that is transmitted into the medium is therefore 1− |ρ|2.

11-5 AJW, EEE3055F, UCT 2012

Example Calculation

Calculate the fraction of transmitted and reflected power for the case of a

1mW laser beam normally incident from air (ε = ε0 and µ = µ0) onto glass(εr = 5 µ = µ0).

Solution

The impedances in air is

η1 ≈√µ0ε0

≈√

4π × 10−7

8.85× 10−12≈ 377Ohms

and in the glass is

η2 =

√4π × 10−7

5× 8.85× 10−12≈ 169Ohms

The reflection coefficient is

ρ =η2 − η1η2 + η1

=169− 377

169 + 377= −0.380

and hence |ρ|2 = 0.145. The reflected power is |ρ|2 Pinc = 0.145 mW andthe transmitted power is the balance being 0.855 mW.

11.4 Reflection and Transmission (refraction) at a

dielectric interface - arbitrary incident angle

(ref: Griffiths 1981, and Ramo et al. 1994)

In this section, the relationships between incident, reflected and refracted

waves are studied for a wave incident on a dielectric interface.

Consider a wave incident on a dielectric boundary on the plane z = 0 as

depicted below.

11-6 AJW, EEE3055F, UCT 2012

Medium 1 Medium 2

z

x

Incident wave

Reflected

k+

k−

θ+

θ−k2

θ2

Transmitted

Figure 11.3: Propagation directions of incident, reflected and refracted waves at an interfacebetween two media.

The problem is analysed as follows:

1. Boundary conditions are applied to the case of arbitrary linear polar-ization to establish

a) the fact that the incident, reflected and refracted waves lie in the

same plane of propagation known as the “plane of incidence”; i.e.the vectors k+, k− and k2 all lie in the same plane.

b) Snell’s laws relating the angles of reflection and refraction to theincident angle.

2. The amplitudes of the reflected and refracted waves are analysed byresolving the incident wave into two orthogonal polarizations:

a) Case where the electric field is perpendicular to the plane of inci-

dence, known as transverse electric polarization (or TE polariza-tion).

b) Case where the magnetic field is perpendicular to the plane ofincidence - referred to as transverse magnetic polarization (or TM

polarization).

11-7 AJW, EEE3055F, UCT 2012

11.4.1 Derivation of Snell’s Laws

We consider first, a wave of arbitrary linear polarization incident on the

z = 0 plane at an angle θ. We model the incident, reflected and refractedwaves in phasor notation as

E+(r, t) = E+e−jk+·rejωt H+(r, t) =

1

η1k+ × E+(r, t)

E−(r, t) = E−e−jk−·rejωt H−(r, t) =

1

η1k− × E−(r, t)

E2(r, t) = E2e−jk2·rejωt H2(r, t) =

1

η1k2 × E2(r, t)

NOTE: The term k+ · r = k+k · r is just k+ (the wave number) times thecomponent of r in the k direction. Imagine defining an axis pointing in the

k direction, and let the distance along this axis be ζ. The term e−jk+k·r isequal to e−jk+ζ .The total electric field in medium 1 is the sum of incident and reflectedfields

E1(r, t) = E+e−jk+·rejωt + E−e

−jk−·rejωt

The angles of incidence and reflection are found by imposing the boundaryconditions previously derived in Section 4, i.e.

(a) Dn1 = Dn2 Normal component of electric flux density is continu-

ous

(b) Et1 = Et2 Tangential components of E are equal on either side of

interface.

(c) Bn1 = Bn2 Normal component of magnetic flux density is contin-uous.

(d) Ht1 = Ht2 Tangential components of H are equal on either side

of interface.

11-8 AJW, EEE3055F, UCT 2012

It can be shown that imposition of all boundary conditions results in equa-tions of the form

( )e−jk+·r + ( )e−jk−·r = ( )e−jk2·r at z = 0

where the bracketed quantities ( ) are either parallel, perpendicular or nor-

mal components of the E or H fields.

Consider, for example, the electric field. At the interface plane, the bound-ary condition Et1 = Et2 becomes:

(E+)xe−jk+·rejωt + (E−)xe

−jk−·rejωt = (E2)xe−jk2·rejωt at z = 0

and

(E+)ye−jk+·rejωt + (E−)ye

−jk−·rejωt = (E2)ye−jk2·rejωt at z = 0

where (E+)x is notation for the x component of E+ etc. The other boundary

conditions result in similar equations.

For the constraint

( )e−jk+·r + ( )e−jk−·r = ( )e−jk2·r at z = 0

to hold true for all x, y in the z = 0 plane requires1 that

k+ · r = k− · r = k2 · r at z = 0

or(k+)xx+ (k+)yy = (k−)xx+ (k−)yy = (k2)xx+ (k2)yy

This must hold along the line y = 0 in the xy plane, and so

(k+)x = (k−)x = (k2)x

and similarly it must hold along the line x = 0 in the xy plane, and so

(k+)y = (k−)y = (k2)y

1The only way this can hold true for all positions at z = 0 if the phase factors are equal.

11-9 AJW, EEE3055F, UCT 2012

Thus we can conclude that k+, k− and k2 lie in the same plane (known asthe plane of incidence).

If we orientate the axes such that the plane of incidence is the xz planeas shown in the figure, then the y components of the k vectors are zero, and

the remaining boundary condition

(k+)x = (k−)x = (k2)x

impliesk1 sin θ+ = k1 sin θ− = k2 sin θ2

from which we obtain Snell’s laws of reflection and refraction:

k1 sin θ+ = k1 sin θ− ⇒ θ+ = θ−

and

k1 sin θ+ = k2 sin θ2 ⇒ sin θ2sin θ+

=k1k2

=ω/v1ω/v2

=v2v1

Since for most dielectrics, µ1 ≈ µ2 ≈ µ0, the ratio

sin θ2sin θ+

=v2v1

=c/n2c/n1

=n1n2

≈√ε1√ε2

≈ η2η1

where n1 = c/v1 and n2 = c/v2 are the refractive indexes of the dielectrics.To analyse the magnitudes of reflected and transmitted waves, we need

examine the two cases of a transverse electric (TE) incident wave, and atransverse magnetic (TM) incident wave. An arbitrary polarization can besplit into the sum of these two orthogonal components.

11.4.2 Case Transverse Magnetic (TM) Incident Wave

(also known as “parallel polarization”)

If the electric field is parallel to the incident plane (and the magnetic fieldperpendicular), there exist only Ex , Ez and Hy components as shown in

the illustration below.

11-10 AJW, EEE3055F, UCT 2012

(into)

(out)

CASETransverseMagnetic(TM)

Medium 2Medium 1

z

x

k+

k−

E+

E−

H−

H+

θ+

θ−H2

E2

k2θ2

Figure 11.4: Definitions of vector quantities on either side of an interface for the case of aTM incident wave.

H+ = yE+

η1e−jk+·r E+ = (y× k+)E+e

−jk+·r

H− = −yE−η1e−jk−·r E− = −(y× k−)E−e

−jk−·r

H2 = yE2

η2e−jk2·r E2 = (y × k2)E2e

−jk2·r

Take note of the directions of the vectors indicated: these represent the di-rections in which E and H point at the interface (i.e. at x = 0 and z = 0) at

time t = 0, assuming the constants E+, E−, E2 were all positive. Also, it isnoted that H− is defined as pointing into the page, such that when θ+ = 0,

E− and E+ will point in the same direction (for positive E+, E−, E2), toallow comparison with the derivation in Section 9.1. The relationships be-

tween E+, E− and E2 will now be established by applying the boundaryconditions.

The boundary conditions imposed on the x, y and z components of theelectric and magnetic fields are (see diagram) are

11-11 AJW, EEE3055F, UCT 2012

(a) Dn1 = Dn2 ⇒ ε1(−E+ sin θ+) + ε1(E− sin θ−) = −ε2E2 sin θ2

(b) Et1 = Et2 ⇒ E+ cos θ+ + E− cos θ− = E2+ cos θ2

(c) Bn1 = Bn2 ⇒ nothing useful as Hz = 0

(d) Ht1 = Ht2 ⇒ H+ − H− = H2 ⇒ 1η1E+ − 1

η1E− = 1

η2E2

Thus (d) implies (so does (a) after substituting Snell’s law and some ma-

nipulation)

E+ − E− =η1η2E2

and (b) implies

E+ + E− =cos θ2cos θ+

E2

We solve these two equations to obtain the reflected and transmitted am-

plitudes in terms of the incident amplitude:

E− =η2 cos θ2 − η1 cos θ+η2 cos θ2 + η1 cos θ+

E+

and

E2 =2η2 cos θ+

η2 cos θ2 + η1 cos θ+E+

The reflection and transmission coefficients are defined as

ρTM =E−

E+

=η2 cos θ2 − η1 cos θ+η2 cos θ2 + η1 cos θ+

and

τTM =E2

E+

=2η2 cos θ+

η2 cos θ2 + η1 cos θ+For the special case of the incident angle being zero, i.e. θ+ = θ− = θ2 = 0,we obtain, as in Section 11.3,

ρTM =η2 − η1η2 + η1

and

τTM =2η2

η2 + η1

11-12 AJW, EEE3055F, UCT 2012

The Brewster Angle

It is interesting2 to plot ρ and τ as a function of incidence angle varying

from 0 degrees (normal incidence) to 90 degrees. Is is noted that at acertain angle known as the Brewster angle, ρTM = 0 hence the no signal is

reflected. This angle occurs when the numerator (η2 cos θ2 − η1 cos θ+) = 0,

which, after substituting Snell’s law and ηi =√

µi

εiand some manipulation

we obtain

sin2 θ+ =1− µ2ε1

µ1ε2

1− (ε1ε2 )2

For the case where µ1 ≈ µ2 ≈ µ0, the relationship simplifies to

tan θ+ ≈√ε2ε1

≈ n2n1

It is also noted that the sign of the reflection coefficient changes from nega-tive to positive, implying that the phase of the reflected wave jumps by 180

degrees as the incidence angle crosses the Brewster angle.

11.4.3 Case Transverse Electric (TE) Incident Wave

(also known as “perpendicular polarization”)If the electric field is perpendicular to the incident plane (and the magnetic

field parallel), there exist only Ey and Hx and Hz components as shown inthe illustration below.

2See example plot for “Example: Plane wave air into glass”

11-13 AJW, EEE3055F, UCT 2012

CASETransverseElectric(TE)

Medium 1 Medium 2

(out)

z

x

k2+

θ+

θ−

k+

H+

E+

E−

k−

H−

E2

θ2H2

Figure 11.5: Definitions of vector quantities on either side of an interface for the case of aTE incident wave.

E+ = yE+e−jk+·r H+ = (k+ × y)

E+

η1e−jk+·r

E− = yE−e−jk−·r H− = (k− × y)

E−η1e−jk−·r

E2 = yE2e−jk2·r H2 = (k2 × y)

E2

η2e−jk2·r

The boundary conditions imposed on the x, y and z components of the

electric and magnetic fields are (see diagram) are

(a) Dn1 = Dn2 ⇒ nothing useful as Ez = 0

(b) Et1 = Et2 ⇒ E+ + E− = E2

(c) Bn1 = Bn2 ⇒ µ1H+ sin θ+ + µ1H− sin θ− = µ2H2 sin θ2

⇒ µ1

η1E+ sin θ+ + µ1

η1E− sin θ− = µ2

η2E2 sin θ2

11-14 AJW, EEE3055F, UCT 2012

(d) Ht1 = Ht2 ⇒ −H+ cos θ+ + H− cos θ− = −H2 cos θ2

⇒ − 1η1E+ cos θ+ + 1

η1E− cos θ− = − 1

η2E2 cos θ2

Thus we have from (b) and (and(c) after substituting Snell’s law and some

manipulation)E+ + E− = E2

and from (d)

E+ − E− =η1 cos θ2η2 cos θ+

E2

Solving for E− and E2 we obtain

E− =η2 cos θ+ − η1 cos θ2η2 cos θ+ + η1 cos θ2

E+

and

E2 =2η2 cos θ+

η2 cos θ+ + η1 cos θ2E+

The reflection and transmission coefficients are

ρTE =E−

E+

=η2 cos θ+ − η1 cos θ2η2 cos θ+ + η1 cos θ2

and

τTE =E2

E+

=2η2 cos θ+

η2 cos θ+ + η1 cos θ2

For TE polarization, it can be shown3 that for dielectrics with equal perme-

abilities, there exists no angle for which ρTE = 0. The Brewster angle onlyoccurs for TM polarization. Thus the Brewster angle is sometimes called

the polarization angle, as light with an arbitrary polarization incident atthe Brewster angle is polarized on reflection. The TM components are not

reflected, but the TE components are4.

3Do this as an exercise by analysing the numerator of ρ.4See example plot for “Example: Plane wave air into glass”; compare TM and TE plots.

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11.4.4 Angle of Total Reflection

Total reflection occurs when |ρ| = 1. Analysis shows that for both TE and

TM polarization this occurs only for the case of an EM wave incident froma medium of slower velocity to that of a higher velocity, i.e. for v2 > v1.

The critical incidence angle θ+ = θc occurs at

sin θc =v1v2

≈√ε2ε1

At angles greater than this complete internal reflection occurs. The interface

behaves as a perfect mirror.Examples of the application of total reflection include: (i) use of prisms as

reflectors in optics (ii) light propagates down optic fibres, which are madeof glass, as a consequence of total internal reflection.

11-16 AJW, EEE3055F, UCT 2012

Example 1: Plane Wave from air into glass

Air: εr = 1.0, µr = 1.0 ⇒ v1 = c = 3.0× 108ms−1 η1 = 377 ohms

Glass: εr = 5, µr = 1.0 ⇒ v2 =c√εr= 1.35× 108ms−1 η2 = 169 ohms

The transmitted wave is refracted towards the normal, as illustrated inthe diagram below. The wavelength will be shorter in the glass than in air

by the factor1√εr

=1√5= 0.46

Medium 1

AIR

Medium 2

GLASS

θ+

θ−z

x

Incident wave

Reflected

Transmitted

θ2

Figure 11.6: An incident wave going from air into glass.

11-17 AJW, EEE3055F, UCT 2012

The Brewster angle for TM polarization occurs at θB = 65.9 degrees.NOTE: Compare the reflection coefficient |ρ| as the incident angle changes

from 0 to 90 degrees. For TE polarization, |ρTE| increases smoothly fromabout |ρTE| = 0.4 to |ρTE| = 1. For TM polarization, as θ increases from 0,|ρTM | decreases smoothly and reaches |ρTM | = 0 at θ = θB = 65.9 degrees;

increasing the angle further increases |ρTM | until |ρTM | = 1 at θ = 90degrees.

0 10 20 30 40 50 60 70 80 900

0.2

0.4

0.6

0.8

1Magnitude of Reflection coefficient

Incident angle in degrees

|rho| TM|rho| TE

0 10 20 30 40 50 60 70 80 900

0.1

0.2

0.3

0.4

0.5

0.6

0.7Magnitude of Transmission coefficient


|tau| TM|tau| TE

CASE: Parameters for plane wave from AIR into GLASS.

Brewsterangle

|ρTM |

|ρTE |

|τTE ||τTM |

Figure 11.7:

11-18 AJW, EEE3055F, UCT 2012

The set of figures below show ρTM , τTM , ρTE, τTE again for the air intoglass case, but the signed values; it can be seen that the phase of ρTM flips

by 180 degrees at the Brewster angle.

0 10 20 30 40 50 60 70 80 900

5

10

15

20

25

30Angle of Refraction vs Angle of Incidence


Ref

ract

ion

angl

e (d

eg)

0 10 20 30 40 50 60 70 80 90−0.4

−0.2

0

0.2

0.4

0.6

0.8

1Transverse Magnetic Incident Wave


rhotau

0 10 20 30 40 50 60 70 80 90−1

−0.5

0

0.5

1Transverse Electric Incident Wave


rhotau

Brewsterangle

From Air into Glass

Plots of various parameters for the case of AIR into GLASS

as a function of the incident angle.

τTM

ρTM

τTE

ρTE

θ2 versus θ+

Figure 11.8:

11-19 AJW, EEE3055F, UCT 2012

Example 2: Plane Wave from glass into air

The transmitted wave is refracted away from the normal as it enters the

air, as illustrated in the diagram below. The wavelength will increase as thewave enters by the factor

√5 = 2.2, in the air.

Medium 1 Medium 2

GLASS AIR

θ+

θ−z

x

Incident wave

Reflected

Transmitted

θ2

Figure 11.9: An incident wave going from glass into air.

NOTE:

Brewster angle for TM polarization occurs at θB = 24.1 degrees.Angle of total internal reflection occurs at θc = 26.6 degrees.Reflection coefficient becomes complex for θ > θc and an angle-dependent

phase shift occurs (not plotted here).Below are plotted ρ and τ for 0 < θ < θc.

11-20 AJW, EEE3055F, UCT 2012

0 10 20 30 40 50 60 70 80 900

20

40

60

80

100Angle of Refraction vs Angle of Incidence


Re

fra

ctio

n a

ng

le (

de

g)

0 10 20 30 40 50 60 70 80 90−1

0

1

2

3

4Transverse Magnetic Incident Wave


rhotau

0 10 20 30 40 50 60 70 80 900

0.5

1

1.5

2Transverse Electric Incident Wave


rhotau

angleBrewster

Brewster

From Glass into Air

Plots of various parameters for the case of GLASS into AIR

as a function of the incident angle.

τTM

τTE

ρTE

ρTM

θ2 versus θ+

Figure 11.10:

11-21 AJW, EEE3055F, UCT 2012

11.4.5 Power Considerations

For a wave at an oblique angle, the power density (referred to as the intensity

I) crossing the dielectric interface is given by the component of the Poyntingvector normal to the interface, i.e.

I = P · n Wm−2

where P is the Poynting vector and n = z is the surface normal.

Medium 1 Medium 2

x

P−

P+

zn

P2

θ2

θ+

θ−

Figure 11.11: Poynting vectors on either side of an interface.

For the incident wave

P+ =1

2

E2+

η1k+

and thus

I+ = P+ · n =1

2

E2+

η1cos θ+

For the reflected wave

P− =1

2

E2−η1

k−

and thus

I− = P− · n = −1

2

E2−η1

cos θ−

11-22 AJW, EEE3055F, UCT 2012

For the transmitted wave

P2 =1

2

E22

η2k2

and thus

I2 = P2 · n =1

2

E22

η2cos θ2

By substituting∣∣∣E−

∣∣∣ = |ρ|∣∣∣E+

∣∣∣ and∣∣∣E2

∣∣∣ = |τ |∣∣∣E+

∣∣∣ it can be seen that

|I+| − |I−| = |I2|

which is consistent with the fact that the net energy per second entering a

surface from one side must balance with that leaving from the other side.

Note: It should be understood that

|P+| − |P−| 6= |P2|

unless the incident angle is zero.

11-23 AJW, EEE3055F, UCT 2012

12 Propagation in Conducting Mediaand the Skin Depth

In this chapter the following topics are covered

1. Wave Equation in conducting media

2. Skin depth

3. Skin effect in conducting wires carrying AC currents

References:“Electromagnetics”by IS Grant and WR Phillips.“Introduction to Electrodynamics” by DJ Griffiths

12.1 Wave propagation in a conducting medium

Wave propagation in a conducting medium results in an exponential decayof the wave amplitude as it propagates. The following illustration shows

an exponentially decaying sinusoidal EM wave. Notice that there exists aphase shift between E and B fields in such a medium.In the treatment that follows, the particular case of a propagation in a

highly conductive material (known as a good conductor) is considered.

12.1.1 Conducting media (particularly “good conductors”)

In a conducting medium, free electrons will move under the influence ofthe electric field i.e. Jc = σE. A “good conductor”, is one for which the

conduction current is significantly greater than the displacement current,i.e. |Jc| ≫ |Jd| =

∣∣∂D∂t

∣∣, which allows us to neglect the displacement current

term in Maxwell’s equations.

12-1

Figure 12.1: Exponentially decaying plane wave in a conducting (lossy) medium. Note Eand H are not in phase in a lossy medium.

If all field quantities are sinusoidal of form ejωt, then |Jc| ≫∣∣∂D∂t

∣∣

⇒ |σE| ≫∣∣∣∣ε∂E

∂t

∣∣∣∣ = |jωεE|

⇒ σ ≫ ωε

Thus for a physical material to be regarded as a good conductor, the fre-

quency satisfies

ω ≪ σ

ε

This requirement is satisfied for all metals from DC to 100 GHz (and con-

siderably higher in frequency).

12.1.2 Derivation of the Differential Equation

Consider Maxwell’s equations in a conducting medium for which ρ ≈ 0inside the medium and J = σE ≫ ∂D

∂t . The equations become

12-2 AJW, EEE3055F, UCT 2012

∇ ·D = ρ ≈ 0

∇ ·B = 0

∇×E = −∂B∂t

∇×H = J+∂D

∂t≈ J

Let us derive an equation containing only E. We start by taking the curl of

Faraday’s law

∇×∇×E = ∇× (−∂B∂t ) = − ∂

∂t∇×B

into which we can substitute the “good conductor” approximation

∇×B = µJ = µσE

Additionally, we can use the identity and the approximation ∇ ·E = ρ ≈ 0

∇×∇×E = −∇2E+∇∇ · E ≈ −∇2E

Making these substitutions, we obtain a form of the wave equation governing

propagation in good conductors:

∇2E = µσ∂E

∂t

12.1.3 Plane wave solution inside the conducting medium

Consider a plane wave propagating in z direction. For this case, ∂∂x = 0 and

∂∂y

= 0, and the wave equation simplifies to

∂2E

∂z2= µσ

∂E

∂t

which has a solution of the form

E = E0ej(ωt−βz)e−αz

modelling a wave propagating in the forward direction. The factor e−αz

represents a decaying amplitude in the z direction.

12-3 AJW, EEE3055F, UCT 2012

To obtain expressions for the constants α and β, substitute this solutionback into the wave equation to obtain

LHS =∂

∂z

[∂E

∂z

]=

∂

∂z

∂

∂z

[E0e

j(ωt−βz)e−αz]

=∂

∂z

[(−α− jβ)E0e

j(ωt−βz)e−αz]

=∂

∂z[(−α− jβ)E]

= (−α− jβ)∂E

∂z= (−α− jβ)2E

and

RHS = µσ∂E

∂t= µσ · jωE

Equating LHS = RHS implies

(−α− jβ)2 = jωµσ

α2 − β2 + 2jαβ = jωµσ

Equating real parts givesα2 − β2 = 0

which implies α = β (or α = −β). Equating imaginary parts gives

2αβ = ωµσ

substituting α = β implies2α2 = ωµσ

or

α = β =

√ωµσ

2

The physical sinusoidal signal is

E = ReE0e

j(ωt−βz)e−αz

= E0 cos(ωt− βz) e−αz

12-4 AJW, EEE3055F, UCT 2012

Figure 12.2: Exponentially decaying plane E(z, t = 0) wave in a“good conductor”medium.The E field envelope is E0e

−z/δ.

A plot of one of the components of the electric field is shown in the followingillustration.

Analysis of the H field shows that H is is 45 degrees out of phase withthe E field - see tutorial example.

12.2 Skin Depth

The exponential decay e−αz, may be characterised by the decay constant

1/α, known as the “skin depth”

δ =1

α=

√2

ωµσ

The skin depth is the distance at which the wave in the medium has decayed

to 1/e ≈ 0.707 of its original amplitude. Note δ ∝ 1√ωi.e skin depth reduces

rapidly as frequency increases.

12-5 AJW, EEE3055F, UCT 2012

Figure 12.3:

Example

Consider a sinusoidal 50 Hz plane wave propagating inside copper (which

is a very good conductor) σ = 5.96× 107 µr = 1.00 at 300 K. The skindepth at 50 Hz is

δ =

√2

4π10−7 × 5.9× 107 × 2π50= 0.0092m ≡ 9.2mm

At f = 50 MHz (radio frequency)

δ ≈ 10−5 m ≡ 0.01mm

At f = 10 Gz (radar & microwave communications)

δ = 6.5× 10−7m

At RF and Microwave frequencies, the penetration into metal is small (less

than a micrometre). Significant currents flow on the surface of good con-ductors (i.e. concentrated mainly within skin depth) - we talk of the current

as being a “surface current”.The table shows the skin depth for copper ( σ = 5.96× 107) for a range

of frequencies.

12-6 AJW, EEE3055F, UCT 2012

Frequency δ for Copper

50 Hz 9.2 mm

10 kHz 0.65 mm

100 kHz 0.21 mm

1 MHz 0.065 mm

10 MHz 0.021 mm

10 GHz 0.00065 mm

12.3 Skin Effect in Conducting Wires

At high frequencies, the AC current flowing in a conducting wire is concen-trated in thin outer layer or “skin” of the wire, as is illustrated in the cross

sections below.

Figure 12.4:

This concentration of current density in a thin outer layer causes the

resistance (per metre) to increase as frequency increases. A plot showingthe variation in current density J across the diameter of a 1mm conductor

is shown below [Ramo, Whinnery & van Duzer].

12-7 AJW, EEE3055F, UCT 2012

Figure 12.5:

DC Resistance

The DC resistance of a piece of wire of length l and thickness 2r is

R0 = lσA =

l

σπr2[Ω]

where A = πr2 is the cross sectional area.

Figure 12.6:

12-8 AJW, EEE3055F, UCT 2012

High Frequency AC Resistance

The high frequency resistance of a length of wire of l metres can be crudely

estimated by pretending that the current is uniform within an effectiveconducting area Askin = 2πrδ. With this giant leap of faith, we obtain a

formula for the high frequency resistance

RHF ≈ l

σAskin=

l

σ2πrδ=

1

2πr

√µ0ω

2σ[Ω]

which turns out to be not far out in practice for the case where δ ≪ r.

The high frequency AC resistance can also be express as

RHF ≈ l

σ2πrδ= R0 ×

r

2δ[Ω]

Example

Consider a the difference in resistance per metre at DC and at 50 MHz ofa copper wire of diameter 5 mm. ⇒ r = 2.5× 10−3 m

At DC,R0

l≈ 0.0008 Ωm−1

At 50 MHz,

RHF

l≈

(R0

l

)r

2δ= 8× 10−4 2.5× 10−3

2× 10−5= 0.1Ωm−1

which is 125 times greater, than at DC.

12-9 AJW, EEE3055F, UCT 2012

13 Radiation from Antennas

13.1 Hertzian Dipole

13.2 Dipole Antennas

13.3 Phase array pattern

• field pattern from a linear array.

• mention SAR [not covered 2009]

• explain the antenna pattern in terms of constructive and destructive

interference. [see handwritten handout issued 2009]

13.4 Aperture Antennas [simplified treatment; lab

session]

• horn antenna

13.5 Discussion on why circuits that are small compared

to the wavelength do not radiate

13-1

14 Thermal Radiation from WarmObjects

• why warm objects radiate

• frequency spectrum (watts/Hz)

• good and bad radiators; black body radiators versus metalic objects.

• noise power received by an antenna aimed at a warm surface. (ekTB)

• demonstration of UCT radiomter (35 GHz or 10 GHz)

14-1

eee3055f_notes.2012_V36

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