ECE 546 – Jose Schutt-Aine1 ECE 546 Lecture 02 Review of Electromagnetics Spring 2014 Jose E. Schutt-Aine Electrical & Computer Engineering University.

ECE 546 – Jose Schutt-Aine 1

ECE 546 Lecture 02

Review of ElectromagneticsSpring 2014

Jose E. Schutt-AineElectrical & Computer Engineering

University of [email protected]


Electromagnetic Quantities

E

HD

B

Electric field (Volts/m)

Magnetic field (Amperes/m)

Electric flux density (Coulombs/m2)

Magnetic flux density (Webers/m2)

J

Current density (Amperes/m2)

Charge density (Coulombs/m2)


Faraday’s Law of Induction

Ampère’s Law

Gauss’ Law for electric field

Gauss’ Law for magnetic field

BE

t

H J D

t

D

0B

Maxwell’s Equations


B H

D E

Constitutive Relations

Permittivity e: Farads/m

Permeability m: Henries/m

Free Space

128.85 10 /o F m

74 10 /o H m


Continuity EquationD

H Jt

0D

H J J Dt t

D

0Jt


0E

with E

ElectrostaticsAssume no time dependence 0

t

Since 0, such thatE

where is the scalar potentialE

2we get

Poisson’s Equation 2we get

2 0 Laplace’s Equationif no charge is present


Integral Form of ME

C S

E dl B dSt

C S S

H dl J dS D dSt

S V

D dS dv

0S

B dS


Boundary Conditions

1 2ˆ 0n E E

1 2ˆ Sn H H J

1 2ˆ Sn D D

1 2ˆ Sn B B

n̂

1 1 1 1, , ,E H D B

2 2 2 2, , ,E H D B


Faraday’s Law of Induction

Ampère’s Law

Gauss’ Law for electric field

Gauss’ Law for magnetic field

o

HE

t

o

EH

t

0E

0B

Free Space Solution


oE Ht

2o o

EE E

t t

Wave Equation

22

2o o

EE

t

Wave Equation

can show that2

22o o

HH

t


Wave Equation2 2 2 2

2 2 2 2o o

E E E E

x y z t

separating the components

2 2 2 2

2 2 2 2x x x x

o o

E E E E

x y z t

2 2 2 2

2 2 2 2

y y y yo o

E E E E

x y z t

2 2 2 2

2 2 2 2z z z z

o o

E E E E

x y z t


Wave Equation Plane Wave

0x y

(a) Assume that only Ex exists Ey=Ez=0

2 2

2 2x x

o o

E E

z t

(b) Only z spatial dependence

This situation leads to the plane wave solution

In addition, assume a time-harmonic dependence

j txE e then j

t


Plane Wave Solution

j zx o

j zx o

E E e forward wave

E E e backward wave

solution

where

In the time domain

22

2x

o o x

EE

z

o o propagation constant

( ) cos

( ) cosx o

x o

E t E t z forward traveling wave

E t E t z backward traveling wave

solution


Plane Wave Characteristics

where

In free space

o o propagation constant

1v propagation velocity

813 10 /

o o

v c m s


HE E j H

t

Solution for Magnetic Field

ˆ j zoE xE e

If we assume that

ˆ ˆ ˆ

1 1

0 0x

x y z

H Ej j x y z

E

then ˆ j zoEH y e

ˆ j zoE xE e

If we assume that then ˆ j zoE

H y e

intrinsic impedance of medium


( )P t E t H t

Time-Average Poynting Vector

We can show that

Poynting vector W/m2

time-average Poynting vector W/m2

0 0

1 1( )

T TP P t dt E t H t dt

T T

*1Re

2P E H

where and are the phasors of ( ) and ( ) respectivelyE H E t H t


s: conductivity of material medium (W-1m-1)

HE

t

EH J

t

J E

Material Medium

E j H

H J j E

1H E j E E j j Ej

2 2 1E Ej

1

j

or

since then


Wave in Material Medium2 2 21E E E

j

g is complex propagation constant

2 2 1j

1j jj

: a associated with attenuation of wave

: b associated with propagation of wave


Wave in Material Medium

ˆ ˆz z j zo oE xE e xE e e

1/22

1 12

decaying exponentialSolution:

1/22

1 12

ˆ z j zoEH y e e

j

j

Magnetic field

Complex intrinsic impedance


Wave in Material Medium1/2

22

1 1pv

Phase Velocity:

0

0

1. Perfect dielectric

Special Cases

Wavelength:

1/22

2 21 1

f

air, free space

and


Wave in Material Medium2. Lossy dielectric

Loss tangent: 1

2

2 2

31

8 2j

2

2 21

8

2

2 21

2 8 2


Wave in Material Medium

3. Good conductors

Loss tangent: 1

j j j

f f

1j f

jj


aattenuation

bpropagation

h dp H, E Examples

PEC - 0 0 0 supercond

Good conductor

finite

copper

Poor conductor finite

Ice

Perfectdielectric 0 finite

air

2

2

1f

j

1

f

2

/

12

j

2

/

Material Medium


Radiation - Vector Potential

E j H

H J j E

/D

0B

(1)

(2)

(3)

(4)

Assume time harmonicity ~ j te


Radiation - Vector Potential

0 such thatB A A B

A

: vector potential

0A

0E j A E j A

Using the property: 0any vector


0vector vector

where is the scalar potentialE j A

Since a vector is uniquely defined by its curl and its divergence, we can choose the divergence of A

choose such thatA

0A j Lorentz

condition

Vector Potential


B J j E

A J j j A

2 2A A J A j

2 2A j J A j

2 2A A J

D’Alembert’sequation

Vector Potential


'

, '4 '

j r reG r r

r r

'

'

''

4 '

j r r

V

J r eA r dv

r r

From A, get E and H using Maxwell’s equations

Three-dimensional free-space Green’s function

Vector potential

Vector Potential


ˆ' ( ') ( ') ( ')oJ r zI dl x y z

For infinitesimal antenna, the current density is:

ˆ4

j roI dlA r z e

r

ˆˆˆ cos sinz r

Calculating the vector potential,

In spherical coordinates,

Vector Potential


Resolving into components,

ˆˆˆ cos sin4

j roI dlA r z r e

r

cosˆ

4j ro

r

I dlA z e

r

sin

4j roI dl

A er

Vector Potential


Calculate E and H fields

1H A

1sin 0

sinr

AH A

1 10

sinrAH rA

r r

E and H Fields


1 rAH rAr r

11 sin

4j roI dl

H j er j r

1E H

j

1 1sin

sinrE Hr j

E and H Fields


2

2cos 1 11

4j ro

r

I dlE j e

r j r j

1 1E rH

r r j

2 2

1 1 1sin 1

4j roj I dl

E j er j r r j

E and H Fields


2

1 1sin 1

4j roj I dl

E er j r j r

2

2cos 1 11

4j ro

r

I dlE j e

r j r j

11 sin

4j roI dl

H j er j r

E and H Fields


Far field : 1 orr r 2

1then, 0

r

, whereE H

0rE

sin4

j roj I dlH e

r

Note that:

sin4

j roj I dlE e

r

Far Field Approximation


• Uniform constant phase locus is a plane• Constant magnitude• Independent of q• Does not decay

Characteristics of plane waves

Similarities between infinitesimal antenna far field radiated and plane wave

(a) E and H are perpendicular(b) E and H are related by h(c) E is perpendicular to H

Far Field Approximation


P t E t H t

Time-average Poynting vector or TA power density

1Re *

2P E H

E and H here are PHASORS

22 2ˆ

ˆRe sin2 2 4

oI dlrP H r

r

Poynting Vector


Total power radiated (time-average)

2

0 0P ds

P= 2ˆwith sinds rr d d

22 2 2

0 0sin sin

2 4oI dl

r d dr

P=

23

0

2 sin2 4

oI dld

P=

Time-Average Power


24

3 4oI dl

P=

Poynting Power DensityDirective Gain =

Average Poynting Power Density

over area of sphere with radius r

2Directive Gain =

/ 4

P

r

P

Time-Average Power


2

2

2

2

sin2 4

Directive Gain4

/ 43 4

o

o

I dl

r

I dlr

For infinitesimal antenna,

23Directive Gain sin

2

Directivity


Directivity: gain in direction of maximum value

Radiation resistance:

From 21

2 oRIP we have: 2

2rad

o

RI

P

For infinitesimal antenna:

2

2 2

2 4 2

3 4 3o

rado

I dl dlR

I

Directivity


2280rad

dlR

For free space, 120

The radiation resistance of an antenna is the value of a fictitious resistance that would dissipate an amount of power equal to the radiated power Pr when the current in the resistance is equal to the maximum current along the antenna

(for Hertzian dipole)

A high radiation resistance is a desirable property for an antenna

Radiation Resistance

ECE 546 – Jose Schutt-Aine1 ECE 546 Lecture 02 Review of Electromagnetics Spring 2014 Jose E. Schutt-Aine Electrical & Computer Engineering University.

Documents

jose schuttaine

wave equation plane

wave equation wave equationcan

components ece

henriesmfree space ece

continuity equation

boundary conditions

plane wave solutionin