Waveguide Introduction & Analysis Setup

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Electromagnetics:

Electromagnetic Field Theory

Waveguide Introduction& Analysis Setup

Lecture Outline

•What is a waveguide?

•Governing equations for waveguide analysis.• Setup for TEM, TE, and TM modes

• Setup for analyzing hybrid modes

• Setup for analyzing slab waveguides

Slide 2

Note, this lecture covers just setting up Maxwell’s equations for solution. This lecture does not attempt to obtain solutions.

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Slide 3

What is a Waveguide?

What is a Waveguide?

Slide 4

A waveguide is a structure that confines the propagation of waves to a single path.

They are “pipes” for electromagnetic waves.

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Waveguide Modes

Slide 5

The field inside a waveguide must obey Maxwell’s equations. This limits what field configurations are possible into a discrete set. Each solution is called a “mode.” Each mode looks different and behaves differently inside the waveguide.

Slab Vs. Channel Waveguides

Slide 6

Slab waveguides confine waves in only one transverse direction.

Channel waveguides confine waves in two transverse directions.

ConfinementConfinement

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Map of Waveguides (LI Media)

Slide 7

Transmission Lines

• Contains two or more conductors.• No low‐frequency cutoff.• Thought of more as a circuit clement

• Confines and transports waves.• Supports higher‐order modes.

• Has TEM mode.• Has TE and TM modes.

stripline

coaxial microstrip

slotline

coplanar

“Pipes”

• Has one or less conductors.• Usually what is implied by the label “waveguide.”

Metal Shell Pipes Dielectric Pipes

Inhomogeneous

Homogeneous

• Enclosed by metal.• Does not support TEM mode.• Has a low frequency cutoff.

• Supports TE and TM modes

• Supports TE and TM modes only if one axis is uniform.

• Otherwise supports quasi‐TM and quasi‐TE modes.

rectangular circular

Channel Waveguides

Slab Waveguides

• Composed of a core and a cladding.• Symmetric waveguides have no low‐frequency cutoff.

• Confinement only along one axis.• Supports TE and TM modes.• Interfaces can support surface waves.

• Confinement along two axes.• TE & TM modes only supported in circularly symmetric guides.

dielectric Slab interface

optical Fiber rib

dual‐ridge

no uniform axis(no TE or TM)

Waveguides

Homogeneous Inhomogeneous

• Supports only quasi‐(TEM, TE, & TM) modes.

Single‐Ended

Differential

buried parallel plate

coplanar strips

photonic crystal

shielded pairlarge‐area

parallel plate

uniform axis(has TE and TM)

Notes on Transmission Lines

• Contains two or more conductors• No low frequency cutoff. Works down to DC.• Supports TEM, TE, and TM modes when the dielectric is homogeneous• Supports higher‐order modes, not just TEM.• Serve more as a circuit element than a wave device• Very compact for low frequency signals.• Tend to be lossy at very high frequencies (> 10 GHz) due to skin effect and dielectric loss.

Slide 8

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Notes on Metal Pipe Waveguides

•Contains on a single conductor•Has a low frequency cutoff below which there is no propagation of waves.• Supports TE and TM waves only if dielectric is homogeneous.• Field confined to inside of the waveguide.• Less lossy for very high frequency waves.•Prohibitively large size at low frequencies.

Slide 9

Notes on Dielectric Waveguides

• Does not contain any metals

• Symmetric dielectric waveguides have no low‐frequency cutoff.

• Symmetric waveguides (e.g. slabs & circularly symmetric) support TE and TM modes.

•Most have a low frequency cutoff below which no waves can propagate.

• Hybrid modes still tend to be strongly linearly polarized.

• Optical fibers are dielectric waveguides.• Field extends outside of the core.

Slide 10

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Channel Waveguides for Integrated Optics

Slide 11

Stripe waveguide Diffused waveguide Buried‐strip waveguide

Buried‐rib waveguide Rib waveguide Strip‐loaded waveguide

Channel Waveguides for Radio Frequencies

Slide 12

Coaxial Cable

Twisted Pair Transmission LineIsolated Wire

Shielded‐Pair Transmission Line

Rectangular Waveguide

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Channel Waveguides for Printed Circuits

Slide 13

Transmission lines are metallic structures that guide electromagnetic waves from DC to very high frequencies.

Microstrip

Stripline Slot Line

Parallel‐Plate Transmission Line

Coplanar Line

Structures Supporting Surface Waves

Slide 14

Surface‐Plasmon Polariton (SPP)

Dyakonov Surface Wave Bloch Surface Wave

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Notes on Waveguides

•Waveguides support an infinite number of discrete modes

• The modes have some amplitude profile that just accumulates phase as it propagates.

•Modes have cutoff frequencies, below which they are not supported and decay very quickly.

Slide 15

Slide 16

Governing Equations for Waveguide

Analysis

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Steps for Waveguide Analysis

Slide 17

1. Draw the waveguide2. Assume a form of the solution. Outer regions must decay exponentially or be equal to

zero.3. Substitute solution into Maxwell’s equations.4. Simplify equations based on the geometry of the waveguide.5. Manipulate equations into a differential equation to solve. This is called the governing

equation.6. Solve the governing equation in each homogeneous region of the waveguide.7. “Connect” the solutions in each region using boundary conditions.8. Calculate the overall field solution.9. Use the field solution to calculate the waveguide parameters such as , Z0, and the

profile of the fields.

Various Wave Equations

Slide 18

E j H

H j E

1. Maxwell’s Curl Equations

E

E j H Hj

2. Wave Equation in General Media

1 2

H j E

Ej E

j

E E

3. Wave Equation in LHI Media

1 2

2

E E

E E

E

2 2

2 2 0

E k E

E k E

4. Wave Equation Decouples

2 2

2 2

2 2

0

0

0

x x

y y

z z

E k E

E k E

E k E

We can solve these equations independently.wave numberk

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Expand Maxwell’s Equations

Slide 19

Maxwell’s equations are used to analyze waveguides.

E j H

H j E

The two curl equations expand into a set of six coupled partial differential equations.

yzx

x zy

y xz

EEj H

y z

E Ej H

z xE E

j Hx y

yzx

x zy

y xz

HHj E

y z

H Hj E

z xH H

j Ex y

There are six field components to solve for: Ex, Ey, Ez, Hx, Hy, and Hz.

Yikes!!

General Form of Solution for Waveguides

Slide 20

A mode in a waveguide has the following general mathematical form.

0, , , j zE x y z E x y e

0 ,E x y

x

y

z

phase constant

complex amplitude,mode shape

accumulation of phase in z direction

j ze

This means we can solve the problem by just analyzing the cross section in the x-yplane. This reduces to a two‐dimensional problem.

3D 2D

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Animation of a Waveguide Mode

Slide 21

Assume the Form of the Solution

Slide 22

For a waveguide uniform in the z direction, the solution will have the form

0 0, , , , , ,j z j zE x y z E x y e H x y z H x y e

If we substitute this solution into our six equations, we get

0,0, 0,

0,0, 0,

0, 0,0,

zy x

zx y

y xz

Ej E j H

y

Ej E j H

xE E

j Hx y

0,0, 0,

0,0, 0,

0, 0,0,

zy x

zx y

y xz

Hj H j E

y

Hj H j E

xH H

j Ex y

Things are a little more simple, but we still have six field components to solve for.

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Slide 23

Setup for TEM, TE, and TM Modes

Existence Conditions for TEM, TE, and TM Modes

Slide 24

TEM modes only exist in transmission lines with two or more conductors embedded in a homogeneous fill.

TE and TM modes only exist in waveguides with a homogeneous fill or in waveguides with a uniform axis like slabs and circularly symmetric guides.

No TEMSupports TEM

Supports TE and TM No TE or TM

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Goal of Following Derivation

Slide 25

0,0, 0,

0,0, 0,

0, 0,0,

zy x

zx y

y xz

Ej E j H

y

Ej E j H

xE E

j Hx y

0,0, 0,

0,0, 0,

0, 0,0,

zy x

zx y

y xz

Hj H j E

y

Hj H j E

xH H

j Ex y

0, 0,0, 2

0, 0,0, 2

z zx

c

z zy

c

E HjH

k y x

E HjH

k x y

0, 0,0, 2

0, 0,0, 2

z zx

c

z zy

c

E HjE

k x y

E HjE

k y x

0,

0,

?

?z

z

E

H

We now only need to find E0,z and H0,z.

2 2

2 2

0

0

z z

z z

E k E

H k H

Reduce the Number of Terms to Solve (1 of 2)

Slide 26

0,0, 0,

0,0, 0,

0, 0,0,

Eq. 1

Eq. 1

Eq. 1

zy x

zx y

y xz

Ej E j H a

y

Ej E j H b

xE E

j H cx y

0,0, 0,

0,0, 0,

0, 0,0,

Eq. 1

Eq. 1

Eq. 1

zy x

zx y

y xz

Hj H j E d

y

Hj H j E e

xH H

j E fx y

Step 1 – Solve Eq. (1e) for E0,y.

0,0, 0,

1 zy x

HE j H

j x

Step 2 – Substitute this expression into Eq. (1a) to eliminate E0,y.

0, 0,0, 0,

1z zx x

E Hj j H j H

y j x

Step 3 – Recall that k2 = 2 and solve this new expression for H0,x.

0, 0,

0, 2 2

z zx

E HjH

y xk

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Reduce the Number of Terms to Solve (2 of 2)

Slide 27

Step 4 – Derive three more similar equations.

Solve Eq. (1d) for E0,x, substitute that expression into Eq. (1b) and solve for H0,y.

0, 0,

0, 2 2

z zy

E HjH

x yk

Solve Eq. (1b) for H0,y, substitute that expression into Eq. (1d) and solve for E0,x.

0, 0,

0, 2 2

z zx

E HjE

x yk

Solve Eq. (1a) for H0,x, substitute that expression into Eq. (1e) and solve for E0,y.

0, 0,

0, 2 2

z zy

E HjE

y xk

Reduced Set of Equations

Slide 28

Step 5 – Define the cutoff wave number kc as

2 2 2ck k

We now have all of the transverse field components expressed in terms of the longitudinal components.

0, 0,0, 2

c

0, 0,0, 2

c

z zx

z zy

E HjH

k y x

E HjH

k x y

0, 0,0, 2

c

0, 0,0, 2

c

z zx

z zy

E HjE

k x y

E HjE

k y x

Now all that we have to do is determine E0,z and H0,z. The remaining field components can be calculated from just these two terms.

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How Do We Find E0,z and H0,z?

Slide 29

Recall that in LHI media, our wave equation simplified to

1 2

2 2

2 2

2 2

2 2

0

0

0

0

x x

y y

z z

E E

E k E

E k E

E k E

E k E

1 2

2 2

2 2

2 2

2 2

0

0

0

0

x x

y y

z z

H H

H k H

H k H

H k H

H k H

Substituting our solution into the bottom equations above gives

2 20, c 0, 0z zE k E 2 2

0, c 0, 0z zH k H

Solution Categories

Slide 30

If E0,z = H0,z = 0, we have a transverse electromagnetic (TEM) solution because all of the field components are transverse to the direction of propagation. Analysis reduces to an electrostatics problem.

If E0,z = 0 and H0,z ≠ 0, we obtain a transverse electric (TE) solution because the electric field has no longitudinal component.

If E0,z ≠ 0 and H0,z = 0, we obtain a transverse magnetic (TM) solution because the magnetic field has no longitudinal component.

If E0,z ≠ 0 and H0,z ≠ 0, we have a hybrid solution that must be analyzed differently.

E0,z H0,z Solution

0 0 TEM

0 TE

0 TM

Hybrid

2 20, c 0,

2 20, c 0,

0

0

z z

z z

E k E

H k H

In LHI media, E0,z and H0,z are solved independently. This means solutions can be obtained in any combination. This is the origin of TEM, TE, TM, and hybrid modes.

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TEM Analysis (1 of 3)

Slide 31

For TEM waves, we have E0,z = H0,z = 0. Under this condition, Maxwell’s equations reduce to

0,zE

y

0, 0,

0,0,

Eq. 1y x

zx

j E j H a

Ej E

x

0,

0, 0,0,

Eq. 1y

y xz

j H b

E Ej H

x y

Eq. 1c

0,zH

y

0, 0,

0,0,

Eq. 1y x

zx

j H j E d

Hj H

x

0,

0, 0,0,

Eq. 1y

y xz

j E e

H Hj E

x y

Eq. 1 f

0, 0,

0, 0,

0, 0,

Eq. 2

Eq. 2

0 Eq. 2

y x

x y

y x

j E j H a

j E j H b

E Ec

x y

0, 0,

0, 0,

0, 0,

Eq. 2

Eq. 2

0 Eq. 2

y x

x y

y x

j H j E d

j H j E e

H Hf

x y


Slide 32

0, 0,

0, 0,

0, 0,

Eq. 2

Eq. 2

0 Eq. 2

y x

x y

y x

j E j H a

j E j H b

E Ec

x y

0, 0,

0, 0,

0, 0,

Eq. 2

Eq. 2

0 Eq. 2

y x

x y

y x

j H j E d

j H j E e

H Hf

x y

0, 0, y xH E

Solve Eq. (2d) for H0,y.

0, 0,

2 20, 0,

2 20, 0,

x x

x x

x x

j E j E

E E

E k E

Substitute H0,y into Eq. (2b).

After dropping E0,z and H0,z, we get

k This tells us that for TEM

Previously, we defined .2 2 2ck k

If = k, then kc = 0 implying that there is no cutoff frequency for the TEM mode.

In summary, for the TEM mode we have

c and 0 no cutoffk k

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Slide 33

In LHI media, recall that the wave equation was

But for the TEM mode, kc = 0.

The wave equation reduces to Laplace’s equation from electrostatics.

2 20, c 0, 0xy xyE k E

2 20, c 0,xy xyE k E

20,

0

0xyE

Alternate Derivation of TEM Analysis

Slide 34

The TEM mode in a transmission line has no cutoff frequency (kc = 0).This means that it can be analyzed as 0 and the problem reduces to an electrostatics problem.

Derivation

Maxwell’s equations for electrostatics

0 Eq. 3

0 Eq. 3

Eq. 3

Eq. 3

E a

D b

D E c

E V d

Substitute Eq. (3c) into Eq. (3b).

0 Eq. 4E

Substitute Eq. (3d) into Eq. (4).

0V

For isotropic dielectrics

0V

For homogeneous dielectrics

2 0V

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TE Analysis in LHI Media

Slide 35

We are free to set E0,z = 0 and H0,z ≠ 0. This means we do not have to solve for E0,z. We only have to obtain a solution for H0,z.

2 20, c 0, 0z zE k E 2 2

0, c 0, 0z zH k H An added benefit of this solution approach is that H0,z is tangential to all boundaries in a waveguide.

For TE analysis, the other field components are calculated just from H0,z.

From this, we can calculate the impedance.

0,2

0, cTE

0,0,2c

z

x

zy

HjE k y k

ZHjH

k y

0,0, 2

c

0,0, 2

c

zx

zy

HjH

k x

HjH

k y

0,0, 2

c

0,0, 2

c

zx

zy

HjE

k y

HjE

k x

We must still determine by solving the wave equation.

TM Analysis in LHI Media

Slide 36

We are free to set E0,z ≠ 0 and H0,z = 0. This means we do not have to solve for H0,z. We only have to obtain a solution for E0,z.

2 2 2 20, c 0, 0, c 0,0 0z z z zE k E H k H An added benefit of this solution approach is that

E0,z is tangential to all boundaries in a waveguide.

For TM analysis, the other field components are calculated just from E0,z.

0,0, 2

c

0,0, 2

c

zx

zy

EjH

k y

EjH

k x

0,0, 2

c

0,0, 2

c

zx

zy

EjE

k x

EjE

k y

From this, we can calculate the impedance.

0,2

0, cTM

0,0,2c

z

x

zy

EjE k x

ZEjH k

k x

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Slide 37

Setup for AnalyzingHybrid Modes

Hybrid Mode Analysis (1 of 3)

Slide 38

To setup a solution to Maxwell’s equations, we back up to Maxwell’s equations in linear and isotropic media (i.e. can be inhomogeneous).

0,0, 0,

0,0, 0,

0, 0,0,

Eq. 1a

Eq. 1b

Eq. 1c

zy x

zx y

y xz

Ej E j H

y

Ej E j H

xE E

j Hx y

0,0, 0,

0,0, 0,

0, 0,0,

Eq. 2a

Eq. 2b

Eq. 2c

zy x

zx y

y xz

Hj H j E

y

Hj H j E

xH H

j Ex y

Solve Eq. (1c) for H0,z and solve Eq. (2c) for E0,z.

0, 0,0,

1 Eq. 3ay x

z

E EH

j x y

0, 0,

0,

1 Eq. 3by x

z

H HE

j x y

Substitute Eq. (3a) into Eqs. (2a) and (2b), & substitute Eq. (3b) into Eqs. (1a) and (1b).

0, 0,0, 0,

0, 0,0, 0,

1 1 Eq. 4a

1 1 Eq. 4b

y xy x

y xx y

H HH E

x x y

H HH E

y x y

0, 0,0, 0,

0, 0,0, 0,

1 1 Eq. 5a

1 1 Eq. 5b

y xy x

y xx y

E EE H

x x y

E EE H

y x y

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Slide 39

We can write our four remaining equations more compactly as

0, 0,

0, 0,

1 1 1 1

Eq. 61 1 1 1

x x

y y

x y x x H E

H E

y y y x

0, 0,

0, 0,

1 1 1 1

Eq. 71 1 1 1

x x

y y

E Hx y x xE H

y y y x

Full Wave Analysis

Solve Eq. (7) for the magnetic field components.

0, 0,

0, 0,

1 1 1 1

1= Eq. 8

1 1 1 1

x x

y y

H Ex y x xH E

y y y x

Solve Eq. (8) into Eq. (6) to arrive at the final wave equation to be solved.

0, 0,2

0, 0,

1 1 1 1 1 1 1 10

01 1 1 1 1 1 1 1

x x

y y

x y x x E Ex y x xE E

y y y x y y y x

Yikes!! This is typically solved numerically on a computer.


Slide 40

Quasi‐LP AnalysisRecognizing that the hybrid modes tend to be strongly linearly polarized, we can make a simplifying approximation that the cross coupling between E0,x and E0,y is weak and can be neglected. Under this condition, our governing equation separates into two independent equations, one for each LP mode.

2 22

2 22

1 1 1 1 1 1

1 1 1 1 1 1

xx

yy

x x y y x y x y

y y x x y x y x

xx xy

yx0, 0,2

0, 0,

0

0x x

y yyy

E E

E E

20, 0,

20, 0,

0

0xx x x

yy y y

E E

E E

Quasi‐LP AnalysisThese final two equations simplify even more for homogeneous dielectrics and for slab waveguides. They are still typically solved numerically on a computer.

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Slide 41

Setup for Analyzing Slab Waveguides

Geometry and Solution

Slide 42

z

x

y

0, , j zE x y z E ex

Amplitude Profile

Wave oscillations

phase constant

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LI Slab Waveguide Analysis (1 of 3)

Slide 43

Given this geometry

0y

0,zE

y

0, 0,

0,0, 0,

0, 0,

Eq. 1

Eq. 1

y x

zx y

y x

j E j H a

Ej E j H b

x

E E

x y

0, Eq. 1zj H c

0,zH

y

0, 0,

0,0, 0,

0, 0,

Eq. 1

Eq. 1

y x

zx y

y x

j H j E d

Hj H j E e

x

H H

x y

0, Eq. 1zj E f


Slide 44

Maxwell’s equations have decoupled into two independent sets of equations.

0,0,

0, 0

0

,

0,

0

,

,

E

Eq. 2

Eq

q. 2

.

2

zx

y x

yz

y

j E j H a

Ej H c

x

Ej E j H b

x

0,

0, 0,

0,0,

0, 0,

Eq. 2

Eq.

q

2

E . 2

z

y x

yz

x y

Hj H j E e

x

j H j E d

Hj E f

x

0,0, 0,

0, 0,

0,0,

Eq. 3

Eq. 3

Eq. 3

zx y

y x

yz

Hj H j E a

xj E j H b

Ej H c

x

0,0, 0,

0, 0,

0,0,

Eq. 3

Eq. 3

Eq. 3

zx y

y x

yz

Ej E j H d

xj H j E e

Hj E f

x

TE Mode (i.e. E0,z = 0) TM Mode (i.e. H0,z = 0)

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Slide 45

0, 20,

0, 0,

0,0,

10 Eq. 4

Eq. 4

1 Eq. 4

yc y

x y

yz

Ek E a

x x

H E b

EH c

j x

0, 20,

0, 0,

0,0,

10 Eq. 5

Eq. 5

1 Eq. 5

yc y

x y

yz

Hk H a

x x

E H b

HE c

j x

TE Mode

TM Mode

We can derive wave equations by substituting the 2nd and 3rd equations into the 1st.

Typical Modes in a Slab Waveguide

Slide 46

TE Modes

TM Modes

ncore = 2.0

nclad = 1.5

ncore = 2.0

nclad = 1.5

01.8

01.8

Effective refractive indices

Effective refractive indices

x

yz

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Remarks About Slab Waveguide Analysis

•Waves are confined in only one transverse direction.•Waves are free to spread out in the uniform transverse direction•Propagation within the slab can be restricted to a single direction without loss of generality.•Maxwell’s equations rigorously decouple into two distinct modes.•No approximations are necessary

Slide 47

Summary of This Lecture• General waveguide concepts

• Waveguides, modes

• Identify what types of modes are supported• TEM, TE, TM, or hybrid

• Analysis setup

Slide 48

TEM

2 0V

TE

2 20, c 0, 0z zH k H

TM

2 20, c 0, 0z zE k E

Hybrid

0, 0,2

0, 0,

1 1 1 1 1 1 1 10

01 1 1 1 1 1 1 1

x x

y y

x y x x E Ex y x xE E

y y y x y y y x

Slab

0, 20,

0, 20,

1TE: 0

1TM: 0

yc y

yc y

Ek E

x x

Hk H

x x

47

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Waveguide Introduction & Analysis Setup

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