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ELEC 351 Notes Set #15Assignment #7

Problem 8.3 Normal incidenceProblem 8.16 Normal incidenceProblem 8.27 Oblique incidenceProblem 8.28 Oblique incidenceProblem 8.40 Waveguide modesProblem 8.41 Waveguide modes

Do this assignment by November 16.

The final exam in ELEC 351 is December 13, 2018 from 2:00 to 5:00.

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Figures from: http://www.antenna-theory.com/tutorial/waveguides/waveguide.php

Rectangular Waveguide

Magic Tee = 3 dB Splitter

3 dB Splitter

Apply a signal to arm 3, and half the power emerges from arm 1 and half from arm 2.

Apply a signal to arm 4 and half the power emerges from arm 1 and half from arm 2.

Sums and Differences

Apply signal #1 to arm 1 and signal #2 to arm 2.

Arm 3: signal #1 plus signal #2 Arm 4: signal #1 minus signal #2.

Diplexer = separate signals at two frequencies

WikipediaA diplexer is a passive device that implements frequency-domain multiplexing. Two ports (e.g., L and H) are multiplexed onto a third port (e.g., S). The signals on ports L and H occupy disjoint frequency bands. Consequently, the signals on L and H can coexist on port S without interfering with each other.L port= the lower frequencyH port= the higher frequencyS port= the sum of the two frequencies

Television diplexer consisting of a high-pass filter (left) and a low-pass filter (right). The antenna cable is connected on the back to the screw terminals to the left of center.Designed and built by Arnold Reinhold

Waveguide diplexer: separate signals at two frequencies

“Waveguide diplexer links point to point systems”http://www.mwrf.com/passive-components/waveguide-diplexer-links-point-point-systems

Output channel 3:18.7 to 19.1 GHz

Output channel 2: 17.7 to 18.1 GHz

Input: both channelsChannel 2: 17.7 to 18.1 GHzChannel 3: 18.7 to 19.1 GHz

Waveguide Diplexer: 12.05 and 13.90 GHz

Irene Ortiz de Saracho Pantoja, , Ku Waveguide Diplexer Design For Satellite Communication, Universidad Politechnica de Madrid, 2015

12.05 GHz

13.9 GHzInput: both 12.05

GHz and 13.9 GHz

Orthomode Transducer (OMT) Combine two linearly polarized signals into one signal with

a horizontal and a vertical component. Two input ports in rectangular waveguideLinearly polarized signals of equal amplitude and 90 degrees out of phase.

One output port in circular waveguide which will feed a circular horn.The output port is circularly polarized by adding the signals at the two input ports with a 90 degree phase shift between them.

Source: Widipediahttps://en.wikipedia.org/wiki/Orthomode_transducer

OMT designed by PhD student Mohamed Abdelaal

Port 1: common portSquare waveguide,Circular polarization

Port 2Rectangular waveguideHorizontalpolarization

Port 3Rectangular waveguideVerticalpolarization

Mohamed Abdelaziz Mohamed Abdelaal,“Design and Analysis of Microwave Devices Based on Ridged Gap Waveguide Technology for 5G Applications”, ECE Departmet, Concordia University, April 2018.

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Standard Waveguide SizesD.M. Pozar, Microwave Engineering, 3rd edition, Wiley, 2005.

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Rectangular and Circular Waveguide

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General Solution of Maxwell’s Equations for Transmission Lines and Waveguide

Convention: ba >

( ) ( ) ( ) zjzz

zjyy

zjxx eyxeaeyxeaeyxeaE βββ −−− ++= ,ˆ,ˆ,ˆ

( ) ( ) ( ) zjzz

zjyy

zjxx eyxhaeyxhaeyxhaH βββ −−− ++= ,ˆ,ˆ,ˆ

( ) zjzz eyxeE β−= ,

Ulaby 7th edition Section 8.6

𝑘𝑘 = 𝜔𝜔 𝜇𝜇𝜇𝜇 is the phase constant in the dielectric filling the waveguide.𝛽𝛽 is the phase constant for waves travelling in the waveguide.

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TEM, TE and TM Modes

TEM field = transverse electric and magnetic field, hence zero axial components: 0=zE 0=zHand

TE field = transverse electric field, hence zero axial component of the electric field:

TM field = transverse magnetic field, hence zero axial component of the magnetic field:

0=zE

0=zH

( ) ( ) ( ) zjzz

zjyy

zjxx eyxeaeyxeaeyxeaE βββ −−− ++= ,ˆ,ˆ,ˆ

( ) ( ) ( ) zjzz

zjyy

zjxx eyxhaeyxhaeyxhaH βββ −−− ++= ,ˆ,ˆ,ˆ

( ) zjyy eyxeE β−= ,( ) zj

xx eyxeE β−= , and

( ) zjyy eyxhH β−= ,( ) zj

xx eyxhH β−= , and

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Rectangular Waveguides

( ) ( ) ( ) zjzz

zjyy

zjxx eyxhaeyxhaeyxhaH βββ −−− ++= ,ˆ,ˆ,ˆ( ) ( ) ( ) zj

zzzj

yyzj

xx eyxeaeyxeaeyxeaE βββ −−− ++= ,ˆ,ˆ,ˆ

•TE modes have , solve for•TM modes have , solve for

0=zE0=zH

( ) zjzz eyxhH β−= ,

zjzz eyxeE β−= ),(

We will show that if we know the axial components, we can find the transverse components from Maxwell’s curl equations:

𝑘𝑘𝑐𝑐2 = 𝑘𝑘2 − 𝛽𝛽2 where 𝑘𝑘 = 𝜔𝜔 𝜇𝜇𝜇𝜇 and 𝛽𝛽 is the phase constant for waves travelling in the waveguide.

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All the Field Components

βjz→

∂∂

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Relationship of the Transverse Components to the Axial Components

If we know the axial components, then we can find the transverse components.

If we know and , then we can find

We can prove this from the Maxwell curl equations: zE zH xE yE xH yH

HjE ωµ−=×∇ ( )EjH ωεσ +=×∇and

zjyy eeE β−=

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Maxwell’s Curl EquationsWrite out all three components of

And all three components of

To show that:

HjE ωµ−=×∇

( )EjH ωεσ +=×∇

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Find the Transverse ComponentsAssume that we know the axial components, and , then

find the transverse components,zE zH

yHxHxE yE

Use k for the wave number in the material filling the waveguide:

Transverse in terms of axial:

Ulaby 7th edition page 384

Repeat the derivation for 𝐻𝐻𝑥𝑥 on the previous page to get 𝐻𝐻𝑦𝑦, 𝐸𝐸𝑥𝑥 and 𝐸𝐸𝑦𝑦:

𝐸𝐸𝑥𝑥 =−𝑗𝑗𝑘𝑘𝑐𝑐2

𝛽𝛽𝜕𝜕𝐸𝐸𝑧𝑧𝜕𝜕𝜕𝜕

+ 𝜔𝜔𝜇𝜇𝜕𝜕𝐻𝐻𝑧𝑧𝜕𝜕𝜕𝜕

𝐸𝐸𝑦𝑦 =𝑗𝑗𝑘𝑘𝑐𝑐2

−𝛽𝛽𝜕𝜕𝐸𝐸𝑧𝑧𝜕𝜕𝜕𝜕

+ 𝜔𝜔𝜇𝜇𝜕𝜕𝐻𝐻𝑧𝑧𝜕𝜕𝜕𝜕

𝐻𝐻𝑥𝑥 =𝑗𝑗𝑘𝑘𝑐𝑐2

𝜔𝜔𝜇𝜇𝜕𝜕𝐸𝐸𝑧𝑧𝜕𝜕𝜕𝜕

− 𝛽𝛽𝜕𝜕𝐻𝐻𝑧𝑧𝜕𝜕𝜕𝜕

𝐻𝐻𝑦𝑦 =−𝑗𝑗𝑘𝑘𝑐𝑐2

𝜔𝜔𝜇𝜇𝜕𝜕𝐸𝐸𝑧𝑧𝜕𝜕𝜕𝜕

+ 𝛽𝛽𝜕𝜕𝐻𝐻𝑧𝑧𝜕𝜕𝜕𝜕

If we can find 𝐸𝐸𝑧𝑧 and 𝐻𝐻𝑧𝑧, then we can find all the other field components.TE Modes: Assume 𝐸𝐸𝑧𝑧 = 0 and find 𝐻𝐻𝑧𝑧TM Modes: Assume 𝐻𝐻𝑧𝑧 = 0 and find 𝐸𝐸𝑧𝑧

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Convention: ba >

Rectangular Waveguide: TE Modes

• The axial component of the electric field is zero:• Find the axial component of the magnetic field

0=zEzH

( ) ( ) zjyy

zjxx eyxeaeyxeaE ββ −− += ,ˆ,ˆ

( ) ( ) ( ) zjzz

zjyy

zjxx eyxhaeyxhaeyxhaH βββ −−− ++= ,ˆ,ˆ,ˆ

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Wave Equation for the TE Modes

( )yxhz ,

( ) ( ) ( ) zjzz

zjyy

zjxx eyxhaeyxhaeyxhaH βββ −−− ++= ,ˆ,ˆ,ˆ

Find

where

Look at the z component:

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The field ℎ𝑧𝑧(𝜕𝜕, 𝜕𝜕) must satisfy this wave equation!

To find ℎ𝑧𝑧(𝜕𝜕,𝜕𝜕), we must satisfy the wave equation and satisfy the boundary condition that 𝐸𝐸𝑡𝑡𝑡𝑡𝑡𝑡 = 0 at the surfaces of the waveguide walls.

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Separation of Variables (ELEC365)

Dispersion equation:where and µεω=k

andLet where 𝑘𝑘𝑥𝑥 and 𝑘𝑘𝑦𝑦 are constants.

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Separation of Variables Solution

The field ℎ𝑧𝑧(𝜕𝜕, 𝜕𝜕) satisfies the wave equation.

What values of 𝑘𝑘𝑥𝑥 and 𝑘𝑘𝑦𝑦 satisfy the boundary conditions?

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Boundary Conditions

For the TE solution, the axial component 𝐸𝐸𝑧𝑧 is zero everywhere.

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Find and in terms of : xE yE zH

( ) zjzz eyxhH β−= ,

leads to

where

∂∂−

=y

Hk

jE z

cx ωµ2

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Satisfy the Boundary Conditions

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Waveguide Modes

Find the Phase Constant mnβ

The value of the phase constant depends on the waveguide mode:

Dispersion equation:where and µεω=k

222ckk −=β

2222yx kkk −−=β

2222

,

−

−=

bn

amknm

ππβ 222

,

−

−=

bn

amknm

ππβ

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Cutoff Frequency for Mode m,n

mnβFor mode m,n to propagate, must be a real number:

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,2

+

=

bn

amf mnc

ππµεπ

The limiting frequency is called the “cutoff frequency” mncf ,

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, 21

+

=

bn

amf mnc

ππµεπ

22

, 21

+

=

bn

amf mnc µε

If the frequency is greater than the “cutoff frequency”, then mode m,n propagates.

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Below Cutoff Waveguide

If the frequency is less than the “cutoff frequency”, then mode m,n attenuates.

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Propagation or Attenuation?22

, 21

+

=

bn

amf mnc µε

For frequency greater than the cutoff frequency :

• is real

•So the waveguide mode propagates with phase factor

f mncf ,

nm,βzj nme ,β−

For frequency less than the cutoff frequency :

• is imaginary:

•So the waveguide mode attenuates with attenuation constant 𝛼𝛼𝑚𝑚,𝑡𝑡

f mncf ,

nm,β mnnm jαβ −=,

( ) zzjjzj nmnmnm eee ,,, ααβ −−−− ==

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Dominant Mode22

, 21

+

=

bn

amf mnc µε

2

10,1

21

=

afc µε

ba >

Lowest cutoff frequency: m=1,n=0

Convention:

Next higher cutoff frequency: m=0,n=1 or m=2,n=0?2

01,1

21

=

bfc µε

Below , no waveguide mode propagates and the waveguide is said to be “cutoff”.

10,cf

Between and the next higher cutoff frequency, only one waveguide mode propagates, which is the TE10 mode, and TE10 is said to be the “dominant mode”.

Above the next higher cutoff frequency, more than one mode can propagate and the waveguide is “overmoded”.

10,cf

2

20,2

21

=

afc µε

𝑓𝑓𝑐𝑐,10 =1

2𝑎𝑎 𝜇𝜇𝜇𝜇

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Example

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(seventh edition page 368)

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For TE modes (𝐸𝐸𝑧𝑧=0) we have found 𝐻𝐻𝑧𝑧 = 𝐴𝐴 cos

𝑚𝑚𝜋𝜋𝜕𝜕𝑎𝑎

cos𝑛𝑛𝜋𝜋𝜕𝜕𝑏𝑏

𝑒𝑒−𝑗𝑗𝛽𝛽𝑚𝑚,𝑛𝑛𝑧𝑧

so

𝐸𝐸𝑥𝑥 =−𝑗𝑗𝑘𝑘𝑐𝑐2

𝛽𝛽𝑚𝑚𝑡𝑡𝜕𝜕𝐸𝐸𝑧𝑧𝜕𝜕𝜕𝜕

+ 𝜔𝜔𝜇𝜇𝜕𝜕𝐻𝐻𝑧𝑧𝜕𝜕𝜕𝜕

=−𝑗𝑗𝑘𝑘𝑐𝑐2

𝜔𝜔𝜇𝜇𝜕𝜕𝐻𝐻𝑧𝑧𝜕𝜕𝜕𝜕

𝐸𝐸𝑦𝑦 =𝑗𝑗𝑘𝑘𝑐𝑐2

−𝛽𝛽𝑚𝑚𝑡𝑡𝜕𝜕𝐸𝐸𝑧𝑧𝜕𝜕𝜕𝜕

+ 𝜔𝜔𝜇𝜇𝜕𝜕𝐻𝐻𝑧𝑧𝜕𝜕𝜕𝜕

=𝑗𝑗𝑘𝑘𝑐𝑐2

𝜔𝜔𝜇𝜇𝜕𝜕𝐻𝐻𝑧𝑧𝜕𝜕𝜕𝜕

𝐻𝐻𝑥𝑥 =𝑗𝑗𝑘𝑘𝑐𝑐2

𝜔𝜔𝜇𝜇𝜕𝜕𝐸𝐸𝑧𝑧𝜕𝜕𝜕𝜕

− 𝛽𝛽𝑚𝑚𝑡𝑡𝜕𝜕𝐻𝐻𝑧𝑧𝜕𝜕𝜕𝜕

=𝑗𝑗𝑘𝑘𝑐𝑐2

−𝛽𝛽𝑚𝑚𝑡𝑡𝜕𝜕𝐻𝐻𝑧𝑧𝜕𝜕𝜕𝜕

𝐻𝐻𝑦𝑦 =−𝑗𝑗𝑘𝑘𝑐𝑐2

𝜔𝜔𝜇𝜇𝜕𝜕𝐸𝐸𝑧𝑧𝜕𝜕𝜕𝜕

+ 𝛽𝛽𝑚𝑚𝑡𝑡𝜕𝜕𝐻𝐻𝑧𝑧𝜕𝜕𝜕𝜕

=−𝑗𝑗𝑘𝑘𝑐𝑐2

𝛽𝛽𝑚𝑚𝑡𝑡𝜕𝜕𝐻𝐻𝑧𝑧𝜕𝜕𝜕𝜕

Find the transverse field components.Given 𝐻𝐻𝑧𝑧, find 𝐸𝐸𝑥𝑥, 𝐸𝐸𝑦𝑦, 𝐻𝐻𝑥𝑥, and 𝐻𝐻𝑦𝑦 .

Substitute 𝐻𝐻𝑧𝑧 = 𝐴𝐴𝑚𝑚𝑡𝑡 cos

𝑚𝑚𝜋𝜋𝜕𝜕𝑎𝑎

cos𝑛𝑛𝜋𝜋𝜕𝜕𝑏𝑏

𝑒𝑒−𝑗𝑗𝛽𝛽𝑚𝑚𝑛𝑛𝑧𝑧

to find the transverse field components 𝐸𝐸𝑥𝑥, 𝐸𝐸𝑦𝑦, 𝐻𝐻𝑥𝑥, and 𝐻𝐻𝑦𝑦:

Homework: show that

𝐸𝐸𝑥𝑥 = 𝐴𝐴𝑚𝑚𝑡𝑡𝑗𝑗𝜔𝜔𝜇𝜇𝑘𝑘𝑐𝑐2

𝑛𝑛𝜋𝜋𝑏𝑏

cos𝑚𝑚𝜋𝜋𝜕𝜕𝑎𝑎

sin𝑛𝑛𝜋𝜋𝜕𝜕𝑏𝑏

𝑒𝑒−𝑗𝑗𝛽𝛽𝑚𝑚𝑛𝑛𝑧𝑧

𝐸𝐸𝑦𝑦 = 𝐴𝐴𝑚𝑚𝑡𝑡−𝑗𝑗𝜔𝜔𝜇𝜇𝑘𝑘𝑐𝑐2

𝑚𝑚𝜋𝜋𝑎𝑎

sin𝑚𝑚𝜋𝜋𝜕𝜕𝑎𝑎

cos𝑛𝑛𝜋𝜋𝜕𝜕𝑏𝑏

𝑒𝑒−𝑗𝑗𝛽𝛽𝑚𝑚𝑛𝑛𝑧𝑧

𝐻𝐻𝑥𝑥 = 𝐴𝐴𝑚𝑚𝑡𝑡𝑗𝑗𝛽𝛽𝑚𝑚𝑡𝑡

𝑘𝑘𝑐𝑐2𝑚𝑚𝜋𝜋𝑎𝑎

sin𝑚𝑚𝜋𝜋𝜕𝜕𝑎𝑎

cos𝑛𝑛𝜋𝜋𝜕𝜕𝑏𝑏

𝑒𝑒−𝑗𝑗𝛽𝛽𝑚𝑚𝑛𝑛𝑧𝑧

𝐻𝐻𝑦𝑦 = 𝐴𝐴𝑚𝑚𝑡𝑡𝑗𝑗𝛽𝛽𝑚𝑚𝑡𝑡

𝑘𝑘𝑐𝑐2𝑛𝑛𝜋𝜋𝑏𝑏

cos𝑚𝑚𝜋𝜋𝜕𝜕𝑎𝑎

sin𝑛𝑛𝜋𝜋𝜕𝜕𝑏𝑏

𝑒𝑒−𝑗𝑗𝛽𝛽𝑚𝑚𝑛𝑛𝑧𝑧

Ulaby 7th edition page 388

Wave Impedance for TE Modes

Define the “wave impedance” as

𝑍𝑍𝑇𝑇𝑇𝑇 =𝐸𝐸𝑥𝑥𝐻𝐻𝑦𝑦

=

−𝑗𝑗𝑘𝑘𝑐𝑐2

𝜔𝜔𝜇𝜇 𝜕𝜕𝐻𝐻𝑧𝑧𝜕𝜕𝜕𝜕−𝑗𝑗𝑘𝑘𝑐𝑐2

𝛽𝛽 𝜕𝜕𝐻𝐻𝑧𝑧𝜕𝜕𝜕𝜕

=𝜔𝜔𝜇𝜇𝛽𝛽

For mode m, n the phase constant is

𝛽𝛽𝑚𝑚𝑡𝑡 = 𝑘𝑘2 −𝑚𝑚𝜋𝜋𝑎𝑎

2−

𝑛𝑛𝜋𝜋𝑏𝑏

2

So the wave impedance can be calculated as 𝑍𝑍𝑇𝑇𝑇𝑇,𝑚𝑚𝑡𝑡 =

𝜔𝜔𝜇𝜇𝛽𝛽

=𝜔𝜔𝜇𝜇

𝑘𝑘2 − 𝑚𝑚𝜋𝜋𝑎𝑎

2− 𝑛𝑛𝜋𝜋

𝑏𝑏2

In general: 𝐻𝐻𝑧𝑧 = 𝐴𝐴𝑚𝑚𝑡𝑡 cos

𝑚𝑚𝜋𝜋𝜕𝜕𝑎𝑎

cos𝑛𝑛𝜋𝜋𝜕𝜕𝑏𝑏

𝑒𝑒−𝑗𝑗𝛽𝛽𝑚𝑚𝑛𝑛𝑧𝑧

𝐸𝐸𝑥𝑥 = 𝐴𝐴𝑚𝑚𝑡𝑡𝑗𝑗𝜔𝜔𝜇𝜇𝑘𝑘𝑐𝑐2

𝑛𝑛𝜋𝜋𝑏𝑏

cos𝑚𝑚𝜋𝜋𝜕𝜕𝑎𝑎

sin𝑛𝑛𝜋𝜋𝜕𝜕𝑏𝑏

𝑒𝑒−𝑗𝑗𝛽𝛽𝑚𝑚𝑛𝑛𝑧𝑧

𝐸𝐸𝑦𝑦 = 𝐴𝐴𝑚𝑚𝑡𝑡−𝑗𝑗𝜔𝜔𝜇𝜇𝑘𝑘𝑐𝑐2

𝑚𝑚𝜋𝜋𝑎𝑎

sin𝑚𝑚𝜋𝜋𝜕𝜕𝑎𝑎

cos𝑛𝑛𝜋𝜋𝜕𝜕𝑏𝑏

𝑒𝑒−𝑗𝑗𝛽𝛽𝑚𝑚𝑛𝑛𝑧𝑧

𝐻𝐻𝑥𝑥 = 𝐴𝐴𝑚𝑚𝑡𝑡𝑗𝑗𝛽𝛽𝑚𝑚𝑡𝑡𝑘𝑘𝑐𝑐2

𝑚𝑚𝜋𝜋𝑎𝑎

sin𝑚𝑚𝜋𝜋𝜕𝜕𝑎𝑎

cos𝑛𝑛𝜋𝜋𝜕𝜕𝑏𝑏

𝑒𝑒−𝑗𝑗𝛽𝛽𝑚𝑚𝑛𝑛𝑧𝑧

𝐻𝐻𝑦𝑦 = 𝐴𝐴𝑚𝑚𝑡𝑡𝑗𝑗𝛽𝛽𝑚𝑚𝑡𝑡𝑘𝑘𝑐𝑐2

𝑛𝑛𝜋𝜋𝑏𝑏

cos𝑚𝑚𝜋𝜋𝜕𝜕𝑎𝑎

sin𝑛𝑛𝜋𝜋𝜕𝜕𝑏𝑏

𝑒𝑒−𝑗𝑗𝛽𝛽𝑚𝑚𝑛𝑛𝑧𝑧

Field Configuration in the Dominant Mode

Dominant mode, m=1,n=0: 𝐻𝐻𝑧𝑧 = 𝐴𝐴10 cos

𝜋𝜋𝜕𝜕𝑎𝑎𝑒𝑒−𝑗𝑗𝛽𝛽10𝑧𝑧

𝐸𝐸𝑥𝑥 = 0

𝐸𝐸𝑦𝑦 = 𝐴𝐴10−𝑗𝑗𝜔𝜔𝜇𝜇𝑘𝑘𝑐𝑐2

𝜋𝜋𝑎𝑎

sin𝜋𝜋𝜕𝜕𝑎𝑎𝑒𝑒−𝑗𝑗𝛽𝛽10𝑧𝑧

𝐻𝐻𝑥𝑥 = 𝐴𝐴10𝑗𝑗𝛽𝛽10𝑘𝑘𝑐𝑐2

𝜋𝜋𝑎𝑎

sin𝜋𝜋𝜕𝜕𝑎𝑎𝑒𝑒−𝑗𝑗𝛽𝛽10𝑧𝑧

𝐻𝐻𝑦𝑦 = 0

46

𝐸𝐸𝑦𝑦 = 𝐴𝐴10−𝑗𝑗𝜔𝜔𝜇𝜇𝑘𝑘𝑐𝑐2

𝜋𝜋𝑎𝑎

sin𝜋𝜋𝜕𝜕𝑎𝑎𝑒𝑒−𝑗𝑗𝛽𝛽10𝑧𝑧

47

Power Flow in the Dominant Mode

byax ≤≤≤≤ 0,0 dxdyasd zˆ=

zjx e

axAajH 10cos10 βπ

πβ −=

zj

cy e

axA

akjE 10sin2

βπωµπ −−=

( )∫∫ ⋅×= dxdyaaHaEP zxxyy ˆˆˆ21 *

10

( )∫∫ ⋅−= dxdyaaHEP zzxy ˆˆ21 *

10

( )∫∫ −

−= − dxdye

axAaje

axA

akjP zjzj

c

1cossin21

1010 10210

ββ ππβπωµπ

𝑃𝑃10 = 𝐴𝐴 2 𝜔𝜔𝜔𝜔𝑡𝑡3𝑏𝑏

4𝜋𝜋2𝛽𝛽10 watts Homework: do

the integration!

48

7. Design a rectangular waveguide such that the operating frequency of 10.4 GHz is 20% above the cutoff frequency for the dominant mode. Also, the cutoff frequency of the next higher waveguide mode must be 10% above the operating frequency. The waveguide is filled with air.

7.1) What is the cutoff frequency for the TE10 mode?

7.2) What is the “a” dimension?

7.3) What is the “b” dimension?

7.4) Find the phase constant for the dominant mode at the operating frequency.

7.5) Find the wave impedance for the dominant mode at the operating frequency.

7.6) How many modes can propagate in the waveguide at 20 GHz?

Question from an old final exam:

49

7.1) What is the cutoff frequency for the TE10 mode?We are told that the “operating frequency of 10.4 GHz is 20% above the cutoff frequency for the dominant mode”, so 10.4 = 1.2𝑓𝑓𝑐𝑐,10, where the cutoff frequency is𝑓𝑓𝑐𝑐,10 = 10.4

1.2= 8.667 GHz.

7.2) What is the “a” dimension?

𝑓𝑓𝑐𝑐,𝑚𝑚𝑡𝑡 = 12𝜋𝜋 𝜔𝜔𝜇𝜇

𝑚𝑚𝜋𝜋𝑡𝑡

2+ 𝑡𝑡𝜋𝜋

𝑏𝑏

2and for air-filled waveguide c = 1

𝜔𝜔0𝜇𝜇0= 30 cm/ns

𝑓𝑓𝑐𝑐,10 =302𝜋𝜋

1𝜋𝜋𝑎𝑎

2

+0𝜋𝜋𝑏𝑏

2

=302𝜋𝜋

𝜋𝜋𝑎𝑎

=15𝑎𝑎

15𝑡𝑡

= 8.667

𝑎𝑎 = 158.667

= 1.731 cm

7.3) What is the “b” dimension? “The cutoff frequency of the next higher waveguide mode must be 10% above the operating frequency.”So the next higher cutoff frequency is 1.1x10.4=11.44 GHz.

The next higher cutoff frequency might be 𝑓𝑓𝑐𝑐,01 or might be 𝑓𝑓𝑐𝑐,20

50

𝑓𝑓𝑐𝑐,𝑚𝑚𝑡𝑡 =302𝜋𝜋

𝑚𝑚𝜋𝜋1.731

2+

𝑛𝑛𝜋𝜋𝑏𝑏

2

𝑓𝑓𝑐𝑐,01 =302𝜋𝜋

0𝜋𝜋1.731

2

+1𝜋𝜋𝑏𝑏

2

=302𝜋𝜋

𝜋𝜋𝑏𝑏

=15𝑏𝑏

= 11.44

𝑏𝑏 = 1511.44

= 1.311 cmCheck: what is 𝑓𝑓𝑐𝑐,20?

𝑓𝑓𝑐𝑐,20 = 302𝜋𝜋

2𝜋𝜋1.731

2+ 0𝜋𝜋

𝑏𝑏

2= 30

2𝜋𝜋2𝜋𝜋

1.731= 30

22

1.731= 30

1.731= 17.33 GHz

Since this is higher than 11.44 GHz, choose 𝑏𝑏 = 1.311 cm

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7.4) Find the phase constant for the dominant mode at the operating frequency.

𝛽𝛽,𝑚𝑚𝑡𝑡 = 𝑘𝑘2 − 𝑚𝑚𝜋𝜋𝑡𝑡

2− 𝑡𝑡𝜋𝜋

𝑏𝑏

2

Where 𝑘𝑘 = 𝜔𝜔 𝜇𝜇𝜇𝜇 = 2𝜋𝜋𝜋𝜋𝑐𝑐

= 2𝜋𝜋𝑥𝑥10.430

= 2.178 rad/cm (with 𝑐𝑐 = 1𝜔𝜔𝜇𝜇

= 30 cm/ns)

𝛽𝛽,10 = 𝑘𝑘2 − 1𝜋𝜋𝑡𝑡

2− 0𝜋𝜋

𝑏𝑏

2= 𝑘𝑘2 − 𝜋𝜋

𝑡𝑡

2= 2.1782 − 𝜋𝜋

1.731

2= 1.024 rad/cm

7.5) Find the wave impedance for the dominant mode at the operating frequency. 𝑍𝑍𝑇𝑇𝑇𝑇 = 𝜔𝜔𝜔𝜔

𝛽𝛽= 2𝜋𝜋𝑥𝑥10.4𝑥𝑥109𝑥𝑥4𝜋𝜋𝑥𝑥10−7

102.4= 682.0 ohms

7.6) How many modes can propagate in the waveguide at 20 GHz?

𝑓𝑓𝑐𝑐,𝑚𝑚𝑡𝑡 =302𝜋𝜋

𝑚𝑚𝜋𝜋1.731

2+

𝑛𝑛𝜋𝜋1.311

2= 15

𝑚𝑚1.731

2+

𝑛𝑛1.311

2

𝑓𝑓𝑐𝑐,01 = 11.44 GHz, propagates at 20 GHz𝑓𝑓𝑐𝑐,20 = 17.33 GHz, propagates at 20 GHz

𝑓𝑓𝑐𝑐,02 = 15 01.731

2+ 2

1.311

2= 22.88 GHz, does NOT propagate

𝑓𝑓𝑐𝑐,11 = 15 11.731

2+ 1

1.311

2= 14.35 GHz propagates

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7.5) Find the wave impedance for the dominant mode at the operating frequency.Continued:

𝑓𝑓𝑐𝑐,21 = 15 21.731

2+ 1

1.311

2= 20.76 GHz does NOT propagate

𝑓𝑓𝑐𝑐,12 = 15 11.731

2+ 2

1.311

2= 24.47 GHz does NOT propagate