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Page 1: ELEC 351 Notes Set #15 - Encstrueman/elec351/ELEC351...Separation of Variables (ELEC365) Dispersion equation: where and k =Ο‰ ¡Ρ Let and w here π‘˜π‘˜ π‘₯π‘₯ and π‘˜π‘˜ 𝑦𝑦

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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

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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.

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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

https://en.wikipedia.org/wiki/Electronic_component
https://en.wikipedia.org/wiki/Frequency-division_multiplexing
https://en.wikipedia.org/wiki/Port_(circuit_theory)
https://commons.wikimedia.org/wiki/User:ArnoldReinhold
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Waveguide diplexer: separate signals at two frequencies

β€œWaveguide 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

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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

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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

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OMT designed by PhD student Mohamed Abdelaal

Port 1: common portSquare waveguide,Circular polarization

Port 2Rectangular waveguideHorizontalpolarization

Port 3Rectangular waveguideVerticalpolarization

Mohamed Abdelaziz Mohamed Abdelaal,β€œDesign 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 Ξ²Ξ²Ξ² βˆ’βˆ’βˆ’ ++= ,Λ†,Λ†,Λ†

( ) ( ) ( ) zjzz

zjyy

zjxx eyxhaeyxhaeyxhaH Ξ²Ξ²Ξ² βˆ’βˆ’βˆ’ ++= ,Λ†,Λ†,Λ†

( ) zjzz eyxeE Ξ²βˆ’= ,

Ulaby 7th edition Section 8.6

π‘˜π‘˜ = πœ”πœ” πœ‡πœ‡πœ‡πœ‡ 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 Ξ²Ξ²Ξ² βˆ’βˆ’βˆ’ ++= ,Λ†,Λ†,Λ†

( ) ( ) ( ) zjzz

zjyy

zjxx eyxhaeyxhaeyxhaH Ξ²Ξ²Ξ² βˆ’βˆ’βˆ’ ++= ,Λ†,Λ†,Λ†

( ) zjyy eyxeE Ξ²βˆ’= ,( ) zj

xx eyxeE Ξ²βˆ’= , and

( ) zjyy eyxhH Ξ²βˆ’= ,( ) zj

xx eyxhH Ξ²βˆ’= , and

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Rectangular Waveguides

( ) ( ) ( ) zjzz

zjyy

zjxx eyxhaeyxhaeyxhaH Ξ²Ξ²Ξ² βˆ’βˆ’βˆ’ ++= ,Λ†,Λ†,Λ†( ) ( ) ( ) zj

zzzj

yyzj

xx eyxeaeyxeaeyxeaE Ξ²Ξ²Ξ² βˆ’βˆ’βˆ’ ++= ,Λ†,Λ†,Λ†

β€’TE modes have , solve forβ€’TM modes have , solve for

0=zE0=zH

( ) zjzz eyxhH Ξ²βˆ’= ,

zjzz eyxeE Ξ²βˆ’= ),(

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

π‘˜π‘˜π‘π‘2 = π‘˜π‘˜2 βˆ’ 𝛽𝛽2 where π‘˜π‘˜ = πœ”πœ” πœ‡πœ‡πœ‡πœ‡ and 𝛽𝛽 is the phase constant for waves travelling in the waveguide.

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All the Field Components

βjz→

βˆ‚βˆ‚

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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 Ο‰Β΅βˆ’=Γ—βˆ‡ ( )EjH ωΡσ +=Γ—βˆ‡and

zjyy eeE Ξ²βˆ’=

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Maxwell’s Curl EquationsWrite out all three components of

And all three components of

To show that:

HjE Ο‰Β΅βˆ’=Γ—βˆ‡

( )EjH ωΡσ +=Γ—βˆ‡

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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:

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Transverse in terms of axial:

Ulaby 7th edition page 384

Repeat the derivation for 𝐻𝐻π‘₯π‘₯ on the previous page to get 𝐻𝐻𝑦𝑦, 𝐸𝐸π‘₯π‘₯ and 𝐸𝐸𝑦𝑦:

𝐸𝐸π‘₯π‘₯ =βˆ’π‘—π‘—π‘˜π‘˜π‘π‘2

π›½π›½πœ•πœ•πΈπΈπ‘§π‘§πœ•πœ•πœ•πœ•

+ πœ”πœ”πœ‡πœ‡πœ•πœ•π»π»π‘§π‘§πœ•πœ•πœ•πœ•

𝐸𝐸𝑦𝑦 =π‘—π‘—π‘˜π‘˜π‘π‘2

βˆ’π›½π›½πœ•πœ•πΈπΈπ‘§π‘§πœ•πœ•πœ•πœ•

+ πœ”πœ”πœ‡πœ‡πœ•πœ•π»π»π‘§π‘§πœ•πœ•πœ•πœ•

𝐻𝐻π‘₯π‘₯ =π‘—π‘—π‘˜π‘˜π‘π‘2

πœ”πœ”πœ‡πœ‡πœ•πœ•πΈπΈπ‘§π‘§πœ•πœ•πœ•πœ•

βˆ’ π›½π›½πœ•πœ•π»π»π‘§π‘§πœ•πœ•πœ•πœ•

𝐻𝐻𝑦𝑦 =βˆ’π‘—π‘—π‘˜π‘˜π‘π‘2

πœ”πœ”πœ‡πœ‡πœ•πœ•πΈπΈπ‘§π‘§πœ•πœ•πœ•πœ•

+ π›½π›½πœ•πœ•π»π»π‘§π‘§πœ•πœ•πœ•πœ•

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

β€’ The axial component of the electric field is zero:β€’ Find the axial component of the magnetic field

0=zEzH

( ) ( ) zjyy

zjxx eyxeaeyxeaE Ξ²Ξ² βˆ’βˆ’ += ,Λ†,Λ†

( ) ( ) ( ) zjzz

zjyy

zjxx eyxhaeyxhaeyxhaH Ξ²Ξ²Ξ² βˆ’βˆ’βˆ’ ++= ,Λ†,Λ†,Λ†

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Wave Equation for the TE Modes

( )yxhz ,

( ) ( ) ( ) zjzz

zjyy

zjxx eyxhaeyxhaeyxhaH Ξ²Ξ²Ξ² βˆ’βˆ’βˆ’ ++= ,Λ†,Λ†,Λ†

Find

where

Look at the z component:

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The field β„Žπ‘§π‘§(πœ•πœ•, πœ•πœ•) must satisfy this wave equation!

To find β„Žπ‘§π‘§(πœ•πœ•,πœ•πœ•), we must satisfy the wave equation and satisfy the boundary condition that 𝐸𝐸𝑑𝑑𝑑𝑑𝑑𝑑 = 0 at the surfaces of the waveguide walls.

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Separation of Variables (ELEC365)

Dispersion equation:where and ¡Ρω=k

andLet where π‘˜π‘˜π‘₯π‘₯ and π‘˜π‘˜π‘¦π‘¦ are constants.

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Separation of Variables Solution

The field β„Žπ‘§π‘§(πœ•πœ•, πœ•πœ•) satisfies the wave equation.

What values of π‘˜π‘˜π‘₯π‘₯ and π‘˜π‘˜π‘¦π‘¦ 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 Ξ²βˆ’= ,

leads to

where

βˆ‚βˆ‚βˆ’

=y

Hk

jE z

cx ω¡2

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Satisfy the Boundary Conditions

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Waveguide Modes

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Find the Phase Constant mnΞ²

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

Dispersion equation:where and ¡Ρω=k

222ckk βˆ’=Ξ²

2222yx kkk βˆ’βˆ’=Ξ²

2222

,

βˆ’

βˆ’=

bn

amknm

ππβ 222

,

βˆ’

βˆ’=

bn

amknm

ππβ

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Cutoff Frequency for Mode m,n

mnΞ²For mode m,n to propagate, must be a real number:

22

,2

+

=

bn

amf mnc

ππ¡Ρπ

The limiting frequency is called the β€œcutoff frequency” mncf ,

22

, 21

+

=

bn

amf mnc

ππ¡Ρπ

22

, 21

+

=

bn

amf mnc ¡Ρ

If the frequency is greater than the β€œcutoff frequency”, then mode m,n propagates.

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Below Cutoff Waveguide

If the frequency is less than the β€œcutoff frequency”, then mode m,n attenuates.

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Propagation or Attenuation?22

, 21

+

=

bn

amf mnc ¡Ρ

For frequency greater than the cutoff frequency :

β€’ is real

β€’So the waveguide mode propagates with phase factor

f mncf ,

nm,Ξ²zj nme ,Ξ²βˆ’

For frequency less than the cutoff frequency :

β€’ is imaginary:

β€’So the waveguide mode attenuates with attenuation constant π›Όπ›Όπ‘šπ‘š,𝑑𝑑

f mncf ,

nm,Ξ² mnnm jΞ±Ξ² βˆ’=,

( ) zzjjzj nmnmnm eee ,,, Ξ±Ξ±Ξ² βˆ’βˆ’βˆ’βˆ’ ==

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Dominant Mode22

, 21

+

=

bn

amf mnc ¡Ρ

2

10,1

21

=

afc ¡Ρ

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 ¡Ρ

Below , no waveguide mode propagates and the waveguide is said to be β€œcutoff”.

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 β€œdominant mode”.

Above the next higher cutoff frequency, more than one mode can propagate and the waveguide is β€œovermoded”.

10,cf

2

20,2

21

=

afc ¡Ρ

𝑓𝑓𝑐𝑐,10 =1

2π‘Žπ‘Ž πœ‡πœ‡πœ‡πœ‡

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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

π‘šπ‘šπœ‹πœ‹πœ•πœ•π‘Žπ‘Ž

cosπ‘›π‘›πœ‹πœ‹πœ•πœ•π‘π‘

π‘’π‘’βˆ’π‘—π‘—π›½π›½π‘šπ‘š,𝑛𝑛𝑧𝑧

so

𝐸𝐸π‘₯π‘₯ =βˆ’π‘—π‘—π‘˜π‘˜π‘π‘2

π›½π›½π‘šπ‘šπ‘‘π‘‘πœ•πœ•πΈπΈπ‘§π‘§πœ•πœ•πœ•πœ•

+ πœ”πœ”πœ‡πœ‡πœ•πœ•π»π»π‘§π‘§πœ•πœ•πœ•πœ•

=βˆ’π‘—π‘—π‘˜π‘˜π‘π‘2

πœ”πœ”πœ‡πœ‡πœ•πœ•π»π»π‘§π‘§πœ•πœ•πœ•πœ•

𝐸𝐸𝑦𝑦 =π‘—π‘—π‘˜π‘˜π‘π‘2

βˆ’π›½π›½π‘šπ‘šπ‘‘π‘‘πœ•πœ•πΈπΈπ‘§π‘§πœ•πœ•πœ•πœ•

+ πœ”πœ”πœ‡πœ‡πœ•πœ•π»π»π‘§π‘§πœ•πœ•πœ•πœ•

=π‘—π‘—π‘˜π‘˜π‘π‘2

πœ”πœ”πœ‡πœ‡πœ•πœ•π»π»π‘§π‘§πœ•πœ•πœ•πœ•

𝐻𝐻π‘₯π‘₯ =π‘—π‘—π‘˜π‘˜π‘π‘2

πœ”πœ”πœ‡πœ‡πœ•πœ•πΈπΈπ‘§π‘§πœ•πœ•πœ•πœ•

βˆ’ π›½π›½π‘šπ‘šπ‘‘π‘‘πœ•πœ•π»π»π‘§π‘§πœ•πœ•πœ•πœ•

=π‘—π‘—π‘˜π‘˜π‘π‘2

βˆ’π›½π›½π‘šπ‘šπ‘‘π‘‘πœ•πœ•π»π»π‘§π‘§πœ•πœ•πœ•πœ•

𝐻𝐻𝑦𝑦 =βˆ’π‘—π‘—π‘˜π‘˜π‘π‘2

πœ”πœ”πœ‡πœ‡πœ•πœ•πΈπΈπ‘§π‘§πœ•πœ•πœ•πœ•

+ π›½π›½π‘šπ‘šπ‘‘π‘‘πœ•πœ•π»π»π‘§π‘§πœ•πœ•πœ•πœ•

=βˆ’π‘—π‘—π‘˜π‘˜π‘π‘2

π›½π›½π‘šπ‘šπ‘‘π‘‘πœ•πœ•π»π»π‘§π‘§πœ•πœ•πœ•πœ•

Find the transverse field components.Given 𝐻𝐻𝑧𝑧, find 𝐸𝐸π‘₯π‘₯, 𝐸𝐸𝑦𝑦, 𝐻𝐻π‘₯π‘₯, and 𝐻𝐻𝑦𝑦 .

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Substitute 𝐻𝐻𝑧𝑧 = π΄π΄π‘šπ‘šπ‘‘π‘‘ cos

π‘šπ‘šπœ‹πœ‹πœ•πœ•π‘Žπ‘Ž

cosπ‘›π‘›πœ‹πœ‹πœ•πœ•π‘π‘

π‘’π‘’βˆ’π‘—π‘—π›½π›½π‘šπ‘šπ‘›π‘›π‘§π‘§

to find the transverse field components 𝐸𝐸π‘₯π‘₯, 𝐸𝐸𝑦𝑦, 𝐻𝐻π‘₯π‘₯, and 𝐻𝐻𝑦𝑦:

Homework: show that

𝐸𝐸π‘₯π‘₯ = π΄π΄π‘šπ‘šπ‘‘π‘‘π‘—π‘—πœ”πœ”πœ‡πœ‡π‘˜π‘˜π‘π‘2

π‘›π‘›πœ‹πœ‹π‘π‘

cosπ‘šπ‘šπœ‹πœ‹πœ•πœ•π‘Žπ‘Ž

sinπ‘›π‘›πœ‹πœ‹πœ•πœ•π‘π‘

π‘’π‘’βˆ’π‘—π‘—π›½π›½π‘šπ‘šπ‘›π‘›π‘§π‘§

𝐸𝐸𝑦𝑦 = π΄π΄π‘šπ‘šπ‘‘π‘‘βˆ’π‘—π‘—πœ”πœ”πœ‡πœ‡π‘˜π‘˜π‘π‘2

π‘šπ‘šπœ‹πœ‹π‘Žπ‘Ž

sinπ‘šπ‘šπœ‹πœ‹πœ•πœ•π‘Žπ‘Ž

cosπ‘›π‘›πœ‹πœ‹πœ•πœ•π‘π‘

π‘’π‘’βˆ’π‘—π‘—π›½π›½π‘šπ‘šπ‘›π‘›π‘§π‘§

𝐻𝐻π‘₯π‘₯ = π΄π΄π‘šπ‘šπ‘‘π‘‘π‘—π‘—π›½π›½π‘šπ‘šπ‘‘π‘‘

π‘˜π‘˜π‘π‘2π‘šπ‘šπœ‹πœ‹π‘Žπ‘Ž

sinπ‘šπ‘šπœ‹πœ‹πœ•πœ•π‘Žπ‘Ž

cosπ‘›π‘›πœ‹πœ‹πœ•πœ•π‘π‘

π‘’π‘’βˆ’π‘—π‘—π›½π›½π‘šπ‘šπ‘›π‘›π‘§π‘§

𝐻𝐻𝑦𝑦 = π΄π΄π‘šπ‘šπ‘‘π‘‘π‘—π‘—π›½π›½π‘šπ‘šπ‘‘π‘‘

π‘˜π‘˜π‘π‘2π‘›π‘›πœ‹πœ‹π‘π‘

cosπ‘šπ‘šπœ‹πœ‹πœ•πœ•π‘Žπ‘Ž

sinπ‘›π‘›πœ‹πœ‹πœ•πœ•π‘π‘

π‘’π‘’βˆ’π‘—π‘—π›½π›½π‘šπ‘šπ‘›π‘›π‘§π‘§

Ulaby 7th edition page 388

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Wave Impedance for TE Modes

Define the β€œwave impedance” as

𝑍𝑍𝑇𝑇𝑇𝑇 =𝐸𝐸π‘₯π‘₯𝐻𝐻𝑦𝑦

=

βˆ’π‘—π‘—π‘˜π‘˜π‘π‘2

πœ”πœ”πœ‡πœ‡ πœ•πœ•π»π»π‘§π‘§πœ•πœ•πœ•πœ•βˆ’π‘—π‘—π‘˜π‘˜π‘π‘2

𝛽𝛽 πœ•πœ•π»π»π‘§π‘§πœ•πœ•πœ•πœ•

=πœ”πœ”πœ‡πœ‡π›½π›½

For mode m, n the phase constant is

π›½π›½π‘šπ‘šπ‘‘π‘‘ = π‘˜π‘˜2 βˆ’π‘šπ‘šπœ‹πœ‹π‘Žπ‘Ž

2βˆ’

π‘›π‘›πœ‹πœ‹π‘π‘

2

So the wave impedance can be calculated as 𝑍𝑍𝑇𝑇𝑇𝑇,π‘šπ‘šπ‘‘π‘‘ =

πœ”πœ”πœ‡πœ‡π›½π›½

=πœ”πœ”πœ‡πœ‡

π‘˜π‘˜2 βˆ’ π‘šπ‘šπœ‹πœ‹π‘Žπ‘Ž

2βˆ’ π‘›π‘›πœ‹πœ‹

𝑏𝑏2

Page 42: ELEC 351 Notes Set #15 - Encstrueman/elec351/ELEC351...Separation of Variables (ELEC365) Dispersion equation: where and k =Ο‰ ¡Ρ Let and w here π‘˜π‘˜ π‘₯π‘₯ and π‘˜π‘˜ 𝑦𝑦

In general: 𝐻𝐻𝑧𝑧 = π΄π΄π‘šπ‘šπ‘‘π‘‘ cos

π‘šπ‘šπœ‹πœ‹πœ•πœ•π‘Žπ‘Ž

cosπ‘›π‘›πœ‹πœ‹πœ•πœ•π‘π‘

π‘’π‘’βˆ’π‘—π‘—π›½π›½π‘šπ‘šπ‘›π‘›π‘§π‘§

𝐸𝐸π‘₯π‘₯ = π΄π΄π‘šπ‘šπ‘‘π‘‘π‘—π‘—πœ”πœ”πœ‡πœ‡π‘˜π‘˜π‘π‘2

π‘›π‘›πœ‹πœ‹π‘π‘

cosπ‘šπ‘šπœ‹πœ‹πœ•πœ•π‘Žπ‘Ž

sinπ‘›π‘›πœ‹πœ‹πœ•πœ•π‘π‘

π‘’π‘’βˆ’π‘—π‘—π›½π›½π‘šπ‘šπ‘›π‘›π‘§π‘§

𝐸𝐸𝑦𝑦 = π΄π΄π‘šπ‘šπ‘‘π‘‘βˆ’π‘—π‘—πœ”πœ”πœ‡πœ‡π‘˜π‘˜π‘π‘2

π‘šπ‘šπœ‹πœ‹π‘Žπ‘Ž

sinπ‘šπ‘šπœ‹πœ‹πœ•πœ•π‘Žπ‘Ž

cosπ‘›π‘›πœ‹πœ‹πœ•πœ•π‘π‘

π‘’π‘’βˆ’π‘—π‘—π›½π›½π‘šπ‘šπ‘›π‘›π‘§π‘§

𝐻𝐻π‘₯π‘₯ = π΄π΄π‘šπ‘šπ‘‘π‘‘π‘—π‘—π›½π›½π‘šπ‘šπ‘‘π‘‘π‘˜π‘˜π‘π‘2

π‘šπ‘šπœ‹πœ‹π‘Žπ‘Ž

sinπ‘šπ‘šπœ‹πœ‹πœ•πœ•π‘Žπ‘Ž

cosπ‘›π‘›πœ‹πœ‹πœ•πœ•π‘π‘

π‘’π‘’βˆ’π‘—π‘—π›½π›½π‘šπ‘šπ‘›π‘›π‘§π‘§

𝐻𝐻𝑦𝑦 = π΄π΄π‘šπ‘šπ‘‘π‘‘π‘—π‘—π›½π›½π‘šπ‘šπ‘‘π‘‘π‘˜π‘˜π‘π‘2

π‘›π‘›πœ‹πœ‹π‘π‘

cosπ‘šπ‘šπœ‹πœ‹πœ•πœ•π‘Žπ‘Ž

sinπ‘›π‘›πœ‹πœ‹πœ•πœ•π‘π‘

π‘’π‘’βˆ’π‘—π‘—π›½π›½π‘šπ‘šπ‘›π‘›π‘§π‘§

Field Configuration in the Dominant Mode

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

πœ‹πœ‹πœ•πœ•π‘Žπ‘Žπ‘’π‘’βˆ’π‘—π‘—π›½π›½10𝑧𝑧

𝐸𝐸π‘₯π‘₯ = 0

𝐸𝐸𝑦𝑦 = 𝐴𝐴10βˆ’π‘—π‘—πœ”πœ”πœ‡πœ‡π‘˜π‘˜π‘π‘2

πœ‹πœ‹π‘Žπ‘Ž

sinπœ‹πœ‹πœ•πœ•π‘Žπ‘Žπ‘’π‘’βˆ’π‘—π‘—π›½π›½10𝑧𝑧

𝐻𝐻π‘₯π‘₯ = 𝐴𝐴10𝑗𝑗𝛽𝛽10π‘˜π‘˜π‘π‘2

πœ‹πœ‹π‘Žπ‘Ž

sinπœ‹πœ‹πœ•πœ•π‘Žπ‘Žπ‘’π‘’βˆ’π‘—π‘—π›½π›½10𝑧𝑧

𝐻𝐻𝑦𝑦 = 0

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𝐸𝐸𝑦𝑦 = 𝐴𝐴10βˆ’π‘—π‘—πœ”πœ”πœ‡πœ‡π‘˜π‘˜π‘π‘2

πœ‹πœ‹π‘Žπ‘Ž

sinπœ‹πœ‹πœ•πœ•π‘Žπ‘Žπ‘’π‘’βˆ’π‘—π‘—π›½π›½10𝑧𝑧

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Power Flow in the Dominant Mode

byax ≀≀≀≀ 0,0 dxdyasd zΛ†=

zjx e

axAajH 10cos10 Ξ²Ο€

πβ βˆ’=

zj

cy e

axA

akjE 10sin2

βπω¡π βˆ’βˆ’=

( )∫∫ β‹…Γ—= dxdyaaHaEP zxxyy Λ†Λ†Λ†21 *

10

( )∫∫ β‹…βˆ’= dxdyaaHEP zzxy Λ†Λ†21 *

10

( )∫∫ βˆ’

βˆ’= βˆ’ dxdye

axAaje

axA

akjP zjzj

c

1cossin21

1010 10210

Ξ²Ξ² ππβπω¡π

𝑃𝑃10 = 𝐴𝐴 2 πœ”πœ”πœ”πœ”π‘‘π‘‘3𝑏𝑏

4πœ‹πœ‹2𝛽𝛽10 watts Homework: do

the integration!

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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 β€œa” dimension?

7.3) What is the β€œb” 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:

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7.1) What is the cutoff frequency for the TE10 mode?We are told that the β€œoperating 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 β€œa” dimension?

𝑓𝑓𝑐𝑐,π‘šπ‘šπ‘‘π‘‘ = 12πœ‹πœ‹ πœ”πœ”πœ‡πœ‡

π‘šπ‘šπœ‹πœ‹π‘‘π‘‘

2+ π‘‘π‘‘πœ‹πœ‹

𝑏𝑏

2and for air-filled waveguide c = 1

πœ”πœ”0πœ‡πœ‡0= 30 cm/ns

𝑓𝑓𝑐𝑐,10 =302πœ‹πœ‹

1πœ‹πœ‹π‘Žπ‘Ž

2

+0πœ‹πœ‹π‘π‘

2

=302πœ‹πœ‹

πœ‹πœ‹π‘Žπ‘Ž

=15π‘Žπ‘Ž

15𝑑𝑑

= 8.667

π‘Žπ‘Ž = 158.667

= 1.731 cm

7.3) What is the β€œb” dimension? β€œThe 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

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𝑓𝑓𝑐𝑐,π‘šπ‘šπ‘‘π‘‘ =302πœ‹πœ‹

π‘šπ‘šπœ‹πœ‹1.731

2+

π‘›π‘›πœ‹πœ‹π‘π‘

2

𝑓𝑓𝑐𝑐,01 =302πœ‹πœ‹

0πœ‹πœ‹1.731

2

+1πœ‹πœ‹π‘π‘

2

=302πœ‹πœ‹

πœ‹πœ‹π‘π‘

=15𝑏𝑏

= 11.44

𝑏𝑏 = 1511.44

= 1.311 cmCheck: what is 𝑓𝑓𝑐𝑐,20?

𝑓𝑓𝑐𝑐,20 = 302πœ‹πœ‹

2πœ‹πœ‹1.731

2+ 0πœ‹πœ‹

𝑏𝑏

2= 30

2πœ‹πœ‹2πœ‹πœ‹

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,π‘šπ‘šπ‘‘π‘‘ = π‘˜π‘˜2 βˆ’ π‘šπ‘šπœ‹πœ‹π‘‘π‘‘

2βˆ’ π‘‘π‘‘πœ‹πœ‹

𝑏𝑏

2

Where π‘˜π‘˜ = πœ”πœ” πœ‡πœ‡πœ‡πœ‡ = 2πœ‹πœ‹πœ‹πœ‹π‘π‘

= 2πœ‹πœ‹π‘₯π‘₯10.430

= 2.178 rad/cm (with 𝑐𝑐 = 1πœ”πœ”πœ‡πœ‡

= 30 cm/ns)

𝛽𝛽,10 = π‘˜π‘˜2 βˆ’ 1πœ‹πœ‹π‘‘π‘‘

2βˆ’ 0πœ‹πœ‹

𝑏𝑏

2= π‘˜π‘˜2 βˆ’ πœ‹πœ‹

𝑑𝑑

2= 2.1782 βˆ’ πœ‹πœ‹

1.731

2= 1.024 rad/cm

7.5) Find the wave impedance for the dominant mode at the operating frequency. 𝑍𝑍𝑇𝑇𝑇𝑇 = πœ”πœ”πœ”πœ”

𝛽𝛽= 2πœ‹πœ‹π‘₯π‘₯10.4π‘₯π‘₯109π‘₯π‘₯4πœ‹πœ‹π‘₯π‘₯10βˆ’7

102.4= 682.0 ohms

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

𝑓𝑓𝑐𝑐,π‘šπ‘šπ‘‘π‘‘ =302πœ‹πœ‹

π‘šπ‘šπœ‹πœ‹1.731

2+

π‘›π‘›πœ‹πœ‹1.311

2= 15

π‘šπ‘š1.731

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

Page 49: ELEC 351 Notes Set #15 - Encstrueman/elec351/ELEC351...Separation of Variables (ELEC365) Dispersion equation: where and k =Ο‰ ¡Ρ Let and w here π‘˜π‘˜ π‘₯π‘₯ and π‘˜π‘˜ 𝑦𝑦

52

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


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