elec 351 notes set #15 - encstrueman/elec351/elec351...separation of variables (elec365) dispersion...
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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
โWaveguide diplexer links point to point systemsโhttp://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,โDesign and Analysis of Microwave Devices Based on Ridged Gap Waveguide Technology for 5G Applicationsโ, 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 Maxwellโs Equations for Transmission Lines and Waveguide
Convention: ba >
( ) ( ) ( ) zjzz
zjyy
zjxx eyxeaeyxeaeyxeaE ฮฒฮฒฮฒ โโโ ++= ,ห,ห,ห
( ) ( ) ( ) zjzz
zjyy
zjxx eyxhaeyxhaeyxhaH ฮฒฮฒฮฒ โโโ ++= ,ห,ห,ห
( ) zjzz eyxeE ฮฒโ= ,
Ulaby 7th edition Section 8.6
๐๐ = ๐๐ ๐๐๐๐ is the phase constant in the dielectric filling the waveguide.๐ฝ๐ฝ 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 ฮฒฮฒฮฒ โโโ ++= ,ห,ห,ห
( ) ( ) ( ) zjzz
zjyy
zjxx eyxhaeyxhaeyxhaH ฮฒฮฒฮฒ โโโ ++= ,ห,ห,ห
( ) zjyy eyxeE ฮฒโ= ,( ) zj
xx eyxeE ฮฒโ= , and
( ) zjyy eyxhH ฮฒโ= ,( ) zj
xx eyxhH ฮฒโ= , and
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Rectangular Waveguides
( ) ( ) ( ) zjzz
zjyy
zjxx eyxhaeyxhaeyxhaH ฮฒฮฒฮฒ โโโ ++= ,ห,ห,ห( ) ( ) ( ) zj
zzzj
yyzj
xx eyxeaeyxeaeyxeaE ฮฒฮฒฮฒ โโโ ++= ,ห,ห,ห
โขTE modes have , solve forโขTM modes have , solve for
0=zE0=zH
( ) zjzz eyxhH ฮฒโ= ,
zjzz eyxeE ฮฒโ= ),(
We will show that if we know the axial components, we can find the transverse components from Maxwellโs curl equations:
๐๐๐๐2 = ๐๐2 โ ๐ฝ๐ฝ2 where ๐๐ = ๐๐ ๐๐๐๐ and ๐ฝ๐ฝ is the phase constant for waves travelling in the waveguide.
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All the Field Components
ฮฒjzโ
โโ
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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 ฯยตโ=รโ ( )EjH ฯฮตฯ +=รโand
zjyy eeE ฮฒโ=
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Maxwellโs Curl EquationsWrite out all three components of
And all three components of
To show that:
HjE ฯยตโ=รโ
( )EjH ฯฮตฯ +=รโ
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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 ๐ป๐ป๐ฅ๐ฅ on the previous page to get ๐ป๐ป๐ฆ๐ฆ, ๐ธ๐ธ๐ฅ๐ฅ and ๐ธ๐ธ๐ฆ๐ฆ:
๐ธ๐ธ๐ฅ๐ฅ =โ๐๐๐๐๐๐2
๐ฝ๐ฝ๐๐๐ธ๐ธ๐ง๐ง๐๐๐๐
+ ๐๐๐๐๐๐๐ป๐ป๐ง๐ง๐๐๐๐
๐ธ๐ธ๐ฆ๐ฆ =๐๐๐๐๐๐2
โ๐ฝ๐ฝ๐๐๐ธ๐ธ๐ง๐ง๐๐๐๐
+ ๐๐๐๐๐๐๐ป๐ป๐ง๐ง๐๐๐๐
๐ป๐ป๐ฅ๐ฅ =๐๐๐๐๐๐2
๐๐๐๐๐๐๐ธ๐ธ๐ง๐ง๐๐๐๐
โ ๐ฝ๐ฝ๐๐๐ป๐ป๐ง๐ง๐๐๐๐
๐ป๐ป๐ฆ๐ฆ =โ๐๐๐๐๐๐2
๐๐๐๐๐๐๐ธ๐ธ๐ง๐ง๐๐๐๐
+ ๐ฝ๐ฝ๐๐๐ป๐ป๐ง๐ง๐๐๐๐
If we can find ๐ธ๐ธ๐ง๐ง and ๐ป๐ป๐ง๐ง, then we can find all the other field components.TE Modes: Assume ๐ธ๐ธ๐ง๐ง = 0 and find ๐ป๐ป๐ง๐งTM Modes: Assume ๐ป๐ป๐ง๐ง = 0 and find ๐ธ๐ธ๐ง๐ง
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Convention: ba >
Rectangular Waveguide: TE Modes
โข The axial component of the electric field is zero:โข Find the axial component of the magnetic field
0=zEzH
( ) ( ) zjyy
zjxx eyxeaeyxeaE ฮฒฮฒ โโ += ,ห,ห
( ) ( ) ( ) zjzz
zjyy
zjxx eyxhaeyxhaeyxhaH ฮฒฮฒฮฒ โโโ ++= ,ห,ห,ห
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Wave Equation for the TE Modes
( )yxhz ,
( ) ( ) ( ) zjzz
zjyy
zjxx eyxhaeyxhaeyxhaH ฮฒฮฒฮฒ โโโ ++= ,ห,ห,ห
Find
where
Look at the z component:
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The field โ๐ง๐ง(๐๐, ๐๐) must satisfy this wave equation!
To find โ๐ง๐ง(๐๐,๐๐), we must satisfy the wave equation and satisfy the boundary condition that ๐ธ๐ธ๐ก๐ก๐ก๐ก๐ก๐ก = 0 at the surfaces of the waveguide walls.
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Separation of Variables (ELEC365)
Dispersion equation:where and ยตฮตฯ=k
andLet where ๐๐๐ฅ๐ฅ and ๐๐๐ฆ๐ฆ are constants.
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Separation of Variables Solution
The field โ๐ง๐ง(๐๐, ๐๐) satisfies the wave equation.
What values of ๐๐๐ฅ๐ฅ and ๐๐๐ฆ๐ฆ satisfy the boundary conditions?
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Boundary Conditions
For the TE solution, the axial component ๐ธ๐ธ๐ง๐ง is zero everywhere.
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Find and in terms of : xE yE zH
( ) zjzz eyxhH ฮฒโ= ,
leads to
where
โโโ
=y
Hk
jE z
cx ฯยต2
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Satisfy the Boundary Conditions
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Waveguide Modes
Find the Phase Constant mnฮฒ
The value of the phase constant depends on the waveguide mode:
Dispersion equation:where and ยตฮตฯ=k
222ckk โ=ฮฒ
2222yx kkk โโ=ฮฒ
2222
,
โ
โ=
bn
amknm
ฯฯฮฒ 222
,
โ
โ=
bn
amknm
ฯฯฮฒ
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Cutoff Frequency for Mode m,n
mnฮฒFor mode m,n to propagate, must be a real number:
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,2
+
=
bn
amf mnc
ฯฯยตฮตฯ
The limiting frequency is called the โcutoff frequencyโ mncf ,
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, 21
+
=
bn
amf mnc
ฯฯยตฮตฯ
22
, 21
+
=
bn
amf mnc ยตฮต
If the frequency is greater than the โcutoff frequencyโ, then mode m,n propagates.
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Below Cutoff Waveguide
If the frequency is less than the โcutoff frequencyโ, then mode m,n attenuates.
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Propagation or Attenuation?22
, 21
+
=
bn
amf mnc ยตฮต
For frequency greater than the cutoff frequency :
โข is real
โขSo the waveguide mode propagates with phase factor
f mncf ,
nm,ฮฒzj nme ,ฮฒโ
For frequency less than the cutoff frequency :
โข is imaginary:
โขSo the waveguide mode attenuates with attenuation constant ๐ผ๐ผ๐๐,๐ก๐ก
f mncf ,
nm,ฮฒ mnnm jฮฑฮฒ โ=,
( ) zzjjzj nmnmnm eee ,,, ฮฑฮฑฮฒ โโโโ ==
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Dominant Mode22
, 21
+
=
bn
amf mnc ยตฮต
2
10,1
21
=
afc ยตฮต
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 ยตฮต
Below , no waveguide mode propagates and the waveguide is said to be โcutoffโ.
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 โdominant modeโ.
Above the next higher cutoff frequency, more than one mode can propagate and the waveguide is โovermodedโ.
10,cf
2
20,2
21
=
afc ยตฮต
๐๐๐๐,10 =1
2๐๐ ๐๐๐๐
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Example
39
40
(seventh edition page 368)
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For TE modes (๐ธ๐ธ๐ง๐ง=0) we have found ๐ป๐ป๐ง๐ง = ๐ด๐ด cos
๐๐๐๐๐๐๐๐
cos๐๐๐๐๐๐๐๐
๐๐โ๐๐๐ฝ๐ฝ๐๐,๐๐๐ง๐ง
so
๐ธ๐ธ๐ฅ๐ฅ =โ๐๐๐๐๐๐2
๐ฝ๐ฝ๐๐๐ก๐ก๐๐๐ธ๐ธ๐ง๐ง๐๐๐๐
+ ๐๐๐๐๐๐๐ป๐ป๐ง๐ง๐๐๐๐
=โ๐๐๐๐๐๐2
๐๐๐๐๐๐๐ป๐ป๐ง๐ง๐๐๐๐
๐ธ๐ธ๐ฆ๐ฆ =๐๐๐๐๐๐2
โ๐ฝ๐ฝ๐๐๐ก๐ก๐๐๐ธ๐ธ๐ง๐ง๐๐๐๐
+ ๐๐๐๐๐๐๐ป๐ป๐ง๐ง๐๐๐๐
=๐๐๐๐๐๐2
๐๐๐๐๐๐๐ป๐ป๐ง๐ง๐๐๐๐
๐ป๐ป๐ฅ๐ฅ =๐๐๐๐๐๐2
๐๐๐๐๐๐๐ธ๐ธ๐ง๐ง๐๐๐๐
โ ๐ฝ๐ฝ๐๐๐ก๐ก๐๐๐ป๐ป๐ง๐ง๐๐๐๐
=๐๐๐๐๐๐2
โ๐ฝ๐ฝ๐๐๐ก๐ก๐๐๐ป๐ป๐ง๐ง๐๐๐๐
๐ป๐ป๐ฆ๐ฆ =โ๐๐๐๐๐๐2
๐๐๐๐๐๐๐ธ๐ธ๐ง๐ง๐๐๐๐
+ ๐ฝ๐ฝ๐๐๐ก๐ก๐๐๐ป๐ป๐ง๐ง๐๐๐๐
=โ๐๐๐๐๐๐2
๐ฝ๐ฝ๐๐๐ก๐ก๐๐๐ป๐ป๐ง๐ง๐๐๐๐
Find the transverse field components.Given ๐ป๐ป๐ง๐ง, find ๐ธ๐ธ๐ฅ๐ฅ, ๐ธ๐ธ๐ฆ๐ฆ, ๐ป๐ป๐ฅ๐ฅ, and ๐ป๐ป๐ฆ๐ฆ .
Substitute ๐ป๐ป๐ง๐ง = ๐ด๐ด๐๐๐ก๐ก cos
๐๐๐๐๐๐๐๐
cos๐๐๐๐๐๐๐๐
๐๐โ๐๐๐ฝ๐ฝ๐๐๐๐๐ง๐ง
to find the transverse field components ๐ธ๐ธ๐ฅ๐ฅ, ๐ธ๐ธ๐ฆ๐ฆ, ๐ป๐ป๐ฅ๐ฅ, and ๐ป๐ป๐ฆ๐ฆ:
Homework: show that
๐ธ๐ธ๐ฅ๐ฅ = ๐ด๐ด๐๐๐ก๐ก๐๐๐๐๐๐๐๐๐๐2
๐๐๐๐๐๐
cos๐๐๐๐๐๐๐๐
sin๐๐๐๐๐๐๐๐
๐๐โ๐๐๐ฝ๐ฝ๐๐๐๐๐ง๐ง
๐ธ๐ธ๐ฆ๐ฆ = ๐ด๐ด๐๐๐ก๐กโ๐๐๐๐๐๐๐๐๐๐2
๐๐๐๐๐๐
sin๐๐๐๐๐๐๐๐
cos๐๐๐๐๐๐๐๐
๐๐โ๐๐๐ฝ๐ฝ๐๐๐๐๐ง๐ง
๐ป๐ป๐ฅ๐ฅ = ๐ด๐ด๐๐๐ก๐ก๐๐๐ฝ๐ฝ๐๐๐ก๐ก
๐๐๐๐2๐๐๐๐๐๐
sin๐๐๐๐๐๐๐๐
cos๐๐๐๐๐๐๐๐
๐๐โ๐๐๐ฝ๐ฝ๐๐๐๐๐ง๐ง
๐ป๐ป๐ฆ๐ฆ = ๐ด๐ด๐๐๐ก๐ก๐๐๐ฝ๐ฝ๐๐๐ก๐ก
๐๐๐๐2๐๐๐๐๐๐
cos๐๐๐๐๐๐๐๐
sin๐๐๐๐๐๐๐๐
๐๐โ๐๐๐ฝ๐ฝ๐๐๐๐๐ง๐ง
Ulaby 7th edition page 388
Wave Impedance for TE Modes
Define the โwave impedanceโ as
๐๐๐๐๐๐ =๐ธ๐ธ๐ฅ๐ฅ๐ป๐ป๐ฆ๐ฆ
=
โ๐๐๐๐๐๐2
๐๐๐๐ ๐๐๐ป๐ป๐ง๐ง๐๐๐๐โ๐๐๐๐๐๐2
๐ฝ๐ฝ ๐๐๐ป๐ป๐ง๐ง๐๐๐๐
=๐๐๐๐๐ฝ๐ฝ
For mode m, n the phase constant is
๐ฝ๐ฝ๐๐๐ก๐ก = ๐๐2 โ๐๐๐๐๐๐
2โ
๐๐๐๐๐๐
2
So the wave impedance can be calculated as ๐๐๐๐๐๐,๐๐๐ก๐ก =
๐๐๐๐๐ฝ๐ฝ
=๐๐๐๐
๐๐2 โ ๐๐๐๐๐๐
2โ ๐๐๐๐
๐๐2
In general: ๐ป๐ป๐ง๐ง = ๐ด๐ด๐๐๐ก๐ก cos
๐๐๐๐๐๐๐๐
cos๐๐๐๐๐๐๐๐
๐๐โ๐๐๐ฝ๐ฝ๐๐๐๐๐ง๐ง
๐ธ๐ธ๐ฅ๐ฅ = ๐ด๐ด๐๐๐ก๐ก๐๐๐๐๐๐๐๐๐๐2
๐๐๐๐๐๐
cos๐๐๐๐๐๐๐๐
sin๐๐๐๐๐๐๐๐
๐๐โ๐๐๐ฝ๐ฝ๐๐๐๐๐ง๐ง
๐ธ๐ธ๐ฆ๐ฆ = ๐ด๐ด๐๐๐ก๐กโ๐๐๐๐๐๐๐๐๐๐2
๐๐๐๐๐๐
sin๐๐๐๐๐๐๐๐
cos๐๐๐๐๐๐๐๐
๐๐โ๐๐๐ฝ๐ฝ๐๐๐๐๐ง๐ง
๐ป๐ป๐ฅ๐ฅ = ๐ด๐ด๐๐๐ก๐ก๐๐๐ฝ๐ฝ๐๐๐ก๐ก๐๐๐๐2
๐๐๐๐๐๐
sin๐๐๐๐๐๐๐๐
cos๐๐๐๐๐๐๐๐
๐๐โ๐๐๐ฝ๐ฝ๐๐๐๐๐ง๐ง
๐ป๐ป๐ฆ๐ฆ = ๐ด๐ด๐๐๐ก๐ก๐๐๐ฝ๐ฝ๐๐๐ก๐ก๐๐๐๐2
๐๐๐๐๐๐
cos๐๐๐๐๐๐๐๐
sin๐๐๐๐๐๐๐๐
๐๐โ๐๐๐ฝ๐ฝ๐๐๐๐๐ง๐ง
Field Configuration in the Dominant Mode
Dominant mode, m=1,n=0: ๐ป๐ป๐ง๐ง = ๐ด๐ด10 cos
๐๐๐๐๐๐๐๐โ๐๐๐ฝ๐ฝ10๐ง๐ง
๐ธ๐ธ๐ฅ๐ฅ = 0
๐ธ๐ธ๐ฆ๐ฆ = ๐ด๐ด10โ๐๐๐๐๐๐๐๐๐๐2
๐๐๐๐
sin๐๐๐๐๐๐๐๐โ๐๐๐ฝ๐ฝ10๐ง๐ง
๐ป๐ป๐ฅ๐ฅ = ๐ด๐ด10๐๐๐ฝ๐ฝ10๐๐๐๐2
๐๐๐๐
sin๐๐๐๐๐๐๐๐โ๐๐๐ฝ๐ฝ10๐ง๐ง
๐ป๐ป๐ฆ๐ฆ = 0
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๐ธ๐ธ๐ฆ๐ฆ = ๐ด๐ด10โ๐๐๐๐๐๐๐๐๐๐2
๐๐๐๐
sin๐๐๐๐๐๐๐๐โ๐๐๐ฝ๐ฝ10๐ง๐ง
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Power Flow in the Dominant Mode
byax โคโคโคโค 0,0 dxdyasd zห=
zjx e
axAajH 10cos10 ฮฒฯ
ฯฮฒ โ=
zj
cy e
axA
akjE 10sin2
ฮฒฯฯยตฯ โโ=
( )โซโซ โ ร= dxdyaaHaEP zxxyy หหห21 *
10
( )โซโซ โ โ= dxdyaaHEP zzxy หห21 *
10
( )โซโซ โ
โ= โ dxdye
axAaje
axA
akjP zjzj
c
1cossin21
1010 10210
ฮฒฮฒ ฯฯฮฒฯฯยตฯ
๐๐10 = ๐ด๐ด 2 ๐๐๐๐๐ก๐ก3๐๐
4๐๐2๐ฝ๐ฝ10 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 โaโ dimension?
7.3) What is the โbโ 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 โoperating frequency of 10.4 GHz is 20% above the cutoff frequency for the dominant modeโ, so 10.4 = 1.2๐๐๐๐,10, where the cutoff frequency is๐๐๐๐,10 = 10.4
1.2= 8.667 GHz.
7.2) What is the โaโ dimension?
๐๐๐๐,๐๐๐ก๐ก = 12๐๐ ๐๐๐๐
๐๐๐๐๐ก๐ก
2+ ๐ก๐ก๐๐
๐๐
2and for air-filled waveguide c = 1
๐๐0๐๐0= 30 cm/ns
๐๐๐๐,10 =302๐๐
1๐๐๐๐
2
+0๐๐๐๐
2
=302๐๐
๐๐๐๐
=15๐๐
15๐ก๐ก
= 8.667
๐๐ = 158.667
= 1.731 cm
7.3) What is the โbโ dimension? โThe cutoff frequency of the next higher waveguide mode must be 10% above the operating frequency.โSo the next higher cutoff frequency is 1.1x10.4=11.44 GHz.
The next higher cutoff frequency might be ๐๐๐๐,01 or might be ๐๐๐๐,20
50
๐๐๐๐,๐๐๐ก๐ก =302๐๐
๐๐๐๐1.731
2+
๐๐๐๐๐๐
2
๐๐๐๐,01 =302๐๐
0๐๐1.731
2
+1๐๐๐๐
2
=302๐๐
๐๐๐๐
=15๐๐
= 11.44
๐๐ = 1511.44
= 1.311 cmCheck: what is ๐๐๐๐,20?
๐๐๐๐,20 = 302๐๐
2๐๐1.731
2+ 0๐๐
๐๐
2= 30
2๐๐2๐๐
1.731= 30
22
1.731= 30
1.731= 17.33 GHz
Since this is higher than 11.44 GHz, choose ๐๐ = 1.311 cm
51
7.4) Find the phase constant for the dominant mode at the operating frequency.
๐ฝ๐ฝ,๐๐๐ก๐ก = ๐๐2 โ ๐๐๐๐๐ก๐ก
2โ ๐ก๐ก๐๐
๐๐
2
Where ๐๐ = ๐๐ ๐๐๐๐ = 2๐๐๐๐๐๐
= 2๐๐๐ฅ๐ฅ10.430
= 2.178 rad/cm (with ๐๐ = 1๐๐๐๐
= 30 cm/ns)
๐ฝ๐ฝ,10 = ๐๐2 โ 1๐๐๐ก๐ก
2โ 0๐๐
๐๐
2= ๐๐2 โ ๐๐
๐ก๐ก
2= 2.1782 โ ๐๐
1.731
2= 1.024 rad/cm
7.5) Find the wave impedance for the dominant mode at the operating frequency. ๐๐๐๐๐๐ = ๐๐๐๐
๐ฝ๐ฝ= 2๐๐๐ฅ๐ฅ10.4๐ฅ๐ฅ109๐ฅ๐ฅ4๐๐๐ฅ๐ฅ10โ7
102.4= 682.0 ohms
7.6) How many modes can propagate in the waveguide at 20 GHz?
๐๐๐๐,๐๐๐ก๐ก =302๐๐
๐๐๐๐1.731
2+
๐๐๐๐1.311
2= 15
๐๐1.731
2+
๐๐1.311
2
๐๐๐๐,01 = 11.44 GHz, propagates at 20 GHz๐๐๐๐,20 = 17.33 GHz, propagates at 20 GHz
๐๐๐๐,02 = 15 01.731
2+ 2
1.311
2= 22.88 GHz, does NOT propagate
๐๐๐๐,11 = 15 11.731
2+ 1
1.311
2= 14.35 GHz propagates
52
7.5) Find the wave impedance for the dominant mode at the operating frequency.Continued:
๐๐๐๐,21 = 15 21.731
2+ 1
1.311
2= 20.76 GHz does NOT propagate
๐๐๐๐,12 = 15 11.731
2+ 2
1.311
2= 24.47 GHz does NOT propagate