Parallel Plate
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ECE450 – QUIZ BREAK Sample problem on Parallel Plate Wave guide Consider a parallel plate wave guide with plate separation a = 5 cm, filled with air, working at a frequency f = 10.0 GHz. a) What is the fundamental mode of propagation and what is its cut-off frequency? As in all parallel plate wave guides, the fundamental mode is the TEM (aka TM 00 ) with cut-off frequency f = 0.0 Hz. b) Find all the higher order modes of propagation. The higher modes of propagation must have a cut-off frequency smaller than the operating frequency. For both TM and TE modes, the cutoff frequency is obtained from TM f c λ c TE f c λ c TEM or TM 00 0.0 Hz ∞ TM 10 3.0 GHz 0.1 [m] TE 10 3.0 GHz 0.1 [m] TM 20 6.0 GHz 0.05 [m] TE 20 6.0 GHz 0.05 [m] TM 30 9.0 GHz 0.0333 [m] TE 30 9.0 GHz 0.0333 [m] c) Determine the guide phase velocity, the group velocity, and the angles of incidence on the plates for all the propagating modes. mode θ v pz v g TEM 90° 3.0×10 8 m/s 3.0×10 8 m/s TM 10 & TE 10 72.55° 3.1425×10 8 m/s 2.86×10 8 m/s TM 20 & TE 20 53.16° 3.75×10 8 m/s 2.4×10 8 m/s TM 30 & TE 30 25.9° 6.86×10 8 m/s 1.31×10 8 m/s 8 1/ 0,1, 2,3, 1 3 10 3 [GHz] 2 2 0.05 1, 2,3, c p p c c v v TM m m f m m a TE λ λ µε × × = = = = = × … … 1 2 2 ; cos cos ( ) 1 ( ) 1 ( ) c c c p pz g p c c f f f f v v v v f f f f λ θ θ λ − = = → = = = − −
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Transcript of Parallel Plate
ECE450 – QUIZ BREAKSample problem on Parallel Plate Wave guide
Consider a parallel plate wave guide with plate separation a = 5 cm, filled with air, working at afrequency f = 10.0 GHz.
a) What is the fundamental mode of propagation and what is its cut-off frequency?
As in all parallel plate wave guides, the fundamental mode is the TEM (aka TM00) withcut-off frequency f = 0.0 Hz.
b) Find all the higher order modes of propagation.
The higher modes of propagation must have a cut-off frequency smaller than theoperating frequency. For both TM and TE modes, the cutoff frequency is obtained from
TM fc λc TE fc λcTEM or TM00 0.0 Hz ∞
TM10 3.0 GHz 0.1 [m] TE10 3.0 GHz 0.1 [m]TM20 6.0 GHz 0.05 [m] TE20 6.0 GHz 0.05 [m]TM30 9.0 GHz 0.0333 [m] TE30 9.0 GHz 0.0333 [m]
c) Determine the guide phase velocity, the group velocity, and the angles of incidence onthe plates for all the propagating modes.
mode θ vpz vg
TEM 90° 3.0×108 m/s 3.0×108 m/sTM10 & TE10 72.55° 3.1425×108 m/s 2.86×108 m/sTM20 & TE20 53.16° 3.75×108 m/s 2.4×108 m/sTM30 & TE30 25.9° 6.86×108 m/s 1.31×108 m/s
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1/
0,1,2,3,1 3 10 3 [GHz] 2 2 0.05 1,2,3,cp
pc
cv
v TMm mf m ma TEλ
λ µε× ×= = = = =×
……
1
22
;
cos cos ( )
1 ( )1 ( )
cc
c
ppz g p c
c
f f ffvv v v f ff f
λθ θλ−= = → =
= = −−
d) Determine the guide impedance for the propagating modes
TM ηTM TE ηTETEM or TM00 376.73 Ω
TM10 359.4 Ω TE10 394.9 ΩTM20 301.5 Ω TE20 470.7 ΩTM30 164.7 Ω TE30 861.7 Ω
e) What modes are evanescent?
All TE and TM modes with cut-off frequency higher than the frequency of operation areevanescent. The cut-off frequency for TE40 and TM40 is fc = 12.0 GHz. Therefore, allmodes with index m ≥ 4 are evanescent.
f) Find the attenuation constant for the first evanescent mode
For TE40 and TM40 we have
e) What happens if the wave guide is filled with a different dielectric with relativepermittivity εr = 9.0?
The phase velocity will be decreased by a factor 3.0 and all the cut-off frequencies willbe lowered. The TE10,0 and TM10,0 modes have cut-off frequency of exactly 10.0 GHz.Therefore, modes with index up to m = 9 propagate, and with m = 10 are at cut-offcondition.
22
1 ( ) ;1 ( )
cTM TEc
f ff fηη η η= − =
−
-1
2 2 2 2 -1
-1
4 251.33 [ m ]0.05138.71 [ m ]
209.58 [ m ]
x
z x x
maj j j
π πβ
β β β β ω µεβ ω µε
= = =
=− − =− − =−= =
