In: Civil Engineering
cut off frequency can be determine by the following formula
where
v is Velocity of propagation in dielectric =
is dielectric which is equal to 4 (assume)
so for mode TE10
V =
V =
f = 0.0025 GHz
For TE01
V =
f = .0037 GHz
For TE02
V =
f = .0075GHz
For TE11
V =
f = .00450 GHz
For TE20
V =
f = .0050 GHz
We were unable to transcribe this image
We were unable to transcribe this image
We were unable to transcribe this image
We were unable to transcribe this image
31081.5108m/s
- C(speedoflight)
We were unable to transcribe this image
1.5 10G1 30 20
31081.5108m/s
1.5*108 , 0 30 20
31081.5108m/s
We were unable to transcribe this image
31081.5108m/s
1.5 10G1 30 20
31081.5108m/s
1.5*108, 2 30 20