HW#6 - Selected solution
9‐7. A standard air‐filled S‐band rectangular waveguide has dimensions and
at mode types can be used to transmit electromagnetic waves having the following wavelengths?
7.21 3.40 . Wh
a) λ = 10 (cm) b) λ = 5 (cm)
Cutoff wavelength for TEmn and TMmn modes:
( ) ( ) (
/) (
/)
( )2
2
2 m
b n a f m
u
c mn c mn
= +
λ
=( ) ( )
cmTE c 14.42
0721 . 0 / 1 : 2
10 2
10
λ
= =( ) ( )
cmTE c 6.8
034 . 0 / 1 : 2
01 2
01
λ
= =( ) ( )
cmTE c 7.21
0721 . 0 / 2 : 2
20 2
20
λ
= =( ) ( ) ( )
cmTM c 6.15
034 . 0 / 1 0721 . 0 / 1 : 2
2 11 2
11 ≈
= +
λ
a) For
λ
=10cm,( ) λc mn >λ TE10
b) For
λ
=5 cm,( ) λc mn >λ TE10, TE01, TE20,TM11
9‐15. An electromagnetic wave is to propagate along an air‐filled rectangular waveguide at the dominant mode. Assume and the usable bandwidth to be between 1.15 and 15%
below the cutoff frequency of the next higher mode.
2.50
0.25
( )
a) Calculate and compare the permissible bandwidth for , 0.50 , and 0.75 . b) Calculate and compare the average powers transmitted along the three guides in part (a) at 7
(GHz) if the maximum electric intensity is 10 (kV/m). Neglect the losses.
( ) ( ) ( )
c GHzf b
n a c m
fc mn c 3 10 6
, / /
8 10
2
2+ = = × =
= 2 2a 2×0.025 Dominant mode: TE10
a) For
( ) ( )
⎜⎜ ( )
⎜ ⎜
⎜ ⎜
⎜ ⎜
⎝
⎛
⎟ =
⎠
⎜ ⎞
⎝ + ⎛
×
= ×
⎟ ⎠
⎜ ⎞
⎝ + ⎛
=
=
=
× =
×
= ×
=
=
b GHz a a
f c
a GHz f c
b GHz f c
a b
c c c
7 . 25 24
. 0 1 1 025 . 0 2
10 1 3
2 12
025 24 . 0 25 . 0 2
10 3 2
, 25 . 0
8 2 2
11 20
8 01
Next higher mode: TE20
Usable band width: 1.15
( )
fc 10 < f <0.85( )
fc 206 . 9 < f < 10 . 2 GHz
Possible band width= 10 . 2 − 6 . 9 = 3 . 3 GHz
For
( ) ( )
⎜⎜ ( )
⎜ ⎜
⎜ ⎜
⎜ ⎜
⎝
⎛
⎟ =
⎠
⎜ ⎞
⎝ + ⎛
=
=
=
=
=
=
b GHz a a
f c
a GHz f c
b GHz f c
a
b
c c c
4 . 13 2 1
12 2 12
, 5 . 0
2 11
20 01
Next higher mode: TE01,TE20
Usable band width:
6 . 9 < f < 10 . 2 GHz
Possible band width= 10 . 2 − 6 . 9 = 3 . 3 GHz
For
( ) ( )
⎜⎜ ( )
⎜ ⎜
⎜ ⎜
⎜ ⎜
⎝
⎛
⎟ =
⎠
⎜ ⎞
⎝ + ⎛
=
=
=
=
=
=
b GHz a a
f c
a GHz f c
b GHz f c
a b
c c c
10 2 1
12 2 8
, 75 . 0
2 11
20 01
Next higher mode: TE01
Usable band width:
6 . 9 < f < 6 . 8 GHz
No possible bandb) Using (9‐101) for f =7 GHz, E0 =10kV/m,
( )
fc 10 =6.9GHzf b f ab
P
avE 1
c85 . 3 4
2
0 2
0
⎟⎟ ⎠ ≈
⎜⎜ ⎞
⎝
− ⎛
= η
For b=0.25a=0.25×0.025, Pav ≈5.33W b=0.5a=0.5×0.025, Pav ≈10.7W