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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

=  

( ) ( )

cm

TE c 14.42

0721 . 0 / 1 : 2

10 2

10

λ

= =  

( ) ( )

cm

TE c 6.8

034 . 0 / 1 : 2

01 2

01

λ

= =  

( ) ( )

cm

TE c 7.21

0721 . 0 / 2 : 2

20 2

20

λ

= =  

( ) ( ) ( )

cm

TM 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   

         

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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 GHz

f 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 20         

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

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

   
(3)

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 band   

b) Using (9‐101) for  f =7 GHz, E0 =10kV/m,

( )

fc 10 =6.9GHz 

f b f ab

P

av

E 1

c

85 . 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 

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