Fundamentals of Electromagnetics with Engineering Applications
Fundamentals of Electromagnetics with Engineering Applications
1st Edition
ISBN: 9780470105757
Author: Stuart M. Wentworth
Publisher: Wiley, John & Sons, Incorporated
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Textbook Question
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Chapter 7, Problem 7.1P

Find the cutoff frequency for the first eight modes of WR430.

Expert Solution & Answer
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To determine

The cutoff frequency of eight modes of WR430 waveguide.

Answer to Problem 7.1P

The cutoff frequency of the eight modes of WR430 the waveguide are 1.374 GHz , 2.747 GHz , 2.747 GHz , 3.07 GHz , 3.07 GHz , 3.885 GHz , 3.885 GHz and 4.121 GHz .

Explanation of Solution

Given:

The dimensions of the waveguide are a=4.3 in and b=2.15 in .

Concept used:

The cutoff frequency of the waveguide with dimensions a and b is shown below.

  fcmn=c2( m a )2+( n b )2 ..... (1)

Here, m and n represents the half-wave variations.

Calculation:

Convert the dimensions of the waveguide in meters as

  a=4.3 in39.37( 1 m 1 in)0.109 m

The value of b in meters is calculated below.

  b=2.15 in39.37( 1 m 1 in)0.0546 m

The first mode of the waveguide is TE10 and its cutoff frequency is calculated as,

  fc 10=c2 ( 1 a )2+ ( 0 b )2=c2a

Substitute 3×108 for c and 0.109 for a in the above equation.

  fc 10=3× 1082( 0.109)=13.74×1081.374×109

Therefore, the cutoff frequency for TE10 mode is 1.374 GHz .

The second mode of the waveguide is TE01 and its cutoff frequency is calculated as,

  fc 01=c2 ( 0 a )2+ ( 1 b )2=c2b

Substitute 3×108 for c and 0.0546 for a in the above equation.

  fc 01=3× 1082( 0.0546)=27.47×1082.747×109

Therefore, the cutoff frequency for TE01 mode is 2.747 GHz .

The third mode of the waveguide is TE20 and its cutoff frequency is calculated as,

  fc 20=c2 ( 2 a )2+ ( 0 b )2=ca

Substitute 3×108 for c and 0.109 for a in the above equation.

  fc 20=3× 1080.109=27.47×1082.747×109

Therefore, the cutoff frequency for TE20 mode is 2.747 GHz .

The fourth mode of the waveguide is TE11 and its cutoff frequency is calculated as,

  fc 11=c2 ( 1 a )2+ ( 1 b )2=c21 a 2 +1 b 2

Substitute 3×108 for c , 0.109 for a and 0.0546 for b in the above equation.

  fc 11=c21 ( 0.109 ) 2 +1 ( 0.0546 ) 2 =30.7×1083.07×109

Therefore, the cutoff frequency for TE11 mode is 3.07 GHz .

The fifth mode of the waveguide is TM11 and its cutoff frequency is calculated as,

  fc 11=c2 ( 1 a )2+ ( 1 b )2=c21 a 2 +1 b 2

Substitute 3×108 for c , 0.109 for a and 0.0546 for b in the above equation.

  fc 11=c21 ( 0.109 ) 2 +1 ( 0.0546 ) 2 =30.7×1083.07×109

Therefore, the cutoff frequency for TM11 mode is 3.07 GHz .

Next mode of the waveguide is TE21 and its cutoff frequency is calculated as,

  fc 21=c2 ( 2 a )2+ ( 1 b )2=c24 a 2 +1 b 2

Substitute 3×108 for c , 0.109 for a and 0.0546 for b in the above equation.

  fc 21=c24 ( 0.109 ) 2 +1 ( 0.0546 ) 2 =38.85×1083.885×109

Therefore, the cutoff frequency for TE21 mode is 3.885 GHz .

Next mode of the waveguide is TM21 and its cutoff frequency is calculated as,

  fc 21=c2 ( 2 a )2+ ( 1 b )2=c24 a 2 +1 b 2

Substitute 3×108 for c , 0.109 for a and 0.0546 for b in the above equation.

  fc 21=c24 ( 0.109 ) 2 +1 ( 0.0546 ) 2 =38.85×1083.885×109

Therefore, the cutoff frequency for TM21 mode is 3.885 GHz .

The eighth modeof the waveguide is TE30 and its cutoff frequency is calculated as,

  fc 30=c2 ( 3 a )2+ ( 1 b )2=3c2a

Substitute 3×108 for c , and 0.109 for a in the above equation.

  fc 30=3( 3× 10 8 )2( 0.109)=41.21×1084.121×109

Therefore, the cutoff frequency for TE30 mode is 4.121 GHz .

Conclusion:

Thus, the cutoff frequency of the eight modes of WR430 the waveguide are 1.734 GHz , 2.747 GHz , 2.747 GHz , 3.07 GHz , 3.07 GHz , 3.885 GHz , 3.885 GHz and 4.121 GHz .

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