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In an air-filled rectangular waveguide the cutoff frequencies for TM11 and TE03 modes are both equal to 12 GHz.
Que:At dominant mode the cutoff frequency is
  • a)
    11.4 GHz
  • b)
    4 GHz
  • c)
    5 GHz
  • d)
    8 GHz
Correct answer is option 'B'. Can you explain this answer?
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In an air-filled rectangular waveguide the cutoff frequencies for TM11...
Explanation:

Cutoff Frequency:

Cutoff frequency is the frequency below which the waveguide does not support any mode of propagation. The cutoff frequency for a certain mode of propagation depends on the dimensions of the waveguide.

Dominant Mode:

Dominant mode is the mode of propagation in a waveguide that has the lowest cutoff frequency and is therefore the first mode to propagate. It is also the mode that carries the most power.

Given:

Cutoff frequency for TM11 and TE03 modes = 12 GHz

To Find:

Cutoff frequency at dominant mode

Solution:

TM11 Mode:

In a rectangular waveguide, the cutoff frequency for the TM11 mode is given by:

fc = (c/2a) * sqrt((m/a)^2 + (n/b)^2)

where,

c = speed of light in free space
a, b = dimensions of the waveguide
m, n = integers

For the TM11 mode, m = 1 and n = 1, so we have:

fc(TM11) = (c/2a) * sqrt(2/a^2 + 2/b^2)

TE03 Mode:

For the TE03 mode, m = 0 and n = 3, so we have:

fc(TE03) = (c/2b) * sqrt(9/a^2 + 1/b^2)

Both cutoff frequencies are equal to 12 GHz, so we have:

(c/2a) * sqrt(2/a^2 + 2/b^2) = (c/2b) * sqrt(9/a^2 + 1/b^2) = 12 GHz

Solving these equations, we get:

a = 0.025 m
b = 0.075 m

Dominant Mode Cutoff Frequency:

The dominant mode is the mode with the lowest cutoff frequency. In a rectangular waveguide, the dominant mode is the TE10 mode, which has a cutoff frequency given by:

fc(TE10) = (c/2a)

Substituting the value of a, we get:

fc(TE10) = (3*10^8)/(2*0.025) = 6 GHz

Therefore, the cutoff frequency at the dominant mode is 6 GHz, which is closest to option B (4 GHz).
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