Numerical 2:for TE11 and TM01 Modes

Find the cutoff frequency for an air-filled circular waveguide having a radius of \(2\,\text{cm}\) for the dominant mode and the \(TM_{01}\) mode.

Given:

$ a=2\,\text{cm} $

$ P'_{11}=1.8412 $

$ P_{01}=2.4049 $

For an air-filled waveguide, the permeability and permittivity are approximately equal to the free-space values:

$ \mu=\mu_0 $

$ \epsilon=\epsilon_0 $

where:

$ \mu_0=4\pi\times10^{-7}\,\text{H/m} $

$ \epsilon_0=8.854\times10^{-12}\,\text{F/m} $

The radius must be converted from centimetres to metres:

$ a=2\,\text{cm} $

$ a=\frac{2}{100}=0.02\,\text{m} $

1. Cutoff Frequency of the Dominant Mode \(TE_{11}\)

The dominant mode of a circular waveguide is \(TE_{11}\). For a TE mode in a circular waveguide, the cutoff frequency is:

$ f_c = \frac{P'_{nm}} {2\pi a\sqrt{\mu\epsilon}} $

For the \(TE_{11}\) mode:

$ P'_{11}=1.8412 $

Therefore:

$ f_{c,11}^{TE} = \frac{1.8412} {2\pi a\sqrt{\mu_0\epsilon_0}} $

Substituting \(a=0.02\,\text{m}\):

$ f_{c,11}^{TE} = \frac{1.8412} { 2\pi(0.02) \sqrt{ (4\pi\times10^{-7}) (8.854\times10^{-12}) } } $

Using:

$ \frac{1}{\sqrt{\mu_0\epsilon_0}} = c $

where \(c\approx3\times10^8\,\text{m/s}\), the equation can be simplified to:

$ f_{c,11}^{TE} = \frac{1.8412c}{2\pi a} $

Substituting the numerical values:

$ f_{c,11}^{TE} = \frac{ 1.8412(3\times10^8) }{ 2\pi(0.02) } $

$ f_{c,11}^{TE} \approx 4.39\times10^9\,\text{Hz} $

Therefore:

$ \boxed{ f_{c,11}^{TE}\approx4.39\,\text{GHz} } $

2. Cutoff Frequency of the \(TM_{01}\) Mode

For a TM mode in a circular waveguide, the cutoff frequency is:

$ f_c = \frac{P_{nm}} {2\pi a\sqrt{\mu\epsilon}} $

For the \(TM_{01}\) mode, the first root of the Bessel function is given as:

$ P_{01}=2.4049 $

Therefore:

$ f_{c,01}^{TM} = \frac{2.4049} {2\pi a\sqrt{\mu_0\epsilon_0}} $

Substituting \(a=0.02\,\text{m}\):

$ f_{c,01}^{TM} = \frac{2.4049} { 2\pi(0.02) \sqrt{ (4\pi\times10^{-7}) (8.854\times10^{-12}) } } $

Using:

$ \frac{1}{\sqrt{\mu_0\epsilon_0}}=c $

we obtain:

$ f_{c,01}^{TM} = \frac{2.4049c}{2\pi a} $

Substituting the numerical values:

$ f_{c,01}^{TM} = \frac{ 2.4049(3\times10^8) }{ 2\pi(0.02) } $

$ f_{c,01}^{TM} \approx 5.74\times10^9\,\text{Hz} $

Therefore:

$ \boxed{ f_{c,01}^{TM}\approx5.74\,\text{GHz} } $

Comparison of the Two Modes

The calculated cutoff frequencies are:

$ f_{c,11}^{TE}\approx4.39\,\text{GHz} $

$ f_{c,01}^{TM}\approx5.74\,\text{GHz} $

Since:

$ 4.39\,\text{GHz}<5.74\,\text{GHz} $

the \(TE_{11}\) mode has the lower cutoff frequency. Therefore, it is the dominant mode of the circular waveguide, while \(TM_{01}\) is the lowest-cutoff TM mode.

$ \boxed{ TE_{11}\text{ is the dominant mode} } $

$ \boxed{ TM_{01}\text{ is the dominant TM mode} } $