Numerical 4: Circular Waveguide Numerical Problem

A circular waveguide operates at a frequency of $11\,\text{GHz}$. The internal diameter of the waveguide is $4.5\,\text{cm}$. Calculate the cutoff frequency, cutoff wavelength, guide wavelength, group velocity, phase velocity, and characteristic impedance for the $TE_{10}$ mode.

Note: $TE_{10}$ is a rectangular-waveguide mode designation. For a circular waveguide, the corresponding dominant mode is $TE_{11}$. Therefore, the calculation below uses the circular-waveguide dominant mode $TE_{11}$.

Given Data

$ f=11\,\text{GHz} $

$ d=4.5\,\text{cm} $

The radius of the circular waveguide is:

$ a=\frac{d}{2} $

$ a=\frac{4.5}{2}=2.25\,\text{cm} $

$ a=2.25\times10^{-2}\,\text{m} $

For the dominant $TE_{11}$ mode, the first root of the derivative of the Bessel function is:

$ P'_{11}=1.8412 $

1. Cutoff Frequency

For a circular waveguide, the cutoff frequency of a TE mode is:

$ f_c= \frac{P'_{nm}c}{2\pi a} $

For the $TE_{11}$ mode:

$ f_c= \frac{1.8412c}{2\pi a} $

$ f_c= \frac{1.8412(3\times10^8)} {2\pi(2.25\times10^{-2})} $

$ \boxed{ f_c\approx3.91\,\text{GHz} } $

Since the operating frequency is:

$ 11\,\text{GHz}>3.91\,\text{GHz} $

the waveguide operates above cutoff and the signal can propagate.

2. Cutoff Wavelength

The cutoff wavelength is related to the cutoff frequency by:

$ \lambda_c=\frac{c}{f_c} $

$ \lambda_c= \frac{3\times10^8} {3.91\times10^9} $

$ \boxed{ \lambda_c\approx0.0767\,\text{m} } $

$ \boxed{ \lambda_c\approx7.67\,\text{cm} } $

3. Free-Space Wavelength

The wavelength corresponding to the operating frequency is:

$ \lambda_0=\frac{c}{f} $

$ \lambda_0= \frac{3\times10^8} {11\times10^9} $

$ \boxed{ \lambda_0\approx0.02727\,\text{m} } $

$ \boxed{ \lambda_0\approx2.727\,\text{cm} } $

4. Guide Wavelength

For a waveguide operating above cutoff, the guide wavelength is:

$ \lambda_g= \frac{\lambda_0} {\sqrt{1-\left(\frac{f_c}{f}\right)^2}} $

$ \lambda_g= \frac{0.02727} {\sqrt{ 1-\left(\frac{3.91}{11}\right)^2 }} $

$ \boxed{ \lambda_g\approx0.0292\,\text{m} } $

$ \boxed{ \lambda_g\approx2.92\,\text{cm} } $

5. Phase Velocity

The phase velocity in a waveguide is:

$ V_p= \frac{c} {\sqrt{1-\left(\frac{f_c}{f}\right)^2}} $

$ V_p= \frac{3\times10^8} {\sqrt{ 1-\left(\frac{3.91}{11}\right)^2 }} $

$ \boxed{ V_p\approx3.21\times10^8\,\text{m/s} } $

6. Group Velocity

The group velocity is:

$ V_g= c\sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } $

$ V_g= 3\times10^8 \sqrt{ 1-\left(\frac{3.91}{11}\right)^2 } $

$ \boxed{ V_g\approx2.80\times10^8\,\text{m/s} } $

The phase and group velocities satisfy:

$ V_pV_g=c^2 $

7. Characteristic Impedance

For a TE mode, the characteristic wave impedance is:

$ Z_{TE} = \frac{\eta} {\sqrt{1-\left(\frac{f_c}{f}\right)^2}} $

For an air-filled waveguide:

$ \eta=\sqrt{\frac{\mu_0}{\epsilon_0}}\approx377\,\Omega $

Therefore:

$ Z_{TE} = \frac{377} {\sqrt{ 1-\left(\frac{3.91}{11}\right)^2 }} $

$ \boxed{ Z_{TE}\approx404\,\Omega } $

Final Answers

Cutoff frequency: $f_c\approx3.91\,\text{GHz}$
Cutoff wavelength: $\lambda_c\approx7.67\,\text{cm}$
Free-space wavelength: $\lambda_0\approx2.727\,\text{cm}$
Guide wavelength: $\lambda_g\approx2.92\,\text{cm}$
Phase velocity: $V_p\approx3.21\times10^8\,\text{m/s}$
Group velocity: $V_g\approx2.80\times10^8\,\text{m/s}$
Characteristic impedance: $Z_{TE}\approx404\,\Omega$

Since $11\,\text{GHz}$ is greater than the cutoff frequency of $3.91\,\text{GHz}$, the signal propagates through the circular waveguide in the dominant $TE_{11}$ mode.