A TE11 wave is propagating through a circular 
waveguide. The diameter of the guide is 10 cm and the 
guide is air-filled. Given X11=1.842. (Nov/Dec 2016) 
Find the cut-off frequency. Find the wavelength λg in 
the guide for a frequency of 3 GHz

To solve this problem, we'll break it down into two parts: first, calculating the cut-off frequency, and then finding the wavelength in the guide for the given frequency of 3 GHz.

### Part 1: Cut-off Frequency

The cut-off frequency \(f_c\) for a TE\(_{mn}\) mode in a circular waveguide is given by the formula:

\[
f_c = \frac{X_{mn}}{2 \pi a \sqrt{\mu \varepsilon}}
\]

Where:
- \(X_{mn}\) is the m-th root of the Bessel function (given as 1.842 for TE\(_{11}\)),
- \(a\) is the radius of the waveguide,
- \(\mu\) is the permeability of the medium (for air, \(\mu \approx \mu_0 = 4 \pi \times 10^{-7} \, \text{H/m}\)),
- \(\varepsilon\) is the permittivity of the medium (for air, \(\varepsilon \approx \varepsilon_0 = 8.854 \times 10^{-12} \, \text{F/m}\)).

First, let's calculate the radius of the waveguide:
- The diameter is given as 10 cm, so the radius is \(a = 5 \, \text{cm} = 0.05 \, \text{m}\).

Now, we can substitute the values into the formula for \(f_c\):

\[
f_c = \frac{1.842}{2 \pi (0.05) \sqrt{(4 \pi \times 10^{-7})(8.854 \times 10^{-12})}}
\]

Simplifying:

\[
f_c = \frac{1.842}{2 \pi (0.05) \times 3 \times 10^8} = \frac{1.842 \times 3 \times 10^8}{2 \pi \times 0.05}
\]

\[
f_c \approx \frac{1.842 \times 3 \times 10^8}{0.314} \approx 1.76 \, \text{GHz}
\]

Thus, the **cut-off frequency** \(f_c \approx 1.76 \, \text{GHz}\).

### Part 2: Wavelength in the Guide at 3 GHz

The wavelength in the waveguide (\(\lambda_g\)) is related to the frequency and the cut-off frequency by the following formula:

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

Where:
- \(\lambda_0 = \frac{c}{f}\) is the free-space wavelength,
- \(c = 3 \times 10^8 \, \text{m/s}\) is the speed of light,
- \(f\) is the operating frequency (given as 3 GHz).

First, calculate the free-space wavelength \(\lambda_0\):

\[
\lambda_0 = \frac{3 \times 10^8}{3 \times 10^9} = 0.1 \, \text{m}
\]

Now, substitute the values into the formula for \(\lambda_g\):

\[
\lambda_g = \frac{0.1}{\sqrt{1 - \left(\frac{1.76}{3}\right)^2}} = \frac{0.1}{\sqrt{1 - 0.3456}} = \frac{0.1}{\sqrt{0.6544}}
\]

\[
\lambda_g \approx \frac{0.1}{0.8089} \approx 0.1236 \, \text{m}
\]

Thus, the **wavelength in the guide** \(\lambda_g \approx 0.1236 \, \text{m} = 12.36 \, \text{cm}\).

### Final Answers:
- The **cut-off frequency** is approximately **1.76 GHz**.
- The **wavelength in the guide** at 3 GHz is approximately **12.36 cm**.

Would you like further details on any step or clarification?

Here are 5 related questions:
1. What happens if the operating frequency is below the cut-off frequency?
2. How does the mode (TE\(_{mn}\)) affect the cut-off frequency in different waveguides?
3. What is the significance of the Bessel function root \(X_{mn}\) in waveguide analysis?
4. How would the cut-off frequency change if the waveguide were filled with a dielectric material?
5. What is the group velocity of the wave in the waveguide at 3 GHz?

**Tip**: Always check whether the operating frequency is above the cut-off frequency to ensure wave propagation in a waveguide.

A TE11 wave is propagating through a circular waveguide. The diameter of the guide is 10 cm and the guide is air-filled. Given X11=1.842. Find the cut-off frequency. Find the wavelength λg in the guide for a frequency of 3 GHz.

Learn how to calculate the cut-off frequency and wavelength in a circular waveguide for a TE11 wave. This detailed guide covers electromagnetic wave theory and wave propagation, with step-by-step solutions for the cut-off frequency and wavelength in a 10 cm air-filled guide at 3 GHz. Ideal for undergraduate electrical engineering students.

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