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For a rectangular waveguide of dimensions: a √3 × a (in cm) , if cut-off frequency of TE10 mode is 2GHz. What is the cut-off frequency for TM11 in waveguide?
- a)1 GHz
- b)3.46 GHz
- c)4.0 GHz
- d)6.0 GHz
Correct answer is option 'C'. Can you explain this answer?
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For a rectangular waveguide of dimensions: a √3 × a (in cm)...
Solution:
Given, dimensions of rectangular waveguide: a = 3a (in cm)
Cut-off frequency of TE10 mode is 2 GHz.
We know that the cut-off frequency for any mode in waveguide is given by:
$$f_{co} = \frac{1}{2\pi\sqrt{\epsilon_r}}\cdot\frac{c}{\sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}}$$
where, m and n are the mode numbers, a and b are the dimensions of the waveguide, c is the speed of light and εr is the relative permittivity of the waveguide material.
For TE10 mode, m = 1 and n = 0.
So, the cut-off frequency for TE10 mode is given by:
$$f_{co} = \frac{1}{2\pi\sqrt{\epsilon_r}}\cdot\frac{c}{\frac{m}{a}}$$
$$2\text{ GHz} = \frac{1}{2\pi\sqrt{\epsilon_r}}\cdot\frac{3\times 10^8\text{ m/s}}{\frac{1}{3a}}$$
Solving this equation, we get:
$$\epsilon_r = \frac{9}{4\pi^2\times 2^2}$$
For TM11 mode, m = 1 and n = 1.
So, the cut-off frequency for TM11 mode is given by:
$$f_{co} = \frac{1}{2\pi\sqrt{\epsilon_r}}\cdot\frac{c}{\sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}}$$
$$f_{co} = \frac{1}{2\pi\sqrt{\epsilon_r}}\cdot\frac{3\times 10^8\text{ m/s}}{\sqrt{\left(\frac{1}{a}\right)^2 + \left(\frac{1}{3a}\right)^2}}$$
Solving this equation using the value of εr obtained earlier, we get:
$$f_{co} = 4.0\text{ GHz}$$
Therefore, the cut-off frequency for TM11 mode in the given rectangular waveguide is 4.0 GHz.
Given, dimensions of rectangular waveguide: a = 3a (in cm)
Cut-off frequency of TE10 mode is 2 GHz.
We know that the cut-off frequency for any mode in waveguide is given by:
$$f_{co} = \frac{1}{2\pi\sqrt{\epsilon_r}}\cdot\frac{c}{\sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}}$$
where, m and n are the mode numbers, a and b are the dimensions of the waveguide, c is the speed of light and εr is the relative permittivity of the waveguide material.
For TE10 mode, m = 1 and n = 0.
So, the cut-off frequency for TE10 mode is given by:
$$f_{co} = \frac{1}{2\pi\sqrt{\epsilon_r}}\cdot\frac{c}{\frac{m}{a}}$$
$$2\text{ GHz} = \frac{1}{2\pi\sqrt{\epsilon_r}}\cdot\frac{3\times 10^8\text{ m/s}}{\frac{1}{3a}}$$
Solving this equation, we get:
$$\epsilon_r = \frac{9}{4\pi^2\times 2^2}$$
For TM11 mode, m = 1 and n = 1.
So, the cut-off frequency for TM11 mode is given by:
$$f_{co} = \frac{1}{2\pi\sqrt{\epsilon_r}}\cdot\frac{c}{\sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}}$$
$$f_{co} = \frac{1}{2\pi\sqrt{\epsilon_r}}\cdot\frac{3\times 10^8\text{ m/s}}{\sqrt{\left(\frac{1}{a}\right)^2 + \left(\frac{1}{3a}\right)^2}}$$
Solving this equation using the value of εr obtained earlier, we get:
$$f_{co} = 4.0\text{ GHz}$$
Therefore, the cut-off frequency for TM11 mode in the given rectangular waveguide is 4.0 GHz.
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For a rectangular waveguide of dimensions: a √3 × a (in cm) , if cut-off frequency of TE10 mode is 2GHz. What is the cut-off frequency for TM11 in waveguide?a)1 GHzb)3.46 GHzc)4.0 GHzd)6.0 GHzCorrect answer is option 'C'. Can you explain this answer?
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