Question

If in a rectangular waveguide for which \(a=2b\), the cutoff frequency for TE\(_{01}\) mode is 12 GHz, the cutoff frequency for TM\(_{11}\) is

• A. \(3\sqrt{5}\) GHz
• B. \(6\sqrt{5}\) GHz
• C. \(3\sqrt{12}\) GHz
• D. \(6\sqrt{12}\) GHz

Hint:

Cutoff frequency in rectangular waveguide: \(f_{c,mn} = \frac{c}{2\pi\sqrt{\mu\epsilon}} \sqrt{(\frac{m\pi}{a})^2 + (\frac{n\pi}{b})^2}\) (general form) or \(f_{c,mn} = \frac{c}{2} \sqrt{(\frac{m}{a})^2 + (\frac{n}{b})^2}\) if medium is vacuum/air.
For TE modes, \(m,n \ge 0\) (not both zero). For TM modes, \(m,n \ge 1\).
Use the given information for one mode to find a relationship between \(c, a, b\), then use it for the other mode.

Answer 1

The Correct Option is B

The cutoff frequency \(f_{c,mn}\) for TE\(_{mn}\) or TM\(_{mn}\) modes in a rectangular waveguide with dimensions \(a\) (width) and \(b\) (height) is given by: 

\( f_{c,mn} = \frac{c}{2} \sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2} \) where \(c\) is the speed of light, and \(m\) and \(n\) are mode indices (integers).

For TE modes, \(m, n\) can be 0, 1, 2, ... but not both \(m=0\) and \(n=0\) simultaneously. For TM modes, \(m, n\) must be 1, 2, 3, ... (neither \(m\) nor \(n\) can be zero).

Given: \(a=2b\). The cutoff frequency for TE\(_{01}\) mode is 12 GHz.

For TE\(_{01}\) mode, \(m=0\), \(n=1\). Using the formula for cutoff frequency:

For TE\(_{01}\) mode: \( f_{c,01} = \frac{c}{2} \sqrt{\left(\frac{0}{a}\right)^2 + \left(\frac{1}{b}\right)^2} = \frac{c}{2} \sqrt{\frac{1}{b^2}} = \frac{c}{2b} \)

We are given \(f_{c,01} = 12 \text{ GHz}\). So, \( \frac{c}{2b} = 12 \text{ GHz} \). This gives us a relation for \(c/b\): \( c/b = 24 \text{ GHz} \).

Now, we need to find the cutoff frequency for TM\(_{11}\) mode.

For TM\(_{11}\) mode, \(m=1\), \(n=1\). Using the formula:

For TM\(_{11}\) mode: \( f_{c,11} = \frac{c}{2} \sqrt{\left(\frac{1}{a}\right)^2 + \left(\frac{1}{b}\right)^2} \)

Substitute \(a = 2b\):

\( f_{c,11} = \frac{c}{2} \sqrt{\left(\frac{1}{2b}\right)^2 + \left(\frac{1}{b}\right)^2} = \frac{c}{2} \sqrt{\frac{1}{4b^2} + \frac{1}{b^2}} \)

Now simplifying:

\( f_{c,11} = \frac{c}{2} \sqrt{\frac{1+4}{4b^2}} = \frac{c}{2} \sqrt{\frac{5}{4b^2}} = \frac{c}{2} \frac{\sqrt{5}}{2b} = \frac{c\sqrt{5}}{4b} \)

We can write this as:

\( f_{c,11} = \left( \frac{c}{2b} \right) \frac{\sqrt{5}}{2} \)

We know \( \frac{c}{2b} = 12 \text{ GHz} \).

So:

\( f_{c,11} = (12 \text{ GHz}) \times \frac{\sqrt{5}}{2} = 6\sqrt{5} \text{ GHz} \)

This matches option (b).

The final answer is: \( \boxed{6\sqrt{5} \text{ GHz}} \)