Question

The cutoff frequency (in GHz) for the dominant \(TE_{10}\) mode of an air-filled rectangular waveguide of inner dimension \(0.28\) inch \(\times\) \(0.14\) inch is .
(rounded off to two decimal places)

Hint:

For TE10 mode the cutoff frequency depends only on the broad wall dimension a, through fc = c/(2a).

Answer 1

Correct Answer: 21.09

Step 1: Recall the cutoff frequency formula for TE10 mode.
For a rectangular waveguide with broad wall dimension \(a\) and narrow wall dimension \(b\), the cutoff frequency of the dominant \(TE_{10}\) mode depends only on the broad dimension \(a\). It is given by
\[ f_c=\frac{c}{2a} \]
where \(c\) is the speed of light in free space, taken as \(3\times10^{8}\) m/s.

Step 2: Identify the broad dimension.
The waveguide has inner dimensions \(0.28\) inch \(\times\) \(0.14\) inch. The broad wall dimension is the larger one, so
\[ a=0.28\text{ inch} \]

Step 3: Convert the dimension to metres.
Since \(1\) inch \(=0.0254\) m,
\[ a=0.28\times0.0254=0.007112\text{ m} \]

Step 4: Substitute into the cutoff frequency formula.
\[ f_c=\frac{3\times10^{8}}{2\times0.007112}=\frac{3\times10^{8}}{0.014224} \]

Step 5: Evaluate the expression.
\[ f_c\approx2.1091\times10^{10}\text{ Hz}=21.09\text{ GHz} \]

Step 6: Final answer.
The cutoff frequency for the dominant \(TE_{10}\) mode is \[ \boxed{21.09\text{ GHz}} \]