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: Note what decides the cutoff for TE10.
In a rectangular guide, the TE10 mode is the one with the lowest cutoff, and its cutoff frequency uses only the wider side $a$ of the cross section, not the narrower side $b$. The formula is $f_c=\dfrac{c}{2a}$.

Step 2: Pick out the wider side.
The cross section is $0.28$ inch by $0.14$ inch, so $a=0.28$ inch is the wide side.

Step 3: Change inches to metres.
One inch is $0.0254$ m, so
\[ a=0.28\times0.0254=7.112\times10^{-3}\text{ m} \]

Step 4: Plug numbers into the formula.
Taking $c=3\times10^{8}$ m/s,
\[ f_c=\frac{3\times10^{8}}{2\times7.112\times10^{-3}}=\frac{3\times10^{8}}{1.4224\times10^{-2}} \]

Step 5: Simplify the division.
\[ f_c\approx2.109\times10^{10}\text{ Hz}=21.09\text{ GHz} \]
\[ \boxed{21.09\text{ GHz}} \]