Dominant-mode cutoff of a rectangular waveguide
BlinkNBuild practice problem · GATE standard, authored and verified in-house
An air-filled rectangular waveguide has internal dimensions $a = 2.286\ \text{cm}$ and $b = 1.016\ \text{cm}$ (standard WR-90).
The cutoff frequency of the dominant mode, in GHz, is ________.
Show the step-by-step derivation
Answer
6.56 GHz (accept 6.5 to 6.6) GHz (accept 6.5 to 6.6)
Step-by-step derivation
- The general cutoff frequency for the $\text{TE}_{mn}$ mode in a rectangular guide is $$f_{c,mn} = \frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}.$$
- The dominant mode is the one with the lowest cutoff. Since $a > b$, the smallest non-trivial choice is $m = 1$, $n = 0$, giving $\text{TE}_{10}$.
- With $n = 0$ the second term vanishes and the expression collapses to $$f_{c,10} = \frac{c}{2a}.$$ Notice that $b$ does not appear at all - it sets power handling and the cutoff of higher modes, not this one.
- Substitute $c = 3 \times 10^{8}\ \text{m/s}$ and $a = 2.286\ \text{cm} = 0.02286\ \text{m}$: $$f_{c,10} = \frac{3 \times 10^{8}}{2 \times 0.02286} = \frac{3 \times 10^{8}}{0.04572}.$$
- Evaluate: $$f_{c,10} = 6.562 \times 10^{9}\ \text{Hz} = \mathbf{6.56\ GHz}.$$
- Why WR-90 is used at X-band. The next mode up, $\text{TE}_{20}$, cuts off at $2f_{c,10} = 13.1$ GHz. Between those two frequencies only $\text{TE}_{10}$ propagates, and the usable single-mode band is quoted as $8.2$ to $12.4$ GHz - inside that window with margin at both ends.
The trap this question is built around
Including $b$ in the calculation. For $\text{TE}_{10}$ the $n/b$ term is zero, so the broad wall alone sets the cutoff. Using both dimensions gives the $\text{TE}_{11}$ cutoff, which is a different and much higher mode.