A rectangular hollow metal wave guide of internal cross-section of 7.366 cm × 3.556 cm carries a 3 GHz signal in the TE _{10}-mode. Calculate the maximum power handling capability of the waveguide assuming the maximum permissible electric field inside the waveguide to be 30 kV/vm.
Question 4.5.8: A rectangular hollow metal wave guide of internal cross-sect...
\begin{aligned}f_{c} &=\frac{V}{2 a}=\frac{3 \times 10^{8} \times 10^{2}}{2 \times 7.366}=2.036 Ghz \\f_{c} &=\frac{2.036}{3}=0.679 \\\left(\frac{f_{c}}{f}\right)^{2} &=0.46 \\\eta &=\frac{\eta_{0}}{\sqrt{1-\left(\frac{f_{c}}{f}\right)^{2}}}=\frac{120 \pi}{\sqrt{1-0.46}}=513.42 \\P_{\max _{d}} &=\frac{E_{\max }^{2}}{2 \eta}=8.766 \times 10^{9} w / m ^{2} \\P_{\max } &=P_{\operatorname{maxd}} X \text { Area }=23 mw\end{aligned}
Related Answered Questions
For rectangular waveguide.
TE _{10} \text {...
Given, a = 5 cm
TE _{10}, f=30 GHz
...
Given, b = 3 cm, a = 6 cm, a = 2b, Lowest T...
Time delay =T=\frac{\bar{\beta}}{\omega} z...
a = 2.29, b = 1.02 cm, c = 10 GHz
(a) λ = 20 = 4.5...
Given that wave is propagating in +2 direction.
[l...
f_{c}=\frac{V_{o}}{2 \pi} \sqrt{\left(\frac...
(i) \lambda_{ g } \text { [guided wavelengt...