Question

A hallow rectangular waveguide has its axis along z-direction. Its internal dimensions in x and y directions are a = 10 cm and b = 7 cm respectively. the amplitude of the electric field intesity in x -direction $E_x=500V/m$ and operating frequency is $30\%$ higher than the cut off frequency, then the power (in W) transmitted in $T{E}_{11}$ mode is _______ .

• A. 0.37 W
• B. 1.1 W
• C. 23 mW
• D. 552 mW

Answer 1

The correct option is D 552 mW

For $T{E}_{\text{mn}}$ mode, power transmitted

$P_{\text{avg}}=\frac{1}{8}\frac{\text{ab}|E{|}^2}{{\eta }_{TE}}\text{where}|E{|}^2=E_{x_0}^2+E_{y_0}^2\text{and}\frac{E_{y_0}}{E_{x_0}}=\frac{mb}{na}$

In TE ${}_{11}$ mode:

$E_{y_0}=\frac{b}{a}{E}_{x_0}=\frac{7}{10}\times 500=350Vm$

${\eta }_{T{E}_{11}}=\frac{{\eta }_0}{\sqrt{1-{\left(\frac{f_{C_{11}}}{f}\right)}^2}}\text{where}f=1.3f_{c_{11}}$

${\eta }_{T{E}_{11}}=\frac{120\pi }{\sqrt{1-{\left(\frac{f_{C_{11}}}{1.3f_{c_{11}}}\right)}^2}}=589.99\Omega$

$P_{av{g}_{T{E}_{11}}}=\frac{ab|E{|}^2}{8{\eta }_{TE}}=\frac{ab|E_{x_0}^2-E_{y_0}^2{|}^2}{8{\eta }_{T{E}_{11}}}$

$=\frac{10\times 7\times {10}^{-4}({500}^2+{350}^2)}{8\times 589.99}=0.552Watt.$