Question 4.5.2: A rectangular hollow metal waveguide has dimensions a = 2.29...

A rectangular hollow metal waveguide has dimensions a = 2.29 cm and b = 1.02 cm. Microwave power at 10 GHz is transmitted through the waveguide in the TE _{10} mode.

(a) Calculate the cut-off wavelength and the guide wavelength for this mode.
(b) What are the other (TE or TM) modes that can propagate through the waveguide?
(c) If
a = b = 2.29 cm, what are the modes which can propagate through the waveguide?

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a = 2.29, b = 1.02 cm, c = 10 GHz
(a)
λ = 20 = 4.58 cm

\begin{aligned}&\lambda=\frac{c}{f}=\frac{3 \times 10^{8}}{10 \times 10^{9}}=3 cm \\&f_{c}=\frac{V_{0}}{\lambda_{C}}=6.55 GHz \\&\bar{\lambda}=\frac{\lambda}{\sqrt{1-\left(\frac{f_{c}}{f}\right)^{2}}}=3.97 cm\end{aligned}

(b) for TE _{10}, f_{c_{1}}=\frac{V_{0}}{2 a}=6.55 GHz

for TE _{01}, f_{c_{2}}=\frac{V_{0}}{2 b}=14.7 GHz,

f c_{2}>f \text { so only } TE _{10} mode can propagate through the given rectangular waveguide.

(c) a = b = 2.29 cm
f = 10 GHz

for TE _{10}, f_{c_{10}}=\frac{V_{0}}{2 a}=\frac{3 \times 10^{10}}{2 \times 2.29}=6.55 GHz

for TE _{01} f_{c_{01}}=\frac{V_{0}}{2 b}=\frac{3 \times 10^{8}}{2 \times 2.29}=6.55 GHz

for TE _{11} \text { or } TM _{11}

 

f_{C_{11}}=\frac{V_{0}}{2 a} \sqrt{1^{2}+1^{2}}=\frac{V_{0}}{\sqrt{2 a}}=9.226 GHz

for TE _{20} \text { and } TE _{02}

 

f_{C_{20}}=f_{C_{02}}=\frac{V_{0}}{a}=\frac{3 \times 10^{10}}{2.29}=13.1 GHz

So, only  TE _{10}, TE _{01}, TE _{11} \text { and } TM _{11} are the possible mode that propagates through the rectangle wave guide.

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