Answer to Question #222528 in Electrical Engineering for Johny

Question

Assume a rectangular waveguide with dimensions a=2 cm and b=1cm. The
waveguide is filled with air and the operating frequency is 50MHz. Find the
propagation constant, guide wavelength and cut off frequency for TE10,
TM11 and TE20 modes.

Answer 1

a)Propagation constant is given by the formula below

$\gamma=\sqrt{({\dfrac{m\pi}{a}})^2+({\dfrac{n\pi}{b}})^2-\omega^2\mu\epsilon}$

a=0.02m,b=0.01m

For TE10 mode, m=1,n=0

f=50MHz

$\gamma=\sqrt{({\dfrac{1\pi}{0.02}})^2+({\dfrac{0\pi}{0.01}})^2-\dfrac{(2\pi\times{50\times{10^6}})^2}{(3\times{10^8})^2}}\\ \gamma=157.1$ For TM11 mode

m=1,n=1

$\gamma=\sqrt{({\dfrac{1\pi}{0.02}})^2+({\dfrac{1\pi}{0.01}})^2-\dfrac{(2\pi\times{50\times{10^6}})^2}{(3\times{10^8})^2}}\\ \gamma=351.2$ For TE20 mode, m=2, n=0

$\gamma=\sqrt{({\dfrac{2\pi}{0.02}})^2+({\dfrac{0\pi}{0.01}})^2-\dfrac{(2\pi\times{50\times{10^6}})^2}{(3\times{10^8})^2}}\\ \gamma=314.2$ The guide wavelength cannot be calculated without cutoff frequency.

$f_c=\dfrac{u'}{2}\sqrt{(\dfrac{m}{a})^2+(\dfrac{n}{b})^2}$

For TE10 mode,m=1,n=0

$f_c=\dfrac{3\times{10^8}}{2}\sqrt{(\dfrac{1}{0.02})^2+(\dfrac{0}{0.01})^2}=7.5GHz$

for TM11 mode,m=n=1

$f_c=\dfrac{3\times{10^8}}{2}\sqrt{(\dfrac{1}{0.02})^2+(\dfrac{1}{0.01})^2}=16.8GHz$for TE20 mode, m=2, n=0

$f_c=\dfrac{3\times{10^8}}{2}\sqrt{(\dfrac{2}{0.02})^2+(\dfrac{0}{0.01})^2}=15MHz$

guide wave is given by

$\lambda_g=\dfrac{c}{\sqrt{f^2-f_c^2}}$

Guide wavelength exist for only TE20 mode

$\lambda_g=\dfrac{3\times10^8}{\sqrt{(50\times{10^6})^2-(15\times{10^6})^2}}=6.29m$