Question #221891

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

Expert's answer

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}}\\\\\n\\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}}\\\\\n\\gamma=351.2" For TE20, 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}}\\\\\n\\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"

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