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# Answer to Question #221891 in Electrical Engineering for Ubaidullah

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

1
2021-08-03T04:07:01-0400

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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