A short length of cable is a good antenna for frequencies that match the frequen

Question

A short length of cable is a good antenna for frequencies that match the frequency of the fundamental mode of the cable. What should the length of an (air-filled) coaxial cable that terminates in a short circuit circuit at both ends if the cable is to be an antenna for a wave of frequency 88 MHz? What should the length of an (air-filled) coaxial cable that terminates in a short circuit circuit at both ends if the cable is to be an antenna for a wave of frequency 108 MHz? What should the length of an (air-filled) coaxial cable that has one open-circuited end and one close- circuited end if the cable is to be an antenna for a wave of frequency 88 MHz? What should the length of an (air-filled) coaxial cable that has one open-circuited end and one close- circuited end if the cable is to be an antenna for a wave of frequency 108 MHz?

Explanation / Answer

Here ,

for the wavelengths ,

wavelength = 2 * L

hence , L = wavelength/2

L is the length of the cable

A)

for frequency = 88 MHz

L = (3 *10^8/(88 *10^6))/2

L = 1.705 m

B)

for frequency = 108 MHz

L = (3 *10^8/(108 *10^6))/2

L = 1.38 m

C)

for open circuit and close circuit

wavelength = 4 *L

L = wavelength/4

for frequency = 88 MHz

L = (3 *10^8/(88 *10^6))/4

L = 0.852 m

D)

for frequency = 108 MHz

L = (3 *10^8/(108 *10^6))/4

L = 0.694 m

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