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 which 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 which has one open-circuited end anti one close-circuited end if the cable is to be an antenna for a wave of frequency 108 MHz? What is the complex reflection coefficient at a load circuit if the load circuit is driven at the resonaut frequency of the circuit and the char

Explanation / Answer

A) Consider the values of permeability and permittivity has no change in this medium so the velocity of the wave considered to be as 'c'.

frequency = 88MHz = 88*106 Hz

wavelength = (3*108 )/(88*106 ) = 3.409 m

Length of a closed column for fundamental mode of vibration L = 2 * wavelength   { we know that wavelength = L/2 }

   L = 2 * 3.409 = 6.818 m

B) Similarly for frequency = 108MHz

     Wavelength = c / frequency = 2.778 m

     Length of the coaxial cable = 2 * 2.778 = 5.556 m

C) For one side open cable, fundamental mode wavelength = L / 4

    Length = wavelength * 4

    L = 3.409 * 4 = 13.636 m

D) L = 2.778 * 4 = 11.112 m

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