Which expression represents the free-space path loss Lp in decibels for a radio link at distance R with wavelength λ?

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Master the AN/PRC-160 and AN/PRC-163 Radio Operations Test. Utilize flashcards and multiple choice questions, each with hints and explanations. Ace your exam!

Multiple Choice

Which expression represents the free-space path loss Lp in decibels for a radio link at distance R with wavelength λ?

Explanation:
In free space, the signal spreads over the surface of a sphere, so the received power falls off with the square of the distance and with the wavelength. From the Friis transmission equation, Pr = Pt Gt Gr (λ/(4πR))^2. The path loss in decibels is Lp(dB) = 10 log10(Pt/Pr). With unity gains (isotropic antennas), this becomes Lp(dB) = 10 log10((4πR/λ)^2) = 20 log10(4πR/λ). So the correct expression shows path loss grows as R increases and as λ decreases. The other forms either invert the ratio, use a 10 instead of 20, or multiply by λ inside the log, which would imply the wrong dependence on distance and wavelength.

In free space, the signal spreads over the surface of a sphere, so the received power falls off with the square of the distance and with the wavelength. From the Friis transmission equation, Pr = Pt Gt Gr (λ/(4πR))^2. The path loss in decibels is Lp(dB) = 10 log10(Pt/Pr). With unity gains (isotropic antennas), this becomes Lp(dB) = 10 log10((4πR/λ)^2) = 20 log10(4πR/λ). So the correct expression shows path loss grows as R increases and as λ decreases.

The other forms either invert the ratio, use a 10 instead of 20, or multiply by λ inside the log, which would imply the wrong dependence on distance and wavelength.

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