8. A received signal of -35 dBm exists in a channel with a total noise power of 52 dBrn. What is the signal-to-noise (S/N) ratio in decibels?
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The formula to convert dBm to watts is: P (watts) = 10^(P (dBm)/10) Given: Received signal power = -35 dBm Noise power = 52 dBm Converting the received signal power: P_signal (watts) = 10^(-35/10) Converting the noise power: P_noise (watts) = 10^(52/10) Show more…
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