Question: The input to a communications receiver is r(t) = 5 sin(20t + 7 /4 ) + (t) + n(t) where i(t) = 0.2 cos (60t)
The input to a communications receiver is
r(t) = 5 sin(20πt + 7 /4 π) + ¡(t) + n(t)
where
i(t) = 0.2 cos (60πt)
and n(t) is noise having standard deviation σn = 0.1. The transmitted signal is 10 cos(20πt). Determine the SNR at the receiver input and the delay from the transmitted signal to the receiver input.
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We can write so that the receiver input is The signal power is P T 5 2 2 125 and ... View full answer
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