We have an amplifier with a 1 MHz bandwidth and a 20 micro-V input signal. The i

Question

We have an amplifier with a 1 MHz bandwidth and a 20 micro-V input signal. The input noise = 1 micro-V, output signal = 6.0 mV, and output noise = 1 mV. Find (a) SNR at the input Pin = Pn(in)= K(T)BW= 1.38*10-23(290)1MHz= 4*10-15W (b) SNR at the output (c) Noise figure NF= 10 log(400/36)= 10.458dB I know the answers are (a) 400 (b) 36 I am having a hard time finding the input power (Pin) to calculate the SNR for part a. Once I have that the rest is easy. Included is the work I have done on it. Any help is greatly appreciated!!

Explanation / Answer

a) SNR = P(signal) / P(noise) = [V(signal) / V(noise) ] ^2 = (20/1)^2 = 400

b) similarly at output SNR = (6/1)^2 = 36

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