Answer to Question #251644 in Electric Circuits for RJBP

Question

Given the signal power of 2.3μw and the noise power of 0.7μw, calculate the signal-to-noise-ratio.

Answer 1

Solution;

Given;

Signal Power,Ps=$2.3\mu W$

Noise power,$P_n$ =$0.7\mu W$

Signal-to-noiss ratio as a logarithmic function is given as;

$SNR_{dB}=10log\frac{P_s}{P_n}$

$SNR_{dB}=10\log\frac{2.3}{0.7}$

$SNR_{dB}=5.17dB$