A system used to process electronic signals can handle volta
Solution
Signal Vsignal = 5Volts
noise, Vnoise = 0.001 Volts
SNR = V/Vnoise = 5/0.001 = 5000
SNR in dB = 20 log10(Vsignal /Vnoise) = 20 log 5000
= 73.98 = 74 dB

Signal Vsignal = 5Volts
noise, Vnoise = 0.001 Volts
SNR = V/Vnoise = 5/0.001 = 5000
SNR in dB = 20 log10(Vsignal /Vnoise) = 20 log 5000
= 73.98 = 74 dB
