For a 2 Vpp signal that fits within the ADC’s input range, find the SQNR for: a) an 8-bit ADC with an input range of 1V to 6V b) a 9-bit ADC with an input range of 0V to 4V

SNR of ADC

If No of bit = n

So,

SNR = 20 log × rms Signal / rms Noise  

SNR = 6.02n + 1.75 db

SNR only depends on No pf bit

A)No of bit (n) = 8

SNR = 6.02 × 8 + 1.75

SNR = 49.92db  

Or

SNR = 50db

B)No of bit (n) = 9

SNR = 6.02 × 9 + 1.75

SNR = 55.94db

Or

SNR = 56db

If we increase a bit by one than SNR increase by 6db

If n(↑) by 1 bit SNR increase by 6db

it was a fall down when it happens

explanation:

answer:

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~batmans wife dun dun