I need help in calculating the SNR of a PCM signal-

I came across a formule for calculating the SNR of PCM signal, considering the Quantization error as the source of noise- the formulae is:

SNR= (3*P*(2^(2N))) / (Mmax^2)
where P= modulating signal power
N= number of bits used in encoding
Mmax= the maximum amplitude of the quantized signal

But in certain problems, all the informations required to get the SNR is not available. Please help me in solving this problem:
Find the OUTPUT S/N of a PCM sampled at 8000 samples/second and using a 6 bits/word for transmission

 

Find the OUTPUT S/N of a PCM sampled at 8000 samples/second and using a 6 bits/word for transmission

Theoretical SNR is 36dB for a 6 bit sample depth. Simply add or subtract 6dB for each bit difference .
So 16 bit SNR =96dB. 8 bit signal SNR = 48dB, and so on.
Note these are theoretical limits. In practice the effective amount of SNR will depend on how much distortion you can put up with. Small signal inputs will be less well resolved than large ones, without some 'help'.

Which is why the effective SNR of low resolution sample depths, like the 8 bit used in telephony, can be greatly improved by the application of μ-law or A-law algorithms.
Analogue compression/expansion techniques can also help make significant perceived improvements in SNR of PCM siganls. It all depends on your application.

 

hello rogs,
Thanks for your reply,

But to just confirm whether i have understood your explanations correctly, The constant of 6dB against each sample bits, is it obtained as follows:

According to the formulae:
SNR= (3*P*(2^(2N))) / (Mmax^2).

Let us assume the modulating signal to be a sinusoidal AmSinWT
Hence Power P= (Am^2)/2
Mmax= Am
Hence SNR= 1.5 * (2^2N).

Converting into dB
SNRdb= 10 log1.5 +10* 2Nlog2= 1.76+6.02*N

Hence according to your explanation, if 1.76 is neglected
SNRdb= 6*N (approx 6.02 to 6)
But will this calculation be valid for any modulating signal??

Thanks in advance.