I want to estimate the SNR of a bipolar signal. I know that each bit increases the SNR at 6 dB. Do i have to subtract one bit for sign, so that for bipolar signals the SNR is (n-1) x 6 dB ?


Your question appears to be about the signal to quantization noise ratio for uniform quantization. You have to count each bit, but that generally doesn't mean that the SNR is $6N$ dB, where $N$ is the number of bits. You normally have a constant offset:

$$\text{SNR}_{\text{dB}}\approx c+6N\tag{1}$$

where the constant $c$ depends on the Crest factor of the signal.

To give you an example, assume that the signal is in the range $[-x_p,x_p]$, and assume uniform quantization with $N$ bits. Then the quantization step size is


If we further assume that the quantization error is uniformly distributed in the interval $[-q/2,q/2]$, the quantization noise power is


With $x_{\text{RMS}}$ the RMS value of the signal, the signal to quantization noise ratio is given by


where $C_x$ is the signal's Crest factor. From $(4)$, the SNR in dB is

$$\text{SNR}_{\text{dB}}=10\log_{10}\left(\frac{3}{C^2_x}\right)+20N\log_{10}2=10\log_{10}\left(\frac{3}{C^2_x}\right)+6.02\cdot N\tag{5}$$

which has the form of Eq. $(1)$.


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