Quantization SNR of bipolar signal 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

$$q=\frac{2x_p}{2^N}\tag{2}$$

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

$$P_n=\frac{1}{q}\int_{-q/2}^{q/2}\xi^2d\xi=\frac{q^2}{12}=\frac13\frac{x_p^2}{2^{2N}}\tag{3}$$

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

$$\text{SNR}=\frac{x^2_{\text{RMS}}}{P_n}=3\frac{x^2_{\text{RMS}}}{x^2_p}\cdot2^{2N}=\frac{3}{C^2_x}\cdot2^{2N}\tag{4}$$

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. $(1)$.