7 How to find the SNR for sampling quantization?

Suppose we have a message \(m\left ( t\right ) \) that is sampled. Assume we have \(n\) bits to use for encoding the sample levels. Hence there are \(2^{n}\) levels of quantizations. We want to find the ration of the signal to the noise power. Noise here is generated due to quantization (i.e. due to the rounding off values of \(m\left ( t\right ) \) during sampling).

This is the algorithm:

Input: \(n\), the number of bits for encoding, \(m_{p}\) absolute maximum value of the message \(m\left ( t\right ) \), the pdf \(f_{X}\left ( t\right ) \) of the message \(m\left ( t\right ) \) is \(m\left ( t\right ) \) is random message or \(m\left ( t\right ) \) function if it is deterministic (such as \(\cos \left ( t\right ) \))

1. Find the quantization step size \(S=\frac {2m_{p}}{2^{2}}\)
2. Find \(P_{av}\) of the error is \(\frac {1}{12}S^{2}\) where \(S\) is the step size found in (1), hence \(P_{av}=\frac {1}{12}S^{2}=\frac {1}{12}\left ( \frac {2m_{p}}{2^{2}}\right ) ^{2}=\frac {m_{p}^{2}}{3\times 2^{2n}}\)
3. If \(m\left ( t\right ) \) is deterministic find \(p_{av}=\left \langle m^{2}\left ( t\right ) \right \rangle =\frac {1}{T}\int _{0}^{T}\left \vert m\left ( t\right ) \right \vert ^{2}dt\)
4. If \(m\left ( t\right ) \) is random, find \(p_{av}=E\left ( m\left ( t\right ) \right ) ={\int }m^{2}\left ( t\right ) f_{X}\left ( t\right ) dt\), this is called the second moment of the pdf
5. \(SNR=\frac {E\left ( m\left ( t\right ) \right ) }{\frac {m_{p}^{2}}{3\times 2^{2n}}}\)

Hence find \(SNR\) for noise quantisation comes down to finding the power in the message \(m\left ( t\right ) \).

Examples: For sinusoidal message \(m\left ( t\right ) \), \(SNR_{db}=6n+1.761\). For random \(m\left ( t\right ) \) with PDF which is uniform distributed \(SNR_{db}=6n\), for random \(m\left ( t\right ) \) which is AWGN. Do this later