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# Signal to Quantization noise ratio derivation

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Let $m(t)$= $A_m$ cos 2$\pi$ $f_mt$ be the signal which is undergoing quantization.

Then for this signal, the  power will be  $A_m^{2}$/2

i.e $P_s$ =$A_m^{2}/2$

Suppose if we are quantizing a signal with n-bits , then the step size(i.e difference between quantization steps) will be

$q$=  $\frac {V_{max}-V_{min}}{2^{n}}$$\frac {2A_m}{2^{n}}$

The maximum quantization error can be q/2. So it is usually assumed that quantization noise is uniformly distributed from -q/2 to q/2. Hence the probability density function is a rectangle with height 1/q and extends it’s base from -q/2 to +q/2. Note that the area under PDF curve should be equal to 1.

Quantization noise power is calculated as follows

$P_q$ = $\frac{1}{q}\int_{-q/2}^{q/2}x^2dx$ = $\frac{q^2}{12}$

Signal to quantization noise ratio is a ratio of signal power $P_s$ to the quantization noise power $P_q$

SQNR= $\frac{P_s}{P_q}$ =1.5 * $2^{2n}$

$SQNR_{dB}$= 10 log {1.5 * $2^{2n}$}

$SQNR_{dB}$ =1.76 +6.02n

For any doubts regarding derivation, please use comment section.

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