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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.

 

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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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