The dynamic range of a signal V is the ratio of the maximum to the minimum, expr

Question

The dynamic range of a signal V is the ratio of the maximum to the minimum, expressed in decibels. The dynamic range expected in a signal is to some extent an expression of the signal quality. It also dictates the number of bits per sample needed in order to reduce the quantization noise down to an acceptable level; e.g., we may like to reduce the noise to at least an order of magnitude below Vmin.

Suppose the dynamic range for a signal is 60 dB. Can we use 10 bits for this signal? Can we use 16 bits?

Explanation / Answer

Answer is as follows :

The range is mapped to ?2(N-1) . . . 2 (N-1)-1. Vmax is mapped to top value, ? 2(N-1).

In fact, whole range Vmax down to (Vmax ? q/2) is mapped to that, where q is the quantization interval. The largest negative signal, ?Vmax is mapped to ?2 (N-1). Therefore q = (2 ? Vmax)/(2N ), since there are 2N intervals.

The dynamic range is Vmax/Vmin, where Vmin is the smallest positive voltage we can see that is not masked by the noise. Since the dynamic range is 60 dB, we have 20 log10(Vmax/Vmin) = 60 so Vmin = Vmax/1000.

At 10 bits, the quantization noise, equal to q/2 == half a quantization interval q, is q/2 = (2 ? Vmax/2N )/2 = Vmax/(210), or in other words Vmax/1024. So this is not sufficient intensity resolution.

At 16 bits, the noise is Vmax/(216) = Vmax /(64?1024), which is more than an order of magnitude smaller than Vmin so is fine.

if there is any query please ask in comments.....

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