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Explain Quantization with the equation.
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Solution:

QUANTIZER: It is defined as the process of converting discrete time continuous amplitude signal into discrete time discrete amplitude signal.

Quantization is an approximation process The quantizer converts each sample to one of the values that are closest to it from among a pre-selected set of discrete amplitudes.

The encoder represents each one of these quantized samples by an R -bit code word. This bit stream travels on the channel and reaches the receiving end.

To perform quantization we consider standard amplitude levels (representation levels) which are finite.

$\rightarrow$ let us denote standard amplitude levels with,

$ \mathrm{L}=2^{\mathrm{n}} \text { (representation levels) }\\ $

Where $\mathrm{n} \rightarrow$ no. of binary digits (bits) transmitted per sample.

$\rightarrow$ let us consider $\mathrm{n}=3$

$\mathrm{L}=2^{\mathrm{n}}=2^3=8$ levels,

$\rightarrow$ Quantization is a process of rounding off the sample values to their nearest quantization levels ( standard representation levels/reference levels).

Step size $=\mathrm{R} / \mathrm{L}$

$\Rightarrow$ in this examples $\Delta=8 / 8=1$

$\rightarrow$ Quantized sample value,

$ \mathrm{T}_{\mathrm{s}} \rightarrow 1.5 ; 2 \mathrm{~T}_{\mathrm{s}} \rightarrow 3.5 ; 3 \mathrm{~T}_{\mathrm{s}} \rightarrow 2.5 ; 4 \mathrm{~T}_{\mathrm{s}} \rightarrow 0.5 ; 5 \mathrm{~T}_{\mathrm{s}} \rightarrow 0.5 ; 6 \mathrm{~T}_{\mathrm{s}} \rightarrow 2.5 ; 7 \mathrm{~T}_{\mathrm{s}} \rightarrow 3.5 $

$\rightarrow$ Code numbers

$ \mathrm{T}_{\mathrm{s}} \rightarrow 5 ; 2 \mathrm{~T}_{\mathrm{s}} \rightarrow 7 ; 3 \mathrm{~T}_{\mathrm{s}} \rightarrow 6 ; 4 \mathrm{~T}_{\mathrm{s}} \rightarrow 4 ; 5 \mathrm{~T}_{\mathrm{s}} \rightarrow 3 ; 6 \mathrm{~T}_{\mathrm{s}} \rightarrow 1 ; 7 \mathrm{~T}_{\mathrm{s}} \rightarrow 0 $

$\rightarrow$ Binary representation,

$ \mathrm{T}_{\mathrm{s}} \rightarrow 101 ; 2 \mathrm{~T}_{\mathrm{s}} \rightarrow 111 ; 3 \mathrm{~T}_{\mathrm{s}} \rightarrow 110 ; 4 \mathrm{~T}_{\mathrm{s}} \rightarrow 100 ; 5 \mathrm{~T}_{\mathrm{s}} \rightarrow 011 ; 6 \mathrm{~T}_{\mathrm{s}} \rightarrow 001 ; 7 \mathrm{~T}_{\mathrm{s}} \rightarrow 000 $

$\rightarrow \mathrm{o} / \mathrm{p}$ digital signal generated at,

(Bit rate) $-R_b=$ No. of samples/sec no. of bits/samples $=n \times f_s$

$ \begin{aligned}\\ &=7 \times 3 \\\\ &=21 \mathrm{bps}\\ \end{aligned}\\ $

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