The discrete-time signal x(n) = 6.35 cos(π/10)n is quantized with a resolution (a) = 0....

(a)

Write the expression of the resolution of the quantizer.

\(\Delta=\frac{x_{\max }-x_{\min }}{N}\)

Here, \(N\) is the number of bits.

Substitute \(6.35 \mathrm{~V}\) for \(x_{\max },-6.35 \mathrm{~V}\) for \(x_{\min }\) and \(0.1\) for \(\Delta\).

\(0.1=\frac{6.35-(-6.35)}{2^{N}-1}\)

\(2^{N}-1=\frac{12.7}{0.1}\)

\(2^{N}=128\)

\(N=7\) bits

Thus, the number of bits required in the A/D convertor are 7 .

(b)

Write the expression of the resolution of the quantizer.

\(\Delta=\frac{x_{\max }-x_{\min }}{N}\)

Here, \(N\) is the number of bits.

Substitute \(6.35 \mathrm{~V}\) for \(x_{\max },-6.35 \mathrm{~V}\) for \(x_{\min }\) and \(0.02\) for \(\Delta\).

\(0.02=\frac{6.35-(-6.35)}{2^{N}-1}\)

\(2^{N}-1=\frac{12.7}{0.02}\)

\(2^{N}=636\)

\(N=10\) bits

Thus, the number of bits required in the A/D convertor are 10 .