Question

An input signal \(\rm x(t) = 2 + 5 \sin{(100πt)}\) is sampled with a sampling frequency of \(\rm 400\ Hz\) and applied to the system whose transfer function is represented by

\(\rm \frac{{Y\left( z \right)}}{{X\left( z \right)}} = \frac{1}{N}\left( {\frac{{1 - {z^{ - N}}}}{{1 - {z^{ - 1}}}}} \right)\)

where, \(\rm N\) represents the number of samples per cycle. The output \(\rm y[n]\) of the system under steady state is

1. \(\rm 0\)
2. \(\rm 1\)
3. \(\rm 2\)
4. \(\rm 5\)

Answer 1

Correct Answer - Option 1 : \(\rm 0\)

 

Concept:

Final value theorem,

\(Y(∞) = \displaystyle\lim _{Z \rightarrow 1} ( 1 - Z^{-1}) Y(Z)\)

Calculation:

We have:

\(\frac{Y(z)}{X(z)} = \frac{1}{N} \left( \frac{1 - Z^{-N}}{1 - Z^{-1}} \right)\)....(1)

and x(t) = 2 + 5sin(100πt)

sampling frequency, f_s = 400 Hz

put t = nT_s, the output of the sampling process is,

x(nT_s) = 2 + sin(100πnT_s)

x(nT_s) = 2 + 5 sin\(\left( 100 π n \times \frac{1}{400} \right)\)

x(nT_s) = 2 + 5\(\sin \left(\dfrac{n π}{4} \right)\)

where,

ω₀ = π/4

\(N = \frac{2\pi}{\omega_0} = \frac{2\pi}{\frac{\pi}{4}} =8\)

The z-transform of x(n) is,

\(\rm ZT[x(n)] = ZT \left[ 2 + 5 \sin \left( \frac{\pi n}{4} \right) \right]\)

\(\rm X(Z) = 2 + \frac{5Z \sin \left( \frac{\pi} {4} \right)}{Z^2 - 2Z \cos \left( \frac{\pi}{4} \right) + 1}\)

\(\rm X(Z) = 2 + \frac{\frac{5Z}{\sqrt 2}}{Z^2 - \frac{2z}{\sqrt Z } + 1}\)

\(X(Z) = 2+ \frac{2.5 \sqrt 2 Z}{Z^2 - \sqrt 2 Z + 1}\)

From equation (i)

\(\frac{Y(z)}{X(z)} = \frac{1}{N} \left( \frac{1 - Z^{-N}}{1 - Z^{-1}} \right)\)

\(\rm Y(Z) = \frac{1}{8} \left( \frac{1 - Z^{-8}}{1 - Z^{-1}} \right)X(Z)\)

\(\rm Y(Z) = \frac{1}{8} \left( \frac{1 - Z^{-8}}{1 - Z^{-1}} \right) \left[ 2+ \frac{2.5 \sqrt 2 Z}{Z^2 - \sqrt 2 Z + 1} \right]\)

using final value theorem,

\(Y(∞) = \displaystyle\lim _{Z \rightarrow 1} ( 1 - Z^{-1}) Y(Z)\)

\(Y(∞) = \displaystyle\lim _{Z \rightarrow 1} ( 1 - Z^{-1}) \frac{1}{8} \left( \frac{1 - Z^{-N}}{1 - Z^{-1}} \right) \left[ 2+ \frac{2.5 \sqrt 2 Z}{Z^2 - \sqrt 2 Z + 1} \right]\)

\(Y(∞) = \displaystyle\lim _{Z \rightarrow 1} \frac{1}{8}( 1 - Z^{-8}) \left[ 2+ \frac{2.5 \sqrt 2 Z}{Z^2 - \sqrt 2 Z + 1} \right]\)

y(∞) = 0