A realization of a stable discrete-time system is shown in the fi
A realization of a stable discrete-time system is shown in the figure. If the system is excited by a unit step sequence input x[n], the response y[n] is
A. <span class="math-tex">\(4{\left( { - \frac{1}{3}} \right)^n}u\left[ n \right] - 5{\left( { - \frac{2}{3}} \right)^n}u\left[ n \right]\)</span>
B. <span class="math-tex">\(5{\left( { - \frac{2}{3}} \right)^n}u\left[ n \right] - 3{\left( { - \frac{1}{3}} \right)^n}u\left[ n \right]\)</span>
C. <span class="math-tex">\(5{\left( {\frac{1}{3}} \right)^n}u\left[ n \right] - 5{\left( {\frac{2}{3}} \right)^n}u\left[ n \right]\)</span>
D. <span class="math-tex">\(5{\left( {\frac{2}{3}} \right)^n}u\left[ n \right] - 5{\left( {\frac{1}{3}} \right)^n}u\left[ n \right]\)</span>
Please scroll down to see the correct answer and solution guide.
Right Answer is: C
SOLUTION
Calculation:
Given network is shown below:
where 
Now,
\(Y\left[ z \right] = - \frac{5}{3}.{z^{ - 1}}{Y_1}\left[ z \right] + \frac{5}{3}{z^{ - 1}}{Y_1}\left[ z \right]\)
\( Y\left[ z \right] = \left[ { - \frac{5}{3}{z^{ - 1}} + \frac{5}{3}{z^{ - 2}}} \right]{Y_1}\left[ z \right]\) ----1)
\(X\left[ z \right] + {z^{ - 1}}.{Y_1}\left[ z \right] + \left( { - \frac{2}{9}} \right)\left( {{z^{ - 2}}} \right){Y_1}\left[ z \right] = {Y_1}\left[ z \right]\)
\( X\left[ z \right] = \left[ {1 - {z^{ - 1}} + \frac{2}{9}{z^{ - 2}}} \right]{Y_1}\left[ z \right]\) ----2)
Dividing equation 1) & 2) we get:
\(\frac{{Y\left[ z \right]}}{{X\left[ z \right]}} = \frac{{\frac{5}{3}\left[ {1 - z} \right]}}{{{z^2} - z + \frac{2}{9}}}\) ----3)
Since x(n) = u(n)
\(X\left[ z \right] = \frac{1}{{1 - {z^{ - 1}}}}\)
Putting this in Equation-(3), we get;
\(Y\left[ z \right] = \left( {\frac{5}{3}} \right)\frac{{\left( { - z} \right)}}{{\left( {z - \frac{1}{3}} \right)\left( {z - \frac{2}{3}} \right)}}\)
\(Y\left[ z \right] = \frac{{5z}}{{z - \frac{1}{3}}} - \frac{{5z}}{{z - \frac{2}{3}}}\) (Using Partial Fractions)
\( {Since\;{\alpha ^n}u\left[ n \right]\mathop \to \limits^{Z.T} \frac{z}{{z - \alpha }}} \)
By taking the Inverse Laplace Transform;
\(y\left[ n \right] = \left( 5 \right){\left( {\frac{1}{3}} \right)^n}u\left[ n \right] - \left( 5 \right){\left( {\frac{2}{3}} \right)^n}u\left( n \right)\)
Hence Option (3) is Correct.
