For the discrete-time system shown in the figure, the poles of th

For the discrete-time system shown in the figure, the poles of the system transfer function are located at

A. 2, 3

B. \(\frac{1}{2},3\)

C. \(\frac{1}{2},\frac{1}{3}\)

D. \(2,\frac{1}{3}\)

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Right Answer is: C

SOLUTION

\(\begin{array}{l} \frac{{{\rm{Y}}\left( {\rm{z}} \right)}}{{{\rm{X}}\left( {\rm{z}} \right)}} = {\rm{H}}\left( {\rm{z}} \right) = \frac{1}{{1 - \frac{5}{6}{{\rm{z}}^{ - 1}} + \frac{{{{\rm{z}}^{ - 2}}}}{6}}} = \frac{{{{\rm{z}}^2}}}{{{{\rm{z}}^2} - \frac{5}{6}{\rm{z}} + \frac{1}{6}}}\\ = \frac{{{{\rm{z}}^2}}}{{\left( {{\rm{z}} - \frac{1}{2}} \right)\left( {{\rm{z}} - \frac{1}{3}} \right)}} \end{array}\)

So, poles are \({\rm{z}} = \frac{1}{2},{\rm{z}} = \frac{1}{3}\)

For the discrete-time system shown in the figure, the poles of th

Right Answer is: C

SOLUTION

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