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GATE 2015 EC – Question 54

Networks, Signals and Systems · Discrete-time Signals · 2 marks · Multiple choice

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

A block diagram with input $X[n]$ and output $Y[n]$. The output goes through a delay $Z^{-1}$ and a gain of $\frac{5}{6}$ back to the second adder, and through a second delay $Z^{-1}$ and a gain of $-\frac{1}{6}$ back to the first adder.
  1. 2, 3
  2. $\frac{1}{2}$, 3
  3. $\frac{1}{2}$, $\frac{1}{3}$
  4. 2, $\frac{1}{3}$

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Correct answer: (C) $\frac{1}{2}$, $\frac{1}{3}$

Explanation

Reading the feedback gains from the diagram, the difference equation is $y[n] = x[n] + \frac{5}{6}y[n-1] - \frac{1}{6}y[n-2]$. The poles satisfy $z^2 - \frac{5}{6}z + \frac{1}{6} = 0$, which factors as $\left(z - \frac{1}{2}\right)\left(z - \frac{1}{3}\right) = 0$. The poles are at $\frac{1}{2}$ and $\frac{1}{3}$.