Topic: Signals and Systems
What are poles and zeros of a system having following transfer function?
$H(Z)=\frac{(1-Z^{-2})}{(1+1.3Z^{-1}+0.36Z^{-2})}$
- A. $Zeros=-1, 1$; $Poles=-0.4, -0.9$
- B. $Zeros=-1, 1$; $Poles=0.4, 0.9$
- C. $Zeros=1$; $Poles=0.4, 0.9$
- D. $Zeros=1, 1$; $Poles=0.4, 0.9$
Answer: A. $Zeros=-1, 1$; $Poles=-0.4, -0.9$
Explanation: Multiply the numerator and denominator by $Z²$:
$H(Z) = \frac{Z²-1}{Z²+1.3Z+0.36} = \frac{(Z-1)(Z+1)}{(Z+0.9)(Z+0.4)}$.
Zeros are obtained by equating the numerator to zero: $Z²-1 = 0 \Rightarrow Z = 1, -1$.
Poles are obtained by equating the denominator to zero: $Z²+1.3Z+0.36 = 0 \Rightarrow Z = -0.9, -0.4$.