Topic: Signals and Systems
For the following function, poles and zeros will be:
$x(t)=\delta(t)-\frac{4}{3}e^{-t}u(t)+\frac{1}{3}e^{2t}u(t)$
- A. Poles at -1 and 2, Zero at -1
- B. Poles at -1 and 2, Zero at 1
- C. Poles at -1 and -2, Zero at -1
- D. Poles at 1 and -2, Zero at -1
Answer: B. Poles at -1 and 2, Zero at 1
Explanation: Taking the Laplace transform of the given function:
$X(s) = 1 - \frac{4/3}{s+1} + \frac{1/3}{s-2}$
$X(s) = \frac{3(s+1)(s-2) - 4(s-2) + (s+1)}{3(s+1)(s-2)}$
$X(s) = \frac{3(s^2 - s - 2) - 4s + 8 + s + 1}{3(s+1)(s-2)}$
$X(s) = \frac{3s^2 - 3s - 6 - 3s + 9}{3(s+1)(s-2)} = \frac{3s^2 - 6s + 3}{3(s+1)(s-2)} = \frac{3(s-1)^2}{3(s+1)(s-2)}$
Simplifying, $X(s) = \frac{(s-1)^2}{(s+1)(s-2)}$.
This gives a double zero at $s = 1$, and poles at $s = -1$ and $s = 2$. Option (b) is the best matching choice.