Question

Find x(∞) if X(z) = \(\frac{z+3}{(z+1)(z+2)}\).

(a) ∞

(b) -1

(c) 1

(d) 0

I had been asked this question during an interview for a job.

The above asked question is from Properties of Z-Transforms in division Z-Transform and Digital Filtering of Signals and Systems

Answer 1

Right choice is (d) 0

Easiest explanation: Given X(z) = \(\frac{z+3}{(z+1)(z+2)} = \frac{z[1+(3/z)]}{z^2 [1+(1/z)][1+(2/z)]} = \frac{1}{z} \frac{1+(3/z)}{[1+(1/z)][1+(2/z)]}\)

The initial value theorem of Z-transform states that

If x(n) ↔ X(z), then x(0) = Ltz→∞⁡ X(z)

x(0) = Ltz→∞⁡ X(z) = \(Lt_{z→∞} \frac{1}{z} \frac{[1+(3/z)]}{[1+(1/z)][1+(2/z)]}\) = 0.