What is the Z-transform H(z), ROC, and the stability of the system for h[n] = (1/2)^n u[n]?

What is the Z-transform H(z), ROC, and the stability of the system for h[n] = (1/2)^n u[n]?

• A. H(z) = 1 / (1 − (1/2) z^{−1}); ROC |z| > 1/2; stable since pole at z = 1/2 lies inside unit circle.
• B. H(z) = 1 / (1 − (1/2) z^{−1}); ROC |z| > 2; unstable.
• C. H(z) = 1 / (1 − z^{−1}/2); ROC |z| < 1/2; unstable.
• D. H(z) = 1 / (1 − z^{−1}/2); ROC |z| > 2; unstable.

Answer: (A)

The Z-transform of a causal, right-sided sequence like h[n] = (1/2)^n u[n] is a geometric series in z^{-1}. Its ratio is (1/2) z^{-1}, so the series converges when |(1/2) z^{-1}| < 1, i.e. |z| > 1/2. Summing the geometric series gives H(z) = 1 / [1 − (1/2) z^{-1}]. The pole occurs at z = 1/2, which lies inside the unit circle. For a causal system, the ROC is outside the pole, so |z| > 1/2. Because the unit circle lies in the ROC, the system is BIBO stable; the impulse response is absolutely summable: ∑_{n=0}^∞ (1/2)^n = 2. Thus the best match is H(z) = 1 / (1 − (1/2) z^{-1}); ROC |z| > 1/2; stable since the pole at z = 1/2 lies inside the unit circle.

The Z-transform of a causal, right-sided sequence like h[n] = (1/2)^n u[n] is a geometric series in z^{-1}. Its ratio is (1/2) z^{-1}, so the series converges when |(1/2) z^{-1}| < 1, i.e. |z| > 1/2. Summing the geometric series gives H(z) = 1 / [1 − (1/2) z^{-1}]. The pole occurs at z = 1/2, which lies inside the unit circle. For a causal system, the ROC is outside the pole, so |z| > 1/2. Because the unit circle lies in the ROC, the system is BIBO stable; the impulse response is absolutely summable: ∑_{n=0}^∞ (1/2)^n = 2.

Thus the best match is H(z) = 1 / (1 − (1/2) z^{-1}); ROC |z| > 1/2; stable since the pole at z = 1/2 lies inside the unit circle.