Digital Signal Processing Basics
Which of the following correctly gives the poles and the stability of the causal system with transfer function H(z) = 1/(1 − 1.3z⁻¹ + 0.4z⁻²)?
- AThe poles are z = −0.5 and z = −0.8, and the system is stable
- BThe poles are z = 0.5 and z = 0.8, and the system is stable
- CThe poles are z = 1.25 and z = 2, and the system is unstable
Show answer and explanation
Answer: B. The poles are z = 0.5 and z = 0.8, and the system is stable
Multiplying the numerator and denominator by z² gives H(z) = z²/(z² − 1.3z + 0.4), and the poles follow from z² − 1.3z + 0.4 = (z − 0.5)(z − 0.8) = 0 as z = 0.5 and 0.8. A causal system is stable when all its poles lie inside the unit circle (|z| < 1), so this system is stable. Solving the denominator as a polynomial in z⁻¹, 1 − 1.3p + 0.4p² = 0 (p = z⁻¹), gives p = 1.25, 2, but these are the reciprocals of the poles, not the poles themselves.
- A. The signs are wrong. The roots of z² − 1.3z + 0.4 = 0 are the positive values 0.5 and 0.8.
- B. Correct. (z − 0.5)(z − 0.8) = z² − 1.3z + 0.4, and both poles lie inside the unit circle.
- C. These solve the equation in z⁻¹ and are the reciprocals of the poles (1/0.8 = 1.25, 1/0.5 = 2).