Exercise · ZW-POLES-ZEROS-AND-STABILITY-006 · ★★☆ Intermediate
A causal system has \(H(z)=1/(1-1.2z^{-1})\). Is it stable?
Choose one option.
Show answer
- A. No, because the pole is at \(z=1.2\)
Explanation
The pole is at z=1.2, whose modulus exceeds one, so the system is unstable. Having a single pole does not guarantee stability.
Review the concept: Poles, zeros and stability
The poles and zeros of a discrete-time transfer function \(H(z)\) are the values of \(z\) where \(H(z)\) vanishes or diverges.
\[|p_k| < 1 \quad \forall k\]
More practice problems: poles, zeros and stability
- ZW-POLES-ZEROS-AND-STABILITY-001 A system has H(z)=1/(1-0.6z^-1). Where is its pole?
- ZW-POLES-ZEROS-AND-STABILITY-002 What is the zero of H(z)=1-0.4z^-1?
- ZW-POLES-ZEROS-AND-STABILITY-003 A causal system has a single pole at z=0.9. Is it stable?
- ZW-POLES-ZEROS-AND-STABILITY-004 What condition must a pole of a causal system satisfy for the system to be stable?
- ZW-POLES-ZEROS-AND-STABILITY-005 A system has H(z)=1/(1+0.5z^-1). Where is its pole?
- ZW-POLES-ZEROS-AND-STABILITY-007 Where are the zeros of H(z)=(1-0.3z^-1)(1+0.7z^-1)?
- ZW-POLES-ZEROS-AND-STABILITY-008 A system has H(z)=(1-0.5z^-1)/(1-0.8z^-1). What are its pole and zero?
- ZW-POLES-ZEROS-AND-STABILITY-009 A causal system has poles at z=0.5 and z=-1.5. Is it stable?
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