Z-transform and digital filters · Definition
Poles, zeros and stability
Before you start you need to know
Once the transfer function is defined, the key question is where its behavior concentrates in the z-plane. Poles and zeros show how the system responds at different frequencies and whether its outputs stay bounded.
Definition
The poles and zeros of a discrete-time transfer function \(H(z)\) are the values of \(z\) where \(H(z)\) vanishes or diverges.
For a rational system H(z)=N(z)/D(z), the zeros are the roots of N(z) and the poles are the roots of D(z). A causal system with all poles inside the unit circle has an absolutely summable impulse response, which is equivalent to BIBO stability. Poles on or outside the unit circle produce unbounded responses.
Consider \(H(z)=(1-0.5z^{-1})/(1-0.8z^{-1})\). Its zero is at \(z=0.5\) and its pole at \(z=0.8\), both inside the unit circle, so the system is stable. Evaluating on the circle \(|z|=1\) gives the corresponding frequency response.
A common mistake is forgetting that poles are read in powers of \(z\) rather than \(z^{-1}\): in \(H(z)=1/(1-0.8z^{-1})\) the pole is at \(z=0.8\), but in \(1/(1-0.8z)\) it would be at \(z=1.25\) and the system would be unstable. Always rewrite the function with positive powers of \(z\) before locating the poles.
Think about it
Can a causal system with all poles inside the unit circle have a zero outside it and still be stable?
Yes. Zeros do not affect stability; only the poles determine whether the impulse response is absolutely summable.
Exercises 10
- A system has \(H(z)=1/(1-0.6z^{-1})\). Where is its pole?
- What is the zero of \(H(z)=1-0.4z^{-1}\)?
- A causal system has a single pole at \(z=0.9\). Is it stable?
- What condition must a pole of a causal system satisfy for the system to be stable?
- A system has \(H(z)=1/(1+0.5z^{-1})\). Where is its pole?
- A causal system has \(H(z)=1/(1-1.2z^{-1})\). Is it stable?
- Where are the zeros of \(H(z)=(1-0.3z^{-1})(1+0.7z^{-1})\)?
- A system has \(H(z)=(1-0.5z^{-1})/(1-0.8z^{-1})\). What are its pole and zero?
- A causal system has poles at \(z=0.5\) and \(z=-1.5\). Is it stable?
- A causal system has \(H(z)=(1+z^{-1})/(1-1.1z^{-1}+0.3z^{-2})\). Is it stable and what are its poles?
References
- Alan V. Oppenheim, Alan S. Willsky, Signals and Systems, 2nd edition, Pearson, 1996.
- Alan V. Oppenheim, Ronald W. Schafer, Discrete-Time Signal Processing, 3rd edition, Pearson, 2010.