Z-transform and digital filters · Definition

Poles, zeros and stability

signal processingz-transform and digital filters

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.

\[|p_k| < 1 \quad \forall k\]

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

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.