# I can't wrap my mind around the concept of filtering and the unit circle ? any help? [closed]

Can someone elaborate this theory where there is a link between the stability of a filter and the zeros or nulls being on the unit circle?

• Hi Abby! That sounds like every signals and systems textbook. Have you one of these? – Marcus Müller Mar 10 at 17:47

BIBO stability of LTI systems implies that their impulse response is absolutely summable, that is, $$$$\sum_{n=-\infty}^{+\infty}|h(n)| < +\infty$$$$

That exact same relationship is a sufficient condition for the Fourier Transform of the impulse response - the so-called Frequency Response - to converge.

Convergence of the frequency response means that it takes finite values, that is, $$|H(e^{j\omega})|<+\infty \Longrightarrow \sum_{n=-\infty}^{+\infty}|h(n)|<+\infty$$

The frequency response is the evaluation of the Z-Transform on the unit circle: $$H(e^{j\omega})= H(z)\Big|_{z=e^{j\omega}}$$

Moreover, a pole is a point on the complex plane where the Z-Transform's magnitude $$|H(z)|$$ takes an infinite value: if $$z=z_p$$ is a pole, then $$|H(z_p)| = +\infty$$. Given that the Fourier Transform is a "version" of the Z-Transform on the unit circle, if there is a pole on the unit circle, the frequency response does not converge. If it does not converge, the system is not BIBO-stable.

The last two statements hold because $$$$\sum_{n=-\infty}^{+\infty}|h(n)| < +\infty$$$$ does not hold.

• The last sentence is not clear ... " A frequency response ... take an infinite value.", could you elaborate ? – Abby_DSP Mar 11 at 10:32
• I have edited my answer. – GKH Mar 11 at 11:18
• Is it better now? – GKH Mar 11 at 20:21

If you're familiar with LaPlace transforms, you can see the Z transform by analogy. The unit circle is equivalent to the jw axis, with zero frequency at 1+j0 and the Nyquist rate at -1+j0.

We get the Fourier Transform of a signal at the unit circle. If the ROC does not include the unit circle, that means that the Fourier transform does not converge which means that the system is unstable. Also, please read bores signal processing basics website and Alan Oppenheim. Its explained really well there.