I am not trying to design a filter. I just want to understand the theory of laplace transform and s-domain.I read an example from the book The Scientist and Engineer's Guide to Digital Signal Processing

I know that I can change the value of the resistor, the capacitor and the inductor . Respectively , I can change the position of the zeros and the poles. So how to design the frequency response from the position of the zeros and the poles?
For example , if I want to design a notch filter with the narrow stop band at Pi/4. How to choose the positions of the zeros and the poles? If the zeros and poles are moving around. How would them affect the frequency response(the fourier transform along the jw axies)?

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  • \$\begingroup\$ What do you mean how do you design the frequency response? Frequency response is the magnitude of the transfer function along the \$j\omega\$ axis. \$\endgroup\$
    – Matt Young
    Feb 16, 2016 at 3:52
  • \$\begingroup\$ If I want to design a notch filter with the narrow stop band at Pi/4. How to choose the positions of the zeros and the poles? \$\endgroup\$
    – Frank
    Feb 16, 2016 at 4:10
  • \$\begingroup\$ A notch-filter is simply a band-stop filter with a narrow stop-band (high Q factor). You haven't really specified what your requirements are, hence it would be hard to provide an answer. However, if you go to wikipedia, you'll see a very simple circuit using passive components. With more stringent requirements, you're looking at using op-amps. \$\endgroup\$ Feb 16, 2016 at 4:44
  • \$\begingroup\$ I'm flagging this question as Too Broad because I'm currently taking an entire university course on this. It's a huge topic. \$\endgroup\$
    – Greg d'Eon
    Feb 16, 2016 at 10:51

1 Answer 1


The poles and zeros of the Laplace TF may be plotted on the complex s-plane, and their locations give some insight into the time domain and frequency domain characteristics of the system.

For the frequency domain, note that the vertical axis is the \$\small j\omega\$ axis, and each point on this axis corresponds to a particular input frequency, \$\small \omega \: rad \:s^{-1}\$. It turns out that, if you choose one such point and draw vectors to each of the zeros and poles, the gain at that particular frequency will be the product of all the zero vector lengths divided by the product of all the pole vector lengths.

Now start at the origin and work your way up the \$\small j\omega \$ axis (we normally just consider positive frequencies). As \$\small \omega \$ gets closer to \$\small z_1 \$ the corresponding vector length reduces until, when \$\small \omega =z_1\$, the vector length is zero and hence the overall gain is also zero. Conversely, for \$\small \omega \$ values that are quite distant from the pole/zero pairs, the pole and zero vector lengths are approximately the same, hence the overall gain is unity.

Given a desired frequency response characteristic, you can place the poles and zeros at appropriate locations, but note that there will be practical constraints that prevent a completely free choice!


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