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I'm a bit confused about the idea of poles. In some of my courses such as digital signal processing, we are taught that in filter design, you place a pole at a frequency you want to enhance and a zero where you want to cause a dip.

In other courses in which we deal with Bode plots, I notice that a pole causes a -20dB/decade slope decrease?

Why is there an incosistency?

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  • \$\begingroup\$ Have a look at low pass wrong pole location to start. Do you follow the discussion there?$$\mid\: H(s)\mid =\frac{\mid N(s)\mid}{\mid D(s)\mid}=\frac{\mid N(j\omega)\mid_{s=j\omega}}{\mid D(j\omega)\mid_{s=j\omega}}=\frac{\scr{N}(\omega)}{\scr{D}(\omega)}=\scr{G}(\omega)$$ Given \$\scr{G}(\omega)\$, the job is to find the \$H(s)\$ that achieves it. For example, the roots of \$\mid D(j\omega)\mid^2=0\$ will have twice the roots of \$D(s)\$, whose roots are the poles. The zeros are the roots of \$N(s)\$. With associated implications. \$\endgroup\$
    – jonk
    Nov 20, 2018 at 0:07

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Here's a bit of a crash course in poles: -

enter image description here

The upper pictures show a bode plot with frequency peaking in various amounts. The lower left picture attempts to show a 3D view of bode plot and poles and the lower right picture is the conventional pole zero diagram.

In other courses in which we deal with Bode plots, I notice that a pole causes a -20dB/decade slope decrease? Why is there an incosistency?

There is no inconsistency; a single pole causes a peak in the spectrum and, due to the mathematical relationship (1/distance) between the pole centre and more distant positions on the jw axis there is a slope that results.

For a 2nd order low pass filter (example) there are two poles and the reciprocal-of-distance relationship leads directly to the ability calculate any point on the jw axis: -

enter image description here

Hope this helps. Related question and another related question and yet another related question. And another.

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  • \$\begingroup\$ Oh wow. That 3-D diagram made so much sense. Thank you so much. Just one more thing, in regards to the resonant frequency. I know that for a second-order cannconical system, there is a value for the resonant frequency based on the damping ratio. Now, does that mean that only one frequency in a system can ever cause a resonance? If so, where does the entire pole stuff come into then? Or is there a damping ratio associated with each pole? \$\endgroup\$
    – AlfroJang80
    Nov 20, 2018 at 12:50
  • \$\begingroup\$ Both poles in a 2nd order filter have the same damping ratio as shown on the horizontal axis on the bottom right image. The value of resonant frequency is (or can be) independent of damping ratio. \$\endgroup\$
    – Andy aka
    Nov 20, 2018 at 12:55
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Resonant poles with low damping (i.e., poles that are close to the imaginary axis in the \$s\$ domain, or close to the unit circle in the \$z\$ domain) cause peaking at frequencies close to the pole. This is probably what you picked up on in DSP class. Any pole causes that 20dB/decade/pole amplitude decrease once you get past the pole frequency.

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