How does one deduct the number of poles from a Bode plot? Is it further possible to decide where the poles are? I have not understood this concept even though there is so much focus on Bode plots in engineering.

E.g. for the Bode plot below?

enter image description here

  • 1
    \$\begingroup\$ It has one pole because the phase change is a maximum of 90 degrees and the slope of the amplitude response is 20 dB per decade. \$\endgroup\$
    – Andy aka
    Mar 26, 2020 at 13:24
  • \$\begingroup\$ ......and the single pole is at the frequecy, where the phase is -45 deg. Do you know the reason why we speak about a "pole" (something like "infinite")? \$\endgroup\$
    – LvW
    Mar 26, 2020 at 13:36
  • \$\begingroup\$ @Andyaka Thanks! So what are the rules for 2 or 3 poles? \$\endgroup\$
    – Lamar
    Mar 26, 2020 at 14:41
  • \$\begingroup\$ 2 poles, the mag plot drops off 40dB/dec, 3 poles 60dB/dec. One thing though, if in addition to "poles", you also have "zeroes", those will counteract the effects of the poles. There's some good resources online, just wanted to point that out. \$\endgroup\$
    – Big6
    Mar 26, 2020 at 14:48

1 Answer 1


First you must understand that we have different types of poles and zeroes: you have simple poles and simple zeroes, DC poles and DC zeroes, and you have higher order poles and zeroes.

Simple poles lead to a slope of -20 dB/dec in the magnitude plot and a slope of -45 degrees/dec in the phase plot. Respectively, simple zeroes lead to a slope of +20dB/dec and +45 degrees/dec.

DC poles and zeroes will affect the slope starting very small frequencies, in the example below the slope starts at zero, however when you have a DC pole the slope will starting going down -20 dB/dec starting from the gain, DC zeroes would start with a slope of +20 dB/dec. Phase plots are not affected by DC poles and zeroes, nor by the gain.

For higher order poles and zeroes you simply multiply the slope in either plot by n, it being the order of the pole or the order of the zero.

So, if you're dressing the transfer function from a magnitude plot, first locate all the frequencies at which the slope changes: it's a pole if the slope decreases (or a negative zero), or a zero if the slope increases. Then you find the gain in dB from the magnitude axis.

For example:

enter image description here

This is a simple bode plot similar to the one you presented. You can notice one change in the slope from 0 to -20 dB/dec indicating one simple pole at a frequency w = 30 rad/s. There are no other changes in the slope so there are no more poles nor zeroes. Now you find the gain by looking at the y-magnitude axis and can estimate that it starts at around 10 dB, so the gain is around 3.33.

The transfer function looks like this:

$$H(s) = \frac{3.33}{\frac{s}{30}+1}$$

If you're dressing the transfer function from a phase plot:

Method: You locate where the change in slope starts, then find the midpoint between the beginning and end of this slope, the frequency at the midpoint is the frequency of a pole or a zero. However, I use this method when I have fairly simple plots. To apply it on more complicated plots you need to practice well, check below for a good place to start practicing!

Good Reference for practicing and more complicated examples with detailed explanation: Click here.

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    \$\begingroup\$ Just for the sake of exactness (and to avoid misinterpretations): You wrote "...simple poles have a slope of -20dB/dec". No, the poles do not "have" any slope. Better and more accurat: A single pole will lead to a slope of app. -20dB/dec - sufficiently far above the pole frequency (as can be seen in your diagrams). \$\endgroup\$
    – LvW
    Mar 26, 2020 at 16:58
  • \$\begingroup\$ Edited, thank you! @LvW \$\endgroup\$
    – OmarAI
    Mar 26, 2020 at 17:21

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