# Identify filter type from Pole-Zero plot

Thread Starter

#### Sagan820

Joined Jul 10, 2016
19
I was wondering if I could get some help understanding the pole-zero plot of a filter. I did some search on Google and it seemed that tracing the ratio of all the zero vectors to all the pole vectors around a unit circle is the most popular method. Is there a faster method? For example, what symmetry does each one have etc... I can recognize low pass filters quickly because they don't have zeroes and the poles are symmetrical but what about other filters? Thanks in advance!

P.S. I'm only interested in Butterworth filters at the moment since that's the only type of filters we've covered this term. Also the professor is interested in us knowing the type of filters and not other characteristics such as loss rate or gain.

#### Papabravo

Joined Feb 24, 2006
19,577
You should review the derivation of the Butterworth Response in a reference such as: Van Valkenburg, M. E., Analog Filter Design.

https://www.amazon.com/Analog-Filter-Design-M-E-Valkenburg/dp/0030592461/ref=sr_1_1?ie=UTF8&qid=1496885444&sr=8-1&keywords=Analog+filter+Design+,+Van+Valkenburg

From there you will see how the poles of a Butterworth filter are organized around the unit circle. A filter of odd order has a single pole on the negative real axis and the remaining poles are complex conjugate pairs that divide up the 180° spanned by the left half plane. the filters of even order lack the pole on the negative real axis but still have the conjugate pairs that divide up the left half plane.

Thread Starter

#### Sagan820

Joined Jul 10, 2016
19
Hi Papabravo,
Thank you for replying. Could you please clarify this for me? For a second order butterworth low pass there are two poles evenly spaced in the left hemisphere of the unit circle. If I start at w = 0 and go around the circle the response would be gradually increasing as w->the poles then decreasing as w passes the second pole. But this seems like a bandpass. Where did I go wrong here? Thanks!

#### Papabravo

Joined Feb 24, 2006
19,577
Hi Papabravo,
Thank you for replying. Could you please clarify this for me? For a second order butterworth low pass there are two poles evenly spaced in the left hemisphere of the unit circle. If I start at w = 0 and go around the circle the response would be gradually increasing as w->the poles then decreasing as w passes the second pole. But this seems like a bandpass. Where did I go wrong here? Thanks!
In the complex plane ω = 0 is at the origin. ω is not an angle, it is a frequency and is equal to 2πf. You should also note that in the complex plane the point at ∞ is a single point. It might be helpful to review the rules for constructing a root locus to get some idea about what the pole zero plot is telling you.

http://www.rfcafe.com/references/electrical/butter-poles.htm
https://en.wikipedia.org/wiki/Butterworth_filter

The unit circle, where the poles are located, represents the corner frequency. As you increase the frequency from 0 to the corner frequency the response starts to drop off because the denominator of the transfer function starts to increase in magnitude. Increasing the frequency further takes you to the zero(s) at infinity.

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#### MrAl

Joined Jun 17, 2014
9,633
Hi Papabravo,
Thank you for replying. Could you please clarify this for me? For a second order butterworth low pass there are two poles evenly spaced in the left hemisphere of the unit circle. If I start at w = 0 and go around the circle the response would be gradually increasing as w->the poles then decreasing as w passes the second pole. But this seems like a bandpass. Where did I go wrong here? Thanks!

Hello there,

Are you calculating the gain wrong? Start with a simple second order LP filter:
1/(s^2+sqrt(2)*s+1)

Find the two poles, calculate the gain from those two poles by varying w as i think you were doing. See if you get a better result. The result will in fact be a LP filter so if you dont get that then you did something wrong in the calculation of the gain

If you show your work and maybe a pole placement drawing and show how you are getting the gain from that graphic we can see where you went wrong. Strictly speaking, i dont think you can say that the two poles are "evenly spaced" around the left half of the circle but they are symmetrical with respect to the x axis which is maybe what you really meant.

#### Papabravo

Joined Feb 24, 2006
19,577
Hello there,

Are you calculating the gain wrong? Start with a simple second order LP filter:
1/(s^2+sqrt(2)*s+1)

Find the two poles, calculate the gain from those two poles by varying w as i think you were doing. See if you get a better result. The result will in fact be a LP filter so if you dont get that then you did something wrong in the calculation of the gain

If you show your work and maybe a pole placement drawing and show how you are getting the gain from that graphic we can see where you went wrong. Strictly speaking, i dont think you can say that the two poles are "evenly spaced" around the left half of the circle but they are symmetrical with respect to the x axis which is maybe what you really meant.
In the two pole case they are located at:

$$- sin(\frac{\pi}{4}) + j \math cos(\frac{\pi}{4})\;\text and\math\;- sin(-\frac{\pi}{4}) - j \math cos(-\frac{\pi}{4})$$

with angles measured relative to the negative real axis

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#### MrAl

Joined Jun 17, 2014
9,633
In the two pole case they are located at:

$$- sin(\frac{\pi}{4}) + j \math cos(\frac{\pi}{4})\;\text and\math\;- sin(-\frac{\pi}{4}) - j \math cos(-\frac{\pi}{4})$$

with angles measured relative to the negative real axis

Hi there Papa,

Thanks, but your point was?

The insinuation by the OP was that they were equally spaced around the unit circuit in the LHP, but they are not. Just because the angle is the same in those two formulas does not mean they are equally spaced around half of the unit circle

Actually, there is a transfer function that has equally spaced poles but it's not a Butterworth:
1/(s^2+sqrt(3)*s+1)

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#### Papabravo

Joined Feb 24, 2006
19,577
Don't know what you are talking about MrAl. The complex conjugate poles of the even order Butterworth Polynomials are most certainly equally spaced. I guess you must have learned different Mathematics than I did. In the case of an odd order filter the pole on the real axis does violate the "equally spaced" program.

#### MrAl

Joined Jun 17, 2014
9,633
Don't know what you are talking about MrAl. The complex conjugate poles of the even order Butterworth Polynomials are most certainly equally spaced. I guess you must have learned different Mathematics than I did. In the case of an odd order filter the pole on the real axis does violate the "equally spaced" program.
Hello again,

Hee hee, thanks for the humor there as i do like to see some lightening up in discussion as that keeps them more human i think. We are not machines so i dont think we should act like such with only facts and data but throw in a little levity now and then too. I like it and have always like comedy.

Yeah, that's it, i use a different mathematics then you I get mine from another universe where they actually read the subject material before they reply [just joking around here]

The bottom line is that if you believe that the poles are equally spaced around the left HALF plane then why didnt' you just plot them and then maybe show the diagram? That would reveal that they are definitely not spaced equally around the left half plane as stated by the OP.. I think it takes at least a third order filter to see that, but with a second order filter the poles are at unequal placements, and that is even evident from your reply which actually specifies where they actually are. All we have to do is plot your solutions and we see right away that they are not spaced equally around the left half plane. If we start at 90 degrees and travel counter clockwise, we hit the first pole at 45 degrees from 90 degrees, which is 90+45=135 degrees, or in simpler terms, 45 degrees from the x axis. If we continue to travel around the circle we hit the next pole at 180+45 degrees, but again in simpler terms this one is at -45 degrees. Now using my math from the alternate universe, the difference between +45 and -45 degrees is 45-(-45)=90 degrees, but the difference between 90 degrees and 135 degrees is just 45 degrees, and the difference beween -45 degrees and -90 degrees (the lower pole) is also 45 degrees, so we have spacing of 45, 90, 45, and since those are not the same, the spacing is unequal in the left half plane.
So just what is an equal spacing in the left half plane then?
Well, if we start at 90 and go 60 degrees, we arrive at 150 degrees, then go another 60 degrees we get to 210 degrees, then go another 60 degrees we get to 270 degrees which is the same as -90 degrees. So on our way from 90 degrees to -90 degrees we had spacings of:
60, 60, 60
which are all equal, so we can say they are equally spaced, as long as we dont count the fact that there are no poles at 90 and -90 degrees. If we decide that we require that too, then we have to go to a higher order filter that would have poles at 90 and -90 degrees also.

NOW, if instead we look at the ENTIRE PLANE around the ENTIRE circle, that's ENTIRELY different. Then we see spacing that is more even for many different filter orders, if not perhaps all of them if we allow the 2n Butterworth poles. But restricting our search to the left half plane only, we only have some filter orders that will fit the 'evenly spaced' specification.

If this still isnt clear, i'll post a plot. Ahh what the heck, here's a plot anyway

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#### Bordodynov

Joined May 20, 2015
3,025
Use LTpice
Example:

#### MrAl

Joined Jun 17, 2014
9,633
Hi,

That's not a bad idea really if they are allowed to do it that way. They can create the transfer function from the poles and then just graph it.

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