I would like to understand electrically with little math (transfer function) involved about poles and zeroes in a control system (feedback in a DC-DC buck converter, to be specific).

I have been told by math users on this site that I will not understand the concept of poles and zeroes without using transfer functions or math, but I request you all to help me understand this electrically.

What does a having pole/zero in a system mean, and how does this pole/zero impact the system electrically?

How does having a pole decrease the gain in the Bode plot, and how does having a zero in the Bode plot make the slope of the line flat?

Request you to provide answers in simple and basic electrical terms as much as possible. I don't know if this is appropriate, but please excuse if so, "Please explain it to me like a five year old" (maybe an analogy).

  • 1
    \$\begingroup\$ Well, @Newbie, that is an impossible task, for any intuition people give you on control systems using electronics/electricity and no math will just be an intuition. You won't really know its properties and that will lead to you misusing the names/concepts. \$\endgroup\$
    – jDAQ
    Nov 21, 2019 at 5:43
  • \$\begingroup\$ I suggest you really try to learn the "mathematical" concept itself. \$\endgroup\$
    – jDAQ
    Nov 21, 2019 at 5:43
  • \$\begingroup\$ but, in a linear system, the complex poles (in \$s\$) of a ~transfer function~ system will each give you a "description" of how a mode ("output waveform") will behave. The position of the pole will tell you both how that mode grows/decay and also if it will oscillate (and at what frequency). \$\endgroup\$
    – jDAQ
    Nov 21, 2019 at 5:45
  • \$\begingroup\$ It's not one of those intuitive concepts, unfortunately. \$\endgroup\$
    – DKNguyen
    Nov 21, 2019 at 5:48

1 Answer 1


From your questions I understand that you want to understand how the feedback in buck converter works and chosen. For the analysis of the feedback network you should try understand three simple circuit and bode plots in general. (RC, LC and integrator) The first is to get familiar and the LC and integrator are the basic parts of the buck converter.

So first lets look at the the RC circuit which is a first order system.

obtained from https://www.robomart.com/blog/rc-filter-circuits-introduction

Now lets take a look at the transfer function (gain of the system):

enter image description here Where s = jw

So analyzing the equation in frequency response : we see that there is pole (S is in the de denominator). So, the gain decreases as the frequency increases. More specifically, the gain decreases with a factor of 10 for every 10-fold in frequency. You can look into the definition of decibels that this corresponds to a decrease of 20dB/ decade.

Now, lets look at the circuit more intuitively. So at low frequencies the impedance of the capacitor is very high and most of the voltage is across the capacitor (this is Vout). As the frequency increases, the impedance of the capacitor decreases and thus the voltage across the capacitor decreases and Vout decreases. So, this corresponds to the decreasing magnitude of the frequency response.

now lets look at a second order system.

obtained from http://sim.okawa-denshi.jp/en/RLClowkeisan.html

So now we see in the denominator instead of S we see S^2. If you look at the frequency response now the gain decreases with 40db/ decade after the corner frequency, because we have two poles instead of one.

for this case, it is a bit less intuitive: But lets try again , at low frequency most of the voltage will be across the capacitor as the inductor has low impedance. As frequency increases the capacitor impedance decreases but at the same time the inductor impedance increases, so this can somehow explain the second order dynamic. But important here is : that in a LC filter the gain decreases by 40 dB after the corner frequency.

now lets look at the buck converter :

enter image description here

Now look at the output we see the same LC filter- stage.

So if we want to design a feedback network, we can check the frequency response of all the stages. The whole point of looking this in the bode is that we can then easily add the gain of all the stages (filter, pwm and the switch). We can then add a feedback stage which alters the gain of the system in the frequency range of interest. This is to much to cover here, but this should give you a basic idea.

  • \$\begingroup\$ Thank you for this answer. Just one small query I have regarding this buck converter feedback loop. "A switch mode power supply is essentially a sampled-data system, therefore the theoretical maximum bandwidth is one half the switching frequency. Practically the phase and transport lag there make it impossible to close the loop there, so 1/5 to 1/10th the switching frequency is a good rule of thumb." In the quote, could you please tell me what does it mean to close the loop and what effect will it have if I dont close the loop \$\endgroup\$
    – user220456
    Nov 21, 2019 at 7:58
  • \$\begingroup\$ Can you please help with my above comment \$\endgroup\$
    – user220456
    Nov 21, 2019 at 8:17
  • \$\begingroup\$ The term "closing the loop" refers to connecting feedback up, which your Buck converter isn't going to do much without. If you are sampling at only twice the switching frequency the signals are going to be changing alot in between your samples, so your control will be lousy. See here for closed and open-loop systems electronics-tutorials.ws/systems/closed-loop-system.html \$\endgroup\$
    – jramsay42
    Nov 21, 2019 at 8:51

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