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I am trying to understand the feedback compensation techniques of DC-DC Buck converters. And I drift to more and more small questions which make me want to understand the basics of poles/zeroes/frequency

In this link the answer given by user Kaz, in the last para, "One way in which amplifiers with multiple poles are stabilized is with the help of capacitors which create a "dominant pole" whose frequency roll-off is so great that the poles at higher frequencies basically do not matter (the gain is squashed at those frequencies). The amplifier basically "looks" like a single pole one."

I can't able to understand what he is saying in the last para. I know, it is somewhat easy. But i am not able grasp

In general, can someone clarify me the concept of pole, dominant pole with filters/capacitors? Like, if someone says, there is a pole in the feedback loop, or two poles in the loop, how to understand it electrically? one more question regarding this, i understand that if there is a pole, it means that oscillations will be present in the system. But the oscillations will be at what frequency? Will it be at the resonant frequency of the part or the entire systems?

Experts who find mistakes in my question, please let me know where I am wrong.

I also find a single pole and double pole loops. How do they arise?

It would be better if someone can provide me an analogy regarding these poles/frequency for good understanding.

Thanks.

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  • \$\begingroup\$ Do you understand what a bode plot is? Did you fully comprehend this answer to one of your earlier questions? \$\endgroup\$ – Andy aka Aug 1 '19 at 8:23
  • \$\begingroup\$ Yes. I understand the bode plot \$\endgroup\$ – Newbie Aug 1 '19 at 8:24
  • \$\begingroup\$ Does this answer help? \$\endgroup\$ – Andy aka Aug 1 '19 at 8:26
  • \$\begingroup\$ I tried to understand ur answer but I'm unable to understand it mathematically. So, I started to understand your answer in the electrical sense. Since, the question was not related to electrical, I wanted to ask it in a different way. I am trying to understand the concept with less mathematics and more of how this (Poles/Zeroes) means in the electrical domain. I also understand that, if the system has a pole, then it will oscillate infinitely. if it doesn't, output will be zero. But i want to understand at what frequency will the system oscillate? And what does single/double pole mean? \$\endgroup\$ – Newbie Aug 1 '19 at 8:30
  • \$\begingroup\$ Just to correct misunderstanding: a system that has a pole will not oscillate. A system that has two poles may oscillate but certainly won't definitely oscillate. \$\endgroup\$ – Andy aka Aug 1 '19 at 8:34
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You keep trying to understand it "more electrically", but the problem is that pole dominance, and the pole concept itself, is strictly related to Laplace Transform (LT) theory, which is a strong mathematical theory.

Poles are not an "electrical concept" in itself. It is a concept that can be applied to any linear system, or any system amenable to be linearized usefully for the application at hand. They can be electrical systems, but also mechanical, thermal or systems in whatever field of application you may imagine where the LT techniques can be applied.

Chu already gave you a very simplified view of the concept in the time domain for an electrical system. I'll try to dumb down the things some more, but keep in mind that if you don't want to grok some math, it is really not possible to understand the concept in all its depth.

First of all, poles are related to how fast a system can react when it is excited, and the reaction time is related to how the system can store and release energy during and after the excitation.

An (idealized) system which cannot store energy has no reaction time, it reacts immediately to any stimulus it is subjected to (think of a pure resistive circuit, where resistors are ideal). Reaction time for electrical circuits, therefore, depends on elements that can store energy: reactive elements, that is capacitors and inductors (if we keep the things simple and don't consider mutual induction and/or distributed circuits).

Grossly simplifying, each independent reactive element contributes a pole to the circuit, and the number of poles of a system is called the order of the system. So a circuit with one cap and no inductors is a 1st order system. A circuit with a cap and an inductor is a 2nd order system (or two independent caps), and so on.

What does it mean "independent" element? Too complex for you, if you don't want to grok the math, so let's skip it. Very broadly, it means that the level of energy storage of one element is completely independent from the level of the other elements.

OK, now what do the poles do to a system? They describe the "reaction time***s***" of the system. Note the plural! A system that is excited momentarily (imagine a fast electrical impulse) reacts with a response that evolve in time and then dies off (if the system is stable).

This "impulse response" is made up of different components, each of which decays with a different time constant. Every pole contributes a different component with a different time constant. The time constant is a measure of how fast a component decays: the bigger the time constant, the slower the component decay (usually a component is considered zero after 5 time constants have elapsed, actually the decay lasts forever theoretically since it is exponential -- see Chu's answer for a simple math expression).

The impulse response of a stable system dies off after a while and the time it takes to die off gives you an estimate of the reaction time of the system: the time a system needs to stabilize into a new condition when the inputs are varied.

A dominant pole is a pole whose time constant is much "slower", i.e. bigger, than all the other time constants of the circuit, therefore the corresponding component is still observable after all the other, faster decaying, components have died off.

In other words, a system with a dominant pole behaves approximately like a 1st order system, i.e. a system that has a single pole.

In the end, the concept of pole dominance is a way to simplify a system: if a dominant pole exists, with some caveats, the system can be thought as a 1st order system, i.e. the simplest system imaginable that has a non-trivial time evolution.

The problem is that not every system has a dominant pole, so it can't be simplified that way. A technique used sometimes to simplify things is to add a compensating network in the circuit, i.e. a network containing at least one reactive element, in order to introduce a dominant pole in the system and thus allowing the simplification.

Of course, if you want to introduce a new pole in the system that has to be dominant, its time constant must be bigger of any existing pole (unless you do pole-zero cancellation, but that is a still nastier and more advanced technique), therefore you end-up slowing down the system. So it's often a kind of trade-off: you slow down a fast system in order to make it simpler, i.e. more manageable by the rest of the control chain. In the process you lose something (e.g. fast response or gain) in order to gain something else (e.g. stability or accuracy).

Note that my answer contains lots of hand-waving and gross conceptual simplifications, but everything I've said can be justified and expanded mathematically to give the exact definition and interpretation of what is a pole and what is a dominant pole. The problem is, as I've already said, that you can't escape the heavy math then.

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  • \$\begingroup\$ Thank you very much for the answer which I can understand and get some real good clarity. Very grateful to you, sir. \$\endgroup\$ – Newbie Aug 1 '19 at 13:05
  • \$\begingroup\$ In the S-plane, we have poles on the left and right side. What do they mean ? Can you please clarify with the location of poles and the freq wrto s plane? \$\endgroup\$ – Newbie Aug 1 '19 at 13:54
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Dominance is most easily explained in the time domain.

Consider a transfer function: \$\small G(s)=\large \frac{5}{(s+1)(s+5)}\$, with a unit step applied, giving the step response: \$\small C(s)=\large \frac{5}{s(s+1)(s+5)}\$.

Expanding this function in partial fractions:

$$\small C(s)= \frac{1}{s}-\frac{1.25}{s+1}+\frac{0.25}{s+5}$$

And the time response is:

$$\small c(t)=1-1.25e^{-t/1}+0.25e^{-t/0.2} $$

So the transient has two terms, one with a time constant, \$\small \tau=1\$ sec, and the other with a time constant, \$\small \tau=0.2\$ sec. But the faster transient has much smaller residue: (\$\small 0.25\$) compared to the slower transient (\$\small 1.25\$). If you sketch the step response you'll see that it is dominated by the slower transient.

This illustrates the general rule that poles that are closer to the s-plane origin are more dominant than those that are more distant. In this particular case, the pole that is five times faster is five times smaller in magnitude.

To relate this to the frequency domain, we note that the effects of the faster poles (higher 3dB frequencies or resonant frequencies) are more apparent close to \$\small t=0\$, and the slower poles (lower 3dB frequencies or resonant frequencies) are more apparent at larger values of \$\small t\$.

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  • \$\begingroup\$ Sorry, I am not able to understand. Can you tell me in more electrical terms? Like how when we say Single pole, Double pole and all, how does it relate electrically? Like what cause two poles? What does two poles mean in electrical terms? And oscillation occurs at the poles, what will be frequency of oscillation? Can you please relate my questions electrically to get an intuitive understanding for clarity. \$\endgroup\$ – Newbie Aug 1 '19 at 8:02
  • \$\begingroup\$ I just addressed the headline question. Relating it to specific electrical networks is a more extensive answer that I do not have time to do presently, but I may extend the answer later. \$\endgroup\$ – Chu Aug 1 '19 at 8:07
  • \$\begingroup\$ Ok. Thanks. But I want to understand this concept more electrically. \$\endgroup\$ – Newbie Aug 1 '19 at 8:14

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