I've been working with transfer functions and control models for a while, but I never really have gotten an intuitive understanding of what the poles and zeros of a system really represent.

I was wondering if anyone would be kind enough to offer an explanation of what they physically and mathematically represent. Let's take an RL load system for example.

Edit: I just thought I'd bring up a follow up question, in linear differential equations we use the solution: $$ x(t) = e^{st} $$ Where the time-domain solution of the differential equation is characterised by a sum of exponential and oscillatory components. How would this relate to poles and zeros?

  • 2
    \$\begingroup\$ I think you need to start by telling us what you feel you have as strengths. How fluent are you with non-homogeneous differential equations? Would you be able to instantly tell the difference between a homogeneous one (poles) and a non-homogeneous one (may include zeros)? How fluent are you with complex analysis? Etc. What's your background? What do you already know? I certainly have been known to write a lot. I may be even be the worst example of it here. But even I don't want to try and write an answer without knowing more about you, first. And that's saying something. ;) \$\endgroup\$
    – jonk
    Commented Dec 10, 2022 at 0:21
  • \$\begingroup\$ Hi there, my background is electrical engineering. I'm a postgraduate research student. I can tell the difference between a homogenous differential equation and a non-homogeneous one. As far as complex numbers go: I understand phasors, and the algebra of complex numbers - just the basics you need to get by in electrical engineering. \$\endgroup\$ Commented Dec 10, 2022 at 0:26
  • \$\begingroup\$ Are you saying that you've had little to no mathematical exposure to complex analysis but are well aware of non-homogeneous differential equations? But if I said, \$\frac{\text{d}}{\text{d}\: t^2}y + 10\frac{\text{d}}{\text{d}\: t}y + 15 y=5\frac{\text{d}}{\text{d}\: t}v+3\$, you wouldn't immediately be able to write this as \$\frac{5s+3}{s^2+10s+15}\$? \$\endgroup\$
    – jonk
    Commented Dec 10, 2022 at 0:42
  • \$\begingroup\$ Hi there Jonk! Yes I would by just taking the Laplace transform of both sides, and rearranging input / output. I've studies about non-homogeneous differential equations in the past - but it's been quite a while! \$\endgroup\$ Commented Dec 10, 2022 at 1:38
  • \$\begingroup\$ output/input* sorry! \$\endgroup\$ Commented Dec 10, 2022 at 1:44

3 Answers 3


Short answer: absolutely positively, poles arise from things that integrate in the physical world. Examples of this are innumerable.

In electronics, the canonical circuit elements that cause integration are inductors, \$i_L = \frac 1 L \int v_l dt\$, and capacitors, \$v_c = \frac 1 C \int i_C dt\$.

In mechanical systems, velocity integrates to position, and acceleration integrates to velocity. So good ol' Newton's law, where \$F = ma\$ implies a double integration to position: \$x = \frac 1 m \iint F dt\$.

Etc. In fact, if you go further down the dynamic systems rabbit hole, sooner or later you'll run into state-space notation, which turns all of your states into a vector, and says that

$$\begin{aligned} \mathbf x(t) &= \int {\mathbf A \mathbf x(t) + \mathbf B \mathbf u(t)\ dt}\\ \mathbf y(t) &= \mathbf C \mathbf x(t) + \mathbf D \mathbf u(t) \end{aligned}.$$ This notation clarifies a lot, not least of which is that there is one mode (just take "mode" as meaning "pole" here) per element in \$\mathbf x\$, and clearly there's one integration going on per element in \$\mathbf x\$.

Note that there's not a 1:1 correspondence between "integrating things" and modes or poles -- but if there's no pole-zero cancellation the pole count in a transfer function will equal the integrator count -- and even with pole-zero cancellations, in state-space form the modes will always be apparent.

Zeros are smokier. In a lot of ways, you can think of a zero as "that which bypasses an integrator". I.e., a \$1^{st}\$-order RC low-pass filter might have a transfer function of \$\frac{\omega_0}{s + \omega_0}\$. The equivalent high pass filter would have a transfer function of \$1 - \frac{\omega_0}{s + \omega_0}\$ -- after simplification, this is \$\frac{s}{s + \omega_0}\$, and presto, a zero has appeared.

  • 1
    \$\begingroup\$ Hi there Tim, thanks for the reply. Could you tell me more about what you mean by 'modes' and 'nodes'? I've never really heard these terms before. So is what you're saying that in the state space form, there is only one integration per state of the system? \$\endgroup\$ Commented Dec 10, 2022 at 1:47
  • 2
    \$\begingroup\$ The word "node" was a typo -- it's been fixed. Yes, look at the equation for the state-space system. When you integrate a vector quantity you're integrating each element in the vector. \$\endgroup\$
    – TimWescott
    Commented Dec 10, 2022 at 1:57

Assuming you have learned EE basic math.

A transfer function is defined by a ratios of quadratic equations that can be solved in the form as follows;

for i = 1 to n for numerator and d for denominator. 
The max. power of i defines the "order of magnitude" 
 of the equation or transfer function on top or bottom. 

The total i is the number of reactive components. The poles and zeros are used to define the breakpoints where the impedance of the reactor equals the matching resistance that causes the Bode Plot to slope up or down by an order = 6 dB/octave or 20 dB/decade.

A reactor in electronics is defined as either an L inductor or a C capacitor in reactance in ohms as a function of s=jω. In mechanics, it can be a spring + mass and lossy viscous damper as R. In other fields, reactors represent anything that stores energy and resistance which consumes energy (heat).

$$H(s)=_\text{Product sum}\sum_i{_{(n,d)} \dfrac{s-z_i}{s-p_i}}$$

$$s=jω, ~~~ ω=2\pi f$$

Simple functions

  • Poles tend to define attenuation in frequency rolloff as the denominator(s) for frequency s increase.
  • Zeros on top define gain which can be a high pass filter or bandpass on the rising side.

For more detail Read the 13 pages from MIT on Zeros/Poles



enter image description here



Let's look at two 1st order systems:


simulate this circuit – Schematic created using CircuitLab

Then the KCL for these are (left column for the above left-side schematic and right column for the above right-side schematic):

$$ \begin{array}{c|c} \begin{array}{rl} \frac{v_{_\text{OUT}}}{R_1}+\frac{v_{_\text{OUT}}}{R_2}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{OUT}}&=\frac{v_{_\text{IN}}}{R_2} \\\\ \frac{\text{d}}{\text{d}t}v_{_\text{OUT}}+\frac1{\left(R_1\vert\vert R_2\right)C_1}v_{_\text{OUT}}&=\frac1{R_2\,C_1}v_{_\text{IN}} \\\\ \left[D+\frac1{\left(R_1\vert\vert R_2\right)C_1}\right]v_{_\text{OUT}}&=\frac1{R_2\,C_1}v_{_\text{IN}} \end{array} & \begin{array}{rl} \frac{v_{_\text{OUT}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{OUT}}+C_2\frac{\text{d}}{\text{d}t}v_{_\text{OUT}}&=C_2\frac{\text{d}}{\text{d}t}v_{_\text{IN}} \\\\ \frac{\text{d}}{\text{d}t}v_{_\text{OUT}}+\frac1{R_1\left(C_1+ C_2\right)}v_{_\text{OUT}}&=\frac{C_2}{C_1+C_2}\frac{\text{d}}{\text{d}t}v_{_\text{IN}} \\\\ \left[D+\frac1{R_1\left(C_1+ C_2\right)}\right]v_{_\text{OUT}}&=\frac{C_2}{C_1+C_2}\:D\:v_{_\text{IN}} \end{array} \end{array} $$

(Note: Here, \$D=\frac{\text{d}}{\text{d}t}\$, the differential operator with respect to time. Not to be confused with another differential operator use where \$D=\text{d}\$ and implies the product-rule applied with respect to all variables of interest and not just time.)

Now, neither of these are homogeneous.

Suppose they were homogeneous:

$$ \begin{array}{c|c} \begin{array}{rl} {\hphantom{\frac{v_{_\text{OUT}}}{R_1}+\frac{v_{_\text{OUT}}}{R_2}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{OUT}}}}\llap{\left[D+\frac1{\left(R_1\vert\vert R_2\right)C_1}\right]v_{_\text{OUT}}}&={\hphantom{\frac1{R_2\,C_1}v_{_\text{IN}}}}\llap{0} \end{array} & \begin{array}{rl} \hphantom{\frac{v_{_\text{OUT}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{OUT}}+C_2\frac{\text{d}}{\text{d}t}v_{_\text{OUT}}}\llap{\left[D+\frac1{R_1\left(C_1+ C_2\right)}\right]v_{_\text{OUT}}}&=\hphantom{\frac{C_2}{C_1+C_2}\:D\:v_{_\text{IN}}}\llap{0} \end{array} \end{array} $$

Then this means that the brackets surround what's called an annihilator. An annihilator of the form \$\left[D-r\right]\$ represents a solution term of the form \$A\, e^{r t}\$.

(We'll have to return to the concept of annihilators when getting back onto the topic of non-homogeneous equations and zeroes.)

Note that there's no longer a time-dependent input source voltage. Instead, I just set that side to zero (so it's grounded.) So there's only a need to work out the value of \$A\$ based upon initial conditions. Other than that the problems are both solved.

Returning to the earlier non-homogeneous equations, we can re-arrange them so that I can move on to the point about poles:

$$ \begin{array}{c|c} \begin{array}{rl} {\hphantom{\frac{v_{_\text{OUT}}}{R_1}+\frac{v_{_\text{OUT}}}{R_2}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{OUT}}}}\llap{\frac{R_1}{R_1+R_2}\cdot \frac{\frac1{\left(R_1\vert\vert R_2\right)\,C_1}}{D+\frac1{\left(R_1\vert\vert R_2\right)\,C_1}}}&={\hphantom{\frac1{R_2\,C_1}v_{_\text{IN}}}}\llap{\frac{v_{_\text{OUT}}}{v_{_\text{IN}}}} \end{array} & \begin{array}{rl} \hphantom{\frac{v_{_\text{OUT}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{OUT}}+C_2\frac{\text{d}}{\text{d}t}v_{_\text{OUT}}}\llap{\frac{C_2}{C_1+C_2}\cdot\frac{D}{D+\frac1{R_1\left(C_1+ C_2\right)}}}&=\hphantom{\frac{C_2}{C_1+C_2}\:D\:v_{_\text{IN}}}\llap{\frac{v_{_\text{OUT}}}{v_{_\text{IN}}}} \end{array} \end{array} $$

I'd earlier written in comments, "It's poles directly define the homogeneous response." Here, you can now see why I said that. (Note the homogeneous response in the denominator?)

The actual equations developed for the two simple circuits also each involve a nonhomogeneity that must be dealt with. (I've stayed in the time domain, though I'm sure by now you can readily see a distinct similarity with Laplace forms.)

But at least the meaning of pole may have been slightly expanded. They are the homogeneous response.


Just setting \$v_{_\text{IN}}=0\:\text{V}\$ might make the solutions a lot easier. But it's not the case where nonhomogeneities remain because \$v_{_\text{IN}}\ne 0\:\text{V}\$ (and is likely some function of time.)

This also means that the bracketed factors shown earlier are no longer annihilators. (We'll need to fix that problem.)

Obviously, any transfer function zeros must somehow be related to these nonhomogeneities. In fact, the nonhomogeneities/zeroes in the transfer function are directly related to the input signals and external forces acting on the system.

To see that obvious fact, let's return to the earlier equations:

$$ \begin{array}{c|c} \begin{array}{rl} \left[D+\frac1{\left(R_1\vert\vert R_2\right)C_1}\right]v_{_\text{OUT}}&=\frac1{R_2\,C_1}v_{_\text{IN}} \end{array} & \begin{array}{rl} \left[D+\frac1{R_1\left(C_1+ C_2\right)}\right]v_{_\text{OUT}}&=\frac{C_2}{C_1+C_2}\:D\:v_{_\text{IN}} \end{array} \end{array} $$

Clear enough.

Let's also assume that \$v_{_\text{IN}}=V_{_0}\cos\left(\omega\,t\right)\$.

If we knew of an annihilator for \$v_{_\text{IN}}\$ then we could simply apply it to both the left and right sides (to get \$0\$ on the right side.) Then we'd just solve the resulting homogeneous equation.

It turns out that \$\left[D^2+\omega^2\right]\$ is just such an annihilator for \$v_{_\text{IN}}=V_{_0}\cos\left(\omega\,t\right)\$.

(For a glimpse as to how distinct zeroes in the numerator can be converted into gain factors applied to the homogeneous response, see [here](https://electronics.stackexchange.com/a/632840/38098. I'm leaving remaining discussion on the idea of zeroes until I hear more from you and have the time to consider and reply.)


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