Pole is defined as the value at which value of a function becomes infinity.

Let's say I have one LTI system with transfer function H(s)=1/s+1.

How the system will behave when s=-1.

From Bode plot I can see at pole the slope of Bode plot changes at pole frequency.

Can someone clear this thing?

  • \$\begingroup\$ Evaluating H(s) at s = -1 is actually evaluating at an imaginary frequency. The roots of the denominator are the break frequencies where Real = Imag \$\endgroup\$
    – sstobbe
    Commented Jun 23, 2017 at 14:28
  • \$\begingroup\$ What do you mean by a system being 'at its pole'? Also: 'how the system behaves when s=-1', is really vague. \$\endgroup\$
    – Chu
    Commented Jun 23, 2017 at 14:32
  • \$\begingroup\$ The question is VERY vague. What - in your view - is a pole??? There are two possible answers: (1) A complex transfer function H(s) has a "pole" (in your example: s=-1). Mathematically the function will be infinite at s=-1. However, this is a pure ficticious consideration. (2) Another, more realistic view, is the question: What happens with the magnitude/phase response of the function H(s) for a FREQUENCY wo (pole frequency) which is identical to the magnitude of the vector to the (complex) pole location (in your case w=1)? See the figures provided by Andy aka. \$\endgroup\$
    – LvW
    Commented Jun 23, 2017 at 16:19

3 Answers 3


This picture might help: -

enter image description here

Along the top of the picture there are three bode plot examples of the magnitude response for a 2nd order low pass filter. These are just examples that show how the damping ratio (\$\zeta\$) affects the peak of the response.

Bottom left shows the fuller picture where you can see the bode plot and pole zero plot together. Finally, bottom right is the conventional pole zero diagram (as viewed from above in the previous diagram).

So, if you have a pole at -1, that pole exists along the \$\sigma\$ axis and is at a frequency where jw = 0. Because the \$\sigma\$ axis is concerned with damping (al la \$\zeta\$ or its inverse Q/2) the further to the left you travel the more damping there is.

What happens to an electrical system at its pole?

A lot of systems will begin to turn oscillatory as the pole advances towards and aligns with the jw axis. If the pole advances further then almost certainly, a system will become unstable and oscillate.

Can someone clear this thing?

Hopefully this will help or maybe trigger a related question that I should be able to clarify.


How the system will behave when s=-1.

In other words, what will be the output of the system when the input is \$e^{-t}\$?

We usually think of the \$s\$ domain transfer function as

$$H(s) = \frac{Y(s)}{X(s)}$$

where \$Y(s)\$ and \$X(s)\$ are the Laplace transforms of the output and input signals respectively.

But, it's also true that if \$x(t) = e^{st}\$, then \$y(t) = H(s)e^{st}\$. This may not be obvious so here's a quick demonstration. Consider the canonical RC low-pass filter differential equation:

$$\dot{y}(t) + \frac{1}{RC}y(t) = \frac{1}{RC}x(t)$$

Let \$x(t) = e^{st}\$ and then assume that \$y(t) = Ae^{st}\$ (where \$A\$ is a complex constant) such that

$$\dot{y}(t) = sAe^{st} = sy(t)$$

and then the differential equation becomes

$$\left( s + \frac{1}{RC} \right)y(t) = \frac{1}{RC}e^{st}$$

Solving for \$y(t)\$ yields

$$\Rightarrow y(t) = \frac{1}{1 + sRC}e^{st} = H(s)e^{st}$$

This is valid for all \$s\$ except for the pole \$s_o = -\frac{1}{RC}\$. Since the denominator blows up there, we must go back to the differential equation and see what happens. For \$s = -\frac{1}{RC}\$, try instead \$y(t) = Ate^{-t/RC}\$ such that

$$\dot{y}(t) = Ae^{-t/RC} - \frac{1}{RC}Ate^{-t/RC} = \left(\frac{1}{t}-\frac{1}{RC} \right)y(t)$$

and then the differential equation becomes

$$\left(\frac{1}{t}-\frac{1}{RC} +\frac{1}{RC} \right)Ate^{-t/RC} = Ae^{-t/RC} = \frac{1}{RC}x(t) = \frac{1}{RC}e^{-t/RC}$$

Thus \$A = \frac{1}{RC}\$ and so

$$y(t) = \frac{t}{RC}e^{-t/RC}$$

which is of a different form than the input (due to the multiplication by \$t\$). Further, though the output goes to zero as \$t \rightarrow \infty\$, the ratio of the output to the input goes to infinity as we should expect since \$H\left(-\frac{1}{RC}\right)\$ is 'infinite' (undefined actually).

Note: we can't actually have in input \$x(t) = e^{-t/RC}\$ (it is unbounded in the past) but an \$x(t) = u(t)e^{-t/RC}\$ is a bounded input signal. I'll leave the system response to that input as an exercise.


The H transform graph would be a set of two 3D contour graphs, one for the modulo of H and one for its angle.

As you may recall, the permanent response of a system is represented by the graph of H over the y axis, namely, s = j.w.

The modulo of H is indeed infinity for s=-1, but that point is not at the modulo of H graph that is usually analyzed, which is a graph for s=j.w.


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