Let's say we have a system (for example an integrator) whose step response has a negative slope as shown by the MATLAB code below:

clc;clear ;close all
den=[R*C 0]
subplot 211
title('pole zero map for pure integrator, with -1 in numerator')
subplot 212
pzmap(sys2)%plotting poles and zeros
title('pole zero map for integrator, with +1 in numerator')
subplot 211
title('step response for pure integrator, with -1 in numerator')
subplot 212
title('step response for integrator, with +1 in numerator')

sys1 is an ideal integrator and we know that ideal integrators are stable systems.

But what if we replace -1 in the numerator with 1 (sys2): then the step response slope will be positive: is sys2 unstable because of the positive slope?

  • 1
    $\begingroup$ What do you mean by "pure" integrator, an ideal integrator? Anyway, changing the sign of a transfer function will definitely not change its stability properties. $\endgroup$
    – Matt L.
    Commented Sep 7, 2022 at 7:47
  • 1
    $\begingroup$ An ideal integrator with a transfer function of $1/s$ is only marginally stable since it has pole at $s=0$ and infinite gain at DC. That's why the step response of the integrator grow to inifinity $\endgroup$
    – Hilmar
    Commented Sep 7, 2022 at 11:12
  • $\begingroup$ @MattL. please check updated question $\endgroup$
    – DSP_CS
    Commented Sep 7, 2022 at 11:20

1 Answer 1


An ideal integrator is not BIBO-stable, i.e., there are bounded input signals which result in an unbounded output. A step at the input is such a signal.

Clearly, changing the sign of the transfer function doesn't make any difference. In both cases, the system is not BIBO-stable.

As mentioned in a comment, the ideal integrator is only marginally stable because it has a pole on the imaginary axis. This makes it BIBO-unstable. Any DC at the input will result in an output signal with linearly increasing magnitude.


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