While trying to understand control engineering from first principles I came across the following which I cannot yet explain intuitively or mathematically.

What is the relationship, between a step input and an integrator?

Why are they identical to each other?

I kept on seeing \$1/s\$ being used to both represent a step input and an integrator.

  • The Laplace transform of the unit-step function is \$1/s\$.
  • An integrator symbol is also \$1/s\$.

Step Function:

enter image description here

Integrator Block:

enter image description here

Multiplication by s in Frequency (Laplace) domain is differentiation in time.

Dividing by s in Frequency (Laplace) domain is equivalent to integration in time.

Is a step input equivalent to integrating in the time domain, or is it purely coincidental that they both have a spectrum that falls as frequency increases?

Why the Laplace transform of the integral is 1/s?

\$\int\$ in Time Domain = \$ 1/s\$ in Freq Domain


\$\mathscr{L} \{1/s\} = 1\$


If I am understanding the answers correctly, there is not relationship between a step INPUT and an integrator, but there is a relationship between a step FUNCTION and integrator, as explained below.

  • 2
    \$\begingroup\$ The step response is the integral of the impulse response. \$\endgroup\$
    – jonk
    Nov 24, 2018 at 19:33
  • \$\begingroup\$ To answer the question you just added to the bottom: yes. \$\endgroup\$
    – Hearth
    Nov 24, 2018 at 20:27
  • \$\begingroup\$ And the step input signal is the differential of a ramp input signal LOL. \$\endgroup\$
    – Andy aka
    Nov 24, 2018 at 21:52
  • \$\begingroup\$ Just to cap it off.... The impulse response is \$h\left(t\right)=\mathscr{L}^{-1}\{H\left(s\right)\}\$, which is the derivative of the step response, \$h \left( t \right)= \frac{ \text{d} }{ \text{d} t }y_{ \gamma } \left( t \right)\$. So: \$y_{ \gamma } \left(t\right)=\int h \left( t \right) \: \text{d}t\$. \$\endgroup\$
    – jonk
    Nov 24, 2018 at 21:58
  • \$\begingroup\$ Just let's remind everyone what this question is: What is the relationship if any, between a step input and an integrator?. The key word here is INPUT and not output. \$\endgroup\$
    – Andy aka
    Nov 25, 2018 at 9:39

2 Answers 2


Consider what happens when an integrator gets a unit impulse as its input. What is the output waveform? What's the Laplace transform of a unit impulse?

  • \$\begingroup\$ When an integrator gets a unit impulse, according to my simulink the output rises from 0 to 1 over a 1 second duration and then remains constant at 1. The Laplace transform of a unit impulse is 1, but I don't intuitively see any connection. \$\endgroup\$
    – rrz0
    Nov 24, 2018 at 19:21
  • 2
    \$\begingroup\$ A discrete impulse is different from a really-o truly-o Dirac delta functional, which is an "impulse" in continuous-time signal processing. \$\delta(t)\$ has no width, is zero everywhere except for \$t=0\$, and \$\int_{0^-}^{0^+} \delta(t) dt = 1\$. If you haven't seen it before and it's boggling your mind -- relax, you have company. \$\endgroup\$
    – TimWescott
    Nov 24, 2018 at 20:44
  • 2
    \$\begingroup\$ If you do the math to answer the question "what is the impulse response of an integrator?" then you will be the width of a Dirac impulse away from understanding the answer to your question. \$\endgroup\$
    – TimWescott
    Nov 24, 2018 at 20:50
  • 1
    \$\begingroup\$ The Dirac impulse is indeed the relationship. This contradicts the answer of Andy aka does it not? \$\endgroup\$
    – rrz0
    Nov 24, 2018 at 21:38
  • 1
    \$\begingroup\$ @Rrz0 I never let contradicting Andy stand in the way of my posting the correct answer. \$\endgroup\$
    – Neil_UK
    Nov 25, 2018 at 5:38

enter image description here

What is the relationship if any, between a step INPUT and an integrator?

For the purpose of reminding folk what this question is about I have edited the quote above (from the OP) to highlight the word INPUT. Folk seem to be reading response (or output) instead and that is somewhat baffling.

Consider this:

  • White noise has a spectrum of "x" at all frequencies i.e. it is spectrally flat
  • A perfect amplifier with a gain of "x" has a transfer function of "x" at all frequencies.

Does anyone get in a muddle about this? Do they have a relationship?

So, a unit step has a spectrum that falls as frequency increases and an integrator also has a transfer function that happens to do the same. Should this be a big deal?

And a final reminder about the question asked: -

  • 4
    \$\begingroup\$ While it's true that there doesn't have to be a relationship between two things, it can often be enlightening to consider why two things look similar. I don't disagree with your answer, but I think it might be too dismissive; this is, after all, the type of question that leads to great insight in mathematics. Honestly, after reading your answer, now I find myself thinking about whether there is some mathematical relationship between a perfect amplifier and white noise. \$\endgroup\$
    – Hearth
    Nov 24, 2018 at 18:48
  • \$\begingroup\$ @Felthry I insist you make this an answer!! \$\endgroup\$
    – Andy aka
    Nov 24, 2018 at 18:49
  • \$\begingroup\$ I would, but it doesn't answer the question! And I'm not currently able to write it up into a form that does so. \$\endgroup\$
    – Hearth
    Nov 24, 2018 at 18:51
  • 2
    \$\begingroup\$ Thanks for the answer. It feels as though I am still missing something fundamental, which I can't yet explain. It could be because I don't fully understand your last sentence. @Felthry I think that would actually be very helpful. \$\endgroup\$
    – rrz0
    Nov 24, 2018 at 18:57
  • \$\begingroup\$ @Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent? \$\endgroup\$
    – rrz0
    Nov 24, 2018 at 21:16

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