If we have an impulse response of a circuit which is u(t) and if one has the input and wants to find the output, we use convolution of the input and the impulse response to find the output, that is to my knowledge the use of convolution. If one wants to find the steady-state response to the sinusoidal input such as $5\cos(2t)$, why should we use convolution. $$\mathcal{L}(u(t)* 5\cos(2t))=\mathcal{L}(u(t)) \mathcal{L}(5\cos(2t))$$

Does the steady case- response in this case mean the output? Second, if we have two cascaded circuits and to find the impulse response of the whole circuit, why do we have to convolve them. I would appreciate any answers.


2 Answers 2


Convolution simply gives you the output of a system given an input. In the case of a periodic input, you'll end up with a periodic output. This implies that the steady state response to this input will be a periodic "steady state" response. I put steady state in quotes because it's not really steady in terms of DC analysis. It is steady though in terms of the frequency domain.

To answer your main question succinctly: No (but almost), the steady state response means the output after the initial transient has settled out.

Taking some quotes from wikipedia may make it more clear:
"steady state is an equilibrium condition of a circuit or network that occurs as the effects of transients are no longer important"
"A transient event is a short-lived burst of energy in a system caused by a sudden change of state."

The burst of energy referred to here would be the u(t) that gives the change in state. The periodic function is actually part of the steady state nature of the system and it will affect the transient response and the periodic steady state of the system, but it itself isn't the transient event. It's periodic and so forever in the past and forever in the future will continue to do it's thing which in the frequency domain is definitely steady.

Let me know if that's still confusing. There's some nice pictures on this page showing it better than I ever can in words: https://ccrma.stanford.edu/~jos/fp/Transient_Response_Steady_State.html


The reason convolution is necessary is because the definition of a Laplace transform involves an integral. You can factor constants in and out of it, but you cannot simply distribute the Laplace operator over multiplication. Convolution solves the problem by inserting a dummy variable that (hopefully) goes away, but makes you deal with the fact that Laplace transforms are integrals.

Convolution usually is a part of finding the inverse of the Laplace transform, when you want to go back to the time domain.

  • \$\begingroup\$ The symbol in the LHS refers to convolution. The laplace of two convolved functions is equal to the multiplication of the individual laplace transformed functions. \$\endgroup\$
    – user29568
    Commented May 8, 2014 at 10:03
  • \$\begingroup\$ Noted, answer fixed. I haven't seen that notation in a few years. \$\endgroup\$
    – user39962
    Commented May 8, 2014 at 10:08

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