Why is the output of an LTI system the convolution of the input funtion and the impulse response? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T07:02:45Z http://mathoverflow.net/feeds/question/68428 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68428/why-is-the-output-of-an-lti-system-the-convolution-of-the-input-funtion-and-the-i Why is the output of an LTI system the convolution of the input funtion and the impulse response? NW Patrick 2011-06-21T19:50:01Z 2011-06-22T00:15:22Z <p>I am looking at the description of LTI systems in the time domain.</p> <p>Intuitively, I'd have guessed it would be the composition of the input function and some "system function". $$y(t) = f(x(t)) = (f\circ x)(t)$$ Where $x(t)$ is the input, $y(t)$ output and $f(x)$ a "system function".</p> <p>Why is it not that way? Could such a "system function" be found for, say, an R-C-Circuit?</p> <p>The actual output function y(t), is defined as $$y(t) = (h * x)(t)$$ Where $h(t)$ is the response to a dirac impulse. This is hard to grasp for me. Why is it so? I have looked at various explanations, drawings of rectangles becoming infinitely narrow, which I sort of understood, but it is still "hard to grasp"! I am looking for a simple explanation in one or two sentences here.</p> <p><a href="http://en.wikipedia.org/wiki/LTI_system_theory" rel="nofollow">http://en.wikipedia.org/wiki/LTI_system_theory</a></p> http://mathoverflow.net/questions/68428/why-is-the-output-of-an-lti-system-the-convolution-of-the-input-funtion-and-the-i/68453#68453 Answer by Steve for Why is the output of an LTI system the convolution of the input funtion and the impulse response? Steve 2011-06-21T23:11:50Z 2011-06-21T23:11:50Z <p>One way to approach this through the frequency domain. (Assume all the conditions about continuity of the system to make the following work.) Let $L$ denote the input-output operator. For each $f\in\mathbb R$, let $e_f(t)=exp(2\pi ift)$. Then it is not hard to see that $Le_f=H(f)e_f$ for some constant $H(f)$, the frequency response at $f$. (The $e_f$'s are eigenfunctions, if you like.) Then using the inverse Fourier transform, $$x(t)=\int \hat x(f) e^{2\pi ift} df$$ and the linearity of $L$ to obtain $$(Lx)(t)=\int \hat x(f) H(f) e^{2\pi ift} df,$$ which is the inverse transform of the product $\hat x H$. So $Lx$ must be the convolution of $x$ and $h$, the inverse Fourier transform of $H$. This is one reason why the Fourier transform and its relatives work so well with linear time-invariant systems, tying the impulse response to the frequency response. Many systems, especially in signal processing, are specified by the frequency response.</p> <p>Of course, one can also approach this from the time domain using approximation by step functions etc.</p> <p>The system function for an RC circuit is easy to find either in the time domain (solving an ODE) or in the frequency domain (using the impedances of the components and simple algebra.)</p> http://mathoverflow.net/questions/68428/why-is-the-output-of-an-lti-system-the-convolution-of-the-input-funtion-and-the-i/68463#68463 Answer by Sridhar Ramesh for Why is the output of an LTI system the convolution of the input funtion and the impulse response? Sridhar Ramesh 2011-06-22T00:08:56Z 2011-06-22T00:15:22Z <p>Well, it's just as simple as this: the output at any moment reflects the effect of the input at just that moment, plus the lingering effect after one second of the input from one second before, plus the lingering effect after two seconds of the input from two seconds before, plus the lingering effect after three seconds of the input from three seconds ago, etc. [And half a second and 2.8 seconds and so on as well...]. That is, the sum (or, rather, integral) over all t of "The input from t seconds ago" * "The amount of lingering effect a unit of input contributes after t seconds". That is, precisely a convolution of the input signal and the impulse response.</p>