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I am looking at the description of LTI systems in the time domain.

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".

Why is it not that way? Could such a "system function" be found for, say, an R-C-Circuit?

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.

http://en.wikipedia.org/wiki/LTI_system_theory

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    $\begingroup$ I think that the key difference between the transformations $x \mapsto f \circ x$ and $x \mapsto h * x$ is that the output from the former depends only on the instantaneous behaviour of $x$—one can determine the response *right now* by knowing only the input *right now*—whereas the former allows the behaviour of $x$ at all (past) times to have an effect on the present response. The latter behaviour is what one expects out of a real-world circuit. Of course, this doesn't say why convolution (as opposed to any other integral transform—they all exhibit this sort of behaviour) is the ‘right’ $\endgroup$
    – LSpice
    Commented Jun 21, 2011 at 20:47
  • $\begingroup$ thing to do. For that, I rather like Terry Tao's explanation (mathoverflow.net/questions/5892/what-is-convolution-intuitively/…) of convolution as a ‘blurring’ process; but my physics is too weak to know whether one can draw any sort of sensible connection between the optics he describes there and the circuit behaviour in which you're interested. $\endgroup$
    – LSpice
    Commented Jun 21, 2011 at 20:48

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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.

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  • $\begingroup$ This (together with khanacademy.org/video/… & a variety of other documents) helped me actually understand what convolution is. -- Thanks a lot! -- // Because an R-C-Circuit can only be described by dirac- or step-function response or an ODE in the time domain, my quest for a "system function" f(x) must be futile, it seems. Correct? $\endgroup$
    – PFiver
    Commented Jun 22, 2011 at 1:05
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    $\begingroup$ @NW Patrick, as indicated in my comment above, any response described by a ‘system function’ yields an output $y$ for each input $x$ such that $y(t)$ depends only on $x(t)$. For example, the constant signal $x : t \mapsto 1$ will produce the same response at time $t = 1$ as the ‘ramp’ signal $y : t \mapsto t$. Unless you've got a very special system, this just won't happen. $\endgroup$
    – LSpice
    Commented Jun 22, 2011 at 1:17
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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.

Of course, one can also approach this from the time domain using approximation by step functions etc.

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.)

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  • $\begingroup$ Thanks a lot. The answer is not that useful to me because I am not familiar with the term "input-output operator". I find $f$ a confusing choice for the name of a constant here. Furthermore I'd prefer to look at the problem without changing to the frequency domain. As to the "system function": How would it look like? $\endgroup$
    – PFiver
    Commented Jun 22, 2011 at 0:17
  • $\begingroup$ The input-output operator is just a fancy name for the function that maps the input to the output. The use of $f$, which stands for frequency, is fairly standard in this area. One very important reason to work frequency domain is important is that it is the analogue of diagonalizing an operator. The convolution is fairly messy to compute in time but in the frequency domain, it is just a multiplication operator. Also, in many applications, it is the frequency domain that is important. For example, an ideal lowpass filter would be specified by $H(f)=1$ for $|f|\leq f_0$ and $0$ elsewhere. $\endgroup$
    – Steve
    Commented Jun 22, 2011 at 1:29
  • $\begingroup$ All right. I was too locked in to use $f$ as the $f$unction letter. Also, our prof has been using $\omega$, the circular frequency, which equals to $2\pi f$, mostly. -- I am aware of the advantages of doing things in the frequency domain but I'd nevertheless like to see the mess I'm avoiding, just once, to have an better understanding of what I'm doing $\endgroup$
    – PFiver
    Commented Jun 22, 2011 at 13:14

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