How Does Convolution Work for Analog Signals?

[N.Saravanan]

Dear Experts,
For convolution to work in any input signal we should be able to represent the input signal in terms of appropriately scaled and shifted unit impulses. This one holds good for discrete time system in which the input signal can be represented as sum of scaled shifted unit impulses. But is it possible to represent an analog input signal as sum of scaled and shifted unit impulse. If so how? Why I ask is unlike in discrete system for which the unit impulse has no width, the practical unit impulse in analog system has negligible width. The unit impulse signal raises to value 1 from 0 in a very short time interval and falls back to zero again. Sum of scaled and shifted unit impulses repeat this action at a faster rate. So if we represent an analog input signal by scaled and shifted unit impulses the representation is actually a signal which touches the zero axis at intermediate intervals. But the original input analog signal need not touch the zero axis. So won't the signal approximation in continuous time produce a distorted input signal. So if we convolve this distorted zero touching input signal will we get the actual response of a system to any input? Kindly please explain the concept.

Thank You,
N.Saravanan.

 

[turin]

I'm not sure that I understand your concern. Suppose that your transfer function has some maximum characteristic frequency (some maximum imaginary pole value, say). Then, I think that if you send the impulses at a much higher frequency than this, the system will smooth them out for you (basically like a low pass filter - every system is a low pass filter at some point). So, even if the input is "jagged", the system cannot resolve the undesired jaggedness anyway, and the output is practically the same as if the ideal analog signal was presented at the input.

That's more of a practical, rather than theoretical, answer to what I think is your concern. There is also a theoretical answer, but I think you were looking for something practical?

 

[elibj123]

In analog signals sum of impulses is actually integration.

Actually using the basic convolution identity you can write:

$$x(t)=\int^{\infty}_{-\infty}x(\tau)\delta(t-\tau)d\tau$$

Which is the equivalent of superposition of discrete impulses:

$$x[k]=\sum^{\infty}_{n=-\infty}x[n]\delta[k-n]$$

The integration in the continuous case smoothes the infinite jumps of the impulse because they are being damped by the infinitesimal length of dt. This is eactly why:

$$\int^{\infty}_{-\infty}\delta(t)dt=1$$

And not infinity or zero.

Actually the representation helps you understand why convolution works for LTI systems:

if h(t) is the response to d(t), then the response to d(t-k) will be h(t-k),
so:

$$y(t)=H{x(t)}=H{\int^{\infty}_{-\infty}x(\tau)\delta(t-\tau)d\tau}=<br /> \int^{\infty}_{-\infty}x(\tau)H{\delta(t-\tau)}d\tau=\int^{\infty}_{-\infty}x(\tau)h(t-\tau)d\tau=(x*h)(t)$$