# Convolution Sums

1. Feb 1, 2005

### BlkDaemon

I'm an EE student currently taking a Systems & Signals class. I've been searching high and low for information about convolution sums and convolution integrals. (Currently using the Haykin/Van Veen text).

Here's my problem: I'm not grokking how the original input signal morphs into the output signal. My understanding is that the input signal is multiplied by an impulse sequence, and then the result is considered overall input into an additional system, which yields the desired output signal.

Is that right? Am I missing something? And how does one hack that out mathematically? And *then* where does the actual summation come into play?

BlackDaemon

2. Feb 3, 2005

### cyeokpeng

I am not getting your question.

When an input signal x(t) is multiplied by an impulse sequence, we get sampling characteristics of the original analog signal x(t), i.e. a digitized signal at the sampling frequency of your impulse sequence. But what is your question?

3. Feb 3, 2005

### da_willem

When you know the systems response on an impulse you can find out the response to an arbitrary input by taking the convolution of the input with the impuls response.

Eg. when the system just outputs the input, the impulse resonse is an impulse so we get (input x(t), the output y(t))

$$y(t)= x(t)*\delta(t)=\int_{-\infty}^{+\infty} x(t')\delta(t-t')dt'=x(t)$$

For an arbitrary response h(t):

$$y(t)= x(t)*h(t)=\int_{-\infty}^{+\infty} x(t')h(t-t')dt'$$

When dealing with discrete signals the convolution is a sum:

$$y[n]= x[n]*h[n]=\sum_{k=-\infty}^{\infty} x[k]h[n-k]$$

4. Feb 3, 2005

### BlkDaemon

Close, but not quite....

First, thanks for responding.

I understand the theory. Believe me when I say I've spent a lot of time staring at the formulas you shared in your post. My problem is that I can't figure out mathematically how to do it. Maybe it's really simple, and I'm just not getting it.

Consider this bunch of arbitrary values from, let's say, a discrete time function.

say x(1) = 4
and h(x) = x^2

wouldn't y(1) = 16?

Or is this waaaaaay oversimplifying the issue? In the examples I've looked at, I don't understand how they get the terms they end up summing. Should I provide a more concrete example?

Daemon

5. Feb 4, 2005

### da_willem

You're welcome. And in the example I don't see why this would be incorrect. I'll reply more elaborate later today, and maybe you could provide a more concrete example where your idea of thins does not work out...

6. Feb 19, 2005

### Mr Jake

generally you would need to know the impulse response of the filter, so if the impulse response is described as h(k) = x^2, then the impulse response would be
h(k) = [0, 1, 4, 9, 16], now if you have a finite impulse response you would have a limited number of values for h(k).
if all the inputs are known then you can easily calculate the outputs, for example if the input where:
x(n) = [1,2,3,4,5,6,7,8,9]
using the convolution sum
y(0) = 0*1=0
y(1)=0*2+1*1=1
y(2)=0*3+1*2+4*1=6
y(3)=0*4+1*3+4*2+9*6=65
etc...

oh and if you want to describe this filter in the z-domain then it is simpliy
H(z)=z^-2+4*z^-3+9*z^-4+16*z^-5

7. Feb 21, 2005

### mcfetridges

I am also taking a Signals Class right now. When studying for my midterm I stumbled upon this site http://www.jhu.edu/~signals/convolve/ . I found that the actual integral way of convolution is very easy but in some cases such as the convolution of general shapes you have to do it graphically, i.e the flip,shift and integrate. I have some good examples of convolution so if you still stuck email me at sdm8@ualberta.ca

8. Mar 16, 2005

### Antiphon

Take the impluse response of your signal (what the system does if you "ping" it)
and reverse it along time. Take your complicated input signal and lay it on top
of the backwards impulse response. At each point of the two signals, multiply
them together and add. This number you get will be the system output AT ONE TIME T.

You then shift the input signal over one time increment and repeat to get the next
time increment's output.

This explanation is easiest to see in the discrete case, but the continuous case is
the same with the integrals that others have shown above.

There is no "second system" to consider.