# Convolution Help

#### jp1390

Joined Aug 22, 2011
45
Okay, so after reading my textbook a billion times and scouring previously made thread topics and online resources, I still find myself stumped as to what the application of Convolution is.

For instance, using the Laplace transform, we are able to take a time signal and convert it into its frequency domain counterpart to avoid differential equations, but what do we get with Convolution?

I keep on hearing that Convolution in the time domain is like multiplication in the frequency domain. That's great, but what value does using Convolution have?

I have seen a lot of animations of a square pulse travelling over two signals overlapped, but I don't know its significance.

Thanks,
JP

#### steveb

Joined Jul 3, 2008
2,436
... what value does using Convolution have?
Value is in the mind of the beholder, but there are many applications for covolution calculations.

In the context of what you are talking about, convolution gives you a choice to do calculations in an easy way, in the time domain. The output of a system is a convolution of the input with the impulse response of that system. Hence, if you know the impulse response, the output can be calculated in a straightforward way, either with formulas, or numerically by computer.

Yes, you might well prefer to do you calculations with transforms, to and from the frequency domain, but some real-world problems are better solved in the time domain due to numerical issues.

In any event, good understanding of linear system theory, including Laplace and z-transforms, is critically dependent on the conceptual understanding of convolution. So, coming to grips with the concepts is well worth the time, as you will carry this understanding with you for life, whether or not you actually need it for a particular calculation.

#### t_n_k

Joined Mar 6, 2009
5,455
That's great, but what value does using Convolution have?
Enhancement of medical imaging techniques for one.

#### jp1390

Joined Aug 22, 2011
45
Thank you for the responses!

Okay, so what I am getting at is that this convolution is what the frequency domain calls frequency response H(w)?

Where:

$$Y(w) = X(w)H(w)$$ in the frequency domain and

$$y(t) = x(tau)*h(t-tau)$$ in the time domain?

My question is, are these methods equivalent say for an electrical system in finding output responses from inputs and known frequency response characteristics? For instance, if I had an input x(t) and converted to frequency domain to X(w) and I knew the frequency response H(w) and found Y(w), then converted back to y(t) in the time domain, could I have received the same result if I convoluted x(t) with h(t)?

Also, why is there a delay (t - tau) for convolution? I understand that the one function is sweeping through the other to find overlaps, but with that delay shift, it seems like they would never arrive at the same time.

#### steveb

Joined Jul 3, 2008
2,436
Thank you for the responses!

Okay, so what I am getting at is that this convolution is what the frequency domain calls frequency response H(w)?
Not precisely. The corresponding entity to H(w) in the time domain is h(t). As you said before, there is a correspondence of convolution in the time domain and multiplication in the frequency domain. I think you understand this, but the terminology is important.

My question is, are these methods equivalent say for an electrical system in finding output responses from inputs and known frequency response characteristics? For instance, if I had an input x(t) and converted to frequency domain to X(w) and I knew the frequency response H(w) and found Y(w), then converted back to y(t) in the time domain, could I have received the same result if I convoluted x(t) with h(t)?
Yes, at least in principle. In practice, you can run into numerical issues and one method may be preferred over the other depending on the data you have. But, mathematically, they are exactly the same.

Also, why is there a delay (t - tau) for convolution? I understand that the one function is sweeping through the other to find overlaps, but with that delay shift, it seems like they would never arrive at the same time.
Understanding this requires several steps. I can outline the steps, but you will likely need to consult other sources to fill in the details, if you are not already up to speed on them (perhaps you already are).

1. The first concept to understand is the meaning of an impulse function in both continuous time (Dirac delta function) and discrete time (Kronecker delta function). The continuous time case is a little more challenging to come to grips with because it is the limit of a very narrow and tall pulse with an area of one, at time t=0. The discrete time function is simply the value of 1 at n=0, and zero for all other n, integers from minus infinity to plus infinity.

2. The impulse function is indicated as $$\delta (t)$$ and $$\delta [n]$$ when the pulse occurs at time zero (t=0 or n=0). A shifted impulse function is indicated as $$\delta (t-\tau)$$ and $$\delta [n-k]$$ when it occurs at a later time.

3. An input signal x(t) or x[n] can be represented as a weighted integral or as a weighted sum of delayed impulse functions, respectively.

4. In a linear system the output can be considered to be the integral or the summation (respectively) of the responses to all of the delayed and scaled impulse functions. This is valid because of the principle of superposition.

Once you understand all of the above, it's clear why the impulse response functions are shifted, and why they do indeed "arrive" at different times. Each piece of the input signal (represented as a scaled and delayed impulse function) starts at a different time and triggers an impulse response function in the output. The total output is then the total sum of all of these overlapping impulse response functions, with each one beginning and ending at different times. Note that some impulse response functions go on forever and never end, but causality requires that they start at a particular time and can not exist previous to the triggering input impulse.

Yes, they did. Hopefully, my answers made sense.