1. ## Convolution in depth

If we have two functions: $f(t)$ and $g(t)$.

When taking the convolution we replace t with $\tau$ let's say for $f(t)$, so we didn't change it.
The fun ends when for the other function we write $g(t-\tau)$. If we have on the abscissa now $\tau$, so when we add the minus sign inside then we reflect the function across the $x = 0$.

Now we must move the function to the far left. So if I understand correctly we write $\infty$ instead of $t$ when integrated, because we move it to the left. Because if we have function $h(x+\infty)$ then we move it to the $\infty$ in the left direction.

But what about $\tau$?

2. ## Re: Convolution in depth

You are taking two functions

$f(\tau),~g(t-\tau)$

and integrating their product for $\tau \in (-\infty, \infty)$

what is $g(t-\tau)=g((-\tau)+t)$ ?

It is the time reversed version of $g(\tau)$ that has been shifted to the left by $t$

So to obtain the convolution of $f(t),~g(t)$ at time $t$ what you do is

i) take $g$, flip it with respect to the time axis

ii) shift it to the left by $t$ units

iii) multiply it times $f$ unaltered

iv) integrate that product across all time

In this process $\tau$ is the variable of integration, and that integration is performed across all $\tau \in (-\infty, \infty)$

Further, the delay $t$, can also take any value $t \in (-\infty, \infty)$

3. ## Re: Convolution in depth

Originally Posted by romsek

iv) integrate that product across all time
I think, you made a mistake here. We integrate across all $\tau$

So if we move it by t units to the left, how do we know how much, because the two functions can overlap.

4. ## Re: Convolution in depth

Originally Posted by Nforce
I think, you made a mistake here. We integrate across all $\tau$

So if we move it by t units to the left, how do we know how much, because the two functions can overlap.
$\tau$ is a time variable.

The functions certainly will overlap, otherwise their product would be zero.

I'm not sure I really understand what you are asking.

5. ## Re: Convolution in depth

Yes, we start with the product zero, then it's something nonzero and finally again zero. (if the functions are not overlaping).

The correct result is something nonzero.

I am asking how do we choose correctly t, such that time shift will be exactly at the start of first function and second function.

6. ## Re: Convolution in depth

Originally Posted by Nforce
Yes, we start with the product zero, then it's something nonzero and finally again zero. (if the functions are not overlaping).

The correct result is something nonzero.

I am asking how do we choose correctly t, such that time shift will be exactly at the start of first function and second function.
why?

suppose your functions have no starting point?

Can you give an example of the sort of convolution you are trying to do?

7. ## Re: Convolution in depth

I am just visualizing like this: (From wikipedia)

On wikipedia, it's written that we start sliding at $t$ $-\infty$ and we go to $\infty$.

But if we insert $-\infty$ here $g(t-\tau)$. We get $g(-\infty - \tau)$. Where we move to the right not to the left.

8. ## Re: Convolution in depth

Originally Posted by Nforce
I am just visualizing like this: (From wikipedia)

On wikipedia, it's written that we start sliding at $t$ $-\infty$ and we go to $\infty$.

But if we insert $-\infty$ here $g(t-\tau)$. We get $g(-\infty - \tau)$. Where we move to the right not to the left.
what you need to do is let $g(\tau) = rect(\tau)$ as shown and write down $g(t-\tau)$ for some examples of $t$

These will all be $rect()$ functions but now centered at $t$ instead of $0$.

It's not until $t=-1$ that $f(\tau)$ and $g(t-\tau)$ overlap and start to produce a non-zero integral.

Then on the other side at $t=1$, $g(t-\tau)$ slides past $f(\tau)$ and the integral resumes being $0$ for $t>1$

9. ## Re: Convolution in depth

Yes, I see that from the picture.

But forget the picture at the moment, and try to answer this:

On wikipedia, it's written that we start sliding at $t -\infty$ and we go to $\infty$.

But if we insert $-\infty$ here $g(t-\tau)$. We get $g(-\infty - \tau$). Where we move to the right not to the left.
Because this is the part I don't understand.

10. ## Re: Convolution in depth

Do you understand what am I asking?

11. ## Re: Convolution in depth

Originally Posted by Nforce
Do you understand what am I asking?
No, not really.

I think you are having trouble understanding the difference in the roles that $t$ and $\tau$ play in this whole thing but I'm not sure.

When you say $g(-\infty -\tau)$ "moves to the left" I really have no idea what you mean.

12. ## Re: Convolution in depth

We are integrating between $-\infty$ and $\infty$. We insert $-\infty$ instead of $t$. When inserting the value, the graph of the function moves to the right. In to infinity to the right to be exact.

13. ## Re: Convolution in depth

Originally Posted by Nforce
We are integrating between $-\infty$ and $\infty$. We insert $-\infty$ instead of $t$. When inserting the value, the graph of the function moves to the right. In to infinity to the right to be exact.
In the case of their demo $g(t-\tau)$ is a $rect()$ function centered at $t$

It is a function of $\tau$ with a parameter $t$

When finding the convolution like they show you plunk down a $rect()$ centered at $t$ and integrate the product over all $\tau$ This only has non-zero value where the functions overlap.

to obtain the convolution as a function of time $t$, you repeat this process for all $t \in \mathbb{R}$

What they show in the demo is the $rect()$ function moving to the right as $t$ increases from $-\infty$ to $\infty$

The confusion might stem from the fact that they are using the same axis to display both $t$ and $\tau$ on.

Maybe just work some examples by hand until you are comfortable with what's going on.