# Convolution Integral Properties

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1. Feb 13, 2016

### roam

1. The problem statement, all variables and given/known data

Either by using the properties of convolution or directly from the definition, show that:

If

$$F(t)=\int^t_{-\infty} f (\tau) d \tau$$

then

$$(F * g) (t) = \int^t_{-\infty} (f * g) (\tau) d \tau$$

2. Relevant equations

The convolution of $f$ with $g$ is given by:

$$(f*g)(t) = \int^{\infty}_{-\infty} f (\tau) g(t- \tau) d \tau$$

3. The attempt at a solution

From definition:

$$(F * g) (t) = \int^{\infty}_{-\infty} \left( \left( \int^t_{-\infty} f (\tau) d \tau \right) . g(t-\tau) \right) d \tau$$

Switching the order of integration:

$$= \int^{t}_{-\infty} \left( \int^{\infty}_{-\infty} f (\tau) d \tau \right) d \tau . \int^{\infty}_{-\infty} g (t - \tau) d \tau$$

So, from here how can I show that this is equal to $\int^t_{-\infty} (f*g) (\tau) d \tau$?

Because I must show that the expression above is equal to:

$$(F * g)(t) = \int^t_{-\infty} (f*g) (\tau) d \tau = \int^t_{-\infty} \left( \int^{\infty}_{-\infty} f (\tau) g(t-\tau) d \tau \right) d \tau$$

What properties of the convolution integral do I need to use?

Any help would be greatly appreciated.

2. Feb 13, 2016

### Staff: Mentor

From definition I get:

$$(F * g) (t) = \int^{\infty}_{-\infty} F (\tau) g(t- \tau) d \tau = \int^{\infty}_{-\infty} \left( \left( \int^\tau_{-\infty} f (\sigma) d \sigma \right) g(t-\tau) \right) d \tau$$

3. Feb 13, 2016

### roam

Did you change the dummy variable $\tau$ to $\sigma$? So shouldn't we then change all the $\tau$'s:

$$\int^{\infty}_{-\infty} \left( \left( \int^\sigma_{-\infty} f (\sigma) d \sigma \right) g(t-\sigma) \right) d \sigma$$

How does changing the variable help?

4. Feb 13, 2016

### Staff: Mentor

That's not what I wrote. I only replaced the expressions by their definitions. The 'dummy variable' I've changed because the variable of the outer integral is the upper bound for the variable of the inside integral. (Maybe ψ is better to see.)

$$(F * g) (t) = \int^{\infty}_{-\infty} F (\psi) g(t- \psi) d \psi = \int^{\infty}_{-\infty} \left( \left( \int^{\psi}_{-\infty} f (\sigma) d \sigma \right) g(t-\psi) \right) d \psi$$

5. Feb 13, 2016

### roam

Thank you so much, it makes perfect sense now.

6. Feb 14, 2016

### roam

I have one more question:

$$\int^{\infty}_{-\infty} \left( \left( \int^{\tau}_{-\infty} f (\sigma) d \sigma \right) g(t-\tau) \right) d \tau =\int^{\infty}_{-\infty} f (\tau) g (t - \tau) d \tau = (f*g) (t)$$

How do we write this as $\int^t_{-\infty} (f*g) (\tau) d \tau$? Because I think ultimately this is what the question asks.

Thank you.

7. Feb 14, 2016

### Samy_A

The last integral should be $$\int^{\infty}_{-\infty} F(\tau) g (t - \tau) d \tau =F*g(t)$$
so you are back where you started.

To get $f*g$ in the expression given by fresh_42, $$\int^{\infty}_{-\infty} \left( \left( \int^{\tau}_{-\infty} f (\sigma) d \sigma \right) g(t-\tau) \right) d \tau$$ you may try a substitution, and then switching the order of integration.
Your goal is to get to $$\int^t_{-\infty} (f * g) (\tau) d \tau$$
For that you need to get the $t$ in the integration limit, so a substitution should transform $\sigma=\tau$ into $\psi=t$.
A substitution achieving this is replacing $\sigma$ by $\psi=\sigma-\tau+t$.

Last edited: Feb 14, 2016
8. Feb 15, 2016

### roam

Could you please explain why you chose the substitution $\psi=\sigma-\tau+t$? I'm really confused here as I'm not sure how it would help.

With this substitution the expression of fresh_42 becomes:

$$(F*g) (t) = \int^{\infty}_{-\infty} \left( \left( \int^{t}_{-\infty} f (\underbrace{\sigma-\tau+t}_{\psi}) d \sigma \right) g(t-\tau) \right) d \tau$$

I am not sure what properties we need to manipulate the expression above to get to:

$$\int^t_{-\infty} \left( \int^{\infty}_{-\infty} f (\tau) g(t- \tau) d \tau \right) d \tau = \int^t_{-\infty} (f * g) (\tau) d \tau$$

Any explanation would be really appreciated.

9. Feb 15, 2016

### Samy_A

The upper integration limit for $\sigma$ is $\tau$. In your expected end result, there is a $t$ as upper integration limit.
So my suggested substitution is purely opportunistic: if we set $\psi=\sigma-\tau+t$, then if $\sigma=\tau$, $\psi=t$. That's all really. I'm not saying that I was sure before doing the whole computation that that would help (it really does), but certainly it is a substitution worth the effort.
This is not correct, you get
$$(F*g) (t) = \int^{\infty}_{-\infty} \left( \left( \int^{t}_{-\infty} f (\underbrace{\psi+\tau-t}_{\sigma}) d \psi \right) g(t-\tau) \right) d \tau$$
If you want to rename the integration variable from $\psi$ to $\sigma$, fine, but then you get
$$(F*g) (t) = \int^{\infty}_{-\infty} \left( \left( \int^{t}_{-\infty} f (\sigma+\tau-t) d \sigma \right) g(t-\tau) \right) d \tau$$

In the expected end result, the integral with $t$ as upper limit is the outside integral. So switch the integration order to get that upper limit where you want it to be.
Then look at what you get. If you don't see it immediately (I didn't), a last substitution $\mu=t-\tau$ will help.

Last edited: Feb 15, 2016
10. Feb 15, 2016

### Samy_A

Notice that (not a hint, just a remark):
$\displaystyle F(t)=\int^t_{-\infty} f (\tau) d \tau$ implies $F'(t)=f(t)$.
$\displaystyle (F * g) (t) = \int^t_{-\infty} (f * g) (\tau) d \tau$ implies $(F * g)' (t) =f*g(t)=F'*g(t)$.

So what you are proving in this exercise is the following interesting property of the convolution:
$$(F*g)'(t)=F'*g(t)$$

(provided the functions are sufficiently nice: https://en.wikipedia.org/wiki/Convolution#Differentiation)

Last edited: Feb 15, 2016
11. Feb 15, 2016

### roam

Thank you for pointing out the property showing the relationship with differentiation.

But how does the $u=t-\tau$ substitution help at the end?

Because even if we switch the order of integration, the integral with t as the upper limit does not operate on the second term:

$$(F*g) (t) = \int^{t}_{-\infty} \left( \int^{\infty}_{-\infty} f (\psi+\tau-t) d \tau \right) d \psi . \int^{\infty}_{-\infty} g(t-\tau) d \tau$$

$$=\int^{t}_{-\infty} \left( \int^{\infty}_{-\infty} f (\tau) d \tau \right) d \psi . \int^{\infty}_{-\infty} g(t-\tau) d \tau$$

12. Feb 15, 2016

### Samy_A

This is not correct.
From $$(F*g) (t) = \int^{\infty}_{-\infty} \left( \left( \int^{t}_{-\infty} f (\psi+\tau-t) d \psi \right) g(t-\tau) \right) d \tau$$
switching the order of integration gives:
$$(F*g) (t) = \int^{t}_{-\infty} \left( \int^{\infty}_{-\infty} f (\psi+\tau-t) g(t-\tau) d \tau \right)d \psi$$
When you first integrate over $\tau$, all the terms involving $\tau$ must stay in the interior (first evaluated) integral.

The interior integral should look familiar, possibly after substituting $\mu=t-\tau$.

Last edited: Feb 15, 2016
13. Feb 15, 2016

### roam

Oops, thank you. After the $\mu=t-\tau$ substitution:

$$(F*g) (t) = \int^{t}_{-\infty} \left( \int^{\infty}_{-\infty} f (\psi - \mu) g(\mu) d \mu \right)d \psi$$

And it follows from commutativity:

$$\int^{t}_{-\infty} (g*f) (\psi) d \psi = \int^{t}_{-\infty} (f*g) (\psi) d \psi$$

Is it okay then to change the dummy variable from $\psi$ to $\tau$? Because in the argument of f in the first equation we have $\sigma(=\sigma-\tau+t-t+\tau)$, so we can choose $\sigma$ to be any variable?

14. Feb 15, 2016

### Samy_A

Yes, the name of the variable you integrate over is totally inconsequential. The only element that you have to keep as is in this exercise is the $t$ (because it appears outside the integral too). All the others: play with their names as much as you like (as long as when you replace a variable name by another, you do it everywhere in the integral, including the integral limits).

15. Feb 16, 2016

### roam

Thank you very much for your help.