Convolution Integral Properties

  • #1
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Homework Statement



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

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

Homework Equations



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

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

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? :confused:

Any help would be greatly appreciated.
 

Answers and Replies

  • #2
15,756
14,024

Homework Statement



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

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

Homework Equations



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

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

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? :confused:

Any help would be greatly appreciated.
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
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12
From definition I get:

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

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
15,756
14,024
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?

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
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12
Thank you so much, it makes perfect sense now.
 
  • #6
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I have one more question:

Since we finally had:

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

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
Samy_A
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I have one more question:

Since we finally had:

$$\int^{\infty}_{-\infty} \left( \left( \int^{\tau}_{-\infty} f (\sigma) d \sigma \right) g(t-\tau) \right) d \tau =\int^{\infty}_{-\infty} f (\tau) g (t - \tau) d \tau$$
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:
  • #8
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12
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##.

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
Samy_A
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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.
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.
With this substitution the expression of fresh_42 becomes:

$$(F*g) (t) = \int^{\infty}_{-\infty} \left( \left( \int^{t}_{-\infty} f (\underbrace{\sigma-\tau+t}_{\psi}) d \sigma \right) g(t-\tau) \right) d \tau$$
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.
 
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  • #10
Samy_A
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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)
 
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  • #11
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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
Samy_A
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Thank you for pointing out the property of 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:

$$(F*g) (t) = \int^{t}_{-\infty} \left( \int^{\infty}_{-\infty} f (\psi+\tau-t) d \tau \right) d \psi . \int^{\infty}_{-\infty} g(t-\tau) d \tau$$
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##.
 
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  • #13
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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
Samy_A
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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?
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
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Thank you very much for your help.
 

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