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Fourier transform/Convolution theorem

  • Thread starter Firepanda
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  • #1
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Ok, so first we need to find h(u).

By letting

h(u) = Integral -1 to 1 of (1/2)*g(u-x) dx

Then we can change the limits about by setting u = 2x

so now we have:


h(u) = Integral -2 to 2 of (1/4) du

so h(u) = 1

and I find the fourier transform of this between -2 and 2 and I don't get sin^2(k) / k^2

Can anyone help me here?

Thanks
 

Answers and Replies

  • #2
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That's confussing to read. k is being used as the transform variable which is unusual. I'll use "z" below. How about I re-word it:

If I have the function f(x), then the Fourier transform of f(x), which will be a function of the Fourier integral variable "z" say and not k is:

[tex]F\left\{f(x)\right\}=\int_{-\infty}^{\infty} f(x)e^{izx}dx=\hat{f}(z)[/tex]

and if I have another function h(x) defined by:

[tex]h(x)=\int_{-\infty}^{\infty} g(x-y)f(y)dy[/tex]

then by the convolution theorem, the Fourier transform of h(x) which will be a function of z will be:

[tex]F\left\{h\right\}=\hat{g} \hat{f}[/tex]

So for your g,

[tex]F\left\{g\right\}=\int_{-\infty}^{\infty}g(x)e^{izx}dx=1/2\int_{-1}^{1}e^{izx}dx[/tex]

Ok, now you have the transform of g which is a function of z but if you have to, call it a function of k then. Now use the convolution theorem to find the transform of g*g.
 
  • #3
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Hi Jackmell, thanks for the reply

I believe you're doing the 2nd part? Verifying the solution via the convolution theorem.

What I need first is to show the result of the transform of h(u), thus I need to find what h(u) is.
 
  • #4
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What I need first is to show the result of the transform of h(u), thus I need to find what h(u) is.
I don't think that's gonna' happen dude. Don't see how anyway. Maybe I'm wrong but I think the way I said is the way to go. Maybe others can confirm the matter for us though.

Edit:

Ok, I might be wrong about that. How about we consider the double integral:

[tex]\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} g(x)g(u-x)e^{iwu}dudx=\int_{-\infty}^{\infty}g(x)\int_{-\infty}^{\infty} g(u-x)e^{iwu}dudx=\int_{-\infty}^{\infty}g(x)F\left\{g(u-x)\right\}dx[/tex]

Then maybe use some shifting theorem to finish it off.
 
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  • #5
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Hi Firepanda! :smile:

Let's pick u=10.
Since we have x between -1 and +1 in your integral (before your substitution).
What would f(u-x)=f(10-x) be?

In other words, I think something went wrong with your substitution.
h(u)≠1.
 
  • #6
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Hi Firepanda! :smile:

Let's pick u=10.
Since we have x between -1 and +1 in your integral (before your substitution).
What would f(u-x)=f(10-x) be?

In other words, I think something went wrong with your substitution.
h(u)≠1.
hmm, do you mean what the range of f(10-x) would be?
 
  • #7
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hmm, do you mean what the range of f(10-x) would be?
Yes...
 
  • #8
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Ok, sorry I'm not too confident with all of this

would it be 9<x<11

so the range is still 2, just shifted. So am I right in thinking my subsitution should be.. something like u = x+1 or u = x-1?
 
  • #9
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Sorry. I said f(u-x), but I meant g(u-x), since that is the one you need for h(u).

What is the value of g(10-x) if 10-x is between 9 and 11?
 
  • #10
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Sorry. I said f(u-x), but I meant g(u-x), since that is the one you need for h(u).

What is the value of g(10-x) if 10-x is between 9 and 11?
0 I assume by the definition of g
 
  • #11
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0 I assume by the definition of g
Yes. So what will h(10) be?

As for your substitution, I recommend y=u-x.
 
  • #12
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h(10) = 0.5

so my integral is

h(u) = Integral -(y+1) to (y-1) of (1/4) du ?
 
  • #13
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h(10) = 0.5
Let's revisit your formula for h(u).
[tex]h(u)=\int_{-\infty}^\infty g(u-x)g(x) dx[/tex]
You already reduced this to:
[tex]h(u)=\int_{-1}^1 g(u-x) {1 \over 2} dx[/tex]

For u is 10, you said that g(u-x) would be 0 (for all x).
I'm afraid the integral won't be 0.5 then...

so my integral is

h(u) = Integral -(y+1) to (y-1) of (1/4) du ?
Hmm, in the integral for h(u) there is no integration with respect to u.
It couldn't, because then you'd never get a function of u.

After substitution the integration should be with respect to y.
 
  • #14
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Let's revisit your formula for h(u).
[tex]h(u)=\int_{-\infty}^\infty g(u-x)g(x) dx[/tex]
You already reduced this to:
[tex]h(u)=\int_{-1}^1 g(u-x) {1 \over 2} dx[/tex]

For u is 10, you said that g(u-x) would be 0 (for all x).
I'm afraid the integral won't be 0.5 then...



Hmm, in the integral for h(u) there is no integration with respect to u.
It couldn't, because then you'd never get a function of u.

After substitution the integration should be with respect to y.
Ah shoot, so my reduction is wrong too?

So I need the integral to be 0.5 when u = 10, perhaps I multiply it all by 0.25, or change my limits?
 
  • #15
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Ah shoot, so my reduction is wrong too?

So I need the integral to be 0.5 when u = 10, perhaps I multiply it all by 0.25, or change my limits?
You have:
[tex]h(u)=\int_{-1}^1 g(u-x) {1 \over 2} dx[/tex]

For u is 10, you said that g(u-x) would be 0 (for all x).
This means:
[tex]h(10)=\int_{-1}^1 0 \cdot {1 \over 2} dx = 0[/tex]
 
  • #16
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You have:
[tex]h(u)=\int_{-1}^1 g(u-x) {1 \over 2} dx[/tex]

For u is 10, you said that g(u-x) would be 0 (for all x).
This means:
[tex]h(10)=\int_{-1}^1 0 \cdot {1 \over 2} dx = 0[/tex]
and that integral is equal to 2, which is why I said we may need to multiply it by 0.25, or change the limits?

Sorry it's late, and I'm trying my best! :)

I just need to get the integral before I go to bed then I'm sure I can do it!
 
  • #17
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and that integral is equal to 2, which is why I said we may need to multiply it by 0.25, or change the limits?
Huh? :confused:
Which integral is equal to 2?
What would you need to multiply by 0.25 and why?

When you make a substitution, say y=u-x, then yes, you need to change the limits.
But you need to make a proper substitution wrt y.


Sorry it's late, and I'm trying my best! :)

I just need to get the integral before I go to bed then I'm sure I can do it!
I'll be off to bed soon too.
 
  • #18
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Huh? :confused:
Which integral is equal to 2?
What would you need to multiply by 0.25 and why?

When you make a substitution, say y=u-x, then yes, you need to change the limits.
But you need to make a proper substitution wrt y.




I'll be off to bed soon too.
NO no no noooo

wait please, we can get this!

I was just trying to make it easier to get to 0.5, the integral of h(u) in your earlier post.... oh wait sorry, I though the integral of 0 was x, I realise now its a constant!

Forget what I was saying before.

y = u - x and so i am doing the integral of

h(u) = int from u+1 to u-1 0.5*g(y) dy

correct?
 
  • #19
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The anti-derivative of 0 is a constant, but the integral between any 2 boundaries is 0.

Yes, you have the right integral for h(u).
Now you need to consider on which part of your range g(y) is zero, and on which part it is (1/2).
 
  • #20
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The anti-derivative of 0 is a constant, but the integral between any 2 boundaries is 0.

Yes, you have the right integral for h(u).
Now you need to consider on which part of your range g(y) is zero, and on which part it is (1/2).
g(y) is 0.5 between u-1< 0 < u+1 no?
 
  • #21
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No. Only between -1 and +1.
 
  • #22
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Does that change my integral of h(u)?

Or is that the integral range I use for when I calclulate my transform?
 
  • #23
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So integrating between these values I get:

h(u) = int u-1 to u+1 of 1/4

= 0.25u + 0.25 - 0.25u + 0.25

= 0.5?

Bah, that doesnt work as the fourier transfrm of that between -1 and 1 is sink / k
 
  • #24
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For large positive and negative values of u, h(u)=0.
Not sure why you are bent on giving h(u) some value for all u.

And I'm sorry, but I'm off to bed now. :zzz:
(Perhaps you need some sleep too.)
 
  • #25
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For large positive and negative values of u, h(u)=0.
Not sure why you are bent on giving h(u) some value for all u.

And I'm sorry, but I'm off to bed now. :zzz:
(Perhaps you need some sleep too.)
Ok, thanks anyway

I still can't do it, I cant get any integral that doesn't give me a value of something other than 1 or 0.5..
 

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