Fourier transform/Convolution theorem

In summary: 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... something with itI see. I didn't do the calculation myself. I just trusted you when you said it was 1.So, if you multiply by 0.25 we get h(10)=0.5. That would work, but let's make the limits of integration symmetric around u.h(u)=\int_{-u+1}^{u-1} g(u
  • #1
Firepanda
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0
2v3lzlx.png


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
 
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  • #2
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
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
Firepanda said:
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
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
I like Serena said:
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
Firepanda said:
hmm, do you mean what the range of f(10-x) would be?

Yes...
 
  • #8
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
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
I like Serena said:
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
Firepanda said:
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
h(10) = 0.5

so my integral is

h(u) = Integral -(y+1) to (y-1) of (1/4) du ?
 
  • #13
Firepanda said:
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...

Firepanda said:
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
I like Serena said:
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
Firepanda said:
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
I like Serena said:
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
Firepanda said:
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.


Firepanda said:
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
I like Serena said:
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 realize 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
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
I like Serena said:
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
No. Only between -1 and +1.
 
  • #22
Does that change my integral of h(u)?

Or is that the integral range I use for when I calclulate my transform?
 
  • #23
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 doesn't work as the Fourier transfrm of that between -1 and 1 is sink / k
 
  • #24
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
I like Serena said:
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 can't get any integral that doesn't give me a value of something other than 1 or 0.5..
 
  • #26
Perhaps you should try drawing the graph on paper for some valules of u.
 

What is the Fourier transform?

The Fourier transform is a mathematical operation that decomposes a function into its constituent frequencies. It is used to analyze and represent signals in the frequency domain, which can provide insights into the underlying components of the signal.

What is the Convolution theorem?

The Convolution theorem states that the convolution of two functions in the time domain is equivalent to the multiplication of their Fourier transforms in the frequency domain. This allows for simpler calculations and analysis of signals.

How is the Fourier transform calculated?

The Fourier transform of a function can be calculated using a mathematical formula, which involves integration of the function over all possible values. There are also various numerical methods and algorithms that can be used to calculate the Fourier transform.

What is the relationship between Fourier transform and convolution?

The Fourier transform and convolution are closely related as convolution can be viewed as a combination of multiplication and integration, which are both operations used in the Fourier transform. The Convolution theorem further establishes the relationship between the two, making them essential concepts in signal processing and analysis.

What are the applications of Fourier transform and convolution theorem?

The Fourier transform and convolution theorem have numerous applications in various fields such as signal processing, image and audio compression, digital filtering, and data analysis. They are also used in solving differential equations and understanding the behavior of complex systems.

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