Fourier Transform Scaling Property help

  • Thread starter Uan
  • Start date
  • #1
Uan
14
0

Main Question or Discussion Point

Hi,

I'm following the proof of the "Scaling Property of the Fourier Transform" from here:

http://www.thefouriertransform.com/transform/properties.php

...but don't understand how they went from the integral to the right hand term here:

scalingP2.gif


The definition of the Fourier Trasform they use is this:

fourierDefinition.gif


Thanks,
Uan
 

Answers and Replies

  • #2
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
14,367
5,962
I don't understand, where there is a problem, because it's simply the definition of the Fouier transform. The full calculation goes as follows. We start with the definition of the Fourier transform:
[tex]\mathcal{F}\{g(c t)\}=\int_{\mathbb{R}} \mathrm{d t} g(t) \exp[-2 \pi \mathrm{i} f t].[/tex]
Now we substitute [itex]u=c t[/itex], which gives
[tex]\mathcal{F}\{g(c t)\}=\frac{1}{|c|} \int_{\mathbb{R}} \mathrm{d} t \exp[-2 \pi \mathrm{i} u (f/c)]=\frac{1}{|c|} G(f/c),[/tex]
where
[tex]G(f)=\mathcal{F}\{g(t)\}(f).[/tex]
 
  • #3
Uan
14
0
I don't understand, where there is a problem, because it's simply the definition of the Fouier transform. The full calculation goes as follows. We start with the definition of the Fourier transform:
[tex]\mathcal{F}\{g(c t)\}=\int_{\mathbb{R}} \mathrm{d t} g(t) \exp[-2 \pi \mathrm{i} f t].[/tex]
Now we substitute [itex]u=c t[/itex], which gives
[tex]\mathcal{F}\{g(c t)\}=\frac{1}{|c|} \int_{\mathbb{R}} \mathrm{d} t \ exp[-2 \pi \mathrm{i} u (f/c)]=\frac{1}{|c|} G(f/c),[/tex]
where
[tex]G(f)=\mathcal{F}\{g(t)\}(f).[/tex]
Hi vanhees71,

I get the substitution but I don't see how they went from the integral to the next bit. To me, the form of the integral doesn't quite match the definition in my first post, it has got g(u) but the exponential has exp(-i*2*pi*f*(u/c)), so this is where I get hung up...

Needs to be [STRIKE]either[/STRIKE] g(u/c) [STRIKE][or exp(-i*2*pi*f*u)^(1/c) ? ] (sorry forget this)[/STRIKE] or something to be able to apply the definition. Feels like I'm missing something simple.

One more thing, are you missing g(ct) from your first integral and g(u) on your second integral?

[tex]\mathcal{F}\{g(c t)\}=\int_{\mathbb{R}} \mathrm{d t} g(c t) \exp[-2 \pi \mathrm{i} f t].[/tex]

[tex]\mathcal{F}\{g(c t)\}=\frac{1}{|c|} \int_{\mathbb{R}} \mathrm{d} t \ g(u) exp[-2 \pi \mathrm{i} u (f/c)]=\frac{1}{|c|} G(f/c),[/tex]
 
Last edited:
  • #4
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,394
179
Hi vanhees71,

I get the substitution but I don't see how they went from the integral to the next bit. To me, the form of the integral doesn't quite match the definition in my first post, it has got g(u) but the exponential has exp(-i*2*pi*f*(u/c)), so this is where I get hung up...
Start with the definition of the Fourier transform:
$$G(f) = \int_{-\infty}^{\infty} g(u) \exp[-2\pi i f u] du$$
Now substitute ##f/c## in place of ##f##:
$$G(f/c) = \int_{-\infty}^{\infty} g(u) \exp[-2\pi i (f/c) u] du$$
Multiply by ##1/c##:
$$\frac{1}{c} G(f/c) = \int_{-\infty}^{\infty} \frac{g(u)}{c} \exp[-2\pi i (f/c) u] du$$
which is the second equality in your displayed equation.
 
  • #5
Uan
14
0
Start with the definition of the Fourier transform:
$$G(f) = \int_{-\infty}^{\infty} g(u) \exp[-2\pi i f u] du$$
Now substitute ##f/c## in place of ##f##:
$$G(f/c) = \int_{-\infty}^{\infty} g(u) \exp[-2\pi i (f/c) u] du$$
Multiply by ##1/c##:
$$\frac{1}{c} G(f/c) = \int_{-\infty}^{\infty} \frac{g(u)}{c} \exp[-2\pi i (f/c) u] du$$
which is the second equality in your displayed equation.
Cheers jbunniii, just what I was after! :thumbs:
 

Related Threads for: Fourier Transform Scaling Property help

Replies
4
Views
543
Replies
2
Views
4K
  • Last Post
Replies
15
Views
2K
Replies
4
Views
2K
Replies
3
Views
2K
  • Last Post
Replies
5
Views
20K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
560
Top