I am trying to prove the below identity where $f_c(x)=f(cx)$ such that c is a positive number.

$F_c(\alpha)=\frac 1 c F(\frac\alpha c)$

F above represents the Fourier transformed $f(x)$. I attempted this by representing $f(x)$ as a Fourier series such that we can represent its values in the frequency domain:

$f(x) = a_0 + \sum_{n=1}^\infty a_n \cos(n\pi x ) + \sum_{n=1}^\infty b_n \cos(n\pi x ) $

Because the coefficients are defined over your signal period, the coefficients do not change when writing f(cx). You can then prove that:

$f(cx) = a_0 + \sum_{n=1}^\infty a_n \cos(n\pi xc ) + \sum_{n=1}^\infty b_n \cos(n\pi xc ) $

This gives us an integral we can solve when plugging back into the Fourier transform:

$F_c(\alpha)=\int_{-\infty}^\infty \left(a_0 + \sum_{n=1}^\infty a_n \cos(n\pi xc ) + \sum_{n=1}^\infty b_n \cos(n\pi xc )e^{-i\alpha xc}\right)$

Euler's number can of course be represented as $\cos(\alpha xc)+i\sin(\alpha cx)$

For the sake of brevity, I'll apply this integral to only the $a_0$ term to represent my dilemma.

$F_c(\alpha)=a_0\int_{-\infty}^\infty (\cos(\alpha xc)+i\sin(\alpha cx))dx+...$

$F_c(\alpha)=\frac{a_0}{\alpha c}(\sin(\alpha xc)]^\infty _{-\infty}-\cos(\alpha xc)]^\infty _{-\infty} ) $

And here lies my dilemma. The above equation for f(x) is:

$F(\alpha)=\frac{a_0}{\alpha}(\sin(\alpha x)|^\infty _{-\infty}-\cos(\alpha x)|^\infty _{-\infty} ) $

Since $\sin(\alpha xc)|^\infty _{-\infty}\ne \sin(\frac{\alpha x}{c})|^\infty _{-\infty}$ I don't know how to prove the identity.


1 Answer 1


To be clear, this is the scaling property of the Fourier Transform which is specifically given as:

$$\mathscr{F}\{x(at)\} = \frac{1}{a}X(\omega/a)$$

With $a$ as a positive real number, and the Fourier Transform of $x(t)$ as:

$$\mathscr{F}\{x(t)\} = X(\omega)$$

This is quite intuitive: If we played a recording 10 times slower ($a=10$), all the frequencies would be 10 times lower.

Instead of using the Fourier Series, use the Fourier Transform formula directly:

$$X(\omega) = \int_{-\infty}^{\infty} x(t) e^{j \omega t} dt$$

Keep these things in mind that will also help:

$$d (at) = a dt$$

$$\frac{1}{a}X(\omega/a) = \int_{-\infty}^{\infty} x(at) e^{j (\omega/a) t} d(at)$$

If the above formulas aren't clear, then write out the Fourier Transform using its formula for the Fourier Transform of $x(t)$ (which I gave) and then again for the Fourier transform of $x(at)$, and $X(\omega/a)$, all of which can be found by substitution with the basic formula for the Fourier Transform provided.

  • $\begingroup$ Thank you for your answer! Your math makes since but I don't follow how you got $\omega /a$ in Euler's number. Shouldn't it be $e^{j\omega at}$? $\endgroup$
    – John Smith
    Mar 19, 2022 at 23:39
  • $\begingroup$ Consider if in general a $X(\omega) = e^{j \omega t}$, what would $X(\omega/a)$ be? Just think through each form like that and the right substitutions and manipulations will make sense. $\endgroup$ Mar 19, 2022 at 23:50
  • 1
    $\begingroup$ Ohhhhh it took me awhile to see that you're right. It's a u substitution. If you say that u = cx or u/c=x, you can plug in everything for u and get the identity! Thanks for the answer! $\endgroup$
    – John Smith
    Mar 20, 2022 at 0:57
  • $\begingroup$ Yes I had assumed you would get more out of it if you dug through those details (and that this might be a HW problem, so can't give the full solution -that would be no fun). Glad you got it! $\endgroup$ Mar 20, 2022 at 1:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.