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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.

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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.

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  • $\begingroup$ 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 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 at 23:50
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    $\begingroup$ 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 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 at 1:07

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