How do you prove that the bandwidth of a signal is inversely proportional to the length of the signal?

Question

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.

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.