Why are the coefficients in this proof multiplied by T?

61 Views Asked by At

Online Proof of Fourier Transform

The link above shows the full proof of the Fourier Transform using the limiting case of the Fourier Series. I don't understand why this was done in the proof however:

$$ x(t) = \sum\limits_{n = - \infty }^{ + \infty } {{c_n}{e^{jn{\omega _o}t}}} $$ $$ c_n = {1 \over T}\int_T {x(t){e^{ - jn\omega_0t}}dt} $$ $$ T{c_n} = \int_T {x(t){e^{ - jn\omega_0t}}dt} $$

As $T \rightarrow \infty$ the fundamental frequency, $\omega_0=2\pi/T$, becomes extremely small and the quantity $n\omega_0$ becomes a continuous quantity that can take on any value (since $n$ has a range of $\pm \infty$) so we define a new variable ω=nω0; we also let $X(\omega)=Tc_n$. Making these substitutions in the previous equation yields the analysis equation for the Fourier Transform (also called the Forward Fourier Transform).

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

Wouldn't multiplying coefficients by T, result in the Fourier Transform being scaled by T according to the proof? If so, the Fourier transform does not accurately describe the 'amount' of frequencies in a function as they too are scaled by T.