Fourier transform as additive synthesis

Now consider an arbitrary signal $X[n]$ that repeats every $N$ samples. (Previously we had assumed that $X[n]$ could be obtained as a sum of sinusoids, but we haven't yet found out whether any periodic $X[n]$ can be obtained that way.) Let

$$Y[k] = {\cal FT}\left \{ X[n] \right \} (k)$$

$$= {{\left [ {U^{-k}} \right ]} ^ {0}} X[0] + {{\left [ {U^... ...X[1] + \cdots + {{\left [ {U^{-k}} \right ]} ^ {N-1}} X[N-1]$$

be the Fourier transform for $k = 0, ..., N-1$. Looking hard at $Y[k]$ we see that it is a sum of complex sinusoids, with complex amplitudes $X[n]$ and frequencies $-n\omega$ for $n=0,\ldots,N-1$. In other words, $Y[k]$ can be considered as a waveform in its own right, whose $m$th component has strength $X[-m]$. We can also find the amplitude of the partials of $Y[k]$ using the Fourier transform on $Y[k]$. Equating the two expressions for the partial amplitudes gives:

$${1 \over N} {\cal FT} \left \{ Y[k] \right \} (m) = X[-m]$$

(The expression $X[-m]$ makes sense because $X$ is a periodic signal). This means in turn that $X[-m]$ can be obtained by summing sinusoids with amplitudes $Y[-k]/N$. The same analysis starting with $X[-m]$ shows that $X[m]$ is obtained by summing sinusoids using $Y[k]/N$ as their amplitudes. So now we know that any periodic $X[n]$ can indeed be obtained as a sum of sinusoids. Furthermore, we know how to reconstruct a signal from its Fourier transform, if we know its value for the integers $k=0, \ldots, N-1$.