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 every periodic $X[n]$ can be obtained that way.) Let $Y[k]$ denote the Fourier transform of $X$ for $k = 0, ..., N-1$:

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

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

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

In the second version we rearranged the exponents to show that $Y[k]$ 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 express the amplitude of the partials of $Y[k]$ using its own Fourier transform. Equating the two 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 $Y[k]/N$. Replacing $m$ with its negative gives:

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

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

This shows that any periodic $X[n]$ can indeed be obtained as a sum of sinusoids. Further, it explicitly shows how to reconstruct $X[n]$ from its Fourier transform $Y[k]$, if we know its value for the integers $k=0, \ldots, N-1$.