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Fourier analysis of periodic signals

Suppose $X[n]$ is a complex-valued signal with period $N$, a positive integer. (We are using complex-valued signals rather than real-valued ones because the mathematics will turn out simpler.) The values of $X[n]$ for $n=0,\ldots,N-1$ (one period) determine $X[n]$ for all integer values of $n$.

Suppose further that $X[n]$ can be written as a sum of sinusoids of frequency $0$, $2\pi/N$, $4\pi/N$, $\ldots$, $2(N-1)\pi/N$. These are the partials, starting with the zeroth, for a signal of period $N$. We stop after the $N$th partial because the next one would have frequency $2\pi $, equivalent to frequency $0$, which is already on the list.

Given the values of $X$, we wish to find the complex amplitudes of the partials. Suppose we want the $k$th partial, where $0 \leq k < N$. The frequency of the partial is $\omega=2\pi k / N$. We can get its amplitude by modulating $X$ downward $\omega $ radians per sample in frequency, so that the $k$th partial is modulated to frequency zero. Then we pass the signal through a low-pass filter with such a low cutoff frequency that nothing but the zero-frequency partial remains. Such a filter will essentially average the $N$ samples of its periodic input. In summary, to measure a sinusoidal component of a periodic signal, modulate it down to DC and then average over one period.

Let $U$ be the unit-magnitude complex number with argument $2\pi/N$ (the fundamental frequency of a signal with period $N$):

\begin{displaymath} U = \cos(\omega) + i \sin(\omega) \end{displaymath}

The $k$th partial of the signal $X[n]$ is of the form:

\begin{displaymath} {P_k}[n] = {A_k}{{\left [ {U^k} \right ]} ^ {n}} \end{displaymath}

where ${A_K}$ is the complex amplitude of the partial, and the frequency of the partial is:

\begin{displaymath} \angle({U^k}) = k \angle(U) = k\omega \end{displaymath}

We're assuming for the moment that the signal $X[n]$ can actually be written as a sum of the $n$ partials, or in other words:

\begin{displaymath} X[n] = {P_0}[n] + {P_1}[n] + \cdots + {P_{N-1}}[n] \end{displaymath}


\begin{displaymath} = {A_0}{{\left [ {U^0} \right ]} ^ {n}} + {A_1}{{\left [ {... ...n}} + \cdots + {A_{N-1}}{{\left [ {U^{N-1}} \right ]} ^ {n}} \end{displaymath}

By the heterodyne-filtering argument above, we expect to be able to measure each $A_k$ by multiplying by the sinusoid of frequency $-k\omega$ and averaging over a period:

\begin{displaymath} {A_k} = {1\over N} \left ( {{\left [ {U^{-k}} \right ]} ^ ... ...dots + {{\left [ {U^{-k}} \right ]} ^ {N-1}} X[N-1] \right ) \end{displaymath}

This is such a useful formula that it gets its own notation. The Fourier transform of a signal $X[n]$, over $N$ samples, is defined as:

\begin{displaymath} {\cal FT}\left \{ X[n] \right \} (k) = {V ^ {0}} X[0] + {V ^ {1}} X[1] + \cdots + {V ^ {N-1}} X[N-1] \end{displaymath}

where $V = {U^{-k}}$. The Fourier transform is a function of the variable $k$. So far $k$ has taken integer values but the formula makes sense for any value of $k$ if we define $V$ more carefully as:

\begin{displaymath} V = \cos(-k\omega) + i\sin(-k\omega) \end{displaymath}

where, as defined above, $\omega=2\pi/N$ is the angular frequency associated with the period $N$.


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Next: Fourier transform as additive Up: Fourier analysis and resynthesis Previous: Fourier analysis and resynthesis   Contents   Index
Miller Puckette 2005-07-11