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Fourier analysis of periodic signals

Suppose $X[n]$ is a complex-valued signal that repeats every $N$ samples. (We are continuing to use complex-valued signals rather than real-valued ones to simplify the mathematics.) Because of the period $N,ドル the values of $X[n]$ for $n=0,\ldots,N-1$ determine $X[n]$ for all integer values of $n$.

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

Given the values of $X,ドル we wish to find the complex amplitudes of the partials. Suppose we want the $k$th partial, where 0ドル \leq k < N$. The frequency of this partial is 2ドル\pi k / N$. We can find its complex amplitude by modulating $X$ downward 2ドル\pi k / N$ 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. We can do this in effect by averaging over a huge number of samples; but since the signal repeats every $N$ samples, this huge average is the same as the average of the first $N$ samples. In short, to measure a sinusoidal component of a periodic signal, modulate it down to DC and then average over one period.

Let $\omega=2\pi/N$ be the fundamental frequency for the period $N,ドル and let $U$ be the unit-magnitude complex number with argument $\omega $:

\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] = {A_0}{{\left [ {U^0} \right ]} ^ {n}} + {A_1}{{\l... ...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:

[画像:\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],ドル 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,ドル equal to $N$ times the amplitude of the input's $k$th partial. So far $k$ has taken integer values but the formula makes sense for any value of $k$ if we define $V$ more generally as:

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

where, as before, $\omega=2\pi/N$ is the (angular) fundamental frequency associated with the period $N$.

Subsections

next up previous contents index
Next: Periodicity of the Fourier Up: Fourier analysis and resynthesis Previous: Fourier analysis and resynthesis Contents Index

Miller Puckette 2006年12月30日