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Fourier Series (FS) and Relation to DFT

In continuous time, a periodic signal $ x(t)$, with period $ P$ seconds,B.2 may be expanded into a Fourier series with coefficients given by

$\displaystyle X(\omega_k) \isdef \frac{1}{P}\int_0^P x(t) e^{-j\omega_k t} dt, \quad k=0,\pm1,\pm2,\dots \protect$ (B.5)

where $ \omega_k \isdef 2\pi k/P$ is the $ k$th harmonic frequency (rad/sec). The generally complex value $ X(\omega_k)$ is called the $ k$th Fourier series coefficient. The normalization by $ 1/P$ is optional, but often included to make the Fourier series coefficients independent of the fundamental frequency $ 1/P$, and thereby depend only on the shape of one period of the time waveform.

Relation of the DFT to Fourier Series

We now show that the DFT of a sampled signal $ x(n)$ (of length $ N$), is proportional to the Fourier series coefficients of the continuous periodic signal obtained by repeating and interpolating $ x$. More precisely, the DFT of the $ N$ samples comprising one period equals $ N$ times the Fourier series coefficients. To avoid aliasing upon sampling, the continuous-time signal must be bandlimited to less than half the sampling rate (see Appendix D); this implies that at most $ N$ complex harmonic components can be nonzero in the original continuous-time signal.

If $ x(t)$ is bandlimited to $ \omega T\in(-\pi,\pi)$, it can be sampled at intervals of $ T$ seconds without aliasing (see §D.2). One way to sample a signal inside an integral expression such as Eq.$ \,$(B.5) is to multiply it by a continuous-time impulse train

$\displaystyle \Psi_T(t) \isdef T\sum_{n=-\infty}^\infty \delta(t-nT) \protect$ (B.6)

where $ \delta(t)$ is the continuous-time impulse signal defined in Eq.$ \,$(B.3).

We wish to find the continuous-time Fourier series of the sampled periodic signal $ x(nT)$. Thus, we replace $ x(t)$ in Eq.$ \,$(B.5) by

$\displaystyle x_s(t) \isdef x(t)\cdot \Psi_T(t). $

By the sifting property of delta functions (Eq.$ \,$(B.4)), the Fourier series of $ x_s$ isB.3

\begin{eqnarray*} X_s(\omega_k) = \frac{1}{P} \int_0^P x_s(t) e^{-j\omega_k t} d... ...1}{P} \sum_{n=0}^{\lceil P/T\rceil-1} x(nT) e^{-j\omega_k nT} T. \end{eqnarray*}

If the sampling interval $ T$ is chosen so that it divides the signal period $ P$, then the number of samples under the integral is an integer $ N\isdef P/T$, and we obtain

\begin{eqnarray*} X_s(\omega_k) &=& \frac{T}{P} \sum_{n=0}^{N-1} x(nT) e^{-j\o... ...{1}{N}\hbox{\sc DFT}_{N,k}(x_p), \quad k=0,\pm 1, \pm 2, \dots \end{eqnarray*}

where $ x_p\isdef [x(0),x(T),\dots,x((N-1)T)]$. Thus, $ X_s(\omega_k)=X(\omega_k)$ for all $ k$ at which the bandlimited periodic signal $ x(t)$ has a nonzero harmonic. When $ N$ is odd, $ X(\omega_k)$ can be nonzero for $ k\in[-(N-1)/2,(N-1)/2]$, while for $ N$ even, the maximum nonzero harmonic-number range is $ k\in[-N/2+1,N/2-1]$.

In summary,

$\textstyle \parbox{0.8\textwidth}{% WHY IS THIS NEEDED??? \emph{the Fourier ser... ...he DFT length, and $N$\ is also the number of samples in each period of $x$.}}$

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