Audio Processing

Sampling Theorem

Let $ x(t)$ denote any continuous-time signal having a continuous Fourier transform

$displaystyle X(jomega)isdef int_{-infty}^infty x(t) e^{-jomega t} dt. $

Let

$displaystyle x_d(n) isdef x(nT), quad n=ldots,-2,-1,0,1,2,ldots, $

denote the samples of $ x(t)$ at uniform intervals of $ T$ seconds. Then $ x(t)$ can be exactly reconstructed from its samples $ x_d(n)$ if $ X(jomega)=0$ for all $ vertomegavertgeqpi/T$.D.3

Proof: From the continuous-time aliasing theorem (§D.2), we have that the discrete-time spectrum $ X_d(e^{jtheta})$ can be written in terms of the continuous-time spectrum $ X(jomega)$ as

$displaystyle X_d(e^{jomega_d T}) = frac{1}{T} sum_{m=-infty}^infty X[j(omega_d +mOmega_s )] $

where $ omega_d in(-pi/T,pi/T)$ is the ``digital frequency'' variable. If $ X(jomega)=0$ for all $ vertomegavertgeqOmega_s /2$, then the above infinite sum reduces to one term, the $ m=0$ term, and we have

$displaystyle X_d(e^{jomega_d T}) = frac{1}{T} X(jomega_d ), quad omega_d inleft(-frac{pi}{T},frac{pi}{T}right) $

At this point, we can see that the spectrum of the sampled signal $ x(nT)$ coincides with the nonzero spectrum of the continuous-time signal $ x(t)$. In other words, the DTFT of $ x(nT)$ is equal to the FT of $ x(t)$ between plus and minus half the sampling rate, and the FT is zero outside that range. This makes it clear that spectral information is preserved, so it should now be possible to go from the samples back to the continuous waveform without error, which we now pursue.

To reconstruct $ x(t)$ from