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Continuous-Time Aliasing Theorem

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

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


$\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, and denote its Discrete-Time Fourier Transform (DTFT) by

$\displaystyle X_d(e^{j\theta})\isdef \sum_{n=-\infty}^\infty x_d(n) e^{-j\theta n}.

Then the spectrum $ X_d$ of the sampled signal $ x_d$ is related to the spectrum $ X$ of the original continuous-time signal $ x$ by

$\displaystyle X_d(e^{j\theta}) = \frac{1}{T} \sum_{m=-\infty}^\infty X\left[j\left(\frac{\theta}{T}
+ m\frac{2\pi}{T}\right)\right].

The terms in the above sum for $ m\neq 0$ are called aliasing terms. They are said to alias into the base band $ [-\pi/T,\pi/T]$ . Note that the summation of a spectrum with aliasing components involves addition of complex numbers; therefore, aliasing components can be removed only if both their amplitude and phase are known.

Proof: Writing $ x(t)$ as an inverse FT gives

$\displaystyle x(t) = \frac{1}{2\pi}\int_{-\infty}^\infty X(j\omega) e^{j\omega t} d\omega.

Writing $ x_d(n)$ as an inverse DTFT gives

$\displaystyle x_d(n) = \frac{1}{2\pi}\int_{-\pi}^\pi X_d(e^{j\theta}) e^{j \theta t} d\theta

where $ \theta \isdef 2\pi \omega_d T$ denotes the normalized discrete-time frequency variable.

The inverse FT can be broken up into a sum of finite integrals, each of length $ \Omega_s \isdef 2\pi f_s= 2\pi/T$ , as follows:

x(t) &=& \frac{1}{2\pi}\int_{-\infty}^\infty X(j\omega) e^{j\omega t} d\omega \\
&=& \frac{1}{2\pi}\sum_{m=-\infty}^\infty \int_{(2m-1)\pi/T)}^{(2m+1)\pi/T}
X(j\omega) e^{j\omega t} d\omega \qquad \mbox{(let $\omega\leftarrow m\Omega_s $)}\\
&=& \frac{1}{2\pi}\sum_{m=-\infty}^\infty \int_{-\Omega_s /2}^{\Omega_s /2}
X\left(j\omega + j m\Omega_s \right) e^{j\omega t} e^{j \Omega_s m t} d\omega \\
&=& \frac{1}{2\pi}\int_{-\Omega_s /2}^{\Omega_s /2} e^{j\omega t}
\sum_{m=-\infty}^\infty X\left(j\omega + j m\Omega_s \right) e^{j\Omega_s m t} d\omega

Let us now sample this representation for $ x(t)$ at $ t=nT$ to obtain $ x_d(n) \isdef x(nT)$ , and we have

x_d(n) &=& \frac{1}{2\pi}\int_{-\Omega_s /2}^{\Omega_s /2} e^{j\omega nT}
\sum_{m=-\infty}^\infty X\left(j\omega + j m\Omega_s \right) e^{j\Omega_s m nT} d\omega \\
&=& \frac{1}{2\pi}\int_{-\Omega_s /2}^{\Omega_s /2} e^{j\omega nT}
\sum_{m=-\infty}^\infty X\left(j\omega + j m\Omega_s \right)
e^{j(2\pi/T) m nT} d\omega \\
&=& \frac{1}{2\pi}\int_{-\Omega_s /2}^{\Omega_s /2} e^{j\omega nT}
\sum_{m=-\infty}^\infty X\left(j\omega + j m\Omega_s \right) d\omega

since $ n$ and $ m$ are integers. Normalizing frequency as $ \theta^\prime = \omega T$ yields

$\displaystyle x_d(n) = \frac{1}{2\pi}\int_{-\pi}{\pi} e^{j\theta^\prime n}
\frac{1}{T}\sum_{m=-\infty}^\infty X\left[j\left(\frac{\theta^\prime }{T}
+ m\frac{2\pi}{T}\right) \right] d\theta^\prime .

Since this is formally the inverse DTFT of $ X_d(e^{j\theta^\prime })$ written in terms of $ X(j\theta^\prime /T)$ , the result follows.
$ \Box$

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``Mathematics of the Discrete Fourier Transform (DFT), with Audio Applications --- Second Edition'', by Julius O. Smith III, W3K Publishing, 2007, ISBN 978-0-9745607-4-8.
Copyright © 2021-02-23 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University