Continuous-Time Aliasing Theorem

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

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

Let

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

denote the samples of $x(t)$ at uniform intervals of $T$ seconds, and denote its Discrete-Time Fourier Transform (DTFT) by

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

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

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

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

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

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

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

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:

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

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

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

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

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

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.
$\Box$