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

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

denote the samples of $x(t)$ at uniform intervals of $T_s$ 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

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

The terms in the above sum for $m\neq 0$ are called aliasing terms. They are said to alias into the base band $[-\pi/T_s,\pi/T_s]$. 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 $\theta \isdef 2\pi \omega_d T_s$ denotes the normalized discrete-time frequency variable.

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

$$\begin{eqnarray*} x(t) &=& \frac{1}{2\pi}\int_{-\infty}^\infty X(j\omega) e^{j\o... ...y X\left(j\omega + j m\Omega_s \right) e^{j\Omega_s m t} d\omega \end{eqnarray*}$$

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

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

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

$$x_d(n) = \frac{1}{2\pi}\int_{-\pi}{\pi} e^{j\theta^\prime n} \fr... ...frac{\theta^\prime }{T_s} + m\frac{2\pi}{T_s}\right) \right] d\theta^\prime .$$

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