Impulse-train Sampling
One type of sampling that satisfies the Sampling Theorem is called impulse-train sampling. This type of sampling is achieved by the use of a periodic impulse train multiplied by a continuous time signal, $ x(t)\! $. The periodic impulse train, $ p(t)\! $ is referred to as the sampling function, the period, $ T\! $, is referred to as the sampling period, and the fundamental frequency of $ p(t)\! $, $ \omega_s = \frac{2\pi}{T}\! $, is the sampling frequency. We define $ x_p(t)\! $ by the equation,
Graphically, this equation looks as follows,
$ x(t)\! $ ----------> x --------> $ x_p(t)\! $ ^ | |
a $ p(t) = \sum^{\infty}_{n = -\infty} \delta(t - nT)\! $
By using linearity and the sifting property, $ x_p(t)\! $ can be represented as follows,
$ x_p(t) = x(t)p(t)\! $
$ = x(t)\sum^{\infty}_{n = -\infty} \delta(t - nT)\! $
$ =\sum^{\infty}_{n = -\infty}x(t)\delta(t - nT)\! $
$ =\sum^{\infty}_{n = -\infty}x(nT)\delta(t - nT)\! $
Now, in the time domain, $ x_p(t)\! $ looks like a group of shifted deltas with magnitude equal to the value of $ x(t)\! $ at that time, $ nT\! $, in the original function. In the frequency domain, $ X_p(\omega)\! $ looks like shifted copies of the original $ X(\omega)\! $ that repeat every $ \omega_s\! $, except that the magnitude of the copies is $ \frac{1}{T}\! $ of the magnitude of the original $ X(\omega)\! $.
Why does $ X_p(\omega)\! $ look like copies of the original $ X(\omega)\! $?
This answer can be found simply by using the Fourier Transform of the $ X_p(\omega)\! $.
$ X_p(\omeg) = F(x(t)p(t))\! $
$ = \frac{1}{2\pi}X(\omega) * P(\omega)\! $
$ = \frac{1}{2\pi}X(\omega) * \sum^{\infty}_{k = -\infty}2\pi a_k \delta(\omega - \omega_s), a_k = \frac{1}{T}\! $
$ = \sum^{\infty}_{k = -\infty}\frac{1}{T}X(\omega - k\omega_s)\! $
From the above equation, it is obvious that $ X_p(\omega)\! $ is simply shifted copies of the original function (as can be seen by the $ X(\omega - k\omega_s)\! $) that are divided by $ T\! $ (as can be seen by $ \frac{1}{T}\! $).
