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)\! $
                              ^
                              |
                              |
                  $ 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 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(\omega) = 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 1/T.

How to recover $ x(t) $

In order to recover the original function, $ x_p(t) $, we can simply low-pass filter $ x_p(t) $ as long as the filter,

with some $ \omega_c $ satisfying, $ \omega_m < \omega_c < \omega_s - \omega_m $. Also, the low-pass filter must have a gain of $ T $. This can be represented graphically as shown below,

                                           filter

where $ x_r(t) $ represents the recovered original function.

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