Chapter 7
In chapter 7 we're going to learn how to represent a CT sgnal using samples.
Sampling
What is it? Sampling is a process of measuring a CT signal x(t) at some specific values of time t.
For example we can sample a continuous time signal x(t) at point t-1, t-2 and t-3. The sampled signal can represented by the formula $ y[n] = x(nT)\, $
The loss of information while sampling occurs depends on the signal and the frequency of the sample.
Given $ y[n] = x(nT)\, $, it is impossible to recontruct the signal back, since ther is loss of information when converting into a DT signal, but one can approximate it.
There is few types of interpolation. One of the common interpolation would be zero order interpolation, and for better approximation, Kth order interpolation can be used.
Examples of zero order interpolation can be found here and examples of first order interpolation can be found here.
Sampling Theorem
Let $ x(t)\, $ be a random signal with frequency $ \omega \, $
Let the maximum frequency of the signal, $ \omega_m \, $ be a non-negative number.
Consider $ x(t)\, $ as a signal with Fourier transform of $ X(\omega)=0 $ when $ |\omega|>\omega_m $. $ x(t)\, $ is band limited
Now, consider the samples $ x(nT)\, $, with $ n = ... -2, -1, 0, 1, 2 ... \, $
If $ T<\frac{1}{2}(\frac{2\pi}{\omega_m}) \, $
then $ x(t)\, $ can be uniquely recovered from its samples.
Let us assume that $ T\, $ is the sampling period of the sampling process. We can safely say that $ \frac{2 \pi}{T} = \omega_s \, $ where $ \omega_s \, $ is the sampling frequency used tos ample the countinuous time signal.
Thus as $ T<\frac{1}{2}(\frac{2\pi}{\omega_m}) \, $
$ 2 \omega_m<\frac{2 \pi}{T}\, $
$ 2 \omega_m<\omega_s\, $
The sampling frequency must be twice the frequency of the maximum frequency of the signal. The minimum frequency that must be used by the sampling process is called Nyquist Rate.
