Tuesday, 17 May 2011

Understanding the Effect of Sampling on Fourier Domain

This solution is answer to an excercise 7.23 of the Book Openheim, Signal and Systems.
Fig1 : Block Diagram of System

As we can see from the diagram
\[ x_p(t) = x(t) \times p(t) \]
in Fourier domain, it can be written as
\[ X_p(j\omega) = \frac{1}{2\pi}X(j\omega) * P(j\omega) \]
Now calculation the fourier transform of p(t) is little tricky. It is a periodic signal with period 2T. Let us write it as the same form as section 7.1.1
\[ p(t) = \sum_{n=-\infty}^{n=\infty}{(-1)^n \delta(t-n\Delta)} \]
Its fourier transform is given as
\[P(j\omega)= \sum_{n=-\infty}^{n=\infty}{(-1)^n e^{(-n\Delta\omega)}}\]
Now we will simplify it using the fact that (Eq 1)
\[ \sum_{n=-\infty}^{n=\infty}{e^{(-n\Delta\omega)}}= \frac{2\pi}{\Delta} \sum_{n=-\infty}^{n=\infty}{\delta\left(\omega-n\frac{2\pi}{\Delta}\right)}\]
Now ,
 \[P(j\omega)= \sum_{n=-\infty}^{n=\infty}{(-1)^n e^{(-n\Delta\omega)}}\] \[P(j\omega)= \sum_{m=-\infty}^{m=\infty}{(-1)^{2m} e^{(-2m\Delta\omega)}+(-1)^{2m+1} e^{(-2(m+1)\Delta\omega)}}\]  \[P(j\omega)= \sum_{m=-\infty}^{m=\infty}{e^{(-m(2\Delta)\omega)}- e^{(-1\Delta\omega)}e^{(-m(2\Delta)\omega)}}\]\[P(j\omega)= \sum_{m=-\infty}^{m=\infty}{e^{(-m(2\Delta)\omega)}}-e^{(-1\Delta\omega)} \sum_{m=-\infty}^{m=\infty}{  e^{(-m(2\Delta)\omega)}}\]
using the fact above stated (Eq 1)
\[P(j\omega)=\frac{2\pi}{2\Delta} \sum_{m=-\infty}^{m=\infty}{\delta\left(\omega-m\frac{2\pi}{2\Delta}\right)}-e^{(-1\Delta\omega)} \sum_{m=-\infty}^{m=\infty}{  \delta\left(\omega-m\frac{2\pi}{2\Delta}\right)}\]\[P(j\omega)= \frac{2\pi}{2\Delta} \left( 1-e^{(-1\Delta\omega)}\right) \sum_{m=-\infty}^{m=\infty}{\delta\left(\omega-m\frac{2\pi}{2\Delta}\right)}\].
 So it is basically scaled version of impulse train.

Now remaining is to convolve this P(jw) with X(jw) (Refer section 7.1.1). Which results in repitition of X(jw) at 1/2delta interval.
Fig 2: Effect of sampling on Fourier Transform