2.1 lecture 1, Monday January 25, 2010

This lecture was general review of signals. Difference between continuous, discrete and digital signals is given. We obtain discrete signal from continuous by sampling. Digital signal is obtained from discrete by quantization.

Then review was given of unit step function, delta function. Some examples shown. important ones:

1.

\(u[n]=\left \{ \begin{array} [c]{c}1\ \ n\geq 0\\ 0\ else \end{array} \right . \)

2.

\(\delta \lbrack n]=\left \{ \begin{array} [c]{c}1\ \ n=0\\ 0\ else \end{array} \right . \)

3.

\(x[n]={\sum \limits _{k=-\infty }^{n}}\delta \left ( k\right ) =\left \{ \begin{array} [c]{c}1\ \ n\geq 0\\ 0\ else \end{array} \right . =u\left ( n\right ) \)

4.

\(x[n]={\sum \limits _{k=0}^{\infty }}\delta \left ( n-k\right ) =\left \{ \begin{array} [c]{c}1\ \ n\geq 0\\ 0\ else \end{array} \right . =u\left ( n\right ) \)

5.

\(x[n]= \sum \limits _{k=-\infty }^{\infty } \delta (n-k) = \left \{ \begin{array} [c]{c}1\, n\geq 0\\ 1\, n<0 \end{array} \right \} =1\)

6.

\(u[n]-u[n-1]=\delta \lbrack n]\)

To find if a discrete function is periodic for a given frequency.

First method:

From \(e^{j\omega _{0}n}=e^{j\omega _{0}\left ( n+N\right ) }=e^{j\omega _{0}n}e^{j\omega _{0}N}\), where \(N\) is the period. Hence we need to have \(e^{j\omega _{0}N}=1\) for periodic, or \(\omega _{0}N=2\pi k\) for some integer \(k.\) Therefore, the condition for periodicity is that \(\omega _{0}N=2\pi k\) or \(\frac{\omega _{0}}{2\pi }=\frac{k}{N}\)

make sure \(k,N\) are relatively prime. If this is true, then \(N\) is the fundamental period. Notice that \(\frac{\omega _{0}}{2\pi }=f\), the frequency is samples per second.

2.1.1 Example 1

given \(\omega _{0}=\frac{3\pi }{5}\) find if \(\cos \left ( \omega _{0}n\right ) \) is periodic. First note that \(\cos \left ( \omega _{0}n\right ) =\cos \left ( 2\pi fn\right ) \), hence \(f=\frac{\omega _{0}}{2\pi }=\frac{3\pi }{10\pi }=\frac{3}{10}.\) Now \(\cos \left ( \omega _{0}\left ( n+N\right ) \right ) =\cos \left ( 2\pi f\left ( n+N\right ) \right ) =\cos \left ( 2\pi \frac{3}{10}\left ( n+N\right ) \right ) =\cos \left ( 2\pi \frac{3}{10}n+2\pi \frac{3}{10}N\right ) \), Hence if we set \(N=10\), then \(2\pi \frac{3}{10}N\) will be an integer an integer multiple of \(2\pi \)

Hence \(N=10\) is the period and this is periodic.

Second method: Once we find that \(f\) is rational, we can stop and say it is periodic. To find the period, make \(f\) to be lowest relatively prime numbers. Hence the period is the denominator\(\,.\) So, in this example, \(f=\frac{3}{10}\), we see it is periodic right away since \(f\) is rational. so period is \(10\). This second method is faster.

2.1.2 Example 2

First method

given \(\omega _{0}=3\) find if \(\cos \left ( \omega _{0}n\right ) \) is periodic. We see that \(f=\frac{3}{2\pi }\).

\(\cos \left ( \omega _{0}\left ( n+N\right ) \right ) =\cos \left ( 2\pi f\left ( n+N\right ) \right ) =\cos \left ( 2\pi \frac{3}{2\pi }\left ( n+N\right ) \right ) =\cos \left ( 2\pi \frac{3}{2\pi }n+2\pi \frac{3}{2\pi }N\right ) \)

\(=\cos \left ( 2\pi \frac{3}{2\pi }n+3N\right ) \). We see that we can’t find an integer \(N\) to make \(3N\) be a multiple of \(2\pi .\) Hence not periodic.

Second method:

Since \(f=\frac{3}{2\pi }\) is not rational, we stop. Not periodic.