Lecture 2 - Notes

January 6, 2016

Signal Representation

A sinusoid using 6 data points.

$$x(t) = \sum_{k=1}^{6} A_k \sin{(\omega_k t +\phi_k)}$$

$k$	$\omega_k$	$A_k$	$\phi_k$
1	1	0.2388	0.1897
2	2	...	...
3	$2\sqrt{3}$	...	...
4	8	...	...
5	10	...	...
6	12	...	...

Complex Numbers

Let $z_0$ be a complex number with polar coordinates $(r_0,\Theta_0)$ and cartesian coordinates $(x_0,y_0)$. Then,

$$\begin{align} x_0 &= \mathbb{Re}\{z_0\} = r_0 \cos{\Theta_0} \newline y_0 &= \mathbb{Im}\{z_0\} = r_0 \sin{\Theta_0} \end{align}$$

If we plot $r_0 = 2$ and $\Theta_0 = \frac{\pi}{4}$,

Periodic Properties of Discreet Time Signals

We can represent a discreet signal as,

$$x[n] = e^{j \omega_0 n}$$ and we know that,

$$e^{j (\omega_0 + 2\pi) n} = e^{j \omega_0 n}$$

Because the signals are periodic we choose an interval length of $2\pi$, and either $[ -\pi, \pi)$ or $[0,2\pi)$

Condition for Periodicity

A signal $x[n]$ is periodic with period $N$ if,

$$e^{j \omega_0 n} = e^{j \omega_0 (m+N)}$$

for all $m \in \mathbb{R}$.

$$\begin{align} e^{j \omega_0 n} = e^{j(\omega_0n + 2k\pi)} &= e^{j \omega_0 (m + N)} \newline \omega_0n + 2k\pi &= \omega_0(m+N) \newline 2k \pi &= \omega_0 n \end{align}$$

Example

Determine the fundamental period of,

$$x[n] = 2 \cos{\left(\frac{\pi}{4}n\right)} + \sin{\left(\frac{\pi}{8}n\right)} - 2 \cos{\left( \frac{\pi}{2}n + \frac{\pi}{6} \right)}$$

Solution

$N$ will be $\text{lcm}{(N_1,N_2,N_3)}$ where,

$$x[n] = 2 \underset{x_1}\cos{\left(\frac{\pi}{4}n\right)} + \underset{x_2}\sin{\left(\frac{\pi}{8}n\right)} - 2 \underset{x_3}\cos{\left( \frac{\pi}{2}n + \frac{\pi}{6} \right)}$$

So for $N_1$, since $x[n] = x[n + N_1]$ and $\omega_0 n = \omega_0 n + 2k\pi$

$$\begin{align} x_1[n + N_1] &= x_1[n] \newline 2 \cos{\left(\frac{\pi}{4}\left(n + N_1\right)\right)} &= 2 \cos{\left(\frac{\pi}{4}n\right)} \newline 2 \cos{\left(\frac{\pi}{4}\left(n + N_1\right)\right)} &= 2 \cos{\left(\frac{\pi}{4}n + 2k\pi\right)} \newline \frac{\pi}{4} (n + N_1) &= \frac{\pi}{4}n + 2k\pi \newline n + N_1 &= n + 8k \newline N_1 &= 8k \newline \end{align}$$

For $N_2$,

$$\begin{align} x_2[n + N_2] &= x_2[n] \newline \end{align}$$

and so on.