Lecture 12 - Notes

January 29, 2016

Eigenfunctions - Continued

definition: Given a system $y[n] = H\left(e^{j \omega}\right) e^{j \omega n}$ with an impulse response $h[n]$, the frequency response is,

$$\begin{align} H\left(e^{j \omega}\right) = \sum_{n = -\infty}^{\infty} h[n] e^{-j \omega n} \end{align}$$

Given a difference equation,

$$\begin{align} \sum_{i=0}^{M} b_i x[n - i] = \sum_{j=0}^{N} a_j y[n-j] \end{align}$$

the frequency response is,

$$\begin{align} H\left(e^{j \omega}\right) = \frac{\sum_{i=0}^{M} b_i e^{-j \omega i}}{\sum_{k=0}^{N} a_k e^{-j \omega k}} \end{align}$$

Examples

Is $3^n$ an eigenfunction?

Yes. Since $3^n = \left(3e^{j0}\right)^n$.

Is $\cos{\omega_0 n}$ an eigenfunction?

No.

Sinusoidal System Responses

Suppose, $x[n] = A \cos{\omega_0 n + \varphi}$, then

$$\begin{align} x[n] &= A \cos{\omega_0 n + \varphi} \\\\ &= \frac{A}{2}\left( e^{j(\omega_0 n + \varphi)} + e^{-j(\omega_0 n + \varphi)} \right) \\\\ &= \frac{A}{2}\left( e^{j\omega_0 n}e^{j\varphi} + e^{-j\omega_0 n}e^{j\varphi} \right) \\\\ &= \frac{A}{2} e^{j\omega_0 n}e^{j\varphi} + \frac{A}{2} e^{j\omega_0 n}e^{-j\varphi} \\\\ &\xrightarrow{S} \frac{A}{2} e^{j\varphi} H\left(e^{j \omega_0}\right) e^{j\omega_0 n} + \frac{A}{2} e^{-j\varphi} H\left(e^{-j \omega_0}\right) e^{j\omega_0 n} \\\\ \end{align}$$

This gives us,

$$\begin{align} y[n] &= \frac{A}{2} e^{j\varphi} H\left(e^{j \omega_0}\right) e^{j\omega_0 n} + \frac{A}{2} e^{-j\varphi} H\left(e^{-j \omega_0}\right) e^{j\omega_0 n} \\\\ &= A \left|H\left( e^{j \omega_0} \right)\right|\cos{\left(\omega_0 n + \varphi - \theta_0\right)} \end{align}$$