Lecture 15 - Notes

February 17, 2016

Properties of Fourier Transforms

Symmetry

For an arbitrary $x[n]$,

$$\begin{align} x[n] = \underbrace{x_\text{even}[n]}_{X_R\left(e^{j\omega}\right)} + \underbrace{x_\text{odd}[n]}_{X_I\left(e^{j\omega}\right)} \end{align}$$

If,

$$\begin{gather} \left| X\left(e^{j\omega}\right) \right| = \left| X\left(e^{-j\omega}\right) \right| \\ \end{gather}$$

then $x[n]$ is even. If,

$$\begin{align} \angle X\left(e^{j\omega}\right) = \angle X\left(e^{-j\omega}\right) \end{align}$$

then $x[n]$ is odd.

Time and Frequency Shifting

To shift in the time domain,

$$\begin{align} x[n - n_d] \overset F \longleftrightarrow e^{-jwn_d} X \left( e^{j\omega} \right) \end{align}$$

To shift in the frequency domain,

Frequency Differentiation

To differentiate in the frequency domain,

$$\begin{align} nx[n] \overset F \longleftrightarrow j \frac{d X \left( e^{j\omega} \right)}{d\omega} \end{align}$$

Example

Find $X \left( e^{j\omega} \right)$ of,

$$\begin{align} x[n] = (n - 1)\left( \frac{1}{3} \right)^{\left| n \right|} \end{align}$$

Solution

Let,

$$\begin{align} x_1[n] &= \left( \frac{1}{3} \right)^{\left| n \right|} \\\\ &= \begin{cases} \left( \frac{1}{3} \right)^{ n} & n \ge 0 \\\\ \left( \frac{1}{3} \right)^{-n} & n \lt 0 \\\\ \end{cases} \end{align}$$

so,

$$\begin{align} X_1 \left( e^{j\omega} \right) &= \sum_{n = -\infty}^{\infty} \left( \frac{1}{3} \right)^{|n|} e^{-j\omega n} \\\\ &= \sum_{n = -\infty}^{0} \left( \frac{1}{3} \right)^{-n} e^{-j\omega n} + \sum_{n = 0}^{\infty} \left( \frac{1}{3} \right)^{n} e^{-j\omega n} - 1 \\\\ &= \sum_{p = 0}^{\infty} \left( \frac{1}{3} \right)^{p} e^{j\omega p} + \sum_{n = 0}^{\infty} \left( \frac{1}{3} \right)^{n} e^{-j\omega n} - 1 \\\\ &= \frac{1}{1 - \frac{1}{3} e^{j\omega}} + \frac{1}{1 - \frac{1}{3} e^{-j\omega}} - 1 \\\\ &= \frac{4}{5 - 3 \cos{\omega}} \end{align}$$

now,

$$\begin{align} x[n] &= n x_1[n] - x_1[n] \\\\ X \left( e^{j\omega} \right) &= j \frac{d X_1 \left( e^{j\omega} \right)}{d\omega} - X_1 \left( e^{j\omega} \right) \\\\ &~~\vdots \\\\ &= \frac{4}{5 - 3 \cos{\omega}} - j \frac{12 \sin{\omega}}{\left(5 - 3 \cos{\omega} \right)^2} \end{align}$$

Convolution

To transform convolution in the time domain into the frequency domain,

$$\begin{align} x_1[n] &\overset F \longleftrightarrow X_1 \left( e^{j\omega} \right) \\\\ x_2[n] &\overset F \longleftrightarrow X_2 \left( e^{j\omega} \right) \\\\ &~~\downarrow \\\\ x_1[n] * x_2[n] &\overset F \longleftrightarrow X_1 \left( e^{j\omega} \right)X_2 \left( e^{j\omega} \right) \\\\ \end{align}$$