Discrete-time Fourier Transform with Example

A slecture by ECE student Jacob Holtman

Partly based on the ECE438 Fall 2014 lecture material of Prof. Mireille Boutin.

## Definition of Discrete Time Fourier Transform (DTFT)

$X(\omega) = \sum_{k=-\infty}^{\infty}x[n]e^{-j\omega k}$

## Definition of Inverse Discrete Time Fourier Transform (iDTFT)

$x[n] = \frac{1}{2\pi}\int\limits_{-\pi}^{\pi} X(\omega)e^{j\omega n}d\omega$

X(ω) is seen to be periodic with a period of to see this ω is replaced with ω + 2kπ where k is an integer

$X(\omega + 2k\pi) = \sum_{n=-\infty}^{\infty}x[n]e^{-j({\color{blue}\omega + 2k\pi})n}$

Using the multiplicative rule of exponential the ω and 2kπ are split into two different exponential

$X(\omega + 2k\pi) = \sum_{n=-\infty}^{\infty}x[n]{\color{red}e^{-j\omega n}}{\color{blue}e^{2k\pi n}}$

given that n and k are integers k and so ej2kπn = 1 for all k, from Euler's identity and so

$X(\omega + 2k\pi) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} = X(\omega)$

so X(ω + 2kπ) = X(ω) for all ω

to find the DTFT of a complex exponential

$x[n] = e^{j\omega_0 n}$

The first step is to replace x[n] with the exponential in the DTFT equation

$X(\omega) = \sum_{n=-\infty}^{\infty}{\color{blue}e^{j\omega_0 n}}e^{-j\omega n}$

if ω = ω0 then the exponential is always 1 and the sum is divergent.

instead of finding the DTFT of a complex exponential it is easier to make an educated guess for what X(ω) is and checking to see if the equation holds and the initial guess is

2πδ(ω − ω0)

given that ω is periodic, as seen above, ω is between 0 and

in the iDTFT the replacement looks like

$\frac{1}{2\pi}\int\limits_{-\pi}^{\pi} {\color{blue}2\pi\delta(\omega-\omega_0)}e^{j\omega n}d\omega$

Because integration is a sum of infinitely small parts and delta functions are only equal to 1 at a distinct value the integration becomes the value of the equation at the delta offset point which is ω = ω0 so

$x[n] = {\color{red}\frac{1}{2\pi}2\pi\delta(\omega_0-\omega_0)}e^{j\omega_0 n}$

which simplifies to

$x[n] = e^{j\omega_0 n}$

since the iDTFT reproduces the DTFT then the DTFT of x[n] is 2πδ(ω − ω0)