**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 2π to see this ω is replaced with ω + 2*k*π 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 2*k*π 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 *e*^{ − j2kπn} = 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 + 2*k*π) = *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 = ω_{0} 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 − ω_{0})

given that ω is periodic, as seen above, ω is between 0 and 2π

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 = ω_{0} 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 − ω_{0})

## Questions and comments

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