Difference between revisions of "Discrete-time Fourier transform (DTFT) Slecture by Jacob Holtman" - Rhea

Revision as of 20:25, 29 September 2014

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(\omega) $ is seen to be periodic with a period of $ 2\pi $ to see this $ \omega $ is replaced with $ \omega + 2k\pi $ 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 $ \omega $ and $ 2k\pi $ 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^{-j2k\pi n} = 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(\omega + 2k\pi) = X(\omega) $ for all $ \omega $

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 $ \omega $ = $ \omega_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(\omega) $ is and checking to see if the equation holds and the initial guess is

$ 2\pi\delta(\omega-\omega_0) $

given that $ \omega $ is periodic, as seen above, $ \omega $ is between 0 and $ 2\pi $

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 $ \omega = \omega_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 $ 2\pi\delta(\omega - \omega_0) $

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