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==Definition of Inverse Discrete Time Fourier Transform (iDTFT)== | ==Definition of Inverse Discrete Time Fourier Transform (iDTFT)== | ||
| − | <math>x[n] = \frac{1}{2\pi}\int\limits_{ | + | <math>x[n] = \frac{1}{2\pi}\int\limits_{-\pi}^{\pi} X(\omega)e^{j\omega n}d\omega </math> |
| + | ---- | ||
<math>X(\omega) </math> is seen to be periodic with a period of <math>2\pi</math> to see this <math>\omega</math> is replaced with <math>\omega + 2k\pi</math> where k is an integer | <math>X(\omega) </math> is seen to be periodic with a period of <math>2\pi</math> to see this <math>\omega</math> is replaced with <math>\omega + 2k\pi</math> where k is an integer | ||
| − | <math>X(\omega + | + | <math>X(\omega + 2k\pi) = \sum_{n=-\infty}^{\infty}x[n]e^{-j({\color{blue}\omega + 2k\pi})n} </math> |
Using the multiplicative rule of exponential the <math>\omega</math> and <math>2k\pi</math> are split into two different exponential | Using the multiplicative rule of exponential the <math>\omega</math> and <math>2k\pi</math> are split into two different exponential | ||
| − | <math>X(\omega + | + | <math>X(\omega + 2k\pi) = \sum_{n=-\infty}^{\infty}x[n]{\color{red}e^{-j\omega n}}{\color{blue}e^{2k\pi n}} </math> |
| − | given that n and k are integers k and so <math>e^{-j2k\pi n} = 1 </math> from Euler's identity and so | + | given that n and k are integers k and so <math>e^{-j2k\pi n} = 1 </math> for all k, from Euler's identity and so |
| − | <math>X(\omega + | + | <math>X(\omega + 2k\pi) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} = X(\omega) </math> |
| − | so <math>X(\omega + | + | so <math>X(\omega + 2k\pi) = X(\omega) </math> for all <math>\omega</math> |
---- | ---- | ||
| + | to find the DTFT of a complex exponential | ||
| + | |||
| + | <math>x[n] = e^{j\omega_0 n} </math> | ||
| + | |||
| + | The first step is to replace x[n] with the exponential in the DTFT equation | ||
| + | |||
| + | <math>X(\omega) = \sum_{n=-\infty}^{\infty}{\color{blue}e^{j\omega_0 n}}e^{-j\omega n} </math> | ||
| + | |||
| + | if <math>\omega</math> = <math>\omega_0</math> 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 <math>X(\omega)</math> is and checking to see if the equation holds and the initial guess is | ||
| + | |||
| + | <math>2\pi\delta(\omega-\omega_0)</math> | ||
| + | |||
| + | given that <math>\omega</math> is periodic, as seen above, <math>\omega</math> is between 0 and <math>2\pi</math> | ||
| + | |||
| + | in the iDTFT the replacement looks like | ||
| + | |||
| + | <math>\frac{1}{2\pi}\int\limits_{-\pi}^{\pi} {\color{blue}2\pi\delta(\omega-\omega_0)}e^{j\omega n}d\omega </math> | ||
| + | |||
| + | 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 <math>\omega = \omega_0</math> so | ||
| + | <math>x[n] = {color{red}\frac{1}{2\pi}2\pi\delta(\omega_0}-\omega_0)}e^{j\omega_0 n} </math> | ||
| + | which simplifies to | ||
| + | <math>x[n] = e^{j\omega_0 n}</math> | ||
---- | ---- | ||
---- | ---- | ||
Revision as of 20:23, 29 September 2014
Discrete-time Fourier transform
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(\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} $
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