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| − | + | <font size="4">'''Discrete-time Fourier Transform with Example''' </font> | |
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| − | + | A [https://www.projectrhea.org/learning/slectures.php slecture] by [[ECE]] student Jacob Holtman | |
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| − | + | Partly based on the [[2014 Fall ECE 438 Boutin|ECE438 Fall 2014 lecture]] material of [[User:Mboutin|Prof. Mireille Boutin]]. | |
| − | + | </center> | |
| − | Partly based on the [[ | + | |
| − | </center> | + | |
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| − | ==Definition of Inverse Discrete Time Fourier Transform (iDTFT)== | + | == Definition of Discrete Time Fourier Transform (DTFT) == |
| − | <math>x[n] = \frac{1}{2\pi}\int\limits_{-\pi}^{\pi} X(\omega)e^{j\omega n}d\omega </math> | + | |
| + | <math>X(\omega) = \sum_{k=-\infty}^{\infty}x[n]e^{-j\omega k} </math> | ||
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| + | == Definition of Inverse Discrete Time Fourier Transform (iDTFT) == | ||
| + | |||
| + | <math>x[n] = \frac{1}{2\pi}\int\limits_{-\pi}^{\pi} X(\omega)e^{j\omega n}d\omega </math> | ||
---- | ---- | ||
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| − | < | + | <span class="texhtml">''X''(ω)</span> is seen to be periodic with a period of <span class="texhtml">2π</span> to see this <span class="texhtml">ω</span> is replaced with <span class="texhtml">ω + 2''k''π</span> where k is an integer |
| − | + | <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 <span class="texhtml">ω</span> and <span class="texhtml">2''k''π</span> are split into two different exponential |
| − | + | <math>X(\omega + 2k\pi) = \sum_{n=-\infty}^{\infty}x[n]{\color{red}e^{-j\omega n}}{\color{blue}e^{2k\pi n}} </math> | |
| − | <math>X(\omega + 2k\pi) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} = X(\omega) </math> | + | given that n and k are integers k and so <span class="texhtml">''e''<sup> − ''j''2''k''π''n''</sup> = 1</span> for all k, from Euler's identity and so |
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| + | <math>X(\omega + 2k\pi) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} = X(\omega) </math> | ||
| + | |||
| + | so <span class="texhtml">''X''(ω + 2''k''π) = ''X''(ω)</span> for all <span class="texhtml">ω</span> | ||
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to find the DTFT of a complex exponential | to find the DTFT of a complex exponential | ||
| − | <math>x[n] = e^{j\omega_0 n} </math> | + | <math>x[n] = e^{j\omega_0 n} </math> |
| − | The first step is to replace x[n] with the exponential in the DTFT equation | + | 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> | + | <math>X(\omega) = \sum_{n=-\infty}^{\infty}{\color{blue}e^{j\omega_0 n}}e^{-j\omega n} </math> |
| − | if < | + | if <span class="texhtml">ω</span> = <span class="texhtml">ω<sub>0</sub></span> 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 < | + | instead of finding the DTFT of a complex exponential it is easier to make an educated guess for what <span class="texhtml">''X''(ω)</span> is and checking to see if the equation holds and the initial guess is |
| − | < | + | <span class="texhtml">2πδ(ω − ω<sub>0</sub>)</span> |
| − | given that < | + | given that <span class="texhtml">ω</span> is periodic, as seen above, <span class="texhtml">ω</span> is between 0 and <span class="texhtml">2π</span> |
| − | in the iDTFT the replacement looks like | + | 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> | + | <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 < | + | 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 <span class="texhtml">ω = ω<sub>0</sub></span> so |
| − | <math>x[n] = {color{red}\frac{1}{2\pi}2\pi\delta(\omega_0 | + | <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 | which simplifies to | ||
| − | <math>x[n] = e^{j\omega_0 n}</math> | + | <math>x[n] = e^{j\omega_0 n}</math> |
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| + | since the iDTFT reproduces the DTFT then the DTFT of x[n] is <span class="texhtml">2πδ(ω − ω<sub>0</sub>)</span> | ||
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| − | + | == [[Holtman_Slecture_Review|Questions and comments]] == | |
| − | ==[[ | + | |
| + | If you have any questions, comments, etc. please post them on [[Holtman_Slecture_Review|this page]]. | ||
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| − | [[ | + | [[2014_Fall_ECE_438_Boutin_digital_signal_processing_slectures|Back to ECE438 slectures, Fall 2014]] |
| + | |||
| + | [[Category:Slecture]] [[Category:ECE438Fall2014Boutin]] [[Category:ECE]] [[Category:ECE438]] [[Category:Signal_processing]] [[Category:Discrete-time_Fourier_transform]] | ||
Latest revision as of 20:02, 16 March 2015
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 + 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 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 + 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 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 = 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)
Questions and comments
If you have any questions, comments, etc. please post them on this page.
