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− | + | <font size="4">'''Discrete-time Fourier transform''' </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> | + | |
+ | '''Failed to parse (lexing error): X(\omega) := \sum_{k=-\infty}^{\infty}x[n]e^{-j\omega k} ''' | ||
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+ | <br> | ||
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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 |
+ | |||
+ | <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-\omega_0)}e^{j\omega_0 n} </math> | + | <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> |
+ | |||
+ | 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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---- | ---- | ||
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− | If you have any questions, comments, etc. please post them on [[ | + | (create a question page and put a link below) |
+ | |||
+ | == [[Slecture title of slecture review|Questions and comments]] == | ||
+ | |||
+ | If you have any questions, comments, etc. please post them on [[Slecture title of slecture review|this page]]. | ||
+ | |||
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− | [[ | + | |
+ | [[2014 Fall ECE 438 Boutin|Back to ECE438, Fall 2014]] | ||
+ | |||
+ | [[Category:Slecture]] [[Category:ECE438Fall2014Boutin]] [[Category:ECE]] [[Category:ECE438]] [[Category:Signal_processing]] |
Revision as of 20:40, 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)
Failed to parse (lexing error): 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)
(create a question page and put a link below)
Questions and comments
If you have any questions, comments, etc. please post them on this page.