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To show the relationship between the Fourier Transform of frequency <math>\omega</math> versus frequency <math>f</math> (in hertz) we start with the definitions: | To show the relationship between the Fourier Transform of frequency <math>\omega</math> versus frequency <math>f</math> (in hertz) we start with the definitions: | ||
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<math>X(w)=\int\limits_{-\infty}^{\infty} x(t)e^{-jwt} dt \qquad \qquad \qquad \qquad X(f)=\int\limits_{-\infty}^{\infty}x(t)e^{-j2\pi ft} dt </math> | <math>X(w)=\int\limits_{-\infty}^{\infty} x(t)e^{-jwt} dt \qquad \qquad \qquad \qquad X(f)=\int\limits_{-\infty}^{\infty}x(t)e^{-j2\pi ft} dt </math> | ||
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[1].Mireille Boutin, "ECE438 Digital Signal Processing with Applications," Purdue University August 26,2009 | [1].Mireille Boutin, "ECE438 Digital Signal Processing with Applications," Purdue University August 26,2009 | ||
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| + | ==[[Fourier_Transform_as_a_FUnction_of_Frequency_w_versus_Frequency_f_in_Hertz_review|Questions and comments]]== | ||
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| − | + | [[2014_Fall_ECE_438_Boutin_digital_signal_processing_slectures|Back to ECE438 slectures, Fall 2014]] | |
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Latest revision as of 09:50, 14 March 2015
Fourier Transform as a Function of Frequency w Versus Frequency f (in Hertz)
A slecture by ECE student Randall Cochran
Partly based on the ECE438 Fall 2014 lecture material of Prof. Mireille Boutin.
To show the relationship between the Fourier Transform of frequency $ \omega $ versus frequency $ f $ (in hertz) we start with the definitions:
$ X(w)=\int\limits_{-\infty}^{\infty} x(t)e^{-jwt} dt \qquad \qquad \qquad \qquad X(f)=\int\limits_{-\infty}^{\infty}x(t)e^{-j2\pi ft} dt $
now we let $ \omega = 2\pi f $
$ X(2\pi f)=\int\limits_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt $
making $ X(2\pi f) = X(f) $
Examples of the relationship can be shown by starting with known CTFT pairs:
Example 1.
$ x(t)= e^{j\omega_o t} \qquad \qquad X(\omega ) = 2\pi \delta (\omega - \omega_o ) $
Again we will let $ \omega = 2\pi f $ in our Fourier Transform $ X(f) $ , and we will use the scaling property of the Dirac$ \delta $ Function: $ c\delta (ct) = \delta (t) $
$ \begin{align} \\ X({\color{red}2\pi f}) & = 2\pi \delta ({\color{red}2\pi f} - ({\color{red}2\pi f_o}))\\ & = {\color{red}2\pi} \delta ({\color{red}2\pi}(f - f_o )\\ & = \delta (f - f_o ) \end{align} $
And previously it was shown that $ X(2\pi f) = X(f) $ completing the change of variables.
Example 2.
$ x(t) = sin(\omega t) \qquad \qquad X(\omega ) = \frac{\pi }{j}\ [\delta (\omega - \omega_o ) - \delta (\omega + \omega_o )] $
As earlier we will let $ \omega = 2\pi f $ in our Fourier Transform $ X(f) $ , and we will use the scaling property of the Dirac$ \delta $ Function: $ c\delta (ct) = \delta (t) $
$ \begin{align} \\ X({\color{red}2\pi f}) & = \frac{\pi }{j}\ [\delta ({\color{red}2\pi f} - {\color{red}2\pi f_o } ) - \delta ({\color{red}2\pi f} + {\color{red}2\pi f_o })]\\ & = \frac{\pi }{j}\ ( [ \delta ({\color{red}2\pi}(f - f_o ) - \delta({\color{red}2\pi} (f + f_o ))\\ & = \frac{\pi }{j}\ {\color{red}\frac{1}{2\pi }} ({\color{red}2\pi } [ \delta ( {\color{red}2\pi }(f - f_o ) - \delta({\color{red}2\pi} (f + f_o ))\\ & = \frac{1}{2j}\ ( \delta (f - f_o ) - \delta(f + f_o ))\\ \end{align} $
References
[1].Mireille Boutin, "ECE438 Digital Signal Processing with Applications," Purdue University August 26,2009
Questions and comments
If you have any questions, comments, etc. please post them on this page.
