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As earlier we will let <math>\omega = 2\pi f</math> in our Fourier Transform <math>X(f)</math> , and we will use the scaling property of the Dirac<math>\delta</math> Function: <math>c\delta (ct) = \delta (t) </math> | As earlier we will let <math>\omega = 2\pi f</math> in our Fourier Transform <math>X(f)</math> , and we will use the scaling property of the Dirac<math>\delta</math> Function: <math>c\delta (ct) = \delta (t) </math> | ||
+ | <math> \begin{align} \\ | ||
+ | |||
+ | X({\color{red}2\pi f}) & = \frac{\pi }{j}\ [\delta (\omega - \omega_o ) - \delta (\omega + \omega_o )] | ||
+ | \\ | ||
+ | & = {\color{red}2\pi} \delta ({\color{red}2\pi}(f - f_o )\\ | ||
+ | & = \delta (f - f_o ) | ||
+ | \end{align} | ||
+ | </math> | ||
Revision as of 13:21, 18 September 2014
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 (\omega - \omega_o ) - \delta (\omega + \omega_o )] \\ & = {\color{red}2\pi} \delta ({\color{red}2\pi}(f - f_o )\\ & = \delta (f - f_o ) \end{align} $
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