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Example 1. <math>x(t)= e^{j\omega_o t} \qquad \qquad X(f) = 2\pi \delta (\omega - \omega_o )</math> | Example 1. <math>x(t)= e^{j\omega_o t} \qquad \qquad X(f) = 2\pi \delta (\omega - \omega_o )</math> | ||
− | Again we will let <math>\omega | + | Again we will let <math>\omega = 2\pi f</math> in our Fourier Transform <math>X(f)</math> |
<math> X(f) = 2\pi \delta (2\pi f - 2\pi f_o )</math> | <math> X(f) = 2\pi \delta (2\pi f - 2\pi f_o )</math> |
Revision as of 12:29, 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(f) = 2\pi \delta (\omega - \omega_o ) $
Again we will let $ \omega = 2\pi f $ in our Fourier Transform $ X(f) $
$ X(f) = 2\pi \delta (2\pi f - 2\pi f_o ) $
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