Difference between revisions of "Fourier Transform as a FUnction of Frequency w versus Frequency f (in Hertz)" - Rhea

Revision as of 12:37, 18 September 2014

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) $

$ X(2\pi f) = 2\pi \delta (2\pi f - 2\pi f_o ) $

,math>\begin{align} \\

f(x) &= \int\limits_{0}^{2\pi} sin^2(\theta) \ d\theta \\

&= \int\limits_{0}^{2\pi} \big(1-cos^2(\theta) \big)\ d\theta \\ 

&= \int\limits_{0}^{2\pi} \bigg( 1-\Big(\frac{1}{2} +
\frac{1}{2} cos(2\theta)\Big) \bigg)\ d\theta \\ 

&= \int\limits_{0}^{2\pi} \frac{1}{2}\ d\theta -

\frac{1}{2} \int\limits_{0}^{2\pi} cos(2\theta) \ d\theta \\

&= \pi \end{align} </math>

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