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

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 (2\pi f - 2\pi f_o ) - \delta (2\pi f + 2\pi f_o )] \\ & = {\color{red}2\pi} \delta ({\color{red}2\pi}(f - f_o )\\ & = \delta (f - f_o ) \end{align} $

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