## Example of Computation of inverse Fourier transform (CT signals)

Problem: (From Oppenheim/Wisllsky, 4.22 b.) Find $F^{-1} (cos(4w+\frac{\pi}{3})).$ ($F^{-1}$ meaning the inverse Fourier Transform of said function.)

Solution: First, observe that $cos(4w+\frac{\pi}{3}) = \frac{1}{2} (e^{j4w} e^{j\frac{\pi}{3}} + e^{-j4w} e^{-j\frac{\pi}{3}})$. First, let's try to apply the formula for the inverse Fourier transform directly:

$F^{-1} (cos(4w+\frac{\pi}{3})) = \int_{-\infty}^\infty cos(4w+\frac{\pi}{3}) e^{jwt} dt$

$= \frac{1}{2\pi} \int_{-\infty}^\infty \frac{1}{2} (e^{j4w} e^{j\frac{\pi}{3}} + e^{-j4w} e^{-j\frac{\pi}{3}}) e^{jwt} dt$

$= \frac{1}{2\pi} \int_{-\infty}^\infty \frac{1}{2} (e^{j4w} e^{j\frac{\pi}{3}} + e^{-j4w} e^{-j\frac{\pi}{3}}) e^{jwt} dt$

$= \frac{1}{2\pi} \int_{-\infty}^\infty \frac{1}{2} e^{j(t+4)w} e^{j\frac{\pi}{3}} + e^{j(t-4)w} e^{-j\frac{\pi}{3}} dt$

Noting that integrating this is a very difficult undertaking, let's try another approach. Say, for instance, that we were to find x(t) such that $F(x(t)) = e^{-jwt_0}$. We'll take the highly-educated guess that $x(t) = \delta(t-t_0)$:

$F(\delta(t-t_0)) = \int_{-\infty}^\infty \delta(t-t_0) e^{-jwt} dt = e^{-jwt_0}$

So, we can conclude that since $F(\delta(t-t_0)) = e^{-jwt_0}$, $F^{-1}(e^{-jwt_0}) = \delta(t-t_0)$. Applying this to our problem at hand:

$F^{-1} (cos(4w+\frac{\pi}{3})) = F^{-1} ( \frac{1}{2} (e^{j4w} e^{j\frac{\pi}{3}} + e^{-j4w} e^{-j\frac{\pi}{3}}))$

$= \frac{1}{2} (F^{-1} (e^{j4w}) e^{j\frac{\pi}{3}} + F^{-1} (e^{-j4w}) e^{-j\frac{\pi}{3}})$

$= \frac{1}{2} ( e^{j\frac{\pi}{3}} \delta(t + 4) + e^{-j\frac{\pi}{3}} \delta(t-4) )$

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