Example of Computation of inverse Fourier transform (CT signals)

A practice problem on CT Fourier transform


Inverse Fourier Transform

$ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega $



$ X(\omega) = \pi\delta(\omega - 4\pi)(2-3j) + \pi\delta(\omega + 4\pi)(2+3j) $



$ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}[\pi\delta(\omega - 4\pi)(2-3j) + \pi\delta(\omega + 4\pi)(2+3j)]e^{j\omega t}d\omega $

$ =\frac{2-3j}{2}\int_{-\infty}^{\infty}\delta(\omega - 4\pi)e^{j\omega t}d\omega + \frac{2+3j}{2}\int_{-\infty}^{\infty}\delta(\omega + 4\pi)e^{j\omega t}d\omega $

$ =\frac{2-3j}{2}e^{j4\pi t} + \frac{2+3j}{2}e^{-j4\pi t} $

$ =e^{j4\pi t}-\frac{3j}{2}e^{j4\pi t} + e^{-j4\pi t}+\frac{3j}{2}e^{-j4\pi t} $

$ =\frac{3}{2j}e^{j4\pi t}-\frac{3}{2j}e^{-j4\pi t}+e^{j4\pi t} + e^{-j4\pi t} $

$ = \frac{3(e^{j4\pi t} - e^{-j4\pi t})}{2j}+\frac{2(e^{j4\pi t} + e^{-j4\pi t})}{2} $

$ =3sin(4\pi t) + 2 cos(4\pi t) $


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