Examples of the Inverse Fourier Transform

Electrical Circuit Analysis > Fourier Transform

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Example 1: Obtain the inverse Fourier transform of:
(a) $ F(\omega)=\frac{10 j \omega+4}{(j \omega)^{2}+6 j \omega+8} $
(b) $ G(\omega)=\frac{\omega^{2}+21}{\omega^{2}+9} $

Solution:
(a) To avoid complex algebra, we can replace $ j \omega $ with $ s $ for the moment. Using partial fraction expansion,

$$F(s)=\frac{10 s+4}{s^{2}+6 s+8}=\frac{10 s+4}{(s+4)(s+2)}=\frac{A}{s+4}+\frac{B}{s+2}$$

where

$$\begin{array}{l}A=\left.(s+4) F(s)\right|_{s=-4}=\left.\frac{10 s+4}{(s+2)}\right|_{s=-4}=\frac{-36}{-2}=18 \\B=\left.(s+2) F(s)\right|_{s=-2}=\left.\frac{10 s+4}{(s+4)}\right|_{s=-2}=\frac{-16}{2}=-8\end{array}$$

Substituting $ A=18 $ and $ B=-8 $ in $ F(s) $ and $ s $ with $ j \omega $ gives

$$F(j \omega)=\frac{18}{j \omega+4}+\frac{-8}{j \omega+2}$$

we obtain the inverse transform as

$$f(t)=\left(18 e^{-4 t}-8 e^{-2 t}\right) u(t)$$

(b) We simplify $ G(\omega) $ as

$$G(\omega)=\frac{\omega^{2}+21}{\omega^{2}+9}=1+\frac{12}{\omega^{2}+9}$$

the inverse transform is obtained as

$$g(t)=\delta(t)+2 e^{-3|t|}$$

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