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I want to find the inverse Fourier transform of $H(jw)=1/(a+jw)$. We know from the Fourier table that $$ F(e^{-at}) = 1/(a+jw). $$ So that $$ h(t)=e^{-at}. $$

But can we get $h(t)$ directly using inverse Fourier transform formula as below? $$ h(t) = \int_0^\infty H(jw)e^{jwt}dw\,. $$

Please help me. Thanks very much.

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  • $\begingroup$ I'm sorry I made some mistakes..The lower limit should be −∞ and the integral should be divided by 2π. $\endgroup$
    – Albert
    May 20 '15 at 15:39
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Use contour integration.

Case 1: $t>0$

Observe the integral over $C$, where $C$ is comprised of the real line plus an infinite semi-circle, $C_{\infty}$, enclosing the upper-half plane of the complex $\omega $ plane.

$$\frac{1}{2\pi}\oint_C\frac{e^{i\omega t}}{a+i\omega}d\omega=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{i\omega t}}{a+i\omega}d\omega+\int_{C_{\infty}}\frac{e^{i\omega t}}{a+i\omega}d\omega$$

Use of Jordan's Lemma shows that the integration over $C_{\infty}$ vanishes. Thus, we have

$$\frac{1}{2\pi}\oint_C\frac{e^{i\omega t}}{a+i\omega}d\omega=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{i\omega t}}{a+i\omega}d\omega$$

The contour integral is equal to $2\pi i$ times the (only) residue at $\omega =ia$. The residue is straightforward to compute and given by $-i e^{-at}$.

Thus, for $t>0$,

$$H(\omega)=e^{-at}$$


Case 2: $t<0$

Observe the integral over $C$, where $C$ is comprised of the real line plus an infinite semi-circle, $C_{\infty}$, enclosing the lower-half plane of the complex $\omega $ plane.

$$\frac{1}{2\pi}\oint_C\frac{e^{i\omega t}}{a+i\omega}d\omega=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{i\omega t}}{a+i\omega}d\omega+\int_{C_{\infty}}\frac{e^{i\omega t}}{a+i\omega}d\omega$$

Use of Jordan's Lemma shows that the integration over $C_{\infty}$ vanishes. Thus, we have

$$\frac{1}{2\pi}\oint_C\frac{e^{i\omega t}}{a+i\omega}d\omega=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{i\omega t}}{a+i\omega}d\omega$$

The contour integral is equal to $2\pi i$ times the residue in the lower-half plane. However, there is no singularity in the lower-half plane and the residue is trivially zero.

Thus, for $t<0$,

$$H(\omega)=0$$

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  • $\begingroup$ There is something wrong with my previous comment.. My question is: Consider the frequency spectrum H(jw) is 1/(a+jw) only when |w|< W; and when |w|> W H(jw)=0. Can we still use inverse fourier formula to get h(t)? In other words, can we get the expression of h(t) by calculating the integral below? $$ h(t) = \frac{1}{2π}\int_{-W}^W \frac{e^{jwt}}{a+jw}dw\,. $$ $\endgroup$
    – Albert
    May 20 '15 at 21:23
  • $\begingroup$ The new truncated integral can be viewed as the FT of $1/(a+iω)$ times a pulse function of unit height that extends from $−W$ to $W$ . Then recall that the FT of the product is the convolution of the Fourier Transforms. We have the FT of $1/(a+iω)$ and the FT of a pulse function is a sinc function. Can you convolve these? $\endgroup$
    – Mark Viola
    May 21 '15 at 18:19

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