Inverse Fourier Transform

Question

I've got a problem where I need to find the IFT of

$$F(\omega) = \frac{1 + i\omega}{6-\omega^2+5i\omega}$$

I've been trying to solve it through partial fractions, but that gives us

$$\frac{1 + i\omega}{6-\omega^2+5i\omega} = \frac{1 + i\omega}{(-\omega + 3i)(\omega-2i)} = \frac{i(\omega - i)}{(-\omega+3i)(\omega-2i)}= \frac{A(\omega-2i) + B(-\omega+3i)}{(-\omega + 3i)(\omega-2i)} $$

and I'm not entirely certain how to solve for A and B here.

Substituting it into the IFT formula give us

$$\frac 1{2\pi} \int_{-\infty}^\infty \frac{1 + i\omega}{6-\omega^2+5i\omega} \cdot e^{i\omega t} \ d\omega$$

or any combination of the above factorizations, but I have no idea how to even begin integrating something like this.

Anything, even a hint at what I should do in such a situation like this would help.

Thanks a lot for taking a look!

Answer 1

By using partial fractions we have \begin{align} \frac{i\omega +1}{6-\omega^2+5i\omega}&=\frac{i\omega+1}{(i\omega+2)(i\omega+3)}\\ &=\frac{A}{i\omega+3}+\frac{B}{i\omega+2} \end{align} Where $A$ and $B$ are constants such that \begin{align} A(i\omega+2)+B(i\omega+3)&=i\omega+1\\ (A+B)i\omega+2A+3B&=i\omega+1 \end{align} Then, solving the linear system of equations \begin{align} A+B&=1\\ 2A+3B&=1 \end{align} we get $\color{red}{A=2}$ and $\color{red}{B=-1}$. Hence \begin{align} \frac{i\omega +1}{6-\omega^2+5i\omega}&=\frac{2}{i\omega+3}-\frac{1}{i\omega+2} \end{align} From here you can apply the inverse FT: \begin{align} \mathscr{F}^{-1}\left\{\frac{i\omega +1}{6-\omega^2+5i\omega}\right\}&=2\mathscr{F}^{-1}\left\{\frac{1}{i\omega+3}\right\}-\mathscr{F}^{-1}\left\{\frac{1}{i\omega+2}\right\}\\[5pt] &=\color{blue}{\boxed{\left(2e^{-3t}-e^{-2t}\right)u(t)}} \end{align}

Answer 2

We can evaluate the integral using contour integration. First, let

$$F(\omega) = \frac{1 + i\omega}{6-\omega^2+5i\omega} $$

Then, the inverse Fourier Transform $\mathscr{F}^{-1}\{F\}(t)$ is given by

$$\begin{align} \mathscr{F}^{-1}\{F\}(t)&=\frac{1}{2\pi}\int_{-\infty}^\infty F(\omega)e^{i\omega t}\,d\omega\\\\ &=\frac{1}{2\pi}\int_{-\infty}^\infty \frac{1 + i\omega}{6-\omega^2+5i\omega}e^{i\omega t}\,d\omega\\\\ &=-\frac{1}{2\pi}\int_{-\infty}^\infty \frac{1 + i\omega}{(\omega-3i)(\omega-2i)}e^{i\omega t}\,d\omega\\\\ \end{align}$$

We now analyze the closed-contour integral $I(t)$ as given by

$$I(t)=-\frac1{2\pi}\oint_C \frac{1 + iz}{(z-3i)(z-2i)}e^{iz t}\,dz$$

where $C$ is the contour comprised of the line segment along the real axis from $-R$ to $R$ and a semi-circular arc $C_R$, with radius R, centered at the origin in the upper-half plane (lower-half plane) for $t>0$ ($t<0$).

Note that the integrand has poles at $z =3i$ and $z=2i$. Therefore, from the residue theorem we have

$$I(t)= \begin{cases} 2e^{-3t}-e^{-2t}&, t>0\\\\\ 0&,t<0 \end{cases}$$

Then, we have

$$\begin{align}I(t)&=-\frac1{2\pi}\int_{-R}^R\frac{1 + i\omega}{(\omega-3i)(\omega-2i)}e^{i\omega t}\,d\omega+\int_0^{\pm \pi}\frac{1 + iRe^{i\phi}}{(Re^{i\phi}-3i)(Re^{i\phi}-2i)}e^{iRe^{i\phi} t}\,iRe^{i\phi}\,d\phi\\\\&=\begin{cases} 2e^{-3t}-e^{-2t}&, t>0 \tag 1\\\\\ 0&,t<0 \end{cases} \end{align}$$

The first integral on the right-hand side of $(1)$ approaches $\mathscr{F}^{-1}\{F\}(t)$ as $R\to \infty$, while the second integral on the right-hand side of $(1)$ goes to $0$.

Putting it all together, we arrive at

$$\mathscr{F}^{-1}\{F\}(t)\begin{cases} 2e^{-3t}-e^{-2t}&, t>0 \\\\\ 0&,t<0 \end{cases}$$