# Inverse Fourier Transform of transfer functions

I want to find the inverse fourier transform of the following transfer function : $$H(iw) = \frac{10 + (iw)}{4 - w^2 + 4(iw)}$$ So my first idea was to replace $iw$ with $s$. Then convert this into some euler formula. This gives me : $$h(t) = \frac{10 + s}{4-w^2 + 4s}$$ But I can't really factor the denominator since there are 2 different variables. So how exactly do I proceed?

I know the inverse fourier transform formula is $0.5(pi)$ * integral of $h(t)*e^(st)$ from negative to positive infinitiy.

You are right, $s=j\omega$, so $$\omega = \frac{s}{j}= \frac{js}{j^2} = -j s$$ Substituting that into your transfer function $$H(j\omega) = \frac{10+j\omega}{4-\omega^2+4j\omega} = \frac{10+s}{4-\left(\frac{s}{j}\right)^2+4s} = \frac{10+s}{4+4s+s^2},$$ since $\frac{1}{j^2} = \frac{1}{-1}=-1$.

In order to find $h(t)$, you need to calculate $$h(t) = L^{-1}\{H(s)\} = \frac{1}{2\pi j}\lim_{T\to\infty}\int_{s=-\gamma-jT}^{\gamma+jT}H(s)e^{st}\; ds.$$

You can do this by using Cauchy's residue theorem, or else make it easier for yourself and use tables. In this case, $$4+4s+s^2 = (s+2)^2,$$ so with $s'=s+2=s-(-2)$, $$H(s) = \frac{s'+8}{(s')^2}=\frac{1}{s'}+\frac{8}{(s')^2},$$ and the rest should be trivial.

## Update

We have \begin{align} L\{ a f(t) + b g(t) \} &= a\,F(s)+b\,G(s) & \text{(linearity)}\\ L\{1\} &= \frac{1}{s} & \text{(constant)}\\ L\{t\} &= \frac{1}{s^2} & \text{(first order)}\\ L\{t^n\} &= \frac{n!}{s^n} & \text{(n-th order)}\\ L\{e^{kt}f(t)\} &= F(s-k) & \text{(s-plane shift)} \end{align} so setting $F(s) = s^{-1}$, then $f(t)=1$ and $$L^{-1} \left\{ \frac{1}{s-(-2)} \right\} = e^{-2t},$$ and setting $F(s) = s^{-2}$ for the next term, then $f(t)=t$ and $$L^{-1} \left\{ \frac{8}{(s-(-2))^2} \right\} = 8t\,e^{-2t},$$ so you end up with $$h(t) = L^{-1}\{H(s)\} = (1+8t)e^{-2t}$$

• Can you help me continue from there as well? – Jonathan Mar 2 '16 at 10:56
• the change of variable is not really useful! – R Djorane Mar 2 '16 at 12:59
• Not useful? $L\{e^{kt}f(t)\} = F(s-k)$. Look at Chu's answer, and you'll notice the $e^{-2t}$ factor. – Pål-Kristian Engstad Mar 2 '16 at 18:57

Write $H(s)$ as $$H(s)= \frac{10}{s^2+4s+4}\:+\:\frac{s}{s^2+4s+4}$$ The time response for the first term is easily found from the Laplace Transform tables $\small (\zeta=1$, $\omega_n \small =2)$; then differentiate this and divide by 10 for the time response of the second term.

This gives: $$h(t)=e^{-2t}(1+8t)$$