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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.

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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 $$ h(t) = L^{-1}\{H(s)\} = (1+8t)e^{-2t} $$

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  • \$\begingroup\$ Can you help me continue from there as well? \$\endgroup\$
    – user91567
    Mar 2, 2016 at 10:56
  • \$\begingroup\$ the change of variable is not really useful! \$\endgroup\$
    – R Djorane
    Mar 2, 2016 at 12:59
  • \$\begingroup\$ Not useful? \$L\{e^{kt}f(t)\} = F(s-k)\$. Look at Chu's answer, and you'll notice the \$e^{-2t}\$ factor. \$\endgroup\$ Mar 2, 2016 at 18:57
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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: $$h(t)=e^{-2t}(1+8t)$$

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