# Inverse Laplace Transform

A system given by $$\frac{s-1}{(s+1)(s-2)}$$ has to be inverse transformed so that it is anticausal and nonstable. The answer given is $$h(t)=-\frac{1}{3}(2e^{-t}+e^{2t})u(-t)$$

Why the minus sign at the beginning?

First you have to remember a Laplace transform property:

$$e^{a t} u(t) \longleftrightarrow \frac{1}{s-a} ~~~,~~~ \mathcal{Re}\{s\} > \mathcal{Re}\{a\} \tag{1}$$

$$-e^{a t} u(-t) \longleftrightarrow \frac{1}{s-a} ~~~,~~~ \mathcal{Re}\{s\} < \mathcal{Re}\{a\} \tag{2}$$

This property states that a given Laplace transform $$H(s)$$ can correspond to multiple inverse transforms, depending on the region of convergence ROC.

Therefore given $$H(s) = \frac{1}{s-a}$$ , you can find two possible inverses as $$x(t) = e^{at} u(t)$$ or $$x(t) = -e^{at} u(-t)$$ depending on whether the ROC is to the left or right of the pole location. Note that one of them is causal and the complementary is anti-causal.

Now, given your transfer function

$$H(s) = \frac{s-1}{(s+1)(s-2)} = \frac{2/3}{s+1} + \frac{1/3}{s-2} , \tag{3}$$

it has two poles at $$s = -1$$ and $$s=2$$. There will be three possible ROC's with three different inverses :

\begin{align} \mathcal{Re}\{s\} < \mathcal{Re}\{-1\} & \implies h(t) = -\frac{2}{3} e^{-t}u(-t) - \frac{1}{3} e^{2t} u(-t) \tag{4} \\ \mathcal{Re}\{-1\} < \mathcal{Re}\{s\} < \mathcal{Re}\{2\} &\implies h(t) = \frac{2}{3} e^{-t}u(t) - \frac{1}{3} e^{2t} u(-t) \tag{5} \\ \mathcal{Re}\{s\} > \mathcal{Re}\{2\} & \implies h(t) = \frac{2}{3} e^{-t}u(t) + \frac{1}{3} e^{2t} u(t) \tag{6} \end{align}

The impulse response in (4) is anti-causal and unstable.

The impulse response in (5) is non-causal and stable.

The impulse response in (6) is causal and unstable.

• Thanks for explaining in detail! But can you please tell me why there are two minus signs in (4)? – John Sep 13 '19 at 5:10
• in case (4), the chosen ROC is: Re{s} < {-1} ... And then accoring to (2), both terms of the partial fraction expansion of (3), will be anti-causal and negated; i.e., 1/(s+1) will be $-e^{-t}u(-t)$ and 1/(s-2) will be $-e^{2t}u(-t)$ – Fat32 Sep 13 '19 at 22:19