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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?

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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.

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  • $\begingroup$ Thanks for explaining in detail! But can you please tell me why there are two minus signs in (4)? $\endgroup$ – John Sep 13 at 5:10
  • $\begingroup$ 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 $-e^{2t}u(-t)$ $\endgroup$ – Fat32 Sep 13 at 22:19

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