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In book signals and systems 2 edition a question is given which is as follows:

$$ x(t)=e^{-3(t+1)}u(t+1) $$

and we are asked to find the unilateral Laplace Transform of the signal. The method that is given in the solution manual is as follows:

Using Table 9.2 and time shifting property we get:

$$ X_2(s) = \frac{e^s}{s+3} $$

Now I am given a question which is as follows:

$$ e^{-2t}u(t-1) $$ and asked to find the Laplace Transform. Now can I apply the method as used above for unilateral Laplace Transform and get:

$$ \frac{e^{-s}}{s+2} \rightarrow A $$

Or does that method only holds true for unilateral Laplace Transforms? Because the answer marked A is wrong when I use this method. Also tell me when can I apply the property?

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If you have written the function correctly then its Laplace transform could be found very similary to your first example:

Given $$x(t) = e^{-2 t} u(t-1)$$ its Laplace transform could be found as follows. First denote the signal

$$x_0(t) = e^{-2} e^{-2t} u(t) $$

then its obvious that $$x(t) =x_0(t-1) $$

Using the tables and properties to conclude:

$$X(s) = e^{-s} X_0(s) $$ $$X(s) = e^{-s} \frac{ e^{-2} }{s + 2} = \frac{ e^{-(s+2)} }{s + 2} $$

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