# confused about time shifting property of Laplace Transform

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?

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:

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