Question

The inverse Laplace transform of

$X\left(s\right)=\frac{2+2s{e}^{-2s}+4e^{-4s}}{(s^2+4s+3)},Re\left(s\right)>-1$ is of the form

$x\left(t\right)=(e^{-t}+e^{-3t}u(t)+[-e^{-t-a}-3e^{-3(t-a)}\left]u\right(t-a)+2[e^{-t-2a}-e^{-3(t-2a)}\left]u\right(t-2a)$
value of a is ______

• A. 2

Answer 1

The correct option is A 2

X(s) is a sum of $X_1\left(s\right)+X_2\left(s\right)e^{-2s}+X_3\left(s\right)e^{-4s}$

Where, $X_1\left(s\right)=\frac{2}{s^2+4s+3}$

$X_2\left(s\right)=\frac{2s}{s^2+4s+3}$

$X_3\left(s\right)=\frac{4}{s^2+4s+3}$

If $x_1\left(t\right)⟷X_1\left(s\right)$

$x_2\left(t\right)⟷X_2\left(s\right)$

$x_3\left(t\right)⟷X_3\left(s\right)$

Then by using linearity and time shifting property we obtain

$x\left(t\right)=x_1\left(t\right)+x_2(t-2)+x_3(t-4)$

Comparing this x(t) with the expression given in question

$a=2$