Question

7. Find the inverse Laplace Transform of X(so2 with ROC-1< Rels) 1.

Answer 1

Given the Laplace transform is

$X(s)=\frac{-5s-7}{(s&plus;1)(s-1)(s&plus;2)}$

Let's split the above expression into partial fractions.

$\frac{-5s-7}{(s&plus;1)(s-1)(s&plus;2)}=\frac{A}{s&plus;1}&plus;\frac{B}{s-1}&plus;\frac{C}{s&plus;2}$

$\Rightarrow -5s-7=A(s-1)(s&plus;2)&plus;B(s&plus;1)(s&plus;2)&plus;C(s&plus;1)(s-1)$

$\Rightarrow -5s-7=A(s^2&plus;s-2)&plus;B(s^2&plus;3s&plus;2)&plus;C(s^2-1)$

$\Rightarrow -5s-7=s^2(A&plus;B&plus;C)&plus;s(A&plus;3B)&plus;(-2A&plus;2B-C)$

Comparing the coefficients on both sides

$\\A&plus;B&plus;C=0 \\A&plus;3B=-5 \\-2A&plus;2B-C=-7$

Solving the above set of three equations

$A=1;\ B=-2\ \&\ C=1$

So,

$X(s)=\frac{1}{s&plus;1}&plus;\frac{-2}{s-1}&plus;\frac{1}{s&plus;2}=X_1(s)&plus;X_2(s)&plus;X_3(s)$

Laplace transform, that is expressed in partial fractions as $X(s)=\sum_{j=1}^m\frac{A_j}{s&plus;a_j}$ ,

For each $\frac{A_j}{s&plus;a_j}$ ,

if the ROC is right to the pole at $s=-a_i$, the inverse Laplace transform is  $A_ie^{-a_it}u(t)$

if the ROC is left to the pole at $s=-a_i$ , the inverse Laplace transform is $-A_ie^{-a_it}u(-t)$

In the given problem, we have the region of convergence (ROC) at $-1<\text{Re}\left \{ s \right \}<1$

For

$X_1(s)=\frac{1}{s&plus;1}$, the pole is at $s=-1$.

So, the ROC is to the right of the poles. Hence the inverse Laplace transform is

$x_1(t)=e^{-t}u(t)$

For

$X_2(s)=\frac{-2}{s-1}$, the pole is at $s=1$ .

So, the ROC is to the left of the poles. Hence the inverse Laplace transform is

$x_2(t)=2e^{t}u(-t)$

For

$X_3(s)=\frac{1}{s&plus;2}$, the pole is at $s=-2$ .

So, the ROC is to the right of the poles. Hence the inverse Laplace transform is

$x_3(t)=e^{-2t}u(t)$

So, the inverse Laplace transform of $X(s)$ is

$x(t)=x_1(t)&plus;x_2(t)&plus;x_3(t)$

$\Rightarrow x(t)=e^{-t}u(t)&plus;2e^{t}u(-t)&plus;e^{-2t}u(t)$

Hence, the inverse Laplace transform of $X(s)=\frac{-5s-7}{(s&plus;1)(s-1)(s&plus;2)}$ with region of convergence (ROC) $-1<\text{Re}\left \{ s \right \}<1$ is

$x(t)=e^{-t}u(t)&plus;2e^{t}u(-t)&plus;e^{-2t}u(t)$