Question

In Problems 1-10, determine the inverse Laplace transform of the given function.

$\frac{\mathbf{3}\mathbf{s}\mathbf{-}\mathbf{15}}{\mathbf{2}{\mathbf{s}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{s}\mathbf{+}\mathbf{10}}$

Short answer

The inverse laplace transform of the given function is$\frac{3}{2}{\mathrm{e}}^{\mathrm{t}}\cos 2\mathrm{t}-3{\mathrm{e}}^{\mathrm{t}}\sin 2\mathrm{t}$.

Step-by-step solution

Determining the inverse laplace transform

• For a given transfer function H, the Inverse Laplace Transform takes the output Y(s) and determines what X(s) it is in terms of (s).
• Consider a function F(s), if there is a function f(t) that is continuous on $\left[0,\infty \right]$and satisfies $\mathcal{L}\left\{f\right\}=F$then we say that f(t) is the inverse Laplace transform of F(s) and employ the notation.
• localid="1662725842956" $\mathrm{f}={\mathcal{L}}^{-1}\left\{\mathrm{F}\right\}$
  
• localid="1662725849590" ${\mathcal{L}}^{-1}\left\{\frac{\mathrm{n}!}{{(\mathrm{s}-\mathrm{a})}^{\mathrm{n}+1}}\right\}={\mathrm{e}}^{\mathrm{at}}{\mathrm{t}}^{\mathrm{n}},\mathrm{n}=1,2,\dots$
  

Find inverse laplace transform for the given function

The given function is $\frac{3s-15}{2s^2-4s+10}$.

Simplify $\frac{3s-15}{2s^2-4s+10}$as:

$\begin{array}{rcl}\frac{3s-15}{2s^2-4s+10}& =& \frac{3(s-5)}{2(s^2-2s+5)}\\ & =& \frac{3}{2}\left(\frac{(s-5)}{(s^2-2s+5)}\right)\\ & =& \frac{3}{2}\left(\frac{s-5}{{(s-1)}^2+{\left(2\right)}^2}\right)\\ & =& \frac{3}{2}\left(\frac{s-1-4}{{(s-1)}^2+{\left(2\right)}^2}\right)\\ & & \end{array}$

Further simplify the equation as follows:

$\begin{array}{rcl}\frac{3s-15}{2s^2-4s+10}& =& \frac{3}{2}\left(\frac{(s-1)-4}{{(s-1)}^2+{\left(2\right)}^2}\right)\\ & =& \frac{3}{2}\left(\frac{s-1}{{(s-1)}^2+{\left(2\right)}^2}\right)-3\left(\frac{2}{{(s-1)}^2+{\left(2\right)}^2}\right)\\ & & \end{array}$

Find the inverse laplace transform of

$\frac{3\mathrm{s}-15}{2{\mathrm{s}}^2-4\mathrm{s}+10}=\frac{3}{2}\left(\frac{\mathrm{s}-1}{{(\mathrm{s}-1)}^2+{\left(2\right)}^2}\right)-3\left(\frac{2}{{(\mathrm{s}-1)}^2+{\left(2\right)}^2}\right)$using

${\mathcal{L}}^{-1}\left\{\frac{\mathrm{b}}{{(\mathrm{s}-\mathrm{a})}^2+{\left(\mathrm{b}\right)}^2}\right\}={\mathrm{e}}^{\mathrm{at}}\mathrm{sinbt}$and ${\mathcal{L}}^{-1}\left\{\frac{\mathrm{b}}{{(\mathrm{s}-\mathrm{a})}^2+{\left(\mathrm{b}\right)}^2}\right\}={\mathrm{e}}^{\mathrm{at}}\mathrm{cosbt}$as:

$\begin{array}{rcl}{\mathcal{L}}^{-1}\left\{\frac{3\mathrm{s}-15}{2{\mathrm{s}}^2-4\mathrm{s}+10}\right\}& =& {\mathcal{L}}^{-1}\left\{\frac{3}{2}\left(\frac{\mathrm{s}-1}{{(\mathrm{s}-1)}^2+{\left(2\right)}^2}\right)-3\left(\frac{2}{{(\mathrm{s}-1)}^2+{\left(2\right)}^2}\right)\right\}\\ & =& {\mathcal{L}}^{-1}\left\{\frac{3}{2}\left(\frac{\mathrm{s}-1}{{(\mathrm{s}-1)}^2+{\left(2\right)}^2}\right)\right\}-{\mathcal{L}}^{-1}\left\{3\left(\frac{2}{{(\mathrm{s}-1)}^2+{\left(2\right)}^2}\right)\right\}\\ & =& \frac{3}{2}{\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}-1}{{(\mathrm{s}-1)}^2+{\left(2\right)}^2}\right\}-3{\mathcal{L}}^{-1}\left\{\frac{2}{{(\mathrm{s}-1)}^2+{\left(2\right)}^2}\right\}\\ & =& \frac{3}{2}{\mathrm{e}}^{\mathrm{t}}\cos 2\mathrm{t}-3{\mathrm{e}}^{\mathrm{t}}\sin 2\mathrm{t}\end{array}$

Therefore, the inverse laplace transform of the given function is$\frac{3}{2}{\mathrm{e}}^{\mathrm{t}}\cos 2\mathrm{t}-3{\mathrm{e}}^{\mathrm{t}}\sin 2\mathrm{t}$