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I am trying to figure out the inverse laplace transform of $\dfrac{6s -19}{s^2 - 6s + 13}$. Looking at my table of Laplace Transforms in my textbook, it seems that either I must break up this fraction using partial fractions into linear terms or I need to use some trick that I'm missing. I know that the denominator cannot be broken up further into linear terms, since if I try to solve $s^2 - 6s + 13$ I get complex roots, so I'm unsure what to do. Any help is appreciated.

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    $\begingroup$ try to complete square. $\endgroup$ – cand Jun 9 '15 at 4:14
  • $\begingroup$ I didn't think of that; good idea. Thanks. $\endgroup$ – letsmakemuffinstogether Jun 9 '15 at 4:15
  • $\begingroup$ Okay I am having a problem with the partial fraction decomposition. Should I be doing it like this: $\dfrac{A}{s-3} + \dfrac{B}{s-3} + \dfrac{C}{4}$? $\endgroup$ – letsmakemuffinstogether Jun 9 '15 at 4:55
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Hint. One may recall that

$$ \begin{align} &\mathcal{L}^{-1}\left(\frac {s-a}{ \left( s-a \right) ^{2}+{b}^{2}}\right)=e^{at}\cos(bt)\\\\ &\mathcal{L}^{-1}\left(\frac {b}{ \left( s-a \right) ^{2}+{b}^{2}}\right)=e^{at}\sin(bt). \end{align} $$ Then one may write $$ \frac{6s -19}{s^2 - 6s + 13}=6\frac {(s-3)}{ \left( s-3 \right) ^{2}+{2}^{2}}-\frac12\frac {2}{ \left( s-3 \right) ^{2}+{2}^{2}} $$ observing that $$ \mathcal{L}^{-1}(\alpha f+\beta g)=\alpha \mathcal{L}^{-1}( f)+\beta \mathcal{L}^{-1}(g). $$

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  • $\begingroup$ Thank you however I tried the answer: $6e^{3t}cos(2t) - e^(3t)sin(2t)$ but it didn't work. $\endgroup$ – letsmakemuffinstogether Jun 9 '15 at 5:41
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    $\begingroup$ what did you try? maybe you should have focused on $6e^{3t}\cos(2t)-\frac12 e^{3t}\sin(2t)$. $\endgroup$ – Math-fun Jun 9 '15 at 6:10
  • $\begingroup$ @letsmakemuffinstogether You can check that the right answer is $\displaystyle 6e^{3t}\cos(2t)-\frac12 e^{3t}\sin(2t)$. Thanks. $\endgroup$ – Olivier Oloa Jun 10 '15 at 4:17
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Given:

$$Y(s)=\frac{6s-19}{s^2-6s+13}$$

APPROACH #1:

Completing the square of the denominator gives $$Y(s)=\frac{6s-19}{(s-3)^2+4}=\frac{6s}{(s-3)^2+4}-\frac{19}{(s-3)^2+4}$$ Algebraic manipulation of each term gives an equivalent equation in a more appropriate form (reasons behind this step will become apparent shortly) $$Y(s)=6\frac{s-3+3}{(s-3)^2+4}-\frac{19}{(s-3)^2+4}=6\frac{s-3}{(s-3)^2+4}-\frac{1}{(s-3)^2+4}$$ or $$Y(s)=6\frac{s-3}{(s-3)^2+4}-\frac{1}{2}\frac{2}{(s-3)^2+2^2}$$

From the transform tables, we use the pairs:

  • $e^{at}cos(bt)\leftrightarrow \frac{s-a}{(s-a)^2+b^2}$
  • $e^{at}sin(bt)\leftrightarrow \frac{b}{(s-a)^2+b^2}$

Identifying: $a=3$ and $b=2$, we get

$$ y(t) = \mathcal{L^{-1}} \left \{ Y(s) \right \} = 6e^{3t}cos(2t)-\frac{1}{2}e^{3t}sin(2t)$$


APPROACH #2:

Invoking the idea of partial fraction decomposition, we write $$\frac{6s-19}{s^2-6s+13}=\frac{A}{s-3-2i}+\frac{B}{s-3+2i}$$ or after multiplying by $s^2-6s+13$ $$(s-3+2i)A+(s-3-2i)B=6s-19$$

  • If $s=3-2i$: $$0+(3-2i-3-2i)B=6(3-2i)-19\Rightarrow B=3-\frac{1}{4}i$$

  • If $s=3+2i$: $$(3+2i-3+2i)A+0=6(3+2i)-19\Rightarrow A=3+\frac{1}{4}i$$

Therefore $$Y(s)=\frac{3+\frac{1}{4}i}{s-3-2i}+\frac{3-\frac{1}{4}i}{s-3+2i}$$

From the transform tables, we use the pair:

  • $\alpha e^{(a+jb)t}\leftrightarrow \frac{\alpha}{s-(a+jb)}$

to invert the terms.

Identifying: $a=3$ and $b=2$, we get

$$ y(t) = \mathcal{L^{-1}} \left \{ Y(s) \right \} = (3+\frac{1}{4}i)e^{\left (3+2i \right )t}+(3-\frac{1}{4}i)e^{\left (3-2i \right )t}$$

To show that this is equivalent to the one found above, we use little complex algebra: \begin{align} y(t) &= (3+\frac{1}{4}i)e^{\left (3+2i \right )t}+(3-\frac{1}{4}i)e^{\left (3-2i \right )t} \\ & = e^{3t}\left [ (3+\frac{1}{4}i)e^{2it}+(3-\frac{1}{4}i)e^{-2it} \right ] \\ & = e^{3t}\left [ 3(e^{2it}+e^{-2it})+\frac{1}{4}i(e^{2it}-e^{-2it})\right ]\\ & = e^{3t}\left [ 3(2cos(2t))+\frac{1}{4}i(2isin(2t))\right ] \\ & = e^{3t}\left [ 6cost(2t)-\frac{1}{2}sin(2t)\right ] \\ & = 6\;e^{3t}cost(2t)-\frac{1}{2}\;e^{3t}sin(2t) \end{align}

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