inverse laplace using partial fractions and completing square

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Try this way:

$$\frac{1}{(s+1)(s^2+s+1)} = \frac{A}{s+1}+\frac{Bs+C}{s^2+s+1}$$

Notice that the second term has complex solutions, so it's a good idea if you try to transform it in something like $\frac{s+a}{(s+a)^2+b^2}$ and $\frac{b}{(s+a)^2+b^2}$

The solution is:

$$e^{-t}-\frac{e^{-t/2} (\sqrt3 \cos(\frac{\sqrt3 t}{2})-\sin(\frac{\sqrt3 t}{2}))}{\sqrt3}$$


$1)\frac{s+a}{(s+a)^2+b^2} \to e^{-a}\cos(bt)$
$2)\frac{b}{(s+a)^2+b^2} \to e^{-a}\sin(bt)$

Example: If $B=1$ and $C=1$

$$\frac{s+1}{s^2+s+1} = \frac{s+1}{(s+\frac12)^2+(1-\frac14)} = $$

$$\frac{s}{(s+\frac12)^2+(\frac34)} + \frac{\sqrt{\frac34}}{(s+\frac12)^2+(\frac34)}*(\sqrt{\frac43}) $$

In this case you have to add $\sqrt{\frac43}$, so the numerator can be equal to $1$, but you need $\sqrt{\frac34}$ to use the second transform (obviously you will add $\sqrt{\frac43}$ to the result of this transformation).

...

In your case is $A=1, B=-1, C=0$, so:

$$\frac{1}{s+1}-\frac{s}{(s+\frac12)^2+\frac34}$$

Now, to use the transformation $1)$ you need something to add to the fraction like that: $$\frac{{\frac12}}{(s+\frac12)^2+\frac34}$$

So:

$$\frac{1}{s+1}-(\frac{s}{(s+\frac12)^2+\frac34} + \frac{{\frac12}}{(s+\frac12)^2+\frac34} - \frac{{\frac12}}{(s+\frac12)^2+\frac34}) $$

So:

$$\frac{1}{s+1}-(\frac{s+{\frac12} }{(s+\frac12)^2+\frac34} - \frac{{\frac12}}{(s+\frac12)^2+\frac34}) $$

That becomes:

$$\frac{1}{s+1}-(\frac{s+{\frac12} }{(s+\frac{1}{2})^2+\frac34} - \frac{{\frac{\sqrt{3}}{2}}}{(s+\frac12)^2+\frac34}*\frac{2}{\sqrt{3}}*\frac12 )$$

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kazekage
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Updated on August 01, 2022

Comments

  • kazekage
    kazekage over 1 year

    what is the inverse Laplace transform of this equation $$\frac{1}{(s+1)(s^2+s+1)}$$ I know that completing the square for the quadratic term is required to avoid complex roots and then I need to use partial fractions. After completing the square the denominator becomes:

    (s+1)((s+0.5)^2+0.75)

    how shall I proceed from here on?

    many thanks

    • Santosh Linkha
      Santosh Linkha over 9 years
      try using convolution theorem
    • kazekage
      kazekage over 9 years
      the convolution is more difficult than the laplace inverse transform
  • kazekage
    kazekage over 9 years
    can you please explain more on the transformation you mentioned in the end?
  • FdT
    FdT over 9 years
    @user84310 Sure
  • kazekage
    kazekage over 9 years
    thank you, but I think in my case A=1, B=7/3 and C=0, so can you please modify your answer to fit these values?
  • FdT
    FdT over 9 years
    @user84310 Ok, now I think that you can continue easily ;)
  • kazekage
    kazekage over 9 years
    Billions of thanks :)