inverse laplace using partial fractions and completing square
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
Updated on August 01, 2022Comments

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 over 9 yearstry using convolution theorem

kazekage over 9 yearsthe convolution is more difficult than the laplace inverse transform


kazekage over 9 yearscan you please explain more on the transformation you mentioned in the end?

FdT over 9 years@user84310 Sure

kazekage over 9 yearsthank 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 over 9 years@user84310 Ok, now I think that you can continue easily ;)

kazekage over 9 yearsBillions of thanks :)