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$$\frac{0.5}{s[(s+0.5)^2 + 0.25]} = \frac1s - \frac{(s+0.5)+0.5}{(s+0.5)^2+0.5^2}$$ The first equation is simplified into the second equation. My question is HOW? I tried partial fractions: $$\frac{0.5}{s[(s+0.5)^2+0.25]} = \frac As + \frac B{(s+0.5)^2+0.25}$$ I get $A=1$ and $B=-2$, which does NOT equal the solution given here. Somebody pls help me.

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Unfortunately, that is not the correct method of partial fraction decomposition for this paticular problem.

The quadratic $(s+\frac{1}{2})^2+\frac{1}{4}$ expands as $s^2+s+\frac{1}{2}$; this is irreducible in the real numbers, because $b^2-4ac=1-4\cdot1\cdot\frac{1}{2}=-1<0$ (where we write our quadratic as $as^2+bs+c$). So, the term that you include in the decomposition for this part should be of the form $$ \frac{Bs+C}{s^2+s+\frac{1}{2}}, $$ so that your overall goal is to find $A,B,C\in\mathbb{R}$ such that $$ \frac{\frac{1}{2}}{s((s+\frac{1}{2})^2+\frac{1}{4})}=\frac{A}{s}+\frac{Bs+C}{s^2+s+\frac{1}{2}}. $$ Try solving for $A,B,C$ here, and see if you get what you were hoping for.

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  • $\begingroup$ +1: In other words, one should have a numerator as a linear expression if the denominator is quadratic and as constant if the denominator is linear (and so on...) $\endgroup$ – Abhimanyu Arora Jan 19 '14 at 20:35
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Assume for some constant $A,B$ and $C$ we have: $$\frac{0.5}{s\left((s+0.5)^2+0.25\right)}=\frac{A}{s}+\frac{Bs+C}{\left((s+0.5)^2+0.25\right)}$$ Now try to find the constants.

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Try:

$$\dfrac{A}{s} + \dfrac{Bs+C}{(s+.5)^2 + .5^2}$$

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$$\frac{\frac12}{s\left(\left(s+\frac12\right)^2+\frac14\right)}=\frac As+\frac{Bs+C}{\left(s+\frac12\right)^2+\frac14}\iff$$

$$\iff 1=2A\left(\left(s+\frac12\right)^2+\frac14\right)+2(Bs+C)s\iff$$

$$\iff1=A\left(2s^2+2s+1\right)+2(Bs+C)s$$

The above is a polynomial identity in $\;s\;$ , so it must have same respective coefficients in both sides and the remains upon substitution. For example

$$s=0\implies 1=A$$

$$\text{Quadratic coefficient in both sides}\implies 0=2A+2B\implies B=-1$$

and etc.

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  • $\begingroup$ how did you get the 0=2A+2B? Thanks. $\endgroup$ – user122415 Jan 19 '14 at 20:43
  • $\begingroup$ Exactly what's written there, @user122415: compare the quadratic coefficient in both sides in the third line's equality. $\endgroup$ – DonAntonio Jan 19 '14 at 20:46
  • $\begingroup$ ok so solve for A and B and you end up with 1=2s^2+2s+1-2s^2+2Cs. I solve this for C and get 1=c(2s+1)+1. C=0. But if you plug it back in the partial fractions you don't get the original answer. Where did i mess up. $\endgroup$ – user122415 Jan 19 '14 at 20:50
  • $\begingroup$ @user122415 You make lots of mistakes, and perhaps if you used latex to write mathematics you could see this more easily:$$\color{red}1=\color{green}{2s^2}+2s+\color{red}1-\color{green}{2s^2}+2Cs\implies 2s(C+1)=0\implies C=-1$$ $\endgroup$ – DonAntonio Jan 19 '14 at 21:11

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