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Okay so I have an equation in my book which is as follows.. $$ \frac {a}{s(s+a)} $$ it says "using partial fractions this can be expanded to $$ \frac {1}{s} + \frac {-1}{s+a} $$

My usual method would be to cross multiply and do something like this $$ \frac {a}{s(s+a)} = \frac {A(s+a)}{s(s+a)} + \frac {B(s)}{s(s+a)} $$

Then cancel off the denominators and solve..

$$ a = A(s+a) + B(s) $$

usually though the a would be some constant but here I have no values to play around with.. how has he done it in the book?

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  • $\begingroup$ Sometimes you have $s+a$ in the denominator; sometimes $s+1$. Which is correct? $\endgroup$ – Empy2 Nov 3 '13 at 14:33
  • $\begingroup$ Sorry I copied it down wrongly.. I have redone it now $\endgroup$ – Shasam Nov 3 '13 at 14:38
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We have:

$$\dfrac{a}{s(s+a)} = \dfrac{A}{s}+\dfrac{B}{s+a}$$

So,

$$a = A(s+a) + Bs = (A+B)s + A a$$

we have $A = 1, B = -1$

Final result:

$$\dfrac{a}{s(s+1)} = \dfrac{1}{s}-\dfrac{1}{s+a}$$

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  • $\begingroup$ Ahh right I get it.. Making A+B = 0 and making A = 1.. Leaves us with a = 0 + 1*a, which means a=a.. Thanks! $\endgroup$ – Shasam Nov 3 '13 at 14:49
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    $\begingroup$ You got it, but I look at it slightly different. We see the LHS is $a$ by itself, right off that tells us what $Aa$ is, so $A = 1$. Now I need for the $(A+B)s$ to cancel, and already know $A$, so $B = -1$. Clear? Regards $\endgroup$ – Amzoti Nov 3 '13 at 14:50
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Let $s=0$, so $a=Aa$. Let $s=1$, so $a=A(a+1)+B$
Solve for $A$ and $B$.

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The equality you have is $$ a=(A-B)s+aA. $$ This suggests taking $A=B=1$.

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  • $\begingroup$ Sorry i totally copied it down wrong.. I have changed it now $\endgroup$ – Shasam Nov 3 '13 at 14:41

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