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$$\frac{e^{-as}}{s(1-e^{-as})}=\frac{A}{s}+\frac{B}{1-e^{-as}}$$

Multiply the whole thing by $s(1-e^{-as})$ $$e^{-as}=A(1-e^{-as})+Bs$$

I distribute and try to find A and B by matching coefficients.

I get $A=1, -1$ and $B=0$. Now what?

Turns out by a little trial and error, I get: $$-\frac{1}{s}+\frac{1}{s(1-e^{-as})}$$

John Lou
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most venerable sir
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  • Partial fractions is a decomposition theorem for rational functions (quotients of polynomials), not for general expressions involving quotients. Particularly, there's no guarantee that your particular function can be decomposed in the form of the first line, and your calculation shows that no values of $A$ and $B$ make the first line true. – Andrew D. Hwang Apr 30 '17 at 23:14

1 Answers1

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Add and substract $1$ upstairs to have:

$$\frac{e^{as}}{s (1-e^{as})} = \frac{1 - 1 +e^{as}}{s (1-e^{as})} = -\frac{1}{s} + \frac{1}{s (1-e^{as})}$$

Dmoreno
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    Note that the point of this was the cancellation $\frac{-1+e^{as}}{1-e^{as}}=-1$. – Ian Apr 30 '17 at 23:02