cambrassk3

2022-07-03

How do you solve B and C for $\frac{s-1}{s+1}\frac{s}{{s}^{2}+1}=\frac{A}{s+1}+\frac{Bs+C}{{s}^{2}+1}$?
$A={\frac{{s}^{2}-s}{{s}^{2}+1}|}_{s=-1}=\frac{1-\left(-1\right)}{1+1}=1$

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Charlize Manning

Expert

We have
$\begin{array}{}\text{(1)}& \frac{s-1}{s+1}\frac{s}{{s}^{2}+1}=\frac{A}{s+1}+\frac{Bs+C}{{s}^{2}+1}\end{array}$
Multiplying $\left(1\right)$ by $s$ and making $s\to \mathrm{\infty }$ gives
$1=A+B$
from which $B=0$
Making $s=0$ in $\left(1\right)$ gives
$0=A+C$
from which $C=-1$

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Banguizb

Expert

Cross-multiplying on the right hand side gives
$A{s}^{2}+A+B{s}^{2}+Bs+Cs+C$ on the numerator. Plug A = 1
$\left(1+B\right){s}^{2}+\left(B+C\right)s+\left(1+C\right)={s}^{2}-s$. Equate the coefficients
$1+B=1\to B=0$
$B+C=-1\to C=-1$

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