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I am trying to solve for H1, I was able to get the Coefficients for B,C, and D. Yet, I have forgotten how to solve for A.

All I can remember is that one must take the derivative of both sides. After that, I can not remember and I rather not use a system of linear equations.

Can anyone refresh my memory on using the Derivative approach?

This was edited to include the work to find A.

Worked out Solution

James Hayek
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2 Answers2

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So far you have:

$$ s+2 = As^2(s+1)(s+3) + B(s+1)(s+3) + Cs^2(s+3) + Ds^2(s+1)\\ B=\dfrac23, C=\dfrac12, D=\dfrac{1}{18}$$

If you substitute in another value, say $s=1$, you get

$$ 3 = A(2)(4) + B(2)(4) + C(4) + D(2)\\ 3=8A+\dfrac{16}{3}+2+\dfrac{1}{9}\\ 8A=3-2-\dfrac{48}{9}-\dfrac{1}{9}\\ 8A=\dfrac{9-49}{9}=\dfrac{-40}{9}\\ A=\dfrac{-5}{9}$$

Frentos
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If you take the derivative of $s^2p(s)$ you will still have a factor of $s$. So starting from $$\begin{align} s+2&=As(s+1)(s+3)+B(s+1)(s+3)+s^2(\text{stuff})&&\text{apply }\frac{d}{ds}\\ 1&=A(s+1)(s+3)+As(\text{stuff})+B(s+1)+B(s+3)+s(\text{stuff})&&\text{evaluate at }s=0\\ 1&=A(1)(3)+B(1)+B(3)\\ 1&=3A+4B\end{align}$$ Since you know $B$, you can find $A$.

But before I would use that differentiation technique, starting from $$s+2=As(s+1)(s+3)+B(s+1)(s+3)+Cs^2(s+3)+Ds^2(s+1)$$ the coefficient of $s^3$ must be the same on either side: $$0=A+C+D$$ so since you know $C$ and $D$, you can find $A$.

2'5 9'2
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  • Thank you guys. Thanks Alex.jordon. My mistake was not taking the derivative of the entire right hand side of the equation. I have worked out the problem. I will post below. Thanks again! – James Hayek Jan 26 '16 at 03:23