1. ## partial fraction expansion

Could some give a little help.

$\frac{1}{\left(s^2+0.76s+1\right)\left(s^2+1.84s+1 \right)}=\frac{\text{A1s}+\text{A2}}{s^2+0.76s+1}+ \frac{\text{B1s}+\text{B2}}{s^2+1.84s+1}$

2. Originally Posted by jut
Could some give a little help.

$\frac{1}{\left(s^2+0.76s+1\right)\left(s^2+1.84s+1 \right)}=\frac{\text{A1s}+\text{A2}}{s^2+0.76s+1}+ \frac{\text{B1s}+\text{B2}}{s^2+1.84s+1}$

$1 = (A+Bs)(s^2+1.84s+1) +(C+Ds)(s^2+0.76s+1)$

Now expand the right side. Can you take it fron here?

3. Ok, I get...

$1=s^2 (1.84 A+B+0.76 C+D)+s (A+1.84 B+C+0.76 D)+s^3 (A+C)+B+D$

But how can I find the 4 constants??

4. the coefficients of the previous three terms must be 0, and the sum of B and D must be 1. that is, we get a linear equation system with four unknowns.

5. Originally Posted by jut
Ok, I get...

$1=s^2 (1.84 A+B+0.76 C+D)+s (A+1.84 B+C+0.76 D)+s^3 (A+C)+B+D$

But how can I find the 4 constants??
In order that two polynomials be equal for all values of the variable coefficients of the same powers must be equal.
Here you have
$0 s^3+ 0s^2+ 0s+ 1$ $=s^2 (1.84 A+B+0.76 C+D)+s (A+1.84 B+C+0.76 D)+s^3 (A+C)+B+D$
so you must have 1.84A+ B+ 0.76C+ D= 0, A+1.84B+ C+ 0.76D= 0, A+ C= 0, and B+ D= 1. That gives you four equations to solve for A, B, C, and D.