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Math Help - partial fraction expansion

  1. #1
    jut
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    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}
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  2. #2
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by jut View Post
    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?
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  3. #3
    jut
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    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??
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  4. #4
    Senior Member Shanks's Avatar
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    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.
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  5. #5
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    Quote Originally Posted by jut View Post
    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.
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