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Thread: partial fraction

  1. #1
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    partial fraction

    can anybody help solve this?

    x= 1/(s^2(s+2))

    i need to find the partial fractions of the RHS of the equation.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by al2308 View Post
    can anybody help solve this?

    x= 1/(s^2(s+2))

    i need to find the partial fractions of the RHS of the equation.
    $\displaystyle \frac 1{s^2(s + 2)} = \frac As + \frac B{s^2} + \frac C{s + 2}$

    can you continue?
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  3. #3
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    I got that far and then i got A=1/2, thats where i'm stuck from
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  4. #4
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    (No partial fractions foundation.)

    $\displaystyle \begin{aligned}
    \frac{1}
    {{s^2 (s + 2)}} &= \frac{4}
    {{4s^2 (s + 2)}} \hfill \\
    &= \frac{{4 - s^2 + s^2 }}
    {{4s^2 (s + 2)}} \hfill \\
    &= \frac{{(2 - s)(s + 2) + s^2 }}
    {{4s^2 (s + 2)}} \hfill \\
    &= \frac{1}
    {4}\left( {\frac{{2 - s}}
    {{s^2 }} + \frac{1}
    {{s + 2}}} \right).
    \end{aligned}$
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jhevon View Post
    $\displaystyle \frac 1{s^2(s + 2)} = \frac As + \frac B{s^2} + \frac C{s + 2}$
    just to finish this method up, since Krizalid showed you another.

    $\displaystyle \Rightarrow 1 = As(s + 2) + B(s + 2) + Cs^2$ .............(1)

    plug in $\displaystyle s = 0$ in (1), we get:

    $\displaystyle 1 = 2B \implies \boxed{B = \frac 12}$

    plug in $\displaystyle s = -2$ in (1), we get:

    $\displaystyle 1 = 4C \implies \boxed{C = \frac 14}$

    can we plug anything else in to eliminate B and C but not A? Nah, we will have to do A the hard way, but it will be easier considering we know B and C

    expand (1):

    $\displaystyle 1 = As^2 + 2As + Bs + 2B + Cs^2$

    group like powers of $\displaystyle s$:

    $\displaystyle 1 = (A + C)s^2 + (2A + B)s + 2B$

    now the coefficients of each power of $\displaystyle s$ on the right must be the same as the coefficient of that same power on the left.

    thus we have the system:

    $\displaystyle A + C = 0$

    $\displaystyle 2A + B = 0$

    $\displaystyle 2B = 1$

    and we could have easily found each variable from this system to begin with (but i wanted to show you the two methods, you usually need only one). but now let's just find A. we can find A by plugging in the value we know for C in the $\displaystyle A + C = 0$ equation
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