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Math Help - Laplace and Equating Coefficients

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    Laplace and Equating Coefficients

    I have a couple of questions I'm hoping someone can help me with. I'm using partial fractions to transfer a couple of equation from F(s) to f(t).

    First question is:

    F(s) = 4/s^2(s+1)(s+2)

    By equating coefficients this becomes: 4/s^2(s+1)(s+2) = A/s + B/s^2 + C/(s+1) D/(s+2)

    By multiplying by the numerator the equation becomes 4= As(s+1)(s+2) + B(s+1)(s+2) + Cs^2(s+2) + Ds^2(s+1)

    Getting B,C & D is enough enough by substituting s=0,-1&-2. The answer I have for getting A I just don't understand, it says equate coefficients of s so that:

    0=2A + 3B

    I just don't see where this comes from.

    Second question is to do with a closed loop control system. Using a standard control loop equation I have G(s)/1+ G(s).H(s)

    This then becomes: (s/s+2) / [1+ 4.(1/s+2)] I have the answer as 2/s+10. I'm useless at working out fractions like this so any help would be appreciated.

    I'm sure it's really basic stuff but I'm going number blind at this point. Thanks for your help.
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  2. #2
    Senior Member BAdhi's Avatar
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    Re: Laplace and Equating Coefficients

    Quote Originally Posted by Hime View Post
    By multiplying by the numerator the equation becomes 4= As(s+1)(s+2) + B(s+1)(s+2) + Cs^2(s+2) + Ds^2(s+1)

    Getting B,C & D is enough enough by substituting s=0,-1&-2. The answer I have for getting A I just don't understand, it says equate coefficients of s so that:

    0=2A + 3B

    I just don't see where this comes from.
    first you've mentioned that you get,

    4=As(s+1)(s+2)+B(s+1)(s+2)+Cs^2(s+2)+Ds^2(s+1)

    by expanding you'll get,

    4=A(s^3+3s^2+2s)+B(s^2+3s+2)+C(s^3+2s^2)+D(s^3+s^2  )

    not lets make some adjustments,

    4=(A+C+D)s^3+(3A+B+2C+D)s^2+(2A+3B)s+2B

    now equate the coefficient of s you'll get,

    0=2A+3B

    just like that you can also equate coefficients of  s^3 , s^2,... This is another way of finding constants on partial fractions (of course there are much easier methods)
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    Re: Laplace and Equating Coefficients

    Thanks a bunch mate, most appreciated.

    Do you have any idea how to solve the second question?
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    Re: Laplace and Equating Coefficients

    Quote Originally Posted by Hime View Post
    I have a couple of questions I'm hoping someone can help me with. I'm using partial fractions to transfer a couple of equation from F(s) to f(t).

    First question is:

    F(s) = 4/s^2(s+1)(s+2)

    By equating coefficients this becomes: 4/s^2(s+1)(s+2) = A/s + B/s^2 + C/(s+1) D/(s+2)

    By multiplying by the numerator the equation becomes 4= As(s+1)(s+2) + B(s+1)(s+2) + Cs^2(s+2) + Ds^2(s+1)

    Getting B,C & D is enough enough by substituting s=0,-1&-2. The answer I have for getting A I just don't understand, it says equate coefficients of s so that:

    0=2A + 3B

    I just don't see where this comes from.

    Second question is to do with a closed loop control system. Using a standard control loop equation I have G(s)/1+ G(s).H(s)

    This then becomes: (s/s+2) / [1+ 4.(1/s+2)] I have the answer as 2/s+10. I'm useless at working out fractions like this so any help would be appreciated.

    I'm sure it's really basic stuff but I'm going number blind at this point. Thanks for your help.
    Equating coefficients is the long and complicated method. It's easier to note that this equation: \displaystyle \begin{align*} 4 \equiv A\,s(s + 1)(s + 2) + B(s + 1)(s + 2) + C\,s^2(s + 2) + D\,s^2(s + 1) \end{align*} holds true no matter the value of \displaystyle \begin{align*}x  \end{align*}, so by using some inspired guesswork, we can see that if we let \displaystyle \begin{align*} s = 0 \end{align*} we have

    \displaystyle \begin{align*} A(0)(0 + 1)(0 + 2) + B(0 + 1)(0 + 2) + C(0)^2(0 + 2) + D(0)^2(0 + 1) &= 4 \\ 2B &= 4 \\ B &= 2 \end{align*}

    By letting \displaystyle \begin{align*} s = -1 \end{align*} we have

    \displaystyle \begin{align*} A(-1)(-1 + 1)(-1 + 2) + B(-1 + 1)(-1 + 2) + C(-1)^2(-1 + 2) + D(-1)^2(-1 + 1) &= 4 \\ C &= 4 \end{align*}

    By letting \displaystyle \begin{align*} s = -2 \end{align*} we have

    \displaystyle \begin{align*} A(-2)(-2 + 1)(-2 + 2) + B(-2 + 1)(-2 + 2) + C(-2)^2(-2 + 2) + D(-2)^2(-2 + 1) &= 4 \\ -4D &= 4 \\ D &= -1 \end{align*}

    By letting \displaystyle \begin{align*} B = 2, C = 4, D = -1  \end{align*} and \displaystyle \begin{align*} s = 1 \end{align*} we have

    \displaystyle \begin{align*} A(1)(1 + 1)(1 + 2) + 2(1 + 1)(1 + 2) + 4(1)^2(1 + 2) - 1(1)^2(1 + 1) &= 4 \\ 6A + 12 + 12 - 2 &= 4 \\ 6A + 22 &= 4 \\ 6A &= -18 \\ A &= -3 \end{align*}
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