# Laplace and Equating Coefficients

• May 20th 2012, 12:22 PM
Hime
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.
• May 20th 2012, 08:42 PM
Re: Laplace and Equating Coefficients
Quote:

Originally Posted by Hime
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,

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

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

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)
• May 21st 2012, 05:37 AM
Hime
Re: Laplace and Equating Coefficients
Thanks a bunch mate, most appreciated.

Do you have any idea how to solve the second question?
• May 21st 2012, 06:04 AM
Prove It
Re: Laplace and Equating Coefficients
Quote:

Originally Posted by Hime
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*}