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Algebraic Expansion with complex numbers

  1. Aug 18, 2012 #1
    Hi all,

    I am a bit rusty and have hit a snag with decomposition of partial fractions, I am taking an Engineering course dealing with Laplace transforms. The example is:

    F(s)= 3 / s(s2+2s+5)

    Now I get that there are complex roots in the denominator and that there are conjugate complex roots (s+1±2i) giving:

    F(s)= 3 / s(s+1-2i)(s+1-2i)

    So partial fraction decomposition would give:

    F(s) = K1/s + K2/(s+1+2i) + K3/(s+1-2i)

    Ok, so solving for K1= 3/5 is easy and I get that, but when it comes to K2 which involves substituting (-1-2i) in for 's' and then expanding, I can't seem to get the answer. What I did was:

    K2=3[STRIKE](s+1+2i)[/STRIKE]/s(s+1-2i)[STRIKE](s+1+2i)[/STRIKE]
    K2=3/s(s+1-2i)
    K2=3/(-1-2i)[(-1-2i)+1-2i)]

    It is the algebraic expansion on the denominator that is getting me. When I do it I get:

    (-1-2i)[(-1-2i)+1-2i)]
    = (-1-2i)(0-4i)
    =-1(-4i)-2i(-4i)
    =4i+8i2
    =8+4i
    =4(2+i)

    However my textbook gets -3/20 (2+i) for the final partial fraction... I'm lost any help would be greatly appreciated.
     
  2. jcsd
  3. Aug 18, 2012 #2

    lewando

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    4i+8i2 = 4i-8 = [-8+4i, edit]
     
    Last edited: Aug 18, 2012
  4. Aug 18, 2012 #3
    Ah yes, simple error... my bad. But it still doesn't match the answer in my text :(
     
  5. Aug 19, 2012 #4

    eumyang

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    That's because you haven't simplified K2 completely. With lewando's correction the denominator simplifies to -8+4i, so K2 is:
    [itex]K_2 = \frac{3}{-8 + 4i}[/itex]
    Multiply the top and bottom by the conjugate of the denominator, and eventually, you will get:
    [itex]K_2 = -\frac{3}{20}\left( 2 + i \right)[/itex]
     
  6. Aug 19, 2012 #5
    Ok, thank you for your time, I needed to simplify my fraction to get the complex term out of the denominator didn't I?
     
  7. Aug 19, 2012 #6

    eumyang

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    Yes, the fraction wouldn't be considered simplified otherwise.
     
  8. Aug 19, 2012 #7
    It seems simpler to me to directly find the residues and then sum them to obtain the inverse Laplace, given that all poles here are simple. I do not know if you have any knowledge of the inverse formula, but it is certainly easier to apply here.
     
  9. Aug 19, 2012 #8
    Hi Millennial,

    So say that the residues are (took me a while to get this :uhh::)

    K1=-0.4615
    K2=0.2308-j2.0110
    K3=0.2308+j2.0110

    I can make this

    g(t)= -0.4615e(-1.4t)+[(0.2308-j2.0110) e(-(1-j0.7)t)+(0.2308+j2.0110) e(-(1+j0.7)t) ] just by 'inspecting' the residues ??? As opposed to breaking the partial fractions down and then doing inverse Laplace transform?

    And I am little confused with the method to express it as trig identities:

    The formulae: cosθ=(e+e-jθ)/2

    and: sinθ=(e-e-jθ)/2j

    I get a bit stuck using this to convert from exponential form to trig form...
     
  10. Aug 19, 2012 #9

    LCKurtz

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    Of course, you can avoid all the complex number stuff by using the partial fraction expansion$$
    \frac 3 {s(s^2+2s+5)}= \frac A s + \frac{Bs+C}{s^2+2s+5}$$and writing ##s^2+2s+5=(s+1)^2+4## when taking the inverse Laplace Transform.
     
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