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Damped System

  1. Sep 8, 2014 #1
    I'm trying to do some refreshing of differential equations featuring damped systems. Specifically, I have a question regarding the differential equation solution to an under damped system involving complex roots.

    Referring to the attached pdf, an under damped system will yield a complex conjugate pair of roots. I am curious as to why the basic real solution features a sine term (refer to second attachment). If I remember Euler's formula correctly, the sine term is always imaginary and is not featured in the real solution. However, this document states otherwise... I believe I have a fundamental misconception regarding this topic.

    Any idea why the negative imaginary conjugate yields a sine term in the time domain?
     

    Attached Files:

  2. jcsd
  3. Sep 8, 2014 #2

    HallsofIvy

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    If you have complex numbers are solutions to your characteristic equations, say, a+ bi and a- bi, then the solution to the corresponding differential equation is of the form [tex]Ae^{(a+ bi)t}+ Be^{(a- bi)t}= Ae^{at}e^{ibt}+ Be^{at}e^{-ibt}= e^{at}(Ae^{ibt}+ Be^{-ibt})[/tex][tex]= e^{at}(Acos(bt)+ Ai sin(bt)+ Bcos(bt)- Bsin(bt))= e^{at}((A+ B)cos(bt)+ i(A- B)sin(bt))[/tex]

    There is, in fact, an "i" multiplying A- B. HOWEVER, "A" and "B" themselves are complex numbers, not real numbers. As long as we have only real number values for the initial or boundary values, because this has to solve a problem involving only real numbers, we must have A+ B real and A- B imaginary (so that i(A- B) is real). That is the same as saying that A and B must be conjugate complex numbers.

    (If you had a differential equation in which the initial values or boundary values involved complex numbers themselves, then you could not assume the coefficients are real- but in that case, you would be better off leaving the solutions as [itex]e^{(a+ bi)t}[/itex] and [itex]e^{(a- bi)t}[/itex].)
     
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