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Math Help - Complex number - Oscillating system

  1. #1
    Junior Member
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    Complex number - Oscillating system

    "An oscillation in a system is given by  x = \frac{4}{100} e^{- \frac{1}{100} t} \sin{(12t)} . Write this in the form

     x = {\rm Re} \left( c e^{\alpha + i \beta} \right) ."

    So I have:

    \begin{aligned}<br />
   x  & = \frac{4}{100}e^{\frac{-1}{100}t}\sin(12t)\\<br />
      & = {\rm Re}\left(ce^{\alpha+i\beta}\right)\\<br />
      & = {\rm Re}\left(ce^{\alpha}e^{i\beta}\right)\\<br />
      & = {\rm Re}\left(ce^{\alpha}\left[\cos\beta+i\sin\beta\right]\right)\\<br />
      & = ce^{\alpha}\cos\beta\\<br />
 \frac{4}{100}e^{\frac{-1}{100}t}\sin(12t) & =  ce^{\alpha}\sin\left(\beta-\frac{\pi}{2}\right)<br />
\end{aligned}

    By comparing respective "components", I propose that

     c = \frac{4}{100}, \; \alpha = \frac{-1}{100} t , \; \text{ and } 12t = \beta - \frac{\pi}{2} , \; \text{ So, } \beta = 12t + \frac{\pi}{2}

    However, I'm not sure of the logically justification behind this. Why is OK to compare the respective components like this? For instance, is it impossible that one could transform both the exponential and the sine parts, so that they have different parameters, but their product remains unchanged, thereby introducing multiple solutions to the problem?
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  2. #2
    Member alunw's Avatar
    Joined
    May 2009
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    Well, the original function is certainly well-defined, so if you can come up with another expression of this form which works it has to actually be the same function, so does not introduce any more solutions to whatever the problem is.
    In fact the expression is not unique. For example you could put alpha equal to -1/100t+k
    where k is any real constant. Then you just need to multiply c by e^-k. But clearly this makes no difference to the value, so the sensible choice is to put k=0.
    But apart from that your reasoning is fine. For example the sin(12t) term forces the function to 0 at intervals of pi/12, to the beta term has to be 12t+ something otherwise you'd get zeros in the wrong places. And since the function is 0 at t=0 that forces the factor of pi/2. But now you get a second chance to do things differently. By adding another constant factor of pi you don't change the magnitude of the sin term, but you do change the sign, so then you'd have to change the sign of the c term as well. And you can add any multiple of 2*pi to beta without changing anything at all.
    So the only thing that really matters is that you can show that your rewritten function is actually the same function as your original one everywhere, and your reasoning would certainly prove that.
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