anonymous
  • anonymous
Why is the "s" variable in laplacean transform complex number ?
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
Not sure i get the question? s=d/dt?
anonymous
  • anonymous
http://upload.wikimedia.org/wikipedia/en/math/8/a/7/8a70463abaee0dc1bbd690ae17fb1276.png http://upload.wikimedia.org/wikipedia/en/math/1/a/e/1ae3de8f169c70efa9f3d47a8cda6289.png why is the s complex number
TuringTest
  • TuringTest
*bookmark, good question

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anonymous
  • anonymous
i doubt that you will get answer here
anonymous
  • anonymous
Check this one: http://www.edaboard.com/thread174416.html#post733852
anonymous
  • anonymous
So, most of the diff. equations (In circuit, system, control theory...) have a solution, in time domain, that include this expression: \[\ e^{st}= e^{t(\sigma+j\omega)}=e^{\sigma t}e^{j\omega}\]. We can see that \[\ e^{\sigma t} \] is the decaying factor (if it is negative) and it determines the amplitude of the signal, where \[\ e^{j\omega}\] is identical to a sinusoidal wave. I suppose we use a complex number because of the possibility that the signal's amplitude will degrade over time (that's where the real part comes from). Suppose sigma being zero, then you could exchange s with only jomega, which is useful, because the AC source usually behaves in this way. It's amplitude doesn't degrade and it changes in a sinusoidal manner. That's more from engineering point of view. I am not satisfied with my explanation and I would also like a better explanation. I know why it is useful because I use it very often (Laplace transform) in EE, but that wasn't the question. Did we just assume s being a complex number because it is convenient or is there some mathematical rule or some underlying principle of why that is?

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