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## anonymous 5 years ago Why is the "s" variable in laplacean transform complex number ?

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1. anonymous

Not sure i get the question? s=d/dt?

2. anonymous
3. TuringTest

*bookmark, good question

4. anonymous

i doubt that you will get answer here

5. anonymous

Check this one: http://www.edaboard.com/thread174416.html#post733852

6. 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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