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i have a decomposition of a square wave signal:

$ y = \frac{4h}{\pi}(\sin(x) + \frac{1}{3}\sin(3x) + \frac{1}{5}\sin(5x) + ...) $

I computed the fundamental wave and 2 harmonic waves:

$ U_{r0} = 27.5e^{j90.8} $ $ U_{r1} = 35e^{j63} $ $ U_{r2} = 38 $

Till here, it is correct. Now i have to show the time function of this square wave and my solution is this one:

$U_r(t) = 27\sin(628t+86.497) + 35\sin(628\cdot 3t+56) + 38.2\sin(628\cdot 5t)$

But when i plot with Wolfram Alpha it does not look like a square wave. Just too less harmonics or did i do something wrong?

enter image description here

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    OK, i flagged it too.2011-05-18

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If I plot $\sin(x)+\sin(3x)/3+\sin(5x)/5$ I get a pretty nice square wave. In your case the +86.497 and +56 introduce phase shifts, which will ruin the wave. Your leading coefficients are also not in the ratio $1:\frac{1}{3}:\frac{1}{5}$ that they should be. Presumably this came out of your calculation of $U_r0, U_r1, U_r2$, which you don't describe. Note that $e^{90.8}$ is a very large number. Where did that come from?

Added: from your circuit, it appears the output voltage is taken from a divider. So $U_o=\frac{R}{R+j\omega L+\frac{1}{j \omega C}}U_i=\frac{j \omega CR}{j \omega CR-\omega^2 CL+1}U_i$ The fact that the filter ratio depends upon the frequency means that a square wave in will not come out a square wave.

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    @joriki: I noticed that it was somewhere around $10^{40}$ :-) But looking at the circuit of the OP I doubt that he cares about the ratio of gravity to electromagnetic force... But in the mean time there is an imaginary unit appearing in the exponent. This changes a lot...2011-05-18