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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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    I can't make any sense out of this. What are those huge exponentials doing there? I presume you meant $U_{r0}, \dotsc$? When you have more than one subscript you need to enclose them in curly braces to group them together. The fundamental and the first two harmonics are already there in your first expansion; there's nothing to compute (other than substituting $h$ if you have a value for it). Your last equation isn't a square wave, but a superposition of three sine waves, but with much higher frequency than your original signal. There must be a major misunderstanding or mistake here.2011-05-18
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    Sorry, forgot the imaginary number j !2011-05-18
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    It is a resonant circuit, so the higher frequencies. Isn't a square wave a summation of sine waves ?2011-05-18
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    It appears you are feeding the circuit with a square wave and calculating the output. The answer is no, the output is not a square wave. The various components are transmitted with different amplitude ratios and different phase shifts, which changes the shape of the wave. Certainly nothing should shift the frequency from 1 to 628.2011-05-18
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    Yes, a square wave is a summation of sine waves. But it is a specific summation of sine waves, the one you give at the start. Even though the circuit is resonant, the output should still be at the same frequency as the input-it is really a filter.2011-05-18
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    Well, y is only a formula from a table book. So it is not an input/output calculation. There is also no frequency shifting.2011-05-18
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    $U_{r0}$, $U_{r1}$,.. that are the harmonic waves and the fundamental wave.2011-05-18
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    I am confused with your notation. In the RHS of your expression for $U_r(t)$ you have $x$, instead of $t$. In terms of the circuit what do $y$ and $x$ correspond to? And $h$, is it just a dimensional factor? Which are the variables for time and for frequency in your question?2011-05-18
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    $h = U_0 = 150V$. Sorry for the x in $U_r$. The y is stated in my exam as an example on how to decompose a square wave. I know it is a bit complicated to describe and it is also written in german.2011-05-18
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    @madmax: In accordance with Ross Millikan's plot and my calculations, the series $\frac{4}{\pi }\sum_{p=1}^{\infty }\frac{1}{2p-1}\sin (\left( 2p-1\right) t)$ is the trigonometric Fourier series of the periodic odd function $f(t)$ defined in $\left[ -\pi ,\pi \right] $ by $$f(t)=\left\{ \begin{array}{ccc} 1 & \text{if} & 02011-05-18
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    But that is just y, isn't it ? Only with a sum.2011-05-18
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    That's my question translated: What is the time function of the output voltage from uR, if you assume that only the fundamental and the first two harmonics are considered.2011-05-18
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    @madmax: Since your question is vague, I voted to close. For instance, you should indicate the numerical values of $\omega,R,L$ and $C$ of your circuit so that one can compute magnitudes and phases. You should also say clearly if the square wave voltage is applyied to the input (voltage $u_q(t)$) and you want to compute the output voltage $u_R(t)$. Is $y\equiv u_q(t)$?2011-05-18
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    @madmax: In the meantime you have commented "But that is just y, isn't it ?". I think so, see my last comment. But it's you that has to confirm, not I.2011-05-18
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    I was thinking that it will be a short answer, so i didn't put all the information in. $y\equiv u_q(t)$2011-05-18
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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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    How can $e^{90.8}$ at all appear in any calculation (which even remotely has something to do with the world around us).2011-05-18
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    @Fabian: This is, to within 20%, the ratio of the fine structure constant to the gravitational coupling constant for a proton and an electron :-)2011-05-18
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    The calculation of $U_{r0},...$ is correct. I have a solution sheet. But i could not find any example on how to make the function of time out of my $U_{r0},...$. So i tried it my way ;-) But it has to look something like that2011-05-18
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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