I've the following function $\text{U}\left(t\right)$ that is defined as posted in the picture:

enter image description here

My book tells me that the Fourier series looks like:

$$\tag{1}U(t)=\sum_{n=1}^{\infty}2 \hat{u} \dfrac{\tau}{T}\dfrac{\sin(\dfrac12 n \omega \tau)}{\dfrac12 n \omega \tau}\cos(n \omega t-\dfrac12 n \omega \tau) \ \ \ \text{with} \ \ \omega:=\dfrac{2 \pi}{T}.$$

Now, I used Mathematica to plot the function given by formula $(1)$ but I got something else (I set some values for the constants in the function and plot it for $t$) so not the thing I was supposed to get.

Where is the mistake in the series?

  • $\begingroup$ would you mind to include the picture directly and describe it properly? $\endgroup$ – tired Nov 20 '16 at 15:25
  • $\begingroup$ I can not I've not enougth + $\endgroup$ – asdasd Nov 20 '16 at 15:26
  • $\begingroup$ i uploaded the picture for you, but now it is your turn to give a proper description $\endgroup$ – tired Nov 20 '16 at 15:29
  • $\begingroup$ What is $\omega$? What is the $x$-coordinate of the first square wave? $\endgroup$ – rogerl Nov 20 '16 at 15:29
  • $\begingroup$ I have done it. But something is missing : picture 2. $\endgroup$ – Jean Marie Nov 20 '16 at 15:29

Understanding that $\omega=2*\pi /T$, then the formula in the book is just missing the constant term ${\hat u\,\frac{\tau }{T}}$.
The correct formula is in fact: $$u(t) = \hat u\,\frac{\tau }{T} + \sum\limits_{1\, \le \,n} {2\,\hat u\,\frac{\tau }{T}\frac{{\sin \frac{1}{2}n\,\omega \,\tau }}{{\frac{1}{2}n\,\omega \,\tau }}} \cos \left( {n\,\omega \,t - \frac{1}{2}n\,\omega \,\tau } \right) $$

with the further understanding that the function actually be: $$ u(t)\quad \left| {\;0 \le t < T} \right.\quad = \left\{ {\begin{array}{*{20}c} {\,\hat u} & {0 \le t < \tau } \\ 0 & {\tau \le t < T} \\ \end{array}} \right. $$ i.e. that the step starts from $t=0$, which from the picture is not so evident

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  • $\begingroup$ So it is right but I've to add: $\hat u\,\frac{\tau }{T}$? $\endgroup$ – asdasd Nov 20 '16 at 15:44
  • $\begingroup$ @asdasd: yes, in fact $\endgroup$ – G Cab Nov 20 '16 at 15:45

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