0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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: $$u(t) = \hat u\,\frac{\tau }{T} + \sum\limits_{1\, \le \,n} {2\,\hat u\,\frac{\tau }{T}\frac{{\sin \frac{1}{2}n\,\omega \,\tau }}{{\frac{1}{2}n\,\omega \,\tau }}} \cos \left( {n\,\omega \,t - \frac{1}{2}n\,\omega \,\tau } \right) $$

p.s.:
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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.