1
$\begingroup$

Hello I have to calculate the Fourier series coefficients for the following function:

$$f(t)=\sum_{n=-\infty}^{+\infty} \Pi(\dfrac{t-nT_o}{T_o/2})$$

where "$\Pi$" indicates the rectangular function.


I know that the Fourier series are in the form:

$f(t)\simeq \frac{a_o}{2}+\sum_{n=1}^{+\infty} [a_n \cos(\dfrac{2n\pi t}{T}) + b_n \sin (\dfrac{2n\pi t}{T})]$

And each coefficient is defined as:

$a_o=\dfrac{2}{T} \int_{-T/2}^{T/2} f(t)$

$a_n = \dfrac{2}{T} \int_{-T/2}^{T/2} f(t)\cos(\dfrac{2n\pi t}{T})$

$b_n= \dfrac{2}{T} \int_{-T/2}^{T/2} f(t)\sin(\dfrac{2n\pi t}{T})$

I have been given the following indications in order to compute these coefficients (if I don't want to use the formulas above):

i) $TF\{\Pi(\dfrac{t}{T/2})\}=\dfrac{2}{\omega}\sin(\dfrac{\omega T}{4})$

ii) $x(t) = \sum_{n=-\infty}^{\infty} x_b (t-nT) = x_b \star \sum_{n=-\infty}^{\infty} \delta(t-nT) $

iii) $TF\{\sum_{n=-\infty}^{\infty} \delta (t-nT) = \sum_{n=-\infty}^{\infty} e^{-jnT\omega} = \dfrac{2\pi}{T} \sum_{n=-\infty}^{\infty} \delta (\omega-n\dfrac{2 \pi n}{T})$

I know how to do it by using the coefficients equations. My question is how to compute the coefficients using indications (i),(ii),(iii).

$\endgroup$

migrated from engineering.stackexchange.com May 1 '15 at 20:03

This question came from our site for professionals and students of engineering.

  • $\begingroup$ TF= Fourier Transform? $\endgroup$ – Paul May 1 '15 at 19:54
  • $\begingroup$ Yes, it is the Fourier Transform. $\endgroup$ – user3780731 May 1 '15 at 19:59

Your Answer

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

Browse other questions tagged or ask your own question.