0
$\begingroup$

Let's say $f(t)$ is a periodic and bounded signal, so it can be represented with Fourier series:

$f(t)= a_0 + \sum_{n=1}^{\infty}a_n\mathbf{cos}(n\omega _0t) + \sum_{n=1}^{\infty}b_n\mathbf{sin}(n\omega _0t)$

If $0 \leq f(t) \leq 1 $, what constraints can be on $a_0$, $a_n$ and $b_n$ ($n \geq 1 $)?

$\endgroup$
  • $\begingroup$ Is $f(t)$ continuous? $\endgroup$ – Mark Jan 12 '17 at 4:30
  • $\begingroup$ Yes. $f(t)$ is continuous $\endgroup$ – regress Jan 12 '17 at 4:35
  • $\begingroup$ Tell us about what you tried. $\endgroup$ – SchrodingersCat Jan 12 '17 at 4:40
  • $\begingroup$ Well I don't know even what I can try. $\endgroup$ – regress Jan 12 '17 at 4:45
2
$\begingroup$

By Parseval equation we get $$ \frac{a_0^2}{2}+\sum_{n=1}^{\infty}(a_n^2+b_n^2)=\frac{1}{\pi}\int_{-\pi}^{\pi}\lvert f(x)\rvert^2\, dx\, \leq \frac{1}{\pi}\int_{-\pi}^{\pi}dx\, =2 $$

$\endgroup$
  • $\begingroup$ Thanks! I've forgotten the Parseval's theorem. $\endgroup$ – regress Jan 12 '17 at 5:09

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.