0

I'm studying for a test next week and came across this question in the past exam papers I've looked back in my notes, but I haven't a notion how to even attempt it. All help is appreciated.

Calculate the Fourier series in complex exponential form, of the following function: $$ f(t)=e^{jk\sin(\frac{2\pi t}{T})}$$

Sepideh Abadpour
  • 1,348
  • 4
  • 13
  • 24
Olivia
  • 1
  • 2
    Assuming $j$ is the imaginary unit, the coefficients are given by $c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t) \exp(-jnt) dt$. Did you attempt this calculation? If so, you should provide your work and explain where you're unable to proceed. – Chester Aug 17 '15 at 18:55
  • You should also give us the formula used for the Fourier transform, because there are several conventions with constant in different places. – Alex M. Aug 17 '15 at 19:01
  • Sorry guys, @Chester. i am not familiar with this concept whatsoever i can do the real Fourier series no problem and understand some of the theory behind it but when it comes to questions i don't have a clue. – Olivia Aug 18 '15 at 17:47
  • @AlexM. the transform i use would be: $$ \hat{f}(t)={1/2pi}{∫}{e^{itx}}$$ our lecturer has a habit of changing the constants when it comes to the exams to stop us learning the maths off. (the limits on that integral are -pi to pi) – Olivia Aug 18 '15 at 18:01
  • I'm sure that what is displayed above is not what you meant. But, then, what did you mean? – Alex M. Aug 18 '15 at 18:03
  • sorry edited it there i am unfamiliar with typing maths supposed to be a dx at the end too – Olivia Aug 18 '15 at 18:05
  • @Olivia, the complex exponential form of Fourier series is: $$f(t) = \sum_{n=-\infty}^{\infty} c_n e^{int}$$ where the $c_n$ are defined as above in my first comment (replace $j$ with $i$). So the series is determined by the $c_n$. Thus, the work is to compute these coefficients, which amounts to calculating the integral that defines that them. – Chester Aug 18 '15 at 19:15
  • ok that doesn't seems too bad, thanks guys I'll let you know if I need more help :) – Olivia Aug 18 '15 at 22:47

0 Answers0