2

I am trying to solve the following question:

enter image description here

I have tried to make an attempt at this question but I am not sure where to begin. My current understanding is to apply $f(x) = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \sin(x)e^{-inx} dx$ to find the $c_n$ coefficient for the general formula of a complex Fourier series $\sum_{n = -\infty}^{\infty} c_ne^{inx}$.

Is this the correct approach to this question?

Any help, clarification or suggestions will be greatly appreciated so I begin to solve this question. Thank you!

EDIT: attempted to evaluate $\int sin(x)e^{-ix} dx$, as a start (dont worry i did not forget about $e^{-inx}$ to obtain $\frac{-ix}{2} - \frac{1}{4} e^{-2ix} + c$ but still not sure if this is correct.

lohboys
  • 437

2 Answers2

1

$$\hat{f}(n) = \frac{1}{2\pi}\int_{0}^{2\pi} f(x) e^{-inx} dx$$

$$= \frac{1}{2\pi} \int_{0}^{\pi} \sin(x) e^{-inx} dx$$

Now writing $\sin x = \frac{e^{ix} -e^{-ix}}{2i}$, we can obtain $\hat{f}(n)$.

Once we have that we can write the Fourier series as:

$$\sum_{n=-\infty}^{\infty} \hat{f}(n) e^{inx}$$

Vizag
  • 3,257
  • so i have to evaluate $ \frac{1}{2 \Pi} \int_{0}^{\Pi} \frac{e^{ix} - e^{-ix}}{2i} e^{-inx} dx$ ? – lohboys May 26 '19 at 04:46
  • i obtained $\frac{-ix}{2} - \frac{1}{4} e^{-2ix} + c$ is this correct? – lohboys May 26 '19 at 05:10
  • 2
    @lohboys I found $\dfrac{1}{2\pi}\dfrac{e^{-in\pi}+1}{1-n^2}$ – Nosrati May 26 '19 at 08:43
  • 1
    Yes that integration will give you the Fourier Series. – Vizag May 26 '19 at 10:49
  • @Nosrati i've tried to work that integration again but don't seem to obtain your answer. Could you show me how you obtained your answer? – lohboys May 26 '19 at 23:19
  • 2
    @lohboys $$ c_n =\dfrac{1}{2\pi}\int_0^\pi\dfrac{e^{ix}-e^{-ix}}{2i}e^{-inx}dx =\dfrac{1}{4\pi i}\left(\dfrac{e^{ix(1-n)}}{i(1-n)}-\dfrac{e^{ix(-1-n)}}{i(-1-n)}\right)_0^\pi =\dfrac{1}{4\pi i}\left(\dfrac{e^{i\pi(1-n)}-1}{i(1-n)}+\dfrac{e^{-i\pi(1+n)}-1}{i(1+n)}\right)_0^\pi =\dfrac{1}{2\pi}\dfrac{e^{-in\pi}+1}{1-n^2} $$ – Nosrati May 26 '19 at 23:46
  • @Nosrati thank you! – lohboys May 28 '19 at 06:32
  • @Nosrati may I ask how you went from the 2nd to the 3rd step (integrating) and then from the third to fourth step? Im stuck with the integrating in the 2nd step – lohboys May 30 '19 at 06:20
0

An alternate solutions is using fourier relations as f.g -> F * G muliplication goes to convolution. Your function is sin multiplied by a top hat function convolved by a comb. The transform will thus be the two deltas from the sin function convolved by a sinc times by a comb (with a phase shift to compensate for the offset of the comb).

Paul Childs
  • 811