Given the $2\pi$-periodic function $f(t)=t^2$ such that $-\pi \le t \le \pi$,

I want to determine the coefficients $f_k$ of the fourier series of this signal.

Therefore I use

$$f_k = \frac{1}{2\pi} \int_{-\pi}^{\pi} t^2 e^{-ikt}\,dt$$

Is that true? Because I'm looking at the answers right now and I don't see the minus sign in the integral. Maybe there is a reason for this that I don't know?

Shortly, the 'correct anwer' according to my textbook should be $$f_k = \frac{1}{2\pi} \int_{-\pi}^{\pi} t^2 e^{ikt}\,dt$$ (and this worked out of course).

  • 1
    $\begingroup$ The $f_k$ are typically with a minus sign as in $e^{-ikt}$, but to some extent it doesn't matter as long as you are consistent when you invert and use other formulae. The $f_k$ are given by the inner product of $t \mapsto t^2$ with the basis function $t \mapsto \frac{1}{2 \pi}e^{ikt}$, the minus sign comes from the conjugation in the inner product. $\endgroup$ – copper.hat Jun 30 '13 at 19:46
  • $\begingroup$ In light of what @copper.hat said, what is your textbook's definition of a forward Fourier series/transform? Does it use $e^{-j\omega t}$ or $e^{j\omega t}$? $\endgroup$ – AnonSubmitter85 Jun 30 '13 at 20:20
  • $\begingroup$ $e^{-jw_ot}$. But the answer has been written by the prof $\endgroup$ – Applied mathematician Jun 30 '13 at 20:22
  • $\begingroup$ Note that $t^2 e^{\pm ikt} = t^2 \cos(kt) \pm i t^2\sin(kt)$. Upon integrating, the second term vanishes regardless of choice of sign. I'm guessing your professor made a typo, which incidentally doesn't affect the answer. $\endgroup$ – dls Jun 30 '13 at 20:36

You write $$ f(t) = t^2 = \sum_{n \in \mathbb Z} c_n e^{int} $$ where $$ c_k = \frac 1{2\pi} \int_{-\pi}^{\pi} f(t) e^{-ikt} \, dt. $$ The reason why this makes sense is that if you make the substitution, assuming $$ f(t) = \sum_{n \in \mathbb Z} c_n e^{int} $$ for some coefficients $c_n \in \mathbb C$, then you have $$ \int_{-\pi}^{\pi} \left( \sum_{n \in \mathbb Z} c_n e^{int} \right) e^{-ikt} \, dt = \sum_{n \in \mathbb Z} c_n \int_{-\pi}^{\pi} e^{int} e^{-ikt} \, dt = 2\pi c_k, $$ because $$ \int_{-\pi}^{\pi} e^{i(n-k)t} \, dt = \begin{cases} 2\pi & \text{ if } n=k \\ 0 & \text{ otherwise }. \end{cases} $$ (I didn't justify the switch of the summation and integral but you can still understand that your minus sign needs to be there by looking at that.)

You're saying you got the answer right by putting a minus sign where it shouldn't be, so I expect a sign error in your computations. Can we see the details? Or maybe you can look them up yourself.

(I am VERY confident that this sign should be there ; the idea behind Fourier analysis is all about using the inner product : $$ \langle f,g \rangle = \int_{-\pi}^{\pi} f(t) \overline{g(t)} \, dt $$ and the statement that "Fourier analysis works" is just that $\{ e^{int} \, | \, n \in \mathbb Z \}$ is an Hilbert basis of the $L^2$ integrable functions over $[-\pi,\pi]$ under this inner product, hence the $-$ in the decomposition because $\overline{e^{int}} = e^{-int}$.)

Hope that helps,

  • $\begingroup$ I dont know anything about Hilbert spaces :) $\endgroup$ – Applied mathematician Jun 30 '13 at 20:05
  • $\begingroup$ But thanks for this, its understandable now $\endgroup$ – Applied mathematician Jun 30 '13 at 20:22
  • $\begingroup$ @MSK : It's okay if you don't, but at least the mathematical community reading the question will not doubt the presence of this sign if they know what I am talking about. Perhaps that should make you more confident in the presence of the sign too. $\endgroup$ – Patrick Da Silva Jun 30 '13 at 20:26

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