3
$\begingroup$

Define a 1-periodic function on $\mathbb{R}$ by:

$f(x) :=$ $\left\{\begin{matrix} 1 & if & 0<x<\frac{1}{10}\\ 0 & if & \frac{1}{10}<x<1 \end{matrix}\right.$

with Fourier series $f(x) = \sum_{-\infty}^{\infty}c_k e^{2\pi ikx}$.

I'm trying to find the graphs of the following two Fourier series:

$\sum_{-\infty}^{\infty}(c_k)^2 e^{2\pi ikx}$ and $\sum_{-\infty}^{\infty}c_{3k} e^{2\pi ikx}$.

Basically my strategy with the others has been to find a way to get these Fourier series into the form $a\sum_{-\infty}^{\infty}c_k e^{2\pi ik(bx)}$ for $a,b\in\mathbb{C}$. But I haven't been able to do it with these, hopefully someone can help, thanks.

Improve this question
$\endgroup$
0

2 Answers 2

Reset to default
4
$\begingroup$

This doesn't require any complicated computations! Look up some tables of identities about the Fourier Series. Which operation would you have to apply to $f$ so that its Fourier series would become $\sum_{-\infty}^{\infty}(c_k)^2 e^{2\pi ikx}$?

For example, we know that derivating in the time domain is equivalent to multiplying the Fourier series coefficients by $i k$ ; so we know that the graph of the Fourier series $\sum_{-\infty}^{\infty} ik (c_k) e^{2\pi ikx}$ is the graph of the derivative of $f$.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Ok so I found an identity which says that $F(x)=\int_0^1 f(t)f(x-t)dt$ has coefficients $(c_k)^2$, and I think F(x) comes out to be a 1-periodic function zero everywhere but for $\frac{1}{10}<x<\frac{2}{10}$ where it is given by the function $\frac{-3}{2}x + \frac{2}{10}$. Does this look correct? $\endgroup$
    – Zaubertrank
    Apr 9, 2012 at 0:10
  • $\begingroup$ Right approach but the $F(x)$ you find (which is the convolution of $f$ by itself) is not correct. Maybe you have forgotten to take into account that $f$ is periodic when computing your integral. The interpretation of your integral is: $F(a)$ is the area, over on period, of the area common to the graph of $f$ and the graph of $f$ reflected along the $x = \frac{a}{2}$ axis. Visualize $f$ and its reflected version sliding on top of each other, and how the area of the intersection evolves. First no overlap, then a maximum overlap when $a = \frac{1}{10}$, then no overlap. $\endgroup$
    – pichenettes
    Apr 9, 2012 at 0:28
  • $\begingroup$ What about for the other one, c_3k, how should I approach that one? $\endgroup$
    – Zaubertrank
    Apr 9, 2012 at 14:25
  • $\begingroup$ Start from the expression giving $c_k$ as an integral of $f(x)$ and a complex exponential. Which transformation do you have to apply to $f$ to get the $c_{3k}$ coefficient instead? $\endgroup$
    – pichenettes
    Apr 9, 2012 at 14:38
  • $\begingroup$ Well we have $c_k = \int_0^{\frac{1}{10}}e^{-2\pi ikx}dx$. Thus what I would like to do is multiply $f(x)$ by $e^{-4\pi ikx}$, but I can't include $k$ in my transformation of $f(x)$, thus I'm not sure what to do.. $\endgroup$
    – Zaubertrank
    Apr 9, 2012 at 16:32
2
$\begingroup$

Is it cheating to inverse the Fourier transform to obtain an expression for the $c_k$?

  1. Find the graph of the inverse function (compute the inverse Fourier transform)

  2. Transform the inverse function according to the transformation of the $c_k$

  3. Compute the Fourier transform of this result, yielding the same expression.

So, if you could find the graph of $c_{3k}$ and $(c_k)^2$, you could find the graph of those expressions by taking the Fourier transform.

My immediate instinct was to identify the time-domain transformation and to analytically work out how this changes the frequencies and the amplitudes. Clearly the $c_{3k}$ option is going to see a shift in the frequencies, and no change in amplitude where $(c_k)^2$ is going to see a shift in the amplitude but no shift in the frequencies.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Ya I think it's cheating to do that, supposedly this is doable without knowing c_k explicitly. $\endgroup$
    – Zaubertrank
    Apr 8, 2012 at 22:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.