I am given a signal \$x[n]\$ that have the following properties:

  1. Real and odd
  2. Period of \$N=8\$ and Fourier coefficients \$a_k\$
  3. \$a_9 = 6j\$
  4. The sum of \$|x[n]|^2\$ from \$n=0\$ to \$n=7\$ is \$576\$.

I want to solve for \$a_k\$ and \$x[n]\$. What I have are the following: $$x[n] = \sum_{k=0}^{N-1}\alpha_k e^{jkn(\frac{2\pi}{N})}$$ $$a_k = \frac{1}{N}\sum_{n=0}^{N-1}x[n]e^{-jkn(\frac{2\pi}{N})}$$ $$\sum_{n=0}^{7}|x[n]|^2 = 576$$ I expanded the series for \$a_k\$ and ended up with something that looks like this: $$a_k = \frac{1}{8}(x[0] + x[1]e^{-jk(\frac{2\pi}{8})} + x[2]e^{-jk(\frac{2\pi}{4})}+..)$$ However, before I proceed any further, I know there must be a technique I should be using to simplify this problem, especially using the fact that this is an odd function. However, an odd function will simplify terms be helping me cancel out terms on either side of the number line. However, in this case, since I'm only summing up on the positive side, I'm not sure how to simplify this equation. How should I proceed and how do I best use the extra information given?


1 Answer 1


Since the function is periodic, it is also defined for negative arguments, e.g. \$x[-1]=x[7]\$ or \$x[-2]=x[6]\$, etc. That said, the notion of being odd makes sense.

Note that also the \$a_k\$ are periodic (in partiular \$a_9=a_1\$), and you get the following constraints:

  • being odd implies \$a_k\$ are imaginary
  • being real implies \$a_k=a^*_{-k}=a^*_{N-k}\$

Btw. for real DFTs, one normally takes advantage of the latter, see e.g. the FFTW docs: http://www.fftw.org/fftw3_doc/One_002dDimensional-DFTs-of-Real-Data.html#One_002dDimensional-DFTs-of-Real-Data

[Edited after comment below, thanks!]

  • \$\begingroup\$ But how would the negative co-efficients help me in the summation since I'm only summing over a range of positive values of \$n\$, not even \$k\$. A little confused on that part. Also, why is \$a_k = 0\$ for 0, 2, etc? Is it only provable that it is 0 for k = 0,8, etc? \$\endgroup\$
    – user91567
    Feb 13, 2017 at 6:59
  • \$\begingroup\$ Sure, sorry, my mistake! Symmetry (odd or even) and being purely real or imaginary are Fourrier-Inverses. (This 2k argument came from a different memory layout, sorry again; also fixed a capital N). But again: if you know some negative coefficient, you also know some positive, as there is periodicity. Hint: do you want to prove that your solution is uniqe, give the set of all solutions, or what one solution suffice? \$\endgroup\$
    – magnustron
    Feb 13, 2017 at 8:03

Your Answer

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