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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?

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    Are you sure that $a_9$ is given? If so, due to periodicity, it's just the same as $a_1$. What you can use is the fact that the Fourier coefficients must be purely imaginary and odd, because $x[n]$ is real-valued and odd. 'Odd' means that $x[n]=-x[N-n]$, from which it follows that $x[0]=x[4]=0$. The same goes for $a_k$. You might also find Parseval's Theorem useful, relating the sum of squares of $x[n]$ and $a_k$. – 2017-02-09
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    Yes, $a_9$ is given. So, the $x[1]e^{...} = 6j$? And, just to double check, coefficients must be on the opposite axis but same parity? As in, if $x[n]$ is imaginary and even, should the Fourier coefficients be real and even? And why does $x[0]$ equal 0? Is that purely because it is odd? Lastly, in my case it should actually be $x[0] = x[8] = 0$ right? – 2017-02-09
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    @MattL. are you still there? – 2017-02-13
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    I am. Have a look at my answer. – 2017-02-13

1 Answers 1

1

From what is given we can conclude the following:

  1. $x[n]$ is real $\Longrightarrow$ $a_k=a^*_{N-k}$
  2. $x[n]$ is odd $\Longrightarrow$ $a_k=-a_{N-k}$
  3. $\sum_n|x[n]|^2=576\quad\Longrightarrow\quad N\sum_k|a_k|^2=576\quad$ (Parseval)
  4. $a_9=a_1=6j$

From $1$ and $2$, we can see that $a_k$ are purely imaginary and odd. Consequently, we have $a_7=-a_1=-6j$. From 3 we know that

$$\sum_k|a_k|^2=\frac{576}{N}=72$$

Now since $|a_1|^2+|a_7|^2=2|a_1|^2=72$, we can conclude that all other coefficients $a_k$ must be zero. This gives us the solution

$$\begin{align}x[n]&=\sum_{k=0}^{N-1}a_ke^{j2\pi nk/N}\\&=a_1e^{j2\pi n/N}+a_7e^{j2\pi n7/N}\\&=a_1\left(e^{j2\pi n/N}-e^{-j2\pi n/N}\right)\\&=-2\text{ Im}\{a_1\}\sin\left(\frac{2\pi n}{N}\right)\\&=-12\sin\left(\frac{2\pi n}{N}\right)\end{align}$$

x =

   -0.00000
   -8.48528
  -12.00000
   -8.48528
   -0.00000
    8.48528
   12.00000
    8.48528