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Following information is given about a signal $x[n]$

  1. $x[n]$ is real and even signal
  2. $x[n]$ has a period $N=10$ and Fourier coefficients $a_k$
  3. $a_{11}=5$
  4. $\frac1 {10}\sum_{n=0}^9 |x[n]|^2=50$


How can we obtain $x[n]$ from these information?
I know the first info makes the Fourier coefficients real and even, third info makes the Fourier coefficient $a_1=a_{11}=5$(since Fourier coeffs are also periodic with N=10) and fourth info is Parsevals theorem.

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    $\begingroup$ Is it a homework of some kind? $\endgroup$
    – jojek
    Aug 4 '14 at 15:31
  • $\begingroup$ No, Book exercise, I have the final answer $\endgroup$
    – mahes
    Aug 4 '14 at 15:39
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As you indicated, Fourier coefficients are even from first point. This tells you that $a_9 = a_1$, so $a_9 = 5$.

Then from Parseval's theorem, you get

$$ \begin{align} \frac{1}{10} \sum_{n=0}^9 \left|x[n]\right|^2 &= \sum_{k=0}^9 \left|a_k\right|^2 \\ &= 50 \end{align} $$

Since $a_1^2 + a_9^2 = 50$, the other coefficients must be 0.

Then $x[n]$ can be computed as:

$$ \begin{align} x[n] &= \sum_{k=0}^9 a_k \exp(2\pi j n k / 10) \\ &= a_1 \exp(2\pi j n / 10) + a_9 \exp(2\pi j n \cdot 9 /10) \\ &= a_1 \exp(2\pi j n / 10) + a_9 \exp(-2\pi j n /10) \\ &= a_1\left[\exp(2\pi j n / 10) + \exp(-2\pi j n /10)\right] \\ &= 2 a_1 \cos(2\pi n/10) \\ &= 10 \cos(2\pi n/10) \end{align} $$

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  • $\begingroup$ The final answer is $x[n]=10\cos(\frac{n\pi}{5})$ $\endgroup$
    – mahes
    Aug 4 '14 at 17:21
  • $\begingroup$ How does $a_9$ become $a_1$. Isn't it $a_1=a_{11}$ $\endgroup$
    – mahes
    Aug 4 '14 at 17:23
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    $\begingroup$ By symmetry (even Fourier coefficients), $a_{10} = a_0$, $a_9 = a_1$, $a_8 = a_2$, ... You also have $a_{11} = a_1$ from periodicity. $\endgroup$
    – SleuthEye
    Aug 4 '14 at 17:26
  • $\begingroup$ since $a_k$s are even shouldn't $a_0$ be non_zero $\endgroup$
    – mahes
    Aug 4 '14 at 17:29
  • $\begingroup$ If $a_k$s were odd then $a_0$ would have to be 0. But since $a_k$s are even, $a_0$ can be anything including 0. $\endgroup$
    – SleuthEye
    Aug 4 '14 at 17:32

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