# Synthesis discrete time signal from fourier coefficients

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.

• Is it a homework of some kind? – jojek Aug 4 '14 at 15:31
• No, Book exercise, I have the final answer – user2332665 Aug 4 '14 at 15:39

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}

• The final answer is $x[n]=10\cos(\frac{n\pi}{5})$ – user2332665 Aug 4 '14 at 17:21
• How does $a_9$ become $a_1$. Isn't it $a_1=a_{11}$ – user2332665 Aug 4 '14 at 17:23
• 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. – SleuthEye Aug 4 '14 at 17:26
• since $a_k$s are even shouldn't $a_0$ be non_zero – user2332665 Aug 4 '14 at 17:29
• 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. – SleuthEye Aug 4 '14 at 17:32