Suppose a periodic signal $x[n] \xrightarrow{Fourier Series} a_k$ has the following properties:
$a)$ Period of $x[n]$ is $8$.
$b) x[n]$ is real
$c) a_3 = 2j $
$d) \sum\limits ^{7}_{n=0} |x[n]|^2= 64$.
Find the signal $x[n]$
How do i find the signal x[n]?From here ,i just know these,the formula below, from the $a)$~$d)$
$x[n]=\sum\limits_{k}a_ke^{k \times j2\pi f n}=\sum\limits_{k \neq 3}a_ke^{k \times j2\pi f n}+2j\times e^{3 \times 2 \pi f n}=x[-n]$,and $64=(2j)^2+\sum\limits_{k\neq 3}(a_k)^2$,however,i still can't know what is the signal $x[n]$
I don't know how to do it, but here are a few thoughts:
You have 8 numbers to find: $x[0]$, ..., $x[7]$. The Fourier coefficients are defined as $$a_k=\sum_{n=0}^7 x[n]e^{-2i\pi\frac{kn}8}$$ If we find those coefficients, then we can figure out the original signal, by the inverse Fourier transform $$x[n]=\frac 1 8\sum_{k=0}^7 a_ke^{2i\pi\frac{kn}8}$$ Now since $x[0]$, ..., $x[7]$ are real, it's easy to verify that $$a_k=a_{8-k}^\star$$ where $z^\star$ is the complex conjugate of $z$. So we only need to find the first 5 coefficients $a_0$, ..., $a_4$, and we will derive the other 3.
Of those 5 numbers, you already know $a_3$. Still 4 to go.
Equation (d) only gives you one constraint on the Fourier coefficients (Plancherel theorem). So you're down to 3 coefficients. I'm not sure where to go from here. It seems that the problem is under-constrained, in that there are an infinite number of solutions.