Let $x(n)=\sin (2 \pi k n / N), n=0,1, \ldots, N-1$, where $2 k \neq N$ and $0<k \leq N-1$. Then the circular convolution of $\{x(n)\}$ with itself is

- $N \cos (4 \pi k n / N)$
- $N \sin (4 \pi k n / N)$
- $-N \cos (2 \pi k n / N) / 2$
- $-N \sin (2 \pi k n / N) / 2$
- None of the above