# Circular convolution of a non causal signal

I know how we compute the $$N$$ point circular convolution of a two causal signals, but what about a signal such as $$\{1,-1,2,1\}$$ where, the position of 2 is the $$0^{th}$$ index and the other sequence is $$\{2, -1\}$$ which we can assume to be causal, what about the 4 point circular convolution. According to me it is

$$\begin{bmatrix}1&1&2&-1 \\-1&1&1&2\\2&-1&1&1\\1&2&-1&1 \end{bmatrix} \begin{bmatrix} 2\\-1\\0\\0\end{bmatrix} =\begin{bmatrix} 1\\-3\\5\\0\end{bmatrix}$$ With the position of 5 being the zeroth index because only then the 2 from the first signal got multiplied with the 2 of the second signal, giving off the zero position. But now I am confused, as to how to arrange the other indices. Can anyone help me out?

• What do you mean by "causal signal"? – MBaz Dec 6 '18 at 17:48
• One whose starting index is the first index of the array. – Himanshu Sharma Dec 7 '18 at 3:23
• Interesting. I had only ever seen "causal" in the context of systems, not signals. – MBaz Dec 7 '18 at 3:26

For an $$N$$-point circular convolution you can think of each signal as being periodically extended with period $$N$$. For your example with $$N=4$$ that would mean that the two sequences are
2 1 1 -1 and 2 -1 0 0
where both now start at index $$n=0$$. The result of the cyclic convolution is
5 0 1 -3
which is just a cyclic shift (by $$2$$) of the (correct) result that you obtained.