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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?

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    $\begingroup$ What do you mean by "causal signal"? $\endgroup$
    – MBaz
    Commented Dec 6, 2018 at 17:48
  • $\begingroup$ One whose starting index is the first index of the array. $\endgroup$
    – Himanshu Sharma
    Commented Dec 7, 2018 at 3:23
  • $\begingroup$ Interesting. I had only ever seen "causal" in the context of systems, not signals. $\endgroup$
    – MBaz
    Commented Dec 7, 2018 at 3:26

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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.

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