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/* p(n|n-1)=A^2*p(n-1|n-1)+q */ state->p[0][0] = state->A[0][0] * state->p[0][0] + state->A[0][1] * state->p[1][0] + state->q[0];
还有q应该是2乘2的矩阵,这里为什么是二维向量?
The text was updated successfully, but these errors were encountered:
确实,他少乘了一个A的转置,但是对于你的第二个问题,一般来说Q和R矩阵都是对角阵,副对角线上的两个值都为零,因为两个误差值一般是独立的,所以他用二维向量来表示是可以的
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还有q应该是2乘2的矩阵,这里为什么是二维向量?
The text was updated successfully, but these errors were encountered: