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Table 5 Computational complexity of KF algorithm

From: Family of state space least mean power of two-based algorithms

Step

Operation

Multiplication

Additions

Predict

\(\bar {\mathbf {x}}[k]_{n\times 1} = \mathbf {A}[k-1]_{n\times n}\hat {\mathbf {x}}[k-1]_{n\times 1}\)

n 2

n 2n

 

\(\bar {\mathbf {P}}[k]_{n\times n} = \mathbf {A}[k-1]_{n\times n}\mathbf {P}[k-1]_{n\times n}\mathbf {A}^{T}[k-1]_{n\times n} + \mathbf {Q}[k]_{n\times n}\)

2n 3

2n 3n 2

Update

\({\mathbf {\varepsilon }}[k]_{m\times 1} = \mathbf {y}[k]_{m\times 1} -\mathbf {C}[k]_{m\times n}\bar {\mathbf {x}}[k]_{n\times 1}\)

mn

nm

 

\(\mathbf {S}[k]_{m\times m} =\mathbf {C}[k]_{m\times n}\bar {\mathbf {P}}[k]_{n\times n}\mathbf {C}^{T}[k]_{n\times m} + \mathbf {R}[k]_{m\times m}\)

m n 2+m 2 n

m 2 n+m n 2m n

 

\(\mathbf {K}[k]_{n\times m} = \bar {\mathbf {P}}[k]_{n\times n}\mathbf {C}^{T}[k]_{n\times m}\mathbf {S}^{-1}[k]_{m\times m}\)

m 3+m 2 n+m n 2

m 3+m 2 n

 

\(\hat {\mathbf {x}}[k]_{n\times 1} =\bar {\mathbf {x}}[k]_{n\times 1} + \mathbf {K}[k]_{n\times m}{\mathbf {\varepsilon }}[k]_{m\times 1}\)

mn

nm

 

\(\mathbf {P}[k]_{n\times n} = (\mathbf {I}_{n\times n}-\mathbf {K}[k]_{n\times m}\mathbf {C}[k]_{m\times n})\bar {\mathbf {P}}[k]_{n\times n}\)

m n 2+n 3

m n 2+n 3n 2

Total for the KF algorithm

 

m 3+2m 2 n+3m n 2+2m n+3n 3+n 2

m 3+2m 2 n+3m n 2m n+3n 3+n 2n