242 6 Process Identification
Varying exponential forgetting In this case is λ
2
= 1 and exponential forget-
ting λ
1
is given as
λ
1
(k)=1− kγ(k)
2
(k) (6.100)
where the constant k is a small positive number (for example 0.001).
In this case the algorithm works as follows: if the identified process
changes, the prediction error
2
will increase and results in a decrease
of λ
1
. Older data will be forgotten more quickly. If
2
decreases, the pro-
cess is well identified, λ
1
approaches the its upper limit and the rate of
forgetting will be slower.
Constant trace In this case are λ
i
chosen in such a way that the trace of the
covariance matrix will be constant
trP (k +1)=trP (k)=ng (6.101)
where n is the number of identified parameters and g =0.1 − 4 is initial
gain.
This modification is suitable for estimation of time-varying parameters.
Constant gain In this case is λ
1
=1,λ
2
= 0 and the covariance matrix is
given as
P (k +1)=P (k)=P (0) (6.102)
This algorithm can be used for identification of a small number of param-
eters (≤ 3) if the signal-to-noise ratio is small. Convergence of parameters
is usually smaller but the algorithm can be easily implemented.
Also used are combinations of the above mentioned methods, e. g. constant
trace with exponential forgetting. These implementations are suitable for es-
timation of time-varying parameters if initial estimates are poor.
There is one drawback of recursive methods with exponential forgetting.
If there are no new data (z(k +1)=z(k)) it can happen that the covariance
matrix increases and is no longer positive definite. The algorithm will break
down (so called bursting effect). In general it is recommended to check the
trace of the covariance matrix.
A method has been developed to improve stability that forgets only in that
direction from which some new information comes. The formulas describing
this method are as follows:
r(k)=z(k +1)
T
P (k)z(k + 1) (6.103)
L(k +1)=
P (k)z(k +1)
1+r(k)
(6.104)
β(k)=
$
λ
1
(k) −
1−λ
1
(k)
r(k)
if r(k) > 0
1ifr(k)=0
(6.105)
P (k +1)=P (k) −
P (k)z(k +1)z(k +1)
T
P (k)
β(k)
−1
+ r(k)
(6.106)