Environmental Engineering Reference
In-Depth Information
θ
1
ϕ
1
θ
2
ϕ
2
θ
m
ϕ
m
y
=
+
+...
(4.96)
and
y
=
y
+
ε
.
(4.97)
The problem is then to find the optimal parameter vector θ which minimizes
ε
=
ε
2
E
(
.
It is assumed that ϕ
k
are not functions of
)
θ so (4.96) is a LIP model. If not,
θ would imply changes in the bases and the space would not be linear.
changes of
Letting
θ
θ
1
θ
m
T
T
=[
...
]
ϕ
=[
ϕ
1
...
ϕ
m
]
,
(4.98)
ϕ
T
θ
y
ϕ
T
θ
y
=
=
+
ε
.
(4.99)
The ϕ
k
are functions of the plant measured variables. For example, in a plant
having as inputs
u
1
(
t
)
and
u
2
(
t
)
and as output
y
(
t
)
some of the bases might be
log
ϕ
j
(
t
)=−
y
(
t
−
3
)
ϕ
l
(
t
)=
u
1
(
t
−
1
)
u
2
(
t
−
1
)
ϕ
r
(
t
)=
(
u
2
(
t
−
3
)).
(4.100)
See also (4.45) in Example 2 in Section 4.2.
Clearly this form makes it possible to introduce nonlinear terms representing
plant phenomenology in LIP models. This is an important property in the case of
soft sensors.
Then, from (4.94) and (4.96):
θ
1
ϕ
1
θ
2
ϕ
2
θ
m
ϕ
m
y
−
y
,
ϕ
k
=
y
−(
+
+...+
),
ϕ
k
=
0;
(4.101)
θ
1
ϕ
1
θ
2
ϕ
2
θ
m
ϕ
m
y
,
ϕ
k
=
+
+...+
,
ϕ
k
.
(4.102)
m
, there is a system of
m
equations with
m
unknown θ
k
=
, ...,
Therefore, for
k
1
given by
θ
1
θ
m
y
,
ϕ
1
=
ϕ
1
,
ϕ
1
+...+
ϕ
m
,
ϕ
1
,
.
θ
1
θ
m
y
,
ϕ
k
=
ϕ
1
,
ϕ
k
+...+
ϕ
m
,
ϕ
k
,
(4.103)
.
θ
1
θ
m
y
,
ϕ
m
=
ϕ
1
,
ϕ
m
+...+
ϕ
m
,
ϕ
m
.
Hence the optimal parameter vector,
θ
θ
1
θ
k
θ
m
T
=[
...
...
]
,
(4.104)
2
2
is given by
which minimizes the squared error norm
e
=
y
−
y
P
xx
θ
=
P
yx
,
(4.105)
where
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