Environmental Engineering Reference
In-Depth Information
10.2.1 Fault Detection
The following hypothesis testing problem is the basis for the development of fault
detection algorithms.
Data: A set of independent random vectors R
1
¼f
r
ð
1
Þ;
r
ð
2
Þ;
...
;
r
ð
k
Þg
, where
k denotes the present time instant, characterized by the probability density function
p
h
ð
r
ð
i
ÞÞ
. The latter depends on a parameter vector h that takes value h
0
in fault free
mode and h
1
in faulty mode, with h
1
6¼
h
0
. Typically h will be the mean or the
variance of the probability distribution in the sequel.
Problem: choose between the following two hypotheses:
H
0
L
ð
r
ð
i
ÞÞ ¼
p
h
0
ð
r
ð
i
ÞÞ
for i
¼
1
;
...
;
k
H
1
L
ð
r
ð
i
ÞÞ ¼
p
h
0
ð
r
ð
i
ÞÞ
for i
¼
1
;
...
;
k
0
1
¼
p
h
1
ð
r
ð
i
ÞÞ
for i
¼
k
0
;
...
;
k
where k
0
is the fault occurrence time which is actually unknown, and L
ð
r
ð
k
ÞÞ
denotes the probability law of r(k).
By Niemann Pearson's lemma, the relevant test to decide between the two
hypotheses is based on the likelihood ratio [
8
]. In our specific setting, assuming
that k
0
is known, this can be written as:
K
k
0
ð
R
1
Þ¼
Q
k
0
1
p
h
0
ð
r
ð
i
ÞÞ
Q
i
¼
k
0
p
h
1
ð
r
ð
i
ÞÞ
Q
i
¼
1
p
h
0
ð
r
ð
i
ÞÞ
i
¼
1
¼
Y
k
p
h
1
ð
r
ð
i
ÞÞ
p
h
0
ð
r
ð
i
ÞÞ
i
¼
k
0
r
ð
i
Þ;
i
¼
1
;
...
;
k
where
the
mutual
independence
of
the
samples
has
been
accounted for. The test can be stated as follows:
Accept hypothesis H
0
when K
k
0
ð
R
1
Þ
k
a
, otherwise accept H
1
, where k
a
is a
user defined threshold that depends of the acceptable probability of false alarm a.
Taking the natural logarithm of both sides of the above expression yields an
equivalent test which is often easier to implement, particularly with distributions
belonging to the exponential family, like the Gaussian distribution. The following
notation will be used to denote the resulting log-likelihood ratio:
S
k
0
ð
R
1
Þ¼
ln K
k
0
ð
R
1
Þ¼
X
p
h
0
ð
r
ð
i
ÞÞ
¼
X
k
k
ln
p
h
1
ð
r
ð
i
ÞÞ
s
ð
i
Þ
ð
10
:
1
Þ
i
¼
k
0
i
¼
k
0
where s(i) denotes the log-likelihood ratio for the i-th sample. The test obviously
becomes:
Accept
S
k
0
ð
R
1
Þ
h
a
,
hypothesis H
0
when
otherwise
accept H
1
,
where
h
a
¼
ln k
a
.
