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In-Depth Information
•
μ
Q
1
(
)
=
(μ
close
-
to
4
(
.
), μ
big
(
))
=
(
.
,
)
With Mamdani's method:
y
Min
3
5
y
Min
0
5
y
(see the figure above, upper right corner)
μ
Q
2
(
y
)
=
Min
(μ
small
(
3
.
5
), μ
small
(
y
))
=
Min
(
0
.
65
,
1
−
y
)
(see lower right corner
of previous figure)
Hence,
μ
Q
∗
(
y
)
=
Max
(μ
Q
1
(
y
), μ
Q
2
(
y
))
=
Max
(
Min
(
0
.
5
,
y
),
Min
(
0
.
65
,
1
−
y
))
(as shown in following figure).
1
μ
∗
0.65
Q
0.5
0
0.35
0.5
1
•
With Larsen's Method:
μ
Q
1
(
y
)
=
0
.
5
y
(left of following figure)
μ
Q
2
(
y
)
=
0
.
65
(
1
−
y
)
(middle of following figure)
Hence,
μ
Q
∗
(
y
)
=
Max
(
0
.
5
y
,
0
.
65
(
1
−
y
))
(right of following figure)
μ
big
μ
small
1
1
1
0.65
0.65
μ
Q
*
0.5
0.5
μ
Q
*
μ
Q
*
0
0.5
1
0
0.35
1
0
0.56
1
Seventh Step: A More Complex Example with Numerical Inputs and Conse-
quences
Let us consider the case:
With Larsen'smethod and translating the and in the antecedents also by
T
=
Prod
.
•
Rule
R
1
is represented by
J
L
(
T
(μ
P
11
(
x
1
), μ
P
12
(
x
2
)), μ
y
1
(
y
))
μ
P
11
(
x
1
)
·
μ
P
12
(
x
2
),
y
=
y
1
=
μ
P
11
(
x
1
)
·
μ
P
12
(
x
2
)
·
μ
y
1
(
y
)
=
0
,
y
=
y
1
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