Information Technology Reference
In-Depth Information
μ
Q
∗
=
+
+
=
.
×
+
Theareabelow
is A
rectangle(1)
rectangle(2)
triangle (3)
0
3
7
4
×
0
.
4
3
×
0
.
7
+
=
5. Since the area of the rectangle with base
[
0
,
3
]
is 0
.
3
×
3
=
0
.
9,
2
it is
y
0
>
3. Hence,
y
0
3
y
10
dy
0
.
3
dy
+
=
A
/
2
=
2
.
5
.
0
3
10
y
0
5, or
y
0
3
1
Then 3
×
0
.
3
+
ydy
=
2
.
ydy
=
10
(
2
.
5
−
0
.
9
)
=
16. Thus,
3
y
2
2
]
y
0
3
y
0
−
y
0
=
[
=
16
⃒
9
=
32
⃒
41
⃒
y
0
=
6
.
4
.
The defuzzified value that corresponds to
x
0
=
4.
Defuzzifying with the centre of the area we obtained an output for all values
x
0
0
.
3, is
y
0
=
6
.
in
. Let's see it by means of the function CRI with defuzzification made by the
centre of the area.
[
0
,
1
]
1. The graphics, for any input
x
0
1
/
2, is
Theareabelow
μ
Q
∗
is
A
=
10
(
1
−
x
0
)
x
0
+
(
10
−
10
(
1
−
x
0
))(
1
−
x
0
)
+
(
1
−
x
0
−
x
0
)(
10
(
1
−
x
0
)
−
10
x
0
)
2
Search WWH ::
Custom Search