FUZZY LOGIC
Soal:
Sebuah mobil mempunyai kecepatan diantara range
60-180 km/jam. Untuk mengendalikan mobil terhadap suatu rintangan (obstacle),
maka kecepatan mobil dapat diatur dengan memberikan tekanan pada rem (brake).
Tekanan pada rem berkisar antara 0-80 bar. Sedangkan jarak terhadap obstacle
ditentukan dari 100-300 meter.
Disainlah sistem diatas dengan menggunakan
fuzzy logic, dan tentukan berapa bar tekanan pada brake jika kecepatan mobil
adalah 90 km/jam dan jarak terhadap obstacle adalah 240 meter.
Determine using fuzzy logic controller (use
Mamdani method) how
much the pressure of the brake if distance of the two car is 240km
and velocity of the green car is 90km/hr.
Jawab:
Commonly, fuzzy system is like this figure.
Define how much input and output??
2 variable as input are: speed and distance of
obstacle.
1 variable as output is: brake’s pressure
From the system above, we follow the rule of
Fuzzy system as the procedures bellow:
1.
Fuzzification/
Membership Function
Speed Input.
Divide the Speed range
into 3 categories of “low”, “middle” and “high”. I can divide the speed range
more than three or less, depend on assumption. The definition of “low”,
“middle”, and “high” are depend on my assumption too. I can choose a triangle
or trapezium or gauss or sigmoid or other function as the shape of membership
function. In this case, I choose trap MF as a “low” and a “high”, and tri MF as
a “middle”. This figure bellow shows the complete result of three categories of
3 membership function.
Obstacle’s
Distance Input.
Divide the distance
range into 3 categories of “near”, “middle” and “far”. (According the
assumption, it’s can be divided less or more than 3).
Brake
Pressure Output.
Divide the brake
pressure range into 3 categories of “no”, “light” and “strong”. (According the
assumption, it’s can be divided less or more than 3).
2.
Rules/
Inference
Define a look-up table
from membership function as:
3.
Defuzzification
The speed of 90 will cross the low line at 0.67
and cross the middle line at 0.25. The distance of 240 will cross the middle
line at 0.47 and cross the far line at 0.2.
In this case I use and operation to determine
the value of output. ( Speed and Obstacle à Result is Brake
Pressure).
From the figure above
we can define the table bellow.
No
|
Rule
|
Speed(MF)
|
Operator
|
Obstacle(MF)
|
Sp Op
|
Result
|
Out
|
1
|
Rule 1
|
Mhigh
|
AND
|
Mnear
|
0 AND 0
|
0
|
S
|
2
|
Rule 2
|
Mhigh
|
AND
|
Mmiddle
|
0 AND 0.47
|
0
|
S
|
3
|
Rule 3
|
Mhigh
|
AND
|
Mfar
|
0 AND 0.2
|
0
|
L
|
4
|
Rule 4
|
Mmiddle
|
AND
|
Mnear
|
0.25 AND 0
|
0
|
S
|
5
|
Rule 5
|
Mmiddle
|
AND
|
Mmiddle
|
0.25 AND 0.47
|
0.25
|
L
|
6
|
Rule 6
|
Mmiddle
|
AND
|
Mfar
|
0.25 AND 0.2
|
0.2
|
N
|
7
|
Rule 7
|
Mlow
|
AND
|
Mnear
|
0.67 AND 0
|
0
|
L
|
8
|
Rule 8
|
Mlow
|
AND
|
Mmiddle
|
0.67 AND 0.47
|
0.47
|
N
|
9
|
Rule 9
|
Mlow
|
AND
|
Mfar
|
0.67 AND 0.2
|
0.2
|
N
|
Note : Out are S, L and
N from rule (look up table)
No
|
Rule
|
Sp Op
|
Result
|
Out
|
Res*Out
|
1
|
Rule 1
|
0 AND 0
|
0
|
S
|
0*80
= 0
|
2
|
Rule 2
|
0 AND 0.47
|
0
|
S
|
0*80
= 0
|
3
|
Rule 3
|
0 AND 0.2
|
0
|
L
|
0*40
= 0
|
4
|
Rule 4
|
0.25 AND 0
|
0
|
S
|
0*80
= 0
|
5
|
Rule 5
|
0.25 AND 0.47
|
0.25
|
L
|
0.25*40
= 10
|
6
|
Rule 6
|
0.25 AND 0.2
|
0.2
|
N
|
0.2*0
= 0
|
7
|
Rule 7
|
0.67 AND 0
|
0
|
L
|
0*40
= 0
|
8
|
Rule 8
|
0.67 AND 0.47
|
0.47
|
N
|
0.47*0
= 0
|
9
|
Rule 9
|
0.67 AND 0.2
|
0.2
|
N
|
0.2*0
= 0
|
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