1. Fuzzy Inferencing – Inverted Pendulum Problem
CS 621 Artificial Intelligence Lecture /08/05 Prof. Pushpak Bhattacharyya
Fuzzy Inferencing – Inverted Pendulum Problem

2. Example Situation θ = 0.06 units θ` = 0.09 units Sensors 23.08.05
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3. Fuzzification Read off μ values Profiles: a) Zero (z) b) Small +/-
c) Medium +/-
e) Large +/-
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4. μ values for θ = 0.06 μ zero ( 0.06 ) = 0.4 μ small+ ( 0.06 ) = 1.0
μ med+ ( 0.06 ) = 0.25
μ large+ ( 0.06 ) = 0.0
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5. μ values for θ` = 0.09 μ zero (θ` = 0.09 ) = 0.1
μ small+ (θ` = 0.09 ) = 1.0
μ med+ (θ` = 0.09 ) = 0.08
μ large+ (θ` = 0.09 ) = 0.0
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6. Rule Firing Potentially all rules fire with various degrees.
Matching the LHS.
Computing AND (min).
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7. R1: If θ = Z and θ` = Z then i = Z μz(θ) μz(θ`)
LHS: min(0.4, 0.1) = 0.1
RHS: By complete truth transfer
μz(i) = 0.1
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8. R2: If θ = Z and θ` = s+ then i = s-
LHS: min(μz(θ = .06) , μs+(θ`= 0.009)
= min(0.4, 1.0) = 0.4
so, μs-(i) = 0.4
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9. R3: If θ = Z and θ` = M+ then i = M- LHS: min( μz(θ) , μM+(θ`))
= 0.08
RHS: μM-(i) = 0.08
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10. After Rule Firing for θ = 0.06, θ` = 0.09 R1 Gives μz(i) = 0.1
R2 Gives μs -(i) = 0.4
R3 Gives μM-(i) = 0.08
R4 Gives …………. and so on
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11. Defuzzification To Get Crisp Values
Read off the Y – Axis.
Project on to the X – Axis.
C. Decide the value.
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12. Defuzzification (Contd)
μz(i) = 0.1 gives
i = or
as read from the profile
μs-(i) = 0.4 gives
i =
μM-(i) = 0.08 gives
i = - 0.1
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13. Max of Min. Centroid (uses alpha cut).
Two Strategies
Max of Min.
Centroid (uses alpha cut).
Max of Min give
i = units
send units of current

14. Centroid α - cut in the area below the curve cut at μ value
For R1: A1= Area below α – cut
μz(i) = 0.1
R2: A2= Area below α – cut
μs-(i) = 0.4
R3: A3= Area below α – cut
μM-(i) = 0.08
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15. = X- axis value of the centroid of A1, A2, A3, ……
Required Current
= X- axis value of the centroid of
A1, A2, A3, ……
Required ( XR, YR) values of Centroid
XR = Xc1 × A1 + Xc2 × A2 + Xc3 × A3 +…..
A1+ A2+ A3+……………
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