Intelligent Systems II · TEE · papers

Q3(b) Center of sums defuzzification [10]

Final examination · 17 Apr 2024

Actual historical paper ·100 marks attempted /140 offered · Q1 compulsory + any four of Q2–Q7. Source defects and solution assumptions are explicitly identified.

Question. Q3(b) [10 marks · 2024-04-17]

For the given membership functions, determine the defuzzified crisp output using the Center of Sums method.

Source graph redrawn: A1 corners (1,0),(3,0.5),(7,0.5),(8,0); A2 corners (3,0),(4,0.3),(8,0.3),(9,0).

Answer

Given

A1 has height 0.5 and support [1,8], plateau [3,7]; A2 has height 0.3 and support [3,9], plateau [4,8].

To find

Center of sums of the two clipped output sets.

Formula and why it applies

Center of sums adds each set’s area and moment separately; overlapping portions are counted once per set. This differs from the centroid of the max-aggregated union. Split each trapezoid into triangles and a rectangle.

z_{COS}=\frac{A_1c_1+A_2c_2}{A_1+A_2}=\frac{M_1+M_2}{A_1+A_2}

Solution

Step 1. Area and moment of A1

Piece	Area	Centroid x	Area × centroid
Left triangle [1,3]	½×2×0.5=0.5	(1+3+3)/3=7/3	7/6
Rectangle [3,7]	4×0.5=2	(3+7)/2=5	10
Right triangle [7,8]	½×1×0.5=0.25	(7+7+8)/3=22/3	11/6

A_1=0.5+2+0.25=2.75,\quad M_1=7/6+10+11/6=13,\quad c_1=13/2.75=52/11

Step 2. Area and moment of A2

Piece	Area	Centroid x	Moment
Left triangle [3,4]	0.15	11/3	0.55
Rectangle [4,8]	1.2	6	7.2
Right triangle [8,9]	0.15	25/3	1.25

A_2=0.15+1.2+0.15=1.5,\quad M_2=0.55+7.2+1.25=9,\quad c_2=6

Step 3. Weight centroids by areas

z_{COS}=\frac{13+9}{2.75+1.5}=\frac{22}{4.25}=\frac{88}{17}\approx5.176471

Do not simply average 52/11 and 6: A1 has more area and must receive more weight. Do not subtract overlap for Center of Sums.

Final answer. Center of Sums = 88/17 ≈ 5.176471.