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Unformatted text preview: CSE140 Fall 2006 Homework #2Solutions For most of these problems, there is more than one approach. It has been tried to show you several different approaches in several problems. The A and A forms are equivalent. A s in the textbook are written A here. 1. Problem 2.5 from the textbook: There are several different ways for this problem: S1 and S2 represent the two switches and L represents the Light. 1 means On and 0 means Off. We assume that at the beginning, state of the light is 0 (Off) and two switches are Off too (S1=0, S2=0 and L=1). Based on the problem description we will have the following truth table: S1 S2 L 0 0 0 1 0 1 0 1 1 1 1 0 L=S1S2+S1S2 = S1 xor S2 2. Problem 2.7 (a)(b)(c) By changing the 0s to 1 and 1s to 0 in one of the functions, we will get the other function. Therefore, based on the definition of duality, these functions are dual. OR and AND A B OR(A,B) AND(A,B) 0 0 0 0 1 0 1 0 0 1 1 0 1 1 1 1 NOR and NAND A B NOR(A,B) AND(A,B) 0 0 1 1 1 0 0 1 0 1 0 1 1 1 0 0 CSE140 Fall 2006 Homework #2Solutions XOR and XNOR A B XOR(A,B) XNOR(A,B) 0 0 0 1 1 0 1 0 0 1 1 0 1 1 0 1 (d) (XY+XY) = De Morgans (XY) (XY)= De Morgans (X+Y) (X+Y) = Distributive Law (X+Y)X+(X+Y)Y = Distributive Law XX+YX+XY+YY= Complementarity 0+YX+XY+0= Commutative Law XY+XY 3. Problem 2.19 F(A,B,C,D) = m(0,1,2,7,8,9,10,15) (a) F(A,B,C,D) = m0+m1+m2+m7+m8+m9+m10+m15=...
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