solutions_homework_2_ece15a_2009

solutions_homework_2_ece15a_2009 - Solutions to Homework #2...

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Solutions to Homework #2 ECE 15a Winter 2009 1. x+xy = x.1+xy (2-4D) = x(1+y) (2-11) = x.1 (2-5) = x (2-4D) 2. x y: 0 0 = 0, 0 1 = 1, 1 0 = 1 and 1 1 = 0. (a) x y = (x+y)(xy)’? x, y (x+y) (xy)’ (x+y)(xy)’ x y 0 0 0 1 0 0 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 0 0 0 x y = (x+y)(xy)’ (b) x y = (x+y)(xy)’ = (x+y)(x’+y’) = xx’+xy’+yx’+yy’ = xy’+x’y 3. (a) x (y z) = (x+y) (x+z)? Let x = 1 and y = z = 0: x (y z) = 1 (0 0) = 1 0 = 1 (x+y) (x+z) = (1+0) (1+0) = 1 1 = 0 x (y z) (x+y) (x+z) (b) x+(y z) = (x+y) (x+z)? Let x = 1 and y = z = 0: x+(y z) = 1+(0 0) = 1+0 = 1 (x+y) (x+z) = (1+0) (1+0) = 1 1 = 0 x+(y z) (x+y) (x+z) (c) x (y+z) = (x y)+(x z)? Let x = y = 1 and z = 0: x (y+z) = 1 (1+0) = 1 1= 0 (x y)+(x z)= (1 1)+(1 0)= 0+1 = 1 x (y+z) (x y)+(x z)
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4. ax = bx and ax’ = bx’
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solutions_homework_2_ece15a_2009 - Solutions to Homework #2...

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