HWK #2 Solutions

# HWK #2 Solutions - Mallard ECE 290 Computer Engineering I...

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Mallard ECE 290: Computer Engineering I - Spring 2007 - HWK #2 Sol. .. https://mallard.cites.uiuc.edu/ECE290/material.cgi?SessionID=mding3_. .. 1 of 7 4/29/2007 4:23 PM HWK #2 Solutions Problem 2.1 f(x,y,z) = xyz + x'z' + y'z' Why? "f=1 iff xy=z" means f=1 when xy=z=1 and when xy=z=0. Therefore f = (xy)z +(xy)'z' = xyz + (x'+y')z' = xyz + x'z' + y'z' a. Using only AND, OR, and NOT gates, f may be implemented with using 7 gates: 3 ANDs, 1 OR, and 3 NOTs. We can do better, though. Note that a XNOR b = (a XOR b)' = ab + a'b' Making use of this fact, we can write f = (xy)z +(xy)'z' = ((xy) XOR z)' and implement f using just 3 gates: 1 AND, 1 XOR, and 1 NOT, as shown below. b. Function g can be implemented using 3 copies of the part (b) circuit and then ANDing these circuit outputs. c. Problem 2.2

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Mallard ECE 290: Computer Engineering I - Spring 2007 - HWK #2 Sol. .. https://mallard.cites.uiuc.edu/ECE290/material.cgi?SessionID=mding3_. .. 2 of 7 4/29/2007 4:23 PM a. and b. The equality is proved on the left; its dual is proved on the right. Notice that the proofs are dual: each step of the right proof is the dual of that same step in the left proof. x'y' + x'y + xy (x' + y')(x' + y)(x + y) = x'(y' + y) + xy distributive = (x' + y'y)(x + y) = x'1 + xy complementarity = (x' + 0)(x + y) = x' + xy identity = x'(x + y) = x' + y no-name = x'y c. Unfortunately, there seems no way around proofs involving several steps of distribution, although solutions exist that require only a few distributive steps. Here are two possible solutions. Possible solution 1: (7 steps total, 3 distributive steps) ac' + a'b + b'c + d' = ac' + a’b + (b’ + d’)(c + d’) distributive = ac’ + (a’b + b’ + d’)(a’b + c + d’) distributive = (ac’ + a’b + b’ + d’)(ac’ + a’b + c + d’) distributive
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HWK #2 Solutions - Mallard ECE 290 Computer Engineering I...

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