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
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentMallard 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
noname
= 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
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Staff
 Addition, Mallard ECE, a'b + b'c

Click to edit the document details