CSE370 Homework 3 Solution
1) Given f= (a + b' + c')(a' + c')(a+ c + d)(a + b + c + d')
a) Express f in canonical POS form (use M
notation)
f = Π M(0,1,4,6,7,10,11,14,15)
b) Express f in canonical SOP form (use m
notation)
f = Σm(2,3,5,8,9,12,13)
c) Express f' in canonical POS form (use M
notation)
f' = Π M(2,3,5,8,9,12,13)
d) Express f' in canonical SOP form (use m
notation)
f = Σm(0,1,4,6,7,10,11,14,15)
2) Determine a minimized expression for the following using the rules of boolean algebra.
a) F(A,B,C,D) = Σm(1,2,11,13,14,15) + d(0,3,6,10)
Using don't
cares (0,3,10),
A'B'C'D' +
A'B'C'D + A'B'CD + A'B'CD' + A'BCD' + ABC'D + ABCD + ABCD' + AB'CD + AB'CD'
A'B' (C'D' +
C'D + CD + CD') + CA(BD' + BD + B'D' + B'D) + D(ABC + ABC')
A'B' +
AC + D(AB(C+C'))
A'B' + AC +
ABD
b) G(A,B,C,D) = PM(2,5,6,8,9,10)
* D(4,11,12)
Using don't cares(11,12)
First express the function in SOP
form.
Σm(0,1,3,7,13,14,15) +
D(4,11,12)
A'B'C'D' + A'B'C'D + A'B'CD + A'BC'D'
+ A'BCD + ABC'D' + ABC'D + ABCD + ABCD' + AB'CD
A'B'C'(D + D') + CD(AB + A'B +
AB' + A'B') + AB(CD + C'D + CD' + C'D')
A'B'C' + CD + AB
c) Take the complement
of the function F from (a) and express it in a minimized SOP form.
F' = Σm(4,5,7,8,9,12) + d(0,3,6,10)
Using don't cares(0,6)
A'B'C'D' + A'BC'D' + ABC'D' +
AB'C'D' + A'BC'D + A'BCD + A'BCD' + AB'C'D
C'D'(A'B' + A'B + AB' + AB) + A'B(CD
+ C'D + CD' + C'D') + AB'C'(D + D')
C'D' + A'B + AB'C'
3) a) F(A,B,C,D) = Σm(3,6,9,12)
b) F(A,B,C,D) = Σm(0,1,2,3,4,6,8,9,12)
4)
A2 | A1 | A0 | Y1 | Y0 |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 0 |
0 | 1 | 0 | 0 | 1 |
0 | 1 | 1 | 0 | 1 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 1 | 0 |
Y1 = A2
Y2 = A2'A1
5.
6.
7) a) Z = ( (B Å C) * (A Å B)' ) + ( (A Å B) * (B Å C)' )
b)
A | B | C | Z |
0 | 0 | 0 | 0 |
0 | 0 | 1 | 1 |
0 | 1 | 0 | 0 |
0 | 1 | 1 |
1 |
1 | 0 | 0 | 1 |
1 | 0 | 1 | 0 |
1 | 1 | 0 | 1 |
1 | 1 | 1 | 0 |
c)
Z = Σm(1,3,4,6)
= A'B'C + A'BC + AB'C' + ABC'
d)
= A'C(B' + B) + AC'(B'+B)
= A'C + AC'
= A Å
C
e)