数字电路练习题1

Class 2 √ 0 (0,1) √ (0,2) √ (0,4) √ (0,8) I 1 (1,3) √ (1,5) √ (1,9) √ (2,3) √ (2,10) √ (4,5) √ (4,12) √ (8,10) √ (8,12) √ 2 (3,7) √ (3,11) (5,7) (5,13) (9,11) (9,13) (10,11) (10,14) (12,14) 3 (7,15) (11,15) (13,15) (14,15)
cu
3 3 1 1
1
14
1 1 1 1
Z2
Z2Z3
Z2 = bd + ab' + a'c'd + acd' -> done!!!
Z1 = bd + c'd' + a'cd Z2 = bd + ab' + a'c'd + acd' Z3 = bd + b'cd' + a'bc' + + a'c'd + {acd’'} Verification with K-maps: Z1 = bd + c'd' + a'cd cd 00 01 ab 00 01 11 10 1 1 0 1 11 1 1 10 0 0 0 0 cd ab 00 01 11 10 00 0 0 0 1 Z2 = bd + ab' + a'c'd + acd' 01 11 10 cd ab 1 1 1 1 0 1 1 1 0 0 1 1 00 01 11 10 Z3 = bd + b'cd' + a'bc' + + a'c'd + acd' 00 01 11 10 0 1 0 0 1 1 1 0 0 1 1 0 1 0 1 1
•
Z2
1 1 1* Z2Z3
{SEPI} : Z2 = bd + {ab'} +
Z3 = bd + b'cd' + a'bc' + + a'c'd + {acd'} -> done!!!
2) No dominating columns 3) Dominated rows
c'd ac a'c'd acd' (1,5,9,13) (10,11,14,15) (1,5) (10,14) --01 0-0 1-1- 0-0 0-01 0-1-10 0-D G K L
1) Find
c'd' bc' a'cd, c'd ab' ad ac b'cd' ab'c'd' a'bc' a'c'd acd' abc bd (0,4,8,12) (4,5,12,13) (3,7) (1,5,9,13) (8,9,10,11) (9,11,13,15) (10,11,14,15) (2,10) 8 (4,5) (1,5) (10,14) (14,15) (5,7,13,15)
1 1 1 1 1 1 Z2 1 1 1 1 1* 1* 1* 1 1 1 1 1 1 1 1 1 1 1* 1* Z3 = bd + b'cd' + a'bc' + a'c'd
Page 2 of 9
1
Z3 Z1Z2 Z1Z3 1 1 Z2Z3 1 1
Z1Z3Z3
2) No dominating columns
4 5
2. f(a,b,c,d) = a’c’d’+a’b’c+abd’+abd+ ab’d’ + d(a’bcd + ab’c’d) PI: ab , c’d’ , b’d’ , ad’, a’b’c, a’cd, ab’c’, ac’ , bcd EPI: c’d’ RPI: ac’ f1(a,b,c,d) = ab + c’d’ + b’d’ + a’cd c
3) Eliminate dominated rows
c'd ab' ad ac ab'c'd' a'c'd acd' abc (1,5,9,13) (8,9,10,11) (9,11,13,15) (10,11,14,15) 8 (1,5) (10,14) (14,15) --01 0-0 10-- 0-0 1--1 0-0 1-1- 0-0 1000 --0 0-01 0-1-10 0-111- 0-D E F G I K L M Z2 Z3
0 1 3 2
00
0
01
1
11
3
10
2
1*
7
1*
6
1
4 5
1
7
1
6
01 11
1*
12
1
13 15
d
14
1*
12 13
x
15 14
b 1
10
1
8 9
1*
11 10
1
a
8
1
9
1
11
10
d
d
d
d
1
x d
1
f2(a,b,c,d) = ab + c’d’ + b’d’ + a’b’c c
0 1 3 2
f3(a,b,c,d) = ab + c’d’ + ad’ + a’b’c c
cu
1 8 9 10 11 14 14
3 1 1 3 1 1 1 1 Z2 3 1 1 3 1 1 1 5 1 Z1Z2 1 1 4 1 1 1 Z2Z3 4 1 1
Iteration #2
1) Find SEPI
D E G K L
cu
3 3 3 1 4
1 8
9
10 11 14 14
• • • •
(1,5,9,13) --01 0-0 c'd ab' (8,9,10,11) 10-- 0-0 ac (10,11,14,15) 1-1- 0-0 (1,5) 0-01 0-a'c'd (10,14) 1-10 0-acd' 1 1 1* 1 1 1 1 1 1 1
b)
Using reduction methods of the prime implicant table find all sets of minimum sum of products expressions that covers the given set of functions. Iteration #1
1 1 1 1 0 0 Z1 = bd + c'd' + a'cd
Page 3 of 9
Question 3: a) Find the prime implicants of the following function by applying the Quine-McCluskey method: G(v,w,x,y,z) = Σm(0, 2, 4, 5, 6, 7, 8, 9, 10, 11, 13, 15, 21, 23, 26, 28, 29, 30, 31)
Class 3 0 (0,4,8,12) (0,1,4,5) (0,2,8,10) (0,1,8,9) 1 (1,5,9,13) (4,5,12,13) (8,9,10,11) (8,9,12,13) (1,3,5,7) 2 (3,7,11,15) (5,7,13,15) (9,11,13,15) (10,11,14,15)
Digital Logic Design: Principles and Practices ELG5195 (EACJ5705 ), Carleton CRN: 18371
Assignment #1
Question 1: Use Karnaugh maps to find the prime implicants (PI), essential prime implicants (EPI) and redundant prime implicants (RPI) of the following functions. Provide the Boolean expressions of the minimized functions. 1. g(a,b,c,d) = ∑m(2,3,4,5,13,15) + ∑d(6,8,9,10,11) PI: ad , a’bc’ , bc’d, b’c EPI: ad , a’bc’ , b’c RPI: bc’d f(a,b,c,d) = ad + a’bc’ + b’c cd ab 00
0 4 1 3 2
0
4 5
1
7
1
6 5
1
7
1*
6
0
12 13
x
15
0
14
b
12 13
x
15 14
b 1
10
1
a
8 9
1
11
1
10
1
a
8 9
1
11
x
0 d
1
x d
1
Question 2: a) Find the prime implicants of the following multiple-output function by applying the modified QuineMcCluskey method: Z1 (a,b,c,d) = Σ(0, 3, 4, 5, 7, 8, 12, 13, 15) ; Z2 (a,b,c,d) = Σ(1, 5, 7, 8, 9, 10, 11, 13, 14, 15) ; Z3 (a,b,c,d) = Σ(1, 2, 4, 5, 7, 10, 13, 14, 15)
• • • • • • • • •
• •
•
• • • • • •
• •
•
Z1
3 1* 1 1 1 3 1 1 1 1 4 1* 1 3 1 1 1 3 1 1 3 1 3 4 5 1 1 4 1 1 4 1 1 4 4 3 1 1 1 1* 1 1* Z1 = bd + c'd' + a'cd - done! Z2 = bd + …..
EPI
--00 -00 -10- -00 0-11 -00 --01 0-0 10-- 0-0 1--1 0-0 1-1- 0-0 -010 001000 --0 010- -00-01 0-1-10 0-111- 0--1-1 --EPI: A B C D E F G H I J K L M N
0 3 4 5 7 8 12 13 15 1 5 7 8 9 10 11 13 14 15 1 2 4 5 7 10 13 14 15
Class 0 Class 1 Min Group Dec vwxyz Group vwxyz term 0 (0,2) 000-0 √ 0 0 00000 √ (0,4) 00-00 √ (0,8) 0-000 √ 1 2 00010 √ 4 00100 √ 1 (2,6) 00-10 √ 8 01000 √ (2,10) 0-010 √ (4,5) 0010- √ 2 5 00101 √ (4,6) 001-0 √ 6 00110 √ (8,9) 0100- √ 9 01001 √ (8,10) 010-0 √ 10 01010 √ 2 (5,7) 001-1 √ 3 7 00111 √ (5,13) 0-101 √ 11 01011 √ (5,21) -0101 √ 13 01101 √ (6,7) 0011- √ 21 10101 √ (9,11) 010-1 √ 26 11010 √ (9,13) 01-01 √ 28 11100 √ (10,11) 0101- √ 4 15 01111 √ (10,26) -1010 A 23 10111 √ 3 (7,15) 0-111 √ 29 11101 √ (7,23) -0111 √ 30 11110 √ (11,15) 01-11 √ 5 31 11111 √ (13,15) 011-1 √ (13,29) -1101 √ (21,23) 101-1 √ (21,29) 1-101 √ (26,30) 11-10 B (28,29) 1110- √ (28,30) 111-0 √ 4 (15,31) -1111 √ (23,31) 1-111 √ (29,31) 111-1 √ (30,31) 1111- √ Group 0 Class 2 Dec (0,2,4,6) (0,2,8,10) (0,4,2,6) (0,8,2,10) (4,5,6,7) (4,6,5,7) (8,9,10,11) (8,10,9,11) (5,7,13,15) (5,7,21,23) (5,13,7,23) (5,13,21,29) (5,21,7,23) (5,21,13,29) (9,11,13,15) (9,13,11,15) (7,15,23,31) (7,23,15,31) (13,15,29,31) (13,29,15,31) (21,23,29,31) (21,29,23,31) (28,29,30,31) (28,30,29,31) vwxyz 00--0 0-0-0 00--0 0-0-0 001-001-010-010-0-1-1 -01-1 0-1-1 --101 -01-1 --101 01--1 01--1 --111 --111 -11-1 -11-1 1-1-1 1-1-1 111-111--
a b c d Z1Z2Z3 --00 -00 A 0-0-0-0 -00--01 -1010-1-00--1 --11 -1-1 1--1 1-1000 000 000 0-0 -00 0-0 000 000 000 --0-0 0-0
D B Eห้องสมุดไป่ตู้
N F G
Class 4 (0,1,4,5,8,9,12,13) --0- 000
Class 1
0 1
2
3
4
0 1 2 4 8 3 5 9 10 12 7 11 13 14 15
a b c d Z1Z2Z3 0000 -00 0001 0-0010 000100 -01000 --0 0011 -00 0101 --1001 0-0 1010 0-1100 -00 0111 --1011 0-0 1101 --1110 0-1111 ---
a b c d Z1Z2Z3 000- 000 00-0 000 0-00 -00 √ -000 -00 √ 00-1 000 0-01 0-- K -001 0-0 √ 001- 000 -010 00- H 010- -0- J -100 -00 √ 10-0 0-0 √ 1-00 -00 √ 0-11 -00 C -011 000 01-1 --- √ -101 --- √ 10-1 0-0 √ 1-01 0-0 √ 101- 0-0 √ 1-10 0-- L 11-0 000 -111 √ 1-11 0-0 √ 11-1 --- √ 111- 0-- M