数电chapter2-2

AB CD 00 01 11 10 0 4 12 8 1 1 00 01
1
m1+m3+m9+m11=B’D
1
5
13 15
9
1
11 3 1 7 10
2
111 10
6 1 14 1
m4+m12+m6+m14=BD’
MSB=A ; LSB=D
m0=A’B’C’D’, m1=A’B’C’D, m2=A’B’CD’, m3=A’B’CD
m4=A’BC’D’, m5=A’BC’D, m6=A’BCD’, m7=A’BCD m8=AB’C’D’, m9=AB’C’D, m10=AB’CD’, m11=AB’CD
m12=A’BC’D’, m13=A’BC’D, m14=A’BCD’, m15=A’BCD
AB CD 00 01 11 10 0 4 12 8 1 00 1 1 1
By decoding the binary coordinates, We label the decimal value for each square.
MSB LSB
AB C 00 01 11 10 4 0 6 2 0 1
1 3 7 5
AB CD 00 01 11 10 0 4 12 8 00 01
1 5
0
13
9
Writing 0 in the square correspond to a maxterm.
11 3 0 7 0 15 0 11 0
20 6 10 14 0 10
0
MSB=A ; LSB=D
Simplify a equation use N-variable K-map
Boolean Identity AB+AB’=A A sum of 2 logically adjacent mminterms can be simplified by eliminating one variable , which is the different literal in the two minterms . The resulting product term has m-1 literals.
10 18
22
30
26
MSB=A ; LSB=E
It is a matrix of squares. each square represent a minterm or maxterm from a Boolean equation. N-variable karnaugh map have 2n squares. The binary numeral on the sides of k-map is the variable coordinates.
01
1
1
5
1
13
1
9
2n －1
1
11 3 1 7 1 151 11 1
2 1 6 1 14 1 10 10 1
∑ mi = 1
i=0
MSB=A ; LSB=D
Boolean Identity AB+AB’=A All minterm groups must occur in a power of 2, 2n, n is the number of variables to be eliminated by the group. The resulting product term have m-n literals, which are the common variables to all minterms
1
13
1
9
1
11 3 1 7 1 151 11 1 10
2 1 6 1 14 1 10 1
Common literals
A ∑ m (8,9,10,11,12,13,14,15)= A
MSB=A ; LSB=D
∑ m (0--15)= A+A’=1
m=4, n=4
m -n =0
m0=A’B’C’D’, m1=A’B’C’D, m2=A’B’CD’, m3=A’B’CD
m4=A’BC’D’, m5=A’BC’D, m6=A’BCD’, m7=A’BCD m8=AB’C’D’, m9=AB’C’D, m10=AB’CD’, m11=AB’CD
m12=A’BC’D’, m13=A’BC’D, m14=A’BCD’, m15=A’BCD
AB CD 00 01 11 10 0 4 12 8 1 00 1 1 1
10
2
MSB=A ; LSB=D
ABC DE 000 001 011 010 00 0
01 1
4 12 8
5
7 6
13
15 14
9
11 10
11 3
10 2
ABC DE 100 101 111 110 00 16
01 17 11 19
20 28 24
MSB=A ; LSB=E
21
23
29
31
25
27
AB CD 00 01 11 10 0 4 12 8 00 01
1 5 7 6 13 15 14 9 11 10
11 3 10
2
MSB=A ; LSB=D
Describe a switching function by K-map
F(A,B,C,D)=∑m(1,3,5,12,13,14,15)
AB CD 00 01 11 10 0 4 12 8 00 1 01
2 6
111 10
m1+m3+m9+m11=B’D m=4, n=2 m -n =2
MSB=A ; LSB=D
m0=A’B’C’D’, m2=A’B’CD’ m8=AB’C’D’, m10=AB’CD’
AB CD 00 01 11 10 01 4 12 8 1 00 01
1 5 7 13 15 14 9 11 10
AB CD 00 01 11 10 0 4 12 8 00 1 1 01
1
Common literals
A’
1
5
1
13
9
11 10
∑ m (0,1,2,3,4,5,6,7)= A’ m=4, n=3
11 3 1 7 1 15 10
2 1 6 1 14
MSB=A ; LSB=D
m -n =1
m1=A’B’C’D, m3=A’B’CD, m9=AB’C’D, m11=AB’CD m4=A’BC’D’, m12=ABC’D’, m6=A’BCD’, m14=ABCD’
AB CD 00 01 11 10 0 4 12 8 00 1 1 01
1
m0+m1+m2+m3=A’B’ m4+m5+m6+m7=A’B A’B’+A’B=A’
1
5
1
13
9
11 10
11 3 1 7 1 15 10
2 1 6 1 14
MSB=A ; LSB=D
m=4, n=3
m -n =1
m0=A’B’C’D’, m1=A’B’C’D, m2=A’B’CD’, m3=A’B’CD m4=A’BC’D’, m5=A’BC’D, m6=A’BCD’, m7=A’BCD
m1=A’B’C’D, m9=AB’C’D
AB CD 00 01 11 10 0 4 12 8 00 01
1
Common literals:
B’, C’, D
1
5 7 6
13 15 14
9 11
1
11 3 10
2
m1+m9=B’C’D m=4, n=1 m-1=3
10
MSB=A ; LSB=D
m0=A’B’C’D’, m1=A’B’C’D m2=A’B’CD’, m3=A’B’CD
1 5 7 6 13 15 14 9 11 10
11 3
The decimal number is the subscript of the relative minterm or maxterm. So a direct connection can be made between the minterm or maxterm equation list and the appropriate square in karnaugh map.
Any single minterm or permitted group of minterms is called an implicant of output equation. The group size should be an integer power of 2. A prime implicant is a group of minterms that cannot be covered by any other implicant. Essential prime implicant(EPI) is a prime implicant that contains one or more minterms that are unique; that is ,terms not contained in any other prime implicant.
m1=A’B’C’D, m3=A’B’CD
AB CD 00 01 11 10 0 4 12 8 00 01
1
Common literals
A’ ,B’ ,D
1
5
13 15 14
9 11 10
m1+m2=A’B’ D m=4, n=1 m-1=3