数字电路教学课件：chapter2-2

1、Boolean switching algebraKarnaugh maps000111100101236745ABCMSBLSB00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D000111100101236745ABCMSBLSBIt is a matrix of squares. each square represent a minterm or maxterm from a Boolean equation. 000 001 011 010ABCDE000111100 4 12 8 1 5 1

2、3 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=E100 101 111 110ABCDE0001111016 20 28 24 17 21 29 25 19 23 31 27 18 22 30 26 MSB=A ; LSB=EN-variable karnaugh map have 2n squares.00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=Dnumeral on the sides of k-map is the variable coordinates. By d

3、ecoding the binary coordinates, We label the decimal value for each square. 000111100101236745ABCMSBLSB. 00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=DThe squares correspond to the adjacent minterms are Across the top and down the side of k- map, only one bit change occur be

4、tween adjacent squares for each column and row Logically adjacent Adjacent Symmetricalstack00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=DLogically adjacent 00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11

5、2 6 14 10 MSB=A ; LSB=DLogically adjacent 00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=DLogically adjacent 00011110ABCDE0000 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D00101101000011110ABCDE10016 20

6、 28 24 17 21 29 25 19 23 31 27 18 22 30 26 MSB=A ; LSB=D101111110Describe a switching function by K-map00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=DSimplify a equation use N-variable K-mapA group of adja

7、cent minterms eliminate variable from the final expression 000111100101236745ABCMSBLSBA group of adjacent minterms eliminate variable from the final expression 000111100101236745ABCMSBLSBBoolean Identity A sum of logically adjacent can be simplified by , which is the in the two minterms . The result

8、ing product term has literals.Boolean Identity 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 minterms00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2

9、6 14 10 MSB=A ; LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=Dm0+m1=ABCm2+m3=ABCm0+m1+m2+m3=AB00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=Dm1+m3=ABDm9+m11=ABDm1+m3+m9+m11=BD

10、00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=Dm1+m3+m9+m11=BD00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=Dm0+m2+m8+m9=BDm0=ABCD, m1=ABCD, m2=ABCD, m3=ABCDm4=ABCD, m5=ABCD, m6=ABCD, m7=ABCD00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB

11、=A ; LSB=Dm0=ABCD, m1=ABCD, m2=ABCD, m3=ABCDm4=ABCD, m5=ABCD, m6=ABCD, m7=ABCD00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=Dm0=ABCD, m1=ABCD, m2=ABCD, m3=ABCDm4=ABCD, m5=ABCD, m6=ABCD, m7=ABCDm8=ABCD, m9=

12、ABCD, m10=ABCD, m11=ABCDm12=ABCD, m13=ABCD, m14=ABCD, m15=ABCD00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=DBoolean Identity All mint

13、erm 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 mintermsAny single minterm or permitted group of minterms is called of output equation. The group size should be

14、an integer power of 2. is a group of minterms that cannot be covered by any other implicant. is a prime implicant that contains one or more minterms that are unique; that is ,terms not contained in any other prime implicant.A set of 2n k-map squares are combined to form a prime implicant, if n-varia

15、bles of the equation being simplified have 2n permutations within the set and the remaining m-n variables have the same value within the set. In other words Load the minterms into the k-map by placing a 1 in the appropriate squareLook for all prime implicantsLook for all essential prime implicantsSe

16、lect all EPI and a minimal set of remaining implicants that cover all remaining 1s in the karnaugh mapMore than one equally simplified result is possible when more than one set of remaining prime implicants contain the same number of minterms Examples00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2

17、 6 14 10 MSB=A ; LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ;

18、 LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D00011110ABCD000111100 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D00011110ABC