CS221: Digital Design. Dr. A. Sahu. Indian Institute of Technology Guwahati

Transcription

1 CS221: Digital Design QMLogicMinimization Minimization Dr. A. Sahu DeptofComp.Sc.&Engg. Indian Institute of Technology Guwahati 1

2 Outline Quine McCluskey(QM) Logic Minimization Examples Writing C/C++ program for QM Method 2

3 BCDAdder(TwoLevel) Treat as 9 input & 5 output functions Generate TthTbl Truth Table for each outputs t Solve each function using KMAP/QM Method Only Two Level: No carry Propagation Ci a1 b1 c1 d1 a2 b2 c2 d2 Adder 2 Level a b c d Co 3

4 Quine McCluskeyMethod for Minimization KMAP methods was practical for at most 6 variable functions Larger number of variables: need method that can be applied to computer based minimization Quine McCluskey method For example: m(0,1,2,3,5,7,13,15)

5 QMMethod Method PhaseI:findingPis Pis Tabular methods: Grouping and combining PhaseII:CoversminimalPIs PIs 5

6 QM Method QM Method Mintermsthat differ in one variable s value can be combined. Thus we list our mintermsso that they are in groups with each group having the same number of 1s. So the first step is ordering the minterms according to their number of 1s (0 cube list) only mintermsresiding in adjacent groups have the chance to be combined.):

7 QM Method m (0,1,2,3,5,7,13,15 ) 0_Cube

8 QM Method: Combining Adjacent Compare mintermsof a group with those ofanadjacentonetoform1 adjacent one form 1 cube cubelist. When doing the combining, we put checkmark alongside the mintermsin the 0 cubelistthathavebeencombined. that have been combined.

9 QM Method 1_Cube 0_Cube , , , ,4

10 QM Method: Combining Adjacent Do same combination of comparing adjacent group minterms To form 2 cubes, 3 cubes and so on. Ol Only mintermsof it adjacent groups have the chance of being combined Which have an in the same position.

11 QM Method 1_Cube _Cube 0 00 * 01 * *

12 Q MMethod:CoverPIs Method: PIs : terms left without checkmarks. After identifying our PIs, we list them against the mintermsneeded to be covered m (0,1,2,3,5,7,13,15 ) 0 0 x x 0 x x 1 x 1 x Func

13 QMMethod:Covers Method To find a minimal cover, we first need to find essentialpis To do this we need to find columns that only haveone checkmark in them, the according row will thus show the essential PI. AfteridentifyingessentialPIs,thatare essential that are necessarily part of the cover, we cover any remaining mintermsusing a minimal set of PIs. In this example: F = A B +BD

14 QM Method: Another Example Tabular method to systematically find all prime implicants ƒ(a,b,c,d) BCD)= m(4,5,6,8,9,10,13) + d(0,7,15) Stage 1: Find all prime implicants Step 1: Fill Column 1 with ON set and DC set mintermindices. Group by number of 1's. Cube

15 Quine McCluskeyMethod Step 2:ApplyUnitingTheorem: Cube0 Cube1 Cube2 Compare elements of group x00 * w/ N 1's against those with N+1 1s. 1's 0100 x000 * Differ by one bit implies x adjacent x0 Eliminate variable and place in x * next column. E.g., 0000 vs yields x0 * 0000 vs yields x1 When used in a combination, mark with a check. If cannot be combined, mark with a star. These are the prime implicants. Repeat until no further combinations can be made x x x01 * x111 11x1

16 QuineMcluskeyMethod Step2:ApplyUnitingTheorem: Cube0 Cube1 Cube2 Compare elements of group x00 * 01xx * w/ N 1's against those with N+11's 1s x000 * x1x1 * Differ by one bit implies x adjacent x0 Eliminate variable and place in x * next column. E.g., 0000 vs yields x0 * 0000 vs yields x1 When used in a combination, mark with a check. If cannot be combined, mark with a star. These are the prime implicants. Repeat until no further combinations can be made x x x01 * x111 11x1

17 Finding the Minimum Cover We have so far found all the prime implicants 2 nd step of the Q M procedure is to find the smallest set of prime implicantsto coverthecompleteon complete on set setofthe the function

18 Finding the Minimum Cover This is accomplished through the prime implicant chart Columns are labeled with the mintermindices of the onset Rows are labeled with the mintermscovered byagivenprimeimplicant implicant Example a prime implicant ( 1 1) becomes minterms0101, 0111, 1101, 1111, which are indices of mintermsm5, m7, m13, m15

19 Coverage Table/ Chart 0,4(0 00) ( 000) 0,8( 000) 8,9(100 ) 8,10(10 0) 9,13(1 01) 4,5,6,7(01 ) 5,7,13,15( 1 1) Note: Don t include DCs in coverage table; they don t have covered by the final logic expression! rows = prime implicants columns = ON set elements placean""ifon setelementis an element is covered by the prime implicant

20 Coverage Table/ Chart 0,4(0 00) ( 000) 0,8( 000) 8,9(100 ) 8,10(10 0) 9,13(1 01) 4,5,6,7(01 ) 5,7,13,15( 1 1) rows = prime implicants columns = ON setelements elements place an "" if ON set element is covered by the prime implicant If column has a single, than the implicantassociated associated withthe the row is essential. It must appear in minimum cover

21 Coverage Table/ Chart: Eliminate 0,4(0 00) ( 000) 0,8( 000) 8,9(100 ) 8,10(10 0) 9,13(1 01) 4,5,6,7(01 ) 5,7,13,15( 1 1) rows = prime implicants columns = ON set elements place an "" if ON set element is covered by the prime implicant If column has a single il, than the implicantassociated with the row is essential. It must appear in minimum cover Eliminate all columns covered by essential primes

22 Coverage Table/ Chart: Eliminate 0,4(0 00) ( 000) 0,8( 000) 8,9(100 ) 8,10(10 0) 9,13(1 01) 4,5,6,7(01 ) 5,7,13,15( 1 1) rows = prime implicants columns = ON-set elements place an "" if ON-set element is covered by the prime implicant Find minimum set of rows that cover the remaining columns Eliminate all columns covered by essential primes

23 Coverage Table/ Chart: Eliminate 0,4(0 00) ( 000) 0,8( 000) 8,9(100 ) 8,10(10 0) 9,13(1 01) 4,5,6,7(01 ) 5,7,13,15( 1 1) If all are covered: Write the Implicants F= AB D + AC D + A B

24 Thanks 24