Minimisation of Logic Functions. F = (x + x.y).z + = y.z. F = x.y.z + x.z + y.z
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1 Minimisation of Logic Functions Example : F = (x + x.y).z + = y.z Sum of Products: Gate Gate Gate F = x.y.z + x.z + y.z Gate ~ inputs Gate ~ inputs Gate ~ inputs F consists of 7 literals and products # of literals = combined # of inputs to the first level gates # of products = # of inputs to the second level gate Sum of Minterms ( Canonical Form) : F = x.y.z + (y + y ).x.z + (x + x ).y z = x.y.z + x.y.z + x.y.z + x.y.z + x.y.z Each minterm has to include all the variables. Minterm labelling : Minterm Label x.y.z ~ = x.y.z ~ = 7 x.y.z ~ = x.y.z ~ = 4 x.y.z ~ = F = (,, 4,, 7) Karnauph Map z x y # # # # #6 #7 #4 #
2 Cell labelling examples: cell # binary coordinates x, y, z 7 x, y, z Minterms in F = (,, 4,, 7) correspond to the cells with value. Product of Maxterms ( Canonical Product) : F = logical product of all cells holding zero value = (value of cell #).(value of cell #) ).(value of cell #6). ).(value of cell #4) = (x + y + z).(x + y + z).(x + y + z) = (,, 6) Creation of maxterms examples: Cell # binary cell value Note 6 x + y + z the value is obtained as a sum x + y + z of the complemented literals We can check the canonical product expression. Apply De Morgan rule: F = (x + y + z ) + (x + y + z) + (x + y + z) = x.y.z + x.y.z + x.y.z + x.y.z + x.y.z = (,, 4,, 7) to get the obvious identity (,, 4,, 7) ' = (,, 6) Minimisation using Karnauph map. Example (continued) : A minimal sum should have the smallest # of product terms ( st level gates) the smallest # of literals
3 z x y Karnauph Map Minimisation Procedure. Combine adjacent cells holding to form the largest rectangles possible and simplify the content of these using the rule a.(b+b') = a: # and #7 x.y' #4 and # y'.z' # and #4 y.z Minimum Sum : F = x.y' + y'.z + y.z Example : Find the minimum sum for F = (,, 4,, 6) x y z 6 4 Prime implicants are the largest rectangles eclipsing the cells holding a : Prime Implicant Area Value Cube Dim d A = cell 6 + cell 4 = x.z B = cell + cell + cell 4 + cell = y # of cells eclipsed by a prime implicant = d : area of A = = area of B = = 4 F = x.z + y Minimum sum = sum of prime implicants ( but not necessary all of them).
4 Example : Minimise the prime number detector F = N N N N (,,,, 7,, ) N N N N Prime Implicant Area Value A c+c+c+c7 N.N B c+c N.N.N C c+c N.N.N D c+c N.N.N Distinguished cell is covered by only a single prime implicant. Essential Prime Implicant is a prime implicant covering a distinguished cell. Distinguished cell c c c7 c c Essential Prime Implicant A B A D C Include all essential prime implicants : F = A + B + C + D = N.N + N.N.N + N.N.N + N.N.N 4
5 Example 4: Minimise F = w x y z (,,,, 4,, 7, 4, ) w Prime Implicant Area Value Dimension A c+c+c+c w.x B c+c+c+c4 w.y C c+c+c7+c w.z D c7+c x.y.z E c+c4 w.x.y F = A + B + C + D + E = w.x + w.y + w.z + x.y.z + w.x.y not minimal w x y z Essential Prime Implicant A B E Essential Cell c c4 c4 Include all essential prime implicants F = A + B + E + ( largest prime implicants for the remaining cells)
6 Reduced K- map showing cells covered by the essential prime implicants: w x y z Remaining uncovered cells c7 Largest Prime Implicant C The minimum sum: F = A + B + E + C = w.x + w.y + w.x.y + w.z Example : Five input Karnauph map. F = (, 4, 8,,,, 6, 9,,, 4, 8 ) X 4 X X X X 6
7 Prime Implicants Value Dimension A X.X B X 4.X.X.X C X 4.X.X.X They are all essential so F = A + B + C = X.X + X 4.X.X.X + X 4.X.X.X Alternative (Two Dimensional) Representation of variable Karnauph map: X.X X 4 X X The upper and lower halves of each cell correspond to the adjacent cells in the D Karnauph map. 7
8 Tabulation Method of Reduction (Quine-McCluskey procedure). Example Minimise F = A.B.C.D + A.B. C.D + A.B.C.D + A.B.C..D + A'B'CD' Step : Note: Eliminate as many terms as possible using logical adjacencies. The resulting terms will be prime implicants. Logical adjacency between two n variables minterms can only exist if n- variables appear the same in both minterms and only one variable is complemented. So logically adjacent minterms differ in only one variable. Hence to find adjacent minterms, group them according to the number of complemented variables so one needs to compare only the minterms in adjacent groups. Group Number (# of complemented variables) Minterms Binary representation Minterm label A.B.C.D S A.B.C.D S A.B.C.D S 8 A.B.C.D S 4 A.B.C.D S Grouping of minterms according to the # of complemented variables Group Adjacent minterms Resulting Cube Binary Labels Dimension A.B.C.D S A.B.C.D +A.B.C.D A.B.C.D +A.B C.D X X S +S 8 S + S A.B.C.D+A.B.C.D A.B.C.D +A.B.C.D X X S + S S 8 + S Combining minterms in adjacent groups to form D cubes 8
9 Group Cubes Resulting Cube Binary Labels Dimension S S S +S 8 S + S XX XX (S +S 8 ) + ( S + S ) (S + S ) + (S 8 + S ) Combining cubes in adjacent groups to form D cubes Prime implicants are: Composed of Binary Label Minterms Label Value Dimension P S A.B.C.D P S +S 8 + S + S XX B.D P' S + S + S 8 + S XX B.D redundant Step : Identify the essential prime implicants using the prime implicants table: Prime Implicant Minterms S S S 8 S S P X P X X X X Both the prime implicants are essential so they have to be included: F = P + P = A.B.C.D + B.D 9
10 Example: Minimise F = (, 6, 7, 9,, ) A B C D Step : Identify all prime implicants. Group Number (# of complemented variables) Minterms Binary representation Minterm label A.B.C.D S A.B.C.D A.B.C.D S S 7 A.B.C.D A.B.C.D S 9 S 6 A.B.C.D S Group Adjacent minterms Resulting Cube Binary Labels Dimension S, S X S +S S, S 7 X S + S 7 S 7, S 6 X S 7 + S 6 S, S 9 X S + S 9 S 6, S X S 6 + S There are five prime implicants: Label Value Binary Value P S + S A.B.D P S + S 7 B.C.D P S 7 + S 6 A.B.C P4 S + S 9 A.C.D P S 6 + S A.C.D Step : Identify essential prime implicants from prime the implicants table. Prime Implicant minterms S S 6 S 7 S 9 S S P x x P x x P x x P4 x x P x x
11 Find the essential cells (shown in dark). These identify the minterms ( S and S 9 ) which are covered only by a single prime implicant ( E and D respectively). So E and D are the essential prime implicants. Step : Identify the secondary prime implicants which include most if not all of the remaining minterms. Remove the rows and columns corresponding to the essential prime implicants D and E to show the remaining minterms (S 7 and S ). Prime Implicant minterms S S 6 S 7 S 9 S S P x P x x P x Choose the prime implicant B as it includes all the remaining minterms. F = P + P4 + P = A.C.D + A.C.D + B.C.D
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