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ECE 3110: Introduction to Digital Systems

ECE 3110: Introduction to Digital Systems. Simplifying Sum of Products using Karnaugh Maps. Previous…. Algebra Minimization K-Map. Karnaugh Maps.

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ECE 3110: Introduction to Digital Systems

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  1. ECE 3110: Introduction to Digital Systems Simplifying Sum of Products using Karnaugh Maps

  2. Previous… • Algebra Minimization • K-Map

  3. Karnaugh Maps • Karnaugh Map : a representation of the truth table by a matrix of squares (cells) , where each square corresponds to a minterm ( or a maxterm) of the logic function. • For n-variable function, we need 2^n rows truth table and 2^n squares (cells). • The square number is equivalent to the row number in the truth table • To represent a logic function, the truth table values are copied into their corresponding cells . • The arrangements of the squares help to identify the input variable redundancy ( X.Y.Z+X.Y.Z’=X.Y )

  4. Karnaugh-map usage • Plot 1s corresponding to minterms of function. • Circle largest possible rectangular sets of 1s. • # of 1s in set must be power of 2 • OK to cross edges • Read off product terms, one per circled set. • Variable is 1 ==> include variable • Variable is 0 ==> include complement of variable • Variable is both 0 and 1 ==> variable not included • Circled sets and corresponding product terms are called “prime implicants” • Minimum number of gates and gate inputs

  5. Prime-number detector (again)

  6. When we solved algebraically, we missed one simplification -- the circuit below has three less gate inputs.

  7. Simplifying the Sum of Products (again) • Two main steps :1) Combining/Grouping the 1-cells.2) Writing the product term for each group. • Rules : ( for n-variable function ) • 1) The group size must be a power of 2.2) A set of 2^i cells can be combined if there are ( i ) variables that take all possible combinations within the set and the remaining ( n-i ) variables have the same value within that set. • 3) The corresponding product term for each group contains (n-i) literals: - The variable is complemented if it is 0 in the combined cells - The variable is uncomplemented if it’s 1 in the combined cells - The variable is not included in the product term if it takes both values 0 and 1 within the combined cells

  8. X XY 00 01 11 10 Z 0 2 6 4 0 1 1 1 1 1 3 7 5 1 Z 0 0 0 1 Y Example 1 • The canonical sum is : F=X’.Y’.Z’+X’.Y.Z’+X.Y’.Z’+X.Y’.Z+X.Y.Z’ • Combine cells (0,2,6,4)X=0,1 , Y= 0,1 , Z=0Product Term : Z’ • Combine cells ( 4,5 )Z=0,1 , Y=0 , X=1Product Term : XY’ • F= X.Y’+Z’

  9. Definitions : • A logic function P implies a logic function F if for every input combination for which P=1, then F=1 also. ( P is an implicant of F, P implies F, p=>F, F includes P, or F covers P ) • Any minterm or combination of minterms in the canonical sum expression is animplicant of the output function • A prime implicant is a group of combined minterms that cant be combined with any other minterm or group of minterms (a circled set) • Distinguished 1-cell: is an input combination that is covered by only one prime implicant. (a unique minterm) • Essential prime implicant is a prime implicant which covers one or more distinguished 1-cells (i.e. at least one minterm isn’t contained in any other prime implicant.) • A minimal sum of a logic function is a sum-of-products expression for F such that no sum-of-products expression for F has fewer product terms. • In Example 1 : - X’Y’Z’ , (X’Y’Z’+XYZ’) , XY’ are implicants of F - XY’ , Z’ are prime implicants and essential prime implicants

  10. Prime-Implicant Theorem • A minimum sum is a sum of prime implicants. • Complete sum: sum of all the prime implicants of a logic function. • Complete sum is not always minimal. • Which prime implicants should be included and which should not be included in the minimum sum?

  11. Minimal sum • Essential prime implicants (if available) must be included. • Secondary essential prime implicants must be included. • If a logic function with no essential prime implicants at all. • Trial and error • Branching method

  12. How to get minimal sum • Load the minterms and maxterms into the K-map by placing the 1’s and 0’s in the appropriate cells. • Look for groups of minterms and write the corresponding product terms ( the prime implicants): a- The group size must be a power of 2. b- Find the largest groups of minterms first then find smaller groups of minterms until all groups are found and all 1-cells are covered. • Determine theessential prime implicants. • Select all essential prime implicants and the minimal set of the remaining prime implicants that cover the remaining 1’s. • It’s possible to get more than one equally simplified expressionif more than one set of the remaining prime implicants contains the same number of minterms.

  13. Next: • Simplifying Products of sums (POS) • Other minimization • Read Ch. 4.3.6--4.5 • HW #6

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