ECE 3110: Introduction to Digital Systems

Simplifying Sum of Products

using Karnaugh Maps

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Algebra MinimizationK-Map

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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 )

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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

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Prime-number detector (again)

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

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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

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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’

0

1 3

2

XY

Z

X

Z

1 1

0 0

00 01

0

1 7

6

1

0

11

5

41

1

10

Y

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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 an implicant 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

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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?

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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

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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 the essential 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 expression

if more than one set of the remaining prime implicants contains the same number of minterms.

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Next:

Simplifying Products of sums (POS)Other minimizationRead Ch. 4.3.6--4.5HW #6