Objectives:. Four variables maps. 2. Simplification using prime implicants. 3. "on t care" conditions. 4. Summary.. Four variables Karnaugh maps Minterms A A m m m3 m2 A B C m4 C A B C m2 m8 C C m5 C m3 m9 m7 m5 m C m6 m4 C m C B
The rows and columns are numbered in a gray code sequence. One square represents one minterm with four literals. Two adjacent squares represent one term with 3 literals. Four adjacent squares represent one term with 2 literals. Eight adjacent squares represent one term with literal. Examples: Example : Simplify the Boolean function wx yz m m m3 m2 m4 m5 m7 m6 m2 m8 m3 m9 m5 m m4 m w y z w y z w z y x y z xy z x z F w, x, y, z y w z x z The simplified function:
Example 2: plot the following 4-variable expression on a Karnaugh map cd ab d a b acd Example 3: simplify the following function: cd ab Four corner terms combine to give bd abd c f(a, b, c,d) = c + b d + a bd
Example 4: For the following k-map with four variables, obtain the simplified logic expression: C AC C A Example 5: Use a k-map to simplify Multiply out Fill the terms in k-map: Y A B C Simplify:
efinitions: 2. Simplification using prime implicants Prime implicant (PI): is a product term obtained by combining the maximum possible number of adjacent squares in the map. Essential prime implicant: If a minterm in a square is covered by only one prime implicant that prime implicant is said to be essential, and it must be included in the final expression. Note: All of the prime implicants of a function are generally not needed in forming the minimum sum of products. Procedure for selecting implicants:. Find essential prime implicants. 2. Find a minimum set of prime implicants which cover the remaining s on the map. Example : C C C A C C AC are prime implicants. are not prime implicants.
Example 2: find the minimum solution for the following Karnaugh map. AC Essential PI A Essential PI Essential PI Non essential PI are essential prime implicants, to complete the minimum solution, one of the nonessential prime implicants in needed: The final solution: Example 3: Simply using k-map: 2 3 4 5 6 7 8 9 2 3 4 5 6
Groups Implicants Simplification (terms number), 4, 3, 6 Essential prime implicant 6, 7,, Essential prime implicant 3, 4, 5, 6 Prime implicant 3, 7,, 5 Prime implicant,, 4, 5 Prime implicant 3, 4, 5, 6 Prime implicant 2 essentials and four prime implicants Square 3 can be covered with either prime implicants or. Square 4 can be covered with either prime implicants or. Square 5 can be covered with any one of four prime implicants. o Final solution: two essential PI with any two prime implicants that cover minterms 3, 4, 5: four possible ways: ) 2) 3) 4) 3. "on t care" conditions Some logic circuits can be designed so that there are certain input conditions for which there are no specified output levels (can t happened). A circuit designer is free to make the output for any "don t care condition" either a or a in order to produce the simplest output. Example : For the following truth table, use K-map to minimize the function inputs Output on t Care Condition
C C Z A Example 2: esign a logic circuit that controls an elevator door in a three-story building. : Moving signal: ( : moving), ( : stopped), : Floor indicator signals. Truth Table M F F2 F3 Elevator Circuit Open
F2F3 MF F2F3 MF Open M F F2 F3 Example 3: For the following logical expression, use K- map to minimize the function. cd ab cd ab 4. Summary Looping a pair of adjacent s in a k-map eliminate the variable that appears in complemented and uncomplemented form. Looping a quad (4) of adjacent s eliminates the two variables that appear in complemented and uncomplemented form. Looping an octet (8) of adjacent s eliminates the three variables that appear in complemented and uncomplemented form.
o Looping groups of two pairs: C C C BC BC A BC A B o Looping groups of four (quads): C C B A B
o Looping groups of eight (octet): B C B