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ECE 3110: Introduction to Digital Systems

ECE 3110: Introduction to Digital Systems. Simplifying Sum of Products using Karnaugh Maps. Previous…. Simplifying SOP: Draw K-map Find prime implicants Find distinguished 1-cell Determine essential prime implicants if available

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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… • Simplifying SOP: • Draw K-map • Find prime implicants • Find distinguished 1-cell • Determine essential prime implicants if available • Select all essential prime implicants and the minimal set of the remaining prime implicants that cover the remaining 1’s.

  3. Example 2 • - Combining (0,2) Product term : X’Z’- Combining (2,3) Product term : X’Y- Combining (3,7) Product term :YZ • X’Z’, X’Y, and YZ are prime implicants • X’Z’, YZ are essential prime implicants • X’Y is non-essential prime implicant (redundant) because all its minterms are covered in the other essential prime implicants • F= X’Z’+X’Y+YZ (complete sum) • OR: F = X’Z’+YZ ( the minimal sum of F ) XY X 00 01 11 10 Z 0 2 6 4 0 1 1 0 0 1 3 7 5 1 Z 0 1 1 0 Y

  4. Example 3 • The essential prime implicants:- W’X’- W’Y’- WXY • Cell 7 is not covered by any of the essential prime implicants. Its covered by two non-essential prime implicant. We choosethe one with the less number of variables which is W’Z • F= W’X’+W’Y’+WXY+W’Z W WX • The prime implicants :- Cells(0,1,2,3) : W’X’- Cells(0,1,4,5) : W’Y’- Cells(1,3,5,7) : W’Z- Cells(7,15) : XYZ- Cells(14,15) : WXY 00 01 11 10 YZ 0 4 12 8 00 1 1 0 0 1 1 5 13 9 01 1 1 0 0 Z 3 7 15 11 11 1 1 1 0 Y 2 6 14 10 1 10 1 1 0 1 0 X

  5. eclipse • Given two prime implicants P and Q in a reduced map, P is said to eclipse Q if P covers at least all the 1-cells covered by Q. (P…Q). • Removing Q because P is at least as good as Q.

  6. Exercise • Row W X Y Z F 0 0 0 0 0 1 1 0 0 0 1 1 2 0 0 1 0 1 3 0 0 1 1 0 4 0 1 0 0 1 5 0 1 0 1 1 6 0 1 1 0 1 7 0 1 1 1 0 8 1 0 0 0 1 9 1 0 0 1 1 10 1 0 1 0 0 11 1 0 1 1 0 12 1 1 0 0 1 13 1 1 0 1 1 14 1 1 1 0 1 15 1 1 1 1 0 W WX 00 01 11 10 YZ 0 4 12 8 00 1 5 13 9 01 Z 3 7 15 11 11 Y 2 6 14 10 10 X

  7. Exercise Solution: • Essential prime implicants:- cells (0,1,4,5,8,9,12,13) The product term : Y’- cells (2,6,0,4) The product term : W’Z’- Cells (4,6,12,14) The product term : XZ’ • F=Y’+W’Z’+XZ’ W WX 00 01 11 10 YZ 0 4 12 8 00 1 1 1 1 1 5 13 9 01 1 1 1 1 Z 3 7 15 11 11 0 0 0 0 Y 2 6 14 10 10 1 1 1 0 X

  8. XY X 00 01 11 10 Z 0 2 6 4 0 1 1 0 1 1 3 7 5 1 Z 0 1 1 1 Y Another Example F= • The prime implicants: 1- (0,2) X’Z’ 2- (0,4) Y’Z’ 3- (2,3) X’Y 4- (3,7) YZ 5- (4,5) XY’ 6 -(5,7) XZ • No essential prime implicant! • Two possible minimal sums :1- Using the prime implicants 1,4,and 5 , F= X’Z’+YZ+XY’2- Using the prime implicants 2,3,and 6 , F= Y’Z’+ X’Y+XZ

  9. Yet another Example • - Cells (5,13,7,15) can be combined to form an essential prime implicant. W & Y change X & Z remain constant, X=1,Z=1- The product term : XZ • - Cells (0,8,2,10) can be combined to form an essential prime implicant. W & Y change Z & X remain constant, X=0, Z=0- The product term : X’Z’ • F= XZ + X’Z’ • Note that the corner cells (0,2),(0,8),(8,10),(2,10) can be combined to form the implicants : W’X’Z’ , X’Y’Z’, WX’Z’, X’YZ’ but, they are not prime implicants. W WX 00 01 11 10 YZ 0 4 12 8 00 1 0 0 1 1 5 13 9 01 0 1 1 0 Z 3 7 15 11 11 0 1 1 0 Y 2 6 14 10 10 1 0 0 1 X

  10. Next… • Simplifying PoS • Read Chapter 4.4---4.7

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