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ECE 3110: Introduction to Digital Systems

ECE 3110: Introduction to Digital Systems. Minimization Karnaugh Maps. Minimization. Logic Function minimization : Simplifying the logic function to reduce the number and size of gates. Minimization methods: 1- Using theorems T9,T9’, T10,T10’ 2- Karnaugh map. Algebraic simplification.

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ECE 3110: Introduction to Digital Systems

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  1. ECE 3110: Introduction to Digital Systems Minimization Karnaugh Maps

  2. Minimization • Logic Function minimization : Simplifying the logic function to reduce the number and size of gates. • Minimization methods:1- Using theorems T9,T9’, T10,T10’ 2- Karnaugh map

  3. Algebraic simplification • Starting from a truth table, or a minterm (maxterm) list • Reduce the cost of a 2-level circuit: • Minimize # of first-level gates • Minimize # of inputs of gates • Do not consider the cost of input inverters: assuming that both true and complemented versions of all input variables are available.

  4. Algebraic simplification • T10, X.Y+X.Y’=X T10’: (X+Y).(X+Y’)=X • Generalization of the theorems: • (Given product term).Y+(Given product term).Y’= Given product term • (Given product term+Y).(Given product term+Y’)= Given product term • Reduce number of gates and gate inputs

  5. Prime detector

  6. Algebraic simplification • Theorem T10, • Reduce number of gates and gate inputs

  7. Resulting circuit

  8. 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 )

  9. Two-variable Karnaugh map • Example : F = X.Y’+X.Y • Simplification : F = X(Y+Y’) = X.1 = X • Also proven by T(10) X X ROW X Y F 0 1 Y 0 0 0 0 0 2 0 0 1 1 0 1 0 1 3 1 Y 2 1 0 1 0 1 3 1 1 1

  10. Two-variable Karnaugh map • Example : F = X’.Y’+X’.Y • Simplification : F = X’(Y+Y’) = X’.1 = X’ • Exercise : Can we simplify FX,Y= SX,Y (1,2) ? X X ROW X Y F 0 1 Y 0 0 0 1 0 2 0 1 0 1 0 1 1 1 3 1 Y 2 1 0 0 1 0 3 1 1 0

  11. 3-variable Karnaugh map • Can we simplify FX,Y,Z = Sx,y,z (1,3,5,7) ?

  12. Example : F=X’.Y’.Z’+X’.Y’.Z+X.Y’.Z’+X.Y’.Z Row X Y Z F 0 0 0 0 1 1 0 0 1 1 2 0 1 0 0 3 0 1 1 0 4 1 0 0 1 5 1 0 1 1 6 1 1 0 0 7 1 1 1 0 F = X’.Y’.(Z’+Z)+X.Y’.(Z’+Z)=X’.Y’+X.Y’=(X’+X).Y’ = Y’ Three-variable Karnaugh map X XY 00 01 11 10 Z 0 2 6 4 0 1 0 0 1 1 3 7 5 1 Z 1 0 0 1 Y

  13. 4-variable Karnaugh map

  14. Five-variable Karnaugh map • Five variable K-map is formed using two connected 4-variable maps: V VWX W W 000 001 011 010 100 101 111 110 YZ 0 4 12 8 16 20 28 24 00 1 5 13 9 17 21 29 25 01 Z 3 7 15 11 19 23 31 27 11 Y 2 6 14 10 18 22 30 26 10 X X

  15. Next: • Simplifying SOP

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