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CS151 Introduction to Digital Design

CS151 Introduction to Digital Design. Chapter 2-4-1 Map Simplification. Circuit Optimization. Goal: To obtain the simplest implementation for a given function Optimization is a more formal approach to simplification that is performed using a specific procedure or algorithm

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CS151 Introduction to Digital Design

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  1. CS151Introduction to Digital Design Chapter 2-4-1 Map Simplification

  2. CircuitOptimization • Goal: To obtain the simplest implementation for a given function • Optimization is a more formal approach to simplification that is performed using a specific procedure or algorithm • Optimization requires a cost criterion to measure the simplicity of a circuit • To distinct cost criteria we will use: • Literal cost (L) • Gate input cost (G) • Gate input cost with NOTs (GN) CS 151

  3. B C D D C B B C D B LiteralCost • Literal – a variable or its complement • Literal cost – the number of literal appearances in a Boolean expression corresponding to the logic circuit diagram • Examples: • F = BD + A C + A L = 8 • F = BD + A C + A + AB L = • F = (A + B)(A + D)(B + C + )( + + D) L = • Which solution is best? CS 151

  4. B C D B C B D B D C B GateInputCost • Gate input costs - the number of inputs to the gates in the implementation corresponding exactly to the given equation or equations. (G - inverters not counted, GN - inverters counted) • For SOP and POS equations, it can be found from the equation(s) by finding the sum of: • all literal appearances • the number of terms excluding terms consisting only of a single literal,(G) and • optionally, the number of distinct complemented single literals (GN). • Example: • F = BD + A C + A G = 11, GN = 14 • F = BD + A C + A + AB G = , GN = • F = (A + )(A + D)(B + C + )( + + D) G = , GN = • Which solution is best? CS 151

  5. GN = G + 2 = 9 L = 5 G =L+ 2 = 7 B C A F • L (literal count) counts the AND inputs and the single literal OR input. • G (gate input count) adds the remaining OR gate inputs C B • GN(gate input count with NOTs) adds the inverter inputs CostCriteria(continued) • Example 1: • F = A + B C + CS 151

  6. B A A B C F A B C F A C C B CostCriteria(continued) • Example 2: • F = A B C + • L = 6 G = 8 GN = 11 • F = (A + )( + C)( + B) • L = 6 G = 9 GN = 12 • Same function and sameliteral cost • But first circuit has bettergate input count and bettergate input count with NOTs • Select it! CS 151

  7. Map Simplification • Simplifying Boolean Expressions • Algebraic manipulation • Lacks rules for predicting steps in the process • Difficult to determine if the simplest expression has been reached • Map simplification (Karnough map or K-map) • Straightforward procedure for function of up to 4 variables • Provides a pictorial form of the truth table • Maps for five and six variables can be drawn, but are more cumbersome to use CS 151

  8. Karnaugh Maps (K-map) • A K-map is a collection of squares • Each square represents a minterm • Adjacent squares differ in the value of one variable • Visual diagram of all possible ways a function may be expressed in standard forms • Sum-of-Products • Product-of-Sums • Alternative algebraic expressions for the same function are derived by recognizing patterns of squares select simplest. CS 151

  9. Y’ Y X’ X Two-Variable Map • 2 variables  4 minterms  4 squares. CS 151

  10. K-Map Function Representation • Example: F(X,Y) = XY’ + XY • From the map, we see that F (X,Y) = X. • This can be justified using algebraic manipulations: F(X,Y) = XY’ + XY = X(Y’ +Y) = X.1 = X 1 1 CS 151

  11. 1 1 1 K-Map Function Representation • Example:G(x,y) = m1 + m2 + m3 G(x,y) = m1 + m2 + m3 = X’Y + XY’ + XY From the map, we can see that G = X + Y CS 151

  12. Three-Variable Maps 3 variables 8 minterms (m0 – m7). How can we locate a minterm square on the map?  Use figure (a) OR  use column # and row # from figure (b)E.g. m5 is in row 1 column 01 (5 10 = 101 2) Q. Show the area representing X’? Y’? Z’? CS 151

  13. y 0 3 x 2 1 x 4 7 6 5 z z z Alternative Map Labeling • Map use largely involves: • Entering values into the map, and • Reading off product terms from the map. • Alternate labelings are useful: y y z y x 00 01 11 10 0 3 2 1 0 4 7 6 5 1 x z CS 151

  14. 1 1 1 1 Y Y YZ YZ 00 00 01 01 11 11 10 10 X X 0 0 1 1 X X Z Z Example Functions • By convention, we represent the minterms of F by a "1" in the map and leave the minterms of F’ blank • Example: • Example: • Learn the locations of the 8 indices based on the variable order shown (x, most significantand z, least significant) on themap boundaries 1 1 1 1 CS 151

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