1 / 14

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

zohar
Download Presentation

CS151 Introduction to Digital Design

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related