1 / 69

Combinational Logic (mostly review!)

Combinational Logic (mostly review!). Logic functions, truth tables, and switches NOT, AND, OR, NAND, NOR, XOR, . . . Minimal set Axioms and theorems of Boolean algebra Proofs by re-writing Proofs by perfect induction Gate logic Networks of Boolean functions Time behavior Canonical forms

jeniferb
Download Presentation

Combinational Logic (mostly review!)

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. Combinational Logic (mostly review!) • Logic functions, truth tables, and switches • NOT, AND, OR, NAND, NOR, XOR, . . . • Minimal set • Axioms and theorems of Boolean algebra • Proofs by re-writing • Proofs by perfect induction • Gate logic • Networks of Boolean functions • Time behavior • Canonical forms • Two-level • Incompletely specified functions • Two-level minimization CS 150 - Spring 2008 – Lec #3: Combinational Logic - 1

  2. 0 1 X not X Y not Y X xor Y X = Y X and Y X nand Ynot (X and Y) X or Y X nor Ynot (X or Y) Possible Logic Functions of Two Variables • 16 possible functions of 2 input variables: • 2**(2**n) functions of n inputs X F Y • X Y 16 possible functions (F0–F15)0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 10 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 • 1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 • 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 CS 150 - Spring 2008 – Lec #3: Combinational Logic - 2

  3. Cost of Different Logic Functions • Some are easier, others harder, to implement • Each has a cost associated with the number of switches needed • 0 (F0) and 1 (F15): require 0 switches, directly connect output to low/high • X (F3) and Y (F5): require 0 switches, output is one of inputs • X' (F12) and Y' (F10): require 2 switches for "inverter" or NOT-gate • X nor Y (F4) and X nand Y (F14): require 4 switches • X or Y (F7) and X and Y (F1): require 6 switches • X = Y (F9) and X  Y (F6): require 16 switches • Because NOT, NOR, and NAND are the cheapest they are the functions we implement the most in practice CS 150 - Spring 2008 – Lec #3: Combinational Logic - 3

  4. X Y X nor Y0 0 11 1 0 X Y X nand Y0 0 11 1 0 Minimal Set of Functions • Implement any logic functions from NOT, NOR, and NAND? • For example, implementing X and Yis the same as implementing not (X nand Y) • Do it with only NOR or only NAND • NOT is just a NAND or a NOR with both inputs tied together • and NAND and NOR are "duals", i.e., easy to implement one using the other • Based on the mathematical foundations of logic: Boolean Algebra X nand Y not ( (not X) nor (not Y) ) X nor Y not ( (not X) nand (not Y) ) CS 150 - Spring 2008 – Lec #3: Combinational Logic - 4

  5. Logic Functions and Boolean Algebra • Any logic function that can be expressed as a truth table can be written as an expression in Boolean algebra using the operators: ', +, and • X Y X • Y0 0 00 1 01 0 0 1 1 1 X Y X' X' • Y0 0 1 00 1 1 11 0 0 0 1 1 0 0 X Y X' Y' X • Y X' • Y' ( X • Y ) + ( X' • Y' )0 0 1 1 0 1 10 1 1 0 0 0 01 0 0 1 0 0 0 1 1 0 0 1 0 1 ( X • Y ) + ( X' • Y' ) º X = Y Boolean expression that is true when the variables X and Y have the same value and false, otherwise X, Y are Boolean algebra variables CS 150 - Spring 2008 – Lec #3: Combinational Logic - 7

  6. A S B Cout Cin A B Cin S Cout 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 0 1 0 0 1 0 0 0 1 0 1 1 1 Simple Example • 1-bit binary adder • Inputs: A, B, Carry-in • Outputs: Sum, Carry-out S = A' B' Cin + A' B Cin' + A B' Cin' + A B Cin Cout = A' B Cin + A B' Cin + A B Cin' + A B Cin CS 150 - Spring 2008 – Lec #3: Combinational Logic - 14

  7. Apply the Theorems to Simplify Expressions • Theorems of Boolean algebra can simplify Boolean expressions • e.g., full adder's carry-out function (same rules apply to any function) Cout = A' B Cin + A B' Cin + A B Cin' + A B Cin = A' B Cin + A B' Cin + A B Cin' + A B Cin + A B Cin = A' B Cin + A B Cin + A B' Cin + A B Cin' + A B Cin = (A' + A) B Cin + A B' Cin + A B Cin' + A B Cin = (1) B Cin + A B' Cin + A B Cin' + A B Cin = B Cin + A B' Cin + A B Cin' + A B Cin + A B Cin = B Cin + A B' Cin + A B Cin + A B Cin' + A B Cin = B Cin + A (B' + B) Cin + A B Cin' + A B Cin = B Cin + A (1) Cin + A B Cin' + A B Cin = B Cin + A Cin + A B (Cin' + Cin) = B Cin + A Cin + A B (1) = B Cin + A Cin + A B CS 150 - Spring 2008 – Lec #3: Combinational Logic - 15

  8. From Boolean Expressions to Logic Gates (cont’d) • More than one way to map expressions to gates • e.g., Z = A' • B' • (C + D) = (A' • (B' • (C + D))) T2 T1 use of 3-input gate A Z A B T1 B Z C C T2 D D CS 150 - Spring 2008 – Lec #3: Combinational Logic - 18

  9. Waveform View of Logic Functions • Just a sideways truth table • But note how edges don't line up exactly • It takes time for a gate to switch its output! time change in Y takes time to "propagate" through gates CS 150 - Spring 2008 – Lec #3: Combinational Logic - 19

  10. A B C Z0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 0 Choosing Different Realizations of a Function two-level realization(we don't count NOT gates) multi-level realization(gates with fewer inputs) XOR gate (easier to draw but costlier to build) CS 150 - Spring 2008 – Lec #3: Combinational Logic - 20

  11. Which Realization is Best? • Reduce number of inputs • literal: input variable (complemented or not) • can approximate cost of logic gate as 2 transistors per literal • why not count inverters? • Fewer literals means less transistors • smaller circuits • Fewer inputs implies faster gates • gates are smaller and thus also faster • Fan-ins (# of gate inputs) are limited in some technologies • Reduce number of gates • Fewer gates (and the packages they come in) means smaller circuits • directly influences manufacturing costs CS 150 - Spring 2008 – Lec #3: Combinational Logic - 21

  12. Which is the Best Realization? (cont’d) • Reduce number of levels of gates • Fewer level of gates implies reduced signal propagation delays • Minimum delay configuration typically requires more gates • wider, less deep circuits • How do we explore tradeoffs between increased circuit delay and size? • Automated tools to generate different solutions • Logic minimization: reduce number of gates and complexity • Logic optimization: reduction while trading off against delay CS 150 - Spring 2008 – Lec #3: Combinational Logic - 22

  13. Canonical Forms • Truth table is the unique signature of a Boolean function • Many alternative gate realizations may have the same truth table • Canonical forms • Standard forms for a Boolean expression • Provides a unique algebraic signature CS 150 - Spring 2008 – Lec #3: Combinational Logic - 25

  14. A B C F F'0 0 0 0 1 0 0 1 1 0 0 1 0 0 1 0 1 1 1 0 1 0 0 0 1 1 0 1 1 0 1 1 0 1 0 1 1 1 1 0 + A'BC + AB'C + ABC' + ABC A'B'C Sum-of-Products Canonical Forms • Also known as disjunctive normal form • Also known as minterm expansion F = 001 011 101 110 111 F = F' = A'B'C' + A'BC' + AB'C' CS 150 - Spring 2008 – Lec #3: Combinational Logic - 26

  15. A B C minterms0 0 0 A'B'C' m0 0 0 1 A'B'C m1 0 1 0 A'BC' m2 0 1 1 A'BC m3 1 0 0 AB'C' m4 1 0 1 AB'C m5 1 1 0 ABC' m6 1 1 1 ABC m7 Sum-of-Products Canonical Form (cont’d) • Product term (or minterm) • ANDed product of literals – input combination for which output is true • Each variable appears exactly once, in true or inverted form (but not both) • F in canonical form:F(A, B, C) =m(1,3,5,6,7)= m1 + m3 + m5 + m6 + m7 • = A'B'C + A'BC + AB'C + ABC' + ABC • canonical formminimal formF(A, B, C) = A'B'C + A'BC + AB'C + ABC + ABC' • = (A'B' + A'B + AB' + AB)C + ABC' • = ((A' + A)(B' + B))C + ABC'= C + ABC'= ABC' + C • = AB + C short-hand notation forminterms of 3 variables CS 150 - Spring 2008 – Lec #3: Combinational Logic - 27

  16. Four Alternative Two-level Implementations of F = AB + C A canonical sum-of-productsminimized sum-of-productscanonical product-of-sumsminimized product-of-sums B F1 C F2 F3 F4 CS 150 - Spring 2008 – Lec #3: Combinational Logic - 31

  17. Waveforms for the Four Alternatives • Waveforms are essentially identical • Except for timing hazards (glitches) • Delays almost identical (modeled as a delay per level, not type of gate or number of inputs to gate) CS 150 - Spring 2008 – Lec #3: Combinational Logic - 32

  18. off-set of W on-set of W don't care (DC) set of W these inputs patterns should never be encountered in practice – "don't care" about associated output values, can be exploitedin minimization Incompletely Specified Functions • Example: binary coded decimal increment by 1 • BCD digits encode decimal digits 0 – 9 in bit patterns 0000 – 1001 A B C D W X Y Z0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 1 0 0 1 1 0 1 0 0 0 1 0 0 0 1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1 1 0 0 0 1 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0 1 0 1 0 X X X X 1 0 1 1 X X X X 1 1 0 0 X X X X 1 1 0 1 X X X X 1 1 1 0 X X X X 1 1 1 1 X X X X CS 150 - Spring 2008 – Lec #3: Combinational Logic - 34

  19. Notation for Incompletely Specified Functions • Don't cares and canonical forms • So far, only represented on-set • Also represent don't-care-set • Need two of the three sets (on-set, off-set, dc-set) • Canonical representations of the BCD increment by 1 function: • Z = m0 + m2 + m4 + m6 + m8 + d10 + d11 + d12 + d13 + d14 + d15 • Z =  [ m(0,2,4,6,8) + d(10,11,12,13,14,15) ] • Z = M1 • M3 • M5 • M7 • M9 • D10 • D11 • D12 • D13 • D14 • D15 • Z =  [ M(1,3,5,7,9) • D(10,11,12,13,14,15) ] CS 150 - Spring 2008 – Lec #3: Combinational Logic - 35

  20. B has the same value in both on-set rows– B remains A has a different value in the two rows– A is eliminated The Uniting Theorem • Key tool for simplification: A (B' + B) = A • Essence of simplification: • Find two element subsets of the ON-set where only one variable changes its value – this single varying variable can be eliminated and a single product term used to represent both elements F = A'B'+AB' = (A'+A)B' = B' A B F 0 0 1 0 1 0 1 0 1 1 1 0 CS 150 - Spring 2008 – Lec #3: Combinational Logic - 37

  21. 11 01 0 1 Y 2-cube 1-cube X 10 00 X 111 1111 0111 4-cube 3-cube Y 101 Z Y Z 000 W X 1000 0000 X Boolean Cubes • Visual technique for indentifying when the uniting theorem can be applied • n input variables = n-dimensional "cube" CS 150 - Spring 2008 – Lec #3: Combinational Logic - 38

  22. 11 01 0 1 Y 2-cube 1-cube X 10 00 X 111 3-cube Y 101 Z 000 X Boolean Cubes • Visual technique for indentifying when the uniting theorem can be applied • n input variables = n-dimensional "cube" 1111 0000 CS 150 - Spring 2008 – Lec #3: Combinational Logic - 39

  23. two faces of size 0 (nodes) combine into a face of size 1(line) A varies within face, B does notthis face represents the literal B' 11 01 B 10 00 A Mapping Truth Tables onto Boolean Cubes • Uniting theorem combines two "faces" of a cube into a larger "face" • Example: F A B F 0 0 1 0 1 0 1 0 1 1 1 0 ON-set = solid nodesOFF-set = empty nodesDC-set = 'd nodes CS 150 - Spring 2008 – Lec #3: Combinational Logic - 40

  24. (A'+A)BCin AB(Cin'+Cin) A(B+B')Cin 111 B 101 C 000 A Three Variable Example • Binary full-adder carry-out logic A B Cin Cout 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1 the on-set is completely covered by the combination (OR) of the subcubes of lower dimensionality - note that “111”is covered three times Cout = BCin+AB+ACin CS 150 - Spring 2008 – Lec #3: Combinational Logic - 41

  25. 011 111 110 010 001 B 101 C 100 000 A Higher Dimensional Cubes • Sub-cubes of higher dimension than 2 F(A,B,C) = m(4,5,6,7) on-set forms a squarei.e., a cube of dimension 2 represents an expression in one variable i.e., 3 dimensions – 2 dimensions A is asserted (true) and unchanged B and C vary This subcube represents the literal A CS 150 - Spring 2008 – Lec #3: Combinational Logic - 42

  26. m-Dimensional Cubes in an n-Dimensional Boolean Space • In a 3-cube (three variables): • 0-cube, i.e., a single node, yields a term in 3 literals • 1-cube, i.e., a line of two nodes, yields a term in 2 literals • 2-cube, i.e., a plane of four nodes, yields a term in 1 literal • 3-cube, i.e., a cube of eight nodes, yields a constant term "1" • In general, • m-subcube within an n-cube (m < n) yields a term with n – m literals CS 150 - Spring 2008 – Lec #3: Combinational Logic - 43

  27. A A B F 0 0 1 0 1 0 1 0 1 1 1 0 0 1 B 1 1 0 0 2 1 3 1 0 0 Karnaugh Maps • Flat map of Boolean cube • Wrap–around at edges • Hard to draw and visualize for more than 4 dimensions • Virtually impossible for more than 6 dimensions • Alternative to truth-tables to help visualize adjacencies • Guide to applying the uniting theorem • On-set elements with only one variable changing value are adjacent unlike the situation in a linear truth-table CS 150 - Spring 2008 – Lec #3: Combinational Logic - 44

  28. A A AB 12 8 13 9 0 4 1 5 11 10 00 01 C 0 6 4 7 5 0 2 1 3 D 1 C 15 11 14 10 3 7 2 6 A C B 6 4 7 5 0 2 1 3 B C B Karnaugh Maps (cont’d) • Numbering scheme based on Gray–code • e.g., 00, 01, 11, 10 • Only a single bit changes in code for adjacent map cells 13 = 1101= ABC’D CS 150 - Spring 2008 – Lec #3: Combinational Logic - 45

  29. A 000 010 001 011 110 100 111 101 C B Adjacencies in Karnaugh Maps • Wrap from first to last column • Wrap top row to bottom row 011 111 110 010 001 B 101 C 100 000 A CS 150 - Spring 2008 – Lec #3: Combinational Logic - 46

  30. B’ A 1 1 0 0 AB + ACin + BCin A B 0 0 0 1 1 0 1 1 Cin A 1 0 0 0 0 1 1 1 B C AC + B’C’ + AB’ B Karnaugh Map Examples • F = • Cout = • f(A,B,C) = m(0,4,6,7) obtain thecomplementof the function by covering 0swith subcubes CS 150 - Spring 2008 – Lec #3: Combinational Logic - 47

  31. A A A A 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 1 1 1 1 0 0 0 C C C = AC + B’C’ B B B = BC’ + A’C More Karnaugh Map Examples G(A,B,C) = F(A,B,C) = m(0,4,5,7) F' simply replace 1's with 0's and vice versa F'(A,B,C) =  m(1,2,3,6) CS 150 - Spring 2008 – Lec #3: Combinational Logic - 48

  32. C + A’BD + B’D’ 1111 0111 0 1 0 0 1 0 0 1 C D A 1000 1 1 1 1 1 1 1 1 0000 B C Karnaugh Map: 4-Variable Example • F(A,B,C,D) =m(0,2,3,5,6,7,8,10,11,14,15)F = A D B find the smallest number of the largest possible subcubes to cover the ON-set (fewer terms with fewer inputs per term) CS 150 - Spring 2008 – Lec #3: Combinational Logic - 49

  33. A’D + B’C’D A X 0 X 1 0 0 1 1 D 0 0 0 0 1 1 0 X C B Karnaugh Maps: Don’t Cares • f(A,B,C,D) = m(1,3,5,7,9) + d(6,12,13) • without don't cares • f = CS 150 - Spring 2008 – Lec #3: Combinational Logic - 50

  34. + C'D A'D by using don't care as a "1"a 2-cube can be formed rather than a 1-cube to coverthis node A X 0 X 1 0 0 1 1 D 0 0 0 0 1 1 0 X C B Karnaugh Maps: Don’t Cares (cont’d) • f(A,B,C,D) = m(1,3,5,7,9) + d(6,12,13) • f = A'D + B'C'D without don't cares • f = with don't cares don't cares can be treated as1s or 0sdepending on which is more advantageous CS 150 - Spring 2008 – Lec #3: Combinational Logic - 51

  35. A1 A2B1 B2 P1 P2P4 P8 Design Example: 2x2-bit Multiplier A2 A1 B2 B1 P8 P4 P2 P1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 1 0 0 1 1 0 1 1 0 1 1 0 0 0 0 0 0 0 1 0 0 1 1 1 0 0 1 1 0 1 1 1 0 0 1 block diagram and truth table 4-variable K-map for each of the 4 output functions CS 150 - Spring 2008 – Lec #3: Combinational Logic - 55

  36. A2 A2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 B1 B1 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 B2 B2 A1 A1 A2 A2 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 1 B1 B1 0 1 1 0 1 0 0 0 0 1 0 1 0 1 0 0 B2 B2 A1 A1 Design Example: 2x2-bit Multiplier (cont’d) K-map for P4 K-map for P8 P4 = A2B2B1' + A2A1'B2 P8 = A2A1B2B1 K-map for P2 K-map for P1 P1 = A1B1 P2 = A2'A1B2 + A1B2B1' + A2B2'B1 + A2A1'B1 CS 150 - Spring 2008 – Lec #3: Combinational Logic - 56

  37. Basic Two-Level Logic Simplification Algorithm • Enumerate all the minterms of the function • Enumerate all the primes of the function • Form the “Covering Table” of Primes vs Minterms • Choose the smallest set of primes that cover all the minterms • Every minterm is contained in at least one prime CS 150 - Spring 2008 – Lec #3: Combinational Logic - 61

  38. Ex: F=a’b’d’ + abd’ +a’b ON = (0,2,4,5,6,7,12,14) DC=15 OFF=(1,3,8,9,10,11,13) CS 150 - Spring 2008 – Lec #3: Combinational Logic - 62

  39. Ex: F=a’b’d’ + abd’ +a’b Primes = a’d’,bd’,a’b,bc CS 150 - Spring 2008 – Lec #3: Combinational Logic - 63

  40. Covering Table Rows are minterms Columns are primes “1” in Square x/y if prime y covers minterm x E.g., “1” in Square 6/01xx because prime a’b contains a’bcd’ (minterm 6) CS 150 - Spring 2008 – Lec #3: Combinational Logic - 64

  41. Solving the Covering Table Choose minimum set of columns so that every row is “covered” by a column “Covered” == has a ‘1’ in some chosen column CS 150 - Spring 2008 – Lec #3: Combinational Logic - 65

  42. “Dominated” Rows Row x dominates row y if covering x automatically covers y Primes covering x are a subset of primes covering y Must choose Prime 1 or Prime 2 to cover x But then y automatically covered => x dominates y CS 150 - Spring 2008 – Lec #3: Combinational Logic - 66

  43. “Dominated” Columns Column 1 dominates Column 2 if Column covers every row covered by Column 2 Minterms covered by 1 are a superset of minterms covered by 2 Choose prime 1 = > cover x or y or z Choose prime 2 => cover x or z Always better to choose 1 over 2 => 1 dominates 2 CS 150 - Spring 2008 – Lec #3: Combinational Logic - 67

  44. 1 1 1 • 1 1 1 1 • 1 1 1 1 • …. K-ary Essential Primes Cyclic core Algorithm (again) • While there are dominated rows or columns • Eliminate dominated rows • Eliminate dominated columns 1 1 1 1 1 1 • Solve Cyclic Core by branch-and-bound • Solution is k-ary essential primes + Cyclic Core Solution CS 150 - Spring 2008 – Lec #3: Combinational Logic - 68

  45. Solving the Covering Table Step 1: Eliminate “dominated” rows Rows automatically “covered” by another Ex: row 4 dominated by rows 0 and 2 Cover 0 => automatically cover 4 CS 150 - Spring 2008 – Lec #3: Combinational Logic - 69

  46. Solving the Covering Table Done! (Not usually so lucky) Minimum cover is a’d’+bd’+a’b CS 150 - Spring 2008 – Lec #3: Combinational Logic - 70

  47. What can go wrong? • Too Many primes! • Can have as many as 3n/n (McMullen) • Too Many Minterms! • In general, have about 2n/k • All subproblems are NP-Hard • Is cube c a prime of F? • Find the minimum cover of a covering table • NP-Hard == Computer Scientist-speak for “no fast algorithm exists (we believe)” • Best known algorithm = “try all solutions” • 2**(3**n/n) possible solutions! • n = 4 implies 2**20 (approx 1,000,000) solutions • n = 5 implies 2**49 (approx 10,000,000,000,000,000,000 solutions) • Easy to generate problems with 100 or so variables CS 150 - Spring 2008 – Lec #3: Combinational Logic - 71

  48. What do we do about it? • Inpractice, things are not so bad! • Circuits people actually build usually have “reasonable” number of primes • Covering tables for circuits people actually build are usually easy to solve • => Use algorithms that are in practice efficient for generating primes, covering table, solve covering table • Use heuristics • Don’t necessarily need minimum form • Given one sum-of-products, generate a better form • Iterate until we run out of time/get answer that is good enough CS 150 - Spring 2008 – Lec #3: Combinational Logic - 72

  49. What do we mean by “usually”? Circuits People Actually build Circuits we can prove we do well on Circuits we in practice do well on CS 150 - Spring 2008 – Lec #3: Combinational Logic - 73

  50. What do we mean by “usually”? All 2^2^n n-input circuits Circuits we see… CS 150 - Spring 2008 – Lec #3: Combinational Logic - 74

More Related