CArcMOOC 03.01 - Boolean algebra and Logic Synthesis
0 likes326 views
The document provides an overview of Boolean algebra and logic synthesis, covering definitions, properties, and the formulation of canonical forms. It explains Boolean functions, truth tables, and the methodology for minimizing Boolean expressions through various laws and techniques. Additionally, it emphasizes the importance of canonical forms in representing unique Boolean expressions and the challenges in achieving minimal representations.
CArcMOOC 03.01 - Boolean algebra and Logic Synthesis
- 1. Carc 03.01 alessandro.bogliolo@uniurb.it 03. Logic Networks 03.01. Boolean algebra and logic synthesis • Definitions • Boolean functions • Properties • Canonical forms • Synthesis and minimization Computer Architecture alessandro.bogliolo@uniurb.it
- 2. Carc 03.01 alessandro.bogliolo@uniurb.it Definitions • Boolean set: • Boolean constants: 0, 1 • Boolean variable: • Boolean functions: z=f(x1,x2,...,xn) • Operations: 10,B xyyxz)y,x( :and BBB yxz)y,x( :or BBB x'xzx :not BB 10,x x y xy 0 0 0 0 1 0 1 0 0 1 1 1 x y x+y 0 0 0 0 1 1 1 0 1 1 1 1 x x' 0 1 1 0 BB n :f
- 3. Carc 03.01 alessandro.bogliolo@uniurb.it Boolean functions • Truth table: Table of 2n rows that associates a Boolean value to each configuration of n independent variables There are 22n different functions of n variables • Boolean expression: Expression of Boolean variables, Boolean constants and operators c)ab(cab'cbaf BB n :f a b c f 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 1
- 4. Carc 03.01 alessandro.bogliolo@uniurb.it Properties xxx xxx )zx()yx()zy(x )zx()yx()zy(x z)xy()yz(xxyz z)yx()zy(xzyx xyyx yxxy xx 1 xx 0 00x 11x 0 'xx 1 'xx xxyx x)yx(x 'y'x)'xy( 'y'x)'yx( Idempotent laws Distributive laws Associative laws Commutative laws Identity elements Null laws (forcing elements) Complement laws Absorption laws De Morgan’s laws (duality principle)
- 5. Carc 03.01 alessandro.bogliolo@uniurb.it Idempotent laws xxx xxx Idempotent laws x y xy 0 0 0 0 1 0 1 0 0 1 1 1 x x*x 0 0*0 0 1 1*1 1 x y x+y 0 0 0 0 1 1 1 0 1 1 1 1 x x+x 0 0+0 0 1 1+1 1 By perfect induction:
- 6. Carc 03.01 alessandro.bogliolo@uniurb.it xxyx x)yx(x Absorption laws Absorption laws xx)y(xyxx 11 xyxxyxxx)yx(x By Boolean manipulation:
- 7. Carc 03.01 alessandro.bogliolo@uniurb.it 'y'x)'xy( 'y'x)'yx( De Morgan’s laws De Morgan’s laws By perfect induction: x=0, y=0 (00)’=0’=1 0’+0’=1+1=1 x=0, y=1 (01)’=0’=1 0’+1’=1+0=1 x=1, y=0 (10)’=0’=1 1’+0’=0+1=1 x=1, y=1 (11)’=1’=0 1’+1’=0+0=0 x=0, y=0 (0+0)’=0’=1 0’0’=11=1 x=0, y=1 (0+1)’=1’=0 0’1’=10=0 x=1, y=0 (1+0)’=1’=0 1’0’=01=0 x=1, y=1 (1+1)’=1’=0 1’1’=00=0
- 8. Carc 03.01 alessandro.bogliolo@uniurb.it Canonical forms • There are infinite equivalent Boolean expressions. • The equivalence (i.e., identity) between two expressions can be demonstrated: 1. By perfect induction 2. By Boolean manipulation • Canonical forms associate unique expressions to each function • Checking the equivalence between two functions reduces to a comparison of their canonical representations
- 9. Carc 03.01 alessandro.bogliolo@uniurb.it Canonical forms (Sum of Products) • Literal: independent variable taken either in true or complemented form (e.g., x, x’) • Minterm: Product of all independent variables taken either in true or complemented form • A minterm represents a Boolean function that takes value 1 corresponding to a unique configuration of input variables (e.g., f(a,b,c)=ab’c takes vale 1 for abc=101) • A Boolean function that takes value 1 for M different configurations (that has M 1’s in the truth table) can be expressed as the sum of the M minterms associated with the M 1’s • Any Boolean function can be expressed as a sum of products • A sum of minterms, with fixed variable order, is a canonical form
- 10. Carc 03.01 alessandro.bogliolo@uniurb.it From truth tables to SoPs a b c f 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 1 f = a’b’c’+ a’bc’+ ab’c’+ abc’+ abc 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 = + + + +
- 11. Carc 03.01 alessandro.bogliolo@uniurb.it Boolean minimization Given a Boolean function, find a Boolean expression that represents the function using a minimum number of literals • In general, this is not an easy task • There is a closed-form solution for 2-level SoP expressions • There is no closed-form solution for general multi-level expressions. • Heuristic solutions found by Boolean manipulation f = a’b’c’+ a’bc’+ ab’c’+ abc’+ abc 15 literals f = a’c’+ ab’c’+ ab 7 literals f = a’c’+ ab’c’+ abc’+ abc 11 literals
- 12. Carc 03.01 alessandro.bogliolo@uniurb.it Boolean minimization (example) f = a’b’c’+ a’bc’+ ab’c’+ abc’+ abc 15 literals (SoP) = a’c’(b’+ b) + ab’c’+ ab (c’+ c) 11 literals (distributive) = a’c’+ ab’c’+ ab 7 literals (complement) = a’c’+ a (b’c’+ b) 6 literals (distributive) = a’c’+ a (b’c’+ b c’+ b) 8 literals (absorption) = a’c’+ a ((b’+b) c’+ b) 7 literals (distributive) = a’c’+ a (c’+ b) 5 literals (complement) = a’c’+ a c’+ ab 6 literals (distributive) = (a’+ a) c’+ ab 5 literals (distributive) = c’+ ab 3 literals (complement) Remark: the number of literals doesn’t decrease at every step. This makes the process non trivial