Combinational Logic Circuits Part III Theoretical Foundations


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1 Combinational Logic Circuits Part III Theoretical Foundations
2 Overview Simplifying Boolean Functions Algebraic Manipulation Karnaugh Map Manipulation (simplifying functions of 2, 3, 4 variables) Systematic Approach for Simplifying Functions using Kmaps Implicants, Prime Implicants (PIs), and Essential Prime Implicants Simplifying Functions using Essential and Nonessential PIs Don tcare Conditions and Simplification using Don t Cares Fall 27 Digital Technique by Todor Stefanov, Leiden University 2
3 Boolean Functions as Equations Truth table and Kmap of a Boolean function are unique representations However, representing a Boolean function as an equation can be done in many different ways Canonical and Standard forms Example: F(,,) = + + F2(,,) = + F3(,,) = + F4(,,) = + The corresponding truth tables for F to F4 are identical! Thus, F = F2 = F3 = F4 However, F2 and F3 are simpler than F and F4 is simpler than the others. F F2 F3 F4 How do we simplify Boolean functions? Fall 27 Digital Technique by Todor Stefanov, Leiden University 3
4 Simplifying a Boolean Function hy simplifying Boolean functions? Boolean functions are used to design digital logic circuits Simpler Boolean function can mean cheaper, smaller, faster circuit Three main approaches to simplify Boolean functions: Algebraic Manipulations using the Boolean Algebra as a tool for simplifications Karnaugh Map Manipulations very easy graphical method to simplify Boolean functions it works for functions of up to 4 variables! Algorithmic Techniques used to program a computer to do the simplifications Fall 27 Digital Technique by Todor Stefanov, Leiden University 4
5 Algebraic Manipulation e use basic identities, properties, and theorems of the Boolean Algebra to manipulate and simplify Boolean functions Example: Simplify F = + + F = apply identity 4 = (+ ) +  apply identity 7 = +  apply identity 2 = + Example2: Simplify G = + + F = apply identity 5 = apply identity 4 = ( +) + ( +)  apply identity 7 = +  apply identity 2 = + Fall 27 Digital Technique by Todor Stefanov, Leiden University 5
6 Karnaugh Map Manipulations e can use a Kmap to simplify a Boolean function of 2, 3, or 4 variables as SumOfProducts Procedure: Enter s in the Kmap for each minterm (product term) in the function Group adjacent Kmap cells containing s to obtain a product term with fewer variables The number of cells in a group must be a power of 2 (2, 4, 8, )! Try to group as many as possible cells containing s in a group Such group corresponds to a simpler product term! Try to make as less as possible groups to cover all cells containing s This corresponds to fewer product terms in the simplified function! Do not forget to handle boundary cells for Kmaps of 3 or 4 variables when you do the grouping Important: The result after the simplification may not be unique! Fall 27 Digital Technique by Todor Stefanov, Leiden University 6
7 Simplifying a Boolean Function using 2variable Kmap (examples) Given functions: F(,) = Σm(,) = = + Simplified functions: F(,) = F2(,) = Σm(,3) = = + F2(,) = + F3(,) = Σm(,2,3) = = + + F3(,) = + F4(,) = Σm(,,2,3) = = F4(,) = Fall 27 Digital Technique by Todor Stefanov, Leiden University 7
8 Simplifying a Boolean Function using 3variable Kmap (groupings) minterm m m m 3 m 2 m m m 3 m 2 m 4 m 5 m 7 m 6 m 4 m 5 m 7 m 6 Group of 2 adjacent cells gives product term of two literals. m m m 3 m 2 m m m 3 m 2 m 4 m 5 m 7 m 6 m 4 m 5 m 7 m 6 Group of 4 adjacent cells gives product term of one literal. m m m 3 m 2 m m m 3 m 2 m 4 m 5 m 7 m 6 m 4 m 5 m 7 m 6 Fall 27 Digital Technique by Todor Stefanov, Leiden University 8
9 Simplifying a Boolean Function using 3variable Kmap (examples) Given functions: F(,,) = Σm(,2,4,7) F2(,,) = Σm(2,3,4,5) Fall 27 Digital Technique by Todor Stefanov, Leiden University 9 Simplified functions: Simplification is not possible F2(,,) = + F3(,,) = Σm(,2,4,6) F3(,,) = F4(,,) = Σm(,,2,3,6,7) F4(,,) = +
10 Simplifying a Boolean Function using 3variable Kmap (more examples) Given functions: F5(,,) = Σm(3,4,6,7) F6(,,) = Σm(,2,4,5,6) F7(,,) = Σm(,2,3,5,7) F8(,,) = Σm(,3,4,5,6) Simplified functions: F5(,,) = + Fall 27 Digital Technique by Todor Stefanov, Leiden University F6(,,) = + F7(,,) = + F8(,,) = + + or F8(,,) = + + Not unique solution
11 Simplifying a Boolean Function using 4variable Kmap (grouping examples) Group of 2 adjacent cells gives product term of 3 literals. m m m 3 m 2 m 4 m 5 m 7 m 6 m 2 m 3 m 5 m 4 m 8 m 9 m m Group of 8 adjacent cells gives product term of literal. m m m 3 m 2 m 4 m 5 m 7 m 6 m 2 m 3 m 5 m 4 m 8 m 9 m m Group of 4 adjacent cells gives product term of 2 literals. Group of all cells gives constant one. Fall 27 Digital Technique by Todor Stefanov, Leiden University m m m 3 m 2 m 4 m 5 m 7 m 6 m 2 m 3 m 5 m 4 m 8 m 9 m m m m m 3 m 2 m 4 m 5 m 7 m 6 m 2 m 3 m 5 m 4 m 8 m 9 m m
12 Simplifying a Boolean Function using 4variable Kmap (examples) Given function: F2(,,,) = = Σm(,,2,4,5,6,8, 9,2,3,4) Simplified function: F2(,,,) = = + + Given function: F(,,,) = = Σm(,,2,4,5,7,8,9,,2,3) Simplified function: F(,,,) = = + + Fall 27 Digital Technique by Todor Stefanov, Leiden University 2 Given function: F3(,,,) = Simplified function: F3(,,,) = = + +
13 Simplifying with Kmaps Systematically ou have seen intuitive procedure on how to group cells and simplify Boolean functions! Can we have more systematic procedure? ES, if we introduce the terms: implicant prime implicant essential prime implicant An Implicant I of a function F() is a product term which implies F(), i.e., F() = whenever I = All minterms of a function F are implicants of F All rectangles in a Kmap made up of cells containing s correspond to implicants Fall 27 Digital Technique by Todor Stefanov, Leiden University 3
14 Prime Implicant (PI) An implicant I of F is called a Prime Implicant (PI) if the removal of any literal from I results in a product term that is NOT an implicant of F The above should hold for all literals in I Thus, a prime implicant is not contained in any simpler implicant The set of prime implicants corresponds to all rectangles, in a Kmap, made up of cells containing s that satisfy the following condition: each rectangle is not contained in a larger rectangle Fall 27 Digital Technique by Todor Stefanov, Leiden University 4
15 Example of Prime Implicants (PIs) Consider function F(,,,) whose Kmap is shown at right is not a prime implicant because it is contained in is not a prime implicant because it is contained in Product terms,, are prime implicants. hy? Consider the term and obtain terms by deleting any literal: e get two terms: term and term Both terms are NOT implicants of F Thus, the term is prime implicant Fall 27 Digital Technique by Todor Stefanov, Leiden University 5
16 Essential Prime Implicants (EPIs) If a minterm of function F is included in ONL one prime implicant pi, then pi is an Essential Prime Implicant of F An essential prime implicant MUST appear in all possible SOP expressions of function F To find essential prime implicants: Generate all prime implicants of a function Select those prime implicants that contain at least one that is not covered by any other prime implicant For the previous example, the PIs are,, and ; all of these are essential. Fall 27 Digital Technique by Todor Stefanov, Leiden University 6
17 Essential Prime Implicants (examples) Consider function F(,,,) whose Kmap is shown below: All Prime Implicants are:,,,,,, Essential Prime Implicants are: Fall 27 Digital Technique by Todor Stefanov, Leiden University 7 Consider function F2(,,,) whose Kmap is shown below: All Prime Implicants are:,, Essential Prime Implicants are: and
18 Systematic Procedure for Simplifying Boolean Functions Given : The Kmap of a Boolean function Obtain: The simplest SOP expression for the function. Find all prime implicants (PIs) of the function 2. Select all essential PIs 3. For remaining minterms not included in the essential PIs, select a set of other PIs to cover them, with minimal overlap in the set 4. The resulting simplified function is the logical OR of the product terms selected above Fall 27 Digital Technique by Todor Stefanov, Leiden University 8
19 Example F(,,,) = m(,,2,3,4,5,7,4,5). All prime implicants (PI) are:,,,, Select all essential PIs:,, Select other PIs to cover all s with minimal overlap: Possibilities: or e select because it is simpler. F(,,,) = + ++ Fall 27 Digital Technique by Todor Stefanov, Leiden University 9
20 Other Examples Consider function F(,,,) whose Kmap is shown at right. All prime implicants are:,,,,, Essential prime implicants are:,,, Nonessential prime implicants are:, Simplified function (solution not unique): F = F = Fall 27 Digital Technique by Todor Stefanov, Leiden University 2
21 Other Examples (cont.) Consider function F(,,,) = m(,,2,4,5,,,3,5) whose Kmap is shown at right. All prime implicants are:,,,,,, Essential prime implicants are: Nonessential prime implicants are:,,,,, Simplified function (solution not unique): F = ++ + F = ++ + F = ++ + F = ++ + Fall 27 Digital Technique by Todor Stefanov, Leiden University 2 and are NONoverlapping PIs. and are NONoverlapping PIs.
22 ProductOfSums (POS) Simplification So far, we have considered simplification of a Boolean function expressed in SumOfProducts (SOP) form using a Kmap. Sometimes the ProductOfSums form of a function is simpler than the SOP form. Can we use Kmaps to simplify a Boolean function in ProductOfSums form? Procedure: Use sumofproducts simplification on the zeros of function F in the K map. In this way you will get the simplified complement of F (F ). Find the complement of F which is F, i.e., (F ) = F Recall that the complement of a Boolean function can be obtained by () taking the dual and (2) complementing each literal. OR, using DeMorgan s Theorem. Fall 27 Digital Technique by Todor Stefanov, Leiden University 22
23 POS Simplification Example F = m(,,2,3,4,5,7,4,5) Simplify using zeros: F = + + Complement F to find F, i.e., F = (F ) First get the dual of F : dual(f ) = (+ ) (+ ) ( +++ ) The complement of F (F ) Complement each literal in dual(f ) to get F as POS F = ( +) ( +) (+ + +) Fall 27 Digital Technique by Todor Stefanov, Leiden University 23
24 Don tcare Conditions Sometimes a Boolean function is not specified for some combinations of input values. hy? There may be a combination of input values which will never occur If they do occur, the value of the function is of no concern Such combinations is called don tcare condition The function value for such combinations is called a don'tcare The don tcare function values are usually denoted with x x may be arbitrarily set to or in an implementation Don tcares can be used to further simplify a function Fall 27 Digital Technique by Todor Stefanov, Leiden University 24
25 Simplification using Don tcares Treat don'tcares as if they are s to generate prime implicants in order to produce simple expressions Delete prime implicants that cover only don'tcare minterms Treat the covering of remaining don't care minterms as optional in the selection process they may be covered but it is not necessary Fall 27 Digital Technique by Todor Stefanov, Leiden University 25
26 Example with Don tcare Conditions Consider the following incompletely specified function F that has three don tcare minterms d: A F(A,B,C,D) = m(,3,7,,5) d(a,b,c,d) = m(,2,5) CD AB x C D x x F = CD + A B B CD AB Fall 27 Digital Technique by Todor Stefanov, Leiden University 26 A x C D x x F2 = CD + A D Notice: F and F2 are algebraically not equal. Both include the specified minterms of F, but each includes different don tcare minterms. B
27 Other Examples with Don tcares () Simplify the function G(A,B,C,D) whose Kmap is shown at right. CD AB A C x x x x x x x D B G = A C + AB or G = A C +BD or G = BD + C D CD AB A C x x x x x x x CD AB B A C x x x x x x x CD AB B A C x x x x x x x D D D Fall 27 Digital Technique by Todor Stefanov, Leiden University 27 B
28 Other Examples with Don tcares (2) Simplify the function F(A,B,C,D) whose Kmap is shown at the topright. F = A BC +AB +CD +A C D or F = A BD +AB +CD +A C D The middle two terms are EPIs, while the first and last terms are selected to cover the minterms m, m 4, and m 5. There s a third solution! Can you find it? CD AB Fall 27 Digital Technique by Todor Stefanov, Leiden University D 28 A A C CD AB x x x x D C CD AB A x x x x D C x x x x B B B
29 Algorithmic Techniques for Simplification Simplification of Boolean functions using Kmaps works for functions of up to 4 variables hat do we do for functions with more than 45 variables? ou can code up a minimiser program Use the QuineMcCluskey algorithm Base on (essential) prime implicants e won t discuss these techniques here Search on Internet to find more information about the QuineMcCluskey algorithm Fall 27 Digital Technique by Todor Stefanov, Leiden University 29
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