ENG2410 Digital Design Combinational Logic Circuits


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1 ENG240 Digital Design Combinational Logic Circuits Fall 207 S. Areibi School of Engineering University of Guelph Binary variables Binary Logic Can be 0 or (T or F, low or high) Variables named with single letters in examples Really use words when designing circuits Basic Functions AND OR NOT 4 Resources Chapter #2, Mano Sections 2. Logic Gates 2.2 Boolean Algebra 2.3 Min Terms & Max Terms 2.4 Logic Optimization Logic Signals Binary 0 is represented by a low voltage (range of voltages) Binary is represented by a high voltage (range of voltages) The voltage ranges guard against noise 2 Example of binary signals 5 Week #2 Topics Logic Gates Name Graphical Symbol Algebraic Function Truth Table Binary Logic and Gates Boolean Functions Boolean Algebra (Truth Tables) Basic Identities Standard Forms (Minterms, Maxterms) Circuit Optimization (KMaps) Map Manipulations (Prime Implicants) AND OR NOT NAND NOR A B A B A A B A B F F F F F F = A B or F = F = A + B F = A or F = A F = F = A + B A B F A B F A F 0 0 A B F A B F XOR A B F F = A B A B F Figure. Basic Logic Gates 6 / 45 School of Engineering
2 NOT Operator Switching Circuits (AND Gate) AND Unary Operator Symbol is bar = X Truth table > Inversion AND Gate: Symbol 7 0 Inverter Gate Timing Diagram AND Gate: Timing Timing Diagrams 8 AND Gate OR Gate Symbol is dot = X.Y Or no symbol = XY Truth table > is only if Both X and Y are Switches in series => AND Symbol is + Not addition = X + Y Truth table > is if either Switches in parallel => OR 9 2 School of Engineering 2
3 Switching Circuits (OR Gate) Logic Diagram Boolean Expression OR Logic Diagram OR Gate: Symbol What is the Boolean expression? F = X + Y. 3 6 OR Gate: Timing Boolean Expression Logic Diagram Timing Diagrams Consider function F = XY + XY + X How many gates do we need to implement this function? 4 7 Binary Logic Truth Tables, Boolean Expressions, and Logic Gates AND OR NOT x y z x y z x z z = x y = x y z = x + y z = x = x 5 Boolean Expression Truth Table F = X + Y. X Y Y Y F What is the Truth Table? School of Engineering 3
4 Boolean Expression Truth Table More Inputs F = X + Y. What is the Truth Table? X Y Y Y F Work same way What s output? Boolean Expression Truth Table NAND Gates F = X + Y. What is the Truth Table? X Y Y Y F Very common for discrete logic Boolean Expression Truth Table NOR Gates F = X + Y. What is the Truth Table? X Y Y Y F NOT OR Also common F = X + Y X Y School of Engineering 4
5 Boolean Operator Precedence Design Steps (Simple Example) The order of evaluation in a Boolean expression is:. Parentheses 2. Not 3. And 4. Or Consequence: Parentheses appear around OR expressions Example: F = A(B + C)(C + D) Step #, since we have three inputs then the Truth Table has to have 8 entries (2 3 ) X Y F Mechanically Go From Truth Table to Function Design Steps (Simple Example) Steps of design:. Start with Problem Statement 2. Obtain Truth Table 3. Obtain a algebraic boolean function 4. Draw the circuit diagram Step #2, Pick the entries in the truth table where F is equal. Give Expression: F = X Y X Y F Design Steps (Simple Example) Design Steps (Simple Example) Specification: Design a circuit that has 3 inputs x, y, z and a single output F. F is true when the three inputs are the same and false otherwise. Step #3, Draw a logic diagram that represents the boolean functions. F = X Y X Y F School of Engineering 5
6 Representation: Truth Table Table of Identities Give Boolean Expression? F = X Y + X Y + X Y + X Y + X Y Not Simplified! X Y 2 n rows where n # of variables X Y X Y X Y X Y 3 34 Boolean Functions: Complexity Dual of an Expression There can be different representations for a boolean function Usually we want the simplest (Why?) Fewer gates (Less Area, Less Power Consumption, Faster!) We can use identities to reduce complexity of Boolean expressions. The dual of an expression is obtained by:. Changing AND to OR and OR to AND throughout 2. Changing s to 0 s and 0 s to s For example The dual of X+0 is: X., The dual of X.0 is: 32 X+ 35 Boolean Algebra & Identities Duals There is only one way that a Boolean function can be represented in a truth table. However, when the function is in algebraic form, it can be expressed in a variety of ways. Boolean algebra is a useful tool for simplifying digital circuits. Use identities to manipulate functions Left and right columns are duals School of Engineering 6
7 Single Variable Identities Distributive Identity 4 is well known from ordinary algebra! Identity 5 is the dual of identity Commutative DeMorgan s Theorem Order independent Used a lot NOR equals invert AND NAND equals invert OR Proof? 38 4 Associative Truth Tables for DeMorgan s Independent of order in which we group So can also be written as and X + Y School of Engineering 7
8 Useful Theorems Algebraic Manipulation x y + x y = y x + x y = x x x + y = x Minimization ( ) Absorption ( + y) = x y Simplification x + x y = x + y x x x y + x z + y z = x y + x z Consensus ( x + y) ( x + z) ( y + z) = ( x + y) ( x + z) 43 o o o F = XY + XY + X When a Boolean equation is implemented with logic gates, i. Each term requires a gate, ii. Each variable designates an input to the gate. We define a literalas a single variable within a term that may or may not be complemented. The expression above has THREE termsand EIGHT literals. 46 Example : Boolean Algebraic Proof Algebraic Manipulation A + A B = A Proof Steps A + A B (Absorption Theorem) Justification (identity or theorem) = A + A B X = X = A ( + B) X Y + X = X (Y + ) = A + X = = A X = X How many gates do we need to implement this function? Consider function F = XY + XY + X Cont.. Boolean Algebraic Proof Simplify Function Our primary reason for doing proofs is to learn: F = XY + XY + X Apply F = XY ( + ) + X Careful and efficient use of the identities and theorems of Boolean algebra, and How to choose the appropriate identity or theorem to apply to make forward progress, irrespective of the application. F = XY + X F = XY + Apply Apply X End Result? School of Engineering 8
9 Fewer Gates F = XY + X 2. Cont..Complement of a Function The complement of a function can be derived algebraically by applying DeMorgan s theorem.. Fewer Gates! 2. Fewer Inputs per gate! F = XY + XY F = XY + XY F = ( XY ).( XY) F = ( X+ Y + )( X+ Y+ ) The Duality Principle 3. Cont..Complement of a Function Recall how to obtain the dual of an expression! Theduality principlestates that a Boolean equation remains valid if we take the dual of the expression on both sides of the equal sign. X + XY = X + Y X ( X + Y) = XY Yet another way of deriving the complement of a function is to: Take the dual of the function equation Complement each literal. F = XY + XY F = ( XY ) + ( XY ) Dual ( X + Y + ).( X + Y + ) F = ( X + Y + ).( X + Y + ) Complement of a Function The complement representation for a function F, (F ), is obtained from an interchange of s to 0 s and 0 s to s for the values of F in the truth table. F = XY + XY X Y F F ENG24 Digital Design Cont..Week #2 Combinational Logic Circuits F = X Y 5 School of Engineering 9
10 Resources Standard Forms Chapter #2, Mano Sections 2.3 Standard Forms 2.4 Circuit Optimization (KMaps) Assume we have a circuit with 3 variables X, Y, Definitions: Product Term a subset of the variables appear XY, X, X, XY Min Term is a product term in which all variables appear once (XY, X Y, X Y ) Week #2, #3 Topics Number of Minterms Standard Forms (Minterms, Maxterms) Circuit Optimization (KMaps) Map Manipulations (Prime Implicants) For n variables, there will be 2 n minterms Like binary numbers from 0 to 2 n From Truth Table to Function Definition: Minterm Consider a truth table Can implement F by taking OR of all terms that are F = XY MinTerm Is a Product Term in which ALL variables appear once (complemented or not) F = X Y + X Y F ( optimized ) = Y + X Product Term School of Engineering 0
11 Sum of Minterms (SOM) Sum of Products (SOP) OR all of the minterms of truth table row with a F = X Y In this case m 0 +m 2 +m 5 +m 7 F = XY + XY + XY + XY In this case m + m 3 + m 4 + m 6 m 0 m m 2 m 3 m 4 m 5 m 6 m 7 The sumofminterms form is a standard algebraic expression that is obtained directly from a truth table. The expression obtained contains the maximum number of literals in each term. Simplification of the sumofminterms expression is called sumofproducts Alternative Representation SOM vs. SOP Sum of minterms F = X. Y. + X. Y. + X. Y. + X. Y. In this case m 0 +m 2 +m 5 +m 7 F ( X, Y, ) = m(0,2,5,7) F = X. Y. + X. Y. + X. Y. + X. Y. In this case m + m 3 + m 4 + m 6 F ( X, Y, ) = m(,3,4,6 ) F = X Y After Simplification we get the Sum of products F = X. + X Summary of Properties of Minterms Sum of Products Implementation There are 2 n minterms for a Boolean function with n variables Any Boolean function can be expressed as a logical sum of minterms The complement of a function contains those minterms not included in the original function. A function that includes all the 2 n minterms is equal to logic. Sum of products F = Y + XY + XY We refer to this implementation as a twolevel circuit School of Engineering
12 2level Implementation Minterm related to Maxterm Sum of products has 2 levels of gates Minterm and maxterm with same subscripts are complements m j = Mj Example (Use Demorgans theory) Is there a 3level rep of this circuit? m 3 = XY m = XY = X + Y + = 3 M level vs. 3level Implementation Product of Maxterms We can also express F as AND of all rows that should evaluate to CE can be Also expressed as + C(D+E) What s best? Hard to answer!! More gate delays? But maybe we only have 2input gates F = M M 3 M 4 M 6 F( X, Y, ) = ΠM (,3,4,6) F = ( X + Y + )( X + Y + ) ( X + Y + )( X + Y + ) M 0 M M 2 M 3 M 4 M 5 M 6 M Definition: Maxterm TwoLevel Circuit Optimization Is a Sum Term in which all variables appear once (complemented or not) The complexity of the digital logic gates that implement a Boolean function is directly related to the algebraic expression from which the function is implemented!! Boolean expressions may be simplified by algebraic manipulation (i.e. identities) but it is awkward and not straight forward! School of Engineering 2
13 Karnaugh Maps ThreeVariable Map Graphical depiction of truth table A box for each minterm So 2 variables, 4 boxes 3 variable, 8 boxes And so on Useful for simplification By inspection Algebraic manipulation harder Eight minterms Look at encoding of columns and rows Only one variable changes from box to box Examples Simplification There are implied 0s in other boxes Figure (b) F = m + m 2 + m 3 Adjacent squares (horizontally or vertically) are minterms that vary by single variable Draw rectangles on map to simplify function Illustration next F = XY + XY + XY = X + Y Simplification How to obtain a simplified expression? F = XY + XY + XY = X Y + X ( Y + Y ) = X Y + X () = ( X + X ) ( X + Y ) = X +Y Combine as many s as possible from the map to form rectangle Only adjacent s can be combined vertically or horizontally but not diagonally. The number of s that can be combined is a power of 2, (2 0 =, 2 =2, 2 2 =4, 2 3 =8), School of Engineering 3
14 Example Another Minimization Example X Y F F = XY + XY + XY + XY F = XY + XY Each of the two adjacent pairs of entries can be simplified by eliminating the changing bit. x is eliminated in column 2 y is eliminated in the other pair. F = y z + x z Extract minterms from Expression XY Covering 4 Squares F = XY + XY + XY F = XY + XY + XY + XY F = XY + XY X XY 0 Y 0 0 is F = X Y F = X + X F = Adjacency is Cylindrical Another Example Note that wraps from left edge to right edge. In general, as more squares are combined, we obtain a product term with fewer literals. Overlap is allowed School of Engineering 4
15 4variable map Simplifying a 4Variable Function At limit of Kmap Y WX F( W, X, Y, ) = m(0,,2,4,5,6,8,9,2,3,4 ) W 0 X Y Don t Care (a) D (d) (b) B 0 0 (e) BC 0 0 (c) D (f) BD So far we have dealt with functions that were always either 0 or Sometimes we have some conditions where we don t care what result is Example: dealing with B Only care about first 0 (g) A (h) D (i) C Figure.8 Example Use of Karnaugh Maps Also Wraps (toroidal topology) Mark With an `X In a Kmap, mark don t care with an X Simpler implementations Can select an X either as or 0 Example: F( A, B, C, D) = m(,3,7,,5), d(0,2,5) School of Engineering 5
16 Example F( A, B, C, D) = m(,3,7,,5), d(0,2,5) Finding Simplified Expressions or The procedure for finding simplified expressions:. Determine all essential prime implicants. 2. The final expression is formed from the logical sum of: I. The essential prime implicants, with II. Other prime implicants needed to cover the remaining minterms There may be more than one simplified expression. What would we have if Xs were 0? 9 94 Prime Implicants Prime Implicants: Example Each element in the KMap is an implicant. A prime implicantis a product term obtained by combining the maximum possible number of adjacent squares in the map. A single on a maprepresents a prime implicantif it is not adjacent to any other. Two adjacent sform a prime implicant, provided they are not within a group of four adjacent squares. Four adjacent sform a prime implicantif they are not within a group of eight adjacent squares, and so on. If a mintermin a square is covered by only one prime implicant, that prime implicant is said to be essential. They are found by looking at each square marked with a and checking the number of prime implicantsthat cover it. Those with only one prime implicant are essential. 92 Two essential prime implicants (caused by m 0 and m 5 ) This gives us two terms: x z and xz Finding prime implicants for the remainders results in four expressions: F = xz + x z + yz + wz F = xz + x z + yz + wx F = xz + x z + x y + wz F = xz + x z + x y + wx 95 Implicant/Prime Implicant/Essential on Consider: f = { b d, abc, bd, acd, abc, bcd} Each Element of f on is an Implicant Prime Implicants are: { b d, abc, bd, acd, abc} abc is a Prime Implicant but NOT an Essential Prime Implicant bcd is an Implicant but NOT a Prime Implicant bd is an Essential Prime Implicant Essential Prime Implicants F(A, B,C, D, E) = m(0, 2, 4, 6, 7, 8,0,,2,3,4,6,8,9, 29, 30) Prime Implicants (6,7) X X (0,) X X (2,3) X X (8,9) X X (3,29) X X (4,30) X X (0,2,6,8) X X X X (0,2,4,6,8,0,2,4) X X X X X X X X 93 School of Engineering 6
17 Three Variable Maps Three variable maps exhibit the following characteristics:. One box represents minterm with 3 literals 2. Rectangle of 2 boxes > 2 literals 3. Rectangle of 4 boxes > literal 4. Rectangle of 8 boxes > Logic (on 3variable map) 97 Consensus Theorem Slight Variation XY + X + Y = XY + X The third term is redundant Can just drop What does it mean? For third term to be true, Y & both Then one of the first two terms must be! Overlap is OK. No need to use full m 5  waste of input 98 0 Proof of Consensus Theorem Another Example XY + X + Y = XY + X + Y( X + X ) XY + X + XY + XY XY + XY+ X+ XY XY ( + ) + X( + Y ) XY +X 99 Four adjacent corners can be combined to form the two literal term x z. Four adjacent squares can be combined to form the two literal term x y. The remaining is combined with a single adjacent to obtain the three literal term w y z. F = x z + x y+ w y z 02 School of Engineering 7
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