Digital Logic CPE231

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# Digital Logic CPE231 - PowerPoint PPT Presentation

Digital Logic CPE231. Digital Fundamentals, Ninth Edition By Floyd Instructor: Eng. Tuqa Manasrah. Ch4: Boolean Algebra and Logic Simplification. Apply the basic laws and rules of Boolean algebra Apply DeMorgan's theorems to Boolean expressions

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### Digital Logic CPE231

Digital Fundamentals, Ninth Edition

By Floyd

Instructor: Eng. Tuqa Manasrah

Ch4: Boolean Algebra and Logic Simplification
• Apply the basic laws and rules of Boolean algebra
• Apply DeMorgan's theorems to Boolean expressions
• Describe gate networks with Boolean expressions
• Evaluate Boolean expressions
• Simplify expressions by using the laws and rules of Boolean algebra
• Convert any Boolean expression into a sum-of-products (SOP) form
• Convert any Boolean expression into a product-of-sums (POS) form
• Use a Karnaugh map to simplify Boolean expressions
• Use a Karnaugh map to simplify truth table functions
• Utilize "don't care" conditions to simplify logic functions

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Term

Boolean function

Laws and rules of Boolean algebra
• Boolean algebra is the mathematics of digital systems.
• Boolean function: described by Boolean equation.
• Boolean equation: express logical relationship between binary variables

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Basic laws of Boolean algebra
• Commutative law: the order of variables makes no difference
• X + Y = Y + X
• XY = YX
• Associative law: the result is the same regardless of the grouping of the variables.
• X + (Y + Z) = (X + Y) + Z
• X(YZ) = (XY)Z
• Distributive law: express the process of factoring in which a common variable is factored out the product term.
• XY + XZ = X( Y + Z)
• X + YZ = (X + Y)(X + Z)

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Rules of Boolean algebra

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Continue..
• X + XY = X

= X(1 + Y) …distributive law

= X . 1 … Rule 2

= X … Rule 4

• X + X’Y = X +Y

= (X + XY) + X’Y … Rule 10

= XX + XY + X’y … Rule 7

= XX + XY + XX’ + X’Y …Rule 8

= (X + X’) (X + Y)…Factoring

= 1 . (X+Y) … Rule 6

= X + Y … Rule 4

• (X + Y)(X + Z) = X + YZ

Do it …

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Demorgan’s Theorem
• State the equivalency of NAND & negative-OR:
• State the equivalency of NOR & negative-AND:
• Prove using truth table.

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Examples
• Apply DeMorgans Theorem:
• (AB’+A’B)’
• (XYZ)’
• (X’+Y’+Z’)’
• ((A+B+C)D)’
• ((ABC)’+D+E)’
• (A’B(C+D’)+E)’

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Boolean Analysis of Logic Circuits
• Each logic function can be expressed with three methods:
• Boolean expression
• Logic circuit
• Truth table
• Having a function expressed with one method, you can derive the remaining.

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Example
• The logic circuit shown below implement two logic functions: F1 and F2.
• Analyze the circuit by deriving the Boolean expressions of F1 and F2.
• Once the Boolean expression for a given logic circuit has been determined, a truth table that shows the output for all values of the input variables can be developed. Derive the truth table of the logic functions F1, F2.

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Simplification using Boolean algebra.
• The goal is to reduce a given expression to its simplest form, or to a more convenient one for implementation.
• Use laws, rules and theorems of Boolean algebra to manipulate and simplify an expression

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Examples
• A+AB+AB’C = A
• AB’ + A(B+C)’ + B(B+C)’ = AB’
• ABC’+A’B’C+A’BC+A’B’C’ = ABC’+A’C+A’B’
• (AB)’+(AC)’+A’B’C’ = A’ + B’+ C’
• AB’C(BD+CDE)+AC = A(C’ + B’DE)

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Standard forms for Boolean algebra
• Standardization makes the evaluation, simplification and implementation of Boolean expressions much more systematic and easier.
• All Boolean expressions, regardless of their form, can be converted into either of two forms: sum-of-product, product-of-sum.

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Sum-of-Products (SOP) form
• A product term consists of the product of literals (variables or their complements).
• When two or more products are summed the results is an SOP.
• E.g. AB + ABC , A’B + A’BC’+ AC
• SOP expression can contain a single variable term.
• In SOP a single overbar cannot extend over more than one variable ( )

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Continue…
• Implementation:
• AND/OR
• NAND/NAND
• Conversion from general expressions:
• Apply Boolean algebra techniques.
• E.g. AB + B(CD+EF), ((A+B)’+C)’
• The domain of a Boolean expression is the set of variables contained in the expression either complemented or not.
• SOP may not include the domain of variable in all products. (A’BC + ABD’ + AB’CD)
• A standard SOP expression is a one in which all the variables in the domain appear in each product term. (A’BC + AB’C + ABC’ + A’B’C’)
• A non standard SOP can be converted to the standard one by using the Boolean algebra rule (x + x’ =1).
• Convert wx’y + x’yz’ + wxy’ to the standard sop form.

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The Product-of-Sum (POS) from
• A sum term consists of the sum of literals.
• When two or more sum terms are multiplied the results is an POS.
• E.g. (A+B)(A+B+C), (A’+B)(A’+B+C’)(A+C)
• POS expression can contain a single variable term.
• In POS a single overbar cannot extend over more than one variable ( )

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Continue…
• Implementation:
• OR/AND
• NOR/NOR
• POS may not include the domain of variable in all terms. (A’+B+C)(A+B+D’)(B’+C+D)
• A standard SOP expression is a one in which all the variables in the domain appear in each product term. (A’+B+C)(A+B+C’)(A+B+C’)(A’+B’+C’)
• A non standard POS expression can be converted to the standard form by using the Boolean algebra rule 8 (xx’=0) and rule12((X+Y)(X+Z)= X+YZ).
• Convert (a+b’)(b+c) to the standard POS form.

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Boolean expressions and Truth Table
• A SOP expression is equal to 1 if at least one of the product terms is equal to 1.
• A POS expression is equal to 0 if at least one of the sum terms is equal to 0.
• E.g. determine the truth table for the expressions:
• F1= A’BC’ + AB’C
• F2= (A+B’+C)(A+B+C’)(A’+B’+C’)
• Determine the standard expression from the truth table:
• SOP
• List the binary values of the inputs for which the output is 1.
• Convert the binary values to the corresponding product term.(1  corresponding variable, 0corresponding variable complement)
• POS
• List the binary values of the inputs for which the output is 0.
• Convert the binary values to the corresponding sum term.(0  corresponding variable, 1corresponding variable complement)

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Definition: Minterm
• Product term in which all variables appear once (complemented or not).
• For the variables X, Y and Z example minterms: X’Y’Z’, X’Y’Z, X’YZ’, …., XYZ
• Each minterm represents exactly one combination of the binary variables in a truth table.
• For n variables, there will be 2n minterms.
• Minterms are labeled from minterm 0, to minterm 2n-1

m0 , m1 , m2 , … , m2n-2 , m2n-1

• For n = 3, we have

m0 , m1 , m2 , m3 , m4 , m5 , m6 , m7

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Definition: Maxterm
• Sum term in which all variables appear once (complemented or not)
• For the variables X, Y and Z the maxterms are: X+Y+Z , X+Y+Z’ …. , X’+Y’+Z’
• Minterms and maxterms with the same subscripts are complements:
• Example:

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Standard Form of F:Sum of Minterms
• The same as standard SOP
• OR all of the minterms of truth table for which the function value is 1
• F = m0 + m2 + m5 + m7

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Complement of F
• Not surprisingly, just sum of the other minterms
• In this case

F’ = m1 + m3 + m4 + m6

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Product of Maxterms
• The same as standard POS
• F = m0 + m2 + m5 + m7
• Can express F as AND of all Maxterms of rows that should evaluate to 0

or

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The Karnaugh Map
• A systematic method for simplifying Boolean expressions.
• If properly used, will produce the simplest SOP or POS expression.
• Graphical representation of truth table.
• A box for each minterm
• So 2 variables, 4 boxes
• 3 variable, 8 boxes
• And so on

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K-Map from Truth TableExamples
• There are implied 0s in empty boxes

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Function from K-Map
• Can generate function from K-map

Simplifies to X + Y (in a moment)

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A 3-variable K-Map
• Eight minterms
• Look at encoding of columns and rows

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A 4-variabel K-Map
• At limit of K-map

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• Cells are arranged so that there is only a single variable change between adjacent cells.
• Each cell is adjacent to the cells that are immediately next to it on any of its four sides.
• A cell is not adjacent to the cells that are diagonally touch any of its corners.
• Note that Z’ wraps from left edge to right edge.

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Also Wraps (toroidal topology)

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K-Map SOP minimization
• A minimized SOP expression contains the fewest possible terms with the fewest possible variables per term. And can be implemented with fewer logic gates than a standard expression.
• Mapping
• From truth table
• From standard SOP (Sum of Minterms)
• From a nonstandard SOP
• simplification

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Mapping Examples
• Map the following standard SOP expressions:
• A’B’C + A’BC’ + ABC’ + ABC
• A’BC + AB’C +AB’C’
• A’B’CD + A’BC’D’+ ABC’D + A’B’C’D + AB’CD’
• A’BCD’ + ABCD’ + ABC’D’ + ABCD
• Map the following nonstandard SOP expressions:
• X’ + XY’ + XYZ’
• YZ + X’Z’
• X’Y’ + WX’ + WXY’ + WX’YZ’ + W’X’Y’Z + WX’YZ
• W + Y’Z + WYZ’ + W’XYZ’

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Simplification of SOP expressions
• The minimum SOP expression is obtained by grouping the 1’s and determining the minimum expression from the map.
• Rules of grouping:
• A group must contain a power of 2 cells, i.e. either 1,2,4,8, or 16.
• Each cell must be adjacent to one or more cells in the group, but all cells not need to be adjacent.
• Include the largest possible number of 1’s in a group.
• Each 1 on the map must be included in at least one group.

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Continue..
• Determining the minimum product term of each group:
• For a 3-variable map
• One cell -> 3 literals
• Rectangle of 2 cell s -> 2 literals
• Rectangle of 4 cell s -> 1 literal
• Rectangle of 8 cell s -> Logic 1
• For a 4-variable map
• One cell -> 4 literals
• Rectangle of 2 cell s -> 3 literals
• Rectangle of 4 cell s -> 2 literal
• Rectangle of 8 cell s -> 1 literal
• Rectangle of 16 cell s -> Logic 1

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Examples

Solve this one

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Overlap is OK.

• No need to use full m5-- waste of input

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K-map POS minimization
• A 0 is placed in k-map for each sum term.
• E.g. for A+ B’ + C, a 0 is placed in cell 010.
• Cell with no zeros are for which the expression is 1.
• Mapping procedure:
• Determine the binary value for each sum term in standard SOP.
• In the k-map place zeros in the corresponding cells.

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Simplification of POS expressions
• Group 0s instead of 1s.
• Follow the same rules as grouping 1s.
• Conversion between POS and SOP for the sake of implementation with fewer gates.
• Use the k-map to optimize a SOP expression: (W+X’+Y+Z’)(W’+X+Y’+Z’)(W’+X’+Y’+Z)(W’+X’+Z’)

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Examples:
• Map the following expressions on a k-map:
• A’BC + AB’C + AB’C’
• A’BCD’+ABCD’+ABC’D’+ABCD
• BC+A’C’
• A+ C’D+ACD’+A’BCD’
• (A+B’+C’+D)(A+B+C+D’)(A+B+C+D)(A’+B+C’+D)
• Use the k-map to minimize the following expressions:
• XY’Z+XYZ’+X’YZ+X’YZ’+XY’Z’+XYZ
• W’X’Y’Z’+WX’YZ+WX’Y’Z+W’YZ+WX’Y’Z’
• (X+Y’+Z)(X+Y’+Z’)(X’+Y’+Z)(X’+Y+Z)

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Don’t care conditions
• Some input combinations are not allowed never occurred treated as don’t care.
• For each don’t care term, X is placed in the cell.
• The Xs can be treated as 1s in case of grouping.

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Examples
• Optimize the expression with don’t care conditions:
• F(a,b,c)=∑m(1,2,4), d(a,b,c)= ∑m(0,3,6,7)
• F(a,b,c,d)=∑m(0,1,7,13,15), d(a,b,c,d)= ∑m(2,6,8,9,10)
• F(a,b,c,d)=∑m(2,4,9,12,15), d(a,b,c,d)= ∑m(3,5,6,13)
• F(w,x,y,z)=∏M(3,11,13,15), d(w,x,y,z)= ∑m(0,2,5,8,10,14)

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