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Boolean Algebra (Lecture #4). ECE 331 – Digital System Design. The slides included herein were taken from the materials accompanying Fundamentals of Logic Design, 6 th Edition , by Roth and Kinney, and were used with permission from Cengage Learning. . Converting POS to SOP.
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Boolean Algebra (Lecture #4) ECE 331 – Digital System Design The slides included herein were taken from the materials accompanying Fundamentals of Logic Design, 6th Edition, by Roth and Kinney, and were used with permission from Cengage Learning.
ECE 331 - Digital System Design Converting POS to SOP Given an expression in product-of-sums (POS) form, the corresponding sum-of-products (SOP) expression can be obtained by multiplying out, using the two distributive laws: • X(Y + Z) = XY + XZ (3-1) • (X + Y)(X + Z) = X + YZ (3-2) In addition, the following theorem is very useful for factoring and multiplying out: • (X + Y)(X′ + Z) = XZ + X′Y (3-3)
ECE 331 - Digital System Design Exercise: Convert the following Boolean Expression in POS form to a Boolean Expression in SOP form. (Hint: Multiply out the expression) F = (A'+D')(B+C')(C+D+E') Boolean Algebra
ECE 331 - Digital System Design Exercise: Convert the following Boolean Expression in POS form to a Boolean Expression in SOP form. (Hint: Multiply out the expression) F = (A+B+C')(A+B+D)(A+B+E)(A+D'+E)(A'+C) Boolean Algebra
ECE 331 - Digital System Design Boolean Algebra If we were to multiply out by brute force, we would generate 162 terms, and 158 of these terms would then have to be eliminated to simplify the expression. Instead, we will use the distributive laws to simplify the process.
ECE 331 - Digital System Design Converting SOP to POS The same theorems that are useful for multiplying out expressions are also useful for factoring. By repeatedly applying (3-1), (3-2), and (3-3), an expression in sum-of-products (SOP) form can be converted to an expression in product-of-sums (POS) form. • X(Y + Z) = XY + XZ (3-1) • (X + Y)(X + Z) = X + YZ (3-2) • (X + Y)(X′ + Z) = XZ + X′Y (3-3)
ECE 331 - Digital System Design Exercise: Convert the following Boolean Expression in SOP form to a Boolean Expression in POS form. (Hint: Factor the expression) F = ACE + BCE + ADE + BDE + ACF + BCF + ADF + BDF Boolean Algebra
ECE 331 - Digital System Design Exercise: Convert the following Boolean Expression in SOP form to a Boolean Expression in POS form. (Hint: Factor the expression) F = AC + A'BD' + A'BE + A'C'DE Boolean Algebra
ECE 331 - Digital System Design Boolean Algebra
ECE 331 - Digital System Design Exclusive-OR and Equivalence The following theorems of Boolean Algebra apply to Exclusive-OR and Equivalence (Exclusive-NOR):
ECE 331 - Digital System Design Exercise: Simplify the following Boolean Expression. Boolean Algebra
ECE 331 - Digital System Design Exercise: Simplify the following Boolean Expression. Boolean Algebra
ECE 331 - Digital System Design Exercise: Simplify the following Boolean Expression. F(A,B,C) = (A xor B)' xor C Boolean Algebra
ECE 331 - Digital System Design Consensus Theorem The consensus theorem can be stated as follows: • XY + X'Z + YZ = XY + X'Z (3-20) Dual Form: • (X + Y)(X’ + Z)(Y + Z) = (X + Y)(X’ + Z) (3-21) Consensus theorem proof: XY + X'Z + YZ = XY + X'Z + (X + X')YZ = (XY + XYZ) + (X'Z + X'YZ) = XY(1 + Z) + X'Z(1 + Y) = XY + X'Z
ECE 331 - Digital System Design 1. Combining terms. Use the theorem XY + XY′ = X to combine two terms. 2. Eliminating terms. Use the theorem X + XY = X to eliminate redundant terms if possible; then try to apply the consensus theorem (XY + X′Z + YZ = XY + X′Z) to eliminate any consensus terms. 3. Eliminating literals. Use the theorem X + X’Y = X + Y to eliminate redundant literals. Simple factoring may be necessary before the theorem is applied. 4. Adding redundant terms. Redundant terms can be introduced in several ways such as adding xx′, multiplying by (x + x′), adding yz to xy + x′z, or adding xy to x. When possible, the added terms should be chosen so that they will combine with or eliminate other terms. Simplifying Boolean Expressions
ECE 331 - Digital System Design Example: Boolean Algebra
ECE 331 - Digital System Design Simplify the following Boolean expression using Boolean algebra: F(A,B,C,D) = A'BC'D' + A'B'C'D + A'BC'D + ABC'D + A'BCD + ABCD Boolean Algebra
ECE 331 - Digital System Design Proving Validity of an Equation Often we will need to determine if an equation is valid for all combinations of values of the variables. Several methods can be used to determine if an equation is valid: Construct a truth table and evaluate both sides of the equation for all combinations of values of the variables. (This method is rather tedious if the number of variables is large, and it certainly is not very elegant.) Manipulate one side of the equation by applying various theorems until it is identical with the other side. Reduce both sides of the equation independently to the same expression. 4. It is permissible to perform the same operation on both sides of the equation provided that the operation is reversible.
ECE 331 - Digital System Design Boolean Algebra Example: Show that A'BD' + BCD + ABC' + AB'D = BC'D' + AD + A'BC Solution: Starting with the left side,
ECE 331 - Digital System Design Importance of Boolean Algebra • Boolean Algebra is used to simplify Boolean expressions. • Through application of the Laws and Theorems discussed • Simpler expressions lead to simpler circuit realization, which, generally, reduces cost, area requirements, and power consumption. • The objective of the digital circuit designer is to design and realize optimal digital circuits. • However, in general, there is no easy way to determine when a Boolean expression has been simplified to a minimum number of terms or minimum number of literals.
ECE 331 - Digital System Design Exercise: For the following Boolean expression F(A,B,C) = (A+B+C).(A'+B+C).(A+B'+C).(A+B+C') 1. Using Boolean algebra, simplify the Boolean expression. 2. Derive the Truth table for the simplified expression. 3. From the Truth table, determine an equivalent Boolean expression. 4. Draw the circuit diagram for the “cheapest” expression. Boolean Algebra
ECE 331 - Digital System Design Questions?