ece314_sec4_TA

# ece314_sec4_TA - ECE/CS 314 spring 2007 Section 4...

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ECE/CS 314 spring 2007 Section 4 COMBINATIONAL LOGIC 1) Truth Tables A truth table is a complete representation of a Boolean expression because it explicitly defines the output for every combination of inputs. The corresponding Boolean expression can be written as a Sum of Products (SOP) and can be obtained from the truth table by inspection. For every row where the output = 1, write the value of the inputs (minterm), e. g. for A = 0, B = 1, C = 1 A’·B·C (product) Sum up every row where output = 1 A’·B·C’ A’·B·C Out = A’BC’ + A’BC + AB’C + ABC A·B’·C A·B·C Exercise #1: Sum and Carry-out (from lecture) – what is the SOP for each output? Sum = A’B’Cin + A’B(Cin)’ + AB’(Cin)’ + ABCin Cout = A’BCin + AB’Cin +AB(Cin)’ + ABCin 2) Boolean Simplification As circuit designers, we would like our circuits to be as small and as fast as possible. One way to accomplish this is to minimize the number of operations in a given computation. Algebraic Simplification: Just like middle school algebra, Boolean algebra follows rules that can be used to consolidate and simplify terms in an expression. Some of these rules are listed below. A B C Out 0 0 0 0 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 0 1 1 1 1 A B Cin Sum Cout 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1

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Basic Boolean Algebraic Identities Additive Multiplicative A + 0 = A 0·A = 0 A + 1 = 1 1·A = A A + A = A A·A = A A + A’ = 1 A·A’ = 0 Basic Boolean Algebraic Properties Additive Multiplicative A + B = B + A Commutative A·B = B·A A + (B + C) = (A + B) + C Associative A·(B·C) = (A·B)·C A·(B + C) = A·B + A·C Distributive Boolean Rules for Simplification A + A·B = A A + A’·B = A + B (A + B)·(A + C) = A + B·C DeMorgan’s Theorem (X 1 ·X 2 ·…·X n )’ = X 1 ’+X 2 ’+…+X n (X 1 +X 2 +…+X n )’ = X 1 ’·X 2 ’·…·X n Alright, that was a lot to look at all at once – let’s get some practice using these rules by simplifying the expression A’·B·C’ + A’·B·C + A·B’·C + A·B·C (from our first example). A’·B·C’ + A’·B·C + A·B’·C + A·B·C (Distributive) factor out A’·B from left two terms A’·B· (C + C’) + A·B’·C + A·B·C (Identity) C + C’ = 1 A’·B·(1) + A·B’·C + A·B·C (Distributive) factor out A·C from right two terms (Identity) 1·(A’·B) = A’·B A’·B + A·C·(B + B’) (Identity) B + B’ = 1 A’·B + A·C·(1) (Identity) 1·(A·C) = A·C A’·B + A·C Okay, now it’s your turn. Exercise #2: Using the rules of Boolean logic simplify the expression for Cout that you obtained above. Cout = A’BCin + AB’Cin +AB(Cin)’ + ABCin (Identity) ABCin + ABCin = ABCin Cout = A’BCin + ABCin + AB’Cin + ABCin + AB(Cin)’ + ABCin
(Distributive) factor out B·Cin from left two terms (Distributive) factor out A·Cin from middle two terms (Distributive) factor out A·B from right two terms Cout = (A’+A).BCin + (B’+B).ACin + AB.( (Cin)’+ Cin) (Identity) A’ + A = 1 (Identity) B’ + B = 1 (Identity) (Cin)’ + Cin = 1 Cout = 1. BCin + 1. ACin + AB.1 (Identity) 1. BCin = BCin (Identity) 1. ACin = ACin (Identity) AB.1 = AB Cout = BCin + ACin + AB Karnaugh Map As you can see, algebraic simplification quickly becomes tedious and time consuming. Luckily there is a graphical method of laying out a truth table such that

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ece314_sec4_TA - ECE/CS 314 spring 2007 Section 4...

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