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Chapter 2. Logic Circuits. Variables and Functions. x. =. 0. x. =. 1. (a) Two states of a switch. S. x. (b) Symbol for a switch. Figure 2.1. A binary switch. A closed circuit. S. Battery. Light. x. (a) Simple connection to a battery. S. Power. Light. x. supply.
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Chapter 2. Logic Circuits
Variables and Functions x = 0 x = 1 (a) Two states of a switch S x (b) Symbol for a switch Figure 2.1. A binary switch.
A closed circuit S Battery Light x (a) Simple connection to a battery S Power Light x supply (b) Using a ground connection as the return path Figure 2.2. A light controlled by a switch.
AND and OR logic functions S S Power x1 x2 Light supply (a) The logical AND function (series connection) S x1 Power Light supply S x2 (b) The logical OR function (parallel connection) Figure 2.3. Two basic functions.
S X1 S Power Light X3 supply S X2 Figure 2.4. A series-parallel connection.
Truth Table for two input variables Figure 2.6. A truth table for the AND and OR operations.
Truth Table for three input variables Figure 2.7. Three-input AND and OR operations.
x AND, OR and NOT gates 1 x 2 x 1 × × ¼ × x x x × x x 1 2 n 1 2 x 2 x n (a) AND gates x 1 x 2 x 1 ¼ x + x x + x + + x 1 2 1 2 n x 2 x n (b) OR gates x x (c) NOT gate Figure 2.8. The basic gates.
A logic Circuit x 1 x 2 ( ) . f = x + x x 1 2 3 x 3 Figure 2.9. The function from Figure 2.4.
Boolean Algebra Definition:Boolean Algebra: A mathematical system for formulating logical statements with symbols so that problems can be solved in a manner similar to ordinary algebra. In short, Boolean algebra is the mathematics of digital systems. The basic rules for Boolean addition and multiplication are presented in Table 1-9. Axioms of Boolean Algebra
Laws of Boolean Algebra Commutative Laws • The commutative law of addition for two variables is algebraically expressed as • x + y = y + x • The commutative law of multiplication for two variables is expressed as • xy = yx • In summary, the order in which the variables are ORed or ANDed make no difference.
Associative Laws • The associative law of addition of three variables is expressed as • x+ (y + z) = (x + y) + z • The associative law of multiplication of three variables is expressed as • x(yz) = (xy)z • In summary, ORing or ANDing a grouping of variables produces the same result regardless of the grouping of the variables.
Distributive Law • The distributive law of three variables is expressed as follows: • x (y+z) = xy + xz • This law states that ORing several variables and ANDing the result is equivalent of ANDing the single variable with each of the variables in the grouping, then ORing the result.
Duality • To reflect the principle of duality, the axioms and single-variable theorems are listed in pairs. For example, see 5a and 3a. When x =0, by 5a, the result is 0. When x =1, by 5a, the result is 0, which is also proved by 3a.
Boolean Simplification Example1: F = AB’ + C’D + AB’ + C’D = AB’ + C’D (by identity 5) Example2: F = ABC + ABC’ + A’C = AB(C + C’) + A’C (by identity 13) = AB(1) + A’C (by identity 7) = AB +A’C (by identity 4)
Logic circuit implementation (Before simplification) (After simplification)
DeMorgan’s Theorem (A B)’ = A’ + B’ (1) That is, the complement of the product is equivalent to the sum of the complements. This is true for any number of variables. (A B C … Z)’ = A’ + B’ + C’ + …+ Z’ (A +B)’ = A’ B’ (2) Similarly, the complement of the sum is equivalent to the product of the complements. Similarly, (A + B + C + …+ Z)’ = A’ . B’ . C’ . … . Z’
Methods to complement a function: Interchange 1’s and 0’s for the values of F in the truth table . Use DeMorgan’s theorem on algebraic function • Change F to F’, and F’ to F • Change OR to AND • Change AND to OR • Complement each individual variable Example 1: F = AB+ C’D + B’D Applying DeMorgan’s theorem, F’ = (A’ + B’)(C + D’)(B + D’)
Example 2: Simplify F = (x1 + x3) . (x1’+ x3’) F = x1 x1’+ x1 x3’ + x3 x1’+ x3 x3’ (Distributive Property) x1 x1’ and x3 x3’ = 0 ( Identity 8) F = x1 x3’ + x1’ x3 Example 3: F = x’yz + x’yz’ +xz = x’y(z + z’ ) + xz (Factoring out) = x’y .1 + xz ( By Identity 6) = x’y + xz ( By Identity 4)
Practice Problems: • Find the complement: (xyz)’ • Expand: x + yz • Simplify: a) x’y’ +x’y + xy b) x’y’ + xz + xy + yz’ c) wy + w’yz’ + wxy + w’xy’
Proof of DeMorgan’s theorem using truth table Figure 2.11. Proof of DeMorgan’s theorem in 15a.
Boolean Functions Example1: Prove: (A + B) (A’ + B’) = AB’ + A’B LHS =AA’ +AB’ + BA’ + BB’ (by distributive property) = 0 + AB’ +BA’ +0 = AB’ +A’B = RHS Example2: Prove: AC’ +B’ C’ + AC + B’C = A’B’ +AB +AB’ LHS = A(C +C’) + B’(C+C’) = A.1 + B’.1 = A + B’ RHS = A’B’ +AB +AB’ = A’B’ + A(B + B’) = A’ B’ + A = A + B’ (by identity 11) LHS = RHS
Precedence of Operations NOT, AND, and then OR Example: A.B + A’.B’ 1. Complements 2. AND operation 3. OR operation
Synthesis using gates F = m0 .1 + m1 . 1 + m2 .0 +m3 .1 Figure 2.15. A function to be synthesized.
x 1 x 2 f (a) Canonical sum-of-products x 1 f x 2 (b) Minimal-cost realization Figure 2.16. Two implementations of a function in Figure 2.15.
Sum of Products (SOP) and Product of Sums (POS) Minterm: A product term with all ‘n’ variables in asserted or negated form. Maxterm: Complement of a minterm. Figure 2.17 Three-variable minterms and maxterms.
Terms and Definitions: Minterm: A product term with all ‘n’ variables in asserted or negated form. Maxterm: Complement of a minterm. Sum-of-products (SOP). Product-of-sums (POS). Canonical sum-of-products. Canonical product-of-sums. Analysis - The task of determining the function performed by a system Synthesis - is the reverse task of analysis, and is defiend as the designing of a new system that implements a desired functional behavior.
Figure 2.18. A three-variable function. f = x1’ x2’x3 + x1 x2’ x3’ + x1 x2’ x3 + x1 x2 x3’ f( x1, x2, x3) = ∑(m1 m4, m5, m6) = ∑ m(1, 4, 5, 6) f ’ = x1’ x2’ x3‘+ x1‘x2 x3’ + x1‘x2 x3 + x1x2 x3 f = (x1+ x2 + x3) ( x1+x2‘ +x3) (x1 + x2‘+ x3‘) (x1‘ + x2‘ + x3‘) f( x1, x2, x3) = ∏(m0 m2, m3, m7= ∏ m(0, 2, 3, 7)
x 2 f x 3 x 1 (a) A minimal sum-of-products realization 1 x 3 f x 2 f = x1’ x2’x3 + x1 x2’ x3’ + x1 . x2’ . x3 . + x1 x2 x3’ = x2’x3 (x1 + x1‘) + x1x3‘(x2 + x2‘) = x2’x3 + x1x3‘ f = (x1 +x3) (x2’ + x3‘) (b) A minimal product-of-sums realization Figure 2.19. Two realizations of a function in Figure 2.18.
Example 2.3. f( x1, x2, x3) = ∑ m (2, 3, 4, 6,7) Simplify the function and draw the circuit (synthesis). Steps: 1. Write the canonical sop expression 2. Simplify the function. 3. Draw the circuit. 4. Calculate the cost - Count the number of gates together with input line sto those gates and express as CUs. Example 2.4. Express the function in Example 1 as a POS, repeat the steps and draw the circuit. Example 2.5. f( x1, x2, x3, x4 ) = ∑ m (3,7,9,12,13,14,15)
Section 2.7 NAND and NOR Logic Networks Figure 2.20. NAND and NOR gates.
NAND and INVERT-OR Equivalency and NOR and INVERT-AND Equivalency x 1 x x 1 1 x x 2 2 x 2 x x x x = + (a) 1 2 1 2 x 1 x x 1 1 x x 2 2 x 2 x x x x + = (b) 1 2 1 2 Figure 2.21. DeMorgan’s theorem in terms of logic gates.
x x 1 1 x x 2 2 x x 3 3 x x 4 4 x x 5 5 x 1 x 2 x 3 x 4 x 5 SOP - AND-OR and All NAND implementations Figure 2.22. Using NAND gates to implement a sum-of-products.
x x 1 1 x x 2 2 x x 3 3 x x 4 4 x x 5 5 x 1 x 2 x 3 x 4 x 5 POS - OR-AND and All NOR implementations Figure 2.23. Using NOR gates to implement a product-of sums.
x1 x2 f x3 (a) POS implementation x1 x2 f x3 (b) NOR implementation Figure 2.24 NOR-gate realization of the function in Examples 2.4. and 2.6
x1 f x2 x3 (a) SOP implementation x1 f x2 x3 (b) NAND implementation Figure 2.25. NAND-gate realization of the function in Examples 2.3. and 2.7