Logic and Computer Design Fundamentals Chapter 2 Combinational Logic Circuits Part 1 Gate Circuits and Boolean Equations
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Y = A B z = x + y X = A Note: The statement: 1 + 1 = 2 ( one plus one equals two ) is not the same as 1 + 1 = 1 ( 1 or 1 equals 1 ). Chapter 2 - Part 1 6
Operations are defined on the values "0" and "1" for each operator: AND 0 0 = 0 0 1 = 0 1 0 = 0 1 1 = 1 OR 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 1 NOT 0 = 1 1 = 0 Chapter 2 - Part 1 7
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Switches in parallel => OR Switches in series => AND Normally-closed switch => NOT C Chapter 2 - Part 1 9
B C A D Chapter 2 - Part 1 10
+V +V +V F G = X + Y X X. Y X X X Y Y (a) NOR (b) NAND (c) NOT Transistor of logic functions are called logic gates or just gates Transistor gate circuits can be modeled by switch circuits Chapter 2 - Part 1 11
X Y X Z5 X Y Z 5 X 1 Y Y AND gate OR gate (a) Graphic symbols X 0 0 1 1 X Z 5 X NOT gate or inverter Y 0 1 0 1 (AND) X Y 0 0 0 1 (OR) X1 Y 0 1 1 1 (NOT) X 1 1 0 0 (b) Timing diagram Chapter 2 - Part 1 12
Truth Table = X + X Y Z Equation F = X + Y Z Logic Diagram F Chapter 2 - Part 1 13
An algebraic structure defined on a set of at least two elements, B, together with three binary operators (denoted +, and ) that satisfies the following basic identities: 1. X + 0 = X 2. X. 1 = X 3. X + 1 = 1 4. X. 0 0 = 5. X + X = X 6. X. X = X 7. X + X = 1 8. X. X = 0 9. X = X 10. 12. 14. 16. X + Y = Y + X (X + Y) + Z = X + (Y + Z) X(Y + Z) = XY+ XZ X + Y = X. Y 11. 13. 15. 17. XY = YX (XY) Z = X(YZ) X+ YZ = (X + Y)(X + Z) X. Y = X + Y Commutative Associative Distributive DeMorgan s Chapter 2 - Part 1 14
The identities above are organized into pairs. These pairs have names as follows: 1-4 Existence of 0 and 1 5-6 Idempotence 7-8 Existence of complement 9 Involution 10-11 Commutative Laws 12-13 Associative Laws 14-15 Distributive Laws 16-17 DeMorgan s Laws The dual of an algebraic expression is obtained by interchanging + and and interchanging 0 s and 1 s. The identities appear in dual pairs. When there is only one identity on a line the identity is self-dual, i. e., the dual expression = the original expression. Chapter 2 - Part 1 15
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( X + Y)Z + XY = Y(X + ( X + Y )Z + X Y Z) Chapter 2 - Part 1 20
Chapter 2 - Part 1 21 x y y ( )( ) n Minimizatio y y y x y y y x = + + = ( ) tion Simplifica y x y x y x y x = + + = + ( ) Absorption x y x x x y x x = + = + Consensus z y x z y z y x + = + + ( ) ( )( ) ( ) ( ) z y x z y z y x + + = + + + DeMorgan's Laws x x = + + x x x x x x x x y x = +y
F1 = xyz F2 = x + yz F3 = xyz + x y z F4 = xy + x z + xy x y z F1 F2 F3 F4 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 0 1 1 0 0 1 0 0 0 1 1 0 1 0 1 1 1 0 1 1 1 1 1 0 1 Chapter 2 - Part 1 22
A B + ACD + A BD + AC D + A BCD Chapter 2 - Part 1 23
xyz + xyz Chapter 2 - Part 1 24
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XY X Y X Y X Y Chapter 2 - Part 1 27
X + Y X + Y X + Y X + Y Chapter 2 - Part 1 28
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abcd abcd abcd abcd abcd abcd a + b + c + a + b + c + a + b + c + a + b + c + a + b + c + a + b + c + d d d d d d Chapter 2 - Part 1 31
x y m 0 m 1 m 2 m 3 0 0 1 0 0 0 x y M 0 M 1 M 2 M 3 0 0 0 1 1 1 0 1 0 1 0 0 0 1 1 0 1 1 1 0 0 0 1 0 1 0 1 1 0 1 1 1 0 0 0 1 1 1 1 1 1 0 Chapter 2 - Part 1 32
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x y z index m 1 + m 4 + m 7 = F 1 0 0 0 0 0 + 0 + 0 = 0 0 0 1 1 1 + 0 + 0 = 1 0 1 0 2 0 + 0 + 0 = 0 0 1 1 3 0 + 0 + 0 = 0 1 0 0 4 0 + 1 + 0 = 1 1 0 1 5 0 + 0 + 0 = 0 1 1 0 6 0 + 0 + 0 = 0 1 1 1 7 0 + 0 + 1 = 1 Chapter 2 - Part 1 34
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F1 = (x + y + z) (x + y + z) (x + y + z) ( x + y + z) (x + y + z) x y z i M 0 M 2 M 3 M 5 M 6 = F1 0 0 0 0 0 1 1 1 1 = 0 0 0 1 1 1 1 1 1 1 = 1 0 1 0 2 1 0 1 1 1 = 0 0 1 1 3 1 1 0 1 1 = 0 1 0 0 4 1 1 1 1 1 = 1 1 0 1 5 1 1 1 0 1 = 0 1 1 0 6 1 1 1 1 0 = 0 1 1 1 7 1 1 1 1 1 = 1 Chapter 2 - Part 1 36
F(A,B,C,D) = M 3 M8 M11 M14 Chapter 2 - Part 1 37
v + v f = x + x y f = x( y+ y ) + x y f = xy + xy + x y Chapter 2 - Part 1 38
F = A + B C Chapter 2 - Part 1 39
F = A + B C F(A, B,C) = Σ m (1,4,5,6,7) Chapter 2 - Part 1 40
v v f ( x, y, z ) = x + x y x + x y = (x + x)(x + y) = 1(x + y) = x + y x + y + z z = (x + y + z) x+ y + z ( ) Chapter 2 - Part 1 41
Convert to Product of Maxterms: f(a, B,C) = A C + BC + A B Use x + y z = (x+y) (x+z) with x = (A C + BC), y = A, and z = B to get: f = (A C + BC + A)(A C + BC + B) Then use x + x y = x + y to get: f = (C + BC + A)(AC + C + B) and a second time to get: f = (C + B + A)(A + C + B) Rearrange to standard order, f = (A + B + C)(A + B + C) to give f = M 5 M 2 Chapter 2 - Part 1 42
F(x, y, z) = Σ m (1,3,5,7) F(x, y, z) F(x, y, z) = Σm(0,2,4,6) = ΠM(1,3,5,7) Chapter 2 - Part 1 43
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F(A, B,C) = Σm(1,4,5,6,7) Chapter 2 - Part 1 45
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A B C A B C A B C A B C A B C F A B C F Chapter 2 - Part 1 49
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