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CS 140 Lecture 3 Combinational Logic. Professor CK Cheng CSE Dept. UC San Diego. Part I. Specification Implementation K-maps. Definitions. Literals x i or x i ’ Product Term x 2 x 1 ’x 0 Sum Term x 2 + x 1 ’ + x 0 Minterm of n variables: A product of n variables
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CS 140 Lecture 3Combinational Logic Professor CK Cheng CSE Dept. UC San Diego
Part I. • Specification • Implementation • K-maps
Definitions Literals xi or xi’ Product Term x2x1’x0 Sum Term x2 + x1’ + x0 Minterm of n variables: A product of n variables in which every variable appears exactly once.
Implementation Specification Schematic Diagram Net list, Switching expression Obj min cost Search in solution space (max performance) Cost: wires, gates Literals, product terms, sum terms We want to minimize # of terms, # of literals
Implementation (Optimization) Karnaugh map – 2D truth table
Function can be represented by sum of minterms: f(A,B) = A’B+AB’+AB This is not optimal however! We want to minimize the number of literals and terms. We factor out common terms – A’B+AB’+AB= A’B+AB’+AB+AB =(A’+A)B+A(B’+B)=B+A f(A,B) = A+B
On the K-map however: A = 0 A = 1 AB’ 0 2 0 1 1 1 B = 0 B = 1 1 3 A’B AB f(A,B) = A + B
Another Example f(A,B)=A’B+AB=(A’+A)B=B
On the K-map: A = 0 A= 1 0 2 0 0 1 1 B= 0 B = 1 1 3 AB A’B f(A,B)=B
Using Maxterms f(A,B)=(A+B)(A’+B)=(AA’)+B=0+B=B
Two variable K-maps Id a b f (a, b) 0 0 0 f (0, 0) 1 0 1 f (0, 1) 2 1 0 f (1, 0) 3 1 1 f (1, 1) 2 variables means we have 22 entries and thus we have 2 to the 22 possible functions for 2 bits, which is 16. a b f(a,b)
Three variables K-maps Id a b c f (a,b,c) 0 0 0 0 1 1 0 0 1 0 2 0 1 0 1 3 0 1 1 0 4 1 0 0 1 5 1 0 1 0 6 1 1 0 1 7 1 1 1 0
Corresponding K-map b = 1 Gray code (0,0) (0,1) (1,1) (1,0) 0 2 6 4 c = 0 1 1 1 1 1 3 7 5 c = 1 0 0 0 0 a = 1 f(a,b,c) = c’
Karnaugh Maps (K-Maps) • Boolean expressions can be minimized by combining terms • K-maps minimize equations graphically • PA + PA = P
K-map • Circle 1’s in adjacent squares • In the Boolean expression, include only the literals whose true y(A,B)=A’B’
Another 3-Input example Id a b c f (a,b,c) 0 0 0 0 0 1 0 0 1 0 2 0 1 0 1 3 0 1 1 0 4 1 0 0 1 5 1 0 1 1 6 1 1 0 - 7 1 1 1 1
Corresponding K-map b = 1 (0,0) (0,1) (1,1) (1,0) 0 2 6 4 c = 0 0 1 - 1 1 3 7 5 c = 1 0 0 1 1 a = 1 f(a,b,c) = a + bc’
Yet another example Id a b c f (a,b,c,d) 0 0 0 0 1 1 0 0 1 1 2 0 1 0 - 3 0 1 1 0 4 1 0 0 1 5 1 0 1 1 6 1 1 0 0 7 1 1 1 0
Corresponding K-map b = 1 (0,0) (0,1) (1,1) (1,0) 0 2 6 4 c = 0 1 - 0 1 1 3 7 5 c = 1 1 0 0 1 a = 1 f(a,b,c) = b’