EE2420 – Digital Logic Summer II 2013 Hassan Salamy Ingram School of Engineering Texas State University Set 5: Karnaugh Maps

Test 1  Thursday July 18,

Karnaugh map  The key to finding a minimum cost SOP or POS form is applying the combining property (14a for SOP or 14b for POS)  The Karnaugh map (K-map) provides a systematic (and graphical) way of performing this operation  Minterms can be combined by 14a when they differ in only one variable  f(x,y,z) = xyz+xyz’ = xy(z+z’) = xy(1) = xy  The K-map illustrates this combination graphically 3

Karnaugh map  The K-map is an alternative to a truth table for representing an expression  K-map consists of cells that correspond to rows of the truth table  Each cell corresponds to a minterm  A two variable truth table and the corresponding K-map x1x1 x2x2 f 00m0m0 01m1m1 10m2m2 11m3m3 m3m3 m1m1 m2m2 m0m x1x1 x2x2 4

Karnaugh map m3m3 m1m1 m2m2 m0m x1x1 x2x2 Values for the first variable are listed across the top Values for the second variable are listed down the left side 5

Karnaugh map groupings  Minterms in adjacent squares on the map can be combined since they differ in only one variable  Indicated by looping the corresponding ‘1’s on the map (the ‘1’s must be adjacent)  Looping two ‘1’s together corresponds to eliminating a term and a variable from the output expression => xy+xy’ = x xyf x y f=xy’+xy=x 6

K-map groupings example  Note that the bottom two cells differ in only one variable (x) and the right two cells differ in only one variable (y) xyf x y x y f=x+y 7

K-map groupings example  Draw the K-map and give the minimized logic expression for the following truth table.  Show the groupings made in the K-map xyf

Three variable K-map  A three-variable K-map is constructed by laying 2 two-variable maps side by side  K-maps are always laid out such that adjacent squares only differ by one variable (i.e. by 1 bit in the binary expression of the minterm values) xyzMinterm 000m 0 =x’y’z’ 001m 1 =x’y’z 010m 2 =x’yz’ 011m 3 =x’yz 100m 4 =xy’z’ 101m 5 =xy’z 110m 6 =xyz’ 111m 7 =xyz m3m3 m1m1 m2m2 m0m xy z m5m5 m7m7 m4m4 m6m End cells are ‘adjacent’ 9

Example three-variable K-maps f(x,y,z)=  m(0,1,2,4) xy z =x’y’+x’z’+y’z’ xy z f(x,y,z)=  m(0,1,2,3,4) =x’+y’z’ A grouping of four eliminates 2 variables 10

Guidelines for combining terms  Can combine only adjacent ‘1’s  Can group only in powers of 2 (1,2,4,8, etc.)  Try to form as large a grouping as possible  Do not generate more groups than are necessary to “cover” all the ‘1’s 11

Example groupings xy z xy z xy z xy z f=z’f=yz’+x f=z’+y’ f=y+x’z’ 12

K-map groupings example  Draw the K-map and give the minimized logic expression for the following.  f(a,b,c)=  m(1,2,3,4,5,6)  Show the groupings made in the K-map 13

Four variable K-map  A four-variable K-map is constructing by laying 2 three-variable maps together to create four rows  f(a,b,c,d) m5m5 m1m1 m4m4 m0m ab cd m9m9 m 13 m8m8 m m6m6 m2m2 m7m7 m3m3 m 10 m 14 m 11 m

Four variable K-map  Adjacencies wrap around in the K-map m5m5 m1m1 m4m4 m0m ab cd m9m9 m 13 m8m8 m m6m6 m2m2 m7m7 m3m3 m 10 m 14 m 11 m

Example four-variable K-maps ab cd f(a,b,c,d)=m(2,3,9-11,13) =ac’d+b’c ab cd f(a,b,c,d)=m(3-7,9,11,12-15) =b+cd+ad 16

Example groupings ab cd f(a,b,c,d)=b’+d’ ab cd f(a,b,c,d)=b’d+bd’ 17

Example groupings ab cd f(a,b,c,d)=b’d’+bd ab cd f(a,b,c,d)=b’d+bd’+a’b’ 18

Examples  We will revisit some of the examples we studied in the last lecture.  We will simplify the equations using K-Maps this time instead of algebraic manipulations. 19

Multiplexer circuit sxyf f(s,x,y)=m 2 +m 3 +m 5 +m 7 f(s,x,y)=s’xy’+s’xy+sx’y+sxy f(s,x,y)=s’x(y’+y)+sy(x’+x) f(s,x,y)=s’x+sy 20

Car safety alarm DKSBA A(D,K,S,B)=m(4,5,6,7,14) A(D,K,S,B)=D’KS’B’+D’KS’B+D’KSB’+D’KSB+DKSB’ =D’KS’+D’KS+KSB’ =D’K+KSB’ 21

Three-way light control xyzf f(x,y,z)=m 1 +m 2 +m 4 +m 7 f(x,y,z)=x’y’z+x’yz’+xy’z’+xyz This is the simplest sum-of-products form. 22

Majority Function 23 XYZMajority The output of the majority function is equal to the value for the three inputs which occurs on more inputs. Majority(X,Y,Z) =  m(3,5,6,7) Majority(X,Y,Z) = X’YZ + XY’Z + XYZ’ + XYZ Simplified  Majority(X,Y,Z) = XY + XZ + YZ