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# EE2420 – Digital Logic Summer II 2013 - PowerPoint PPT Presentation

EE2420 – Digital Logic Summer II 2013. Set 5: Karnaugh Maps. Hassan Salamy Ingram School of Engineering Texas State University. Test 1. Thursday July 18, 2013. 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)

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EE2420 – Digital LogicSummer II 2013

Set 5: Karnaugh Maps

Hassan Salamy

Ingram School of Engineering

Texas State University

• Thursday July 18, 2013

• 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

x1

0

1

x2

m0

m2

0

m1

m3

1

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

x1

0

1

x2

m0

m2

0

m1

m3

1

Karnaugh map

Values for the first variable

are listed across the top

Values for the second variable

are listed down the left side

0

1

y

0

1

0

0

1

1

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

f=xy’+xy=x

0

1

y

0

1

0

1

1

1

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)

x

f=x+y

y

Draw the K-map and give the minimized logic expression for the following truth table.

Show the groupings made in the K-map

K-map groupings example

Three variable K-map the following truth table.

• 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)

xy

00

01

11

10

z

m0

m2

m6

m4

0

m1

m3

m7

m5

1

xy the following truth table.

00

01

11

10

z

1

1

0

1

0

1

0

0

0

1

Example three-variable K-maps

f(x,y,z)=Sm(0,1,2,4)

=x’y’+x’z’+y’z’

xy

00

01

11

10

z

f(x,y,z)=Sm(0,1,2,3,4)

1

1

0

1

0

=x’+y’z’

1

1

0

0

1

A grouping of four eliminates 2 variables

Guidelines for combining terms the following truth table.

• 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

xy the following truth table.

00

01

11

10

z

1

1

1

1

0

0

0

0

0

1

Example groupings

xy

00

01

11

10

z

0

1

1

1

0

0

0

1

1

1

f=z’

f=yz’+x

xy

xy

00

01

11

10

00

01

11

10

z

z

1

1

1

1

1

1

1

0

0

0

1

0

0

1

0

1

1

0

1

1

f=y+x’z’

f=z’+y’

f(a,b,c)=Sm(1,2,3,4,5,6)

Show the groupings made in the K-map

K-map groupings example

Four variable K-map the following.

• A four-variable K-map is constructing by laying 2 three-variable maps together to create four rows

• f(a,b,c,d)

ab

00

01

11

10

cd

m0

m4

m12

m8

00

m1

m5

m13

m9

01

m3

m7

m15

m11

11

m2

m6

m14

m10

10

Four variable K-map the following.

• Adjacencies wrap around in the K-map

00

01

11

10

ab

cd

m0

m4

m12

m8

00

m1

m5

m13

m9

01

m3

m7

m15

m11

11

m2

m6

m14

m10

10

ab the following.

00

01

11

10

cd

0

0

0

0

00

0

0

1

1

01

1

0

0

1

11

1

0

0

1

10

Example four-variable K-maps

f(a,b,c,d)=Sm(2,3,9-11,13)

=ac’d+b’c

ab

00

01

11

10

cd

f(a,b,c,d)=Sm(3-7,9,11,12-15)

0

1

1

0

00

0

1

1

1

01

1

1

1

1

11

0

1

1

0

10

ab the following.

00

01

11

10

cd

1

1

1

1

00

1

0

0

1

01

1

0

0

1

11

1

1

1

1

10

Example groupings

ab

00

01

11

10

cd

0

1

1

0

00

1

0

0

1

01

1

0

0

1

11

0

1

1

0

10

f(a,b,c,d)=b’+d’

f(a,b,c,d)=b’d+bd’

Example groupings the following.

ab

ab

00

01

11

10

00

01

11

10

cd

cd

1

0

0

1

1

1

1

0

00

00

0

1

1

0

1

0

0

1

01

01

0

1

1

0

1

0

0

1

11

11

1

0

0

1

1

1

1

0

10

10

f(a,b,c,d)=b’d’+bd

f(a,b,c,d)=b’d+bd’+a’b’

Examples the following.

• 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.

Multiplexer circuit the following.

f(s,x,y)=m2+m3+m5+m7

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

Car safety alarm the following.

A(D,K,S,B)=Sm(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’

Three-way light control the following.

f(x,y,z)=m1+m2+m4+m7

f(x,y,z)=x’y’z+x’yz’+xy’z’+xyz

This is the simplest sum-of-products form.

Majority Function the following.

• 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