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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 logic summer ii 2013
EE2420 – Digital LogicSummer II 2013

Set 5: Karnaugh Maps

Hassan Salamy

Ingram School of Engineering

Texas State University

test 1
Test 1
  • Thursday July 18, 2013
karnaugh map
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
karnaugh map1

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
karnaugh map2

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

karnaugh map groupings

x

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

k map groupings example

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

k map groupings example1
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
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)

xy

00

01

11

10

z

m0

m2

m6

m4

0

m1

m3

m7

m5

1

End cells are ‘adjacent’

example three variable k maps

xy

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
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
example groupings

xy

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’

k map groupings example2
Draw the K-map and give the minimized logic expression for the following.

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
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)

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 map1
Four variable K-map
  • 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

example four variable k maps

ab

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

=b+cd+ad

0

1

1

1

01

1

1

1

1

11

0

1

1

0

10

example groupings1

ab

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 groupings2
Example groupings

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
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.
multiplexer circuit
Multiplexer circuit

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
Car safety alarm

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
Three-way light control

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
Majority Function
  • 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
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