Download
ee2420 digital logic summer ii 2013 n.
Skip this Video
Loading SlideShow in 5 Seconds..
EE2420 – Digital Logic Summer II 2013 PowerPoint Presentation
Download Presentation
EE2420 – Digital Logic Summer II 2013

EE2420 – Digital Logic Summer II 2013

108 Views Download Presentation
Download Presentation

EE2420 – Digital Logic Summer II 2013

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. EE2420 – Digital LogicSummer II 2013 Set 5: Karnaugh Maps Hassan Salamy Ingram School of Engineering Texas State University

  2. Test 1 • Thursday July 18, 2013

  3. 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

  4. 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

  5. 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

  6. 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

  7. 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

  8. 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

  9. 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’

  10. 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

  11. 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

  12. 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’

  13. 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

  14. 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

  15. 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

  16. 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

  17. 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’

  18. 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’

  19. 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.

  20. 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

  21. 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’

  22. 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.

  23. 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