1. Computer Engineering (Logic Circuits) (Karnaugh Map)
Lec. # 6
(Karnaugh Map)
Dr. Tamer Samy Gaafar
Dept. of Computer & Systems Engineering
Faculty of Engineering
Zagazig University

2. Course Web Page http://www.tsgaafar.faculty.zu.edu.eg

3. Lec. # 6
Karnaugh Map (Contd.)

4. Truth Table to K-Map Mapping
Karnaugh Mapping
Truth Table to K-Map Mapping
Digital Electronics
2.2 Intro to NAND & NOR Logic
Four Variable K-Map W X Y Z
FWXYZ
Minterm – 0
Minterm – 1 1
Minterm – 2
Minterm – 3
Minterm – 4
Minterm – 5
Minterm – 6
Minterm – 7
Minterm – 8
Minterm – 9
Minterm – 10
Minterm – 11
Minterm – 12
Minterm – 13
Minterm – 14
Minterm – 15 1 V 1 3 2 4 5 7 6 12 13 15 14 8 9 11 10 1 1 1
Demonstrate how the minterms of a four variable truth table are mapped to a K-Map.
Project Lead The Way, Inc.
Copyright 2009

5. Four Variable K-Map Groupings
Karnaugh Mapping
Digital Electronics
2.2 Intro to NAND & NOR Logic
Four Variable K-Map Groupings
Groups of One – 16 (not shown)
Groups of Two – 32 (not shown)
Groups of Four – 24 (seven shown) 1 1 1 V 1 1 1 1
Four Variable K-Map:
There are 16 groups of one that are not shown because they are obvious.
There are 32 groups of two that are not shown because they are obvious.
There are 24 groups of four. Seven are show, the other 17 are obvious.
Project Lead The Way, Inc.
Copyright 2009

6. Four Variable K-Map Groupings
Karnaugh Mapping
Digital Electronics
2.2 Intro to NAND & NOR Logic
Four Variable K-Map Groupings
Groups of Eight – 8 (two shown) 1 V 1
Four variable K-Map: There are 8 groups of eight. Two are shown, the other 6 are obvious.
Project Lead The Way, Inc.
Copyright 2009

7. Four Variable K-Map Groupings
Karnaugh Mapping
Digital Electronics
2.2 Intro to NAND & NOR Logic
Four Variable K-Map Groupings
Group of Sixteen – 1 V 1
Four variable K-Map, group of 16.
Project Lead The Way, Inc.
Copyright 2009

8. Example : 4 Variable K-Map
Karnaugh Mapping
Example : 4 Variable K-Map
Digital Electronics
2.2 Intro to NAND & NOR Logic
Example:
After labeling and transferring the truth table data into the K-Map, write the simplified sum-of-products (SOP) logic expression for the logic function F3. R S T U F3 1 V
Pause the presentation and allow students to complete the example. The solution is on the next slide.
Project Lead The Way, Inc.
Copyright 2009

9. Example : 4 Variable K-Map
Karnaugh Mapping
Digital Electronics
2.2 Intro to NAND & NOR Logic
Example:
After labeling and transferring the truth-table data into the K-Map, write the simplified sum-of-products (SOP) logic expression for the logic function F3. R S T U F3 1 V 1
Solution:
Here is the solution. If you print handouts, do not print this page.
Project Lead The Way, Inc.
Copyright 2009

10. Steps of graphical method (continued):
Combine adjacent 1’s into group of 2n each such that
Each group contains only 1’s.
The group is not completely a part of a larger group.
Choose the minimum number of the largest sized groups needed to cover all the 1’s.
Each group is represented by an expression which is an intersection of the minterm in the group.
The simplified solution is a logical OR of the expressions of all the groups chosen in steps 3 above.

11. Definitions An implicant is a product term of a function
Any group of 1’s in a K-Map
A prime implicant is a product term obtained by combining the maximum possible number of adjacent 1’s in a k-map
Biggest groups of 1’s
Not all prime implicants are needed!
If a minterm is covered by exactly one prime implicant then this prime implicant is called an essential prime implicant

12. Terminology
An implicant is a product term in an SOP expression (or a sum term in POS expression)
Implicants are always rectangular in shape and the # of 1's covered is a power of 2
A prime implicant is an implicant that is not fully contained in some other larger implicant
B C A 00 01 11 10 1
red  implicant (but not prime) many more are not shown
blue  prime implicants (only two)

13. Essential Prime Implicants
An essential prime implicant is a prime implicant that contains a 1 not included in any other prime implicant
The minimum Boolean expression must use this term
A cover is a collection of implicants that accounts for all valuations in which the function is “on” (e.g. 1)

14. Example Y=1 Consider F(X,Y,Z) = Sm(1,3,4,5,6)
List all implicants, prime implicants and essential prime implicants
Solution:
Implicants: XY’Z’, XZ’, XY’, XY’Z, X’Y’Z, Y’Z, …
P.Is: XY’, XZ’, Y’Z, X’Z
EPIs: X’Z, XZ’
YZ X 00 01 11 10 1
X=1
Z=1
Y=1
YZ X 00 01 11 10 1
X=1
The simplest expression is NOT unique!
Z=1

15. Finding minimum SOP
Find each essential prime implicant and include it in the solution
If any minterms are not yet covered, find minimum number of prime implicants to cover them (minimize overlap).

16. Example Simplify F(A, B, C, D) = ∑ m(0, 1, 2, 4, 5, 10,11,13, 15)
Note:
Only A’C’ is E.P.I
For the remaining minterms:
Choose 1 and 2 (minimize overlap)
For m2, choose either A’B’D’ or B’CD’
F = A’C’ + ABD + AB’C + A’B’D’

17. Examples to illustrate terms
0 0
1 1
1 0 D A
0 1 B C
5 prime implicants: BD, ABC', ACD, A'BC, A'C'D
essential
minimum cover: 4 essential implicants

18. 4 Variable Maps f(A,B,C,D) = m(0,1,2,3,6,8,9,11,13,14)
f = A C' D + B C D' + B' C ' + B' D + A'B'
C D A B 00 01 11 10 1

19. These are not essential!
Another 4 Variable Map
f(A,B,C,D) = m(0,1,2,5,6,7,8,9,10,13,15)
f = B D + A' B C + B' D' + B' C' or
f = B D + A' B C + B' D' + C' D (there are 2 more)
C D A B 00 01 11 10 1
These are not essential!

20. essential prime implicants
Find the Essential Prime Implicants
C D A B 00 01 11 10 1
C D A B 00 01 11 10 1
essential prime implicants
f = C'D' + A'B + B'D' + ACD

21. 1 square = a term with 4 literals 2 square = a term with 3 literal
16 square = a function equal to 1 Z WX Y 01 00 YZ 11 10 X W

22. Simplification of k-Map
Generate all PIs
Find EPIs
If EPIs can cover all minterms, then it is answer. Otherwise choose some non-essential PIs (which has less cost) such that all minterms are cover.

23. Step 1 Generate PIs Blue circle are PIs
They are the largest circle you can drawn on kmap

24. Step 2 Find EPIs Red circle are EPIs
Minterm 5, 14, 11 can only be cover by these 3 red circle

25. Step 3 EPIs cannot cover minterm 7
Choose between green/blue circle to cover minterm 7
Green is chosen as it is larger
Less cost

26. Final
Final result is obtained
x3x4’ + x2’x3 + x1’x3 + x2x3’x4

27. Filling The Map Given F= WX’ + YZ Required: 1) K-Map
2) Simplify F as POS
Solution:
4 Variables K-Map

28. Filling The Map F= WX’ + YZ 0 0 0 1 1 1 10 1 1 1 1 1 Y YZ WX 00 01 X YZ 11 10 X W 1

29. Filling The Map
POS Z WX Y 01 00 YZ 11 10 X W 1

30. Filling The Map Z WX Y 01 00 YZ 11 10 X W
0 0 1

31. Karnaugh Mapping
Don’t Care Conditions
Digital Electronics
2.2 Intro to NAND & NOR Logic
A don’t care condition, marked by (X) in the truth table, indicates a condition where the design doesn’t care if the output is a (0) or a (1).
A don’t care condition can be treated as a (0) or a (1) in a K-Map.
Treating a don’t care as a (0) means that you do not need to group it.
Treating a don’t care as a (1) allows you to make a grouping larger, resulting in a simpler term in the SOP equation.
Project Lead The Way, Inc.
Copyright 2009

32. Don’t Cares
In some cases, the output of the function (1 or 0) is not specified for certain input combinations either because
The input combination never occurs (Example BCD codes), or
We don’t care about the output of this particular combination
Such functions are called incompletely specified functions
Unspecified minterms for these functions are called don’t cares
While minimizing a k-map with don’t care minterms, their values can be selected to be either 1 or 0 depending on what is needed for achieving a minimized output.

33. f(w,x,y,z) = m(0,2,4,5,8,14,15), d(w,x,y,z) = m(7,10,13)
Practice K-map
Find a MSOP (Minimal SOP) for
f(w,x,y,z) = m(0,2,4,5,8,14,15), d(w,x,y,z) = m(7,10,13)

34. Solutions for Practice K-map
Find a MSOP for:
f(w,x,y,z) = m(0,2,4,5,8,14,15),
d(w,x,y,z) = m(7,10,13) Z WX Y 01 00 YZ 11 10 X W 1 1 1 1 X X 1 1 1 X

35. Solutions for Practice K-map
Find a MSOP for:
f(w,x,y,z) = m(0,2,4,5,8,14,15), d(w,x,y,z) = m(7,10,13) Y 1 x X W Z
f(w,x,y,z)= x’z’ + w’xy’ + wxy

36. Example: Don’t Care Conditions
Karnaugh Mapping
Example: Don’t Care Conditions
Digital Electronics
2.2 Intro to NAND & NOR Logic
Example:
After labeling and transferring the truth table data into the K-Map, write the simplified sum-of-products (SOP) logic expression for the logic function F4. Be sure to take advantage of the don’t care conditions. R S T U F4 X 1 V
Pause the presentation and allow students to complete the example. The solution is on the next slide.
Project Lead The Way, Inc.
Copyright 2009

37. Example: Don’t Care Conditions
Karnaugh Mapping
Example: Don’t Care Conditions
Digital Electronics
2.2 Intro to NAND & NOR Logic
Example:
After labeling and transferring the truth table data into the K-Map, write the simplified sum-of-products (SOP) logic expression for the logic function F4. Be sure to take advantage of the don’t care conditions. R S T U F4 X 1 V X 1
Solution:
Here is the solution. If you print handouts, do not print this page.
Project Lead The Way, Inc.
Copyright 2009

38. Example
F = ∑m(1, 3, 7) + ∑d(0, 5) B B A 00 01 11 10 X 1 3 2 4 5 7 6
Circle the x’s that help get bigger groups of 1’s (or 0’s if POS).
Don’t circle the x’s that don’t help. A C

39. Example F = C F = ∑m(1, 3, 7) + ∑d(0, 5)
Circle the x’s that help get bigger groups of 1’s (or 0’s if POS).
Don’t circle the x’s that don’t help. B B A 00 01 11 10 X 1 3 2 4 5 7 6 A
F = C C

40. Example F(A, B, C, D) = ∑ m(1, 3, 7, 11, 15) + ∑ d(0, 2, 5)
Two possible solutions!
Both acceptable.
All 1’s covered

41. Chapter – 4 – Combinational Logic Circuit

42. Combinational Circuits
Output is function of input only
i.e. no feedback
When input changes, output may change (after a delay)
Combinational
Circuits
n inputs
m outputs • • 

43. Combinational Circuits
Analysis
Given a circuit, find out its function
Function may be expressed as:
Boolean function
Truth table
Design
Given a desired function, determine its circuit ? ?

44. (A’+B’)(A’+C’)(B’+C’)
Analysis Procedure
Boolean Expression Approach
ABC
A+B+C
AB'C'+A'BC'+A'B'C
(A’+B’)(A’+C’)(B’+C’)
AB+AC+BC
F1=AB'C'+A'BC'+A'B'C+ABC
F2=AB+AC+BC

45. Analysis Procedure
Truth Table Approach
A B C F1 F2
= 0 1

46. Analysis Procedure Truth Table Approach 1 0 A B C F1 F2 0 0 0 0 0 1
= 0
= 1 1 1 1 1

47. Analysis Procedure Truth Table Approach 1 0 A B C F1 F2 0 0 0 0 0 1 1 1
= 0
= 1 1 1 1 1

48. Analysis Procedure Truth Table Approach 0 1 A B C F1 F2 0 0 0 0 0 1 1 1
= 0
= 1 1 1

49. Analysis Procedure Truth Table Approach 1 0 A B C F1 F2 0 0 0 0 0 1 1 1
= 1
= 0 1 1 1 1

50. Analysis Procedure Truth Table Approach 0 1 A B C F1 F2 0 0 0 0 0 1 1 1
= 1
= 0 1 1

51. Analysis Procedure Truth Table Approach 0 1 A B C F1 F2 0 0 0 0 0 1 1 1
= 1
= 0 1 1

52. Analysis Procedure Truth Table Approach F1=AB'C'+A'BC'+A'B'C+ABC 1
= 1 1 1 1 B 1 A C B 1 A C
F1=AB'C'+A'BC'+A'B'C+ABC
F2=AB+AC+BC

53. Design Procedure Given a problem statement: Example: ?
Determine the number of inputs and outputs
Derive the truth table
Simplify the Boolean expression for each output
Produce the required circuit
Example:
Design a circuit to convert a “BCD” code to “Excess 3” code ?
4-bits
0-9 values
4-bits
Value+3

54. Design Procedure BCD-to-Excess 3 Converter C B A D C B A D w = A+BC+BD1 B A x D C 1 B A x D
A B C D
w x y z
x x x x
w = A+BC+BD
x = B’C+B’D+BC’D’ C 1 B A x D C 1 B A x D
y = C’D’+CD
z = D’

55. Design Procedure BCD-to-Excess 3 Converter w = A + B(C+D)
w x y z
x x x x
w = A + B(C+D)
y = (C+D)’ + CD
x = B’(C+D) + B(C+D)’
z = D’

56. Example
Design a combinational circuit that obtains the square for the numbers from > 7
Solution:
# i/ps = 3
# o/ps = 6 ( 49 )

58. Continue
K-Maps (A, B, C, D, E, F)
Then
Implement the Circuit