1. Prof. Sin-Min Lee Department of Computer Science
CS147 Lecture 6
POS, K-map and Multiplexer
Mid 1 Revision
Prof. Sin-Min Lee
Department of Computer Science

2. Unsimplifying expressions
xy + y’z + xz = (xy  1) + (y’z  1) + (xz  1)
= (xy  (z’ + z)) + (y’z  (x’ + x)) + (xz  (y’ + y))
= (xyz’ + xyz) + (x’y’z + xy’z) + (xy’z + xyz)
= xyz’ + xyz + x’y’z + xy’z
You can also convert the expression to a sum of minterms with Boolean algebra.
Apply the distributive law in reverse to add in missing variables.
Very few people actually do this, but it’s occasionally useful.
In both cases, we’re actually “unsimplifying” our example expression.
The resulting expression is larger than the original one!
But having all the individual minterms makes it easy to combine them together with the K-map.

3. K-maps from truth tables
You can also fill in the K-map directly from a truth table.
The output in row i of the table goes into square mi of the K-map.
Remember that the rightmost columns of the K-map are “switched.”

4. The Sum-of-Products (SOP) Form
When two or more product terms are summed by Boolean addition

5. Conversion of a General Expression to SOP Form
Any logic expression can be change into SOP form by applying Boolean Algebra techniques
Example:
Try This:

6. The Standard SOP Form
Multiply:

7. The Products-of-Sum (POS) Form
When two or more sum terms are multiplied.

8. The Standard POS Form
Rule 12!
Add:

9. Converting SOP to Truth Table
Examine each of the products to determine where
the product is equal to a 1.
Set the remaining row outputs to 0.

10. Converting POS to Truth Table
Opposite process from the SOP expressions.
Each sum term results in a 0.
Set the remaining row outputs to 1.

11. Converting from Truth Table to SOP and POS
Inputs
Output A B C X 1
POS:

12. The 3-Variable K-Map

13. The 4-Variable K-Map

14. K-Map SOP Minimization
A 1 is placed on the K-Map for each product term in the expression.
Each 1 is placed in a cell corresponding to the value of a product term

15. Example: Map the following standard SOP expression on a K-Map:
Solution:

16. Example: Map the following standard SOP expression on a K-Map:
Solution:

17. Exercise:
Map the following standard SOP expression on a K-Map:

18. Answer:

19. K-Map Simplification of SOP Expressions
A group must contain either 1, 2, 4, 8 or 16 cells.
Each cell in group must be adjacent to one or more cells in that same group but all cells in the group do not have to be adjacent to each other
Always include the largest possible number 1s in a group in accordance with rule 1
Each 1 on the map must be included in at least one group. The 1s already in a group can be included in another group as long as the overlapping groups include noncommon 1s
Goal:
To maximize the size of the groups and to minimize the number of groups

20. Example: Group the 1s in each K-Maps

21. Determining the minimum SOP Expression from the Map
Groups the cells that have 1s. Each group of cells containing 1s create one product term composed of all variables that occur in only one form (either uncomplemented or complemented) within the group. Variable that occurs both uncomplemented and complemented within the group are eliminated. These are called contradictory variables.

22. Example: Determine the product term for the K-Map below and write the resulting minimum SOP expression 1

24. Example: Use a K-Map to minimize the following standard SOP expression

25. Example: Use a K-Map to minimize the following standard SOP expression

26. Mapping Directly from a Truth Table

27. Don’t Care (X) Conditions
A situation arises in which input variable combinations are not allowed
Don’t care terms either a 1 or a 0 may be assigned to the output

28. Exercise: Use K-Map to find the minimum SOP from1 2

34. SOP

35. POS

43. MUX
Enable
2n Data
Inputs
Data
Output n
Input
Select

44. (Only one O/P asserted at any time
Remember the 2 – 4 Decoder?
Sel(3) S1
Sel(2)
Sel(1) S0
Sel(0)
Mutually Exclusive
(Only one O/P asserted at any time

45. Z Y X F
F=0 1
F=1
F= X´
F= X 1 X´ X F
Z Y

46. Y X Z F
F=Z 1
F=0
F= Z´
F= 1 Z Z´ 1 F
Y X

47. Y X Z F
X .Y ´=0 1
X⊕Y=0
X⊕Y=1
X .Y ´=1

48. MUX Example (cont.) When A=B=0, F=C When A=0, B=1, F=C1
When A=B=0, F=C
When A=0, B=1, F=C
When A=1, B=0, F=C
When A=B=1, F=C’

49. Implementation using MUXs: Now we implement the output f1
using an 8x1MUX. Selection inputs to the MUX are x2 y1 y2
(that is, if x2y1y2=001, input I1 of the MUX is selected).
The implementation table is tabulated as

52. Combinational circuit implementation using MUX
We can use Multiplexers to express Boolean functions also.
Expressing Boolean functions as MUXs is more efficient than as decoders.
First n-1 variables of the function used as selection inputs; last variable used as data inputs.
If last variable is called Z, then each data input has to be Z, Z’, 0, or 1.