1. ECE 2110: Introduction to Digital Systems PoS minimization Don’t care conditions

2. 2 Previous… Sum-of-Products (SoP) minimization

3. 3 Simplifying the Product of Sums (Principle of Duality: looking on the 0s on a K-map) Two main steps : 1) Combining/Grouping the 0 cells. 2) Writing the sum term for each group. Rules : ( for n-variable function ) 1) The group size must be a power of 2. 2) A set of 2^i cells can be combined if there are ( i ) variables that take all possible combinations within the set and the remaining ( n-i ) variables have the same value within that set. 3) The corresponding sum term for each group contains (n-i) literals: - The variable is complemented if it is 1 in the combined cells - The variable is uncomplemented it it 0 in the combined cells - The variable in not included in the product term if it takes the values 0 and 1 within the combined cells

4. 4 Example F= The prime implicants: - Cells (0,1,8,9) X=0, Y=0 The sum term : X+Y - Cells (8,10,12,14) W=1, Z=0 The sum term : W’+Z The two prime implicants are essential prime implicants and cover all zeros The minimal product of sums : F=(X+Y).(W’+Z) 0 15 4 WX YZ W Z 01 01 0001 00 13 12 0 1 11 9 8 0 0 10 X 3 26 7 11 11 14 15 1 0 10 11 1 0 01 11 10 Y

5. 5 Simplifying Products of Sums PoS- Another method F= 1- Complement the function. (F’) 2- Use K-map to get the minimal sum of the complement function (F’). 3- Complement the minimal sum to get the minimal product Example : In the previous example the function is complemented and represented using K-map : - The essential prime implicants are: X’.Y’, W.Z’ - The minimal SoP : M= X’.Y’+W.Z’ - F = (F’)’=M’ = (X’.Y’+W.Z’)’ = (X+Y).(W’+Z) 0 15 4 WX YZ W Z 10 10 0001 00 13 12 1 0 11 9 8 1 1 10 X 3 26 7 00 00 14 15 0 1 10 11 0 1 01 11 10 Y

6. 6 Minimal PoS vs. Minimal SoP : F= In this same example, the minimal SoP is  F =W’.X+W’.Y+X.Z+Y.Z  The minimal PoS was F=(X+Y).(W’+Z) Which representation has lower cost?  In this case the minimal PoS implementation is cheaper. See note. In general: To find the best realization compare the minimal product and the minimal sum products. 0 15 4 WX YZ W Z 01 01 0001 00 13 12 0 1 11 9 8 0 0 10 X 3 26 7 11 11 14 15 1 0 10 11 1 0 01 11 10 Y

7. 7 Don’t Care Conditions (not in Test2): In some applications, the Boolean function for certain combinations of the input variables is not specified. The corresponding minterms (maxterms) are called “don't care minterms (maxterms)”. In K-map, the “don't care minterms/maxterms” are represented by “d”. Since the output function for those minterms (maxterms) is not specified, those minterms (maxterms) could be combined with the adjacent 1 cells(0-cells) to get a more simplified sum-of- products (product-of-sums) expression.

8. 8 New Rules for circling sets of 1s (Sum-of-Products simplification) Allow d’s to be included when circling sets of 1s, to make the sets as large as possible Do NOT circle any sets that contain only d’s. As usual, cover all 1s, none of 0s.

9. 9 New Rules for circling sets of 0s (Product-of-Sums simplification) Allow d’s to be included when circling sets of 0s, to make the sets as large as possible Do NOT circle any sets that contain only d’s. As usual, cover all 0s, none of 1s.

10. 10 Example Build a logic circuit that determines if a decimal digit is >= 5 Solution:  The decimal digits(0,1,2,...,9) can be represented by 4 bit BCD code.  The logic circuit should have 4 input variables and one output.  There are 16 different input combinations but only 10 of them are used.  The logic function should produce 0 if the number is = 5

11. 11 Example - The Truth table The Truth table for the function is specified as follows: Row W X Y Z F 0 0 0 0 0 0 1 0 0 0 1 0 2 0 0 1 0 0 3 0 0 1 1 0 4 0 1 0 0 0 5 0 1 0 1 1 6 0 1 1 0 1 7 0 1 1 1 1 8 1 0 0 0 1 9 1 0 0 1 1 10 1 0 1 0 d 11 1 0 1 1 d 12 1 1 0 0 d 13 1 1 0 1 d 14 1 1 1 0 d 15 1 1 1 1 d Maxterm list: F=? K-Map?

12. 12 Example - K-Map The Minimal Sum : Combining the 1 cells only the minimal sum is: F = W.X’.Y’+W’.X.Z+W’.X.Y Combining the don't care minterms d (s) with the 1 cells The minimal sum is (with using d (s)): F = W+X.Z+X.Y Exercise: Find the Minimal Product 0 15 4 WX YZ W Z 00 01 0001 00 13 12 d d 11 9 8 1 1 10 X 3 26 7 01 01 14 15 d d 10 11 d d 01 11 10 Y 0 15 4 WX YZ Z 00 01 0001 00 13 12 d d 11 9 8 1 1 10 X 3 26 7 01 01 14 15 d d 10 11 d d 01 11 10 Y W

13. 13 Exercise solution The Minimal Product F= (W+X).(W+Y+Z) Which one is cheaper for this problem: The SoP or the PoS? 0 15 4 WX YZ W Z 00 01 0001 00 13 12 d d 11 9 8 1 1 10 X 3 26 7 01 01 14 15 d d 10 11 d d 01 11 10 Y

14. 14 Example- Implementation The minimal Sum implementation : F = W+X.Z+X.Y The minimal Product implementation : F= (W+X).(W+Y+Z) How to choose? W X Y Z F W X Y Z F W X Y Z F

15. Next Test 2. Chapter 4 only. Closed book. One sheet of Boolean Algebra theorems will be provided. Then, we will start Chapter 6. 15