1. Announcements Project components starting to arrive No assignment due this week – assignment 7 will be posted on Thursday. Final exam date. Let’s leave it where it is (Dec 13 th ). Project reports due to me by email, Sunday Dec 11 th. Individual reports. NO EXTENSIONS

2. Lecture 17 Overview Karnaugh map examples Applications of Combinational logic

3. Karnaugh Maps: Example 2: ABCDO 00001 00011 00101 00110 01000 01011 01100 01110 10001 10011 10100 10111 11000 11011 11100 11111 Draw the table

4. Karnaugh Maps: Example 2: ABCDO 00001 00011 00101 00110 01000 01011 01100 01110 10001 10011 10100 10111 11000 11011 11100 11111 00011110 001101 010100 110110 101110 AB CD Draw the table

5. Karnaugh Maps: Example 2: ABCDO 00001 00011 00101 00110 01000 01011 01100 01110 10001 10011 10100 10111 11000 11011 11100 11111 00011110 001101 010100 110110 101110 AB CD Draw the table Find any isolated cells - 1's with no neighbours

6. Karnaugh Maps: Example 2: ABCDO 00001 00011 00101 00110 01000 01011 01100 01110 10001 10011 10100 10111 11000 11011 11100 11111 00011110 001101 010100 110110 101110 AB CD Draw the table Find any isolated cells Find any 2-cell subcubes which are not adjacent to other 2-cell subcubes. Remember wrapping 00011110 001101 010100 110110 101110 AB CD

7. Karnaugh Maps: Example 2: ABCDO 00001 00011 00101 00110 01000 01011 01100 01110 10001 10011 10100 10111 11000 11011 11100 11111 00011110 001101 010100 110110 101110 AB CD Draw the table Find any isolated cells Find 2-cell subcubes. Find 4-cell subcubes which are not adjacent to other four cell subcubes Find 8-cell subcubes etc etc. Minimal expression is formed by the smallest number of maximal subcubes

8. Karnaugh Maps: Example 2: ABCDO 00001 00011 00101 00110 01000 01011 01100 01110 10001 10011 10100 10111 11000 11011 11100 11111 00011110 001101 010100 110110 101110 AB CD A'·B'·D'

9. Karnaugh Maps: Example 2: ABCDO 00001 00011 00101 00110 01000 01011 01100 01110 10001 10011 10100 10111 11000 11011 11100 11111 B'·C' 00011110 001101 010100 110110 101110 AB CD

10. Karnaugh Maps: Example 2: ABCDO 00001 00011 00101 00110 01000 01011 01100 01110 10001 10011 10100 10111 11000 11011 11100 11111 C'·D 00011110 001101 010100 110110 101110 AB CD

11. Karnaugh Maps: Example 2: ABCDO 00001 00011 00101 00110 01000 01011 01100 01110 10001 10011 10100 10111 11000 11011 11100 11111 A·D 00011110 001101 010100 110110 101110 AB CD

12. Karnaugh Maps: Example 2: ABCDO 00001 00011 00101 00110 01000 01011 01100 01110 10001 10011 10100 10111 11000 11011 11100 11111 So sum-of products realization is: O=A'·B'·D'+B'·C'+C'·D+AD Requires 9 gates 00011110 001101 010100 110110 101110 AB CD

13. Karnaugh Maps: Example 2: ABCDO 00001 00011 00101 00110 01000 01011 01100 01110 10001 10011 10100 10111 11000 11011 11100 11111 So sum-of products realization is: O=A'·B'·D'+B'·C'+C'·D+AD Requires 9 gates A'·B'·D' B'·C' C'·D AD

14. Karnaugh Maps: Your turn ABCDO 00001 00011 00101 00111 01001 01011 01101 01111 10001 10011 10100 10110 11000 11010 11100 11110

15. ABCDO 00001 00011 00101 00111 01001 01011 01101 01111 10001 10011 10100 10110 11000 11010 11100 11110 00011110 00 01 11 10 AB CD Solve the Karnaugh map and draw the circuit

16. Karnaugh Maps: Fill the map ABCDO 00001 00011 00101 00111 01001 01011 01101 01111 10001 10011 10100 10110 11000 11010 11100 11110 00011110 001111 011111 110000 101100 AB CD

17. Karnaugh Maps: Find the subcubes ABCDO 00001 00011 00101 00111 01001 01011 01101 01111 10001 10011 10100 10110 11000 11010 11100 11110 00011110 001111 011111 110000 101100 AB CD

18. Karnaugh Maps: Evaluate ABCDO 00001 00011 00101 00111 01001 01011 01101 01111 10001 10011 10100 10110 11000 11010 11100 11110 00011110 001111 011111 110000 101100 AB CD A' B'C' sum of products realization is A'+B'C'

19. Karnaugh Maps: Evaluate ABCDO 00001 00011 00101 00111 01001 01011 01101 01111 10001 10011 10100 10110 11000 11010 11100 11110 sum of products realization is A'+B'C'

20. Karnaugh Maps: Setting up the Maps DeMorgan's Theorem shows that any logic function can be implemented by using just OR and NOT gates, or by just AND and NOT gates A consequence of this is that any logical expression can be reduced to either a "Sum-of-Products" form or a "Product-of-Sums" form

21. Karnaugh Maps: Boxing zeros for sum of products So far, we have been "boxing the ones" for a sum-of-products It is also possible to realize a sum-of-products by "boxing the zeros" Truth table Karnaugh Map A'·B' A·C So, realization is O=(A'·B'+A·C)' (all inverted as we've boxed zeros) Can convert to a product-of-sums with DeMorgan's Theorem: (A'·B'+A·C)' =(A'·B')'·(A·C)'= (A+B)·(A·C)'=(A+B)·(A'+C')

22. Karnaugh Maps: Boxing zeros for product of sums Knowing this, we can get the same product of sums more directly Truth table Karnaugh Map A+B A'+C' So, realization is O=(A+B)·(A'+C') We have to NOT the inputs, then:

23. Karnaugh Maps So O=(A+B)·(A'+C') This requires 5 gates: Note: this realization, part sum of products, part product of sums: O=(A+B)·(A·C)' only requires 3 gates:

24. Sum of Products (X·Y+W·Z)Product of sums (X+Y)·(W+Z) Box 1'sBox 0's AND operation among inputs which are not changing OR operation among NOTted inputs which are not changing OR operation between boxes AND operation between boxes Result must be NOTted for final answer Choose whether to box 1's or 0's by looking at the map - if you have more 0's, boxing 0's will lead to greater simplification

25. Another Karnough Example: 7 segment displays Truth Table Karnaugh Map for "a" "x" represents a "don't care" condition - the value can be either 0 or 1 Control segments with 4 binary inputs Need a logic circuit for each segment

26. Box the ones for sum-of-products This subcube: B This subcube: A'·C' This subcube: D This subcube: A·C

27. Realization (sum-of-products) is B+D+ A·C+ A'·C'

28. Karnaugh Maps More generally, Karnaugh maps can be used to simplify any logic function (doesn’t have to be electronics!)

29. Final comment on Karnaugh Maps http://puz.com/sw/karnaugh 00011110 000000 011100 110111 100011 AB CD