1. EGR 2131 Unit 3 Gate-Level Minimization
Read Mano & Ciletti, Chapter 3.
Homework #3 and Lab #3 due next week.
Quiz next week.
-Take Quiz #2 and fire up Multisim.
-Handouts: Unit 3 practice sheet, slide xx (K-map procedure).

2. Review: Minimization Techniques
Minimization techniques let us take a digital circuit and reduce it to a simpler but equivalent circuit.
Example: The two circuits show below are equivalent to each other. (In other words, whenever you give them both the same inputs, they’ll produce the same output.)
Given the prevalence of software that can do reduction for us, these manual techniques are less important today than they were 20 years ago. But you should still be familiar with the basic ideas.

3. Review: Two Primary Methods
The two primary manual minimization methods use:
Boolean algebra: a set of rules that let us transform Boolean expressions into equivalent Boolean expressions.
Karnaugh maps (also called K-maps): similar to truth tables.
The Karnaugh-map method provides a step-by-step procedure. If you follow the steps correctly, you’ll get the right answer. The Boolean-algebra method requires more ingenuity on your part, but may be quicker.

4. Karnaugh Maps
A Karnaugh map (or K-map) contains the same information as a truth table, but laid out in a way that reveals patterns that let us read off a simplified Boolean expression.
K-maps can be used for problems with anywhere from 2 to 6 input variables.
But it’s most often used for 3-input and 4-input problems.

5. SOP Or POS?
K-maps can be used to read off either the simplest sum-of-products (SOP) expression for a function or the simplest product-of-sums (POS) expression for that same function.
Recall some examples:
SOP form: F1 = x′y + xyz + y′z′
POS form: F2 = (x+y)(x′+z)(y′+z′+w)
Reading off the simplest SOP expression is slightly easier and requires fewer steps.

6. Two-Variable K-Map A two-variable K-map has two rows and two columns.
Here’s the initial setup of a two-variable map, assuming our inputs are named x and y.
As we’ll see, for a particular problem you’ll insert additional information into some of the map’s squares.

7. Two-Variable K-Map (Cont’d.)
Each square in the map corresponds to one row of a truth table, hence to one minterm.
This map shows which minterm corresponds to each square. But you don’t normally write this information. x y F 1

8. Two-Variable K-Map (Cont’d.)
Here’s another look, with some more information that you don’t need to write yourself.
This marking simply says that y is true in this column (and false in all other columns).
This marking says that x is true in this row (and false in all other rows).

9. Three-Variable K-Map
A three-variable K-map has two rows and four columns.
Here’s the initial setup of a three-variable map, assuming our inputs are named x, y, and z.
Note that the columns are numbered “out of order.” This is intentional, and you must do it this way.

10. Three-Variable K-Map (Cont’d.)
Each square in the map corresponds to one row of a truth table, hence to one minterm.
This map shows which minterm corresponds to each square. But you don’t normally write this yourself. x y z F 1

11. Three-Variable K-Map (Cont’d.)
Here’s another look, with some more information that you don’t need to write yourself.
This marking says that y is true in these columns (and false in in all other columns).
This marking says that z is true in these columns (and false in all other columns).

12. Adjacent Squares
The K-map method relies on our ability to group squares that are adjacent to each other. By “adjacent” we mean side-by-side (such as squares m1 and m3), or above-and-below (such as squares m3 and m7), but not diagonally (such as squares m1 and m7).

13. Why the Funny Numbering?
The “out-of-order” numbering of the columns guarantees that the minterms in adjacent squares differ from each other in exactly one variable.
Examples:
Squares m1 and m3 differ in y but not in x or z.
Squares m3 and m7 differ in x but not in y or z.

14. Wraparound
We regard the right and left edges of the map as wrapping around to touch each other such that:
Squares m0 and m2 are adjacent, differing in y but not in x or z.
Squares m4 and m6 are adjacent, differing in y but not in x or z.
-Hide this slide, and then do Practice Questions 1 and 2.

15. K-Map Procedure For Reading Off Simplified SOP Expression
Set up the K-map, labeling its rows and columns.
From either a truth table or a Boolean expression (preferably in SOP form), place 1s in the appropriate squares.
Circle adjacent 1s in groups of 8, 4, 2, or 1. You want to maximize the size of the circles and minimize the number of circle. Follow this order:
Circle any octet.
Circle any quad that contains one or more 1s that haven’t already been circled, using the minimum number of circles.
Circle any pair that contains one or more 1s that haven’t already been circled, using the minimum number of circles.
Circle any isolated 1s that haven’t already been circled.
Read off the term for each circle by including only those complemented or uncomplemented variables that do not change throughout the circle.
Form the OR sum of the terms generated in Step 4.
Do Practice Question 3.

16. Three-Variable K-Map: Example #1
Using a K-map, find the simplest SOP expression for this truth table. x y z F 1
Do Practice Question 4.
So F = xy′ + x′y

17. Three-Variable K-Map: Example #2
Using a K-map, simplify
F = A′C + A′B + AB′C + BC
Do Practice Question 5.
So F = A′B + C

18. Four-Variable K-Map
A four-variable K-map has four rows and four columns.
Here’s the initial setup of a four-variable map, assuming our inputs are named w, x, y, and z.
Note that columns and rows are numbered “out of order.” You must do it this way.

19. Four-Variable K-Map (Cont’d.)
Each square in the map corresponds to one row of a truth table, hence to one minterm.
This map shows which minterm corresponds to each square. You don’t normally write this yourself. w x y z F 1

20. Four-Variable K-Map (Cont’d.)
Here’s another look, with some more information that you don’t need to write yourself.
This marking says that y is true in these columns (and false in in all other columns).
This marking says that w is true in these rows (and false in all other rows).

21. Why the Funny Numbering?
As with three-variable maps, the “out-of-order” numbering of columns and rows guarantees that the minterms in adjacent squares differ in exactly one variable.
Examples:
Squares m5 and m13 differ in w but not in x, y, or z.
Squares m12 and m13 differ in z but not in w, x, or y.

22. Wraparound
As with three-variable maps, the right and left edges of the map wrap around to touch each other. And so do the top and bottom edges. Examples:
Squares m0 and m8 are adjacent, differing in w but not in x, y, or z.
Squares m4 and m6 are adjacent, differing in y but not in w, x, or z.
-Hide this slide, and then do Practice Questions 6, 7, and 8.

23. Four-Variable K-Map: Example #1w x y z F 1
Using a K-map, find the simplest SOP expression for this truth table.
Do Practice Question 9.
So F = y′ + w′z′ + xz′

24. Four-Variable K-Map: Example #2
Using a K-map, simplify
F = A′B′C′ + B′CD′ + A′BCD′ + AB′C′
Do Practice Question 10.
So F = A′CD′ + B′C′ + B′D′

25. K-Map Procedure For Reading Off Simplified POS Expression
The procedure we’ve been practicing yields a simplified SOP expression. What if you want a simplified POS expression instead?
Instead of writing and circling the 1s from the function’s truth table, write and circle the 0s.
Use the procedure above to read off an expression, which will be the simplified SOP expression for your function’s complement.
Apply generalized DeMorgan’s theorem to the result of Step 2. This will yield an expression for your function, and it will be in POS form.

26. K-Map Procedure For Simplified POS Expression: Example
Find simplified POS expression for
F(A, B, C, D) = ∑(0, 1, 2, 5, 8, 9, 10)
So F′ = AB + CD + BD′
Do Practice Question
So F = (A′+B′)(C′+D′)(B′+D)

27. Don’t-Care Conditions
In some problems we don’t care whether the output of a function is 0 or 1 for some particular combinations of inputs.
This might happen, for instance, if we know that that combination of inputs is physically impossible.
In such cases we’ll put an X rather than a 0 or a 1 in the output column of the truth table.

28. Don’t-Care Conditions: Example
Suppose we’re designing a circuit that controls whether to pump coolant to a machine tool (call this variable P), based on the values of three binary sensors whose signals we’ll call A, B, and C.
Suppose also that it’s physically impossible, based on the machine’s design, for sensors A and C to be on at the same time.
Then we don’t need to worry about the rows in our truth table where A and C are both equal to 1. So we’ll put Xs in those rows of the truth table. See next slide….

29. Don’t-Care Conditions: Example (Cont’d.)
Put 1 in the output column for each combination of sensor readings that should turn the pump on.
Put 0 in the output column for each combination of sensor readings that should turn the pump off.
Put X in the output column for each combination of sensor readings that can never happen. A B C F 1 X

30. Another Example
Suppose our circuit’s inputs come from a thumbwheel whose value can only be set to numbers between 0 (00002) and 9 (10012) .
Then we don’t have to worry about input values of through 11112, because the thumbwheel will never produce such values. So we’ll put Xs in those rows of the truth table. See next slide….

31. Another Example (Cont’d.)w x y z F 1 X
Put 1 in the output column for each input value that should turn your circuit on.
Put 0 in the output column for each input value that should turn your circuit off.
Put X in the output column for each input value that can never happen.

32. Don’t-Care Conditions and K-Maps
When you build a K-map, place an X for each don’t care condition. Then, when you are circling 1s in your map, you can choose to treat each X as either a 0 or a 1—make whichever choice leads to fewer, bigger circles.

33. Don’t-Care Conditions and K-Maps: Example
When you build a K-map, place an X for each don’t care condition. Then, when you are circling 1s in your map, you can choose to treat each X as either a 0 or a 1—make whichever choice leads to fewer, bigger circles.

34. Don’t-Care Conditions and K-Maps: Examplew x y z F X 1
Using a K-map, find the simplest SOP expression for this truth table.
Do Practice Question 11.
So F = w′x′ + yz

35. Universal Gates
NAND gates are sometimes called universal gates because they can be used to produce the other basic Boolean functions. A A A′ AB B
Inverter
AND gate A A
Show how to use Multisim’s logic converter to generate NAND-only logic circuits.
A + B
(A + B)′ B B
OR gate
NOR gate

36. NAND-Only Logic
Because NAND gates are universal gates, and because in most technologies NAND gates are faster and take up less space than AND or OR gates, it’s very common for chips to be manufactured using just NAND gates.
Unfortunately, thinking in terms of NAND is less natural for humans than thinking in terms of AND and OR. Following slides show some tricks to make it easier.
Wakerly: “I won’t go out with you if you’re not clean or not wealthy and also you’re not smart or not friendly” as equivalent to “I’ll go out with you if you’re clean and wealthy, or if you’re smart and friendly.”

37. Two Ways to Draw a NAND Gate: “AND-invert” or “Invert-OR”
DeMorgan’s theorem tells us that the following are equivalent:
Mentally “push” the bubble through from input to output (or vice versa), changing the shape as you do so.

38. Two-Level NAND Implementation
It’s easy to draw a two-level NAND implementation for any SOP expression. Recall that an SOP expression translates directly into a two-level AND-OR circuit such as:
Do Practice Question
1. Insert extra inverter pairs.
2. Then “absorb” the inverters into the gates.

39. NOR Is Also a Universal Gate
What we’ve seen for NAND gates also applies to NOR gates.
First, NOR gates are universal gates and can form all of the basic gates: A A A′
A + B B
Inverter
OR gate A A AB
(AB)′ B B
AND gate
NAND gate

40. Two Ways to Draw a NOR Gate: “OR-invert” or “Invert-AND”
DeMorgan’s theorem also tells us that the following are equivalent:
Again, mentally “push” the bubble through, changing the shape as you do so.

41. NOR-Only Logic
Using the same sort of techniques we saw above for NAND gates, the book shows how to implement Boolean expressions using only NOR gates.
I won’t hold you responsible for this.

42. Summary: Alternative Symbols for Inverter, NAND, and NOR

43. Other Two-Level Implementations
In addition to AND-OR, OR-AND, NAND-NAND, and NOR-NOR implementations, any Boolean expression can also be implemented in several other two-level forms, as discussed in Section 3.7.
While this is interesting from a theoretical standpoint, it’s not that useful in practice, so I won’t hold you responsible for it.

44. Inclusive-OR versus Exclusive-OR in the English Language
Suppose you invite me to a party at your house and I ask you if you have any cake or ice cream in your fridge. Assuming you tell the truth, how would you answer in each of the following cases?
Here we’re using OR in the inclusive sense.
Cake
Ice Cream
Your Answer No
Yes

45. Inclusive-OR versus Exclusive-OR in the English Language (Cont’d)
Suppose you’ve taken your child to a birthday party and it’s time for dessert. You say to the kid, “You can have cake or ice cream. Which one do you want?” Which of the following cases are you allowing?
Here you’re using OR in the exclusive sense.
Cake
Ice Cream
Allowed? No
Yes

46. A X A X
The XOR Gate B B
The XOR gate produces a HIGH output only when the inputs are at opposite logic levels. The truth table is
0 0
0 1
1 0
1 1 1
-XOR gate has only two inputs.
The XOR operation is written as X = A′B + AB′. Alternatively, it can be written with a circled plus sign between the variables as X = A + B.

47. Applying Boolean-Algebra Rules to Expressions with XOR
The “circle-OR” symbol doesn’t appear in any of our Boolean-algebra rules, so if you’re doing a Boolean simplifcation, write out the longer form shown above for XORs.
We’ll do an example later.

48. The XOR Gate Example waveforms: A B X
Notice that the XOR gate will produce a HIGH only when exactly one input is HIGH.
Do as Practice Question 3.

49. Fixed Function Logic
Some common gate configurations are shown.

50. The XNOR operation can be written as X = A′B′ + AB or as X = A + B.
The XNOR Gate B B
The XNOR gate produces a HIGH output only when the inputs are at the same logic level. The truth table is
0 0
0 1
1 0
1 1 1
-XNOR gate has only two inputs.
-As noted earlier, the “circle-OR” symbol doesn’t appear in any of our Boolean rules, so if you’re doing a Boolean derivation, write out the longer form shown above.
The XNOR operation can be written as X = A′B′ + AB or as X = A + B.

51. The XNOR Gate Example waveforms: A B X
Notice that the XNOR gate will produce a HIGH when both inputs are the same. This makes it useful for comparison functions.
Do as Practice Question 4.

52. More Than Two Inputs?
Some textbooks (including ours) say that XOR and XNOR gates can only have two inputs.
Other textbooks say they can have more than two inputs, but there’s some ambiguity in how exactly to define this.
The most common way is to say that XOR output is HIGH if and only if an odd number of inputs are HIGH.
You won’t find XOR or XNOR chips that have more than two inputs, but Multisim has XOR and XNOR gates with more than two inputs.

53. Applications of XOR and XNOR Gates
Three common applications:
Comparators
Controlled inverters
Parity generation and checking

54. Convention for Multi-Bit Strings
When dealing with multi-bit binary strings, we use subscripts to refer to the individual bits in the string.
The least significant bit (LSB) always gets the smallest subscript, which may be either 1 or 0.
Example: In a four-bit string A, the bits may be labeled either A4A3A2A1 or A3A2A1A0 54

55. Application #1: Comparator
A comparator compares two strings of bits to see whether they are equal to each other:
Example: if string A = 0101 and string B = 0100, then A≠ B.
Next slide shows how to build a 4-bit comparator from XNOR gates.

56. Comparator Circuit
Look again at XNOR truth-table, particularly 1st and 4th rows. Do Practice Questions 5, 6.

57. Application #2: Controlled Inverter
A controlled inverter takes an input string and, depending on the logic level on a control line, either
Leaves the string unchanged or
Inverts each bit in the string
Next slide shows how to build an 8-bit controlled inverter from XOR gates.

58. Controlled Inverter
Do as Practice Questions 7, 8.

59. Making Connections to the Altera Board
In Lab 6 you’ll use the o’scope to display a signal being generated on the Altera board.
To do this you must assign the correct pin number in Quartus to send the signal out to a pin on the Altera board’s 14-pin general-purpose I/O connector. Then you must plug in our home-made connector block and use an oscilloscope probe to measure the pin to which you sent your signal.
See next slide for pin numbers.

60. Making Connections to the Altera Board (Cont’d.)
Here are the pin numbers to use when you assign pin numbers in Quartus:
PIN_D9
PIN_E10
PIN_F14
PIN_H14
PIN_H13
PIN_J14
PIN_J10
This one is GROUND.

61. Printing from Our Oscilloscopes
You can print the oscilloscope screen by pressing the PRINT button.
There’s a delay of about 40 seconds before the page will print, so be patient.
Only one oscilloscope can print at a time, or else the printer gets confused and prints hundreds of pages.
Please shout “Printing!” before you press the PRINT button, and make sure that you don’t print while someone else is waiting for their page to print.

62. Example Solution Application #3: Parity Checking
Parity checking is a method of error detection for simple transmission errors involving one bit. A parity bit is an “extra” bit attached to a group of bits to force the total number of 1’s to be either even (even parity) or odd (odd parity).
Example
The ASCII character for “a” is and for “A” is What is the correct bit to append to make both of these have odd parity?
Recall ASCII table on p. 19.
Solution
The ASCII “a” has an odd number of bits that are equal to 1; therefore the parity bit is 0. The ASCII “A” has an even number of bits that are equal to 1; therefore the parity bit is 1.

63. Parity Generators
To implement parity checking, we need circuitry on the sending end that generates the parity bit for each group of bits being sent. This circuitry is called a parity generator.
Next slide shows how to build 4-bit even or odd parity generators.

64. Parity Generators
Do as Practice Questions 9, 10.

65. Parity Checkers
On the receiving end, we need circuitry that checks the data bits and parity bit as they’re received to decide whether an error has occurred during transmission. This circuitry is called a parity checker.
Next slide shows how to build a 4-bit-plus-parity even parity checker.

66. Five-Bit Even Parity Checker

67. A Parity Generator/Checker Chip
74280 Nine-bit Parity Generator/Checker
Most chips we’ve studied have been SSI (small-scale integration) chips containing fewer than 10 gates that are not connected to each other.
The is an MSI (medium-scale integration) chip. Instead of containing a few disconnected gates, it contains about 45 gates connected internally on the chip to perform a specific function.
-Follow links for pin-out and logic symbol.
-Mention presence of and other macrofunctions in Quartus

68. Parity Error Detection System

69. Parity Generator/Checker Chip
The can be used to generate a parity bit or to check an incoming data stream for even or odd parity.
Checker: The can test codes with up to 9 bits. The even output will be HIGH if the data lines have even parity; otherwise it will be LOW. Likewise, the odd output will be HIGH if the data lines have odd parity; otherwise it will be LOW.
Data inputs
S Even
S Odd
Do Practice Question 11.
Generator: To generate even parity, the parity bit is taken from the odd parity output. To generate odd parity, the output is taken from the even parity output.
74280

70. Applying Boolean-Algebra Rules to Expressions with XOR or XNOR
Recall that the “circle-OR” symbol doesn’t appear in any of our Boolean-algebra rules, so if you’re doing a Boolean simplifcation, write out the longer forms shown earlier for XOR and XNOR:
Rewrite X = A + B as X = A′B + AB′.
Rewrite X = A + B as X = A′B′ + AB.
Example: Use Boolean algebra to simplify:
Do as Practice Question 12.

71. Another Example
Example: Use Boolean algebra to simplify:

72. Add new HDL slides here on Sections 3.9, 3.10