1. Instructor: Alexander Stoytchev http://www.ece.iastate.edu/~alexs/classes/ CprE 281: Digital Logic

2. Signed Numbers CprE 281: Digital Logic Iowa State University, Ames, IA Copyright © Alexander Stoytchev

3. Administrative Stuff HW 5 is due today

4. Administrative Stuff No homework is due next week.

5. Administrative Stuff Labs Next Week Mini-Project This one is worth 3% of your grade. Make sure to get all the points.

6. Administrative Stuff Midterm Exam #1 When: Monday Feb 24. Where: This classroom What: Chapter 1 and Chapter 2 plus number systems The exam will be open book and open notes (you can bring up to 3 pages of handwritten notes).

7. Topics for the Midterm Exam Binary Numbers Octal Numbers Hexadecimal Numbers Conversion between the different number systems Truth Tables Boolean Algebra Logic Gates Circuit Synthesis with AND, OR, NOT Circuit Synthesis with NAND, NOR Converting an AND/OR/NOT circuit to NAND circuit Converting an AND/OR/NOT circuit to NOR circuit SOP and POS expressions

8. Topics for the Midterm Exam Mapping a Circuit to Verilog code Mapping Verilog code to a circuit Multiplexers Venn Diagrams K-maps for 2, 3, and 4 variables Minimization of Boolean expressions using theorems Minimization of Boolean expressions with K-maps Incompletely specified functions (with don’t cares) Functions with multiple outputs

9. Quick Review

10. [ Figure 3.1a from the textbook ] Adding two bits (there are four possible cases)

11. [ Figure 3.1b from the textbook ] Adding two bits (the truth table)

12. [ Figure 3.1c from the textbook ] Adding two bits (the logic circuit)

13. [ Figure 3.1c-d from the textbook ] The Half-Adder

14. Bit position i Addition of multibit numbers [ Figure 3.2 from the textbook ]

15. Problem Statement and Truth Table [ Figure 3.3a from the textbook ][ Figure 3.2b from the textbook ]

16. Let’s fill-in the two K-maps [ Figure 3.3a-b from the textbook ]

17. Let’s fill-in the two K-maps [ Figure 3.3a-b from the textbook ]

18. The circuit for the two expressions [ Figure 3.3c from the textbook ]

19. This is called the Full-Adder [ Figure 3.3c from the textbook ]

20. XOR Magic (s i can be implemented in two different ways)

21. HA s c s c c i x i y i c i1+ s i c i x i y i c i1+ s i (a) Block diagram (b) Detailed diagram A decomposed implementation of the full-adder circuit [ Figure 3.4 from the textbook ]

22. HA s c s c c i x i y i c i1+ s i The Full-Adder Abstraction

23. FA c i x i y i c i1+ s i The Full-Adder Abstraction

24. FA We can place the arrows anywhere xixi yiyi sisi c i+1 cici

25. FA x n –1 c n c n1” y n1– s n1– FA x 1 c 2 y 1 s 1 c 1 x 0 y 0 s 0 c 0 MSB positionLSB position n-bit ripple-carry adder [ Figure 3.5 from the textbook ]

26. FA x n –1 c n c n1” y n1– s n1– FA x 1 c 2 y 1 s 1 c 1 x 0 y 0 s 0 c 0 MSB positionLSB position n-bit ripple-carry adder abstraction

27. x n –1 c n y n1– s n1– x 1 y 1 s 1 x 0 y 0 s 0 c 0 n-bit ripple-carry adder abstraction

28. The x and y lines are typically grouped together for better visualization, but the underlying logic remains the same x n –1 c n y n1– s n1– x 1 y 1 s 1 y 0 s 0 c 0 x 0

30. Math Review: Subtraction 39 15 - ??

31. Math Review: Subtraction 39 15 - 24

32. Math Review: Subtraction 82 61 - ?? 48 26 - ?? 32 11 - ??

33. Math Review: Subtraction 82 61 - 21 48 26 - 22 32 11 - 21

35. Math Review: Subtraction 82 64 - ?? 48 29 - ?? 32 13 - ??

36. Math Review: Subtraction 82 64 - 18 48 29 - 19 32 13 - 19

38. The problems in which row are easier to calculate? 82 64 - ?? 48 29 - ?? 32 13 - ?? 82 61 - ?? 48 26 - ?? 32 11 - ??

39. The problems in which row are easier to calculate? 82 64 - 18 48 29 - 19 32 13 - 19 82 61 - 21 48 26 - 22 32 11 - 21 Why?

40. Another Way to Do Subtraction 82 – 64 = 82 + 100 – 100 - 64

41. Another Way to Do Subtraction 82 – 64 = 82 + 100 – 100 - 64 = 82 + (100 – 64) - 100

42. Another Way to Do Subtraction 82 – 64 = 82 + 100 – 100 - 64 = 82 + (100 – 64) - 100 = 82 + (99 + 1 – 64) - 100

43. Another Way to Do Subtraction 82 – 64 = 82 + 100 – 100 - 64 = 82 + (100 – 64) - 100 = 82 + (99 + 1 – 64) - 100 = 82 + (99 – 64) +1 - 100

44. Another Way to Do Subtraction 82 – 64 = 82 + 100 – 100 - 64 = 82 + (100 – 64) - 100 = 82 + (99 + 1 – 64) - 100 = 82 + (99 – 64) +1 - 100 Does not require borrows

45. 9’s Complement (subtract each digit from 9) 99 64 - 35

46. 10’s Complement (subtract each digit from 9 and add 1 to the result) 99 64 - 35 + 1 = 36

47. Another Way to Do Subtraction 82 – 64 = 82 + (99 – 64) +1 - 100

48. Another Way to Do Subtraction 82 – 64 = 82 + (99 – 64) +1 - 100 9’s complement

49. Another Way to Do Subtraction 82 – 64 = 82 + (99 – 64) +1 - 100 = 82 + 35 + 1 - 100 9’s complement

50. Another Way to Do Subtraction 82 – 64 = 82 + (99 – 64) +1 - 100 = 82 + 35 + 1 - 100 9’s complement 10’s complement

51. Another Way to Do Subtraction 82 – 64 = 82 + (99 – 64) +1 - 100 = 82 + 35 + 1 - 100 9’s complement = 82 + 36 - 100 10’s complement // add the first two

52. Another Way to Do Subtraction 82 – 64 = 82 + (99 – 64) +1 - 100 = 82 + 35 + 1 - 100 = 118 - 100 9’s complement = 82 + 36 - 100 = 18 10’s complement // add the first two // delete the leading 1

54. b n1– b 1 b 0 Magnitude MSB (a) Unsigned number b n1– b 1 b 0 Magnitude Sign (b) Signed number b n2– 0 denotes 1 denotes + –MSB Formats for representation of integers [ Figure 3.7 from the textbook ]

55. Sign and magnitude 1’s complement 2’s complement Negative numbers can be represented in following ways

56. Let K be the negative equivalent of an n-bit positive number P. Then, in 1’s complement representation K is obtained by subtracting P from 2 n – 1, namely K = (2 n – 1) – P This means that K can be obtained by inverting all bits of P. 1’s complement

57. Find the 1’s complement of … 0 1 0 0 1 0 0 1 1 1 0 0 1 1

58. Find the 1’s complement of … 0 1 0 0 1 0 0 1 1 1 0 0 1 1 1 0 1 1 0 1 1 0 0 0 1 1 0 0 Just flip 1's to 0's and vice versa.

59. Example of 1’s complement addition + 0 1 1 1 0 1 0 0 1 0 5+ () 2+ () 7+ () +

60. Example of 1’s complement addition [ Figure 3.8 from the textbook ] + 1 1 0 0 1 0 0 0 1 0 2+ () 5–  3-  +

61. Example of 1’s complement addition [ Figure 3.8 from the textbook ] + 0 0 1 0 0 1 1 1 0 1 1 1 0 0 1 1 5+ () 3+ () + 2– 

62. Example of 1’s complement addition [ Figure 3.8 from the textbook ] + 0 1 1 1 1 0 1 1 0 1 1 1 1 0 0 0 5–  7–  + 2– 

63. Let K be the negative equivalent of an n-bit positive number P. Then, in 2’s complement representation K is obtained by subtracting P from 2 n, namely K = 2 n – P 2’s complement

64. For a positive n-bit number P, let K 1 and K 2 denote its 1’s and 2’s complements, respectively. K 1 = (2 n – 1) – P K 2 = 2 n – P Since K 2 = K 1 + 1, it is evident that in a logic circuit the 2’s complement can computed by inverting all bits of P and then adding 1 to the resulting 1’s-complement number. Deriving 2’s complement

65. Find the 2’s complement of … 0 1 0 0 1 0 0 1 1 1 0 0 1 1

66. Find the 2’s complement of … 0 1 0 0 1 0 0 1 1 1 0 0 1 1 1 0 1 11 1 1 0 1 0 0 11 1 0 1

67. Quick Way to find 2’s complement Scan the binary number from right to left Copy all bit that are 0 from right to left Stop at the first 1 Copy that 1 as well Invert all remaining bits

68. [ Table 3.2 from the textbook ] Interpretation of four-bit signed integers

69. Example of 2’s complement addition [ Figure 3.9 from the textbook ] + 0 1 1 1 0 1 0 0 1 0 5+ () 2+ () 7+ () +

70. Example of 2’s complement addition [ Figure 3.9 from the textbook ] + 1 1 0 1 1 0 1 1 0 0 1 0 2+ () 5–  3–  +

71. Example of 2’s complement addition [ Figure 3.9 from the textbook ] + 0 0 1 1 0 1 1 1 1 0 1 ignore 5+ () 3+ () + 2– 

72. Example of 2’s complement addition [ Figure 3.9 from the textbook ] + 1 0 0 1 1 0 1 1 1 1 1 0 1 ignore 5–  7–  + 2– 

74. Example of 2’s complement subtraction – 0 1 0 0 1 0 5+ () 2+ () 3+ () – 1 ignore + 0 0 1 1 0 1 1 1 1 0 [ Figure 3.10 from the textbook ]

75. Example of 2’s complement subtraction [ Figure 3.10 from the textbook ] – 1 0 1 1 0 0 1 0 – 1 ignore + 1 0 0 1 1 0 1 1 1 1 1 0 5–  7–  2+ ()

76. Example of 2’s complement subtraction [ Figure 3.10 from the textbook ] – 0 1 1 1 1 0 5+ () 7+ () – + 0 1 1 1 0 1 0 0 1 02– 

77. Example of 2’s complement subtraction [ Figure 3.10 from the textbook ] – 1 0 1 1 1 1 1 0 – + 1 1 0 1 1 0 1 1 0 0 1 02–  5–  3– 

78. [ Figure 3.11 from the textbook ] Graphical interpretation of four-bit 2’s complement numbers

79. Take Home Message Subtraction can be performed by simply adding the 2’s complement of the second number, regardless of the signs of the two numbers. Thus, the same adder circuit can be used to perform both addition and subtraction !!!

80. s 0 s 1 s n1– x 0 x 1 x n1– c n n-bit adder y 0 y 1 y n1– c 0 Add  Sub control [ Figure 3.12 from the textbook ] Adder/subtractor unit

81. XOR Tricks y control out

82. y 0 y XOR as a repeater

83. y 1 y XOR as an inverter

84. s 0 s 1 s n1– x 0 x 1 x n1– c n n-bit adder y 0 y 1 y n1– c 0 Add  Sub control [ Figure 3.12 from the textbook ] Addition: when control = 0

85. s 0 s 1 s n1– x 0 x 1 x n1– c n n-bit adder y 0 y 1 y n1– c 0 Add  Sub control [ Figure 3.12 from the textbook ] Addition: when control = 0 0 0 00

86. s 0 s 1 s n1– x 0 x 1 x n1– c n n-bit adder y 0 y 1 y n1– c 0 Add  Sub control [ Figure 3.12 from the textbook ] Addition: when control = 0 0 0 00 y n-1 y1y1 y0y0 …

87. s 0 s 1 s n1– x 0 x 1 x n1– c n n-bit adder y 0 y 1 y n1– c 0 Add  Sub control [ Figure 3.12 from the textbook ] Subtraction: when control = 1

88. s 0 s 1 s n1– x 0 x 1 x n1– c n n-bit adder y 0 y 1 y n1– c 0 Add  Sub control [ Figure 3.12 from the textbook ] Subtraction: when control = 1 1 1 11

89. s 0 s 1 s n1– x 0 x 1 x n1– c n n-bit adder y 0 y 1 y n1– c 0 Add  Sub control [ Figure 3.12 from the textbook ] Subtraction: when control = 1 1 1 11 y n-1 y1y1 y0y0 …

90. s 0 s 1 s n1– x 0 x 1 x n1– c n n-bit adder y 0 y 1 y n1– c 0 Add  Sub control [ Figure 3.12 from the textbook ] Subtraction: when control = 1 1 1 11 y n-1 y1y1 y0y0 … 1 carry for the first column!

91. ++ 1 0 1 1 1 0 0 1 0 0 1 0 1 0 0 1 0 1 1 1 0 0 1 0 7+ () 2+ () 9+ () + ++ 0 1 1 1 1 0 0 1 1 1 1 0 0 1 0 1 1 1 1 1 1 0 7+ () 5+ () + 2–  11 c 4 0= c 3 1= c 4 0= c 3 0= c 4 1= c 3 1= c 4 1= c 3 0= 2+ () 7–  5–  + 7–  9–  +2–  Examples of determination of overflow [ Figure 3.13 from the textbook ]

92. ++ 1 0 1 1 1 0 0 1 0 0 1 0 1 0 0 1 0 1 1 1 0 0 1 0 7+ () 2+ () 9+ () + ++ 0 1 1 1 1 0 0 1 1 1 1 0 0 1 0 1 1 1 1 1 1 0 7+ () 5+ () + 2–  11 c 4 0= c 3 1= c 4 0= c 3 0= c 4 1= c 3 1= c 4 1= c 3 0= 2+ () 7–  5–  + 7–  9–  +2–  Examples of determination of overflow [ Figure 3.13 from the textbook ] Overflow occurs only in these two cases.

93. ++ 1 0 1 1 1 0 0 1 0 0 1 0 1 0 0 1 0 1 1 1 0 0 1 0 7+ () 2+ () 9+ () + ++ 0 1 1 1 1 0 0 1 1 1 1 0 0 1 0 1 1 1 1 1 1 0 7+ () 5+ () + 2–  11 c 4 0= c 3 1= c 4 0= c 3 0= c 4 1= c 3 1= c 4 1= c 3 0= 2+ () 7–  5–  + 7–  9–  +2–  Examples of determination of overflow [ Figure 3.13 from the textbook ] Overflow occurs only in these two cases. Overflow = c 3 c 4 + c 3 c 4

94. ++ 1 0 1 1 1 0 0 1 0 0 1 0 1 0 0 1 0 1 1 1 0 0 1 0 7+ () 2+ () 9+ () + ++ 0 1 1 1 1 0 0 1 1 1 1 0 0 1 0 1 1 1 1 1 1 0 7+ () 5+ () + 2–  11 c 4 0= c 3 1= c 4 0= c 3 0= c 4 1= c 3 1= c 4 1= c 3 0= 2+ () 7–  5–  + 7–  9–  +2–  Examples of determination of overflow [ Figure 3.13 from the textbook ] Overflow occurs only in these two cases. Overflow = c 3 c 4 + c 3 c 4 XOR

95. Calculating overflow for 4-bit numbers with only three significant bits

96. Calculating overflow for n-bit numbers with only n-1 significant bits

97. Another way to look at the overflow issue

98. If both numbers that we are adding have the same sign but the sum does not, then we have an overflow.

99. Questions?

100. THE END