1. Instructor: Alexander Stoytchev
CprE 281: Digital Logic
Instructor: Alexander Stoytchev

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

3. Administrative Stuff HW5 is out It is due on Monday Oct 2 @ 4pm.
Please write clearly on the first page (in block capital letters) the following three things:
Your First and Last Name
Your Student ID Number
Your Lab Section Letter
Also, please staple all of your pages together.

4. Administrative Stuff Labs Next Week Mini-Project
This one is worth 3% of your grade.
Make sure to get all the points.
2017_Fall_281/labs/Project-Mini/

5. Quick Review

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

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

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

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

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

11. Analogy with addition in base 10
c3 c2 c1 c0
x2 x1 x0
y2 y1 y0
s2 s1 s0 +

12. Analogy with addition in base 10 +

13. Analogy with addition in base 10
carry +

14. Analogy with addition in base 10
c3 c2 c1 c0
x2 x1 x0
y2 y1 y0
s2 s1 s0 +

15. Problem Statement and Truth Table
[ Figure 3.2b from the textbook ]
[ Figure 3.3a 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. (si can be implemented in two different ways)
XOR Magic
(si can be implemented in two different ways)

21. A decomposed implementation of the full-adder circuitHA c x i HA c c y i + 1 i
(a) Block diagram c i s i x i y i c i + 1
(b) Detailed diagram
[ Figure 3.4 from the textbook ]

22. A decomposed implementation of the full-adder circuitHA c x i HA c c y i + 1 i
(a) Block diagram c i s i x i y i c HA i + 1 HA
(b) Detailed diagram
[ Figure 3.4 from the textbook ]

23. The Full-Adder AbstractionHA c x i c i 1 + HA c y i

24. The Full-Adder AbstractionFA x i c i 1 + y i

25. We can place the arrows anywherexi yi
ci+1 FA ci si

26. n-bit ripple-carry adderx y x y x y
n – 1 n – 1 1 1 c 1 c c n FA c n ” 1 2 FA FA c s s s n – 1 1
MSB position
LSB position
[ Figure 3.5 from the textbook ]

27. n-bit ripple-carry adder abstractionx y x y x y
n – 1 n – 1 1 1 c 1 c c n FA c c n ” 1 2 FA FA s s s n – 1 1
MSB position
LSB position

28. n-bit ripple-carry adder abstractionx y x y x y
n – 1 n – 1 1 1 c c n s s s n – 1 1

29. 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 – s

31. Math Review: Subtraction39 15 - ??

32. Math Review: Subtraction39 15 - 24

33. Math Review: Subtraction82 61 - ?? 48 26 - ?? 32 11 - ??

34. Math Review: Subtraction82 61 - 21 48 26 - 22 32 11 - 21

36. Math Review: Subtraction82 64 - ?? 48 29 - ?? 32 13 - ??

37. Math Review: Subtraction82 64 - 18 48 29 - 19 32 13 - 19

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

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

41. Another Way to Do Subtraction
82 – 64 = –

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

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

44. Another Way to Do Subtraction
82 – 64 = –
= (100 – 64) - 100
= ( – 64) - 100
= (99 – 64)

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

46. 9’s Complement (subtract each digit from 9)99 64 - 35

47. 10’s Complement (subtract each digit from 9 and add 1 to the result)99 64 -
= 36

48. Another Way to Do Subtraction
82 – 64 = (99 – 64)

49. Another Way to Do Subtraction
9’s complement
82 – 64 = (99 – 64)

50. Another Way to Do Subtraction
9’s complement
82 – 64 = (99 – 64) =

51. Another Way to Do Subtraction
9’s complement
82 – 64 = (99 – 64)
10’s complement =

52. Another Way to Do Subtraction
9’s complement
82 – 64 = (99 – 64)
10’s complement = =

53. Another Way to Do Subtraction
9’s complement
82 – 64 = (99 – 64)
10’s complement = =
// Add the first two. =

54. Another Way to Do Subtraction
9’s complement
82 – 64 = (99 – 64)
10’s complement = =
// Add the first two. =
// Just delete the leading 1.
// No need to subtract 100.
= 18

56. Formats for representation of integersb b b n – 1 1
Magnitude
MSB
(a) Unsigned number b b b b n – 1 n – 2 1
Magnitude
Sign
0 denotes +
1 denotes –
MSB
(b) Signed number
[ Figure 3.7 from the textbook ]

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

58. 1’s complement
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 2n – 1, namely
K = (2n – 1) – P
This means that K can be obtained by inverting all bits of P.

59. Find the 1’s complement of …

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

61. A) Example of 1’s complement addition( + 5 ) + ( + 2 ) + ( + 7 )
[ Figure 3.8 from the textbook ]

62. A) Example of 1’s complement addition( + 5 ) + ( + 2 ) + ( + 7 )

63. B) Example of 1’s complement addition( - 5 ) + ( + 2 ) + ( - 3 )
[ Figure 3.8 from the textbook ]

64. B) Example of 1’s complement addition( - 5 ) + ( + 2 ) + ( - 3 )

65. C) Example of 1’s complement addition( + 5 ) + ( – 2 ) + ( + 3 ) 1
[ Figure 3.8 from the textbook ]

66. C) Example of 1’s complement addition( + 5 ) + ( – 2 ) + ( + 3 ) 1

67. C) Example of 1’s complement addition( + 5 ) + ( – 2 ) +
But this is 2! ( + 3 ) 1

68. C) Example of 1’s complement addition( + 5 ) + ( – 2 ) + ( + 3 ) 1 1
We need to perform one
more addition to get the result.

69. C) Example of 1’s complement addition( + 5 ) + ( – 2 ) + ( + 3 ) 1 1
We need to perform one
more addition to get the result.

70. D) Example of 1’s complement addition( – 5 ) + ( – 2 ) + ( – 7 ) 1
[ Figure 3.8 from the textbook ]

71. D) Example of 1’s complement addition( – 5 ) + ( – 2 ) + ( – 7 ) 1

72. D) Example of 1’s complement addition( – 5 ) + ( – 2 ) +
But this is +7! ( – 7 ) 1

73. D) Example of 1’s complement addition( – 5 ) + ( – 2 ) + ( – 7 ) 1 1
We need to perform one
more addition to get the result.

74. D) Example of 1’s complement addition( – 5 ) + ( – 2 ) + ( – 7 ) 1 1
We need to perform one
more addition to get the result.

75. 2’s complement
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 2n , namely
K = 2n – P

76. Deriving 2’s complement
For a positive n-bit number P, let K1 and K2 denote its 1’s and 2’s complements, respectively.
K1 = (2n – 1) – P
K2 = 2n – P
Since K2 = K1 + 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.

77. Find the 2’s complement of …

78. Find the 2’s complement of …
Invert all bits.

79. Find the 2’s complement of … 1 + 1 + 1 + 1 +
Then add 1.

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

81. Find the 2’s complement of …

82. Find the 2’s complement of …
Copy all bits that are 0 from right to left.

83. Find the 2’s complement of …
Stop at the first 1. Copy that 1 as well.

84. Find the 2’s complement of …
Invert all remaining bits.

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

86. Interpretation of four-bit signed integers
The top half is the same in all three representations.
It corresponds to the positive integers.

87. Interpretation of four-bit signed integers
In all three representations the first bit represents the sign.
If that bit is 1, then the number is negative.

88. Interpretation of four-bit signed integers
Notice that in this representation there are two zeros!

89. Interpretation of four-bit signed integers
There are two zeros in this representation as well!

90. Interpretation of four-bit signed integers
In this representation there is one more negative number.

91. The number circle for 2's complement
[ Figure 3.11a from the textbook ]

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

93. B) Example of 2’s complement addition+ 2 ( ) 5 – 3
[ Figure 3.9 from the textbook ]

94. C) Example of 2’s complement addition+ 1
ignore 5 ( ) 3 2 –
[ Figure 3.9 from the textbook ]

95. D) Example of 2’s complement addition+ 1
ignore 5 – ( ) 7 2
[ Figure 3.9 from the textbook ]

97. Naming Ambiguity: 2's Complement
2's complement has two different meanings:
representation for signed integer numbers
algorithm for computing the 2's complement (regardless of the representation of the number)

98. Naming Ambiguity: 2's Complement
2's complement has two different meanings:
representation for signed integer numbers
in 2's complement
algorithm for computing the 2's complement (regardless of the representation of the number)
take the 2's complement

99. Example of 2’s complement subtraction( + 5 ) – ( + 2 ) – + ( + 3 ) 1
ignore
means take the 2's complement
[ Figure 3.10 from the textbook ]

100. Example of 2’s complement subtraction( + 5 ) – ( + 2 ) – + ( + 3 ) 1
ignore
Notice that the minus changes to a plus.
means take the 2's complement
[ Figure 3.10 from the textbook ]

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

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

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

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

105. Example of 2’s complement subtraction( + 5 ) – ( – 2 ) – + ( + 7 )
[ Figure 3.10 from the textbook ]

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

107. Taking the 2’s complement negates the number
decimal
b3 b2 b1 b0
take the 2's complement +7
0111
1001 -7 +6
0110
1010 -6 +5
0101
1011 -5 +4
0100
1100 -4 +3
0011
1101 -3 +2
0010
1110 -2 +1
0001
1111 -1 +0
0000 -8
1000

108. Taking the 2’s complement negates the number
decimal
b3 b2 b1 b0
take the 2's complement +7
0111
1001 -7 +6
0110
1010 -6 +5
0101
1011 -5 +4
0100
1100 -4 +3
0011
1101 -3 +2
0010
1110 -2 +1
0001
1111 -1 +0
0000 -8
1000
This is
the only
exception

109. Taking the 2’s complement negates the number
decimal
b3 b2 b1 b0
take the 2's complement +7
0111
1001 -7 +6
0110
1010 -6 +5
0101
1011 -5 +4
0100
1100 -4 +3
0011
1101 -3 +2
0010
1110 -2 +1
0001
1111 -1 +0
0000 -8
1000
And this
one too.

110. But that exception does not matter( – 5 ) – ( – 8 ) – + ( + 3 )
ignore

111. But that exception does not matter
Add 8

112. But that exception does not matter
Subtract 8

113. 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 !!!

114. Adder/subtractor unity y y n – 1 1
Add
Sub
control x x x n – 1 1 c n
-bit adder c n s s s n – 1 1
[ Figure 3.12 from the textbook ]

115. XOR Tricks y
control
out

116. XOR as a repeater y

117. XOR as an inverter y 1

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

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

120. Addition: when control = 0y y y n – 1 1
Add
Sub
control x x x n – 1 1
yn-1 … y1 y0 c n
-bit adder c n s s s n – 1 1
[ Figure 3.12 from the textbook ]

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

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

123. Subtraction: when control = 1y y y n – 1 1 1
Add
Sub
control 1 1 1 x x x n – 1 1
yn-1 … y1 y0 c n
-bit adder c n s s s n – 1 1
[ Figure 3.12 from the textbook ]

124. Subtraction: when control = 1y y y n – 1 1 1
Add
Sub
control 1 1 1 x x x n – 1 1
yn-1 … y1 y0 c n
-bit adder c n 1
carry for the
first column! s s s n – 1 1
[ Figure 3.12 from the textbook ]

125. Examples of determination of overflow( + 7 ) ( – 7 ) + + + ( + 2 ) + ( + 2 ) ( + 9 ) ( – 5 ) ( + 7 ) ( – 7 ) + + + ( – 2 ) + ( – 2 ) ( + 5 ) 1 ( – 9 ) 1
[ Figure 3.13 from the textbook ]

126. Examples of determination of overflow ( + 7 ) ( – 7 ) + + + ( + 2 ) + ( + 2 ) ( + 9 ) ( – 5 ) ( + 7 ) ( – 7 ) + + + ( – 2 ) + ( – 2 ) ( + 5 ) 1 ( – 9 ) 1
Include the carry bits: c4 c3 c2 c1 c0

127. Examples of determination of overflow ( + 7 ) ( – 7 ) + + + ( + 2 ) + ( + 2 ) ( + 9 ) ( – 5 ) ( + 7 ) ( – 7 ) + + + ( – 2 ) + ( – 2 ) ( + 5 ) 1 ( – 9 ) 1
Include the carry bits: c4 c3 c2 c1 c0

128. Examples of determination of overflowc 4 = 3 1 c 4 = 3 ( + 7 ) ( – 7 ) + + + ( + 2 ) + ( + 2 ) ( + 9 ) ( – 5 ) c 4 1 = 3 c 4 1 = 3 ( + 7 ) ( – 7 ) + + + ( – 2 ) + ( – 2 ) ( + 5 ) 1 ( – 9 ) 1
Include the carry bits: c4 c3 c2 c1 c0

129. Examples of determination of overflowc 4 = 3 1 c 4 = 3 ( + 7 ) ( – 7 ) + + + ( + 2 ) + ( + 2 ) ( + 9 ) ( – 5 ) c 4 1 = 3 c 4 1 = 3 ( + 7 ) ( – 7 ) + + + ( – 2 ) + ( – 2 ) ( + 5 ) 1 ( – 9 ) 1
Overflow occurs only in these two cases.

130. Examples of determination of overflowc 4 = 3 1 c 4 = 3 ( + 7 ) ( – 7 ) + + + ( + 2 ) + ( + 2 ) ( + 9 ) ( – 5 ) c 4 1 = 3 c 4 1 = 3 ( + 7 ) ( – 7 ) + + + ( – 2 ) + ( – 2 ) ( + 5 ) 1 ( – 9 ) 1
Overflow = c3c4 + c3c4

131. Examples of determination of overflowc 4 = 3 1 c 4 = 3 ( + 7 ) ( – 7 ) + + + ( + 2 ) + ( + 2 ) ( + 9 ) ( – 5 ) c 4 1 = 3 c 4 1 = 3 ( + 7 ) ( – 7 ) + + + ( – 2 ) + ( – 2 ) ( + 5 ) 1 ( – 9 ) 1
Overflow = c3c4 + c3c4
XOR

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

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

134. Detecting Overflow x y x y x y c c c FA c FA FA c s s s n – 1 n – 1 1c 1 c c FA c FA FA c n n ” 1 2 s s s n – 1 1

135. Detecting Overflow (with one extra XOR)y x y x y
n – 1 n – 1 1 1 c 1 c c FA c FA FA c n n ” 1 2 s s s n – 1 1
overflow

136. Another way to look at the overflow issue
X= x3 x2 x1 x0
Y= y3 y2 y1 y0
S= s3 s2 s1 s0 +

137. Another way to look at the overflow issue
X= x3 x2 x1 x0
Y= y3 y2 y1 y0
S= s3 s2 s1 s0 +
If both numbers that we are adding have the same sign
but the sum does not, then we have an overflow.

138. Examples of determination of overflow( + 7 ) ( – 7 ) + + + ( + 2 ) + ( + 2 ) ( + 9 ) ( – 5 ) ( + 7 ) ( – 7 ) + + + ( – 2 ) + ( – 2 ) ( + 5 ) 1 ( – 9 ) 1

139. Examples of determination of overflow( + 7 ) ( – 7 ) + + + ( + 2 ) + ( + 2 ) ( + 9 ) ( – 5 ) ( + 7 ) ( – 7 ) + + + ( – 2 ) + ( – 2 ) ( + 5 ) 1 ( – 9 ) 1

140. Examples of determination of overflow3 = y s 1 x 3 1 = y s ( + 7 ) ( – 7 ) + + + ( + 2 ) + ( + 2 ) ( + 9 ) ( – 5 ) x 3 = y 1 s x 3 1 = y s ( + 7 ) ( – 7 ) + + + ( – 2 ) + ( – 2 ) ( + 5 ) 1 ( – 9 ) 1

141. Examples of determination of overflow3 = y s 1 x 3 1 = y s ( + 7 ) ( – 7 ) + + + ( + 2 ) + ( + 2 ) ( + 9 ) ( – 5 ) x 3 = y 1 s x 3 1 = y s ( + 7 ) ( – 7 ) + + + ( – 2 ) + ( – 2 ) ( + 5 ) 1 ( – 9 ) 1
In 2's complement, both +9 and -9 are not representable with 4 bits.

142. Examples of determination of overflow3 = y s 1 x 3 1 = y s ( + 7 ) ( – 7 ) + + + ( + 2 ) + ( + 2 ) ( + 9 ) ( – 5 ) x 3 = y 1 s x 3 1 = y s ( + 7 ) ( – 7 ) + + + ( – 2 ) + ( – 2 ) ( + 5 ) 1 ( – 9 ) 1
Overflow occurs only in these two cases.

143. Examples of determination of overflow3 = y s 1 x 3 1 = y s ( + 7 ) ( – 7 ) + + + ( + 2 ) + ( + 2 ) ( + 9 ) ( – 5 ) x 3 = y 1 s x 3 1 = y s ( + 7 ) ( – 7 ) + + + ( – 2 ) + ( – 2 ) ( + 5 ) 1 ( – 9 ) 1
Overflow = x3 y3 s3 + x3 y3 s3

144. Another way to look at the overflow issue
X= x3 x2 x1 x0
Y= y3 y2 y1 y0
S= s3 s2 s1 s0 +
If both numbers that we are adding have the same sign
but the sum does not, then we have an overflow.
Overflow = x3 y3 s3 + x3 y3 s3

145. Questions?

146. THE END