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Lecture – 09
Fixed-Point Arithmetic
 Addition
 Subtraction
 Multiplication
 Division
Half Adder
X0 Y0 S0 C0
0 0 0 0
0 1 1 0
1 0 1 0
1 1 0 1C0
S0
X0
Y0
000
000
YXC
YXS


Full Adder
1010000
1000




CYCXYXC
CYXS
Full Adder
Serial Binary Adder
 Least expensive
circuit in terms of
hardware cost.
 It adds the numbers bit
by bit and so requires
n clock cycle to
compute the sum of
two n-bit numbers.
 Circuit size
independent of n.
Ripple Carry Adder
 A 1 appearing on the carry in line of a 1-bit adder cause
it to generate a 1 on its carry out line. So, the carry
signal propagate through the adder from right to left.
 The maximum signal propagation delay is nd, where d is
the delay of a full-adder stage.
 The amount of hardware increase linearly with n.
2’s Complement Adder-Subtracter
 When s =0, then X  s = X
 When s=1, then X  s = X
2’s Complement Adder-
Subtracter
Overflow
 When the result of an arithmetic operation exceeds the
standard word size n, overflow occurs.
 Example: let n=8, X=11101011=23510 and Y=00101010=4210
Z = X+Y = 11101011 + 0101010 = 00010101 = 2110
with Cn-1 = C7 = 1.
C7Z = 100010101 = 27710 = 25610 + 2110
 The result of an addition simply wraps around when the
largest number 2n-1 is exceeds.
 For n, the number range for unsigned number is 0 to 2n-1
Overflow
 We can never have overflow on adding a negative and
positive number.
 Example: let n=8 X=11101011=-2110 and
Y=00101010=+4210
Z = X+Y = 00010101 = 2110 and C7 = 1.
So, Cn-1 = 1 does not indicate overflow.
 Overflow in 2’s complement addition can result from adding
1) two positive numbers or
2) two negative numbers.
Overflow
Case 1: Two numbers are positive.
Let n = 4, 7 + 3 = 0111+0011 = 1010 so, cn-2 = 1
Cn-2 =1 indicates that the magnitude of the sum exceeds the n-1 bits
allocated to it.
Case 2: Two numbers are negative.
Let n=4, -7 = 1001, -3 = 1101
so, 1001+1101 = 10110 so, cn-2 = 0
Cn-2 =0 indicates the overflow.
Overflow
Xn-1 Yn-1 Cn-2 Zn-1 v
0 0 0 0 0
0 0 1 0 1
0 1 0 1 0
0 1 1 0 0
1 0 0 1 0
1 0 1 0 0
1 1 0 1 1
1 1 1 1 0
21
211211
021021021 ............:......






nn
nnnnnn
nnnnnn
CCv
CYXCYXv
YYYXXXZZZ
Complete 2’s Complement Adder-Subtracter
Complete 2’s Complement Adder-Subtracter

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Lecture 09

  • 2. Fixed-Point Arithmetic  Addition  Subtraction  Multiplication  Division
  • 3. Half Adder X0 Y0 S0 C0 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1C0 S0 X0 Y0 000 000 YXC YXS  
  • 6. Serial Binary Adder  Least expensive circuit in terms of hardware cost.  It adds the numbers bit by bit and so requires n clock cycle to compute the sum of two n-bit numbers.  Circuit size independent of n.
  • 7. Ripple Carry Adder  A 1 appearing on the carry in line of a 1-bit adder cause it to generate a 1 on its carry out line. So, the carry signal propagate through the adder from right to left.  The maximum signal propagation delay is nd, where d is the delay of a full-adder stage.  The amount of hardware increase linearly with n.
  • 8. 2’s Complement Adder-Subtracter  When s =0, then X  s = X  When s=1, then X  s = X
  • 10. Overflow  When the result of an arithmetic operation exceeds the standard word size n, overflow occurs.  Example: let n=8, X=11101011=23510 and Y=00101010=4210 Z = X+Y = 11101011 + 0101010 = 00010101 = 2110 with Cn-1 = C7 = 1. C7Z = 100010101 = 27710 = 25610 + 2110  The result of an addition simply wraps around when the largest number 2n-1 is exceeds.  For n, the number range for unsigned number is 0 to 2n-1
  • 11. Overflow  We can never have overflow on adding a negative and positive number.  Example: let n=8 X=11101011=-2110 and Y=00101010=+4210 Z = X+Y = 00010101 = 2110 and C7 = 1. So, Cn-1 = 1 does not indicate overflow.  Overflow in 2’s complement addition can result from adding 1) two positive numbers or 2) two negative numbers.
  • 12. Overflow Case 1: Two numbers are positive. Let n = 4, 7 + 3 = 0111+0011 = 1010 so, cn-2 = 1 Cn-2 =1 indicates that the magnitude of the sum exceeds the n-1 bits allocated to it. Case 2: Two numbers are negative. Let n=4, -7 = 1001, -3 = 1101 so, 1001+1101 = 10110 so, cn-2 = 0 Cn-2 =0 indicates the overflow.
  • 13. Overflow Xn-1 Yn-1 Cn-2 Zn-1 v 0 0 0 0 0 0 0 1 0 1 0 1 0 1 0 0 1 1 0 0 1 0 0 1 0 1 0 1 0 0 1 1 0 1 1 1 1 1 1 0 21 211211 021021021 ............:......       nn nnnnnn nnnnnn CCv CYXCYXv YYYXXXZZZ
  • 14. Complete 2’s Complement Adder-Subtracter
  • 15. Complete 2’s Complement Adder-Subtracter