1. Fixed-Point Arithmetics: Part I

2. Binary Representation
Unsigned Magnitude
Signed Magnitude
Two’s Complement

3. Unsigned Magnitude Only positive number are presented No sign bit
For N bits we can represent the signed integers between 0 and 2N -1
Example: 111 -> 7

4. Signed Magnitude
The most significant bit is used to represent the sign: “1” means negative, “0” means positive
For positive numbers the result is the same as unsigned magnitude representation
For N bits we can represent the signed integers between -2(N-1) +1 and +2(N-1)-1

5. Disadvantage of Signed Magnitude
Two encoding for the zero ‘0’.
Addition of two numbers of opposite sign will not yield 0!!
= (85)H + (05) H = (8A) H = -10!!
Arithmetic operations of numbers with unlike sign and with same sign must be handled differently (subtracter needed with unlike signs)
A – B ≠ A + (-B)
Example: 3 – 2
Addition Operation 3 + (-2) Subtraction Operation (3-2)
0011 (+3) (+3)
(-2) (+2)
1100 (-5) WRONG! (+1) CORRECT!

6. Disadvantage of Signed Magnitude
A hardware adder and subtracter needed
Special treatment also for A – B if B > A
Example 3 – 2 & 2 – 3
Subtraction Operation Subtraction Operation 2 - 3
0011 (3) (+2)
(2) (+3)
0001 (1) CORRECT (-7) WRONG
Solution:
Switch order of operands (3 – 2 instead of 2 – 3)
Perform subtraction (Result = 1)
attach the minus sign (Result = -1)

7. Two’s Complement example
1 = −1 = = 1111
2 = −2 = = 1110
6 = −6 = = 1010
8 = 1000 −8 = = 1000
Note that the max positive number that can be represented is 7! Since to get the negative of -8 we need:
-8 = 1000
To bitwise invert : 0111
Add +1 to LSB: +1
Result:
So –(-8) = -8 (abnormal result)  8 cannot be represented
For N bits, the max number is 2(N-1)-1 while the min number is -2(N-1)

8. Geometric Depiction of Twos Complement Integers

9. Benefits of 2’s Complement
One unique representation of the number 0:
0 =
Complement:
Add +1 to LSB: +1
Result: (Ignoring carry bit)
-0 = +0
Negation: take complement and add +1
Extending word length:
For positive number pack with leading zeros
+3 =011
+3 =
For negative numbers pack with leading ones
-4=100
-4 =111000

10. Benefits of 2’s Complement
Subtraction Rule: to subtract ‘b’ from ‘a’, take the 2’s complement of ‘b’ and add it to ‘a’: a – b = a + (-b) a b b a
Signed Mag
2’s
Compare signs bits
Equal?? No
Adder
Yes
If Subtraction enable
Adder: a + b
Yes
Is a > b No
Result
2’s Comp
Sub: a - b
Swap Operand
Sub: a – b
Negate

11. Benefits of 2’s Complement
Overflow Rule: if two numbers with same sign are added and overflow occurs the result number has different sign:
Addition: = 110
= -2
Subtraction: 110 – 011 = 011
= 3
Multiplication: 011 * 010 = 110
3 * = -2

12. What to Do In case of Overflow
Prevention rule: design your code with minimal possible overflows
DSP can be configured to saturate the number at overflow (N=4):
5 x 2 = 0010 Overflow
5 x 2 = 0111 Saturate
Guard bits in accumulator could be useful for intermediate results
Example: 5 x x 4 = 10 – 4 = 6
If accumulator has one extra bit called “guard bit” the final result will not overflow!

13. Multiplication Example
Multiplicand (11 dec)
x Multiplier (13 dec)
Partial products
Note: if multiplier bit is 1 copy
multiplicand (place value)
otherwise zero
Product (143 dec)
Note: need double length result

14. Unsigned Binary Multiplication

15. Execution of Example

16. Flowchart for Unsigned Binary Multiplication

17. Problem: Multiplication & 2’s Complement
Addition and Subtraction can be treated as unsigned integer
1001 (+9) +
0011 (+3)
1100 (+12)
1001 (-7) +
0011 (+3)
1100 (-4)
2’s Complement
Unsigned
What about multiplication??
1010
(10) x
1110
(14)
0000
(140)
1010
(-6) x
1110
(-2)
0000
(-116)
2’s Complement
Unsigned
Wrong Result!!!

18. Problem: Multiplication & 2’s Complement
Addition and Subtraction can be treated as unsigned integer
1001 (+9) +
0011 (+3)
1100 (+12)
1001 (-7) +
0011 (+3)
1100 (-4)
2’s Complement
Unsigned
What about multiplication??
1010
(10) x
1110
(14)
0000
(140)
1010
(-6) x
1110
(-2)
0000
(-116)
2’s Complement
Unsigned
Need to sign extend partial products
Wrong Result!!!

19. Problem: Multiplication & 2’s Complement
Addition and Subtraction can be treated as unsigned integer
1001 (+9) +
0011 (+3)
1100 (+12)
1001 (-7) +
0011 (+3)
1100 (-4)
2’s Complement
Unsigned
What about multiplication??
1010
(10) x
1110
(14)
0000
(140)
1010
(-6) x
1110
(-2)
111010
11010
(44)
2’s Complement
Unsigned
Need to sign extend partial products
Still Wrong Result!!!

20. Multiplying Negative Numbers
This does not work!
Solution 1
Convert to positive if required
Multiply as above
If signs were different, negate answer
Solution 2
Booth’s algorithm

21. Solution 1 Solution 1 is as follows:
If either multiplicand or multiplier is negative convert it to positive
Do the multiplication operation
Check the signs of the multiplicand and multiplier if different negate the result
Adds Hardware
Complexity

22. Solution 2: Booth Algorithm (1951)
Booth starts from the fact that a series can be represented as:
Suppose that we have to perform the following multiplication:
M*30 -> M * ( ) = M * ( )
From series the multiplication can be written as: M * ( ) = M * (25 – 21)
Savings on computations

23. Booth Algorithm:
Multiply the multiplicand by -2n-K when a transition from 0 to 1 (0-1) is encountered going from right to left.
Multiply the multiplicand by 2n+1 whenever an transition from 1-0 is encountered
Add the partial products obtained.

24. Booth Algorithm:
When multiplying 2n-bit numbers, form 22n-bit partial products as follows:
When the first 1 of a block is encountered (1-0), partial product is formed by negating the multiplicand and appropriately shifting the negated multiplicand
When the last 1 of a block is encountered (0-1), partial product is formed by taking the multiplicand and appropriately shifting it
When either 0-0 or 1-1 is encountered, partial product is all zeros
Sign-extend all partial products
Add the partial products to obtain the multiplication result.

25. Booth Algorithm: Example 1
Multiplying: -6 x -2
1010 -6
X 1110
(0) -2
0-0
1-0
(Subtract multiplicand)
000000
1-1
(Shift)
00000
1-1
(Shift)

26. Multiplication: Why MSB is Redundant?
Assume that we have a number represented by 4 bits (N = 4)
2’s complement range is from -8 to +7.
The min/max number obtained from multiplication is
-56/+64  7 bits are enough to represent the result.
In general: 2N-1 bits are needed if NxN multiplication is performed
The additional MSB is a “sign extension bit” and can be removed
Another way to interpret it is that if converted to unsigned the multiplication result will be (N-1) + (N-1) + 1 sign bits giving 2N – 1.