fixed point arithmetics 1

7/29/2019 Fixed Point Arithmetics 1

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Fixed-Point Arithmetics: Part I


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I ra Fu lton School of Engineeri ng

Electri cal Department

EEE404/591Real Time DSP

Binary Representation

Unsigned Magnitude

Signed Magnitude

Twos Complement


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Unsigned Magnitude

Only positive number are presented

No sign bit

For N bits we can represent the signedintegers between 0 and 2N -1

Example: 111 -> 7


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Signed Magnitude

The most significant bit is used torepresent the sign: 1 means negative,

0 means positive

For positive numbers the result is thesame as unsigned magnituderepresentation

For N bits we can represent the signedintegers between -2(N-1) +1 and +2(N-1)-1


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Disadvantage of Signed Magnitude

Two encoding for the zero 0. Addition of two numbers of opposite sign will not yield

0!! -5 + 5 = (85)H + (05) H = (8A) H = -10!!

Arithmetic operations of numbers with unlike sign and

with same sign must be handled differently (subtracterneeded with unlike signs)

AB A + (-B) Example: 3 2Addition Operation 3 + (-2) Subtraction Operation (3-2)

0011 (+3) 0011 (+3)+ 1010 (-2) - 0010 (+2)

------- -------1100 (-5) WRONG! 0001 (+1) CORRECT!


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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 3 - 2 Subtraction Operation 2 - 3

0011 (3) 0010 (+2)- 0010 (2) - 0011 (+3)

------- ------0001 (1) CORRECT 1111 (-7) WRONG

Solution: Switch order of operands (3 2 instead of 2 3) Perform subtraction (Result = 1)

attach the minus sign (Result = -1)


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Twos Complement example

1 = 0001 1 = 1110 + 1 = 1111 2 = 0010 2 = 1101 + 1 = 1110 6 = 0110 6 = 1001 + 1 = 1010 8 = 1000 8 = 0111 + 1 = 1000

Note that the max positive number that can berepresented is 7! Since to get the negative of -8 weneed:

-8 = 1000To bitwise invert : 0111Add +1 to LSB: +1

Result: 1000So(-8) = -8 (abnormal result) 8 cannot be represented

For N bits, the max number is 2(N-1)-1 while the minnumber is -2(N-1)

G t i D i ti f T


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Geometric Depiction of TwosComplement Integers


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Benefits of 2s Complement

One unique representation of the number 0: 0 = 0000 Complement: 1111 Add +1 to LSB: +1 Result: 0000 (Ignoring carry bit)

-0 = +0 Negation: take complement and add +1 Extending word length:

For positive number pack with leading zeros +3 =011

+3 = 000011 For negative numbers pack with leading ones

-4=100 -4 =111000


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Subtraction Rule: to subtract b from a, take the 2s

complement of b and add it to a: a b = a + (-b)

Adder

ab

Result

2s

If Subtractionenable

a b

Compare signs bitsEqual??

Adder: a + b Is a > b

Yes No

Sub: a - b

Yes

Swap OperandSub: a b

Negate

No

2s Comp

Signed Mag


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Overflow Rule: if two numbers with same signare added and overflow occurs the resultnumber has different sign: Addition: 011 + 011 = 110

3 + 3 = -2 Subtraction: 110 011 = 011

-2 - 3 = 3 Multiplication: 011 * 010 = 110

3 * 2 = -2


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What to Do In case of Overflow

Prevention rule: design your code with minimal possibleoverflows

DSP can be configured to saturate the number atoverflow (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 2 - 1 x 4 = 10 4 = 6 If accumulator has one extra bit called guard bit the final

result will not overflow!


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Multiplication Example

1011 Multiplicand (11 dec)

x 1101 Multiplier (13 dec)

1011 Partial products

0000 Note: if multiplier bit is 1 copy 1011 multiplicand (place value)

1011 otherwise zero

10001111 Product (143 dec) Note: need double length result


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Unsigned Binary Multiplication


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Execution of Example


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owc ar or ns gne naryMultiplication


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Problem: Multiplication & 2s Complement

Addition and Subtraction can be treated asunsigned integer

1001 (-7)

+ 0011 (+3)

1100 (-4)

1001 (+9)

+ 0011 (+3)

1100 (+12)

Unsigned

2s Complement

What about multiplication??

1010 (10)

x 1110 (14)

0000

1010

1010

1010

10001100 (140)

Unsigned

1010 (-6)

x 1110 (-2)

0000

1010

1010

1010

10001100 (-116) Wrong Result!!!

2s Complement


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1001 (-7)

+ 0011 (+3)

1100 (-4)

1001 (+9)

+ 0011 (+3)

1100 (+12)

Unsigned

2s Complement


1010 (10)

x 1110 (14)

0000

1010

1010

1010

10001100 (140)

Unsigned

1010 (-6)

x 1110 (-2)

0000

1010

1010

1010

10001100 (-116) Wrong Result!!!

2s Complement

Need to sign extend partial products


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1001 (-7)

+ 0011 (+3)

1100 (-4)

1001 (+9)

+ 0011 (+3)

1100 (+12)

Unsigned

2s Complement


1010 (10)

x 1110 (14)

0000

1010

1010

1010

10001100 (140)

Unsigned

1010 (-6)

x 1110 (-2)

00000000

1111010

111010

11010

000101100 (44) Still Wrong Result!!!

2s Complement

Need to sign extend partial products


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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

Booths algorithm


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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 andmultiplier if different negate the result

Adds HardwareComplexity


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Solution 2: Booth Algorithm (1951)

Knn

K

n

K

nKnnn

22

2

11

2

11

22

1

2

1

2

112222 1

1

2

1

Booth starts from the fact that a seriescan be represented as:

Suppose that we have to perform the followingmultiplication:

M*30 -> M * (00011110) = M * (24+23+22+21)

From series the multiplication can be written as:

M * (00011110) = M * (25 21)Savings on computations


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Booth Algorithm:

Multiply the multiplicand by -2n-Kwhen atransition from 0 to 1 (0-1) is encounteredgoing from right to left.

Multiply the multiplicand by 2n+1whenever an transition from 1-0 isencountered

Add the partial products obtained.


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Booth Algorithm:

When multiplying 2n-bit numbers, form 22n-bit partialproducts as follows:

When the first 1 of a block is encountered (1-0), partialproduct is formed by negating the multiplicand andappropriately shifting the negated multiplicand

When the last 1 of a block is encountered (0-1), partialproduct is formed by taking the multiplicand andappropriately shifting it

When either 0-0 or 1-1 is encountered, partial product is allzeros

Sign-extend all partial products

Add the partial products to obtain the multiplicationresult.


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Booth Algorithm: Example 1

Multiplying: -6 x -2

00001100

0000110

0-000000000

(0)X 1110

1010

1-0000000 1-100000 1-1

-6

-2

(Subtract multiplicand)(Shift)(Shift)


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EEE404/591 R l Ti DSP

Multiplication: Why MSB is Redundant?

Assume that we have a number represented by 4 bits(N = 4) 2s 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.

fixed point arithmetics 1

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