Basic Dividers for Computer Arithmetic - Slides | ECE 645, Study notes of Electrical and Electronics Engineering

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1

Basic Dividers

Lecture 9

2

Required Reading

Parhami,

Chapter 13 Basic Division Schemes

3

Notation

z Dividend z2k- 1 z2k- 2... z 2 z 1 z 0

d Divisor dk- 1 dk- 2... d 1 d 0

q Quotient qk- 1 qk- 2... q 1 q 0

s Remainder sk- 1 sk- 2... s 1 s 0

(s = z - dq)

4

Basic Equations of Division

z = d q + s

sign(s) = sign(z)

| s | < | d |

z > 0

0  s < | d |

z < 0

• | d | < s  0

7

Sequential Integer Division

Justification

s(1)^ = 2 z - qk- 1 (2k^ d)

s(2)^ = 2(2 z - qk- 1 (2k^ d)) - qk- 2 (2k^ d)

s(3)^ = 2(2(2 z - qk- 1 (2k^ d)) - qk- 2 (2k^ d)) - qk- 3 (2k^ d)

s(k)^ = 2(... 2(2(2 z - qk- 1 (2k^ d)) - qk- 2 (2k^ d)) - qk- 3 (2k^ d)...

• q 0 (2k^ d) =

= 2k^ z - (2k^ d) (qk- 1 2 k-^1 + qk- 2 2 k-^2 + qk- 3 2 k-^3 + … + q 020 ) =

= 2k^ z - (2k^ d) q = 2k^ (z - d q) = 2k^ s

8

Fig. 13.2 Examples of sequential division with integer

and fractional operands.

9

Unsigned Fractional Division

zfrac Dividend .z- 1 z- 2... z-(2k-1)z-2k

dfrac Divisor .d- 1 d- 2... d-(k-1) d-k

qfrac Quotient .q- 1 q- 2... q-(k-1) q-k

sfrac Remainder .000…0s-(k+1)... s-(2k-1) s-2k

k bits

10

Integer vs. Fractional Division

For Integers:

z = q d + s  2 - 2k

z 2-2k^ = (q 2-k) (d 2-k) + s (2-2k)

zfrac = qfrac dfrac + sfrac

For Fractions:

where

zfrac = z 2-2k

dfrac = d 2-k

qfrac = q 2-k

sfrac = s 2-2k

13

Sequential Fractional Division

Justification

s(1)^ = 2 zfrac - q- 1 dfrac

s(2)^ = 2(2 zfrac - q- 1 dfrac) - q- 2 dfrac

s(3)^ = 2(2(2 zfrac - q- 1 dfrac) - q- 2 dfrac) - q- 3 dfrac

s(k)^ = 2(... 2(2(2 zfrac - q- 1 dfrac) - q- 2 dfrac) - q- 3 dfrac...

• q-k dfrac =

= 2k^ zfrac - dfrac (q- 1 2 k-^1 + q- 2 2 k-^2 + q- 3 2 k-^3 + … + q-k 20 ) =

= 2k^ zfrac - dfrac 2 k^ (q- 1 2 -^1 + q- 2 2 -^2 + q- 3 2 -^3 + … + q-k 2 - k) =

= 2k^ zfrac - (2k^ dfrac) qfrac = 2k^ (zfrac - dfrac qfrac) = 2k^ sfrac

14

Restoring Unsigned Integer Division

s(0)^ = z

for j = 1 to k

if 2 s(j-1)^ - 2 k^ d > 0

qk-j = 1

s(j)^ = 2 s(j-1)^ - 2 k^ d

else

qk-j = 0

s(j)^ = 2 s(j-1)

15 Fig. 13.5 Shift/subtract sequential restoring divider. 16

Fig. 13.6 Example of restoring unsigned division.

19

Fig. 13.7 Example of nonrestoring unsigned division.

20

Fig. 13.8 Partial remainder variations for restoring and

nonrestoring division.

21 s(j)^ = 2 s(j-1) s(j+1)^ = 2 s(j)^ - 2 k^ d = = 4 s(j-1)^ - 2 k^ d s(j)^ = 2 s(j-1)^ - 2 k^ d s(j+1)^ = 2 s(j)^ + 2k^ d = = 2 (2 s(j-1)^ - 2 k^ d) + 2k^ d = = 4 s(j-1)^ - 2 k^ d

Restoring division Non-Restoring division

Non-Restoring Unsigned Integer Division

Justification

22

Non-Restoring Signed Integer Division

Correction step

z = q d + s

z = (q-1) d + (s+d)

z = q’ d + s’

z = (q+1) d + (s-d)

z = q” d + s”

25

Fig. 13.9 Example of nonrestoring signed division.

26

BSD  2 ’s Complement Conversion

q = (qk- 1 qk- 2... q 1 q 0 )BSD =

= (pk- 1 pk- 2... p 1 p 0 1) 2 ’s complement

where

qi pi

• 1 0

Example:

qBSD 1 - 1 1 1

p

q 2 ’scomp

no overflow if pk- 2 = pk- 1 (qk- 1  qk- 2 )