Sequential Integer Division Basic Equations - Lecture Slides | ECE 645, Study notes of Electrical and Electronics Engineering

Material Type: Notes; Class: Computer Arithmetic; Subject: Electrical & Computer Enginrg; University: George Mason University; Term: Unknown 1989;

Typology: Study notes

Pre 2010

Uploaded on 02/12/2009

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1
Basic Dividers
Lecture 9
2
Notation
z Dividend z2k-1z2k-2 . . . z2z1z0
d Divisor dk-1dk-2 . . . d1d0
q Quotient qk-1qk-2 . . . q1q0
s Remainder sk-1sk-2 . . . s1s0
(s = z - dq)
3
Basic Equations of Division
z = d q + s
sign(s) = sign(z)
| s | < | d |
z > 0
0 s < | d | z < 0
- | d | < s 0
pf3
pf4
pf5
pf8
pf9

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1

Basic Dividers

Lecture 9

2

Notation

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

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

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

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

(s = z - dq)

3

Basic Equations of Division

z = d q + s

sign(s) = sign(z)

| s | < | d |

z > 0

0 ≤ s < | d |

z < 0

  • | d | < s ≤ 0

4

Unsigned Integer Division Overflow

z = zH 2 k^ + zL < d 2k

Condition for no overflow:

z = q d + s < (2k-1) d + d = d 2k

zH < d

5

Sequential Integer Division Basic Equations

s(0)^ = z

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

s(k)^ = 2k^ s

6

Fig. 13.2 Examples of sequential division with integer and fractional operands.

10

Unsigned Fractional Division Overflow

Condition for no overflow:

zfrac < dfrac

11

Sequential Fractional Division Basic Equations

sfrac(0)^ = zfrac

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

s(k)frac = 2k^ sfrac

12

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

13

Restoring Unsigned Integer Division

s(0)^ = z

for j = 1 to k

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

14

Fig. 13.5 Shift/subtract sequential restoring divider.

15

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

Fig. 13.10 Shift-subtract sequential nonrestoring divider.

22

Non-Restoring Signed Integer Division

s(0)^ = z for j = 1 to k if sign(s(j-1)) == sign(d) qk-j = 1 s(j)^ = 2 s(j-1)^ - 2k^ d = 2 s(j-1)^ - qk-j (2k^ d) else qk-j = - s(j)^ = 2 s(j-1)^ + 2k^ d = 2 s(j-1)^ - qk-j (2k^ d) Correction_step q = BSD_2’s_comp_conversion(q);

23

Fig. 13.9 Example of nonrestoring signed division.

24

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”