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U. Glaeser

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FIGURE 9.14 Nonrestoring division.<br />

FIGURE 9.15 Example of nonrestoring division.<br />

Binary SRT Divider<br />

The binary SRT division process is similar to nonrestoring division, but the set of allowable quotient<br />

digits is increased to {1, 0, 1}<br />

and the divisor is restricted to .5 ≤ D < 1. The digit selection and resulting<br />

partial remainder are given for the kth iteration by the following relations:<br />

© 2002 by CRC Press LLC<br />

4. q n-k-1 = -1<br />

P k+1 = 2P k + D<br />

P0 = N = 5<br />

8<br />

D = 7<br />

8<br />

n = 4<br />

LT<br />

1. P 0 = N<br />

k = -1<br />

2. k = k + 1<br />

3.<br />

P k : 0<br />

6.<br />

K : n-1<br />

GE<br />

Q = N/D<br />

GE<br />

LT<br />

5. q n-k-1 = 1<br />

P k+1 = 2P k − D<br />

k = 0, Since P0 > 0, then q3 = 1 and P1 = 2P0 − D = 3<br />

8<br />

k = 1, Since P1 > 0, then q2 = 1 and P2 = 2P1 − D = −1<br />

8<br />

_<br />

k = 2, Since P0 < 0, then q1 = 1 and P3 = 2P2 + D = 5<br />

8<br />

k = 3, Since P0 > 0, then q0 = 1 and P4 = 2P3 − D = 3<br />

8<br />

_<br />

Q = 0 .1 1 1 1 (SIGNED DIGIT FORM)<br />

Q = 0 .1 0 1 1 (BINARY)<br />

If Pk ≥ .5, qn−k−1 = 1 and Pk+1 = 2Pk − D (9.28)<br />

If −.5 < Pk < .5, qn−k−1 = 0 and Pk+1 = 2Pk (9.29)<br />

If Pk ≥ .5, qn−k−1 = 1<br />

and Pk+1 = 2Pk + D (9.30)

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