
Example of Implementation of Div#ion and Square Root 383
7 oSo3
7~
From (7.38) we get
P[j] = 1 + 2-16[j] = 1 + 2-1(1- x[j])
7.40
That is,
P[j] is obtained by complementing x[j] and shifting the fractional part
one bit to the right.
The iteration consists then of
1. p[j]2 = p[j]p[j]
2. x[j + l]=x[j]P[j] 2 S[j + I]=S[j]P[j]
3. P[j +
1] = 1 + 2-1(1- x[j + 1])
Each iteration has three multiplications, but two of them can be performed con-
currently, or in pipelined fashion.
Implementation and Error Issues
The implementation of the square root algorithms and the error analysis are
similar to that for the reciprocal operation. ...