Page 553 - Introduction to Information Optics

538 9. Computing with Optics

Table 9.34
Reduced Truth Table for the Recorded QSD Multiplication [160]

Output literal Minterms a ,«,-,*,
3 122 122
2 Idfo ,2 2dl Td,o72 2dl
! Idl d 20 22 122 Idl d 2o22 122
1 Idl d 2? 1 2? Idl d 2 022 122
2 ld To ,2 2dl 11< 1 T 2 2dl
i ia u)]
3 122 122

N i
addition of the products fr,-a 04° + b {a va^ + ••• + b ia n^^4 " is carry
free if the partial product is permitted to be represented in the original
format. Hence after receding, partial products can be formed in two steps
by digitwise multiplication and addition. Furthermore, by the above
l
analysis, we can conclude that the value of the (/ + j)th digit p\ lj of the
1
(l)
j'th partial product (p = Ab {4 ) of the receded numbers is dependent on
the values of 0,-fly-1 and h {. Therefore, by examining every two consecutive
digits of A and b t in parallel, we can generate the /th partial product in a
single step. In this way, all partial products can be produced simulta-
neously. The reduced minterms for parallel multiplication are summarized
in Table 9.34. Note that the minterms at the right side are digit-by-digit
complement of those at the left side. The partial products are represented
in the original QSD format. When employing this table, it is necessary to
pad one zero trailing the least significant digit of the multiplicand. From
the receding truth table (Table 9.31), one can deduce that the most
significant digit of any receded number is restricted to (T, 0, 1}, According
to the multiplication truth table (Table 9.34), when fl/a/- v = OT or 01, and
/?,. = 2, I, 1, or 2, the resultant product is always 0. Therefore, we do not
need to pad a zero preceding the most significant digit of the multiplicand,
and the partial product Ab { has the same number of digits as recoded A,
Recede only the multiplier B into the small set {2, I, 0, 1, 2). In this
case, when a { is multiplied by b j} the partial product digit is limited to
(2,1,0, 1, 2} and the partial carry digit to {T, 0, 1.}. Similarly, we can form
all the partial products in parallel by examining every pair of consecutive
digits of A and b t at the same time.
When performing nonrecoded QSD multiplication, both the partial-
product digit and the partial-carry digit of a t x b } belong to the set
(2, T, 0, 1, 2); thus, it is impossible for us to obtain the result of Ab t as a
single partial product in parallel by checking two consecutive digits of the
multiplicand Aia^^j) and one digit of the multiplier B(b /). The solution