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8.8 ARITHMETIC AND LOGIC UNITS 367

B A B A
n-1 n-1 . . . B 2 A 2 B 1 A 1 o o
(H) (H) (H) (H) (H) (H) (H) (H)
1 /P/U|\
D(H)-4<
rr/Lj\ 4x
c-\n ) — / '

4 ' \-\ ' 4 X ' 4, ' 4, ' t «,- \ ' 4,' 4 4 -4, '
/
t: D B A E: D B A EE D B A E: [) B A
R R R I— R
L/R —I L/R —I L/R —I UR —I
... EN
LSB
C nout(L) <-C '-'out ""*• n-i *^in o+- «: Co,,< PALU 2 C jn O«-C C m)) PALU, C jn cx-c C out PALU 0 C tn r^^t p
u
J
^~ ,.~m
(L)
MSB C ->C c in C out D->- ->C Cm c out D->C c fn c ou, D-^C c in c out
(L) '"
-C EN -C EN rC EN -C EN
Y Y Y Y Y Y Y Y
<f Y <P Y
EN(L) 1
F,i(H) F 2(H) F^H) F 0(H)
FIGURE 8.37
An rc-bit PALU with operational characteristics given by Figs. 8.34, 8.35, and 8.36, and by Eqs. (8.17).
and (8.18).

that LSB C in = 1 in Fig. 8.37 and that the MSB be reserved for the sign bit as dis-
cussed in Section 8.3. For operation (2), the requirements of F and C ollt are that R =
E © C in = 01102 = 610, E = A®B = AQB = 1001 2 = 9i 0, and that D =
A + B = 1011 = 1110. Operation (3) simply requires that LSB C in = I when A = 0,
for which the requirements of function F are that R = E © C- m = 01102 = 610,
E = B = 1010 2 = lOio, and D = 1 = 1111 = 15i 0 so that D = 0 in Eq. (8.18).
The 2's complement of operation (4) follows Algorithm 2.5 represented by Eq. (2.14).
The 2's complement operation sets LSB C- m = 1, then with B = 1 and A is comple-
mented, R = E © C in = 0110 2 = 610, E = A = 0011 2 = 3 m and D = A =
00112 = 310- Operations (6) through (10) are simple bitwise logic operations for which
F = R = E when R = 1010 2 = lOio, and C out = 1 and L/R = 0 for false carry rejection
when D = 0.
Operation (5) is the comparator operation, A = B, considered either an arithmetic or
logic operation. The requirement for this operation is as follows: If the two operands, A
and B, are equal on a bitwise comparison basis, then F,,_i = 1 for an n-bit PALU with
its LSB C in(L) = 0(L). Or if the operands are not equal, F n-\ = 0. Thus, C out = 0 will
ripple from the LSB stage to the MSB stage and all outputs will be F { = 1. However, if
any one of the bitwise comparisons yields an inequality, the carry C out = 1 will ripple
from that stage to the MSB stage and generate F n-\ = 0 at the MSB stage. Therefore,
operation (5) requires that R = C inE = 0010 2 = 2i 0 , E = A O B = 1001 2 = 9io, and
D = A Q B = 1001, which, when introduced into Eqs. (8.17) and (8.18), yields the results