SMJ320C80 Datasheet, PDF (62/157 Pages) Texas Instruments – DIGITAL SIGNAL PROCESSOR

SMJ320C80
DIGITAL SIGNAL PROCESSOR
SGUS025B – AUGUST 1998 – REVISED JUNE 2002
EALU Boolean functions
EALU operations support all 256 Boolean ALU functions plus the flexibility to add 1 or a carry-in to Boolean sum.
The Boolean function performed by the ALU is:
(F0 & (~A & ~B & ~C)) | (F1 & (A & ~B & ~C)) | (F2 & (~A & B & ~C)) | (F3 & (A & B &
~C)) | (F4 & (~A & ~B & C)) | (F5 & (A & ~B & C)) | (F6 & (~A & B & C)) | (F7 & (A & B
& C)) [+1 | +cin]
Table 27. EALU Boolean Function Codes
d0 BIT
26
25
24
23
22
21
20
19
ALU FUNCTION SIGNAL
F7
F6
F5
F4
F3
F2
F1
F0
PRODUCT TERM
A&B&C
~A & B & C
A & ~B & C
~A & ~B & C
A & B & ~C
~A & B & ~C
A & ~B & ~C
~A & ~B & ~C
EALU arithmetic functions
EALU operations support all 256 arithmetic functions provided by the three-input ALU plus the flexibility to add
1 or a carry-in to the result. The arithmetic function performed by the ALU is:
f(A,B,C) = A & f1(B,C) + f2(B,C) [+1 | cin]
f1(B,C) and f2(B,C) are independent Boolean combinations of the B and C ALU inputs. The ALU function is
specified by selecting the desired f1 and f2 subfunction and then XORing the f1 and f2 code from Table 28 to
create the ALU function code for bits 19–26 of d0. Additional operations such as absolute values and signed
shifts can be performed using d0 bits which control the ALU function based on the sign of one of the inputs.
Table 28. ALU f1(B,C) and f2(B,C) Subfunctions
f1
CODE
00
AA
88
22
A0
0A
80
2A
A8
02
08
A2
8A
20
28
82
f2
CODE
00
FF
CC
33
F0
0F
C0
3F
FC
03
0C
F3
CF
30
3C
C3
SUBFUNCTION
0
–1
B
–B –1
C
–C –1
B&C
–(B & C) – 1
B|C
–(B | C) – 1
B & ~C
–(B & ~C) –1
B | ~C
–(B | ~C) –1
(B & ~C) | ((–B – 1) & C)
(B & C) | ((–B – 1) & ~C)
COMMON USAGE
Zero the term
–1 (All 1s)
B
Negate B
C
Negate C
Force bits in B to 0 where bits in C are 0
Force bits in B to 0 where bits in C are 0 and negate
Force bits in B to 1 where bits in C are 1
Force bits in B to 1 where bits in C are 1 and negate
Force bits in B to 0 where bits in C are 1
Force bits in B to 0 where bits in C are 1 and negate
Force bits in B to 1 where bits in C are 0
Force bits in B to 1 where bits in C are 0 and negate
Choose B if C = all 0s and –B if C = all 1s
Choose B if C = all 1s and –B if C = all 0s
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