TMS320C80GF datasheet(65/171 Pages) TI

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TMS320C80

DIGITAL SIGNAL PROCESSOR

SPRS023B – JULY 1994 – REVISED OCTOBER 1997

POST OFFICE BOX 1443

• HOUSTON, TEXAS 77251–1443

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 are shown below and in Table 27.

(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

ALU FUNCTION SIGNAL

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

CODE

SUBFUNCTION

COMMON USAGE

Zero the term

–1

–1 (All 1s)

–B –1

Negate B

–C –1

Negate C

B & C

Force bits in B to 0 where bits in C are 0

–(B & C) – 1

Force bits in B to 0 where bits in C are 0 and negate

B | C

Force bits in B to 1 where bits in C are 1

–(B | C) – 1

Force bits in B to 1 where bits in C are 1 and negate

B & ~C

Force bits in B to 0 where bits in C are 1

–(B & ~C) –1

Force bits in B to 0 where bits in C are 1 and negate

B | ~C

Force bits in B to 1 where bits in C are 0

–(B | ~C) –1

Force bits in B to 1 where bits in C are 0 and negate

(B & ~C) | ((–B – 1) & C)

Choose B if C = all 0s and –B if C = all 1s

(B & C) | ((–B – 1) & ~C)

Choose B if C = all 1s and –B if C = all 0s

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