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I'm reading on MIPS processors, I try to understand wow they get the logic equation (scheme or second picture) from this truth table

for example for ALU0, I understant the x meaning and understand how they get this small table :

ALUOp | Funct        ||  ALU 
1 0   | 5 4 3 2 1 0  ||  2 1 0
------+--------------++------------
1 x   | x x 0 1 0 1  ||  0 0 1
1 x   | x x 1 0 1 0  ||  1 1 1

but now, the question is : how did they get this logic equation ALU i ??? :

ALU 0 = ALUOp1 (F0 + F3) 

ALU 1= UALOp1\ + F2\

ALU 2 = ALUOp0 + ALUOp1.F1

truth table enter image description here

scheme enter image description here

  • Can you please specify what exactly you don't understand? – Garp Aug 15 '14 at 18:31
  • @Garp I didnt understand how they produced the two equations from the truth table, more exactly, how they get the first equation from the first simplified table. –  Aug 15 '14 at 18:36
  • Okay I will post an answer. – Garp Aug 15 '14 at 18:38

1 Answers1

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You can do the logic by your eyes:

  1. For ALU_0: you see the last 2 rows only are ones, ALUOp is required to be 1, Fun(5) and Fun(4) are constant as 10 in the 2 rows, so they are dont care. The others Fun(3) to Fun(0) are 0101 and 1010, thus Fun(3) OR Fun(0) is sufficient.
  2. For ALU_2: they are ones in the third, fifth and the eighth row, For the third ALuop(0) is sufficient, for the other two you will find that {ALUop(1)} AND {Fun(1)} should be ones. thus the final equation is true.
Garp
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  • you said he others Fun(3) to Fun(0) are 0101 and 1010, thus Fun(3) OR Fun(0) is sufficient. what is means, this is exaclty what I didnt understand. for my logic the equation is `ALU0 = ALU1op1*((f3\ f2 f1\ f0)+ (f3 f2\ f1 f0\))` how from this equation you get `ALU 0 = ALUOp1 (F0 + F3) ` –  Aug 15 '14 at 18:49
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    if Fun(0) or Fun(2) are ones, Operation(0) in the seventh row is one. Or if Fun(1) or Fun(3) are ones, Operation(0) in the eighth row is one. Thus either (Fun(0) OR Fun(3)) or the (Fun(1) OR Fun(2)) is sufficient. – Garp Aug 15 '14 at 18:53