RTN Register Transfer Notation Explain what would happen in

RTN = Register Transfer Notation

Explain what would happen in a Branch instruction if the RegWrite control signal was mistakenly set to 1 instead of 0. Consider all possible case and use RTN to explain the behaviors.

Solution

The control signal set to 1 instead of zero is an classic example of Stuck-at-0 fault. Struck-at-0 are basically functional faults. On a very general level, struck-at-0 represents the system\'s vulnerability to stay at a logic which it is not designed to under normal conditions.

In this model, faults are fixed (0 or 1) value to a net which is an input or an output of a logic gate or a flip-flop in the circuit. If the net is stuck to 0, it is called stuck-at-0 (s-a-0) fault and if the net is stuck to 1, it is called stuck-at-1 (s-a-1) fault. If it is assumed that only one net of the circuit can have a fault at a time, it is called single stuck-at fault model. Without this assumption, it is called multiple stuck-at fault model. This has been observed though fabrication and testing history, that if a chip is verified to be free of single stuck-at faults, then it can be stated with more than 99.9% accuracy that there is no defect in the silicon or the chip is functionally (logical) normal.

Now with regard to the question, follwing behaviours are observed when a RegWrite control signal is set to 1 instead of zero.

a. RegWrite = 0: All R-format instructions, in addition to lw , will not work because these instructions will not be able to write their results to the register file.

b. ALUop0 = 0: beq instruction will not work because the ALU will perform addition instead of subtraction, so the branch outcome may be wrong.

c. ALUop1 = 0: All R-format instructions except add will not work correctly because the ALU will perform add instead of the required ALU operation.

d. Branch = 0: beq will not execute correctly. The branch instruction will always be not taken even when it should be taken.

e. MemRead = 0: lw will not execute correctly because it will not be able to read data from memory.

f. MemWrite = 0: sw will not work correctly because it will not be able to write to the data memory.