第一讲单周期数据通路的设计第二讲单周期控制器的设计第.pptVIP

The control signals setting in the main datapath for the Jump instruction is pretty boring because in most cases, we DON’T CARE. More specifically, control signals ExtOp, ALUSrc, ALUctr are all don’t cares because the ALU is not used at all for the Jump instruction. Control signals MemtoReg and RegDst are don’t are because Jump does not write the register file. That is the reason why we still need to set RegWr to zero. Furthermore, we also need to set MemWr to zero to avoid Data Memroy write. Finally, the control signal Branch is set to zero but Jump is set to 1. +2 = 37 min. (X:17) Complement: So far as we have already seen, almost all instructions use ALU. But this jump instruction is exceptional. which instruction doesnt use ALU? Inside the Instruction Fetch Unit, with Branch set to zero and Jump set to 1, we will not use the output of neither Adder. What we will use is the concatenation of the four most significant bits of the current program counter and the twenty six bits of the target address. With the control signal Jump set to 1, this value will be send to the Program Counter and get written into PC at the next clock tick (points to the Clk bubble). +2 = 39 min. (Y:19) There should be an improvement: When Jump=1, we don’t have to set Branch=0. We can set Branch=x to simply logical design of the control part. But we have to set Jump=0 when Branch=1. Here is a table summarizing the control signals setting for the seven (add, sub, ...) instructions we have looked at. Instead of showing you the exact bit values for the ALU control (ALUctr), I have used the symbolic values here. The first two columns are unique in the sense that they are R-type instrucions and in order to uniquely identify them, we need to look at BOTH the op field as well as the func fiels. Ori, lw, sw, and branch on equal are I-type instructions and Jump is J-type. They all can be uniquely idetified by looking at the opcode field alone. Now let’s take a more careful