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From the textbook : ARM edition of computer organization and design chapter 4 4.

ID: 3859848 • Letter: F

Question

From the textbook : ARM edition of computer organization and design chapter 4

4.5 In this exercise, we examine in detail how an instruction is executed in a

single-cycle datapath. Problems in this exercise refer to a clock cycle in which the

processor fetches the following instruction word: 0xf8014062.

4.5.1 [5] <§4.4> What are the outputs of the sign-extend and the “shift left 2”

unit (near the top of Figure 4.23) for this instruction word?

4.5.2 [10] <§4.4> What are the values of the ALU control unit’s inputs for this

instruction?

4.5.3 [5] <§4.4> What is the new PC address after this instruction is executed?

Highlight the path through which this value is determined

4.5.4 [10] <§4.4> For each mux, show the values of its inputs and outputs during

the execution of this instruction. List values that are register outputs at Reg [Xn].

4.5.5 [10] <§4.4> What are the input values for the ALU and the two add units?

4.5.6 [10] <§4.4> What are the values of all inputs for the registers unit?

Read address Instruction Instruction memory Uncondbranch Branch Mem Read Instudion p1-2 Instruction 9-6) struction po-161 magister 1 Read data 2 register data 2 instruction data Registers Instruction B1-01 2 extend Instruction B1-21 eft 2 ALU Address Read Data

Explanation / Answer

Given instruction word: 0xf8014062 in HEX.

The corresponding value in Binary is:
  
   1111100000000001 0100000001100010

(a)
   Sign-extend takes the lower 16 bits and expands them to be 32 bits,
   outputting 00000000000000000100000001100010.
   The shift left 2 in the upper-left is used for a jump,
   which takes the lower 26 bits and expands them to 28 bits with a left-shift.
   So we have 0000000001010000000110001000 as output.
  
  
(b)
   The ALU control gets the lower 6 bits as it's main input, so 100010.
   It also gets a signal from the main Control indicating the ALU op.
   In this case the main control gets bits 31-26 as it's input, 111110.
   The decimal value of 111110 is 62 in decimal and 3E in HEX.
   This corresponds to “sdc2”, which sends a 00 to the ALU Control.
  
(c)
   New PC = PC + 4.
   Path will be PC to Add (PC+4) to branch Mux to jump Mux to PC.
  
     
             

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