代写COMP 273 Project Mini-MIPS CPU帮做R编程

日期：2024-12-12 05:13

Mini-MIPS CPU

COMP 273 Project

Due: December 8, 2024, on MyCourses

Submission instructions

You can do this project on your own or in a team of 2 students. All work must be your own and must be submitted to myCourses. Include your name(s) and student number(s) in a comment at the top of your Logisim circuit. Submit only one file: project-miniCPU.circ per group. Check your submission by downloading it from myCourses to verify that it was correctly submitted. You will not receive marks for work that is incorrectly submitted.

You must enroll yourself in a group on MyCourses to be able to submit the project. If you are working on your own, you must still enroll yourself in a group on MyCourses.

Purpose

•    Understand how CPUs function.

•    To understand how the CU coordinates the different machines that make the CPU.

Helpful

•    The lectures on Datapath and CU.

•    Optionally you may use your A2 solutions for ALU and RAM (the graded version) in this project.

 

Figure 1 - Mini-MIPS architecture

Overview

Mini-MIPS is a CPU that executes only one instruction at a time. This means it executes the instructions sequentially, one by one.

The Mini-MIPS CPU diagram, Figure 1, is the basic CPU architecture that everyone’s project must adhere to. It represents the basic architecture. The TA will make sure that your CPU looks and behaves as like the above diagram. You are permitted to build the entire CPU on the same circuit page, or you can create subcircuit pages to mimic the above diagram’s look. If you build your CPU on a single page, it is important to label each part of the circuit and to leave empty spaces to separate the parts. This will both help you while debugging and help the TA understand what you have built. If you chose to use subcircuits you will need to take care of the time delays that Logisim adds to subcircuits. The more subcircuits you nest the more delays are added.

The execution of a single instruction requires multiple tricks. The CU, not shown, controls each section of the instruction execution cycle. The execution cycle constitutes of the fowllowing stages:

Stage 0: Loads an instruction from the instruction RAM using the address stored in PC into IR  (notice  RAM and not cache – we are using your A2 RAM to make things easier). PC is incremented to the next address (circuit not shown in diagram). This completes Stage 0.

Stage 1: Uses the Register file to output the values for the ALU’s input registers (not shown in the diagram). Bits from the IR will address the specific Register file registers.

Stage 2: The ALU performs the selected operation updating the status register and ALU output register    (not shown in the diagram) and sending values to intermediate registers (not shown in diagram) for the   multiple data paths that are possible at this point in execution (to prepare for: an overwrite of the PC, or overwrite Data RAM at a specific address, or overwrite a specific register in the Register file).

Stage 3: Completes what was prepped by Stage 2: either write into Data RAM at a particular address or loads a value from Data RAM into the input of the Register file or overwrites the PC, etc.

Stage 4:  Completes the write to Register file from stage 3 (if needed) and then the CU loops back to Stage 0. The CU operates as an infinite loop until the program asks the computer to shutdown.

Your final submission must run by auto ticking at the default Logisim Evolution 1 Hz rate. When you test  your circuit, you can manually click on the clock, but the final submission must auto tick. You can set this up through the menu option Simulate, see Figure 2. Make sure “Auto-propagate” is check marked. Make sure “Auto Tick” is check marked. Make sure “Auto tick frequency” is set to 1 Hz.  If you circuit runs at 1  Hz it will run at higher frequencies as well. But the TA will test your circuit with the default speed.

From the Simulate tab you can start and stop the ticking, see Figure 3. The play and pause buttons can be pressed to start and stop the ticking of the clock. When you pause the clock, you can manually tick the clock. When you press play, it will resume ticking where you left of.

 Figure 2

 Figure 3

Implementation

Using Logisim Evolution create the following Mini-MIPS CPU:

•    To make things easier: everything is a nibble in size. All data area nibble (4-bits). The IR must be able to address a maximum of 8 nibbles.

•    Your circuit must have the following components:

o A clock

o The CPU-parts depicted in figure 1 and described in Overview.

o The Register file only has two registers: R0 and R1.

o The CU controls everything inside the CPU.

o The Status register flags: Sign overflow, zero, and negative.

o Instruction RAM stores 8 nibbles of data.

o Data RAM stores 2 nibbles of data.

o The CU must implement the Fetch and Execute instruction.

o You can add additional components to help you. See the list of legal components.

•    You are permitted to ONLY the following Logisim pre-built elements:

o Clock

o Wire and splitters.

o D-Flip-flop and Register.

o AND, OR, NOT, XOR, Buffer, and controlled buffer gates

o Pins and probes

o Subcircuits (optional)

o Tunnelling (optional)

o Multiplexer

o Decoder

NOTE 1 : The TA must be able to change and see the bits of RAM and registers before and after execution. Provide away for this to happen (this is especially important for those of you using subcircuits).

NOTE 2: Your final circuit must use designs we covered during class. You cannot use any outside (other sourced) circuit designs.

Instructions

•    The Instruction RAM has 8 nibbles. These nibbles will be used to store the machine codes of your assembler instructions. Algorithms must fit within this space and terminate with HALT.

•    The Data RAM has 2 nibbles and will store the program’sdata.

•    To run a program, you will need to first input the instructions into the Instruction RAM, the data in the Data RAM, and set the PC pointing to the first instruction in the Instruction RAM. All other registers are assumed to be zero or overwritable. Then the auto-clock is turned on and the programs runs. Make sure to place the clock on your circuit.

•    LOAD REGISTER, RAM_ADDRESS

o 1st  2 bits : LOAD = 00

o 3rd  bit : REGISTER = 0 for R0, 1 for R1

o 4th  bit : RAM_ADDRESS = 0 for address 0 in RAM, 1 for address 1 in RAM

o Example: LOAD R1, 0  0010

o Example: LOAD R0, 1  0001

•    SAVE REGISTER, RAM_ADDRESS

o 1st  2 bits : SAVE = 01

o 3rd  bit : REGISTER = as in LOAD

o 4th  bit : RAM_ADDRESS = as in LOAD

o Example: SAVE R0, 1  0101

•    ADD REGISTER1 REGISER2

o 1st  2 bits : ADD = 10

o 3rd  bit : REGISTER = as in LOAD

o 4th  bit : REGISTER = as in LOAD

o The solution to the addition is saved in REGISTER1.

o Example: ADD R1 R0  R1 = R1 + R0  1010

o Example: ADD R0 R0  R0 = R0 + R0  1000

•    SUB REGISTER1 REGISTER2

o 1st  2 bits : SUB = 11

o 3rd  bit : REGISTER = as in LOAD

o 4th  bit : REGISTER = as in LOAD

o The solution to the subtraction is saved in REGISTER1.

o Example: SUB R1 R0  R1 = R1 – R0  1110

o Example: SUB R0 R0  R0 = R0 – R0  1100

•    HALT

o All four bits are 1.

o Example: HALT  1111

o This marks the end of the algorithm. The clock’sticking does not affect the circuit anymore. The PC no longer increments.

•    Bonus instruction:

o JUMP

     Use machine code 0000 for jump “j” instruction. PC is updated with whatever is in R0.

•    Your circuit must be able to execute any program set in the RAM with the PC pointing to the first instruction of that algorithm. The PC can be set to any starting address.

Execution

Your CPU circuit must be able to do at least the following algorithms:

1.    Execute a program that loads two numbers, performs an ALU operation, and then saves the solution.

2.    Execute aprogram that has only the HALT instruction.

Note: The TA will test your CPU by first entering a starting address in the PC and loading a program   in your Instruction RAM and data in your Data RAM. Then, they will start the clock and see the result display in the Data RAM.

Note: Your CPU does not need to be optimized, therefore it does not matter the number of clock ticks it needs to execute your instructions.

Marking

•    Bonus points for JUMP +3

•    Maximum 100 points

o +10 Reusing A2 circuits.

o +20 Using one common clock for the circuit

o +20 Complete Datapath implementation (including buses between the different components)

o +10 CPU Registers R0 and R1.

o +10 PC Increment.

o +30 CU and execution of instructions.

•    -20 points for not using the clock.