[SOLVED] Cs2200 project 1 – lc-3300 datapath
5.0
1 customer review
Digital download
Digital download
$25.00
- Availability
- In stock
- Checkout
- One item
Need a hand?
Message us on WhatsApp for payment or download support.
1 Requirements
Gatekeeper restrictions.
• The LC-3300 assembler is written in Python. If you do not have Python 3 or newer installed on your system, you will need to install it before you continue.
2 Project Overview and Description
Project 1 is designed to give you a good feel for exactly how a processor works. In Phase 1, you will design a datapath in CircuitSim to implement a supplied instruction set architecture. You will use the datapath as a tool to determine the control signals needed to execute each instruction. In Phases 2 and 3, you are required to build a simple finite state machine (the “control unit”) to control your computer and actually run programs on it.
3 Phase 1 – Implement the DatapathFigure 1: Datapath for the LC-3300 Processor
In this phase of the project, you must learn the Instruction Set Architecture (ISA) for the processor we will be implementing. Afterwards, we will implement a complete LC-3300 datapath in CircuitSim using what you have just learned.
You must do the following:
1. Learn and understand the LC-3300 ISA. The ISA is fully specified and defined in Appendix A: LC-3300
Instruction Set Architecture. Do not move on until you have fully read and understood the ISA specification. Every single detail will be relevant to implementing your datapath in the next step.
2. Using CircuitSim, implement the LC-3300 datapath. To do this, you will need to use the details of the LC-3300 datapath defined in Appendix A: LC-3300 Instruction Set Architecture. You should model your datapath on Figure 1.
3. Put your name on your CircuitSim data path in a comment box so we know it is your work.
3.1 Hints
3.1.1 Subcircuits
CircuitSim enables you to break create reusable components in the form of subcircuits. We highly recommend that you break parts of your design up into subcircuits. At a minimum, you will want to implement your ALU in a subcircuit. The control unit you implement in Phase 2 is another prime candidate for a subcircuit.
3.1.2 Debugging
As you build the datapath, you should consider adding functionality that will allow you to operate the whole datapath by hand. This will make testing individual operations quite simple. We suggest your datapath include devices that will allow you to put arbitrary values on the bus and to view the current value of the bus. Feel free to add any additional hardware that will help you understand what is going on.
3.1.3 Memory Addresses
Because of CircuitSim limitations, the RAM module is limited to no more than 16 address bits. Therefore, per our ISA, any 32-bit values used as memory addresses will be truncated to 16 bits (with the 16 most significant bits disregarded). If you use the RAM subcircuit we provide, this truncation has already been handled, and you can simply attach the 32-bit value from the MAR (Memory Address Register) to our custom RAM circuit. Otherwise, you will need to truncate the most significant bits of the the address value from the MAR before feeding it into the RAM.
3.1.4 Comparison Logic
The “comparison logic” box in Figure 1 is responsible for performing the comparison logic associated with the BEQ instructions. The comparison logic should read the current value on the bus. When executing BEQ, you should compute A−B using the ALU. While this result of the ALU is being driven on the bus, the comparison logic should read the result A−B and output a single “true” or “false” bit for the condition
A == B.
Your comparison logic should be purely combinational. Feel free to use any CircuitSim components you wish to aid in your implementation.
3.1.5 Register File
3.1.6 Register Select
4 Phase 2 – Implement the Microcontrol Unit
In this phase of the project, you will use CircuitSim to implement the microcontrol unit for the LC-3300 processor. This component is referred to as the “Control Logic” in the images and schematics. The microcontroller will contain all of the signal lines to the various parts of the datapath.
You must do the following:
1. Read and understand the microcontroller logic:
• Please refer to Appendix B: Microcontrol Unit for details.
• Note: You will be required to generate the control signals for each state of the processor in the next phase, so make sure you understand the connections between the datapath and the microcontrol unit before moving on.
2. Implement the Microcontrol Unit using CircuitSim. The appendix contains all of the necessary information. Take note that the input and output signals on the schematics directly match the signals marked in the LC-3300 datapath schematic (see Figure 1).
5 Phase 3 – Microcode and Testing
In this final stage of the project, you will write the microcode control program that will be loaded into the microcontrol unit you implemented in Phase 2. Then, you will hook up the control unit you built in Phase 2 of the project to the datapath you implemented in Phase 1. Finally, you will test your completed computer using a simple test program and ensure that it properly executes.
You must do the following:
1. Open and fill out microcode.xlsx file we’ve provided you (note: the formulas in the provided file will only work with Excel). You will need to mark which control signal is high (that is 1) for each of the states.
2. After you have completed all the microstates, convert the binary strings you just computed into hex and move them into the main ROM. You can just copy and paste the hex column (highlighted yellow) from the spreadsheet directly into the ROM component in Circuitsim. Do the same for the sequencer and condition ROMs.
3. Connect the completed control unit to the datapath you implemented in Phase 1. Using Figures 1 and 2, connect the control signals to their appropriate spots.
4. Finally, it is time to test your completed computer. Use the provided assembler (found in the “assembly” folder) to convert a test program from assembly to hex. For instructions on how to use the assembler and simulator, see README.txt in the “assembly” folder. Note: The simulator does not test your project, it simply provides a model. To test your design, you must load the assembled HEX into CircuitSim. We recommend using test programs that contain a single instruction since you are bound to have a few bugs at this stage of the project. Once you have built confidence, test your processor with the provided pow.s program as a more comprehensive test case.
6 Autograder
• Don’t rename the main subcircuit ”Datapath”
• Name your PC register as “PC”
• Name your IR register as “IR”
• Name your state register as “State”
• Name your 3 ROMs as ”MAIN”, ”SEQ”, and ”CC” respectively
• Reference registers in regfile in lower case with prefixed ‘$’ sign ($zero, $at, $v0…), according to Appendix A: LC-3300 Instruction Set Architecture.
• Use only one clock globally, which means you should not use clock components in any subcircuits besides the main datapath. Instead, you should use an input pin and connect the clock signal to that subcircuit in the main datapath.
• Use only one RAM as memory
• In microcode, use the first row as your first state of FETCH macrostate.
• Don’t change the layout of the microcode Excel sheet
7 Deliverables
To submit your project, you need to upload the following files to Gradescope:
• CircuitSim datapath file (LC-3300.sim)
• Microcode file (microcode.xlsx)
If you are missing one or both of those files, you will get a 0 so make sure that you have uploaded both of them. Always re-download your assignment from Gradescope after submitting to ensure that all necessary files were properly uploaded. If what we download does not work, you will get a 0 regardless of what is on your machine. This project will be demoed.In order to receive full credit, you must sign up for a demo slot and complete the demo.We will announce when demo times are released.
8 Appendix A: LC-3300 Instruction Set Architecture
The LC-3300 is a simple, yet capable computer architecture. The LC-3300 combines attributes of both ARM and the LC-2200 ISA defined in the Ramachandran & Leahy textbook for CS 2200.
The LC-3300 is a word-addressable, 32-bit computer. All addresses refer to words, i.e. the first word (four bytes) in memory occupies address 0x0, the second word, 0x1, etc.
All memory addresses are truncated to 16 bits on access, discarding the 16 most significant bits if the address was stored in a 32-bit register. This provides roughly 64 KB of addressable memory.
8.1 Registers
The LC-3300 has 16 general-purpose registers. While there are no hardware-enforced restraints on the uses of these registers, your code is expected to follow the conventions outlined below.
Table 1: Registers and their Uses
Register Number Name Use Callee Save?
0 $zero Always Zero NA
1 $at Assembler/Target Address NA
2 $v0 Return Value No
3 $a0 Argument 1 No
4 $a1 Argument 2 No
5 $a2 Argument 3 No
6 $t0 Temporary Variable No
7 $t1 Temporary Variable No
8 $t2 Temporary Variable No
9 $s0 Saved Register Yes
10 $s1 Saved Register Yes
11 $s2 Saved Register Yes
12 $k0 Reserved for OS and Traps NA
13 $sp Stack Pointer No
14 $fp Frame Pointer Yes
15 $ra Return Address No
1. Register 0 is always read as zero. Any values written to it are discarded. Note: for the purposes of this project, you must implement the zero register. Regardless of what is written to this register, it should always output zero.
3. Register 2 is where you should store any returned value from a subroutine call.
4. Registers 3 – 5 are used to store function/subroutine arguments. Note: registers 2 through 8 should be placed on the stack if the caller wants to retain those values. These registers are fair game for the callee (subroutine) to trash.
5. Registers 6 – 8 are designated for temporary variables. The caller must save these registers if they want these values to be retained.
7. Register 12 is reserved for handling interrupts. While it should be implemented, it otherwise will not have any special use on this assignment.
8. Register 13 is the everchanging top of the stack; it keeps track of the top of the activation record for a subroutine.
9. Register 14 is the anchor point of the activation frame. It is used to point to the first address on the activation record for the currently executing process.
10. Register 15 is used to store the address a subroutine should return to when it is finished executing.
8.2 Instruction Overview
The LC-3300 supports a variety of instruction forms, only a few of which we will use for this project. The instructions we will implement in this project are summarized below.
Table 2: LC-3300 Instruction Set
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0000 DR SR1 unused SR2
0001 DR SR1 unused SR2
0010 DR SR1 immval20
0011 DR BaseR offset20
0100 SR BaseR offset20
0101 SR1 SR2 offset20
0110 AT RA unused
0111 unused
1000 SR1 SR2 offset20
1001 DR unused PCoffset20
1010 DR SR1 unused 00 SR2
1010 DR SR1 unused 01 SR2
1010 DR SR1 unused 10 SR2
1010 DR SR1 unused 11 SR2
1011 SR unused
ADD
NAND
ADDI
LW
SW
BEQ
JALR
HALT
BGT
LEA
SLL
SRL
ROL
ROR
FABS
8.2.1 Conditional Branching
Branching in the LC-3300 ISA is a bit different than usual. We have a set of branching instructions including BEQ, BGT, and FABS which offer the ability to branch upon a certain condition being met. These instructions use comparison operators, comparing the values of two source registers. If the comparisons are true (for example, with the BGT instruction, if SR1 > SR2), then we will branch to the target destination of incrementedPC + offset20. For FABS, if SR < 0 then we will branch to the series of microstates for negation.
8.3 Detailed Instruction Reference
8.3.1 ADD
Assembler Syntax
ADD DR, SR1, SR2
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0000 DR SR1 unused SR2
Operation
DR = SR1 + SR2;
Description
The ADD instruction obtains the first source operand from the SR1 register. The second source operand is obtained from the SR2 register. The second operand is added to the first source operand, and the result is stored in DR.
8.3.2 NAND
Assembler Syntax
NAND DR, SR1, SR2
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0001 DR SR1 unused SR2
Operation
DR = ~(SR1 & SR2);
Description
The NAND instruction performs a logical NAND (AND NOT) on the source operands obtained from SR1 and SR2. The result is stored in DR.
HINT: A logical NOT can be achieved by performing a NAND with both source operands the same.
For instance,
NAND DR, SR1, SR1…achieves the following logical operation: DR ← SR1.
8.3.3 ADDI
Assembler Syntax
ADDI DR, SR1, immval20
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0010 DR SR1 immval20
Operation
DR = SR1 + SEXT(immval20);
Description
The ADDI instruction obtains the first source operand from the SR1 register. The second source operand is obtained by sign-extending the immval20 field to 32 bits. The resulting operand is added to the first source operand, and the result is stored in DR.
8.3.4 LW
Assembler Syntax
LW DR, offset20(BaseR)
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0011 DR BaseR offset20
Operation
DR = MEM[BaseR + SEXT(offset20)];
Description
An address is computed by sign-extending bits [19:0] to 32 bits and then adding this result to the contents of the register specified by bits [23:20]. The 32-bit word at this address is loaded into DR.
8.3.5 SW
Assembler Syntax
SW SR, offset20(BaseR)
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0100 SR BaseR offset20
Operation
MEM[BaseR + SEXT(offset20)] = SR;
Description
An address is computed by sign-extending bits [19:0] to 32 bits and then adding this result to the contents of the register specified by bits [23:20]. The 32-bit word obtained from register SR is then stored at this address.8.3.6 BEQ
BEQ SR1, SR2, offset20
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0101 SR1 SR2 offset20
Operation
if (SR1 == SR2) {
PC = incrementedPC + offset20
}
Description
A branch is taken if SR1 is equal to SR2. If this is the case, the PC will be set to the sum of the incremented PC (since we have already undergone fetch) and the sign-extended offset[19:0].
8.3.7 JALR
Assembler Syntax
JALR AT, RA
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0110 AT RA unused
Operation
RA = PC;
PC = AT;
Description
First, the incremented PC (address of the instruction + 1) is stored into register RA. Next, the PC is loaded with the value of register AT, and the computer resumes execution at the new PC.
8.3.8 HALT
Assembler Syntax
HALT
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0111 unused
Description
The machine is brought to a halt and executes no further instructions.
8.3.9 BGT
BGT SR1, SR2, offset20
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
1000 SR1 SR2 offset20
Operation
if (SR1 > SR2) {
PC = incrementedPC + offset20
}
Description
A branch is taken if SR1 is greater than SR2. If this is the case, the PC will be set to the sum of the incremented PC (since we have already undergone fetch) and the sign-extended offset[19:0].
8.3.10 LEA
Assembler Syntax
LEA DR, label
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
1001 DR unused PCoffset20
Operation
DR = PC + SEXT(PCoffset20);
Description
An address is computed by sign-extending bits [19:0] to 32 bits and adding this result to the incremented PC (address of instruction + 1). It then stores the computed address into register DR.
8.3.11 SLL
Assembler Syntax
SLL DR, SR1, SR2
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
1010 DR SR1 unused 00 SR2
Operation
DR = SR1 << SR2;
Description
The value stored in SR1 is logically left shifted by the value stored in SR2, and the result is stored in DR.
8.3.12 SRL
SRL DR, SR1, SR2
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
1010 DR SR1 unused 01 SR2
Operation
DR = SR1 >> SR2;
Description
The value stored in SR1 is logically right shifted by the value stored in SR2, and the result is stored in DR.
8.3.13 ROL
Assembler Syntax
ROL DR, SR1, SR2
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
1010 DR SR1 unused 10 SR2
Operation
DR = (SR1 << SR2) | (SR1 >> (32 – SR2));
Description
Bits in SR1 are ”rotated” left by SR2 number of bits using circular shifting. During a left rotation, the bits that are shifted out from the left are brought back in on the right side.
8.3.14 ROR
Assembler Syntax
ROR DR, SR1, SR2
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
1010 DR SR1 unused 11 SR2
Operation
DR = (SR1 >> SR2) | (SR1 << (32 – SR2));
Description
Bits in SR1 are ”rotated” right by SR2 number of bits using circular shifting. During a right rotation, the bits that are shifted out from the right are brought back in on the left side.8.3.15 FABS
Assembler Syntax
FABS SR
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
1011 SR unused
Operation
if (SR < 0) {
SR = ~(SR & SR);
SR += 1;
}
Description
If SR is less than zero, then the microcontroller will redirect to the series of negate microstates.
9 Appendix B: Microcontrol Unit
You will make a microcontrol unit which will drive all of the control signals to various items on the datapath. This Finite State Machine (FSM) can be constructed in a variety of ways. You could implement it with combinational logic and flip flops, or you could hard-wire signals using a single ROM. The single ROM solution will waste a tremendous amount of space since most of the microstates do not depend on the opcode or the conditional test to determine which signals to assert. For example, since the condition line is an input for the address, every microstate would have to have an address for condition = 0 as well as condition = 1, even though this only matters for one particular microstate.
To solve this problem, we will use a three ROM microcontroller. In this arrangement, we will have three ROMs:
• the main ROM, which outputs the control signals,
• the sequencer ROM, which helps to determine which microstate to go at the end of the FETCH state,
• and the condition ROM, which helps determine whether or not to branch during the BEQ and whether or not to negate during the FABS instructions.
Examine the following:Figure 2: Three ROM Microcontrol Unit
As you can see, there are three different locations that the next state can come from: part of the output from the previous state (main ROM), the sequencer ROM, and the condition ROM. The mux controls which of these sources gets through to the state register. If the previous state’s “next state” field determines where to go, neither the OPTest nor ChkCmp signals will be asserted. If the opcode from the IR determines the next state (such as at the end of the FETCH state), the OPTest signal will be asserted. If the comparison circuitry determines the next state (such as in the BEQ instructions), the ChkCmp signal will be asserted. Note that these two signals should never be asserted at the same time since nothing is input into the “11” pin on the MUX.
The sequencer ROM should have one address per instruction, and the condition ROM should have one address for condition true and one for condition false.
Before getting down to specifics you need to determine the control scheme for the datapath. To do this examine each instruction, one by one, and construct a finite state bubble diagram showing exactly what control signals will be set in each state. Also determine what are the conditions necessary to pass from one state to the next. You can experiment by manually controlling your control signals on the bus you’ve created in Phase 1 – Implement the Datapath to make sure that your logic is sound.
Once the finite state bubble diagram is produced, the next step is to encode the contents of the Control Unit ROM. Then you must design and build (in CircuitSim) the Control Unit circuit which will contain the three ROMs, a MUX, and a state register. Your design will be better if it allows you to single step and ensure that it is working properly. Finally, you will load the Control Unit’s ROMs with the hexadecimal generated by your filled out microcode.xlsx.
Note that the input address to the ROM uses bit 0 for the lowest bit of the current state and 5 for the highest bit for the current state.
Table 3: ROM Output SignalsBit Purpose Bit Purpose Bit Purpose Bit Purpose Bit Purpose
0 NextState[0] 6 DrREG 12 LdIR 18 WrMEM 24 OPTest
1 NextState[1] 7 DrMEM 13 LdMAR 19 RegSelLo 25 ChkCmp
2 NextState[2] 8 DrALU 14 LdA 20 RegSelHi
3 NextState[3] 9 DrPC 15 LdB 21 ALU0
4 NextState[4] 10 DrOFF 16 LdCmp 22 ALU1
5 NextState[5] 11 LdPC 17 WrREG 23 ALU2Table 4: Register Selection MapRegSelHi RegSelLo Register
0 0 RX (IR[27:24])
0 1 RY (IR[23:20])
1 0 RZ (IR[3:0])
1 1 unusedTable 5: ALU Function Map (Option 1)ALU2 ALU1 ALU0 IR[5] IR[4] Function
0 0 0 N/A N/A ADD
0 0 1 N/A N/A SUB
0 1 0 N/A N/A NAND
0 1 1 N/A N/A A + 1
1 IR[5] IR[4] 0 0 SLL
1 IR[5] IR[4] 0 1 SRL
1 IR[5] IR[4] 1 0 SRA
1 IR[5] IR[4] 1 1 RORALU2 ALU1 ALU0 Function
0 0 0 ADD
0 0 1 SUB
0 1 0 NAND
0 1 1 A + 1
1 0 0 SHIFT
IR[5] IR[4] Function
0 0 SLL
0 1 SRL
1 0 ROL
1 1 ROR
Table 6: ALU Function Map (Option 2)
NOTE: The ordering of RegSelHi/RegSelLo and ALU2/ALU1/ALU0 are in the opposite order in the microcode excel spreadsheet! Please make sure you are enabling the correct signals in your microcode!