Project 1 Solution

Description

• Requirements

Download the proper version of Brandonsim. A copy of Brandonsim is available under Files on Canvas. In order to run Brandonsim, Java must be installed. If you are a Mac user, you may need to right-click on the JAR le and select \Open” in the menu to bypass Gatekeeper restrictions.

Brandonsim is not perfect and does have small bugs. In certain scenarios, les have been corrupted and students have had to re-do the entire project. Please back up your work using some form of version control, such as a local/private git repository or Dropbox. Do not use public git repositories, it is against the Georgia Tech Honor Code.

The BOB-2200 assembler is written in Python. If you do not have Python 2.6 or newer installed on your system, you will need to install it before you continue.

• 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 Brandonsim 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 nite state machine (the \control unit”) to control your computer and actually run programs on it.

Note: You will need to have a working knowledge of Brandonsim. Make sure that you know how to make basic circuits as well as subcircuits before proceeding. The TAs are always here if you need help.

Project 1 CS 2200 – Systems and Networks

• Phase 1 – Implement the Datapath

BOB-2200 Datapath

LdPC

PC

LdA

A

B

LdB

LdMAR

MAR

LdIR

IR

24

IR[31:0]

ALU:

WrREG

Din

WrMEM

Addr

Din

func

00: ADD

2

01: SUB

Registers

Memory

IR[19:0]

10: NAND

11:A+1

16 x 32 bits

224 x 32 bits

regno

Sign

4

Dout

Dout

Extend

DrPC

DrALU

DrREG

DrMEM

DrOFF

DrCmp

CondType

Comparison

IR[31:28]

OP: 4-bit opcode to control logic

Logic

IR[27:24]

Rx: 4-bit register number to control logic

IR[23:20]

Ry: 4-bit register number to control logic

IR[3:0]

Rz: 4-bit register number to control logic

LdCmp CmpReg

CmpOut – to control logic

Figure 1: Datapath for the BOB-2200 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 BOB-2200 datapath in Brandonsim using what you have just learned.

You must do the following:

1. Learn and understand the BOB-2200 ISA. The ISA is fully speci ed and de ned in Appendix A: BOB-2200 Instruction Set Architecture. Do not move on until you have fully read and understood the ISA speci cation. Every single detail will be relevant to implementing your datapath in the next step.

2. Using Brandonsim, implement the BOB-2200 datapath. To do this, you will need to use the details of the BOB-2200 datapath de ned in Appendix A: BOB-2200 Instruction Set Architecture. You should model your datapath on Figure 1.

3.1 Hints

3.1.1 Subcircuits

Brandonsim enables you to break create reusable components in the form of subcircuits. We highly recom-mend that you break 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.

Project 1 CS 2200 – Systems and Networks

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 Brandonsim limitations, the RAM module is limited to no more than 24 address bits. Therefore, per our ISA, any 32-bit values used as memory addresses will be truncated to 24 bits (with the 8 most signi cant bits disregarded). We recommend you implement this (i.e. truncate the most signi cant bits) before feeding the address value from the MAR (Memory Address Register) to the RAM.

3.1.4 Comparison Logic

The \comparison logic” box in Figure 1 is responsible for performing the comparison logic associated with the BNE and SLT instructions. The comparison logic should read the current value on the bus. When executing BNE or SLT, 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 \true” or \false” value for either the condition A! = B or A < B depending on the CondType signal.

Your comparison logic should be purely combinational. Feel free to use any Brandonsim components you wish to aid in your implementation.

For the SLT instruction, you will need to drive the output of the comparison register to the data bus. Note the tri-state bu er controlled by the DrCmp signal. Keep in mind that the CmpOut signal must be a 1-bit value, but the data bus is 32-bits wide.

3.1.5 Zero Register

Your zero register must be implemented such that writes to it are ine ective, i.e., attempting to write a non-zero value to the zero register will do nothing. Do not forget to do this or you will lose points!

3.1.6 Register Select

From lecture and the textbook, you should be familiar with the \register select” (RegSel) multiplexer. The mux is responsible for feeding the register number from the correct eld in the instruction into the register le. See Table 4 for a list of inputs your mux should have.

• Phase 2 – Implement the Microcontrol Unit

In this phase of the project, you will use Brandonsim to implement the microcontrol unit for the BOB-2200 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.

Project 1 CS 2200 – Systems and Networks

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 Brandonsim. 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 BOB-2200 datapath schematic (see Figure 1).

• Phase 3 – Microcode and Testing

In this nal 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. Fill out the microcode.xlsx (Excel spreadsheet) le that we have provided. You will need to mark which control signal is high (that is 1) for each of the states. We have given you a starting template.

2. After you have completed all the states, convert the binary strings you just computed into hex and move them to the main ROM (the Excel sheet will do this for you; just copy-and-paste the hex column into the Brandonsim ROM). 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 \as-sembly” 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 Brandonsim.

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 con dence, test your processor with the provided pow.s program as a more comprehensive test case.

• Deliverables

Please submit all of the following les in a .tar.gz archive generated by our Make le.

The Make le will work on any Unix or Linux-based machine (on Ubuntu, you may need to sudo apt-get install build-essential if you have never installed the build tools).

Run make submit to automatically package your project into the correct archive format. The generated archive should contain at a minimum the following les:

Brandonsim Datapath File (BOB-2200.circ) Microcode le (microcode.xlsx)

Always re-download your assignment from Canvas after submitting to ensure that all necessary les 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.

Project 1 CS 2200 – Systems and Networks

• Appendix A: BOB-2200 Instruction Set Architecture

The BOB-2200 is a simple, yet capable computer architecture. The BOB-2200 combines attributes of both ARM and the LC-2200 ISA de ned in the Ramachandran & Leahy textbook for CS 2200.

The BOB-2200 is a word-addressable, 32-bit computer. All addresses refer to words, i.e. the rst word (four bytes) in memory occupies address 0x0, the second word, 0x1, etc.

All memory addresses are truncated to 24 bits on access, discarding the 8 most signi cant bits if the address was stored in a 32-bit register. This provides roughly 67 MB of addressable memory.

7.1 Registers

The BOB-2200 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.

2. Register 1 is used to hold the target address of a jump. It may also be used by pseudo-instructions generated by the assembler.

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.

6. Registers 9 – 11 are saved registers. The caller may assume that these registers are never tampered with by the subroutine. If the subroutine needs these registers, then it should place them on the stack and restore them before they jump back to the caller.

Project 1 CS 2200 – Systems and Networks

7. Register 12 is reserved for handling interrupts. While it should be implemented, it otherwise will not have any special use on this assignment.

7. Register 13 is your anchor on the stack. It keeps track of the top of the activation record for a subroutine.

7. Register 14 is used to point to the rst address on the activation record for the currently executing process. Don’t worry about using this register.

7. Register 15 is used to store the address a subroutine should return to when it is nished executing.

7.2 Instruction Overview

The BOB-2200 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: BOB-2200 Instruction Set

31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0

ADD	0000	DR	SR1	unused	SR2
NAND
	0001	DR	SR1	unused	SR2
ADDI
	0010	DR	SR1	immval20
LW
	0011	DR	BaseR	o set20
SW
	0100	SR	BaseR	o set20
BNE
	0101	SR1	SR2	o set20
JALR
	0110	RA	AT	unused
HALT
	0111			unused
SLT
	1000	DR	SR1	unused	SR2
LEA
	1001	DR	unused	o set20

7.2.1 Conditional Branching

Conditional branching in the BOB-2200 ISA is provided via the BNE (\branch if not equal”) instruction. BNE will branch to address “incrementedPC + o set20” only if SR1 and SR2 are not equal

Project 1 CS 2200 – Systems and Networks

7.3 Detailed Instruction Reference

7.3.1 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 rst source operand from the SR1 register. The second source operand is obtained from the SR2 register. The second operand is added to the rst source operand, and the result is stored in DR.

7.3.2 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.

Project 1 CS 2200 – Systems and Networks

7.3.3 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 rst source operand from the SR1 register. The second source operand is obtained by sign-extending the immval20 eld to 32 bits. The resulting operand is added to the rst source operand, and the result is stored in DR.

7.3.4 LW

Assembler Syntax

LW DR, offset20(BaseR)

Encoding

31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0

0011	DR	BaseR	o set20

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 speci ed by bits [23:20]. The 32-bit word at this address is loaded into DR.

Project 1 CS 2200 – Systems and Networks

7.3.5 SW

Assembler Syntax

SW SR, offset20(BaseR)

Encoding

31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0

0100	SR	BaseR	o set20

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 speci ed by bits [23:20]. The 32-bit word obtained from register SR is then stored at this address.

7.3.6 BNE

Assembler Syntax

BNE SR1, SR2, offset20

Encoding

31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0

0101	SR1	SR2		o set20
Operation
if (SR1 != SR2) {
PC = incrementedPC + offset20
}

Description

A branch is taken if SR1 and SR2 are not equal. 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 o set[19:0].

Project 1 CS 2200 – Systems and Networks

7.3.7 JALR

Assembler Syntax

JALR RA, AT

Encoding

31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0

0110	RA	AT	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.

7.3.8 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.

7.3.9 SLT

Assembler Syntax

SW DR, SR1, SR2

Encoding

31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0

1000	DR	SR1	unused	SR2

Operation

if (SR1 < SR2) {

DR=1

} else { DR=0

}

Description

If SR1 is less than SR2, a 1 should be stored into DR. Otherwise a 0 should be stored in DR.

Project 1 CS 2200 – Systems and Networks

7.3.10 LEA

Assembler Syntax

LEA DR, label

Encoding

31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0

1001	DR	unused	PCo set20

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.

Project 1 CS 2200 – Systems and Networks

• 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 ip ops, 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 BNE instruction.

Examine the following:

Figure 2: Three ROM Microcontrol Unit

As you can see, there are three di erent 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” eld determines where to go, neither the OPTest nor ChkComp 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 BNE instruction), 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.

Project 1 CS 2200 – Systems and Networks

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.

Note: Brandonsim has a minimum of two address bits for a ROM (i.e. four addresses), even though only one address bit (two addresses) is needed for the condition ROM. Just ignore the other two addresses. You should design it so that the high address bit for this ROM is permanently set to zero.

Before getting down to speci cs you need to determine the control scheme for the datapath. To do this examine each instruction, one by one, and construct a nite 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 nite state bubble diagram is produced, the next step is to encode the contents of the Control Unit ROM with the provided Excel sheet. Then you must design and build (in Brandonsim) 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 from the Excel sheet.

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 Signals

Bit	Purpose		Bit	Purpose		Bit	Purpose	Bit	Purpose	Bit	Purpose
0	NextState[0]		6	DrREG		12	LdIR	18	WrMEM	24	ChkCmp
1	NextState[1]		7	DrMEM		13	LdMAR	19	RegSelLo	25	DrCmp
2	NextState[2]		8	DrALU		14	LdA	20	RegSelHi	26	CondType
3	NextState[3]		9	DrPC		15	LdB	21	ALULo
4	NextState[4]		10	DrOFF		16	LdCmp	22	ALUHi
5	NextState[5]		11	LdPC		17	WrREG	23	OPTest

Table 4: Register Selection Map

RegSelHi	RegSelLo	Register
0	0	RX (IR[27:24])
0	1	RY (IR[23:20])
1	0	RZ (IR[3:0])
1	1	unused

Table 5: ALU Function Map

ALUHi	ALUlLo	Function
0	0	ADD
0	1	SUB
1	0	NAND
1	1	A + 1