Digital Systems Design
2017
Project 1
Part 1 Computer
Professor Jonathan H. Manton
Interface
A computer needs to have inputs and outputs to be useful.
Numbered from left to right, top to bottom:
1. Current contents of Instruction Pointer (IP) (hexadecimal)
2. Can be blank, or can display a decimal number between -128 and 127 inclusively
3. Available for hardware debugging
4. General Purpose Output: each LED can be turned on or off
5. Reset switch (HIGH to reset CPU)
6. Turbo switch (HIGH for turbo mode)
7. Data In: will be read by the CPU when the Sample Button is released
8. Sample Button (push then release for Data In to be read into CPU)
9. Buttons: CPU will be notified each time a push button is released
In the above list, items in orange are built into the computer hardware, items in purple are outputs
under software control, and those in blue are inputs that can be read by the software. Note though
that the software cannot read in the current values of the switches (Data In) when it pleases.
Rather, the CPU hardware arranges for the values of the switches to be latched into the CPU
whenever the Sample button is released. This latched version is available for the software to read at
any time.
CPU Pins
The Central Processing Unit (CPU) will have the following pins, allowing it to communicate with
external circuitry.
Input Pins
Din (Data In) 8 bits Allows the CPU to read in an 8-bit number. Must be kept stable around
the time when the Sample pin goes from HIGH to LOW (falling edge).
Sample 1 bit On the falling edge, data on the Din pins will be clocked into the CPU.
Btns 3 bits Three buttons. On a falling edge, a button release will be recorded.
Clock 1 bit External clock (square wave, system-wide)
Reset1 bit When HIGH, reset CPU.Minimum pulse duration should be 25 ms.
Turbo 1 bit When HIGH, CPU will run at full speed
Output Pins
Dout (Data Output) 8 bits Used to output a number
Dval (Data Valid) 1 bit HIGH if Dout is valid
GPO (General Purpose Output) 6 bits Additional output lines for CPU
Debug 4 bits Used to pass information out of CPU module
IP (Instruction Pointer) 8 bits Current value of IP register
CPU Architecture
Our CPU is not going to be efficient! Only once we have built our own primitive CPU, can we truly
appreciate more refined CPUs.
Note: If you are not familiar with CPUs then you might not be able to guess from this overview
exactly how a CPU works. More detail will be given later, if not explicitly then at least implicitly,
when the implementation is described in stages.
Registers
The CPU will have access to 32 8-bit wide registers, numbered 0 to 31. Most of them are general
purpose registers. The last four registers though have special purposes.
Register 28 (DINP)holds the latched contents of the Din CPU pins.
The contents of Register 29 (GOUT) will appear on the Dval and GPO output pins of the CPU.
Bit 7 Bit 6 Bit 5 Bit 4 Bit 3 Bit 2 Bit 1 Bit 0
Dval GPO[5] GPO[4] GPO[3] GPO[2] GPO[1] GPO[0]
DVAL
The contents of Register 30 (DOUT) will appear on the Dout output pins of the CPU.
Bit 7 Bit 6 Bit 5 Bit 4 Bit 3 Bit 2 Bit 1 Bit 0
Dout[7] Dout[6] Dout[5] Dout[4] Dout[3] Dout[2] Dout[1] Dout[0]
Register 31 (FLAG) is known as the Flag Register. Certain events will cause it to be modified.
Bit 7 Bit 6 Bit 5 Bit 4 Bit 3 Bit 2 Bit 1 Bit 0
Shift Overflow Sample Btn 2 Btn 1 Btn 0
SHFT OFLW SMPL
If a pushbutton is pressed then released, the corresponding bit of the Flag Register will be set to a 1.
It will remain a 1 until cleared by the programmer. The Overflow bit will be set or cleared after each
arithmetic operation that has the possibility of overflowing, such as addition or multiplication. The
Shift bit will be modified after a shift operation has taken place; it is the value of the bit shifted out.
Program Memory
The CPU will read instructions from the Program Memory, which is a block of memory containing
256 memory cells, each cell being 35 bits wide. This memory is read only and asynchronous1.
Instruction Set
Each instruction is 35 bits wide: such a large width wastes memory but makes the job of both the
programmer and designer easier. The programmer can write in machine code more easily due to the
simple structure, and similarly, when we come to designing the CPU, it will be easier to decode each
instruction. A 35 bit instruction is decomposed as follows.
Bits 34-31 Bits 30-28 Bits 27-18 Bits 17-8 Bits 7-0
Command Group Command Argument 1 Argument 2 Address
Command Group
The first 4 bits of an instruction encode the Command Group.
4b0000 NOP No Operation
4b1000 JMP Jump
4b0100 ATC Atomic Test and Clear
4b0010 MOV Move
4b0001 ACC Accumulate
Command
The next 3 bits of an instruction specify the Command. Their meaning depends on the Command
Group. The following types of JMP are recognised.
3b000 UNC Unconditional Jump
3b010 EQ Jump on Equality
3b100 ULT Jump on Unsigned Less Than
3b101 SLT Jump on Signed Less Than
3b110 ULE Jump on Unsigned Less Than or Equals
3b111 SLE Jump on Signed Less Than or Equals
1 A more realistic computer would use synchronous memory, but then we would have to use every second
clock cycle to fetch the next instruction in from memory, unnecessarily complicating the design. That said, it is
common to use asynchronous memory for controlling FSMs on FPGAs; see Chapter 18 of the textbook.
The ATC Command Group is used to test a particular bit of the Flag Register. The 3-bit Command
specifies which bit (from 0 to 7).
The following types of MOV are recognised.
3b000 PUR Pure Move
3b001 SHL Shift Left
3b010 SHR Shift Right
The following types of ACC are recognised.
3b000 UAD Unsigned Addition
3b001 SAD Signed Addition
3b010 UMT Unsigned Multiplication
3b011 SMT Signed Multiplication
3b100 AND Bitwise And
3b101 OR Bitwise Or
3b110 XOR Bitwise XOR
Arguments 1 and 2
The next 10 bits of an instruction comprise Argument 1, and the 10 bits after those comprise
Argument 2. Arguments describe either an 8-bit number or a location in one of three ways.
Decompose the argument as {a,b} where a is a 2-bit number and b an 8-bit number.If a=00 then the
number is simply b. For example, 10b00_0000_1000 represents the number 8. If a location is
required, it is an error to use a=00. If a=01 then the location is Register b, and if a number is
required, then the number is whatever is in Register b. If a=10 then the location is Register c, where
c is the current value of Register b; this affords an extra level of indirection that can be useful to
have. Similarly, if a number is expected then the number is whatever is in Register c.
Whether a location or a number is expected will be clear from the Command Group.
Argument (2 bits, 8 bits) Location Number
{ 2b00, b } b
{ 2b01, b } Register b (0 to 31) Contents of Register b
{ 2b10, b } Register c, where c = contents
of Register b
Contents of Register c, where c =
contents of Register b
Address
Finally, the remaining 8 bits of an Instruction store an Address that is used for the JMP and ATC
Command Groups.
Remark
There are many instructions CPUs normally have that we have omitted for simplicity. Nevertheless,
this CPU already has more than enough instructions with which to implement an RPN calculator. You
can always add more if you wish though!
Stage 1
Download the Skeleton Project directory, rename it, and use it to develop your project. The top-level
module is in MyComputer.v.
Go to Assignments > Settings > Files and add MyComputer.v, AuxMod.v, CPU.v, ROM.v and CPU.vh
to the project. (The MyComputer.sdc should already be listed there.)
MyComputer.v Top-level design
AuxMod.v Auxilliary Modules
CPU.v CPU module
ROM.v Program Memory (including program to run)
CPU.vh `define definitions
Create an empty module for your CPU in the file CPU.v. You must fill in the correct inputs and
outputs. All inputs and outputs should be unsigned, and use the standard convention for ordering
multiple bits (e.g., input [7:0] Din).
Instantiate the CPU module in the top-level module MyComputer.
Feed the 50 MHz clock into the CPU.
Connect the GPO pins to the six right-most LEDs.
Connect the Debug pins to the remaining LEDs.
Connect the right-most eight switches to Din.
Connect the Turbo switch to the Turbo pin.
Connect an inverted version of the left-most push button to Sample. (The CPU expects 1 to mean
a button has been pressed, but on the DE1-SoC board, pressing a push button produces a 0.)
Connect inverted versions of the remaining three push buttons to Btns.
Create a module Debounce (in AuxMod.v) that takes as input a clock (50 MHz) and a wire, and
returns a debounced version of the signal on that wire. Stage 2 will create the debounce circuitry; for
now, just leave empty the body of the module. Connect the reset switch to the reset pin of the CPU
via an instantiation of this Debounce module.
Create a module Disp2cNum (in AuxMod.v). It takes an 8-bit 2s complement number as input and
displays the number on four 7-segment displays. A subsequent stage will fill in this module; leave it
blank for now. Connect the Dout of the CPU to an instantiation of Disp2cNum, and connect the
output of this Disp2cNum instance to the right-most four 7-segment displays. Add an extra input to
the module, called enable. Connect the Dval CPU pin to enable, so that the CPU can switch off
the display if there is nothing to display.
Create a module DispHex (in AuxMod.v) that displays an 8-bit number in hexadecimal on two 7-
segment displays. Use this to display the output of the CPUs IP pins on the left-most 7-segment
displays. Leave the module blank for now.
Note
The intention is not for you to reverse-engineer any RTL diagrams that are provided. Some of the
writing is very small, possibly illegible. They are there simply to give you something to compare with,
in broad terms, to see if you are on the right track or not. Ultimately, you will need to test that your
circuit works (e.g., by writing test benches). Also, you are encouraged to choose more meaningful
names than I do.
Stage 2
External inputs must be synchronised, to avoid metastability. Furthermore, we are told that the
reset switch needs debouncing, and of a minimum 25 ms duration. (The other switches do not need
debouncing because we will tell anyone that wants to use our computer not to change the switches
around the time when the Sample button is released for else the results may be unpredictable. The
switches still need synchronising though, just in case our friends try to break our computer by
madly pressing buttons and moving switches at the same time)
Synchroniser
Create a module Synchroniser (in AuxMod.v) that takes as input the 50 MHz clock and an input wire,
and by using a double flip-flop synchroniser, produces a synchronised version of the input signal. (If
you are truly paranoid you can use three flip-flops, but Quartus assures me that we can expect less
than one error in 1,000,000,000 years by using just two flip-flops in our design.)
Debounce
Fill in the Debounce module, so that it operates as follows. First, send the input through a
Synchroniser. Then implement a FSM that functions in an equivalent way to the following
description. If the synchronised input remains a 1 for 30 ms then output a 1. If the synchronised
input remains a 0 for 30 ms then output a 0. Otherwise, do not change the output.
Your design must ensure that the shortest period the output can be HIGH for is at least 25 ms
(because this was a stated requirement2 on the CPU Reset pin specification given to you). In
particular, you must ensure there are no glitches on the output. Therefore, you should not use
combinational logic or a latch to produce y. Instead, use a flip-flop.
Make sure that if the synchronised input changes right at 30 ms, that you do not accidentally change
the output to the wrong value (i.e., do not naively make the output equal to the synchronised input
once 30 ms expires, unless you are sure you know it will work).
Hint
Here is my implementation of the Debounce module. Remember, you should not try to reverse-
engineer the RTL, but rather, write the Verilog code yourself and then compare in broad terms
whether the produced RTL is similar to the following. Maybe you have a better design?
2 This requirement is mainly to give you practice; 25 ms is considerably longer than necessary.
Stage 3
To get something up and running, we will implement a CPU that just does the following.
When the Reset pin is HIGH, all the output pins will be set to 0.
When the Reset pin is LOW, the CPU should run, which in this case, just means setting all the GPO
pins to 1.
Simulate your design to see if it works as expected. Make sure to check that the Debounce circuitry
functions correctly.
Stage 4
Fill in the modules Disp2cNum and DispHex.
The following module will display a blank, a negative sign, or a hexadecimal digit. (It should already
be in the AuxMod.v file that you downloaded.)
module SSeg(input [3:0] bin, input neg, input enable, output reg [6:0] segs);
always @(*)
if (enable) begin
if (neg) segs = 7b011_1111;
else begin
case (bin)
0: segs = 7b100_0000;
1: segs = 7b111_1001;
2: segs = 7b010_0100;
3: segs = 7b011_0000;
4: segs = 7b001_1001;
5: segs = 7b001_0010;
6: segs = 7b000_0010;
7: segs = 7b111_1000;
8: segs = 7b000_0000;
9: segs = 7b001_1000;
10: segs = 7b000_1000;
11: segs = 7b000_0011;
12: segs = 7b100_0110;
13: segs = 7b010_0001;
14: segs = 7b000_0110;
15: segs = 7b000_1110;
endcase
end
end
else segs = 7b111_1111;
endmodule
DispHex is straightforward to write: it just requires two instantiations of SSeg.
For Disp2cNum, note that leading zeros should be suppressed, so that -1 is displayed with two
leading blanks followed by -1, rather than -001. If you were writing software, you would use a for
loop. Because this is hardware, you should instead create a module that can be daisy-chained
together, to achieve the same algorithm.
When you believe you have finished, you can test it out by modifying your CPU to include a 23-bit
counter. Every time the CPU is running, on the rising edge of the clock, the counter should increment
by one. When the counter equals zero, the following registers should be incremented by one: Dout,
GPO, IP. Dont forget to enable Dval.
You should observe the following. When the reset switch is on, only 00 is showing (on the left-most
two 7-segment displays). When the reset switch is off, the LEDs should count in binary from 0 to 63,
the left-most two 7-segment displays should count from 00 to FF in hexadecimal, and the remaining
7-segment displays should count from 0 to 127 then from -128 back up to 0.
Note: You can use simulations to verify your design. You may also decide to create a new Project
Directory for each stage. Then when you get to the Lab to use a DE1-SoC board, you can test out
each stage quickly.
Hint
Here is a big hint.
module Disp2cNum(input signed [7:0] x, input enable, output [6:0] H3, H2, H1, H0);
wire neg = (x < 0);wire [7:0] ux = neg ? -x : x;wire [7:0] xo0, xo1, xo2, xo3;wire eno0, eno1, eno2, eno3;// You fill in the rest: create four instances of DispDec. endmodulemodule DispDec(input [7:0] x, input neg, enable, output reg [7:0] xo, output reg eno, output [6:0] segs);wire [3:0] digit;wire n;SSeg converter(digit, n, enable, segs);// You fill in the rest. Only a few lines of code are required. endmoduleIt may be handy to recall that you can find the least-significant decimal digit using the modulo 10 operation, written % 10 in Verilog. And you can divide by 10 using / 10. So if x is your unsigned number, you can first display x % 10 and then pass x / 10 on to the next DispDec. Stage 5 We will start by implementing just a single instruction. We will not even fully decode it properly. We want to be able to move a number into Register 30 so that it will be displayed on the 7-segment display. (Recall that Register 30 is one of the special registers; whatever is stored in Register 30 will be output on the Dout pins of the CPU.) The actual instruction should be written as 35b0010_000_00_xxxxxxxx_01_00011110_00000000 where the xxxxxxxx represents the 8-bit number we want to store in Register 30. Referring to the CPU Architecture Instruction Set, the first 4 bits (0010) specify the MOV command. The next 3 bits (000) show it is a Pure Move command. The first argument is the source: 00_xxxxxxxx. This represents a number because the leading two bits are 00. The destination of the move is the second argument: 01_00011110. The leading 2 bits (01) indicate the destination location is a register, and the following 8 bits (00011110) represent the number 30, hence the destination is Register 30. Writing instructions in binary will give us a good understanding of how CPUs work. However, it quickly becomes tedious. A standard trick is to use constants, as now described. It is important to keep in mind that the substitution occurs at compile time: it has nothing to do with how the CPU actually works. `define MOV 4b0010 `define PUR 3b000 `define NUM 2b00 `define REG 2b01 `define N8 8d0 We can then write {`MOV, `PUR, `NUM, 8h88, `REG, 8d30, `N8} instead of 35b0010_000_00_10001000_01_00011110_00000000. They are identical as far as the Verilog compiler is concerned, but the former is easier to read: we can see immediately that we wish to move the hexadecimal number 88 into Register 30. We also want to be able to slow the clock down so that we can see the computer work on the DE1-SoC board. (In simulations, we want it to run at full speed, hence the reason for implementing a Turbo switch.) Naively, we may think to pass either a slow clock or a fast clock to the CPU module. This is not a good idea though. Instead, the preferred method is to use the Clock Enable pin on flip-flops to slow things down. We will not explicitly tell Quartus to use Clock Enable, but the Verilog code we write will result in the use of Clock Enable. (In fact, the only benefit of Clock Enable is it saves some power; the alternative approach the synthesis tool might take is to feed the output of a flip-flop back to its input via a multiplexer. We could force the issue with a direct_enable attribute, but there is no need. The functionality will be the same.) Step 1 of Stage 5 Copy your Quartus Project Directory if you like, to back it up. Then delete the body of the CPU module since we will be starting again (but using code similar to what you developed already). To slow the computer down when required, implement a counter that resets back to zero when it reaches a particular value. Only start a new instruction cycle when the counter is zero. By changing the value at which the counter resets to zero, the speed of the CPU can be controlled. For the moment, choose a value so that a new instruction cycle starts every 250 ms. Each cycle, simply increment IP by 1, unless the Reset pin is asserted, in which case, set IP to 0. If you have access to a DE1-SoC board, test your design: the left-most two 7-segment displays should count at a rate of 4 numbers per second, or be reset to 0 if the Reset switch is HIGH. If testing via simulation, change the value at which the counter resets, to a small number like eight. Note: The simpler design is having a synchronous Reset. The requirement that the reset pin is HIGH for at least 25 ms is not needed. It was included for practice because some devices really do have a minimum pulse-width requirement. (It is still very important that the Reset pin is synchronised to the 50 MHz clock via the Synchroniser.) Hint For the counter, use something like the following. (You can add the bit widths to the constants, once you figure out how many bits are required.)// Clock circuitry (250 ms cycle)reg [???:0] cnt;localparam CntMax = ???;always @(posedge Clock) cnt <= (cnt == CntMax) ? 0 : cnt + 1; Create a wire for holding the condition for executing the Instruction Cycle. // Synchronise CPU operations to when cnt == 0wire go = !Reset && (cnt == 0); The Instruction Cycle then looks like this.// Instruction Cycle – Instruction Cycle Blockalways @(posedge Clock) begin // Process Instruction if (go) IP <= IP + 8’b1; // Process Reset if (Reset) IP <= 8’b0;endBy putting the if (go) statement first, it is more likely to be used as the Clock Enable pin for the majority of CPU registers, once the rest of the design is implemented. Question Why will Quartus complain if the Instruction Cycle is split over two always blocks, with the first having if (go) and the second having if (Reset)? Step 2 of Stage 5 We need to create the Program Memory, and place a program in there for the CPU to execute. We have choices: synchronous or asynchronous memory, stored inside the FPGA or in external memory attached to the FPGA. We will take the simplest approach. module AsyncROM(input [7:0] addr, output reg [34:0] data);always @(addr) case (addr)0: data = {`MOV, `PUR, `NUM, 8’d 1,`REG, `DOUT, `N8};1: data = {`MOV, `PUR, `NUM, 8’d 3,`REG, `DOUT, `N8};2: data = {`MOV, `PUR, `NUM, 8’d 5,`REG, `DOUT, `N8};3: data = {`MOV, `PUR, `NUM, 8’d 7,`REG, `DOUT, `N8};4: data = {`MOV, `PUR, `NUM, 8’d 9,`REG, `DOUT, `N8};5: data = {`MOV, `PUR, `NUM, 8’d 11, `REG, `DOUT, `N8};6: data = {`MOV, `PUR, `NUM, 8’d 13, `REG, `DOUT, `N8};7: data = {`MOV, `PUR, `NUM, 8’d 15, `REG, `DOUT, `N8};8: data = {`MOV, `PUR, `NUM, 8’d 17, `REG, `DOUT, `N8};9: data = {`MOV, `PUR, `NUM, 8’d 19, `REG, `DOUT, `N8};default: data = 35’b0; // Default instruction is a NOP endcase endmodule Place the above in ROM.v, and in your CPU module, add the following code.// Program Memorywire [34:0] instruction;AsyncROM Pmem(IP, instruction); Test Temporarily include the following lines in your CPU module. initial Dval = 1; always @(*)Dout = instruction[25 -:8]; Then when you run your computer, the display will show 1,3, 5, 7, 9, 11, 13, 15, 17, 19 then remain on 0 until the IP reaches 00 again. Questions How does Pmem(IP, instruction) work? What does the test code do? Why, after showing 19, does the display remain on 0 until the IP reaches 00 again? Are letters or numbers being stored into the program memory? Step 3 of Stage 5 Add the following code to store the CPU Registers. Fix up any compiler errors.// Registersreg [7:0] Reg [0:31];// Use these to Read the Special Registerswire [7:0] Rgout = Reg[29];wire [7:0] Rdout = Reg[30];wire [7:0] Rflag = Reg[31];// Use these to Write to the Flags and Din Registers`define RFLAG Reg[31]`define RDINP Reg[28]// Connect certain registers to the external worldassign Dout = Rdout;assign GPO= Rgout[5:0];// TO DO: Change Laterinitial Dval = 1; Here is a very simple instruction cycle. // Instruction Cycle wire [3:0] cmd_grp = instruction[34:31]; wire [2:0] cmd = instruction[30:28]; wire [1:0] arg1_typ = instruction[27:26]; wire [7:0] arg1 = instruction[25:18]; wire [1:0] arg2_typ = instruction[17:16]; wire [7:0] arg2 = instruction[15:8]; wire [7:0] addr = instruction[7:0];always @(posedge Clock) beginif (go) begin IP <= IP + 8b1;// Default action is to increment IP case (cmd_grp)`MOV: Reg[arg2] <= arg1;// For now, we just assumed a PUR move, with arg1 a number and arg2 a register! endcaseend// Place reset code here end Test This should display 1,3,5 etc when you run the program, due to the instructions loaded into memory in Step 2. Add several more instructions, to get some LEDs to light up. (Move a number into the `GOUT register.) You have built a very simple computer!Stage 6 Implement the Unconditional Jump instruction, and test it out by adding a JMP to your Program Memory so that your program loops after ten or so instructions. Note that by writing IP <= IP+1; before the case (cmd_grp) statement, you can change IP from within the case statement, and that change will take effect: according to the Verilog standard, the last assignment to a variable from within the same block is the one that will take precedence. Turbo Switch You can also implement the Turbo feature now. Add the following inside the CPU module. wire turbo_safe; Synchroniser tbo(Clock, Turbo, turbo_safe); Then make a small change to the definition of go:wire go = !Reset && ((cnt == 0) || turbo_safe); The safe suffix reminds us that we have synchronised the signal. Note Have a look at the RTL for your CPU. This will be the last stage where the RTL is relatively straightforward to take in at a glance. After the next stage, even though only a few more lines of Verilog are added, there will be hundreds of wires everywhere! Questions How does the Turbo feature work? What does it mean to have synchronised the turbo signal? What can go wrong if we had used Turbo instead of turbo_safe?Stage 7 The full repertoire of the MOV Command Group will now be implemented. First create two Verilog functions to help us: get_number and get_location. function [7:0] get_number;input [1:0] arg_type;input [7:0] arg;begin case (arg_type)`REG: get_number = Reg[arg[5:0]];`IND: get_number = Reg[ Reg[arg[5:0]][5:0] ];default: get_number = arg; endcaseend endfunctionfunction [5:0] get_location;input [1:0] arg_type;input [7:0] arg;begin case (arg_type)`REG: get_location = arg[5:0];`IND: get_location = Reg[arg[5:0]][5:0];default: get_location = 0; endcaseend endfunction Then the appropriate part of the case statement can be changed to the following. Outside of the always statement you will also need to declare reg [7:0] cnum;. It will not be synthesised; it just makes the Verilog code easier to read. Since we are treating cnum as a temporary variable that does not get synthesised, we use = to assign values to it. `MOV: begincnum = get_number(arg1_typ, arg1);case (cmd) `SHL: begin`RFLAG[`SHFT] <= cnum[7];cnum = {cnum[6:0], 1’b0}; end `SHR: begin`RFLAG[`SHFT] <= cnum[0];cnum = {1’b0, cnum[7:1]}; endendcaseReg[ get_location(arg2_typ, arg2) ] <= cnum; end Note that very few lines of code were needed to implement a very powerful set of MOV instructions. Test Make sure your original program still executes correctly. Then change some of the instructions to use the new shifting ability of MOV commands, and check that it works. Questions What do get_number and get_location do? o How do they relate to the CPU Instruction Set? What do the instructions `MOV `SHL and `MOV `SHR do exactly? o Why is an assignment made to the Flag Register? How is it possible that cnum is not synthesised? What does the synthesiser do instead? Look at the RTL. Why, with only a few lines of code, is there now a birds nest of wires? Stage 8 Using get_number and get_location, it is not difficult to implement the ACC Command Group. It is expedient to introduce several more temporary variables that do not get synthesised. Here is a (big) hint: maybe you can have a cleaner version though. `ACC: begincnum = get_number(arg2_typ, arg2);cloc = get_location(arg1_typ, arg1);case (cmd) `UAD: word = Reg[ cloc ] + cnum; `SAD: s_word = $signed( Reg[ cloc ] ) + $signed( cnum ); `UMT: word = ; // Fill this in `SMT: s_word = ; // Fill this in `AND: cnum = Reg[ cloc ] & cnum; `OR: cnum = ; // Fill this in `XOR: cnum = ; // Fill this inendcaseif (cmd[2] == 0) if (cmd[0] == 0) begin // Unsigned addition or multiplicationcnum = word[7:0];`RFLAG[`OFLW] <= (word > 255);
end
else begin // Signed addition or multiplication
cnum = s_word[7:0];
`RFLAG[`OFLW] <= (s_word > 127 || s_word < -128); endReg[ cloc ] <= ; // Fill this in end Test Test out your computer with the following program. case (addr)0: data = {`MOV, `PUR, `NUM, 8’d 0, `REG, `DOUT, `N8};4: data = {`ACC, `UAD, `REG, `DOUT, `NUM, 8’d 15, `N8};8:data = {`JMP, `UNC, `N10, `N10, 8’d 4};default: data = 35’b0; endcase Questions Describe exactly how each Instruction works. Why does using the addresses 0, 4 and 8 introduce a delay? When the displayed Dout changes on the board, is the IP 4 or 5? Why?Stage 9 Implement in full the JMP Command Group. Here is a skeleton. `JMP: begin case (cmd) `UNC: cond = 1; `EQ:cond = ( get_number(arg1_typ, arg1) == get_number(arg2_typ, arg2) ); `ULT:cond = `SLT:cond = `ULE:cond = `SLE:cond = default: cond = 0; endcase if (cond) IP <= addr; end Test Test it out with the following program. 0:
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