CPEN 311

Datapath Continued

Updated 2018-01-18

Blocking and Non-blocking Assignments

Blocking Assignments

Blocking: evaluation and assignments are immediate

// Example
always_comb begin
  x = a | b;
  y = a ^ b ^ c;
  z = b & ~c;
end

The “assignments” will hook up the wires

Non-Blocking Assignments

Non-blocking: all assignments deferred until all right-hand sides have been evaluated

// Example
always_ff @(posedge clk) begin
  x <= a | b;
  y <= a ^ b ^ c;
  z <= b & ~c;
end

In terms of the hardware semantics, the registers are all in parallel. Essentially, a, b and c will all be evaluated first, then the x, y, and z are scheduled and evaluated simultaneously at the “end” of the clock cycle.

Note that in both cases, x, y, and z are declared as reg in Verilog. However, in the first blocking code, the assignments won’t be synthesized to to a register. In the second one, the assignments will be synthesized to flip-flops because we need the clock to update the values at the end. Also flip-flops are needed to hold memory.

Rule of Thumb


Lab 2

Reuleaux Triangle: A special triangular round shape that has constant width no matter how to measure it

Goal: Draw a Reauleaux Triangle

Datapath Methods

There are two methods of datapath:

  1. Explicit State Machine / Datapath Method: Design datapath and controller separately (like lab 1)
  2. Implicit Datapath Method: Specify state machine, but for each state, describe data operations to be performed in that state

Explicit State Machine / Datapath Method

  1. Design Datapath
    • Determine all storage elements, operators
    • Create network connecting all the elements and identify input / outputs
    • Describe datapath in Verilog
  2. State Machine (Controller design)
    • Figure out what to do in each cycle
    • Create a state machine that produces output for control signals
    • Describe controller in Verilog
  3. Create top-level design that connects datapath and controller

Tasks

Task 2 - Fill the Screen

Need to turn on each pixel one at a time. Some pseudo code:

for yp = 0 to 119 {
  for xp = 0 to 159 {
    turn on pixel at (xp, yp) with color (xp mod 8)
  }
}

Notice that this will take a lot of cycles. The first approach would be to build a state machine. But there are too many states. So let’s use a for loop.

So consider this following naïve implementation:

integer ix, iy;

always_ff @(posedge(CLOCK_50))
  for (iy = 0; iy < 120; iy = iy + 1) begin
    for (ix = 0; ix < 160; ix + ix + 1) begin
      x <= ix;
      y <= iy;
      color <= ix % 8;
      plot <= 1;
      @ (posedge (CLOCK_50)); // Sneak a clock cycle in here
    end
  end
end

We’re trying to control all pixels in a single clock cycle. So only the last assignment to x. y, etc. are applies. So how to get around this? For lab 2, method 1 is recommended.

A better method utilizes counters using registers.

Taking the Modulo

If we want to take a modulo of 8, notice that 8 is \(2^3\). So all we need to do is take the lower 3 bits in the binary.

Divide by Base 2

Use right shift by \(n\) to divide by \(2^n\).

Drawing a Line

If we want to draw a line from (0, 100) to (100, 0). Possible implementation could be:

for yp = 0 to 119 {
  for xp = 0 to 119 {
    turn on pixel(xp, yp) with color 0;
  }
}

xp = 0; yp = 100;
while (xp != 101) {
  turn on pixle(xp, yp) with color 1;
  xp = xp + 1;
  yp = yp - 1;
}

How will this change our datapath?

We added a subtraction block and a comparator block.

Notice that this only works for perfectly diagonal line.

Drawing a Circle

Insert pseudo for circle

  1. Start with clearing the screen datapath
  2. Figure what needs to be added for datapath
  3. Figure out any extra states

Do we need more storage?

Yes. When drawing a circle using the above described algorithm, we need to store crit