212k views
4 votes
Write Verilog code for one of the following 1 flip-flop "machines" shown below (with inputs X,Y). Use M₁ if you are working on Stations 1,3,5,7,9 and use M₂ for Stations 2,4,6,8.

M₂
X Y Q*
0 0 0 reset
0 1 1 set
1 0 Q' toggle
1 1 Q no change

1 Answer

5 votes

Final answer:

If we want to write Verilog code for the given 1 flip-flop machine (M₂), we can use an always block to describe the behavior of the flip-flop based on the inputs X and Y.

Here's the Verilog code:

```verilog

module flip_flop_M2 (

input wire X,

input wire Y,

output reg Q

);

always (at)(X, Y)

begin

case({X, Y})

2'b00: Q <= 1'b0; // reset

2'b01: Q <= 1'b1; // set

2'b10: Q <= ~Q; // toggle

2'b11: Q <= Q; // no change

endcase

end

endmodule

```

Step-by-step explanation:

1. The `flip_flop_M2` module takes two input signals, X and Y, and has an output signal Q, which represents the state of the flip-flop.

2. The `always (at)(X, Y)` block ensures that the behavior of the flip-flop is sensitive to changes in the X and Y inputs.

3. Inside the `always` block, a `case` statement is used to define the behavior of the flip-flop based on the input values {X, Y}.

4. The cases 2'b00, 2'b01, 2'b10, and 2'b11 represent the four possible input combinations {X, Y} specified in the question.

5. For each case, the value of the output Q is assigned accordingly:

  • - 2'b00 (X=0, Y=0): Q is set to 1'b0, indicating a reset.
  • - 2'b01 (X=0, Y=1): Q is set to 1'b1, indicating a set.
  • - 2'b10 (X=1, Y=0): Q is toggled by assigning it the complement (~) of its current value.
  • - 2'b11 (X=1, Y=1): Q remains unchanged.

6. The `endcase` statement ends the case block, and the `end` statement closes the `always` block.

This Verilog code describes the behavior of the flip-flop machine M₂ based on the given truth table. Make sure to use this code if you are working on Stations 2, 4, 6, or 8.

User Emanuel Hanner
by
8.8k points