FSM with output logic in Verilog - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the state register.

Verilog

reg [1:0] [1];

Drag options to blanks, or click blank then click option'

Ainput

Bstate

Coutput

Dclk

Attempts:

3 left

💡 Hint

Common Mistakes

Using input or output names instead of the state register.

The state register holds the current state of the FSM and is declared as state.

2fill in blank

medium

Complete the code to define the next state logic using a case statement.

Verilog

always @(*) begin
  case (state)
    2'b00: next_state = [1];
    default: next_state = 2'b00;
  endcase
end

Drag options to blanks, or click blank then click option'

A2'b11

B2'b10

C2'b01

D2'b00

Attempts:

3 left

💡 Hint

Common Mistakes

Choosing the current state as next state causing no transition.

The next state after 2'b00 is 2'b01 as per the FSM design.

3fill in blank

hard

Fix the error in the output logic to assign output based on the current state.

Verilog

always @(*) begin
  case ([1])
    2'b00: out = 1'b0;
    2'b01: out = 1'b1;
    default: out = 1'b0;
  endcase
end

Drag options to blanks, or click blank then click option'

Anext_state

Bclk

Cinput

Dstate

Attempts:

3 left

💡 Hint

Common Mistakes

Using next_state or input signals instead of current state.

The output depends on the current state, not the next state or inputs.

4fill in blank

hard

Fill both blanks to complete the synchronous state update logic.

Verilog

always @(posedge [1] or posedge [2]) begin
  if (reset)
    state <= 2'b00;
  else
    state <= next_state;
end

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Aclk

Breset

Cinput

Dout

Attempts:

3 left

💡 Hint

Common Mistakes

Using input or output signals instead of clock or reset.

The state updates on the rising edge of the clock clk or asynchronous reset reset.

5fill in blank

hard

Fill all three blanks to complete the FSM module header and port declarations.

Verilog

module fsm(
  input wire [1],
  input wire [2],
  output reg [3]
);

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Aclk

Breset

Cout

Dstate

Attempts:

3 left

💡 Hint

Common Mistakes

Using internal signals like state as ports.

The FSM module has inputs clk and reset, and output out.

Practice

1. Which part of an FSM with output logic determines the output signals based only on the current state?

easy

A. Input decoder

B. Next state logic

C. State register

D. Output logic in a Moore machine

Solution

1. Step 1: Understand FSM output types

   In FSMs, Moore outputs depend only on the current state, while Mealy outputs depend on state and input.

2. Step 2: Identify output logic role

   The output logic in a Moore machine produces outputs based solely on the current state stored in the state register.

3. Final Answer:

   Output logic in a Moore machine -> Option D

4. Quick Check:

   Moore output = state only [OK]

Hint: Moore outputs depend only on state, not inputs [OK]

Common Mistakes:

• Confusing next state logic with output logic
• Thinking outputs depend on inputs in Moore FSM
• Mixing state register role with output logic

2. Which of the following is the correct Verilog syntax to declare a state register named state with 2 bits?

easy

A. reg [1:0] state;

B. wire [2:0] state;

C. reg state[1:0];

D. state reg [1:0];

Solution

1. Step 1: Recall Verilog register declaration

   Registers are declared with reg keyword followed by bit width in brackets, e.g., reg [1:0] for 2 bits.

2. Step 2: Check each option

   reg [1:0] state; correctly declares a 2-bit register named state. wire [2:0] state; uses wire and wrong width. reg state[1:0]; uses incorrect syntax for bit width. state reg [1:0]; has wrong keyword order.

3. Final Answer:

   reg [1:0] state; -> Option A

4. Quick Check:

   reg [1:0] state; is correct syntax [OK]

Hint: Use reg [width-1:0] name; for registers [OK]

Common Mistakes:

• Using wire instead of reg for state registers
• Incorrect bit width syntax like reg name[1:0]
• Wrong keyword order like state reg

3. Given this Verilog snippet for a Moore FSM output logic:

always @(state) begin
  case(state)
    2'b00: out = 1'b0;
    2'b01: out = 1'b1;
    2'b10: out = 1'b0;
    default: out = 1'b0;
  endcase
end

What is the output out when state is 2'b01?

medium

A. 1'b0

B. Undefined

C. 1'b1

D. 2'b01

Solution

1. Step 1: Check case for state 2'b01

   The case statement sets out = 1'b1 when state is 2'b01.

2. Step 2: Confirm output value

   Since the state matches 2'b01, output out is assigned 1'b1.

3. Final Answer:

   1'b1 -> Option C

4. Quick Check:

   State 01 -> out = 1 [OK]

Hint: Match state in case to find output quickly [OK]

Common Mistakes:

• Confusing output with state value
• Assuming default applies instead of matching case
• Misreading bit values as decimal

4. Identify the error in this FSM output logic code snippet:

always @(posedge clk) begin
  case(state)
    2'b00: out = 1'b0;
    2'b01: out = 1'b1;
    default: out = 1'b0;
  endcase
end

What is the main problem?

medium

A. Output logic should be in a combinational always block, not clocked

B. Missing reset signal for output

C. State values are incorrectly formatted

D. Output out is not declared as reg

Solution

1. Step 1: Understand output logic timing

   Moore FSM output logic is usually combinational and depends only on state, so it should be in an always block sensitive to state changes, not clock edges.

2. Step 2: Analyze given code

   The code uses always @(posedge clk), making output synchronous, which is incorrect for pure output logic in Moore FSM.

3. Final Answer:

   Output logic should be in a combinational always block, not clocked -> Option A

4. Quick Check:

   Output logic = combinational, not clocked [OK]

Hint: Output logic usually uses always @(*) for combinational [OK]

Common Mistakes:

• Using clocked always block for output logic
• Ignoring sensitivity list correctness
• Confusing output and state update blocks

5. You want to design a Mealy FSM with output logic that sets out = 1 only when the current state is 2'b10 and input in is high. Which Verilog code snippet correctly implements this output logic?

hard

A. always @(in) begin\n if (in)\n out = 1'b1;\n else\n out = 1'b0;\nend

B. always @(state or in) begin\n if (state == 2'b10 && in)\n out = 1'b1;\n else\n out = 1'b0;\nend

C. always @(posedge clk) begin\n if (state == 2'b10 && in)\n out <= 1'b1;\n else\n out <= 1'b0;\nend

D. always @(state) begin\n if (state == 2'b10)\n out = 1'b1;\n else\n out = 1'b0;\nend

Solution

1. Step 1: Understand Mealy output logic

   Mealy outputs depend on both current state and inputs, so sensitivity list must include both state and in.

2. Step 2: Check each option

   always @(state or in) begin\n if (state == 2'b10 && in)\n out = 1'b1;\n else\n out = 1'b0;\nend correctly uses always @(state or in) and sets out based on both conditions. always @(state) begin\n if (state == 2'b10)\n out = 1'b1;\n else\n out = 1'b0;\nend ignores input in. always @(posedge clk) begin\n if (state == 2'b10 && in)\n out <= 1'b1;\n else\n out <= 1'b0;\nend uses clocked block which is not typical for Mealy output logic. always @(in) begin\n if (in)\n out = 1'b1;\n else\n out = 1'b0;\nend ignores state.

3. Final Answer:

   always @(state or in) begin if (state == 2'b10 && in) out = 1'b1; else out = 1'b0; end -> Option B

4. Quick Check:

   Mealy output depends on state and input [OK]

Hint: Include both state and input in sensitivity list for Mealy outputs [OK]

Common Mistakes:

• Leaving input out of sensitivity list
• Using clocked always block for Mealy output
• Ignoring state condition in output logic