FSM with output logic in Verilog - Mini Project: Build & Apply

FSM with Output Logic in Verilog

📖 Scenario: You are designing a simple traffic light controller using a finite state machine (FSM). The traffic light cycles through three states: Green, Yellow, and Red. Each state lasts one clock cycle for simplicity.

🎯 Goal: Build a Verilog FSM that changes states on each clock cycle and outputs the correct light signal for each state.

📋 What You'll Learn

Create a state encoding using parameters

Use a register to hold the current state

Implement a synchronous state transition on the clock's rising edge

Add output logic that sets the traffic light signals based on the current state

Print the current state and output signals

💡 Why This Matters

🌍 Real World

Traffic light controllers use FSMs to manage light changes safely and predictably.

💼 Career

Understanding FSMs and output logic is essential for hardware design and embedded systems engineering.

1

Define the FSM states

Create three parameters called GREEN, YELLOW, and RED with values 2'b00, 2'b01, and 2'b10 respectively.

Verilog

// Your code here

Hint

Use parameter keyword to define constant state values.

2

Create the state register and clock input

Declare a reg [1:0] state to hold the current state and an input clk for the clock signal.

Verilog

parameter GREEN = 2'b00;
parameter YELLOW = 2'b01;
parameter RED = 2'b10;

// Your code here

Hint

Use reg [1:0] state; and input clk; declarations.

3

Implement the state transition logic

Write an always @(posedge clk) block that updates state to the next state in the sequence: GREEN → YELLOW → RED → GREEN.

Verilog

parameter GREEN = 2'b00;
parameter YELLOW = 2'b01;
parameter RED = 2'b10;

reg [1:0] state;
input clk;

// Your code here

Hint

Use a case statement inside the always @(posedge clk) block to update state.

4

Add output logic and display the state

Declare three output wires green_light, yellow_light, and red_light. Assign them to 1 when state matches their color, else 0. Then, add a initial block to print the current state and output signals.

Verilog

parameter GREEN = 2'b00;
parameter YELLOW = 2'b01;
parameter RED = 2'b10;

reg [1:0] state;
input clk;

always @(posedge clk) begin
    case (state)
        GREEN: state <= YELLOW;
        YELLOW: state <= RED;
        RED: state <= GREEN;
        default: state <= GREEN;
    endcase
end

// Your code here

Hint

Use conditional assignments for outputs and $monitor to print values.

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