State encoding (binary, one-hot, gray) in Verilog - Mini Project: Build & Apply

State Encoding in Verilog: Binary, One-Hot, and Gray

📖 Scenario: You are designing a simple traffic light controller using Verilog. The controller cycles through three states: Green, Yellow, and Red. To implement this, you will encode these states using three different methods: binary encoding, one-hot encoding, and Gray code encoding.

🎯 Goal: Build a Verilog module that defines the traffic light states using three different state encoding schemes: binary, one-hot, and Gray code. You will declare the states, set a current state variable, and display the encoded values.

📋 What You'll Learn

Create state definitions for Green, Yellow, and Red using binary encoding.

Create state definitions for Green, Yellow, and Red using one-hot encoding.

Create state definitions for Green, Yellow, and Red using Gray code encoding.

Declare a variable current_state and assign it the binary encoded value for Yellow.

Print the encoded values of the states.

💡 Why This Matters

🌍 Real World

State encoding is used in digital circuits like traffic lights, vending machines, and communication protocols to represent different modes or steps clearly and efficiently.

💼 Career

Understanding state encoding is essential for hardware engineers and FPGA/ASIC designers to optimize circuit speed, area, and power consumption.

1

Define states using binary encoding

Create three localparam constants named GREEN_BIN, YELLOW_BIN, and RED_BIN with binary values 2'b00, 2'b01, and 2'b10 respectively.

Verilog

// Define binary encoded states
// Your code here

Hint

Use localparam to define constants. Binary values use the format 2'bXX.

2

Define states using one-hot encoding

Add three localparam constants named GREEN_ONEHOT, YELLOW_ONEHOT, and RED_ONEHOT with one-hot values 3'b001, 3'b010, and 3'b100 respectively.

Verilog

localparam GREEN_BIN = 2'b00;
localparam YELLOW_BIN = 2'b01;
localparam RED_BIN = 2'b10;
// Define one-hot encoded states
// Your code here

Hint

One-hot encoding uses one bit set to 1 and others 0. Use 3 bits here.

3

Define states using Gray code encoding

Add three localparam constants named GREEN_GRAY, YELLOW_GRAY, and RED_GRAY with Gray code values 2'b00, 2'b01, and 2'b11 respectively.

Verilog

localparam GREEN_BIN = 2'b00;
localparam YELLOW_BIN = 2'b01;
localparam RED_BIN = 2'b10;

localparam GREEN_ONEHOT = 3'b001;
localparam YELLOW_ONEHOT = 3'b010;
localparam RED_ONEHOT = 3'b100;
// Define Gray code encoded states
// Your code here

Hint

Gray code changes only one bit between states. Use 2 bits here.

4

Assign current state and display encoded values

Declare a reg [1:0] current_state and assign it the value YELLOW_BIN. Then write $display statements to print the values of GREEN_BIN, YELLOW_BIN, RED_BIN, GREEN_ONEHOT, YELLOW_ONEHOT, RED_ONEHOT, GREEN_GRAY, YELLOW_GRAY, and RED_GRAY.

Verilog

localparam GREEN_BIN = 2'b00;
localparam YELLOW_BIN = 2'b01;
localparam RED_BIN = 2'b10;

localparam GREEN_ONEHOT = 3'b001;
localparam YELLOW_ONEHOT = 3'b010;
localparam RED_ONEHOT = 3'b100;

localparam GREEN_GRAY = 2'b00;
localparam YELLOW_GRAY = 2'b01;
localparam RED_GRAY = 2'b11;

// Declare current_state and assign YELLOW_BIN
// Display all encoded state values
// Your code here

Hint

Use reg [1:0] current_state = YELLOW_BIN; and $display to print values.

Practice

1.

Which state encoding method uses exactly one bit set to 1 for each state, and all other bits are 0?

easy

A. One-hot encoding

B. Binary encoding

C. Gray encoding

D. Thermometer encoding

Solution

1. Step 1: Understand one-hot encoding definition

   One-hot encoding assigns a unique bit to each state, where only one bit is high (1) and the rest are low (0).

2. Step 2: Compare with other encodings

   Binary encoding uses multiple bits to represent states in binary form; Gray encoding changes only one bit between states; Thermometer encoding is different and not the one-hot method.

3. Final Answer:

   One-hot encoding -> Option A

4. Quick Check:

   One-hot = single 1 bit per state [OK]

Hint: One-hot means only one bit is 1 at a time [OK]

Common Mistakes:

• Confusing binary encoding with one-hot
• Thinking Gray encoding sets only one bit
• Mixing thermometer encoding with one-hot

2.

Which of the following is the correct Verilog declaration for a 3-bit binary encoded state register named state?

easy

A. reg state[2:0];

B. reg [3:0] state;

C. reg [2:0] state;

D. wire [2:0] state;

Solution

1. Step 1: Understand Verilog register declaration

   For a 3-bit register, the correct syntax is reg [2:0] name; where bits are indexed from 2 down to 0.

2. Step 2: Check options for correctness

   reg [2:0] state; uses reg [2:0] state; which is correct. reg [3:0] state; declares 4 bits, wire [2:0] state; uses wire (not register), and reg state[2:0]; has incorrect syntax.

3. Final Answer:

   reg [2:0] state; -> Option C

4. Quick Check:

   3-bit reg = reg [2:0] name [OK]

Hint: Use reg [N-1:0] for N-bit registers in Verilog [OK]

Common Mistakes:

• Using wire instead of reg for state storage
• Incorrect bit range syntax
• Confusing array declaration with bit vector

3.

Given the following Verilog snippet using one-hot encoding for 4 states, what is the value of state after the first clock cycle if the initial state is 4'b0001 and the next state logic shifts the 1 bit left by one position?

reg [3:0] state = 4'b0001;
always @(posedge clk) begin
  state <= {state[2:0], state[3]};
end

medium

A. 4'b0001

B. 4'b0100

C. 4'b1000

D. 4'b0010

Solution

1. Step 1: Understand the shift operation in the code

   The assignment state <= {state[2:0], state[3]}; shifts bits left by one, moving bit 0 to bit 1, bit 1 to bit 2, bit 2 to bit 3, and bit 3 to bit 0.

2. Step 2: Apply the shift to initial state 4'b0001

   Initial bits: bit3=0, bit2=0, bit1=0, bit0=1. After shift: bit3=bit2=0, bit2=bit1=0, bit1=bit0=1, bit0=bit3=0 -> 4'b0010.

3. Final Answer:

   4'b0010 -> Option D

4. Quick Check:

   Shift left one bit in one-hot = 4'b0010 [OK]

Hint: Shift bits left, wrap last bit to first [OK]

Common Mistakes:

• Forgetting bit wrap-around
• Confusing bit order in concatenation
• Assuming no change after clock

4.

Identify the error in the following Verilog code snippet that attempts to implement Gray code state encoding for 3 states:

reg [1:0] state;
always @(posedge clk) begin
  case(state)
    2'b00: state <= 2'b01;
    2'b01: state <= 2'b11;
    2'b11: state <= 2'b10;
    2'b10: state <= 2'b01;
    default: state <= 2'b00;
  endcase
end

medium

A. State transitions do not follow Gray code rules

B. State register width is too small for 3 states

C. Missing reset condition for state register

D. Using blocking assignment instead of non-blocking

Solution

1. Step 1: Check state transitions for Gray code property

   Gray code requires only one bit change between consecutive states. Check transitions: 00->01 (1 bit change), 01->11 (1 bit change), 11->10 (1 bit change), 10->01 (2 bits change).

2. Step 2: Identify violation in transitions

   The transition 10->01 changes two bits, violating Gray code rules.

3. Final Answer:

   State transitions do not follow Gray code rules -> Option A

4. Quick Check:

   Gray code = one bit change per transition [OK]

Hint: Gray code changes only one bit between states [OK]

Common Mistakes:

• Ignoring multi-bit changes in transitions
• Assuming register width is incorrect
• Confusing blocking and non-blocking assignments

5.

You need to design a finite state machine (FSM) with 8 states. Which state encoding method is best to minimize the number of flip-flops used, and why?

hard

A. One-hot encoding, because it uses fewer flip-flops

B. Binary encoding, because it uses the minimum number of flip-flops

C. Gray encoding, because it reduces glitches in output

D. Thermometer encoding, because it is simplest to implement

Solution

1. Step 1: Calculate flip-flops needed for each encoding

   For 8 states, binary encoding needs 3 flip-flops (2^3=8). One-hot encoding needs 8 flip-flops (one per state). Gray encoding also uses binary bits, so 3 flip-flops.

2. Step 2: Choose encoding minimizing flip-flops

   Binary encoding uses the fewest flip-flops (3) compared to one-hot (8). Gray encoding is similar to binary but mainly reduces glitches, not flip-flop count.

3. Final Answer:

   Binary encoding, because it uses the minimum number of flip-flops -> Option B

4. Quick Check:

   Binary encoding = minimum flip-flops for many states [OK]

Hint: Binary encoding uses log2(states) flip-flops [OK]

Common Mistakes:

• Choosing one-hot for fewer flip-flops
• Confusing glitch reduction with flip-flop count
• Assuming thermometer encoding is minimal