State encoding (binary, one-hot, gray) in Verilog - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is binary state encoding in Verilog FSM design?

Binary state encoding assigns states using the minimum number of bits needed, representing states as binary numbers (e.g., 00, 01, 10). It is memory efficient but can be slower due to multiple bits changing between states.

beginner

Explain one-hot state encoding.

One-hot encoding uses one flip-flop per state. Only one bit is '1' at a time, and all others are '0'. This makes state decoding simple and fast but uses more hardware resources.

intermediate

What is Gray code state encoding and why is it used?

Gray code encoding assigns states so that only one bit changes between consecutive states. This reduces glitches and switching noise in hardware, useful in some FSM designs.

intermediate

Show a simple Verilog example of a one-hot encoded FSM with 3 states.

```verilog module fsm_one_hot(input clk, reset, input in, output reg out); typedef enum logic [2:0] {S0=3'b001, S1=3'b010, S2=3'b100} state_t; state_t state, next_state; always_ff @(posedge clk or posedge reset) begin if (reset) state <= S0; else state <= next_state; end always_comb begin case(state) S0: next_state = in ? S1 : S0; S1: next_state = in ? S2 : S0; S2: next_state = S0; default: next_state = S0; endcase end always_comb begin out = (state == S2); end endmodule ```

intermediate

Compare advantages and disadvantages of binary, one-hot, and gray state encoding.

• Binary: Uses fewer bits, saves hardware, but can cause glitches due to multiple bit changes.
• One-hot: Simple decoding, fast transitions, but uses more flip-flops.
• Gray: Minimizes glitches by changing one bit at a time, but needs more bits than binary and more complex encoding.

Which state encoding uses one flip-flop per state?

A Binary encoding

B Thermometer encoding

C One-hot encoding

D Gray encoding

One-hot encoding assigns one flip-flop per state, with only one bit set to '1' at a time.

What is the main advantage of Gray code encoding in FSMs?

A Minimizes number of flip-flops

B Only one bit changes between states to reduce glitches

C Simplifies state decoding

D Uses less power than binary encoding

Gray code changes only one bit between consecutive states, reducing glitches and switching noise.

Which encoding is most hardware efficient in terms of flip-flop count?

A Binary encoding

B One-hot encoding

C Gray encoding

D Johnson encoding

Binary encoding uses the minimum number of bits, so it is most hardware efficient.

In a 4-state FSM, how many bits are needed for binary encoding?

A 4 bits

B 1 bit

C 3 bits

D 2 bits

2 bits can represent 4 states (00, 01, 10, 11).

Which encoding typically results in the fastest state transitions?

A One-hot encoding

B Binary encoding

C Gray encoding

D Random encoding

One-hot encoding has simple decoding and only one bit changes, making transitions faster.

Describe the differences between binary, one-hot, and gray state encoding in FSM design.

Think about how many bits change and how many flip-flops each encoding uses.

Write a simple Verilog snippet for a one-hot encoded FSM with three states and explain how the states are represented.

Use a typedef enum with one bit set per state.

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