State encoding (binary, one-hot, gray) in Verilog - Step-by-Step Execution

Concept Flow - State encoding (binary, one-hot, gray)

Define States

↓

Choose Encoding Type

↓

Binary

↓

Assign Codes to States

↓

Use Encoded States in FSM

↓

Simulate and Verify

This flow shows how to define states, pick an encoding type (binary, one-hot, or gray), assign codes, and use them in a finite state machine.

Execution Sample

Verilog

typedef enum logic [1:0] {
  IDLE = 2'b00,
  LOAD = 2'b01,
  EXEC = 2'b11,
  DONE = 2'b10
} state_t;

state_t state, next_state;

Defines a 2-bit binary encoded state machine with four states.

Execution Table

Step	State Name	Binary Code	One-Hot Code	Gray Code
1	IDLE	00	0001	00
2	LOAD	01	0010	01
3	EXEC	11	0100	10
4	DONE	10	1000	11
5	End	-	-	-

💡 All states assigned codes in each encoding scheme.

Variable Tracker

State	Binary Code	One-Hot Code	Gray Code
IDLE	00	0001	00
LOAD	01	0010	01
EXEC	11	0100	10
DONE	10	1000	11

Key Moments - 3 Insights

Why does one-hot encoding use more bits than binary encoding?

One-hot encoding assigns one bit per state, so for N states it uses N bits, unlike binary which uses log2(N) bits. See execution_table rows 1-4 for bit widths.

How does Gray code differ from binary in state transitions?

Gray code changes only one bit between consecutive states, reducing glitches. Look at execution_table rows 1-4 to compare Gray and Binary codes.

Can two states have the same code in these encodings?

No, each state must have a unique code in all encoding schemes to avoid confusion. The execution_table shows unique codes for each state.

Visual Quiz - 3 Questions

Look at the execution_table, what is the one-hot code for the EXEC state?

A 0010

B 1000

C 0100

D 0001

💡 Hint

Check the 'One-Hot Code' column at the row where State Name is EXEC.

At which step does the Gray code for the state change from '01' to '11'?

A Step 1 to 2

B Step 3 to 4

C Step 2 to 3

D Step 4 to 5

💡 Hint

Look at the 'Gray Code' column between rows 3 and 4 in execution_table.

If the FSM had 8 states, how many bits would one-hot encoding require?

A 8 bits

B 3 bits

C 4 bits

D 1 bit

💡 Hint

One-hot uses one bit per state; see key_moments about bit width differences.

Concept Snapshot

State encoding assigns binary codes to FSM states.
Binary uses few bits (log2 states).
One-hot uses one bit per state.
Gray changes one bit between states.
Choose encoding by speed, area, or glitch needs.

Full Transcript

State encoding in Verilog means giving each FSM state a unique code. Binary encoding uses the fewest bits by counting in binary. One-hot encoding uses one bit per state, so only one bit is high at a time. Gray code changes only one bit between states to reduce glitches. We define states, pick an encoding, assign codes, and use them in the FSM. The execution table shows codes for four states in each encoding. One-hot uses more bits but can be faster. Gray code helps with smooth transitions. Each state must have a unique code to avoid errors.