D flip-flop with clock edge in Verilog - Time & Space Complexity

Time Complexity: D flip-flop with clock edge

O(n)

Understanding Time Complexity

We want to understand how the time it takes for a D flip-flop to operate changes as inputs change.

Specifically, how the clock edge triggers affect the execution steps.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

module d_flip_flop(
  input wire clk,
  input wire d,
  output reg q
);

always @(posedge clk) begin
  q <= d;
end
endmodule

This code models a D flip-flop that updates output q on the rising edge of the clock clk.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The flip-flop updates q once every rising clock edge.
• How many times: Once per clock cycle, repeating as many times as the clock ticks.

How Execution Grows With Input

Each clock edge causes one update operation, so the number of operations grows directly with the number of clock cycles.

Input Size (n)	Approx. Operations
10	10 updates
100	100 updates
1000	1000 updates

Pattern observation: The operations increase linearly as the number of clock cycles increases.

Final Time Complexity

Time Complexity: O(n)

This means the work done grows in direct proportion to the number of clock cycles.

Common Mistake

[X] Wrong: "The flip-flop updates multiple times per clock cycle."

[OK] Correct: The flip-flop only updates on the clock's rising edge, so it updates exactly once per cycle, not multiple times.

Interview Connect

Understanding how hardware elements like flip-flops respond to clock signals helps you reason about timing and performance in digital circuits, a useful skill in many technical roles.

Self-Check

"What if we changed the flip-flop to update on both rising and falling clock edges? How would the time complexity change?"

Practice

1.

What does a D flip-flop do in a digital circuit?

easy

A. Stores the input value on the rising edge of the clock

B. Generates a clock signal

C. Performs arithmetic operations

D. Resets the circuit asynchronously

Solution

1. Step 1: Understand the role of a D flip-flop

   A D flip-flop captures and stores the input value (D) at a specific clock event.

2. Step 2: Identify the clock event

   It stores the value on the rising edge (positive edge) of the clock signal.

3. Final Answer:

   Stores the input value on the rising edge of the clock -> Option A

4. Quick Check:

   D flip-flop stores input on clock rising edge [OK]

Hint: Remember: D flip-flop stores data at clock's rising edge [OK]

Common Mistakes:

• Confusing D flip-flop with clock generator
• Thinking it performs calculations
• Assuming it resets asynchronously

2.

Which of the following is the correct Verilog syntax to trigger a D flip-flop on the rising edge of the clock clk?

easy

A. always @(negedge clk)

B. always @(posedge clk)

C. always @(clk)

D. always @(clk posedge)

Solution

1. Step 1: Recall the correct event control syntax

   Verilog uses posedge to detect the rising edge of a signal.

2. Step 2: Match syntax with clock signal

   The correct syntax to trigger on rising edge of clk is always @(posedge clk).

3. Final Answer:

   always @(posedge clk) -> Option B

4. Quick Check:

   Rising edge trigger = posedge [OK]

Hint: Use 'posedge' for rising edge in always block [OK]

Common Mistakes:

• Using negedge instead of posedge
• Omitting posedge keyword
• Incorrect event control syntax

3.

What will be the output q after the following Verilog code runs for one clock cycle if d = 1 and initial q = 0?

module dff(input clk, input d, output reg q);
  always @(posedge clk) begin
    q <= d;
  end
endmodule

medium

A. q remains 0

B. q toggles between 0 and 1

C. q becomes undefined

D. q becomes 1

Solution

1. Step 1: Understand the behavior on posedge clk

   On the rising edge of clk, the flip-flop assigns d to q.

2. Step 2: Apply given inputs

   Since d = 1 and initial q = 0, after the clock edge, q updates to 1.

3. Final Answer:

   q becomes 1 -> Option D

4. Quick Check:

   D input 1 stored at clock edge = q 1 [OK]

Hint: On clock rising edge, q copies d value [OK]

Common Mistakes:

• Assuming q changes immediately without clock edge
• Thinking q toggles automatically
• Ignoring initial value of q

4.

Identify the error in this D flip-flop Verilog code snippet:

module dff(input clk, input d, output reg q);
  always @(clk) begin
    q <= d;
  end
endmodule

medium

A. Missing sensitivity to clock edge, should use posedge

B. Output q should not be reg

C. Input d should be reg type

D. No error, code is correct

Solution

1. Step 1: Check sensitivity list

   The code uses always @(clk), which triggers on any change of clk, not just edges.

2. Step 2: Correct edge sensitivity

   For a D flip-flop, it must trigger only on the rising edge: always @(posedge clk).

3. Final Answer:

   Missing sensitivity to clock edge, should use posedge -> Option A

4. Quick Check:

   Flip-flop triggers on posedge clk only [OK]

Hint: Use posedge in always block for flip-flop [OK]

Common Mistakes:

• Using level sensitivity instead of edge sensitivity
• Confusing reg and wire for outputs
• Assuming inputs must be reg

5.

You want to design a D flip-flop that resets q to 0 asynchronously when rst is high. Which Verilog code snippet correctly implements this behavior?

module dff(input clk, input rst, input d, output reg q);
  always @(posedge clk or posedge rst) begin
    if (rst)
      q <= 0;
    else
      q <= d;
  end
endmodule

hard

A. The sensitivity list is missing negedge rst

B. Reset should be checked only on clock edge, not asynchronously

C. This code correctly implements asynchronous reset

D. q should be assigned with blocking assignment (=) inside always

Solution

1. Step 1: Analyze sensitivity list

   The always block triggers on rising edge of clock or rising edge of reset, enabling asynchronous reset.

2. Step 2: Check reset logic

   If reset is high, q is set to 0 immediately, else q follows d on clock edge.

3. Final Answer:

   This code correctly implements asynchronous reset -> Option C

4. Quick Check:

   Asynchronous reset uses posedge rst in sensitivity list [OK]

Hint: Add posedge rst in sensitivity list for async reset [OK]

Common Mistakes:

• Checking reset only on clock edge (synchronous reset)
• Using blocking assignment for sequential logic
• Missing reset in sensitivity list