D flip-flop with synchronous reset in Verilog - Time & Space Complexity

Time Complexity: D flip-flop with synchronous reset

O(n)

Understanding Time Complexity

We want to understand how the time it takes to run a D flip-flop with synchronous reset changes as input size changes.

Specifically, we ask: how does the number of operations grow when the circuit runs?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

module dff_sync_reset (
  input wire clk,
  input wire reset,
  input wire d,
  output reg q
);
  always @(posedge clk) begin
    if (reset) q <= 0;
    else q <= d;
  end
endmodule

This code models a D flip-flop that updates its output on the clock's rising edge and resets synchronously when reset is high.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The flip-flop updates output on every clock rising edge.
• How many times: Once per clock cycle, repeating indefinitely as the clock runs.

How Execution Grows With Input

Explain the growth pattern intuitively.

Input Size (clock cycles)	Approx. Operations
10	10 updates
100	100 updates
1000	1000 updates

Pattern observation: The number of operations grows directly with the number of clock cycles, increasing steadily.

Final Time Complexity

Time Complexity: O(n)

This means the work done grows linearly with the number of clock cycles the flip-flop runs.

Common Mistake

[X] Wrong: "The flip-flop does many operations inside each clock cycle, so time grows faster than linearly."

[OK] Correct: Each clock cycle triggers only one update operation, so the work grows one-to-one with cycles, not more.

Interview Connect

Understanding how hardware elements like flip-flops behave over time helps you reason about circuit speed and design efficiency, a useful skill in many tech roles.

Self-Check

"What if the flip-flop had an asynchronous reset instead? How would the time complexity change?"

Practice

1. What does a D flip-flop with synchronous reset do when the reset signal is active?

easy

A. It clears the stored data immediately when reset is active.

B. It stores the inverse of the input data.

C. It ignores the reset signal and keeps the data.

D. It clears the stored data only on the clock edge.

Solution

1. Step 1: Understand synchronous reset behavior

   A synchronous reset affects the flip-flop only at the clock edge, not immediately.

2. Step 2: Compare reset types

   Asynchronous reset clears data immediately, but synchronous reset waits for clock edge.

3. Final Answer:

   It clears the stored data only on the clock edge. -> Option D

4. Quick Check:

   Synchronous reset = clear on clock edge [OK]

Hint: Synchronous reset acts only at clock edges [OK]

Common Mistakes:

• Confusing synchronous with asynchronous reset
• Thinking reset clears data immediately
• Assuming reset stores input data

2. Which of the following is the correct Verilog syntax for a D flip-flop with synchronous reset inside an always block?

easy

A. always @(posedge clk or posedge reset) begin if (reset) q <= 0; else q <= d; end

B. always @(negedge clk) begin if (!reset) q <= d; else q <= 0; end

C. always @(posedge clk) begin if (reset) q <= 0; else q <= d; end

D. always @(reset) begin if (reset) q <= 0; else q <= d; end

Solution

1. Step 1: Identify synchronous reset sensitivity list

   Synchronous reset triggers only on clock edge, so only posedge clk is in sensitivity list.

2. Step 2: Check reset condition inside always block

   Reset is checked inside the block, not in sensitivity list, matching synchronous reset behavior.

3. Final Answer:

   always @(posedge clk) begin if (reset) q <= 0; else q <= d; end -> Option C

4. Quick Check:

   Synchronous reset inside posedge clk block [OK]

Hint: Synchronous reset inside posedge clk always block [OK]

Common Mistakes:

• Including reset in sensitivity list (asynchronous reset)
• Using reset in sensitivity list but coding synchronous reset
• Using negedge clk incorrectly

3. Given the following Verilog code, what is the value of q after the second rising edge of clk if reset is 1 at the first clock and 0 at the second?

reg q;
always @(posedge clk) begin
  if (reset) q <= 0;
  else q <= d;
end

medium

A. 1

B. d (input value)

C. 0

D. Undefined

Solution

1. Step 1: Analyze first clock edge with reset=1

   At first posedge clk, reset=1, so q is set to 0.

2. Step 2: Analyze second clock edge with reset=0

   At second posedge clk, reset=0, so q takes value of d input.

3. Final Answer:

   d (input value) -> Option B

4. Quick Check:

   Reset active clears q, then q follows d [OK]

Hint: Reset clears q only at clock edge, then q follows d [OK]

Common Mistakes:

• Assuming reset clears q immediately
• Thinking q stays 0 after reset deactivates
• Confusing asynchronous reset behavior

4. Identify the error in this Verilog code for a D flip-flop with synchronous reset:

always @(posedge clk or posedge reset) begin
  if (reset) q <= 0;
  else q <= d;
end

medium

A. Reset should not be in the sensitivity list for synchronous reset.

B. The assignment operator should be '=' instead of '<='.

C. The clock edge should be negedge, not posedge.

D. The reset condition should be checked after the else statement.

Solution

1. Step 1: Check sensitivity list for synchronous reset

   Synchronous reset triggers only on clock edge, so reset must NOT be in sensitivity list.

2. Step 2: Confirm correct sensitivity list

   Only posedge clk should be in sensitivity list for synchronous reset.

3. Final Answer:

   Reset should not be in the sensitivity list for synchronous reset. -> Option A

4. Quick Check:

   Synchronous reset excludes reset from sensitivity list [OK]

Hint: Synchronous reset excludes reset from sensitivity list [OK]

Common Mistakes:

• Including reset in sensitivity list for synchronous reset
• Using blocking assignment instead of non-blocking
• Checking reset condition incorrectly

5. You want to design a D flip-flop with synchronous reset that sets output q to 0 on reset and otherwise stores input d. Which Verilog code correctly implements this with minimal logic and proper timing?

hard

A. always @(posedge clk) begin if (reset) q <= 0; else q <= d; end

B. always @(posedge clk or posedge reset) begin if (reset) q <= 0; else q <= d; end

C. always @(posedge clk) begin if (reset) q <= d; else q <= 0; end

D. always @(posedge reset) begin if (reset) q <= 0; else q <= d; end

Solution

1. Step 1: Confirm synchronous reset sensitivity list

   For synchronous reset, only posedge clk is in sensitivity list.

2. Step 2: Check reset logic inside always block

   Reset condition sets q to 0 when active; else q follows d input.

3. Step 3: Verify minimal and correct logic

   always @(posedge clk) begin if (reset) q <= 0; else q <= d; end matches synchronous reset behavior with minimal logic and correct timing.

4. Final Answer:

   always @(posedge clk) begin if (reset) q <= 0; else q <= d; end -> Option A

5. Quick Check:

   Synchronous reset inside posedge clk block with correct logic [OK]

Hint: Use posedge clk only and check reset inside block [OK]

Common Mistakes:

• Including reset in sensitivity list (asynchronous reset)
• Using negedge clk incorrectly
• Inverting reset logic