Stimulus process with wait statements in VHDL - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of a stimulus process with sequential wait statements

Consider the following VHDL stimulus process. What will be the value of signal_a after 30 ns?

VHDL

process
begin
  signal_a <= '0';
  wait for 10 ns;
  signal_a <= '1';
  wait for 10 ns;
  signal_a <= '0';
  wait for 10 ns;
  wait;
end process;

A Signal retains previous value before process

B '1'

C Undefined (signal_a is not assigned)

D '0'

💡 Hint

Think about how the signal changes at each wait statement and the total elapsed time.

The process assigns '0' at 0 ns, then waits 10 ns. At 10 ns, it assigns '1', then waits 10 ns. At 20 ns, it assigns '0' again, then waits 10 ns. At 30 ns, the last assignment was '0'.

❓ Predict Output

intermediate

Value of signal after multiple wait statements with different durations

Given this VHDL stimulus process, what is the value of signal_b at 25 ns?

VHDL

process
begin
  signal_b <= '1';
  wait for 5 ns;
  signal_b <= '0';
  wait for 15 ns;
  signal_b <= '1';
  wait for 10 ns;
  wait;
end process;

A '1'

B '0'

C 'U' (undefined)

D Signal retains initial value

💡 Hint

Calculate the time intervals and see which assignment is active at 25 ns.

At 0 ns, signal_b = '1'. After 5 ns, it changes to '0'. After 20 ns (5+15), it changes to '1'. At 25 ns, which is between 20 ns and 30 ns, signal_b is '1'. But wait, 25 ns is after 20 ns assignment to '1'. So the correct value is '1'.

❓ Predict Output

advanced

Output of a stimulus process with wait until and wait for combined

What will be the value of signal_c at 50 ns given this process?

VHDL

process
begin
  signal_c <= '0';
  wait until now = 20 ns;
  signal_c <= '1';
  wait for 20 ns;
  signal_c <= '0';
  wait;
end process;

A '0'

B '1'

C Process will hang and signal_c remains '0'

D Signal_c is 'U' (undefined)

💡 Hint

Consider how wait until and wait for affect the timing of assignments.

The process waits until 20 ns, then assigns '1'. Then it waits 20 ns more (until 40 ns), assigns '0'. At 50 ns, the last assignment was '0'.

❓ Predict Output

advanced

Effect of multiple wait statements on process execution

What is the final value of signal_d after this process completes?

VHDL

process
begin
  signal_d <= '1';
  wait for 10 ns;
  signal_d <= '0';
  wait for 10 ns;
  signal_d <= '1';
  wait for 10 ns;
  signal_d <= '0';
  wait;
end process;

A Process never completes, signal_d remains '1'

B '1'

C '0'

D Signal_d is 'U' (undefined)

💡 Hint

Follow the signal assignments and wait times step by step.

The signal changes at 0 ns to '1', at 10 ns to '0', at 20 ns to '1', and at 30 ns to '0'. The process then waits indefinitely. So the final value is '0'.

🧠 Conceptual

expert

Understanding process suspension with wait statements

In a VHDL stimulus process with multiple wait statements, what happens to the process execution after the last wait; statement?

A The process restarts from the beginning immediately

B The process suspends indefinitely and does not resume unless triggered externally

C The process terminates and cannot be resumed

D The process continues executing the next statements sequentially

💡 Hint

Think about how wait; without conditions affects process behavior.

A wait; statement without any condition causes the process to suspend indefinitely. It will only resume if an external event triggers it (like a signal change if sensitivity list or wait conditions exist). It does not restart or terminate.

Practice

1. What is the main purpose of using wait for statements inside a stimulus process in VHDL?

easy

A. To declare variables inside the process

B. To pause the process and apply inputs step-by-step over time

C. To define the clock period for the design

D. To reset all signals to zero immediately

Solution

1. Step 1: Understand the role of stimulus processes

   Stimulus processes apply test inputs to signals during simulation.

2. Step 2: Identify the function of wait for

   The wait for statement pauses the process for a specified time, allowing signals to hold values before changing again.

3. Final Answer:

   To pause the process and apply inputs step-by-step over time -> Option B

4. Quick Check:

   Stimulus process uses wait for to pause and apply inputs [OK]

Hint: Wait statements pause input changes to simulate time steps [OK]

Common Mistakes:

• Thinking wait for resets signals
• Confusing wait for with variable declaration
• Assuming wait for defines clock period

2. Which of the following is the correct way to end a stimulus process in VHDL to stop it from running indefinitely?

easy

A. wait;

B. stop;

C. end process;

D. wait for 10 ns;

Solution

1. Step 1: Review how stimulus processes terminate

   Stimulus processes often run indefinitely unless explicitly stopped.

2. Step 2: Identify the correct wait statement to stop the process

   The statement wait; causes the process to suspend forever, effectively stopping it.

3. Final Answer:

   wait; -> Option A

4. Quick Check:

   Use wait; to stop stimulus process [OK]

Hint: Use plain wait; to stop stimulus process execution [OK]

Common Mistakes:

• Using wait for with a time instead of wait;
• Trying to use end process; as a stop command
• Using stop; which is not valid VHDL

3. Consider the following VHDL stimulus process snippet:

process
  begin
    signal_a <= '0';
    wait for 10 ns;
    signal_a <= '1';
    wait for 20 ns;
    wait;
end process;

What is the value of signal_a at 25 ns after simulation start?

medium

A. '0'

B. Undefined

C. '1'

D. Signal does not change

Solution

1. Step 1: Trace signal assignments over time

   At 0 ns, signal_a is set to '0'. After 10 ns, it changes to '1'.

2. Step 2: Determine signal value at 25 ns

   At 25 ns, which is after the 10 ns and before the 30 ns (10+20), signal_a remains '1'.

3. Final Answer:

   '1' -> Option C

4. Quick Check:

   Signal changes at 10 ns to '1', stays until 30 ns [OK]

Hint: Add wait times cumulatively to find signal value at given time [OK]

Common Mistakes:

• Assuming signal_a is still '0' at 25 ns
• Ignoring the second wait for 20 ns
• Confusing wait; with wait for

4. Identify the error in the following stimulus process code:

process
begin
  signal_b <= '1';
  wait for 5 ns
  signal_b <= '0';
  wait for 5 ns;
  wait;
end process;

medium

A. Missing semicolon after first wait for statement

B. Signal assignment syntax is incorrect

C. wait; statement is misplaced

D. Process missing sensitivity list

Solution

1. Step 1: Check syntax of wait statements

   The first wait for 5 ns is missing a semicolon at the end.

2. Step 2: Confirm other syntax elements

   Signal assignments and the final wait; are correct. Sensitivity list is not required in a process with wait statements.

3. Final Answer:

   Missing semicolon after first wait for statement -> Option A

4. Quick Check:

   Every wait for must end with semicolon [OK]

Hint: Check semicolons after wait for statements carefully [OK]

Common Mistakes:

• Thinking sensitivity list is required with wait
• Misreading signal assignment syntax
• Assuming wait; must be last line only

5. You want to create a stimulus process that toggles clk every 15 ns and stops after 90 ns. Which code snippet correctly implements this behavior?

hard

A. process
begin
clk <= '0';
wait for 15 ns;
clk <= '1';
wait for 15 ns;
wait;
end process;

B. process
begin
clk <= '0';
wait for 90 ns;
clk <= '1';
wait;
end process;

C. process
begin
for i in 0 to 5 loop
clk <= '0';
wait for 15 ns;
clk <= '1';
wait for 15 ns;
end loop;
wait;
end process;

D. process
variable count : integer := 0;
begin
while count < 6 loop
clk <= not clk;
wait for 15 ns;
count := count + 1;
end loop;
wait;
end process;

Solution

1. Step 1: Understand toggling every 15 ns for 90 ns

   Toggling means changing clk from 0 to 1 or 1 to 0 every 15 ns. For 90 ns total, toggling happens 6 times (90/15).

2. Step 2: Analyze each option's logic

   process variable count : integer := 0; begin while count < 6 loop clk <= not clk; wait for 15 ns; count := count + 1; end loop; wait; end process; uses a loop with a counter to toggle clk 6 times, waiting 15 ns each time, then stops with wait;. Others either toggle once or use incorrect timing.

3. Final Answer:

   variable count : integer := 0; while count < 6 loop clk <= not clk; wait for 15 ns; count := count + 1; end loop; wait; -> Option D

4. Quick Check:

   Loop with wait for 15 ns toggles clk 6 times [OK]

Hint: Use loop with counter and wait for to toggle signals repeatedly [OK]

Common Mistakes:

• Using single wait for without loop for repeated toggling
• Incorrect loop ranges causing wrong toggle count
• Not stopping process after toggling