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VHDLprogramming~10 mins

Process sensitivity list in VHDL - Interactive Code Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to add the correct signal to the process sensitivity list.

VHDL
process([1])
begin
  -- process body
end process;
Drag options to blanks, or click blank then click option'
Adata
Benable
Creset
Dclk
Attempts:
3 left
💡 Hint
Common Mistakes
Including data signals instead of the clock in the sensitivity list.
Leaving the sensitivity list empty.
2fill in blank
medium

Complete the code to include both signals in the sensitivity list.

VHDL
process([1])
begin
  -- process body
end process;
Drag options to blanks, or click blank then click option'
Aclk, reset
Bdata, enable
Cclk, data
Dreset, enable
Attempts:
3 left
💡 Hint
Common Mistakes
Including only the clock or only the reset signal.
Including unrelated signals like enable.
3fill in blank
hard

Fix the error in the sensitivity list by choosing the correct signal.

VHDL
process([1])
begin
  if rising_edge(clk) then
    -- process body
  end if;
end process;
Drag options to blanks, or click blank then click option'
Adata
Breset
Cclk
Denable
Attempts:
3 left
💡 Hint
Common Mistakes
Using a data signal instead of the clock in the sensitivity list.
Omitting the clock signal.
4fill in blank
hard

Fill both blanks to complete the sensitivity list for asynchronous reset and clock.

VHDL
process([1], [2])
begin
  if [1] = '1' then
    -- reset logic
  elsif rising_edge([2]) then
    -- clocked logic
  end if;
end process;
Drag options to blanks, or click blank then click option'
Areset
Bclk
Cenable
Ddata
Attempts:
3 left
💡 Hint
Common Mistakes
Swapping the order of reset and clock signals.
Including unrelated signals like enable or data.
5fill in blank
hard

Fill all three blanks to complete the process sensitivity list and conditions for asynchronous reset and clock edge.

VHDL
process([1], [2], [3])
begin
  if [1] = '1' then
    -- async reset logic
  elsif rising_edge([2]) then
    -- clocked logic
  end if;
end process;
Drag options to blanks, or click blank then click option'
Areset_n
Bclk
Attempts:
3 left
💡 Hint
Common Mistakes
Using active high reset signal name instead of active low.
Mixing up the order of signals in the sensitivity list.

Practice

(1/5)
1. What does the sensitivity list in a VHDL process specify?
easy
A. Which signals trigger the process to run when they change
B. The order in which signals are assigned values
C. The data types of signals inside the process
D. The clock frequency for the process execution

Solution

  1. Step 1: Understand the role of sensitivity list

    The sensitivity list tells VHDL which signals to watch for changes.
  2. Step 2: Identify what triggers the process

    The process runs only when a signal in the sensitivity list changes value.
  3. Final Answer:

    Which signals trigger the process to run when they change -> Option A
  4. Quick Check:

    Sensitivity list = triggers process run [OK]
Hint: Sensitivity list = signals that start the process [OK]
Common Mistakes:
  • Thinking sensitivity list sets signal data types
  • Confusing sensitivity list with signal assignment order
  • Assuming sensitivity list controls clock frequency
2. Which of the following is the correct syntax for a process sensitivity list in VHDL?
easy
A. process {clk, reset}
B. process clk, reset
C. process [clk, reset]
D. process(clk, reset)

Solution

  1. Step 1: Recall VHDL process syntax

    The sensitivity list is enclosed in parentheses after the keyword process.
  2. Step 2: Match correct syntax

    Only the syntax with parentheses and comma-separated signals is correct: process(clk, reset)
  3. Final Answer:

    process(clk, reset) -> Option D
  4. Quick Check:

    Parentheses and commas = correct syntax [OK]
Hint: Use parentheses with commas for sensitivity list [OK]
Common Mistakes:
  • Omitting parentheses around sensitivity list
  • Using square or curly brackets instead of parentheses
  • Separating signals with spaces instead of commas
3. Consider this VHDL process:
process(a, b)
begin
  c <= a and b;
end process;

What happens if signal b changes value?
medium
A. The process does not run because only a triggers it
B. The process runs and updates c
C. The process runs but c remains unchanged
D. The process causes a syntax error

Solution

  1. Step 1: Check sensitivity list signals

    The process sensitivity list includes signals a and b.
  2. Step 2: Understand process triggering

    When b changes, the process runs and updates c with a and b.
  3. Final Answer:

    The process runs and updates c -> Option B
  4. Quick Check:

    Signal in sensitivity list changed = process runs [OK]
Hint: Process runs when any sensitivity signal changes [OK]
Common Mistakes:
  • Thinking only first signal triggers process
  • Assuming process runs but output doesn't update
  • Confusing syntax error with normal behavior
4. Identify the error in this VHDL process:
process(clk)
begin
  if reset = '1' then
    q <= '0';
  elsif rising_edge(clk) then
    q <= d;
  end if;
end process;

Assuming reset is a signal, what is the problem?
medium
A. Reset signal is missing from the sensitivity list
B. Clock signal should not be in the sensitivity list
C. The process should use falling_edge instead of rising_edge
D. The process syntax is correct with no errors

Solution

  1. Step 1: Check signals used inside process

    The process uses reset and clk signals.
  2. Step 2: Verify sensitivity list completeness

    Only clk is in the sensitivity list; reset is missing, which can cause simulation mismatches.
  3. Final Answer:

    Reset signal is missing from the sensitivity list -> Option A
  4. Quick Check:

    All input signals must be in sensitivity list [OK]
Hint: Include all input signals in sensitivity list [OK]
Common Mistakes:
  • Ignoring signals that affect outputs in sensitivity list
  • Thinking clock should be removed from sensitivity list
  • Confusing edge detection with sensitivity list content
5. You have a process sensitive to signals a and b. Inside, you compute c <= a or b;. Later, you add a new signal enable that gates the output: if enable = '1' then c <= a or b; else c <= '0'; end if;
What must you do to the sensitivity list to ensure correct simulation?
hard
A. Remove a and b from the sensitivity list
B. No change needed; sensitivity list stays the same
C. Add enable to the sensitivity list
D. Replace sensitivity list with only enable

Solution

  1. Step 1: Identify all signals affecting outputs

    Now enable controls output c along with a and b.
  2. Step 2: Update sensitivity list accordingly

    To simulate correctly, enable must be added to the sensitivity list with a and b.
  3. Final Answer:

    Add enable to the sensitivity list -> Option C
  4. Quick Check:

    All input signals in sensitivity list = correct simulation [OK]
Hint: Add all signals that affect outputs to sensitivity list [OK]
Common Mistakes:
  • Forgetting to add new control signals to sensitivity list
  • Removing original signals from sensitivity list incorrectly
  • Assuming sensitivity list auto-updates with code changes
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10 questions~5 mineasy hard