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Verilogprogramming~10 mins

Up-down counter with direction control in Verilog - Interactive Code Practice

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

Complete the code to declare a 4-bit register named count.

Verilog
reg [3:0] [1] ;
Drag options to blanks, or click blank then click option'
Acount
Bclk
Cdir
Dreset
Attempts:
3 left
💡 Hint
Common Mistakes
Using clock or reset names instead of the counter variable.
Forgetting to specify the bit width.
2fill in blank
medium

Complete the code to check if the reset signal is active high.

Verilog
if ([1]) begin
  count <= 4'b0000;
end
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Areset
Bclk
Cdir
Denable
Attempts:
3 left
💡 Hint
Common Mistakes
Using clock or direction signals instead of reset.
Not resetting the counter properly.
3fill in blank
hard

Fix the error in the always block sensitivity list to trigger on clock's positive edge.

Verilog
always @([1]) begin
  if (reset) count <= 4'b0000;
  else if (dir) count <= count + 1;
  else count <= count - 1;
end
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Anegedge clk
Bposedge reset
Cclk
Dposedge clk
Attempts:
3 left
💡 Hint
Common Mistakes
Using 'negedge' instead of 'posedge'.
Using the reset signal in sensitivity list incorrectly.
4fill in blank
hard

Fill both blanks to complete the conditional increment and decrement of the counter.

Verilog
if (dir) count <= count [1] 1;
else count <= count [2] 1;
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A+
B-
C*
D/
Attempts:
3 left
💡 Hint
Common Mistakes
Using multiplication or division instead of addition or subtraction.
Mixing up increment and decrement operators.
5fill in blank
hard

Fill all three blanks to complete the module header and port declarations for the up-down counter.

Verilog
module up_down_counter(
  input wire [1],
  input wire [2],
  input wire [3],
  output reg [3:0] count
);
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Aclk
Breset
Cdir
Denable
Attempts:
3 left
💡 Hint
Common Mistakes
Including 'enable' which is not used in this design.
Mixing up the order of inputs.

Practice

(1/5)
1. What does the direction signal control in an up-down counter in Verilog?
easy
A. The reset state of the counter
B. The speed of the clock signal
C. Whether the counter counts up or down
D. The width of the counter output

Solution

  1. Step 1: Understand the role of the direction signal

    The direction signal determines if the counter increments or decrements its value.
  2. Step 2: Relate direction to counting behavior

    If direction is high, the counter counts up; if low, it counts down.
  3. Final Answer:

    Whether the counter counts up or down -> Option C
  4. Quick Check:

    Direction controls count direction = C [OK]
Hint: Direction signal decides count up or down [OK]
Common Mistakes:
  • Confusing direction with clock speed
  • Thinking reset controls counting direction
  • Assuming direction changes output width
2. Which of the following is the correct Verilog syntax to declare a 4-bit up-down counter with direction input?
easy
A. wire [4:0] count; input direction; always @(negedge clk) begin if(reset) count = 0; else if(direction) count = count + 1; else count = count - 1; end
B. reg [3:0] count; input dir; always @(posedge clk) begin if(reset) count <= 0; else if(dir) count <= count + 1; else count <= count - 1; end
C. reg [3:0] count; input dir; always @(posedge clk) begin if(reset) count = 0; else if(dir) count = count + 1; else count = count - 1; end
D. reg [3:0] count; input dir; always @(posedge clk) begin if(reset) count <= 0; else if(dir) count <= count - 1; else count <= count + 1; end

Solution

  1. Step 1: Check variable declarations and types

    Count should be a 4-bit reg, direction input is 'dir'.
  2. Step 2: Verify always block and assignments

    Use posedge clk, non-blocking assignments '<=' for sequential logic, reset sets count to 0, dir controls increment or decrement.
  3. Final Answer:

    Correct syntax with reg, posedge clk, non-blocking assignments, and direction control -> Option B
  4. Quick Check:

    Proper reg and non-blocking assignments = A [OK]
Hint: Use reg, posedge clk, and non-blocking <= for counters [OK]
Common Mistakes:
  • Using wire instead of reg for count
  • Using blocking '=' in sequential always block
  • Incorrect clock edge sensitivity
  • Reversing increment and decrement logic
3. Given this Verilog code snippet for a 3-bit up-down counter with direction control:
reg [2:0] count = 3'b000;
always @(posedge clk) begin
  if(reset) count <= 3'b000;
  else if(dir) count <= count + 1;
  else count <= count - 1;
end

If reset=0, dir=1, and the clock pulses 5 times starting from 0, what is the value of count after 5 pulses?
medium
A. 3'b101
B. 3'b111
C. 3'b100
D. 3'b110

Solution

  1. Step 1: Understand initial conditions and direction

    Count starts at 0, reset is inactive, dir=1 means count up.
  2. Step 2: Calculate count after 5 increments

    Counting up 5 times from 0: 0->1->2->3->4->5, which is 3'b101 in binary.
  3. Final Answer:

    3'b101 -> Option A
  4. Quick Check:

    0 + 5 increments = 5 (binary 101) [OK]
Hint: Count up adds 1 each clock pulse [OK]
Common Mistakes:
  • Forgetting initial count is zero
  • Confusing binary values
  • Counting down instead of up
4. Identify the error in this Verilog code for an up-down counter:
reg [3:0] count;
always @(posedge clk) begin
  if(reset) count = 0;
  else if(dir) count <= count + 1;
  else count <= count - 1;
end
medium
A. Using blocking assignment '=' for reset inside always block
B. Missing sensitivity to negedge clk
C. Count should be declared as wire, not reg
D. Direction signal 'dir' is not declared

Solution

  1. Step 1: Check assignment types in sequential always block

    Sequential logic requires non-blocking assignments '<=' for all assignments inside always @(posedge clk).
  2. Step 2: Identify inconsistent assignment usage

    Reset uses blocking '=' which can cause simulation mismatches; it should use '<=' like other assignments.
  3. Final Answer:

    Using blocking assignment '=' for reset inside always block -> Option A
  4. Quick Check:

    Use '<=' for all sequential assignments [OK]
Hint: Use non-blocking '<=' in sequential always blocks [OK]
Common Mistakes:
  • Mixing blocking and non-blocking assignments
  • Ignoring declaration of signals
  • Wrong sensitivity list for clock
5. You want to design a 4-bit up-down counter with direction control and synchronous reset in Verilog. The counter should wrap around when it reaches max or min values (i.e., counting up from 15 goes to 0, counting down from 0 goes to 15). Which code snippet correctly implements this behavior?
hard
A. always @(posedge clk) begin if(reset) count <= 0; else if(dir) count <= count + 1; else count <= count - 1; end
B. always @(posedge clk) begin if(reset) count <= 0; else if(dir) count <= (count == 4'b0000) ? 4'b1111 : count - 1; else count <= (count == 4'b1111) ? 0 : count + 1; end
C. always @(posedge clk) begin if(reset) count <= 0; else if(dir) count <= count - 1; else count <= count + 1; end
D. always @(posedge clk) begin if(reset) count <= 0; else if(dir) count <= (count == 4'b1111) ? 0 : count + 1; else count <= (count == 4'b0000) ? 4'b1111 : count - 1; end

Solution

  1. Step 1: Understand wrap-around conditions

    When counting up at max (15), reset to 0; when counting down at 0, reset to 15.
  2. Step 2: Check code logic for direction and wrap-around

    always @(posedge clk) begin if(reset) count <= 0; else if(dir) count <= (count == 4'b1111) ? 0 : count + 1; else count <= (count == 4'b0000) ? 4'b1111 : count - 1; end correctly uses conditional operator to wrap count on max and min values based on direction.
  3. Final Answer:

    Code with conditional wrap-around for max and min counts -> Option D
  4. Quick Check:

    Wrap-around logic matches direction and limits [OK]
Hint: Use conditional operator to wrap count at limits [OK]
Common Mistakes:
  • Not wrapping count at max/min values
  • Reversing increment and decrement logic
  • Using incorrect boundary values
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