Window watchdog concept in Embedded C - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to enable the window watchdog timer.

Embedded C

WWDG->CR = [1];

Drag options to blanks, or click blank then click option'

A0x7F

B0x40

C0x80

D0x00

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0x00 disables the watchdog.

Using 0x7F sets the counter but does not enable it.

The control register (CR) is set with 0x80 to activate the watchdog counter.

2fill in blank

medium

Complete the code to set the window value for the watchdog timer.

Embedded C

WWDG->CFR = ([1] << 0);

Drag options to blanks, or click blank then click option'

A0x80

B0x7F

C0x40

D0x3F

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0x7F sets the window too high, disabling the window effect.

Using 0x80 sets a bit outside the window value field.

The window value is set by writing 0x3F to the lower 7 bits of the CFR register.

3fill in blank

hard

Fix the error in the code to refresh the watchdog counter correctly.

Embedded C

WWDG->CR = [1];

Drag options to blanks, or click blank then click option'

A0x7F

B0x80

C0x3F

D0x40

Attempts:

3 left

💡 Hint

Common Mistakes

Writing 0x3F lowers the counter incorrectly.

Writing 0x80 disables the watchdog.

Refreshing the watchdog requires writing 0x7F to the CR register to reset the counter.

4fill in blank

hard

Fill both blanks to configure the prescaler and enable the watchdog interrupt.

Embedded C

WWDG->CFR = ([1] << 7) | (1 << [2]);

Drag options to blanks, or click blank then click option'

A3

B6

C5

D7

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong bit positions disables features.

Confusing prescaler and interrupt bits.

The prescaler is set by shifting 3 to bit 7, and the interrupt enable is bit 6.

5fill in blank

hard

Fill all three blanks to initialize the watchdog with prescaler, window, and refresh.

Embedded C

WWDG->CFR = ([1] << 7) | [2];
WWDG->CR = [3];

Drag options to blanks, or click blank then click option'

A2

B0x3F

C0x7F

D4

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing prescaler and window values.

Using refresh value outside allowed range.

Prescaler set to 4 (shifted to bit 7), window set to 0x3F, and refresh counter set to 0x7F.

Practice

1. What is the main purpose of a window watchdog timer in embedded systems?

easy

A. To increase the processing speed of the microcontroller

B. To reset the system if the software fails to refresh the timer within a specific time window

C. To store data temporarily during program execution

D. To provide power backup during system failure

Solution

1. Step 1: Understand the watchdog timer function

   A watchdog timer monitors if the software is running correctly by expecting periodic refreshes.

2. Step 2: Identify the role of the window watchdog

   The window watchdog resets the system if the refresh happens too early or too late, ensuring software reliability.

3. Final Answer:

   To reset the system if the software fails to refresh the timer within a specific time window -> Option B

4. Quick Check:

   Window watchdog resets system on missed or mistimed refresh [OK]

Hint: Watchdog resets system if refresh is outside allowed time [OK]

Common Mistakes:

• Confusing watchdog with memory storage
• Thinking watchdog speeds up the processor
• Assuming watchdog provides power backup

2. Which of the following is the correct way to refresh a window watchdog timer in embedded C?

easy

A. Call WWDG_Refresh() function within the allowed time window

B. Call WWDG_Refresh() only once at system startup

C. Disable the watchdog timer before refreshing

D. Refresh the watchdog timer continuously without delay

Solution

1. Step 1: Understand watchdog refresh requirements

   The watchdog must be refreshed periodically within a specific time window to avoid reset.

2. Step 2: Identify correct refresh method

   Calling WWDG_Refresh() within the allowed time window is the correct way to refresh.

3. Final Answer:

   Call WWDG_Refresh() function within the allowed time window -> Option A

4. Quick Check:

   Refresh watchdog periodically inside time window [OK]

Hint: Refresh watchdog periodically, not just once or too fast [OK]

Common Mistakes:

• Refreshing only once at startup
• Disabling watchdog before refresh
• Refreshing too frequently causing reset

3. Consider this code snippet using a window watchdog timer:

void main() {
  WWDG_Init();
  while(1) {
    delay(50);  // delay in ms
    WWDG_Refresh();
  }
}

If the allowed refresh window is between 40ms and 60ms, what will happen?

medium

A. The system will reset because refresh is too early

B. The system will reset because refresh is too late

C. The system will run normally without reset

D. The system will freeze due to watchdog error

Solution

1. Step 1: Analyze the refresh timing

   The code delays 50ms before refreshing the watchdog, which is within the 40-60ms window.

2. Step 2: Determine system behavior

   Since refresh happens inside the allowed window, the system will not reset and runs normally.

3. Final Answer:

   The system will run normally without reset -> Option C

4. Quick Check:

   Refresh at 50ms inside 40-60ms window = no reset [OK]

Hint: Check if refresh delay fits inside allowed window [OK]

Common Mistakes:

• Assuming 50ms is too early or too late
• Confusing delay units
• Ignoring the watchdog window limits

4. The following code causes the system to reset unexpectedly. What is the likely error?

void main() {
  WWDG_Init();
  while(1) {
    WWDG_Refresh();
    delay(30);  // delay in ms
  }
}

Assuming the allowed refresh window is 40ms to 60ms.

medium

A. Not calling WWDG_Refresh() frequently enough

B. Delay is too long causing late refresh

C. Missing WWDG_Init() call

D. Refreshing the watchdog too early before the window opens

Solution

1. Step 1: Check refresh timing relative to window

   The code refreshes the watchdog immediately, then delays 30ms, so refresh happens every ~30ms.

2. Step 2: Compare with allowed window

   Since the window starts at 40ms, refreshing at 30ms is too early and causes reset.

3. Final Answer:

   Refreshing the watchdog too early before the window opens -> Option D

4. Quick Check:

   Refresh before 40ms window causes reset [OK]

Hint: Refresh only after window opens, not before [OK]

Common Mistakes:

• Thinking delay is too long
• Ignoring refresh order
• Assuming missing init causes reset here

5. You want to design a system using a window watchdog timer that resets if the refresh happens outside 100ms to 200ms window. Which approach correctly implements this behavior in embedded C?

hard

A. Refresh the watchdog every 150ms

B. Refresh the watchdog every 50ms

C. Refresh the watchdog every 250ms

D. Refresh the watchdog only once at startup

Solution

1. Step 1: Understand the allowed refresh window

   The watchdog must be refreshed between 100ms and 200ms to avoid reset.

2. Step 2: Evaluate each refresh timing option

   Refreshing at 50ms is too early, 150ms is inside window, 250ms is too late, and once at startup is insufficient.

3. Final Answer:

   Refresh the watchdog every 150ms -> Option A

4. Quick Check:

   Refresh inside 100-200ms window avoids reset [OK]

Hint: Refresh watchdog only inside allowed time window [OK]

Common Mistakes:

• Refreshing too early or too late
• Refreshing only once
• Ignoring window timing constraints