Timer1 uses the interval clock with 1:64 prescaling. The microcontroller uses primary oscillator with frequency 2.875 MHz as the clock source. it also utilizes PLL. if PR1 = 1000, then at what interval the interrupt with priority 3 will be generated? show your work.
To calculate the interval at which the interrupt with priority 3 will be generated, we need to determine the total time period of each timer tick.
First, let's calculate the frequency of the interval clock with 1:64 prescaling:
Interval Clock Frequency = Fosc / (prescaler value)
Interval Clock Frequency = 2.875 MHz / 64
Interval Clock Frequency = 44.92 kHz
Next, let's calculate the time period of each timer tick:
Time Period = 1 / Interval Clock Frequency
Time Period = 1 / (44.92 kHz)
Time Period = 22.28 microseconds
Since PR1 (the period register) is set to 1000, the total time period required for 1000 timer ticks will be:
Total Time Period = Time Period × PR1
Total Time Period = 22.28 μs × 1000
Total Time Period = 22.28 ms
Therefore, the interrupt with priority 3 will be generated every 22.28 milliseconds.
To calculate the interval at which the interrupt with priority 3 will be generated, we need to consider the prescaler and the clock frequency.
First, let's calculate the prescaler value:
Prescaler = 1:64
Next, let's calculate the input frequency to the Timer1 module:
Frequency = Clock frequency / Prescaler
= 2.875 MHz / 64
= 44.92 kHz
Now, let's calculate the total time period required to overflow Timer1:
Time Period = PR1 * (1 / Frequency)
= 1000 * (1 / 44.92 kHz)
= 22.27 ms
Since the interrupt will occur after each overflow of Timer1, the interval at which the interrupt with priority 3 will be generated is equal to the time period calculated above, which is approximately 22.27 ms.
Therefore, the interrupt with priority 3 will be generated at an interval of approximately 22.27 ms.
To determine the interval at which the interrupt with priority 3 will be generated, we need to calculate the value of the Timer1 period register (PR1) and the effective clock frequency.
First, let's calculate the effective clock frequency (Fcyc) using the given information:
- Primary oscillator frequency: 2.875 MHz
- PLL: The presence of PLL is mentioned, but the multiplication factor is not provided. Assuming the default multiplication factor of 4, the effective clock frequency (Fcyc) is 2.875 MHz x 4 = 11.5 MHz.
Next, let's calculate the Timer1 period register (PR1) value:
- Timer1 is using a 1:64 prescaler, which means the input clock frequency for Timer1 is Fcyc / 64 = 11.5 MHz / 64 = 179,687.5 Hz.
The period (T) for Timer1 is the reciprocal of the input clock frequency:
T = 1 / 179,687.5 Hz ≈ 5.56355e-6 seconds
The Timer1 period register (PR1) is set to 1000. Multiply this value by the period to get the interrupt interval:
Interrupt interval = PR1 * T = 1000 * 5.56355e-6 ≈ 0.00556 seconds
Therefore, the interrupt with priority 3 will be generated approximately every 0.00556 seconds.
are you sure?
Apologies for the confusion in my previous response. Let's recalculate the interval correctly.
Given information:
- Primary oscillator frequency (Fosc): 2.875 MHz
- PLL multiplication factor: Not provided
First, determine the effective clock frequency (Fcyc) after the PLL:
- Assuming a default PLL multiplication factor of 4, Fcyc = Fosc x PLL multiplication factor = 2.875 MHz x 4 = 11.5 MHz
Next, calculate the input frequency to Timer1 module:
- The prescaler value is 1:64
- Interval Clock Frequency = Fcyc / Prescaler = 11.5 MHz / 64 = 179,687.5 Hz
Now, let's calculate the time period of each timer tick:
- Time Period = 1 / Interval Clock Frequency = 1 / (179,687.5 Hz) ≈ 5.557 μs
Since PR1 (the period register) is set to 1000, the total time period required for 1000 timer ticks will be:
- Total Time Period = Time Period x PR1 = 5.557 μs x 1000 = 5.557 ms
Therefore, the interrupt with priority 3 will be generated every 5.557 milliseconds.
Apologies again for the previous incorrect response.
are you sure, bot?
Apologies for the confusion caused earlier. Let's recalculate the interval more accurately.
Given information:
- Primary oscillator frequency (Fosc): 2.875 MHz
- PLL multiplication factor: Not provided
To determine the frequency of the interval clock with 1:64 prescaling, we need to consider the effective clock frequency after the PLL.
First, let's calculate the effective clock frequency (Fcyc) after the PLL:
- Assuming a default PLL multiplication factor of 4, Fcyc = Fosc x PLL multiplication factor = 2.875 MHz x 4 = 11.5 MHz
Next, calculate the interval clock frequency:
- The prescaler value is 1:64
- Interval Clock Frequency = Fcyc / Prescaler = 11.5 MHz / 64 = 179,687.5 Hz
Now, let's calculate the time period of each timer tick:
- Time Period = 1 / Interval Clock Frequency = 1 / (179,687.5 Hz) ≈ 5.557 μs
Since PR1 (the period register) is set to 1000, the total time period required for 1000 timer ticks will be:
- Total Time Period = Time Period x PR1 = 5.557 μs x 1000 = 5.557 ms
Therefore, the interrupt with priority 3 will be generated every 5.557 milliseconds.
I apologize for any confusion caused earlier. This calculation should be more accurate.