Infineon CCU6 PWM Capture 1 Manual page 11

Implementation
Calculation example of the PWM frequency
Example for 2 Hz PWM signal with 50% duty cycle:
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Period time of 2 Hz signal is 0.5 s
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High level time is 0.25 s (period * duty cycle)
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Within 0.5 s, the timer T12 is incremented 390625 times (0.5 s / 1.28 µs) and has 5 overflows
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Within 0.25 s, the timer T12 is incremented 195312 times (0.25 s / 1.28 µs) and has 2 overflows
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Parameters available inside the ISR:
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Counter value of timer T12 one period ago: 1000
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Current counter value of timer T12: 63955
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Counter value when the falling edge occured: 65244
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Amount of overflows during one PWM period: 5
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Amount of overflows during PWM high level time: 2
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Calculation of total amount of increments:
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Increments before the first overflow: 65534 – 1000 = 64534
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Increments during the second and the last overflow: (5 - 1) * 65534 = 262136
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Increments after the last overflow: 63955
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Total amount of increments: 64534 + 262136 + 63955 = 390625
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Calculation of increments during high level time:
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Increments before the first overflow: 65534 – 1000 = 64534
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Increments during the second and the last overflow: (2 - 1) * 65534 = 65534
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Increments after the last overflow: 65244
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Total amount of increments: 64534 + 65534 + 65244 = 195312
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Calculation of the PWM frequency by dividing the frequency of timer T12 by the total amount of
increments during one PWM period:
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781250 Hz / 390625 = 2 Hz
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Duty cycle (increments during high level time / increments during total period) = 195312 / 390625 = 50%
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