I am trying to calculate the pulse waves for the PWM (Pulse Width Modulation) of a Pure Sine Wave with 60 Hz or 50 Hz.

Each pulse wave duration calculated, either to power on or power off, must be a whole number of 3 μs (microseconds) or greater. The challenge is to closely approximate a pure sine wave within the constraints specified above.

My best attempt is this:
p = power on microseconds (μs) as a whole number of 3 or greater.
n = no power (power off) (μs) as a whole number of 3 or greater.
h = Hz rate, which will be 60 or 50 Hz.
m = microseconds (μs) for a single sine wave cycle at h Hz
t = total microseconds (μs) for the previous pulse calculations

$$h = 60$$ $$m = \frac{1000000}{h} = 16667$$ $$\frac{p}{p+n} ≈ \sin(\frac{p+n+t}{m} \times \frac{\pi}{2})$$

I started with the assumption to turn power on at the minimum 3 μs and then calculated the first power off duration to be 178 μs.

Then I added 1 to the 3 μs to turn power on again for 4 μs and then calculated to have it turned off for 132 μs; I suspect that just adding 1 each time to the power on duration is not correct, but I don't know how to determine when I should increase that duration more rapidly. I want to calculate to sin(0.5) and then use these numbers backwards to zero, and then repeat the full cycle for the negative side of the sine wave.

I know my assumptions are wrong, the total μs exceeds 16667 μs for a single cycle, it came out to be 5676 x 4 = 22704 μs with 214 on-off pulse combinations. I'm close, but not quite there.

This problem has humbled me, I'm absolutely stuck and will appreciate any help. Thank you.


If you want to produce a sine wave as nearly as possible, you need merely vary the on-pulses as multiples of 3 microseconds as a function of sine with time with, e.g., 3 microsecond off-pulses between them. If you want just the full-wave power equivalent, I recommend exploring RMS which effectively reflects the duty cycle requirement I think you are looking for.

  • $\begingroup$ I really appreciate your help. I got lost in the forest, thanks again. $\endgroup$ – Mark Main Jan 12 '19 at 21:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.