# Sine wave analysis

1. Aug 30, 2010

### Willnage

Hi All

I'm in the process of developing a phase control circuit. I need to know the specific time that corresponds to a required % of max power to chop the sine wave of a mains. I know that there is someway of calculating the time interval for a percentage power using integration or some form of wave analysis but my calculusis really rusty. Any help will be greatly appreciated.

Thanks

2. Sep 3, 2010

### klondike

The RMS power of a periodic signal with period of T is defined as:

$$RMS=\sqrt{\frac{1}{T}\int ^{+T/2}_{-T/2}f^{2}(t)dt}$$

You then express the clipped sine wave in several integrating intervals and carry out the simple definite integral. The relation between the clipping time and RMS is then obvious.

The following integral are involved:
In the non-clipping intervals:
$$\int \sin ^{2}\omega t dt =\frac{1}{2}t-\frac{\cos 2\omega t}{4\omega}+C$$

In the clipping interval (clipped at A volts):
$$\int A^{2} dt =A^{2}t+C$$

Your "specific time" determines the integral intervals. Good luck!