Pen-and-paper experiment.

[hamster_nz]

I was just played around with a pen and paper, and did something that others might find interesting to do.

- calculate the power from 0.5V DC into a load (say 1 ohm)
- calculate the power from 0.5V 1Hz square wave (that is +0.5V / -0.5V with 50% duty cycle) into the same load

Add the two signals together (giving a square wave with a DC offset, that alternates between 1V and 0V)
- calculate the average voltage of this signal
- calculate the average power of this signal
- work out the equivalent DC signal that delivers the same power to the same load
- work out the equivalent square wave AC signal that delivers the same power to the same load

For me, it really has the feeling of "oh, that is just blindingly obvious", and "wait, what what? how does that work?" and "So where, if anywhere, does the extra  0.207V DC come from". For me the math is simple, but (at least for me) the implications are very interesting.

Then play around with a few other combinations, maybe changing duty cycles, or maybe graphing the adding of square waves of different frequencies, and perhaps the most interesting the same frequency but with different relative phases (e.g 90 or 180 degrees, pi/2 or pi if you like radians).

[bob91343]

All good stuff, learning the simpler math of Ohm's law.

[jeroen79]

The voltage is squared not only in the waveform.

[hamster_nz]

So DC + one square wave is relatively easy.

I can simple add the power of a 1Hz and 2Hz square wave into a standard load and get the correct answer for the total power

Why then can I not simply add the power of a  1Hz and 3Hz square wave and get the correct answer for total power?

For example:

1V AC square wave at 1 Hz into a 1 Ohm load is 1 Watt.
1V AC square wave at 3 Hz into a 1 Ohm load is 1 Watt.

But (1V AC square wave at 1Hz added to a 1V AC square wave at 3 Hz) into a 1 Ohm load isn't simply 2W.

(yes, I have an answer to why... but just to show that it isn't always super simple)