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I need to calculate the effective value of this signal:

enter image description here

I dont know how to calculate this. Iknow how to calculate the effective value of well known waveforms like square, sin, ..., but this is a mystery for me. I think I should combine it somehow with square waveform, but I don't know how.

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    \$\begingroup\$ Integrate it. Integration of rectangles is dead simple. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 12 '14 at 12:42
  • \$\begingroup\$ The problem is: what is the function. The only information that I have is this picture. Can I read function from the picture somehow? \$\endgroup\$ – depecheSoul Jan 12 '14 at 12:43
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    \$\begingroup\$ Height times width. And watch your signs. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 12 '14 at 12:44
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    \$\begingroup\$ You need to clarify what you mean by "effective value" because that's not how we usually talk about waveforms. Are you interested in this signal's ability to generate heat in a resistor? If so, then integration will not give you the result you want. \$\endgroup\$ – Joe Hass Jan 12 '14 at 12:50
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    \$\begingroup\$ electronics.stackexchange.com/questions/95731/… \$\endgroup\$ – GR Tech Jan 12 '14 at 12:56
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The same of what Andy says above in details and with full equations. In electrical terms, the average (or mean) value of a periodic waveform is also known as the DC level of the waveform, even if it is zero. For example if you will try to charge a battery with a bipolar and symmetrical waveform, then there is no net charge on battery. So the average of the above waveform is

enter image description here

therefore

enter image description here

However, it is not the average current (or voltage) of an AC which determines its power delivering capability. The RMS value for a voltage or current waveform is the effective value from the point of view of power dissipation in a resistive circuit. So the effective or RMS current will be

enter image description here

results

enter image description here

But I have a question here: How we can apply

enter image description here

A practical use of the above is in rectifier circuits, that is shorten the contact time of diode, the power dissipation decreases but RMS current remains the same!

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There are two sections in the waveform that are not 0V then, I assume the waveform repeats. If you are calculating RMS then add the squares of the non-0V sections to get 100 + 4 = 104. Then, divide by baselength of 4 to get 26. Then take the square root to get 5.1V RMS.

For average it's 10 - 2 then divided by 4 = 2V

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