How Do You Calculate Average and RMS Values for a Half-Wave Rectified Signal?

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To calculate the average and RMS values for a half-wave rectified signal, the average potential difference (p.d.) is typically found to be (2/π) times the peak voltage. The RMS value for a half-wave rectified signal differs from the full-wave case and requires integration over the waveform, resulting in a value that is not simply (√2/2) times the peak voltage due to the zero voltage during the negative half-cycle. The heat dissipated in a resistor can be calculated using the average power formula, P_avg = (V_rms)^2 / R. Understanding the definitions of average and RMS values is crucial for accurate calculations. This discussion emphasizes the need for integration to derive these values for half-wave rectified signals.
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Homework Statement


A sinusoidal voltage waveform is rectified so that it becomes a half-wave waveform (i.e. the bottom half has been "chopped off"). Determine:

i) The average p.d.
ii) The RMS p.d.
iii) The heat dissipated in a 200ohm resistor in 100s

Homework Equations


P_{avg} = \frac{(V_{rms})^2}{R}

The Attempt at a Solution


I'm not sure, but I would guess that the average p.d. is the line that cuts the rectified waveform in such a way that the area above and below is the same. I have a vague recollection about it being \frac{2}{\pi} times the peak voltage.
I am also quite uncertain about the RMS, how would the voltage being zero half the time affect this value? I'm guessing it will no longer be \frac{\sqrt2}{2} V_{peak}.

Please could anyone give me some clues on how to work out the RMS and average values?EDIT:Oops sorry I did not mean to post the same thread twice...just ignore the other one!
 
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I don't know of any formula for the half-wave signal. I rather think you will have to find out what "root mean square" means and work it out from the definition - no doubt you will be integrating over a full wavelength (the V = zero part will be easy!).

The average potential will be the same sort of thing, the area divided by the time.
 

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