Firstly, it is important to understand the theory of RMS. If you applied AC voltage to a resistor, it will dissipate energy in the form of heat, just as if the voltage were DC. So, the RMS (root mean square) value is determined based on how much DC voltage would cause an identical amount of heating if it was actually AC voltage on the resistor.Hi everybody,
Can anybody tell me how to calculate the rms value of a chopped sinewave?
Sinewave amplitude = 1V
Sinewave freq. = 30Hz
Chopper freq. = 35KHz
Duty = 65%
Thanks.
Does 65% duty mean that it is on for 65% of the time, or off 65% of the time. I am going to assume it means on. Since the height of the sine has not changed, and the switching occurs as a much higher frequency than the sine, my intuition says that the rms will be 65% of .707 or 0.460 for a 1 volt sine. I base my answer on the observation that the wave will only be able to deliver energy 65% of the time during he same time period as before. That should decrease the energy rate or power to 65% of what it could do before.Can anybody tell me how to calculate the rms value of a chopped sinewave?
Sinewave amplitude = 1V
Sinewave freq. = 30Hz
Chopper freq. = 35KHz
Duty = 65%
It will reduce the power by 65% so the rms value by √0.65 = 0.806, so the multimeter should read 622mV.Hi Ratch,
I thougth that too, but theoretical result and multimeter measure don´t match. First is quite higher.
Thanks,
Cachapo.
Yes, I like your reasoning. I think that may be right.It will reduce the power by 65% so the rms value by √0.65 = 0.806, so the multimeter should read 622mV.
You need a true RMS multimeter to measure the RMS voltage of non sinusoidal waveforms.Hi Ratch,
I thougth that too, but theoretical result and multimeter measure don´t match. First is quite higher.
Thanks,
Cachapo.
What you say is certainly true, but in this case, the shape envelope of the wave has not been altered. The nibbles out of the wave occur so evenly that I don't think the crest factor will change very much.You need a true RMS multimeter to measure the RMS voltage of non sinusoidal waveforms.
I'm hopeless on definitions, but IIRC non-true RMS meters either measure the average rectified voltage or (less likely) the peak voltage and scale for a sine wave. If it measures the average voltage then this will drop by the duty cycle whereas the RMS value drops by the square-root of the duty cycle, and if it measures the peak then this will not drop at all. Either way a non-true RMS meter will be in error.mik3
What you say is certainly true, but in this case, the shape envelope of the wave has not been altered. The nibbles out of the wave occur so evenly that I don't think the crest factor will change very much.
Ratch
Tomorrow, if I get time, I will calculate the RMS and the average value of the given chopped wave by computer methods. Then we can settle the question once and for all.I'm hopeless on definitions, but IIRC non-true RMS meters either measure the average rectified voltage or (less likely) the peak voltage and scale for a sine wave. If it measures the average voltage then this will drop by the duty cycle whereas the RMS value drops by the square-root of the duty cycle, and if it measures the peak then this will not drop at all. Either way a non-true RMS meter will be in error.
You don't need to go to all that trouble, as long as the waves aren't coherent, then if you chop with a duty cycle of D the power is reduced by a factor of D, so the rms amplitude reduced by √D. QED.I'm not sure whether this adds anything to the "mix" but I've attached working on how I have computed RMS values for the function using a Scilab solution. Also included a 'rough' outline of the math method I applied.
Yeah - thanks Tesla23. I take your points. What is interesting is that one can come down to quite low sampling frequencies and the relationship for the RMS value still holds pretty well.You don't need to go to all that trouble, as long as the waves aren't coherent, then if you chop with a duty cycle of D the power is reduced by a factor of D, so the rms amplitude reduced by √D. QED.
In the example given, the chopping frequency is so high that even if the waveforms are harmonically related this will still be very close.