rms value

Thread Starter

cachapo

Joined Aug 24, 2009
12
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.
 

ELECTRONERD

Joined May 26, 2009
1,147
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.
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.

If the peak of the sinewave is 1V, then the RMS value is 0.707 x 1V = ? (I'll let you answer that).

Inversly, if you want to find the peak of a sine wave with a RMS value of 120V, you would do the following: 120V x 1.414 = 170V If you want the peak to peak, you would just multiply the peak answer by two.
 

Ratch

Joined Mar 20, 2007
1,070
cachapo,

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%
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.

Ratch
 

Thread Starter

cachapo

Joined Aug 24, 2009
12
Hi Ratch,

I thougth that too, but theoretical result and multimeter measure don´t match. First is quite higher.

Thanks,

Cachapo.​
 

Tesla23

Joined May 10, 2009
542
Hi Ratch,​


I thougth that too, but theoretical result and multimeter measure don´t match. First is quite higher.​

Thanks,​


Cachapo.​
It will reduce the power by 65% so the rms value by √0.65 = 0.806, so the multimeter should read 622mV.
 

Thread Starter

cachapo

Joined Aug 24, 2009
12
I think Tesla23 may be right, his reason seems good.

I don´t know if my multimeter measures true RMS values...
 

Ratch

Joined Mar 20, 2007
1,070
mik3

You need a true RMS multimeter to measure the RMS voltage of non sinusoidal waveforms.
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
 

Tesla23

Joined May 10, 2009
542
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
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.
 

Ratch

Joined Mar 20, 2007
1,070
Tesla23,

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.
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.

Ratch
 

t_n_k

Joined Mar 6, 2009
5,455
I've run several simulations using PSIM with a 30Hz base wave frequency and 3.5KHz (not 35kHz) switching frequency.

At 65% duty [D] the result looks like that shown in the pdf attachment.

The RMS value is 0.570 V - equating to ...

Vrms(switched)=√D * Vrms(pure sine)

as proposed by Tesla23 (in particular).

Results are quite consistent with this over all simulated D values tried.

The result also appears consistent over a wide range of switching frequencies - BTW with the free (limited) version of PSIM, it was just easier for me to get the simulation looking reasonable at the lower switching frequency. For instance with the switching frequency at 350 Hz the RMS result is the consistent to 3 significant figures for 65% duty. It would be interesting to see just how low a switching frequency will suffice - presumably this relates to the Sampling Theorem in some way.
 

Attachments

t_n_k

Joined Mar 6, 2009
5,455
In fact reducing the switching frequency to 175 Hz (on the same 30Hz base wave) still gives Vrms=0.570 at 65% duty. Remarkable - given the sparse switching regime as shown in the pdf attachment. Even at 20% duty the result only varies slightly from the expected value - 0.3175 vs 0.3162
 

Attachments

Ratch

Joined Mar 20, 2007
1,070
To the Ineffable All,

OK, I got 0.5744562642 effective and 0.4201826677 average for 65%, 0.5049752470 effective and 0.3246755513 average for 50%, and 0.7071067814 effective and 0.6366197507 average for 100% .

I did the average because a lot of analog of meters are average responding. This wave shifts pretty fast, so I don't know if most true RMS meters can track it. Perhaps one of the thermal units would be better for this wave.

I used a chop frequency of 3000 hertz and a sampling freq of 300000. Those are multiples of 100 and should suffice.

Ratch
 

t_n_k

Joined Mar 6, 2009
5,455
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.
 

Attachments

Tesla23

Joined May 10, 2009
542
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.
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.
 

t_n_k

Joined Mar 6, 2009
5,455
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.
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.
 

Ratch

Joined Mar 20, 2007
1,070
To the Ineffable All,

I believe Claude Shannon's sampling theorem says that the sampling frequency should be at least 2 times the highest sinusoidal frequency you wish to recover.

Ratch
 
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