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explaining about RMS

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wondrous

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HI
i have a question about rms
rms = 0.7 vp
where (0.7) come from
is it depend on frequencies 50/60 hz ?
like PWM when we vary duty cycle we obtain different value of volts
but here we vary frequencies
thanks
 

Hi,

0.7 --> more exactly 0.707... = 1/sqrt(2) or 0.5 x sqrt(2)
(only for pure sinewave).

It does not depend on frequency.

It just depends on waveform.

****
PWM:
with DC voltage PWM´d:
V(average) = V_DC x duty_cycle
V(RMS) = V_DC x sqrt(duty_Cycle)

Klaus
 
Vrms = 1/2 (square root of 2) Vpeak applie ONLY to sin waves

if you any other waveshape, you have to do the math as in the link in post 3
 
i have a question about rms; rms = 0.7 vp; where (0.7) come from

Good question.

Consider an arbitrary but periodic waveform. It can be sine or square or any random graph but it should be periodic; because we need that period later.

Any arbitrary periodic graph can be described in terms of moments; let me try to explain.

The first one is the mean: Σx(t) taken over one period. To get the mean you divide the sum by T, the period. x is the voltage at any time t. You need to perform this sum over as many points as you can.

If the waveform is symmetrical (e.g., a sine or a cosine wave), the mean vanishes. In fact, all odd moments do vanish for a symmetrical waveform.

The next moment is the sum of squares: Σx²(t); this is related to the variance or standard deviation and RMS value. RMS value is √((x^2(t))/T)

Whether the waveform is symmetrical or not, the sum of squares is always positive (can be zero only if all the points are zero).

There are higher moments: the third moment is Σx^3(t) is related to skewness (it also vanishes for a symmetrical wave). The fourth moment ∑x^4(t) is related to kurtosis.

In general, they are called generalized mean: generalized mean of the n-th order will be (1/T)Σx^n(t).

Now you see why RMS is so important (the simple mean disappears for a symmetrical wave).

For a sine wave, RMS value is peak value /√2 (or, more simply 0.707 x peak value).

But for a general waveform you must calculate all the moments manually.
 
HI
@ c_mitra I see now there are more and more to learn
and I appreciate every post
 

I see now there are more and more to learn ...

Just like an arbitrary distribution of charges can be described by the moments:

monopole moment:

dipole moment:

quadrupole moment:

Octopole moment:

Wikipedia has a somewhat boring description of the multipole expansion: see, for example, https://en.wikipedia.org/wiki/Multipole_expansion

Coming back to the original question; if you specify the RMS value of a given but random waveform, you get some idea about the waveform.

But if you know all the values of the generalized means (sum over one cycle (x(t))^n) you get to know the complete waveform.

But in electrical engineering, the RMS value is the most important - it is related to the power or energy.

But that is a different area altogether.
 
......
Coming back to the original question; if you specify the RMS value of a given but random waveform, you get some idea about the waveform..

This is incorrect. RMS value does not give any information about the original wave shape.

... But in electrical engineering, the RMS value is the most important - it is related to the power or energy.

But that is a different area altogether.

Actually that is the original and probably only purpose of calculating an RMS value.
 

Coming back to the original question; if you specify the RMS value of a given but random waveform, you get some idea about the waveform.

But if you know all the values of the generalized means (sum over one cycle (x(t))^n) you get to know the complete waveform.

But in electrical engineering, the RMS value is the most important - it is related to the power or energy.

But that is a different area altogether.

many thanks sir
that's helpful
may i ask _ excuse my multiple question _ what is the meaning of true RMS
I see that term in many Measurement topic ?
 

Hi,

True RMS means there is a "true" RMS measurement and calculation...with squaring the signal, averaging and taking the square root.

In opposite there are some (how I call them) "fake RMS" calculations.
Often used:
* peak value measurement ... multiplied with 0.707 (for pure sinewave)
* rectified average measurement .... multiplied with 1.11 (for pure sinewave)
* some odd measurement .... multiplied with just a factor to get the expected output value.

You see the "measurements" and calculations are not true RMS. The measurement has nothing to do with RMS, none of them use "square" and "square root". Thus the output is fake.
Only if you have a known, never changing, reliable waveform .... the the results "seem" to be RMS values.

My personal opinion: It's like measuring the size of your feet to find out how tall you are. For many people you may find a good relationship. But for some people the result simply is wrong.

Klaus
 

..
Coming back to the original question; if you specify the RMS value of a given but random waveform, you get some idea about the waveform.

But if you know all the values of the generalized means (sum over one cycle (x(t))^n) you get to know the complete waveform.
.....

Hey C_Mitra !! Here are your 2 lines again. I may be nit-picking, but you have clearly written that "rms value.... get an idea of waveform" I paraphrase.
Your 2nd line (quoted above) doesn't make any sense, so I ignored it. If you meant that if we know all the points in the signal, then yes of course we know the signal ?!? That's a tautology.

- - - Updated - - -

Yes, you can see a paper by a real mathematician here:
https://www.improbable.com/2016/01/15/facilitating-the-perfect-golden-ratio-eyebrow-new-patent/
Of course this is a joke but the point should not be missed.

Try to stay on topic please. Thanks !
 

HI
i have a question about rms
rms = 0.7 vp
where (0.7) come from
is it depend on frequencies 50/60 hz ?
like PWM when we vary duty cycle we obtain different value of volts
but here we vary frequencies
thanks

It is not dependent on frequency.

Note that 0.707 is only for sine wave. Other waveforms yield different factors.

Actually the factor is the ratio of integrated voltage waveform in one cycle to the total area in one cycle. This is to say the that the factor is the ratio of the area taken up by the voltage waveform over the sum of the area taken up and that not taken up.

- - - Updated - - -

ADD:

Note that the total area is (Vpk * one_period).
 

For a sine wave, both the quantities, integrated voltage waveform in one cycle AND the total area in one cycle are ZERO.

see, for example, https://www.wolframalpha.com/input/?i=integral+of+sin(x)+from+0+to+2*pi

In fact, both these are ZERO for any symmetrical waveform.

No, remember that when we say negative, we imply the direction of flow of current. Current flowing in opposite directions every other half cycle doesn't cancel out. In fact the same heating consequence is produced irrespective of direction.

So when integrating, integrate the absolute values. This would give 2 times what you get for a half cycle but not zero.
 
Last edited:

For a sine wave, both the quantities.......are ZERO for any symmetrical waveform.

Are you implying that the power transferred/ carried by a symmetrical wave (around zero) is always Zero? This is obviously a flawed and misguided statements. And the flaw is self-evident.

Power (and RMS) is an integral over squares. Hence doesn't cancel.

Akanimo made a slight error. Nothing to get so excited about :smile:
 
Last edited:

Note that the total area is (Vpk * one_period). ...

For any symmetrical waveform, the total area over a complete period is always ZERO irrespective of the exact nature of the waveform.

Unless, of course, you are defining area as above. But then could you care to explain the same using a sine wave? (say the peak is 1v, the period is 1s)

- - - Updated - - -

No, remember that when we say negative, we imply the direction of flow of current...

We are talking about the RMS value of a voltage waveform. RMS value does not bother about voltage or current; any distribution (graph in time) does have a RMS value.
 

C_mitra ..... you are forgetting what the 'S' stands for in RMS. It's 'Square'.

Power is proportional to SQUARE of voltage. It's proportional to SQUARE of current. When driving a resistive load where V and I are in phase.

When V and I are not in phase the situation is different, and power too flows back and forth.

Get over it.
 
Last edited:

Hi,

So when integrating, integrate the absolute values. This would give 2 times what you get for a half cycle but not zero.
Integrating the voltages, as well as integrating the absolute_value_of_voltage can never lead to RMS.

--> For RMS you need to integrate square_of_voltage .... which can never be negative.

Integrating the absolute value of voltage gives (ONLY for pure sinewaves) a value (precisely: Pi/4) that has to be multiplied with 1.11 to look like the RMS value.

With any altered waveform, the factor need to be adjusted, too, else the resulting value is wrong.
Without knowing exact waveform an RMS calculation with the use of integrating absolute values is mathematically and physically (generated heat in a resistor) wrong.

Klaus
 

Yes Klaus, thanks.

I just noticed I had made that blunder just before your notification came in.

I had been describing the average voltage all along. RMS is obtained from squaring each sample in a period, summing up all the squared results, dividing by the number of samples and then finding the squareroot of the quotient.

The higher the number of samples, the more accurate the result.
 
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