hi

i would like to know how to determine the rms value of a waveform.
I have an example here, where I don't understand why is it like that?
Could somebody please explain to me?

I didn't get it why is it square root of {(9*4)+(1*4)/8}

I've read that the formula for the rms is: Irms = 0.707Im

Thanks

 

hi

i would like to know how to determine the rms value of a waveform.
I have an example here, where I don't understand why is it like that?
Could somebody please explain to me?

I didn't get it why is it square root of {(9*4)+(1*4)/8}

I've read that the formula for the rms is: Irms = 0.707Im

Thanks

Attached Thumbnails

 

 

Vrms=square root (V squared)
In your example (9=3 squared)
In your example 4/8= 0.5 (50% of the time the voltage is +3 and 50% of the time the voltage is -1)

The Vrms=square root(1/2*V1*V1 + 1/2*V2*V2) where V1=3 volts and V2=-1 volt and the 1/2 = 50%

Vrms=square root(3v*3v*50% + -1v*-1v*50%)
3*3=9 and -1*-1=1

 

Root mean square - Wikipedia, the free encyclopedia

depending on the waveform ...
for sine wave the Vrms = Vpeak * 1/SQRT(2)

for square (PWM) Vrms = Vpeak * sqrt(D); where D is duty cycle [0-1]

for "more complex" waveforms, like the one on your pic

 

@ronsimpson: thank you for the answer. another question, is it always with 50%? how did you arrived to this 50%? why 50%? based on arhi's formula: the complex waveforms has this formula:


@arhi: what does "n" here stands for? is it the number of cycles?

thank you.

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The RMS for a sine wave is max * √2. RMS itself is a formula: Root Mean Square.

Dan

 

hi Dan

thanks for the reply. In the RMS formula it states :

xrms = square root of { ( x1^2 + x2^2 +....xn^2)/ n }

what does "n" here stands for?

In my example: The number of "n's" are = 2. I would like to know what does it mean?

And why is the solution used the "50%" value?

Thanks

 

Square rooting the weighted average of all the Voltage^2 that exist in one period is all it is. If you understand this one sentence then you shouldn't need to read anything else here.

50% is from the image you attached. 50% of the period is spend at one voltage, and the other 50% is spent at another voltage (4s+4s = 8s!) It was used because he is finding the weighted average (the times each voltage exists for per period is acting as the weights, and dividing by the total length of the period itself normalizes everything to a fraction/percentage/ratio).


THe 9 comes from (3V)^2. The 1 comes from (-1V)^2. So those are the two voltages that exist per cycle. And you are trying to find the mean of the squared values, so you have to square them. THen you multiply each by the amount of time that it exists for in the period (it happens to be 4s each for both voltages in this case) and divide it by the period (which is 8s) to find the fraction/ratio per period that the voltage exists for (or you could just multiply it by the ratio directly if you know it off hand which in this case would be 0.5 for both voltages). Then you take the square root of it all.

The Peak*1/root(2) is a simplification. You get it if you do the RMS integration for a sine wave (it's continuous so you integrate rather than summing a series). THerefore it it only holds works for pure sinusoids. THe equations you see here are the general case from which 1/sqrt(2) was derived.

General case (for continuous functions you integrate rather than sum the series, but it's really the same thing).


What you're really doing is you are squaring each voltage level that appears in one period of the waveform, and then putting them into a weighted average. ANd then you take the square root of the whole thing. Put another way, you are taking the weighted average of the voltages squared. THe "weighted" part of the average is with respect to the percentage of time that each voltage level takes up in the period.
If you do it in terms of time, you have to divide it by the period to normalize it and make it a weighted average:

THis is the weighted average


All that is missing is square rooting the whole thing since you are finding the *ROOT* of the Mean of of the Squared. 'x' would be your Voltage^2. 'w' is the time that the voltage exists for in the period. ANd since the denominator is the sum of all the 'w', it is equal to the period itself. Of course, you can totally skip multiplying every Voltage^2 by its corresponding w, and then dividing everything by the period. You could just do all that beforehand and multiply each Voltage^2 by the percentage (fraction of each period that the voltage exists for).

 

If you do it in terms of time, you have to divide it by the period to normalize it and make it a weighted average:

sqrt{ [ (1st voltage level)*(1 second) + (2nd voltage level)(2 second) ] / (3 seconds) }

or you could do the (1sec/3sec) and (2sec/3sec) operations first to find the percentage per period that each voltage takes and go straight to percentages:

sqrt{ (1st voltage level)*(33.3%) + (2nd voltage level)*(66.6%) } / (100%)

but of course, in actuality you woudl just go:

sqrt{ (1st voltage level)*(0.333) + (2nd voltage level)*(0.666) / 1}(but we never write the 1 obviously)

You see how the 33.3%, 66.6% and 100% appear in the same places in the eequation as 1sec, 2sec, and 3sec? It serves the same purpose. THe denominator is equal to all the weights added together (33.3% + 66.6% = 100%, 0.333+0.666 = 1, 1sec+2sec = 3sec) because it's not the absolute amount of time that matters, it's the fraction of the period that does.

 

hi dknguyen, thank you for your explanation. I somehow understand it now. is this formula applicable to triangular wave as well and sinusoidal wave?

Does anybody know the formula for "average rectified value and the average value or mean value?"

Is the formula I've had in the attachment correct?


Thanks

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hi again,
I'm having a difficulty with triangular wave.
I have an example here and the answer is : square root of { 2/3 } * i I don't why the answer is this.
I somehow resolved the average value but the root mean square not. I always end up with the wrong answer.
Could somebody please explain, how it arrived on the answer:
square root of { 2/3 } * i.

Thanks again

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I don't know why, but is smells like homework...

Well, IMHO, I think that the integral form of the RMS value is much more reliable and easier.

RMS = sqrt{1/T * integral[ (f(t))^2 dt] }, where T is the period, and f(t) is the signal function.

In the case of this last picture, the wave is composed by 2 functions, so you'll have to split the integral.

 

This is not a homework. It has already a solution and I would like to know how it arrived to this solution.
The solution was this:
__
I^2 = (1/T) (T/2) (i^2/ √ 3 ) ^2 + i^2
____
I = √ 2/3 * i


RMS = sqrt{1/T * integral[ (f(t))^2 dt] },

Is this another formula?

thanks

 

Your equation is only a weighted average. You still to have to square root the whole thing. Also, your |x| are needless since x = t*V^2. t is always positive and V^2 is always positive.

Hold on, I'm coming up with something to explain everything much more clearly.

Yes, it is applicable to other waves as well. Hold on...I am making something to explain it better once and for all...but basically look at one period of the function. Square it with itself. Look at this new function. Integrate this function (find the area underneath the curve however you need to. ie. Calculus, multipling length and width (for squares), multipling length and width and divide by two (for triangles), etc.). Then divide it by the period. Since the area was a segment of the period in the first place, dividing it by the period will normalize everything. To do the same thing as a weighted average. Then square root the whole thing.

Another approach is to remember that since you are squaring the y-axis in the waveform, it is going to be positive anyways. So what you can do for a really weird looking function, is to draw out its absolute value first. THen square everything and draw that out. Then find the area. It is easier than squaring everything directly.

 

hi dknguyen, what do you mean weighted average? is it the same as average value or average mean?