ELECTRICAL AND ELECTRONICS Instrumentation

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flaxwalter

Junior Member level 2 1) I=2+4sin314t
2) 4 is the amplitude (or peak to peak) current.
3) But rms only takes the half and multiplies it by sqrt(2)/2, wich results 0.7071....
4) Then by superposition: I=2+(4/2)*.7071= 3.4142 A

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Easy peasy Dear Barry, rms of A + rms of B does not give a total rms figure, rms must be added vectorially, unless they are both DC signals, or AC signals of single freq and in phase.

hence square each, add, then take the SQRT, i.e. add as vectors, rms stands for root - mean - squared, i.e. the SQRT of the mean of the squares, this is covered in 6th form mathematics and 1st year university maths, the rms current is that value that gives the same heating effect as a DC current of that value.

Dear Flaxwater, 4 is the peak ( not pk-pk) superposition does not apply here, the answer is 3.46375 amps rms

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Akanimo Hi Barry,

Also take a look at my LTspice simulation result of Post #13. In that simulation I added both currents as you can see in the simulation waveform and calculated the RMS value of the sum. Spice reported 3.464A which is exactly what you get by "squaring/square-rooting".

I'm looking at it as vector addition. Let's analyze the DC part as a sinusoidal representation recognising that the frequency of a DC quantity is zero, we get to see the following:
Putting Idc = 2*sin0 = 2*0 = 0A, we see that Idc is not a sine function.
Putting Idc = 2*cos0 = 2*1 = 2A, we see that Idc is a cosine function.

This means that I = 2 + 4sinx = 2cos0 + 4sinx in electricity.

I hope that's a good one.

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Hi flaxWalter, 4 in the sine wave expression is just the peak value and not peak-to-peak.

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With this question, one thing that I have come to learn the hard way is that DC quantities are cosine functions and not sine functions. No teacher has ever told me that. Sympathetic!

FvM

Super Moderator
Staff member Some explanation are unnecessarily complicated, I think. The keyword is orthogonality.

Sine signals of different frequency (also freq=0, DC) are orthogonal. Their generated power adds linearly, respectively rms values have to be squared.

barry Well, I guess I flunked that test...

Akanimo Some explanation are unnecessarily complicated, I think. The keyword is orthogonality.

Sine signals of different frequency (also freq=0, DC) are orthogonal. Their generated power adds linearly, respectively rms values have to be squared.

I went a step further and I found out that it is not only frequency that determines the RMS value for such combination of waveforms. I found out that differences in phase angles between the waveforms involved play a role too. This makes SQRT(RMSA^2 + RMSB^2) not adequate for determining the RMS value in all cases. This is also the case for combinations of more than two signals.

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This can be confirmed by simulation. I checked with LTspice.

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Easy peasy yes - it is more correct to say the vector addition of two or more uncorrelated signals

where there is a fixed phase difference - e.g. 180 degrees for an obvious case, the rms will be smaller, if the phase difference moves then the vector addition will give an average but not describe the envelope, for unrelated frequencies the vector addition still works

KlausST

Super Moderator
Staff member Hi,

I went a step further and I found out that it is not only frequency that determines the RMS value
I also did a simulation with several frequencies ... and with my results the square and squareroot worked.

Are you sure you did it long enough for the combined waveform to repeat (may be a lot of cycles)
You need an integer number of cycles of the one frequency a d an integer number of the other frequency.
Else you get a non integer cyle of one frequency ... where the RMS is not sqrt(2) of the peak anymore.

Klaus

Akanimo Hi,

I also did a simulation with several frequencies ... and with my results the square and squareroot worked.

Are you sure you did it long enough for the combined waveform to repeat (may be a lot of cycles)
You need an integer number of cycles of the one frequency a d an integer number of the other frequency.
Else you get a non integer cyle of one frequency ... where the RMS is not sqrt(2) of the peak anymore.

Klaus

Actually, this was saying that differences in phase angles between combined waveforms affect the resultant RMS value.  With the same frequency, same amplitudes and at different phase angles between the two combined waveforms, the resultant RMS values turn out to be different.

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Hence the relationship: RMS = SQRT(RMSA^2 + RMSB^2) is for waveforms that are 90 degrees out of phase with each other.

KlausST

Super Moderator
Staff member Hi,

In post#26 .. you refer to FvM's post which clearly states "different frequency".

In post#20 ... I stated that at same frequency there is an exception..

Also in post#27 ... is mentioned "uncorrelated frequencies

And posts#24, #27, #28.....

Don't ignore all this.
For sure adding two identical frequencies result in a rotating vector of same, constant frequency and constant length..
Which is different to when you use different frequencies. (Length of vector changes with time, angular velocity is not constant).

So, your simulation is correct and does not contradict all said before.

Klaus

Akanimo I did not say that the simulation contradicts all the points made previously. I acknowledge the fact that vector addition is correct in all cases as mentioned by Easy peasy. I am just trying to strengthen the point that orthogonality is no determined by frequency but by phase. Two vectors are orthogonal when they are perpendicular and as such 90 degrees out of phase with each other.

RMS = SQRT(RMSA^2 + RMSB^2) is correct but A and B must be 90 degrees out of phase. This goes back to strengthen the point that I made earlier on with my analysis that showed that DC signals are actually cosine functions.

If we look at it, we would see that at any time t the orthogonality can be shown by: Idc + Iac_m*sin(2*pi*f*t) = Idc_m*cos(2*pi*0*t) + Iac_m*sin(2*pi*f*t) = Idc_m*cos(2*pi*0*t) + Iac_m*cos((2*pi*f*t) - 90deg) = Idc_m*sin((2*pi*0*t) + 90deg) + Iac_m*sin(2*pi*f*t).

FvM

Super Moderator
Staff member I am just trying to strengthen the point that orthogonality is not determined by frequency but by phase.
Actually both.
I didn't refer to phase because it's not needed to answer the original question. But it's true that sine and cosine wave of same frequency are orthogonal. Also any harmonic to each other or the DC component.

Your explanation seems to me unnecessary complicated and somehow counter-intuitive. You don't need to refer to phase to explain why signal components of different frequency add up by their power rather than magnitude.

I'm aware that orthogonality is a rather general term, different science disciplines may possibly use it in a different sense.

Easy peasy RMS = SQRT(RMSA^2 + RMSB^2) is correct but A and B must be 90 degrees out of phase

presumably if they are in phase the vector addition works too...? ( simple addition works in this case too - as it does for the 180 deg case )

are you implying that for 45 deg out of phase - vector addition won't work ... ?

Akanimo presumably if they are in phase the vector addition works too...? ( simple addition works in this case too - as it does for the 180 deg case )
Yes, because in phase the forces aid each other and at 180 degrees out of phase the forces are in direct opposition, at all times.

are you implying that for 45 deg out of phase - vector addition won't work ... ?
No, vector addition works in all cases including 45 degrees. I confirmed this in post #30:
… I acknowledge the fact that vector addition is correct in all cases as mentioned by Easy peasy…

RMS = SQRT(RMSA^2 + RMSB^2) is a specific case of vector addition for two sinusoidal functions with one 90 degrees out of phase with the other, Acosx + Bsinx, and so is not suitable for use as a general expression. RMS = SQRT(RMSA^2 + RMSB^2) obviously does not work for two waveforms that are 45 degrees out of phase with each other.

Easy peasy yes, works for un-correlated waveforms ...

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