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RMS of linear combination

  1. Nov 2, 2011 #1
    1. The problem statement, all variables and given/known data

    [tex]
    i_1=10+20sin(377t+(\pi /6))[/tex]
    [tex]i_2=-30[/tex]
    [tex]i_3=10+10sin(100t-(\pi/4))[/tex]
    [tex]i_4=10sin(377t)[/tex]

    [tex]i_T=\sum _i i_i
    [/tex]

    What I am wondering is how you go about calculating the average value for the linear combination of all three functions. Also, how would one calculate the rms for such, given the periods are all different?
     
  2. jcsd
  3. Nov 2, 2011 #2

    The Electrician

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    Gold Member

    The periods are not ALL different, are they? Don't i1 and i4 have the same period?

    Anyway, what you do is integrate a suitable expression for the RMS or average value of the sum i1+i2+i3+i4 for a time such that a whole number of cycles of each of the AC waveforms occurs. A number that would work is 100 cycles of the i1 waveform; that would take the same time as 377 cycles of the i3 waveform.

    The 377 is probably intended to be the radian frequency of the grid voltage in the U.S., which is more exactly 2*PI*60. If you had to use 2*PI*60 (for the radian frequency of i1) to find a length of time where whole numbers of cycles would occur for both the i1 and i3 waveforms, you'd be out of luck since PI is irrational. But, your problem says 377, so you're golden.

    Do you know how to use calculus to get the RMS (and average) value of a waveform?
     
  4. Nov 2, 2011 #3
    Thanks for the response. If it is going for one cycle, then t=0→t=1
    [tex]\bar{i_1}=\int ^{1}_{0}{10+20sin(377t+(\pi/6)}dt[/tex]
    let [tex]u=377t+(\pi/6), \frac{du}{377}=dt[/tex]
    and the lower limit of integration is now pi/6, while the upper is 377+pi/6.
    [tex]\bar{i_1}=\int ^{1}_{0}{10}dt+\frac{20}{377}\int ^{377+\pi /6}_{\pi /6}{sin(u)du}[/tex]
    [tex]\bar{i_1}=10t |^{1}_{0}+\frac{20}{377}(-cos(u))|^{377+\pi /6}_{\pi /6}[/tex]

    Thus the average for this first value is 10. Does it look as though that was executed correctly? Once I have the average for the other 3, do I merely sum them together or do I have to average the sum of the averages?
     
  5. Nov 3, 2011 #4

    NascentOxygen

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    Staff: Mentor

    RMS stands for the square Root of the Mean of the Square of the waveform. So I think you need the integral of 100 + 400.sin2(377t in the above.

    I suspect that you can average each separately over its own period, then add their values together, not bothering with phase nor looking for a common whole period time interval, but to prove this I'd need to work it out both ways and show the results are identical.
     
    Last edited: Nov 3, 2011
  6. Nov 3, 2011 #5
    hmmm it'd be interesting to see if this was the case.
     
  7. Nov 3, 2011 #6
    Since I have made the limits of integration the same for all four functions, then the average value for all of the functions would merely be a sum of all the averages over 0-->1?

    Moreover, when determining the rms, we should ensure that we are integrating over a whole number of periods?

    Once we have the answer for each of the four equations, we can simply multiply its period by the entire period we are analyzing. I just don't see how this is done with all the phase changes and stretches and compresses..
     
  8. Nov 3, 2011 #7

    gneill

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    The average of any sinewave over a complete period is zero. So if the chosen period is a multiple of all periods involved (and you don't even need to know what it is!), then the contribution of all the sinewaves to the average will be nil. This makes finding the average value trivial.

    Things are more tricky for the RMS value because both the positive and negative half-cycles of each sinewave make a contribution. There's not much for it but to do the integration. If you let t run from 0 to [itex]2 \pi[/itex] then you're guaranteed to include full cycles for each of the contributing waves.
     
  9. Nov 3, 2011 #8
    Okay, here's what I have done thus far.
    [tex]i_2+i_3=-20+10sin(100t-45°)[/tex]
    Next, I converted equation 1 and 4 into polar coordinates, and then from polar to rectangular coordinates.
    [tex]i_1=30/\sqrt{2}\angle{30°}\rightarrow{15\sqrt{3/2}i+15/\sqrt{2}j}[/tex]
    [tex]i_4=10/\sqrt{2}\angle 45°\rightarrow{5}i+5j[/tex]
    I then combined these two, converted to polar, and then returned it to sinusoid form.
    [tex]i_{1,4}=23.37i+15.61j\rightarrow{28.1\angle{33.7°}}\rightarrow{39.7sin(337t+33.7°}[/tex]

    Therefore, I now have 2 equations.
    [tex]i_{1,4}=39.7sin(337t+33.7°), i_{2,3}-20+10sin(100t-45°)[/tex]

    Can anyone confirm if that looks okay thus far, before I proceed further? I was a little bit weary regarding my initial conversions as this is a brand new subject for me.
     
    Last edited: Nov 3, 2011
  10. Nov 3, 2011 #9

    gneill

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    I'm not sure why you're inserting the √2 constant during your conversions. Usually that would indicate converting from peak to RMS values, but it's the RMS value that you're trying to find.

    Have a look "en.wikiversity.org/wiki/Phasor_algebra"[/URL] for phasor algebra.
     
    Last edited by a moderator: Apr 26, 2017
  11. Nov 3, 2011 #10
    I did look at it in my textbook. For example, the textbook does this:
    [tex]v_a=50sin(377t+30°)\rightarrow{35.35V\angle{30°}}[/tex]
     
  12. Nov 3, 2011 #11

    gneill

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    Yeah, they're converting directly to RMS for the phasor representation. In the present case I don't think that's wise because you're going to be finding the RMS value of the sum of several waveforms by calculating it with an integration; The method assumes that you're using the peak magnitudes.
     
  13. Nov 3, 2011 #12
    Okay, thank you.
    Revised:
    [tex]i_2+i_3=-20+10sin(100t-45°)[/tex]
    Next, I converted equation 1 and 4 into polar coordinates, and then from polar to rectangular coordinates.
    [tex]i_1=30\angle{30°}\rightarrow{25.98i+15j}[/tex]
    [tex]i_4=10\angle 45°\rightarrow{7.07}i+7.07j[/tex]

    [tex]i_{1,4}=33.05i+22.07j\rightarrow{39.74\angle{33.7°}}\rightarrow{39.7sin(337t+33.7°)}[/tex]

    Therefore,
    [tex]i_{1,4}={39.7sin(337t+33.7°)}, i_2+i_3=-20+10sin(100t-45°)[/tex]
     
  14. Nov 3, 2011 #13

    gneill

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    First I'd collect all the DC terms together and put them aside until integration time. Thus

    DC = 10 - 30 + 10 = -10

    Next I'd combine the same-frequency sinusoids from i1 and i4. You can go directly to phasor rectangular form with just the phase angles. For a sinusoid:
    [tex]A \, sin(\omega t + \phi) \Rightarrow A (cos(\phi) + j \, sin(\phi))[/tex]
    so for i1 you should get
    [tex]20\left(cos\left(\frac{\pi}{6}\right) + j \, sin\left(\frac{\pi}{6}\right) \right) = 20 \left( \frac{\sqrt{3}}{2} + j\frac{1}{2}\right) = 10 \left(\sqrt{3} + j \right)[/tex]
     
  15. Nov 3, 2011 #14
    Okay, also
    [tex]i_4=10, i_{1,4}=10(\sqrt{3}+1)[/tex]
    Thus,
    [tex]i_{1,4}=\sqrt{10\sqrt{3}+11}sin(337t+2.096°)=5.32sin(337t+2.096°)[/tex]
    So we have
    [tex]i_D=-10+\sqrt{10\sqrt{3}+11}sin(337t+2.096°)+10sin(100t-45°)[/tex]

    From here, I need to pick a suitable interval to integrate over such that both sinusoid functions complete full cycles, correct?
     
  16. Nov 3, 2011 #15

    gneill

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    Careful, i1 has a "j" in it! : 10*(√3 + j)
     
  17. Nov 3, 2011 #16
    Whoa, not sure what I did there...... Let me retry that.

    [tex]i_{1.4}=\sqrt{(10+10\sqrt{3})^2+10^2}sin(337t+2.096°)=29.09sin(337t+20.1°)[/tex]

    [tex]i_D=-10+\sqrt{(10+10\sqrt{3})^2+10^2}sin(337t+2.096°)+10sin(100t-45°)[/tex]
     
  18. Nov 3, 2011 #17

    gneill

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    You should also check your angle for the i1+i4 phasor, since your components have changed.
     
  19. Nov 3, 2011 #18
    Oh yes, I had the angle as 20.1°, just forgot to change it in the formula
    [tex]i_D=-10+\sqrt{(10+10\sqrt{3})^2+10^2}sin(337t+20.1°)+10sin(100t-45°)[/tex].

    Okay, now from this point (finally) I can determine the rms voltage?
     
  20. Nov 3, 2011 #19

    gneill

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    Yes, it involves integrating the square of that expression...
    [tex] I_{rms} =\sqrt{\frac{1}{T} \int_0^T (i_D)^2 dt}[/tex]
    In this case the integral is with respect to t, which if you run from 0 to [itex]2 \pi [/itex] you will include some number of full cycles of both sinusoids. The math is going to be both tedious and messy after you square that 3-term expression!
     
  21. Nov 3, 2011 #20
    Okay, before I tackle this question, I have a concern regarding the interval of integration. How exactly do you determine when a full cycle has completed for both sinusoid? For example, if the first limit of integration is 0, how do you determine the upper limit?
     
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