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Sine wave with dc offset

  1. Jan 22, 2012 #1

    D44

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    I have to calculate the RMS and average voltage of a sine wave with a DC offset.

    I have a peak to peak voltage of 11V and an offset which is -6V.

    I have the average voltage as -6V and RMS voltage as 7.15V. I calculated the RMS as follows:

    AC Vrms = 5.5/sqrt(2) = 3.89V

    Vrms = sqrt(3.89^2 + 6^2) = 7.15V

    Does this look correct?
     
  2. jcsd
  3. Jan 22, 2012 #2

    gneill

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    Staff: Mentor

    Check your AC Vrms calculation. [itex] \frac{11}{2}\sqrt{2} \neq \frac{5.5}{\sqrt{2}} [/itex]

    EDIT: Oops. D44's result is correct. See below.
     
    Last edited: Jan 22, 2012
  4. Jan 22, 2012 #3

    D44

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    I thought RMS of a sine wave was Vpeak/sqrt(2)?
     
  5. Jan 22, 2012 #4

    gneill

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    My apologies. You are correct. I blame it on lack of coffee :smile:

    You're result is fine.
     
  6. Jan 22, 2012 #5

    D44

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    I've accidently just clicked on report instead of new reply, so whoever runs this site will wonder why I'm talking about voltages. Sorry about that :/

    No problem. Coffee time for me too I think :) So the results look ok? How is it that I get a positive RMS? The way I see the wave is that the average is -6V, it reaches a peak of -6+5.5 = -0.5V and has a minimum of -6-5.5 = -11.5V. How is it that the RMS could be 7.15V?
     
  7. Jan 22, 2012 #6

    gneill

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    Staff: Mentor

    RMS is always positive. It represents the DC equivalent for power delivered.

    RMS of a function is calculated as the square root of the mean of the square of the function (hence R M S). Your voltage function can be written as:

    [itex] f(\theta) = \frac{V_{pp}}{2}sin(\theta) - 6 [/itex]

    and the RMS value:

    [itex] V_{rms} = \sqrt{\frac{1}{2 \pi} \int_0^{2 \pi} f(\theta)^2 d\theta} [/itex]

    If you perform the integration you should obtain the same, positive value as you did before. Squaring the function ensures that the result must be a positive value.
     
  8. Jan 22, 2012 #7

    D44

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    Thanks, that's great!

    Can I also ask...

    For the half rectified wave I spoke about the other day, again if I was wanting to integrate, given peak current as 11A, Tpulse as 0.02s and Tperiod as 0.035s, I'm thinking that my function would be 5.5sin(50*pi*t). So I integrate this with respect to t? Between the limits of 0 and 0.02? Then divide by 0.035?
     
  9. Jan 22, 2012 #8

    gneill

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    That function also had offsets that you'd need to calculate first and apply. If you specify all the times in milliseconds then you'd have something like:

    [itex] f(t) = A sin(\frac{2 \pi}{35} t - \phi) + Vo [/itex]

    Where [itex] \phi [/itex] is an angular offset for the sine function, and Vo the voltage offset that shifts the base of the pulse up to the 0V level.

    [​IMG]

    [itex] \phi [/itex] can be obtained directly from the Δt value in the above figure. The offset voltage, Vo, requires a bit more work: Note that A + Vo = 11, and Vo = Asin([itex] \phi [/itex]).
     

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  10. Jan 22, 2012 #9

    D44

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    This is the image of the waveform that I have. So I need to find the average and RMS in order to find form factor.
     

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  11. Jan 22, 2012 #10

    gneill

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    Yes. It corresponds to the image that I posted above if you 'chop' the wave off along the horizontal zero line (time axis as shown).
     
  12. Jan 22, 2012 #11

    D44

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    But if the 2 portions of a single period aren't identical, how does this effect the function? That's what's stopping me getting round to the integration part
     

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  13. Jan 22, 2012 #12

    gneill

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    The function does not have to be symmetrical within its period, it only has to be identical from period to period.

    If you write the expression for the full, untruncated-at-zero function, then integrate over only the sections that are greater than zero, then you'll be okay. For the graph that I posted above, that function is:

    [itex] f(t) = A sin(\frac{2 \pi}{35} t - \phi) + Vo [/itex]

    And the integration bounds would go from 0 to 22 (milliseconds) to include just the pulse.
     
  14. Jan 22, 2012 #13

    D44

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    I'm struggling with this. The 2nd pic I put up is the unaltered version of the wave.

    In this case, phi would be 0 and Vo would be 0? So I'm just integrating Asin(((2*pi)/35)t)?
     
  15. Jan 22, 2012 #14

    gneill

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    If you approximate the pulse with a half cycle of a true sinewave then you would have

    [itex] f(t) = 11 sin(\frac{2 \pi}{2 \times 22}t) ~~~~~ 0 \leq t \leq 22 [/itex]

    But this will be an approximation only, one that gets worse as the offset voltage gets larger. This is because while the half-sinewave is everywhere concave from below, a full sinewave alternates from concave to convex in shape at its zero crossings. So the function of a pulse created by rectifying an offset sinewave is not the same thing as stretching a half a sinewave to the same amplitude.
     
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