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Sine wave with dc offset 
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#1
Jan2212, 06:10 AM

P: 70

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
Jan2212, 08:24 AM

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P: 11,614

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. 


#3
Jan2212, 08:31 AM

P: 70

I thought RMS of a sine wave was Vpeak/sqrt(2)?



#4
Jan2212, 08:44 AM

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P: 11,614

Sine wave with dc offset
You're result is fine. 


#5
Jan2212, 08:55 AM

P: 70

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 65.5 = 11.5V. How is it that the RMS could be 7.15V? 


#6
Jan2212, 09:10 AM

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P: 11,614

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. 


#7
Jan2212, 10:10 AM

P: 70

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? 


#8
Jan2212, 11:18 AM

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P: 11,614

[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. [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]). 


#9
Jan2212, 12:07 PM

P: 70

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.



#10
Jan2212, 12:14 PM

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P: 11,614




#11
Jan2212, 12:36 PM

P: 70

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



#12
Jan2212, 12:46 PM

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P: 11,614

If you write the expression for the full, untruncatedatzero 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. 


#13
Jan2212, 01:27 PM

P: 70

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)? 


#14
Jan2212, 01:44 PM

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P: 11,614

[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 halfsinewave 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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