Register to reply

Sine wave with dc offset

by D44
Tags: offset, sine, wave
Share this thread:
D44
#1
Jan22-12, 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?
Phys.Org News Partner Science news on Phys.org
What lit up the universe?
Sheepdogs use just two simple rules to round up large herds of sheep
Animals first flex their muscles
gneill
#2
Jan22-12, 08:24 AM
Mentor
P: 11,682
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.
D44
#3
Jan22-12, 08:31 AM
P: 70
I thought RMS of a sine wave was Vpeak/sqrt(2)?

gneill
#4
Jan22-12, 08:44 AM
Mentor
P: 11,682
Sine wave with dc offset

Quote Quote by D44 View Post
I thought RMS of a sine wave was Vpeak/sqrt(2)?
My apologies. You are correct. I blame it on lack of coffee

You're result is fine.
D44
#5
Jan22-12, 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 -6-5.5 = -11.5V. How is it that the RMS could be 7.15V?
gneill
#6
Jan22-12, 09:10 AM
Mentor
P: 11,682
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.
D44
#7
Jan22-12, 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?
gneill
#8
Jan22-12, 11:18 AM
Mentor
P: 11,682
Quote Quote by D44 View Post
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?
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.



[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]).
Attached Thumbnails
Fig1.gif  
D44
#9
Jan22-12, 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.
Attached Thumbnails
2.JPG  
gneill
#10
Jan22-12, 12:14 PM
Mentor
P: 11,682
Quote Quote by D44 View Post
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.
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).
D44
#11
Jan22-12, 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
Attached Thumbnails
3.JPG  
gneill
#12
Jan22-12, 12:46 PM
Mentor
P: 11,682
Quote Quote by D44 View Post
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
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.
D44
#13
Jan22-12, 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)?
gneill
#14
Jan22-12, 01:44 PM
Mentor
P: 11,682
Quote Quote by D44 View Post
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)?
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.


Register to reply

Related Discussions
High speed sine wave to square wave converter Electrical Engineering 14
Removing a DC Offset from a Square Wave Signal Electrical Engineering 7
Half wave bridge rectifier - Sine wave Amplitude problem Engineering, Comp Sci, & Technology Homework 3
Sine wave resonance using square wave input Electrical Engineering 4
Synthesised Sine wave VS Modified Square Wave inverters Engineering, Comp Sci, & Technology Homework 1