I have some questions about Average Power - Page 1
Forum section: electronics / beginners

author: cvriv
author_rank: Frequent Contributor
timestamp: November 27, 2015, 02:56:34 am
content: Hello people:) So I have some questions about average power. I asked my professor and either I'm not asking him correctly or I am and he's just not understanding me. He keeps saying the same thing over and over and it's not helping me. I think I understand this, I just need to clear this up in my head.So, the book I have says that if I want to calculate the average power for an "AC waveform" that I need to use rms values to do so, and that the average power of an "AC waveform" is the DC equivalent. I know how to calculate rms values and I also know how to calculate half-cycle average values as well, which my professor says are DC equivalent values. Yea. You can probably see where I'm getting confused?So, is the average power calculated with rms values are for full-cycle AC waveforms only? If the rms calculated average power is the DC equivalent, then what if I calculate power using half-cycle average values? Half-cycle average values are DC equivalents, so what would that average power be, because they are slightly different. This is where I'm getting confused.Also, is a half-cycle AC waveform still AC? My professor said yes because the values are still changing. The change isn't bi-directional, but it's still changing. At what point is it not AC anymore? I ask this, because he ran us through the process of converting AC to DC. I know that the process is not prefect and never really perfectly smooths out the waveform. There is always a slight alternation, so slight that it's insignificant, but still on a very very small scale, still alternating. So, at what point do we call it DC?

author: cvriv
author_rank: Frequent Contributor
timestamp: November 27, 2015, 03:09:10 am
content: The book said that rms values are used most of the time. Then what are half-cycles values used for?

author: IanB
author_rank: Super Contributor
timestamp: November 27, 2015, 03:27:49 am
content: Uniform AC waves are symmetric. The negative half cycles are mirror images and look exactly the same as the positive half cycles. So if you calculate the average power over a single half cycle you have calculated the average power for the wave as a whole, no?RMS voltage just means "the DC voltage that would produce the same power in a resistor as this AC wave would produce". You can find out the RMS voltage by integrating the power over a half cycle and finding out what DC voltage that corresponds to.Every uniform cyclic wave can be decomposed into a sum of a DC portion and an AC portion. Pure DC (as from a battery) can be considered 100% DC and 0% AC. Pure AC (as from an alternator) can be considered 100% AC and 0% DC. A DC voltage with some ripple on it is a mixture of DC plus AC. It is neither one or the other. However, it is usually considered DC if for practical purposes the consumer doesn't care about the AC ripple.

author: pmbrunelle
author_rank: Regular Contributor
timestamp: November 27, 2015, 04:31:50 am
content: Go back to the basic definition of power. Power = Voltage x Current.If the voltage and current are changing with time, the above equation is still true (as always), except that the power is changing with time. Could be positive or negative at any moment as energy goes back and forth between the load and supply.So what you need to do is calculate the power at every time index in one cycle. Once you have the power at every time index in the cycle, you can take the average of all these values. Voila, average power! This operation is integration, and how you handle this depends on your math / trigonometry skills, the type of signals (analytical textbook function, or measured data)...***The above method will always work***Now in practice, when you have the RMS values of your voltage and current waveforms, and these waveforms are sinusoidal, and you know your cos(phi), then you don't need to go through the whole integration nonsense (but you can)... but when in doubt, go back to basics!

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 27, 2015, 05:31:23 am
content: If he's telling you the two methods are DC equivalent, then A ~= B ~= C implies A ~= C (using "~=" to mean "equivalent to").For sinusoidal waveforms in linear circuits, they must be equivalent, because all the waveforms are periodic, with the same periodicity; and if it's the same from period to period, it doesn't matter if you take just one period* or an infinity of them (i.e., average over all time).*Specifically, that would be one period of the power waveform. The real component of which is at twice the V/I frequency. Which is where your "half cycle" comes from, too.Tim

author: jitter
author_rank: Frequent Contributor
timestamp: November 27, 2015, 06:57:51 am
content: Quote from: cvriv on November 27, 2015, 02:56:34 amIt's mainly amathematicalthing because if you look at a nice symmetrical sinewave without DC offset, you'll notice that the positive and negative parts are equal, and that only the sign has changed. The equation to calculate RMS uses Vpeak, so the positive half of the waveform. Obviously, the negative half of the sinewave contributes just as much power as the positive half. The equation doesn't need to "know" the other half of the sinewave as it's equal. The RMS calculation first squares the values, so all signs become positive.Of course these formulas only apply to symmetrical waveforms around 0 (no offset).QuoteA signal can (and usually does) consist of ACandDC parts. Both contribute to the power.

author: cvriv
author_rank: Frequent Contributor
timestamp: November 27, 2015, 09:28:38 am
content: I understand a lot of what you guys are saying about symmetry etc. I know that that RMS is a point in magnitude that does not change in time, as long as the AC waveform remains constant. What I want to know is, well the book says:- Average AC power and DC power are equivalents.- AC power calculations arealwaysperformed with RMS values.- Because power is a primary concern with any electrical or electronic system, RMS values are used more often than any other magnitude-related measurements.- When a magnitude related measurement is unidentified, it's RMS.This tells me that RMS values are used most of the time. So im thinking that if average power, calculated with RMS values, is the DC power equivalent, than V-RMS and I-RMS are DC equivalents too, but I'm thinking this not right, because my professor said that the half-cycle average values are DC equivalents. Im so damn confused.Question: What are half-cycle average values used for?

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: November 27, 2015, 09:43:39 am
content: Quote from: cvriv on November 27, 2015, 02:56:34 amThis kinda depends who you are asking. In signal analysis, the DC component is the signal with frequency 0 in the spectrum. However, this is a quite tight definitions, so often people will say that you have 2 signals in power-lines: A DC-offset and a perfect sine. The DC-offset is the value around which the sine oscillates. Yet other people might consider DC anything that is only positive or negative and thus doesn't change sign (I remember in highschool that was the definition thought to us)Quote from: cvriv on November 27, 2015, 09:28:38 amI think you might be getting confused as to what RMS is exactly. The RMS value of a function is a mathematical operation you can calculate for that function: (fromhttps://en.wikipedia.org/wiki/Root_mean_square)Thus, we can calculate this for both current and voltage (for a perfect sine, the RMS value is the amplitude/sqrt(2) ). The reason we say that the RMS values are theDC equivalentis because a DC voltage equal to the RMS value of any non-constant signal over a resistor, will cause a current Irms=Vrms/R to flow which is also equal to the RMS of the current from the AC signal. Say we were to now first calculate the RMS value, and then calculate the amount of energy over a certain amount of time, or we were to first calculate the instantanious power and then get the energy (integrate over t), we will get the same amount of energy, IE: if we were to apply a DC voltage of value Vrms over the load, the energy consumed by the load will be equal to the energy it would consume with the AC signal applied.I realise this will likely be quite confusing, might do a quick recording of this if you don't really understand and are interested.

author: cvriv
author_rank: Frequent Contributor
timestamp: November 27, 2015, 10:33:15 am
content: Still a bit confusing. I asked my professor about RMS values being DC equivalents and he said that only average power. I asked him specifically about V-RMS and I-RMS and that's what he said. He said the V-AVE and I-AVE are DC equivalents. This is whats confusing me. How can average power be a DC equivalent but not V-RMS and I-RMS?I did some reading online and im under the impression now that you use RMS for certain situations and half-cycle average for others. For instance, half-cycle average values are used to make calculations for a half wave rectified waveform.Is RMS only used for pure AC only? Where each alternation is symmetrical and equal in magnitude? If half cycle average used for AC waveforms that are not pure? Like a rectified waveform or a waveform with a DC offset? I think I might be jumping to far ahead of myself. If the book didn't explain all this to me it's either because the book sucks or the book is holding off explaining it to me for whatever reason. This is so damn frustrating. The book doesn't this is great detail at all.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 27, 2015, 10:54:36 am
content: What is meant by "equivalent"?Apply that voltage or current to an ideal resistor, and the resistor dissipates some power, heating up.When averaged over all time, this works for any voltage or current or power waveform.When only DC is applied to a resistor, it also dissipates some power, heating up. When this dissipation is equal to that from an AC waveform, the AC voltage is said to have an RMS equivalent of this voltage.The mathematical function which this process represents is RMS (which can be proven).Vrms is only DC equivalent if the load is a resistor. A resistor dissipates power due to both DC and AC components of a signal, so the DC component of a signal is not equivalent to the Vrms or Irms of that signal. This becomes more important with RLC networks, which can dissipate different amounts of power at different frequencies, or for different waveforms.Tim

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: November 27, 2015, 10:57:08 am
content: I'm going to put a quick clip on youtube in which I attempt to somewhat explain why Pavg is the DC equivalent but not Vrms or Irms (at least from what equivalent seems to mean in your book, but it's confusing).And you can use RMS for any waverform, not just pure AC (as I will demonstrate in that video). Ofcourse, you do need to keep in mind that all of this only applies directly to resistive loads, once you start dealing with capacitive or inductive loads frequency becomes important too. I have a feeling your book does not clearly describe what is meant with "equivalent", as Tim already pointed out.

author: GNU_Ninja
author_rank: Regular Contributor
timestamp: November 27, 2015, 10:58:39 am
content: You may find the link to this PDF useful:http://www.raeng.org.uk/publications/other/8-rmsEssentially, the RMS value has the sameeffectas the DC value. For example, a resistor will dissipate the same amount of power when connected to a 12 V RMS or a 12 VDC source.Hope this helps

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: November 27, 2015, 11:37:57 am
content: Its not great but I recorded this is one take...https://youtu.be/Msm14xJ7IRM

author: cvriv
author_rank: Frequent Contributor
timestamp: November 27, 2015, 01:15:14 pm
content: Quote from: TheUnnamedNewbie on November 27, 2015, 11:37:57 amAwesome! I think you cleared this up for me. I have to watch it again, maybe even a few times, though. Just got off of work and Im tired as hell. Thanks again man!

author: cvriv
author_rank: Frequent Contributor
timestamp: November 27, 2015, 01:15:58 pm
content: Quote from: GNU_Ninja on November 27, 2015, 10:58:39 amIm going to check this PDF out later after I have a nap. LOL. Thanks:)

author: vk6zgo
author_rank: Super Contributor
timestamp: November 27, 2015, 01:24:03 pm
content: Another thing which may help is to grab an old Maths text & look up "Area under the curve".

author: jitter
author_rank: Frequent Contributor
timestamp: November 28, 2015, 08:36:47 am
content: Quote from: cvriv on November 27, 2015, 09:28:38 amDraw yourself a graph with a perfect DC voltage of whatever positive voltage. Colour fill the area between 0 and the curve (actually a straight line in this case). That coloured area is what can contribute to power.If you draw the same graph with the same voltage, but now negative, you'll notice that the area is the same. In other words, the amount of contribution to power is the same.Now draw yourself a symmetrical square wave without offset and half the wave positive and half negative. Use the same peak value as above for the positive and negative waves.If you fill the areas between 0 and the curve, you'll notice that if you add them together, the total area be the same as with the DC wave above, despite half being negative and half positive.Imagine taking the square wave and flipping over all negative parts to the positive side. That's what the rms calculation does and why the terms are squared.With a symmetrical square wave, you're ready, with a different shaped wave, you need to take that shape into account.Draw in your previous square wave graph the sinewave with equal peak values and you'll see that the corners of the square are "chopped off", diminishing the area.But still, area is just area, whatever the shape. Hence the DC-equivalence if you total those areas.The half cycle value is because in the calculations you only use Vpeak, i.e. the value between 0 and the positive peak. Of course, the real signal varies between the positive and negative peak, or the Vpeak-peakvalue. But that negative peak is flipped over to the positive side when it's squared in the calculation. It's a mathematical thing.As said before, this works only for symmetrical shapes around 0.

author: gearhead
author_rank: Contributor
timestamp: November 28, 2015, 09:30:28 am
content: Maybe this will help you understand a little better.it's a couple pages from my Fluke 8050A DMM manual.

author: cvriv
author_rank: Frequent Contributor
timestamp: November 29, 2015, 12:24:41 pm
content: Quote from: TheUnnamedNewbie on November 27, 2015, 11:37:57 amSorry for the late reply. I was very busy this weekend. After watching your video a few times I understand why RMS is used to calculate power. So P-rms is the DC equivalent. Now... my professor said that V-avg and I-avg are the DC equivalents. This right? He did a class where we converted AC to DC and I think we calculated V-avg after rectifying, and then again after filtering. This gave us the final voltage. Is this right?

author: cvriv
author_rank: Frequent Contributor
timestamp: November 29, 2015, 11:20:44 pm
content: So all of this is covered in Chapter 18 of my book. I was working in Chapter 17. This is what happens when your professor jumps around. We did chapter 9, then chapter 17, chapter 18, and then I think we go to chapter 12. I don't why they do this. How can I fully understand the next chapter without having done the previous chapter? My Precalc B book does the same damn thing. Explain a little something and then expects you to do the problems in the back of the section. Not being able to do the problems, I read ahead only to find out that what I need to do the problem for the first section is explained fully over the next 3 sections. Just crap. Wasting my time.