different kinds of mean, arithmetic mean, root mean square (RMS) etc.

Discussion in 'Math' started by PG1995, Oct 1, 2011.

  1. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Hi

    Many people, including me, are most comfortable with arithmetic mean because it makes much sense to them. Think of three persons having 3, 7 and 5 apples. Behind every math concept there is a rational approach and logical reasoning.

    arithmetic mean = (3+7+5) / 3 = 15/3 = 5

    It says that if apples are equally distributed then each person would receive 5 apples. It would make more sense when it is seen in a particular context in 'real' life.

    This question about RMS value came to me when I read about Vrms. Yes, sinusoid is alternating between positive and negative values. In one half cycle it could be +1 and in next half cycle it would be -1. Saying that we have to take root mean square value (RMS) because it keeps alternating between +ve and -ve and gives us 'regular' average to be zero, (+1+(-1))/2 = 0, is little naive. Because one may argue that why not use absolute values of negative numbers such as (+1 + |-1|)/2 = +1.

    There are many different kinds of mean. But we will focus on arithmetic mean and root mean square.

    1: My first question is that in your opinion why there are so many different kinds of mean.
    2: I understand that the problem under discussion would dictate which mean should be used. Under what circumstances RMS is used?
    3: How would you differentiate between arithmetic mean and RMS? Their advantages and disadvantages. If we have values: 7 10 7 0 -3 -1 -2 0. It's RMS would be 5.14. I don't think it represents the 'average' of the values correctly. There is so much deviation.

    Useful links:
    http://en.wikipedia.org/wiki/Mean
    http://www.bcae1.com/voltages.htm
    http://en.wikipedia.org/wiki/Root_mean_square
    http://www.rfcafe.com/references/electrical/square-wave-voltage-conversion.htm
    http://cnx.org/content/m13418/latest/
    http://www.practicalphysics.org/go/Guidance_107.html
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    Actually you have entered into calculus, specifically the area under the curve. This is RMS.

    As for apples, you are talking oranges.
     
  3. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Hi Bill

    Do you mean to say that whenever area under a curve is involved we have to use RMS? I don't think so. Well, the chances are I could be wrong. So, please correct me in that case.

    Suppose 'x' is the amount of money which you deposit in a bank and after 10 years you get the square of the amount deposited. The return which you can get after 10 years can be represented in math language as: y = f(x) = x^2, where 'y' is the return after 10 years and 'x' is the amount deposited.

    If I know the amounts deposited by five persons are (in dollars): 50, 20, 30, 5, 7. After ten years collectively they will get $774.8. If I want to know the average return for each person, then it would be 774.8/5 = $155.

    I didn't use RMS to calculate the average here.

    Regards
    PG
     
  4. Wendy

    Moderator

    Mar 24, 2008
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    Yes, RMS is area under the curve. I'm not a math whiz, but this much I remember. It is the same reason the RMS of a square wave is 1.
     
  5. someonesdad

    Senior Member

    Jul 7, 2009
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    RMS is used for both continuous and discrete data; it's a sum for discrete data and an integral for continuous data. I use such sums a lot with numpy and python.

    When you want the area under a curve, you use a Riemann integral or one of its generalizations.
     
  6. MrChips

    Moderator

    Oct 2, 2009
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    The concept of RMS or root-mean-square comes from the power in question. So for a voltage into a load resistor, one may ask "what is the average power delivered to the load?" To determine this, we integrate the total power over time. Then we take the square root to give us the average voltage. Thus the RMS value can be determined for any voltage waveform, not just sine wave.

    Therefore to calculate the RMS voltage, you have to take the sum of the squares of each voltage reading and then take the square root of the average, hence the expression "root-mean-square".
     
    Last edited: Oct 2, 2011
  7. someonesdad

    Senior Member

    Jul 7, 2009
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    As was pointed out on another forum, the RMS values of current and voltage are useful because it was experimentally determined that their product gives the equivalent heating power of the signal. This allows comparison to DC measurements because the RMS values of a DC waveform are the same as the DC values.

    Here's a tidbit that some folks might not be aware of. The AC-coupled RMS value of a sampled waveform is equal to the population standard deviation of the sampled points. Here's a quickie python script that demonstrates it:

    Code ( (Unknown Language)):
    1.  
    2. from pylab import *
    3.  
    4. # n = points in pulse, m = number of pulses, V = Pulse amplitude
    5. n, m, V = 20, 10, 5.67
    6. x = concatenate([zeros(n), V*ones(n)]*m)  # Make the pulse
    7. x -= average(x)     # Remove DC component
    8. print "RMS =", sqrt(average(x*x))
    9. corr = sqrt((len(x) - 1)/len(x))  # Correction factor for sample sigma --> pop. sigma
    10. print "stddev(corr) =", sqrt(cov(x))*corr
    11.  
    If you want to plot the waveform, just stick in the two lines

    plot(x)
    show()

    at the end.
     
  8. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
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    Thank you, Bill, someonesdad, MrChips.
    1: I don't think we can use arithmetic mean for continuous data. But as you say RMS can also be used for discrete data and for discrete data arithmetic mean can also be used. So, the question pops up, what would be the difference between the two when they are applied on the same data? Please help me to understand it.

    2: Although I haven't seen the derivation Vrms but I would guess that along x-axis we take radian angle along the x-axis and voltage along the y-axis. For a moment think of a more simple case of velocity and time, where time is take along x-axis and velocity is taken along y-axis. The area under the velocity curve gives the distance/displacement (velocity x time). So, when area is calculated under the curve in the case of voltage (assuming radian measure is taken along x-axis), what does that area really represent? Something of sort "voltage x radian"? If that is the case, then I don't think one can Vrms, rather it would be (voltage x radian)rms. Please help me with it.

    3: The same question arises in my mind about DC current with some added confusion! In case of DC what measure/unit do you take along x-axis?

    Thanks a lot for the help and your time.

    Best regards
    PG
     
  9. someonesdad

    Senior Member

    Jul 7, 2009
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    No -- you need to learn to help yourself to understand it. This is something that would take you a few minutes' of simple work with a pencil and a piece of paper. And with the code example I gave above using python, it's even easier.
     
  10. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
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    Hi

    I had been through it before I asked you. I have read even the derivation of pressure exerted by an ideal gas and the speed of gas molecules which uses RMS velocity of gas molecules. But I believe my this particular question wasn't clear enough. Obviously, there is going to be a numerical difference between the two averages. Suppose there are five persons having cash (in dollars): 5, 15, 20, 10, 30. We want average of this money. But which average should we use? Arithmetic Mean = (5+15+20+10+30)/5 = $16.

    $RMS\hspace{0.06in}average=\sqrt{\frac{\left( 5^{2}+15^{2}+20^{2}+10^{2}+30^{2}\right) }{5}}=\allowbreak \sqrt{330}=\$18.2$

    Thank you for the code. Unfortunately, I don't have any programming experience. But these days I'm learning C++.

    Please also have a look on my other queries. Thanks.

    Regards
    PG
     
  11. MrChips

    Moderator

    Oct 2, 2009
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    It depends on what you are averaging. In some instances, it could be one person one vote. In another instance, maybe the heavier guy who can carry a bigger punch should get extra votes. That is were the square comes in handy. With voltages, we are talking about power. Hence taking the square of the voltage is mathematically correct.
     
    Last edited: Oct 3, 2011
  12. Dr_Ram

    New Member

    Sep 14, 2011
    7
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    "We want average of this money. But which average should we use?"
    Actually the answer is hidden in a statement you made in your original post "...problem under discussion would dictate which mean should be used." In other words, which mean must be used is decided by what you want to DO with that mean. In the case of the people with cash, you want to be able to say "Here is a number such that if all 5 people had the same amount of cash, we'd end up with the same total" The desire to have such a figure implies the use of summation and hence the arithmetic mean. If you were talking of a sequence of voltages being fed very quickly to an analog meter, the question "What number will the meter read" ends up (based on the physics of the meter behavior) leading to the arithmetic mean. If your question was "How much heat is produced in the meter coil", then the physics of the situation leads you to RMS. It's all about what you want to use that mean for.
     
    PG1995 likes this.
  13. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    This discussion has an interesting "spin-off" with respect to using a meter to measure RMS values. Certain AC meters are inherently true RMS responding meters. However, if you obtain an old style D'Arsonval moving coil based multi-meter, you'll notice that it may have (at least) an AC voltage range which supposedly reads RMS voltage. How one makes an average (DC mean) responding moving coil meter indicate RMS is an interesting exercise. Generally this meter type reads AC sine wave RMS values with reasonable accuracy. Some simple internal electronics and meter range scaling are required to achieve this. The same can't be said for the same meter's RMS accuracy with respect to non-sinusoidal input voltages or sinusoidal inputs with DC offset.
     
  14. MrChips

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    There is such a thing as a hot-wire ammeter that measures true RMS value of any waveform.
     
  15. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Thank you, MrChips, DrRam, t_n_k.



    For someone interested to see the use of mean square speed in derivation of pressure exerted by an ideal gas can have look on the scanned pages below, the link #4 also make a comparison between different kinds of mean speeds:

    1: http://img717.imageshack.us/img717/6028/page1ml.jpg
    2: http://img834.imageshack.us/img834/4431/page2zd.jpg
    3: http://img31.imageshack.us/img31/4716/page3kv.jpg
    4: http://img713.imageshack.us/img713/1885/page4ny.jpg

    In link #2 you see mean square speed pops up as a result of the process. We don't need to choose if we should use arithmetic mean speed or mean square speed. The method makes its own choice.

    I'm still seeking answers to the questions, particularly #2 and #3, I raised in this post (I have also quoted it above). I'm sorry if you think I should have understand the phenomenon so far. I don't have much background in science but I'm struggling to learn it.

    Could any of you have a link to a simple and straightforward derivation of Vrms of AC current. Perhaps, trying to understand it in a proper context might help me.
     
  16. MrChips

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    Oct 2, 2009
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    PG1995, are you paying attention?

    The RMS value of any voltage or current is the value of a DC voltage or current that will provide the same heating effect of the AC voltage or current.

    To calculate the RMS value, you take a series of equally spaced samples, compute the sum of the squares, divide by the number of samples and take the square root.

    For a continuous random waveform, you take the integral of the square of the voltage or current, divide by the time, and take the square root.
     
  17. Dr_Ram

    New Member

    Sep 14, 2011
    7
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    Let's try a simple Vrms derivation. Let's say you have a signal you plot versus time (volts V on Y-axis, time on X-axis). Draw 11 equally spaced vertical lines representing small-enough time slots Δt. Then the power developed by the voltage in a resistor R in each of these 10 timeslots is
    V1*V1/R, V2*V2/R,..., V10*V10/R (because power is V*I = V*V/R) ---(1)

    Therefore the total heat developed in those 10 timeslots is
    Δt*V1*V1/R + ...+ Δt*V10*V10/R ---(2)
    (because power is the same as heat per unit time).

    Therefore the "average" heat developed per timeslot Δt is the total heat divided by the total number of timeslots = (2) divided by 10 which is
    Δt/R * (V1*V1 + ... + V10*V10)/10 --(3)

    which is the same heat per timeslot that would be developed by a constant voltage Y of value sqrt((V1*V1 + ... + V10*V10)/10) i.e.,
    Δt/R * Y*Y
    i.e,., the power developed by Y would be Y*Y/R.

    Y is by definition the RMS value of the voltage V over the 10 timeslots.
     
    PG1995 likes this.
  18. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    To MrChips, Dr. Ram: Now I understand where I'm having most of the confusion. I will get back to it soon. Dr. Ram, thanks a lot for your recent post.

    Best wishes
    PG
     
    Last edited: Oct 4, 2011
  19. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Perhaps, my question wasn't clear enough, or I'm just missing on something too much obvious. So, I will try to rephrase it.

    When we plot velocity along y-axis and time along x-axis. The result of the definite integral with respect to time is displacement over the specified time interval. Because (velocity x time) is displacement. If we take the average of the displacement resulted from the integration, we get average rate of displacement per unit time over the specified time interval. I hope I have it correct.

    But I think I'm having trouble understanding the RMS because I'm thinking too much in 'graphical' terms.

    Now we plot voltage and time where voltage is taken along y-axis. The integral in this case will give us (voltage x time). Please have a look on the linked image (the same image is also attached to this post):
    http://img708.imageshack.us/img708/7500/imgnpj.jpg

    If you understand what I say in the attachment you could figure out where I'm going wrong.

    Thank you very much for your help, and I have some silly point then please excuse my ignorance.

    Regards
    PG
     
  20. MrChips

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    Oct 2, 2009
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    I think you are stuck in a rut.

    The area under a velocity vs time graph gives distance.
    The area under a current vs time graph gives charge.
    The area under current squared x load vs time gives energy.
    The area under voltage squared divided by load vs time graph gives energy.
    The area under voltage vs time graph gives volts x time. That's it! Period. Done. Fini.
     
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