Integration

Discussion in 'Math' started by Thevenin's Planet, Aug 29, 2011.

  1. Thevenin's Planet

    Thread Starter Active Member

    Nov 14, 2008
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    Good day

    The problem of understanding the intervals over which the combined AC+AC function is greater than or equal to zero volts. Can this be explained? I am told that it is an infinite number of intergration interval opions exits.How are these interval found?
     
  2. russ_hensel

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    Jan 11, 2009
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    Assuming you have the right lower and upper limit now you can adjust either by 2pi ( as long as the difference does not reach 0 ) and do the rms average. That will give you an infinite number of ranges to choose from.
     
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  3. Thevenin's Planet

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    Nov 14, 2008
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    According to what is given,to establish the upper and lower limits the calculation begins with ...Sine=y/hypo.or D.C.offset/Vmax(a.c.),(D.C./12* square root 2)which gives Q1 or +0.558. Is this the beginning of T! or the beginning of the wave point that calculation begins?
    In regard to creating the lower limits the opposite of +0.558 is used,which is -0.558.
    Next, the calculation continue to find the upper limits using Pi +Q1,(Pi + .558 = +3.6995 or 3.7). Now adjusting by 2pi, what exactly are you expressing? For some reason that is not known by me the Pi is added to the sine of 9/17=3.7 radians that produces the upper limits.
    Where are the increments beginning, from Pi ?
     
  4. t_n_k

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    Mar 6, 2009
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    Consider the continuous periodic function

    f(t)=A+Bsin(\frac{2\pi}{T} t)

    which is a DC offset sinusoid with period T. The variable time 't' lies anywhere on the interval -∞<t<∞.

    The mean value is (in purely general terms) given by

    Mean=\frac{1}{T}\int^{t_x+T}_{t_x}f(t)dt

    where -∞< tx <∞ is some arbitrary real time value.

    Performing the integration will always yield Mean=A, irrespective of the value of tx.

    One may further state that this will be true of any continuous periodic function f(t) having period T.
     
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  5. Thevenin's Planet

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    Nov 14, 2008
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    If attachment in the above is used as an example to express F(t)= A+B sin(2pi/T t). Is "A" the DC offset and "B" becomes a.c. peak voltage? If so, the sine of the angular frequency (2pi/T),that is using 60 Hz would be 376.99 rad.Should A + B be added then multiplied by the sin of Angular frequency in radians multiplied by the variable "t" ? How should the variable "t" be chosen ? Taking in consideration that has been said that "t" lies any where on the intervals between opposite extremes.
     
  6. t_n_k

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    The difference in the attachment (which BTW is actually my work from another post / thread) is that the inclusion of a rectifier diode in the original meter circuit (not shown) brings the region of non-conduction from θ2 to θ3 into the analysis. So finding the mean value in this case is different to the analysis in my last post in this thread.

    A lengthy analysis gives the general result for the attached case in post #1 as follows ...

    Mean(f(t))=\frac{A}{2}+\frac{A}{\pi}arcsin(\frac{A}{B})+\frac{B}{\pi}cos([arcsin(\frac{A}{B})])

    With the function f(t) defined as

    f(t)=A+Bsin(\omega t)

    You'll note that the mean value is determined independently of the angular frequency [ω] and time [t] terms.

    With A=9V and B=17V the mean value is 10.69V which is in agreement with the result shown in the attachment.

    I haven't shown the derivation of the relationship but (again) it can be found by suitable integration and manipulation between the arbitrary limits tx & tx+T. T is the period of the sinusoid term in f(t).
     
    Last edited: Sep 1, 2011
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  7. Thevenin's Planet

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    Nov 14, 2008
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    For me to get to understand the equations that is spoken about in Post #6 It's necessary for me to grasp the equation A +B sin (2 pi/Tt) as a p requisite.My knowledge is not as It should be regarding these functions.
    If I use 9 =A and B=17 and 60 Hz.Would "t" be some fraction of an angular frequency(2 pi/T)? In the case 377. Should the period of one cycle is 0.01666 seconds,is the function stating that I use some fraction of 16.6 Ms.multiply the 377 angular frequency,then find the sine of it? And it seems to me add A +B then multiply this sum? Translating this functions can be difficult,It's not common for me to grasp moving things in mathematical notation.
     
    Last edited: Sep 3, 2011
  8. t_n_k

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    The angular frequency ω [unit of radians per second] is defined by the equation

    \omega=2\pi f

    where f is the frequency in Hertz

    So if A=9, B=17, f=60Hz and hence ω=377 rad/sec

    then the aforementioned equation for f(t) becomes

    f(t)=9+17sin(2 \pi 60t)=9+17sin(377t)

    It's now possible to state the value of f(t) for any given time 't', where t is in seconds.

    for instance at t=30 milliseconds=0.03sec

    f(t)=9+17sin(377*0.03)
    f(t)=9+17sin(11.31 [radians])=9+17*(-0.951)=9-16.17=-7.17

    I noted the unit as [radians] for the sine angle term - not degrees.
     
  9. Thevenin's Planet

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    Nov 14, 2008
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    It seems that my understanding regarding the time (t) is not limited to the sine wave Period of 1/60Hz=0.01666 seconds but ,"for any given time", Which gives me the impression of infinity.
    Regarding the result that is given,is -7.17 the instantaneous voltage or have a more explicit meaning?
     
  10. t_n_k

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    Indeed you can find a value for any given t for -∞<t<∞

    This was simply intended to show you how to find the value of the given function f(t) at a certain time t. There is nothing special about the time value I chose as an example case.

    If I may ask - What are you hoping to learn from this discussion? You don't seem to have a clear objective in mind. Perhaps you might consider undertaking a new or refresher course in calculus or perhaps a more basic course in mathematics. You might need to get a better grounding in some of the basic concepts (such as functions) before you try to understand the more subtle issues of the problem under consideration.
     
  11. Thevenin's Planet

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    Nov 14, 2008
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    To understand the function for the mean derivatives.What the outcomes defines.To understand the steps and notation clearer.That is why I ask about -7.17.Is it a instantaneous point in the cycle regarding both voltage? Is it below the offset voltage?A continuous function, you use with a .030 seconds or very close to two cycles with a period of .01666 seconds each. Is that a fact? I hope you are not frustrated,but last night I ponder the equation or the derivatives consider in the aforementioned,I noticed that it is no more than a snap picture of voltages rather then the still pictures. Rambling through some of my past library copies about the subject, that is a.c. superimpose on D.C. it was noticed that the -7.17 voltage is below the zero value,that is below the offset voltage.This is a quote from a copy from a book from the library,"If Vdc is greater than the peak of the sine wave,the combine voltage is a sine wave that never reverses polarity. That is, its a sine wave riding on a D.C. level.The next statement says,"If Vdc is less than the peak value of the sine wave the sine wave will be negative during the a portion of its lower half-cycle.So my slow analyzing
    deduction noticed that in regard to the thumbnail pix in your calculation in the diagram is depicting that the Vdc is less then the ac,therefore ,the sine wave do reverse polarity.Which is in the above case is -7.17 volts.Also,the equation you have given is instantaneous,rather then still picture if you don't mind the metaphors. (It do seems that you have some kind of indifference to diction). That was lack of my attention on my part of thinking.In this particular situation ,"what would I expect on the meter when measuring this type of voltage,"that is a.c. is larger then D.C.? Or can this be measured only on an oscilloscope ? I hope you don't fall in a bad mood regarding this discussion. I am just attempting to get a better understanding about the the subject.,which is the mean function.Hang in there with me, a few stumbles but I would get you to notice what I am trying to get you to show me.
    1/pi Integral (9+17 sin theta)d theta,with +3.7 as the upper limits and -0.558 as the low limits.Is this the same as Mean=1/T Integral f(t)dt ? I am wondering if that is just a instantaneous moment or do I plot out many points for
    a graph ? I hope this gives an explicit insight of my intention. Keep a balance Mind.

     
    Last edited: Sep 6, 2011
  12. t_n_k

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    I haven't given up on the discussion. I've been pondering on other things. I'll try to get a clear understanding of the questions in your last post and come back later with some comments.
     
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  13. t_n_k

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    Returning to our general function

    f(t)=A+Bsin(\omega t)

    Maybe think of this as a time varying voltage function described by a similar equation ...

    v(t)=V_{dc}+V_{max}sin(\omega t)

    So we have a time varying sinusoidal voltage of maximum value Vmax which is offset from zero by a DC value of Vdc.

    The voltage v(t) will never attain reverse polarity [i.e. go below zero] if Vdc is positive and greater than or equal to Vmax. If Vdc is less than Vmax then v(t) will go negative for some part or all of the AC cycle. If Vdc is negative and less than or equal to -Vmax then v(t) will never attain a positive value greater than zero volts.
     
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  14. Thevenin's Planet

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    Nov 14, 2008
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    [​IMG]
    It seem that I did have an inkling of impatience.But the above integral it expresses a time varying situation at a precise moment between two limits but that's all, general principal.
    when I compare it with the function in the thumpnail in post #1 the relation I can comprehen but the sine function of 9/
    17 that also became understandable even though I would not have expected that would have been used.Why did 9+17 sin theta become 17 cosine theta ? Also, I am assuming that delta theta is the same as dt( in the average function) but instead of using seconds it is refering to radian ? Cosine, commomly, I would think of a leading phase by 90 degrees.Another puzzling preception is tx + period,is "t" a function of x ? Also, I notice 1/T=1/2 pi. A fraction of the period. How does that play a part in the function ?
     
    Last edited: Sep 7, 2011
  15. t_n_k

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    Attached is a copy of my general analysis based on time rather than on angular displacements.

    It's rather detailed and requires a reasonable understanding of integration techniques and matters relating to the average value of a time varying function.
     
  16. Thevenin's Planet

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    Nov 14, 2008
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    You might be right about "clear objective of mind", which I think you are off track (in a figure of speech).To remaind you that you stated two Mean equations. Right ? Post #4 and # 6.If It don't take a lot of bits can I invite you to concentrate on the Integral of Post #4,Mean equal the fraction of 1/T followed by a point located on the sine wave graph. I assume that the function after the integral sign(S) f(t)dt is stating literally or in the concept of sine wave behavior,where is a point of the wave(Voltage amplitude is located) in graph mode.Using the upper and lower limits.Which I think is restrictive,but there is situation for such limits.Besides that,The f(x)=dy/dx is the derivative ? Please answer the questions.It seem as though there is a tendecy that you evade question and substitute with equations and not explaining what the number actually represents in the world of sine waves or electronics.I hope ACC have a lot of bit and bytes in regard to the process of your responding.Also are we working withIndefinite Integral or
    definite Inegral?
     
  17. t_n_k

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    Sure - happy to focus on the contents of post#4. Clarity of mind is always good.

    I'll reproduce it [post#4] here for convenience and try to address your questions & comments

    =========================================

    Consider the continuous periodic function

    f(t)=A+Bsin(\frac{2\pi}{T} t)

    which is a DC offset sinusoid with period T. The variable time 't' lies anywhere on the interval -∞<t<∞.

    The mean value is (in purely general terms) given by

    Mean=\frac{1}{T}\int^{t_x+T}_{t_x}f(t)dt

    where -∞< tx <∞ is some arbitrary real time value.

    Performing the integration will always yield Mean=A, irrespective of the value of tx.

    One may further state that this will be true of any continuous periodic function f(t) having period T.

    ===========================================

    1. The function f(t) is meant to be the same in both the cases where it is written above - notwithstanding my comments about the generality of how the mean value of any continuous periodic function might be found. So I would expand the integral to include the specific statement of the mean of f(t) as

    Mean \ of \ f(t)=\frac{1}{T}\int^{t_x+T}_{t_x}(A+Bsin(\omega t))dt

    where ω is the angular frequency in radians/sec and related to the period T of the function f(t) by

    \omega=\frac{2\pi}{T}

    2. You suggest the upper and lower integration limits are restrictive - I disagree, given that I make the point that the time tx (not a function of x - by the way) is any arbitrary time value within the time domain of the function f(t). So tx could be any real time value - whatever you see fit to choose when attempting to complete a solution. Since the integration is averaged over the period T (i.e. by placing the '1/T' in front of the integral sign) the range of the limits must include at least one complete period T of the function - hence the addition of the period T in the upper limit.

    3. You ask whether this is a definite or indefinite integral. The existence of the limits of integration surely mean the former [definite].

    4. You suggest I evade your questions and apparently cover my lack of real understanding by posting a lot of irrelevant equations. The inclusion of the solution I attached as a pdf in my last post is in fact the general solution to the problem material included as the attachment in your original post - which was also my work, & a matter which you failed to acknowledge. In my opinion the attached pdf provided some important insights about how the problem would be solved for any case - not just the specific case originally posted. In my defense I'll say that I've made a reasonable effort to answer your questions. Keep in mind I don't get any financial reward for doing this. It's an interest not an obligation. Your comments might be considered by some as bordering on insensitive. I'm not sure what you meant by your statement - "I hope ACC have a lot of bit and bytes in regard to the process of your responding". Whether or not I understand what I'm submitting and whether it is helpful or not is for you and others to judge.

    5. Since you have objections to my mathematical solutions I won't burden you with the complete solution to the integral. I'll leave that for you to ponder.

    In the interests of maintaining a harmonious relationship with you as a fellow forum member I suggest that ongoing comment / discussion would be better left to continue between yourself and others. I will refrain from making further contributions to this thread. Hopefully others will offer you more practical help. Good luck with your quest.
     
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  18. Thevenin's Planet

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    Nov 14, 2008
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    It seems to me that you are not sure is what is going on in this discusssion ? Some of my humor which was not to ridicule or exprees any kind sarcasm but just expression of irony.Kind of reverse of the literal sense, a enough of bit and bytes which literally i know there is enough of memory yo handle this discussion. In another thought mode I was using the Socratic method of question and answer method which is commomly done,even though many people who use it are not aware of label(s). I hope this view clear the insenstiveness that was produced by misconception.When the expectation of answer then not follow the pattern I assume it was a evasion from the Socratic method.I was in the process of extracting the notation meaning from you regarding function.Attempting to get you to express the notation meaning in actuality or perhaps in graphy picture.A picture can explain something greater than words in 99.99% of the time.Such as the periodic function:
    f(t) A +Bsin(2pi/Tt) which I have found out in this discussion, that it would be a sine wave graph of points and the varying "t or x"is an instanetaneous point on the sine wave.I ponder over this a couple of nights and comprehended the picture of the sine wave graphy.
    Then I did some extra reseaching and found that Intergration is the opposite of differentiation.And when a functionis differentiated,the results is another functionthat expresses the instantanous rate of change or the orginal function.Integration ,in a sense,the reverse.It produces a new function that expresses the cumulative growth of the orginnal function.
    I was trying to decide was f(t)=A+Bsin(2pi/Tt) is the orginal function?
    So I do think this disscuss has been very helpful it must continue all means regardless..
     
  19. Thevenin's Planet

    Thread Starter Active Member

    Nov 14, 2008
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    Howdy
    It seems your firm determination has prevailed.The information you conveyed is very helpful for to observe how waves forms are associated with these functions and attaining specific information. I will missed the opportunity to understand what was actually happening as far as the sine wave graph concerning function manipulations so, I decided to take your advice and review calculus basic.I found a book in the library that would be a good beginning : Calculus for Dummies....Author Mark Ryan.
     
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