To check Periodicity of x(t) = 2cos(t) + 3cos(t/3)

Discussion in 'Homework Help' started by waterineyes, Sep 6, 2014.

  1. waterineyes

    Thread Starter New Member

    Sep 6, 2014
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    My answer which I have got is :

    The function is periodic with Fundamental Period, T = 6(pi)..

    Is that right??
     
  2. waterineyes

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    Sep 6, 2014
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    I am just confirming as My book has given its answer as : T = 2(pi)..
     
  3. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Here is a plot of the waveform.

    Your value of 6pi can clearly be seen as the modulating envelope repeat frequency.

    Your book is referring to the timing between peaks on the underlying 'carrier', which I assume it means by the fundamental.

    In the second diagram I have stripped out the 2 and 3 multiplying factors as they push the peaks up and down but do not alter their timing.
    This allows the carrier to be more clearly identified.

    Remember the x axis is in radian measure.

    You can tell the periodicity because the peaks alternate either side of the x axis although they are vary in height.

    waveform1.jpg
     
    Last edited: Sep 7, 2014
  4. waterineyes

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    Sep 6, 2014
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    I am not getting you, are you saying both are right?? 6(pi) and 2(pi) also? And I am unable to see what carrier you are referring to..
     
  5. studiot

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    You time on forum is short as are your posts so I am not sure in what context you have met this question.
    Perhaps the question is to test if you know the definition of fundamental?


    Your function, x(t) = 2cos(t) + 3cos(t/3) contains two components, both cosine waves, of period 2pi (the first term) and 6pi (the second term).
    You should be able to see this just by looking at the equation.

    Now the fundamental is not necessarily the lowest frequency (longest period) present.
    This is a common misconception.

    The term comes from harmonic analysis and is a technique to analyse any waveform into components.

    The components comprise a fundamental and two sorts of waveforms, although only one sort may be present.

    1)Harmonics which have a frequency of integer multiples of the fundamental, or a period equal to the fundamental period divided by an integer.

    2)Subharmonics which have a frequency equal to the fundamental frequency divided by an integer, or a period equal to the fundamental period multiplied by an integer.

    Now, terms (components) of both 2pi and 6pi period appear in your equation. The first term in your equation has the higher frequency or shorter period and yet is considered the fundamental.

    Are you asking why?
     
    Last edited: Sep 7, 2014
  6. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Yeah it would be better to know more about the 'book' in question here just to see what their general idea is about harmonics.
    Sometimes seeing the actual question the way it was originally worded helps in cases like these, and sometimes we have to look at other questions and examples to see how they handle the other questions that involve the same concepts.
    In other words there are two possibilities:
    1. The fundamental period is 6pi and the book was wrong about stating 2pi.
    2. The fundamental period is 2pi and the book was right about stating 2pi.
     
  7. waterineyes

    Thread Starter New Member

    Sep 6, 2014
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    What do you mean by Fundamental??
     
  8. waterineyes

    Thread Starter New Member

    Sep 6, 2014
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    Now the fundamental is not necessarily the lowest frequency (longest period) present.
    This is a common misconception.

    What does this mean??
     
  9. waterineyes

    Thread Starter New Member

    Sep 6, 2014
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    @studiot , may be you are thinking too much and with your full force, you are trying to prove that Book may be right which is not the case here..

    The book is full of errors..

    cos(t) has 2(pi) and cos(t/3) has 6(pi) as their individual Time Periods.

    So, for overall signal to be periodic, The ratio of both the time periods should be rational which is, yes, rational, in this case and equals 1/3.

    Secondly, Overall Time Period can be found as : LCM{T1 , T2} which is : 6(pi) here and not 2(pi)..

    So, Its Fundamental Periodicity must be 6(pi) here..

    http://www.wolframalpha.com/input/?i=x(t)+=+2cos(t)+++3cos(t/3)+Period

    Check this and don't think too much... :)
     
  10. waterineyes

    Thread Starter New Member

    Sep 6, 2014
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    Do you say this to everyone who uses this website or forum for very first time ??

    Then you should not, I suggest..

    And If I have signed up for this forum two days back, then it is understood that my time and my posts are going to be short here... :)
     
  11. studiot

    AAC Fanatic!

    Nov 9, 2007
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    No but you were asked by two separate responders for more information.

    That is what I meant.

    It means exactly what it says.

    If you include subharmonics in your series they cannot be called fundamental because the subharmonics are never ending.

    That is there is no lower bound to the series so there is no lowest frequency, to call a harmonic.

    However the series of harmonics has a lower bound and therefore a lowest frequency.
    It is this frequency that is the fundamental and this does not affect the fact that the series of harmonics goes on to infinite terms.

    All the infinite terms in both the series of sub harmonics and harmonics are always present, even if their coefficients are zero

    Now 6pi is greater than 2pi so the frequency represented by a period of 6pi is less than the frequency represented by a period of 2pi

    So the 6pi wave is a subharmonic of the 2pi wave.
     
  12. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    The fundamental is the lowest frequency present so that would mean the period is 6*pi unless this book gives more extreme examples, which i doubt very much. Therefore my initial conclusion is that the book is wrong when they declare 2*pi as being the period.
     
  13. studiot

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    So if I pluck a guitar string so it vibrates with one central node and two end nodes is the fundamental present?

    Or if I broadcast a suppressed carrier wireless transmission, is the fundamental present?
     
  14. MrAl

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    Jun 17, 2014
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    Hi,

    If i generate forty three sine waves of different frequencies, is the fundamental present?
    If a guy on the moon paints three pictures of sine waves, is the fundamental present?
    If a tree falls in the woods, is the fundamental present? (he he) :)

    In this thread here we have only two frequencies f1 and f2, we would assume the lowest one is the fundamental otherwise more information would have to be given to explicitly state otherwise. In other words, i believe here we should assume the most typical case not the least typical in the absence of more information. The fundamental being the lowest frequency present is the most typical. By your examples we cant determine anything in this thread or any other thread unless we have that extra information. Going by that we then could not even guess it might be 2pi instead of 6pi.
     
  15. studiot

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    Hey, Mr Al there is no need to mock. The missing fundamental is a recognised phenomenon.

    http://en.wikipedia.org/wiki/Missing_fundamental.

    I think this is exactly right we really need more knowledge of what the book actually said.

    Here is my understand ing of the maths involved in the Wolfram Alpha link.

    A Harmonic analysis of a function y(t) can represent y(t) by a Fourier cosince series

    y(t) = {A_0} + {A_1}\cos ({\omega _f}t + {\phi _1}) + {A_2}\cos (2{\omega _f}t + {\phi _2}) + {A_3}\cos (3{\omega _f}t + {\phi _3}) + .....

    Where

    {\omega _f} = \frac{{2\pi }}{\tau }

    Tau is the fundamental period and omega_f the angular frequency of the fundamental
    and phi_n the phase angles of each component.
    If we consider the original function y(t) as the output of a Fourier cosine analysis and compare coefficients we find immediately that

    {A_0} = {\phi _n} = 0

    That is the fiorst term and all the phase angles phi are zero.

    Now the original function had two terms so trying each one in turn as a proposed second term we have either

    2\cos (t) = {A_1}\cos ({\omega _f}t) and 3\cos \left( {\frac{t}{3}} \right) = {A_3}\cos (3{\omega _f}t)

    Which yields Tau = 2pi and 18 pi ie a contradiction
    or

    2\cos (t) = {A_3}\cos (3{\omega _f}t) and 3\cos \left( {\frac{t}{3}} \right) = {A_1}\cos ({\omega _f}t)

    which yields tau = 6pi as the OP says.
     
  16. MrAl

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    Jun 17, 2014
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    Hi again,

    Oh no i was not mocking at all, i was just trying to illustrate the principle that the simplest solution is probably the best one, and also if we are given a simple set we dont have to explore ideas outside of that set which wont apply anyway. So in a way i was illustrating a sort of Occam's Razor. If you hear hoof beats assume horses not zebras.

    I agree that if we can see the book we might be able to draw a more perfect conclusion, but in the mean time i think we should assume horses :)
    I also like to try to throw in a little comedy here and there to lighten things up a little sometimes. I like comedy almost as much as i like circuits :)

    BTW i do like to read your replies too as i find them interesting. In the case of this thread i think it is very good to think about the subharmonics as well as the 'regular' harmonics.
     
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