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Sinusoidal function. Top percentage of values.

  1. Jun 10, 2013 #1
    1. The problem statement, all variables and given/known data

    For a sinusoidal function, how do you determine the highest value exceeded 10% of the time?
    The pink line in the attached pic indicates that value.
    Just wondering how you actually determine the value for a periodic function?

    [Broken]

    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 10, 2013 #2

    Dick

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    That function is not very sinusoidal. In that case it's simply a numerical problem. If you mean that you have a function like y=sin(x) that is, then just draw a graph, say over the range [0,2pi]. The highest value is 1 and the lowest value is -1, so you want to find the values of x where sin(x)=0.8. That isn't so hard, is it?
     
    Last edited by a moderator: May 6, 2017
  4. Jun 10, 2013 #3
    I was just posting that graph to show what I mean by sound level exceeded 10% of the time I'm actually just trying to do it for a periodic function.
    I'm not sure if whats you're saying answers the question or not. Maybe you misunderstood the question.
    Someone else told me length of interval [pi/2 - pi/10, pi/2 + pi/10] is 10% of [0, 2pi]
    Does this make sense to you?
     
  5. Jun 11, 2013 #4

    Dick

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    Yes, I did misunderstand. I was thinking it was 10% of the max not 10% of the time. 10% of 2*pi is pi/5. Since the max of sin is at pi/2 then I think that "somebody else" is correct.
     
  6. Jun 11, 2013 #5
    Yeah thanks, that gave me the correct answer. If pi/2 - pi/10 is 10% of the time, what is it for 90% of the time?
     
  7. Jun 11, 2013 #6

    Dick

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    Well, x is outside of the range you gave 90% of the time. And y=sin(x) is less than the value of sin(pi/2-pi/10), isn't it?
     
    Last edited: Jun 11, 2013
  8. Jun 11, 2013 #7
    i mean what value is exceed 90% of the time though so it's line on the above graph would be down the bottom somewhere
     
  9. Jun 11, 2013 #8

    Ray Vickson

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    If you draw a horizontal line at ##y = L \; (0 < L < 1)## it cuts the graph of ##y = \sin(x)## at two points, say ##x=a## and ##x=b##, with ##\sin(a) = \sin(b) = L.## If you want ##\sin(x) > L## for 10% of the time, you want ##b-a = 2 \pi/10 = \pi/5,## so you need to solve the equation
    [tex] \sin(a) = \sin(a + \pi/5)[/tex]
    Expanding ##\sin(a + \pi/5)## we get the equation
    [tex] \sin(a) = \sin(a) \cos(\pi/5) + \cos(a) \sin(\pi/5)[/tex]
    Thus
    [tex] [1-\cos(\pi/5)] \sin(a) = \sin(\pi/5) \cos(a) \Rightarrow \tan(a) =
    \frac{\sin(\pi/5)}{1-\cos(\pi/5)}[/tex]
    From this we can find ##a## and can get the level ##L## as
    [tex] L = \sin(a) = \frac{\sin(\pi/5)}{\sqrt{2\left(1-\cos(\pi/5) \right)}}[/tex]
    We have ##a = 1.256637061,## and ##L = .9510565160## is the desired level.
     
  10. Jun 11, 2013 #9

    Dick

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    a=1.256637061 is, not surprisingly, pi/2-pi/10. It's actually kind of obvious if you just look at the graph. Why are you choosing this difficult path to the result?
     
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