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Fourier series and sketch the waveform

  1. Jan 19, 2015 #1

    M P

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    • Member warned about posting with no effort
    1. The problem statement, all variables and given/known data
    Sketch the waveform defined below and explain how you would obtain its Fourier series:
    f(wt) = 0 for 0 ≤wt ≤pi/2 (w=omega)
    f(wt) = Vsin(wt) for pi/2 ≤wt ≤pi
    f(wt) = 0 for pi ≤wt ≤3pi/2
    f(wt) = Vsin(wt) for 3pi/2 ≤wt ≤2pi
    Develop the analysis as far as you are able.


    2. Relevant equations


    3. The attempt at a solution

    Any help how to start this is appreciated..
     
  2. jcsd
  3. Jan 19, 2015 #2

    gneill

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    Staff: Mentor

    You need to make an attempt, else we'll have no choice but to delete your thread. Can you at least make a sketch of the waveform to show us?

    By the way, if you generate characters that are part of a local character set on your computer there's no guarantee that they will render correctly for others:
    Fig1.gif
     
  4. Jan 19, 2015 #3

    M P

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    Yes I will in the evening please dont remove it..
     
  5. Jan 19, 2015 #4

    M P

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  6. Jan 19, 2015 #5

    gneill

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    That link doesn't show anything that resembles the waveform that you described. Wrong link perhaps?
     
  7. Jan 19, 2015 #6

    M P

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    ≤≤
    I attached my thinking regarding to the first equation. f(omega t)= 0 for 0≤ omega t ≤ pi/2
     

    Attached Files:

  8. Jan 19, 2015 #7

    gneill

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    You have a piecewise-continuous function that's defined in several parts. Can you make a sketch of one full period (0 to ##2 \pi## radians). And maybe clarify what those garbled characters are in the function definition. Are they important?
     
  9. Jan 20, 2015 #8

    M P

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    gneill thank you for your patience and help. I am attaching my sketch after your pointers. attaching progress sketch.
     

    Attached Files:

  10. Jan 20, 2015 #9

    gneill

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    Okay, a couple of things before tackling the Fourier series. The first is a minor nit: In your function definition you have used ≤ for both of the segment ends where they transition from one segment to the other. This means that at the points where function changes from one segment to another the function would have two values (one from each segment) so it really wouldn't be a true function.

    The second is that your sketch of the function doesn't look right. For example, the function should be zero until ωt is 90° (the first segment of the function definition) then it transitions to Vsin(ωt). Well, ωt is 90° at that instant, and Vsin(90°) = V, since sin(90°) = 1. In other words, your sketch should show a sharp jump from 0 to V at ωt = 90°. Similarly at 270° there should be another sharp transition but this time from 0 to -V, since sin(270°) = -1.

    After you've sorted out your sketch, see if you can spot the overall function's period and then set up the integrals for the Fourier coefficients.
     
  11. Jan 20, 2015 #10

    M P

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    attempt to a sketch. that is what I understand has to be done. am I getting there ? thank you.
     

    Attached Files:

  12. Jan 20, 2015 #11

    gneill

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    Closer. Note that the transitions from horizontal section to sine curve section should be vertical!
     
  13. Jan 20, 2015 #12

    M P

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    I do not want to make myself look silly but please explain me is this first part shall I integrate over a period and achieve 1/pi x pi/2 and that gives me 1/2 at the begin of the function?
     
  14. Jan 20, 2015 #13

    gneill

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    I'm not sure what you're asking here. My previous comment was to point out that the function makes a sudden transition from 0 to 1 at ωt = 90°, and from 0 to -1 at ωt = 270°. After the transitions the curve follows sin(ωt) until the next zero, where it goes back to a horizontal (zero) value for the next segment.

    Perhaps you should plot a few points around the segment transitions to see.
     
  15. Jan 20, 2015 #14

    M P

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    attempt3.jpg
    a sudden transition from 0 to 1 at pi/2
    same for 270 with minus 1
     
  16. Jan 20, 2015 #15

    gneill

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    f(wt) = 0 for 0 ≤ wt ≤ pi/2 (w=omega) ← at wt = ##\pi/2## f(wt) = 0
    f(wt) = Vsin(wt) for pi/2 ≤ wt ≤pi ← at wt = ##\pi/2## f(wt) = sin(wt) = 1
    f(wt) = 0 for pi ≤ wt ≤3pi/2 ← at wt = ##3 \pi/2## f(wt) = 0
    f(wt) = Vsin(wt) for 3pi/2 ≤ wt ≤2pi ← at wt = ##3 \pi/2## f(wt) = sin(wt) = -1

    Fig1.gif
     
  17. Jan 21, 2015 #16

    M P

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    gneill.png
    Sir I have a question related to the picture above and also connected with thread how a0 is obtained in this particular example? is it something to do with cosine? I am not sure how a0 = 1 is calculated? can I ask for some hint in here?
     
  18. Jan 21, 2015 #17

    gneill

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    I don't see where that expression comes from. How did wt "escape" from its trig function? Can you show more of your work?
     
  19. Jan 21, 2015 #18

    M P

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    this what I am asking I was studying example to try to form some integrals and I did not understand how in this particular example a0 = 1 what is the process..?
     
  20. Jan 21, 2015 #19

    gneill

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    I don't know how they got there since I can't see the full development.

    When I worked through the problem as you've presented it I found an expression for an that yielded a0 = 0. Perhaps they framed the problem differently and thus arrived at a different result.
     
  21. Jan 21, 2015 #20

    M P

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    Can you please clarify something for me 1/∏ ∫ sin (wt) dwt can I "escape" to 1/pi [-cos wdt] and then "escape" to -eg 1/pi[ cos pi - cos pi/2] ??
     
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