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Sine reading prob

Discussion in 'Electronics Homework Help' started by AlanP, Jun 23, 2012.

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  1. AlanP

    AlanP

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    Jun 23, 2012
    Hello;

    Ive recently started messing around with amps during which time Ive received this assignment to read the graph (see att) and convert it into this form here u(t) = U m sin (ωt + Φ ). I can read anything, but I dont know how to get the angle :/ The result is Φ =120 deg. Any help please

    Thanks

    Alan
     

    Attached Files:

    • Sine.PNG
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  2. Harald Kapp

    Harald Kapp Moderator Moderator

    11,522
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    Nov 17, 2011
    The period T of the sine is the time between the 1st and 3rd zero crossing (in the same direction). In the attachment this time is easy to find if you take 10µs for the 1st zero crossing. From the period T you get the frequency F- you should know how. And from the frequency F you get ω.

    One full period is equivalent to a phase angle of 360 °. For Φ =0 ° the zero crossing (rising edge of the sine wave) is at t=0, as you can easily see by graphing sin (ωt + 0 ). Now if the zero crossing is not at t=0 but at t=10µs as in your attachment, You can now find the value for Φ such that (ωt + Φ)=0.

    This result doesn't match the graph in your attachment. If T=60µs, a phase shift of 10µs cannot be equivalent to 120°. Also if the sine is negative at t=0, the phase shift must be negative, too.

    I hope this helps you understanding the issue. I'll be happy to check your results.

    Harald
     
  3. AlanP

    AlanP

    4
    0
    Jun 23, 2012
    Thank you very much for your input mr Harald.
    Ive found some old formula in my books, that states that in this case (and every other case where I would need to find a phase shift angle) the result would be

    Φ=(first_crossing/period)*360*
    By using this here I get 60*. I think that this could be ok. But how would I determine if the sine is negative in t=0? By looking at the graph or?

    Thank you very much

    Alan
     
  4. Harald Kapp

    Harald Kapp Moderator Moderator

    11,522
    2,654
    Nov 17, 2011
    Alan,
    60 ° is the right number.

    This question, excuse me, sounds silly
    Why, yes, of course. In the graph you posted the sine is at approx. -60 for t=0. Or just put the numbers in your equation and evaluate for t=10µs. The result will be 0 only for Φ= -60 °, not for Φ= +60 °.

    Harald

    Harald
     
  5. AlanP

    AlanP

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    0
    Jun 23, 2012
    Kudos Harald. Great help!
     
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