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Oscilloscope trigger homework problem

  1. Jan 14, 2009 #1
    http://img502.imageshack.us/img502/8941/webworks4ez6.jpg" [Broken].

    The first answer is 2.5V, but I'm having trouble getting ωt

    I'm using the equation: V = V0 sin(ωt) --> 2.5V = 5V sin(ωt) --> sin-1(2.5V/5V) = ωt --> ωt = 30 degrees but that's incorrect and so is -30 degrees

    I've just used an oscilloscope for the first time last week, and I don't know much about it. What does changing the trigger slope polarity affect in this case? Apparently the scope still triggers at 2.5V... please shed some light on me!
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 14, 2009 #2
    A positive trigger at 2.5 volts means that scope should trigger on a rising part of the waveform as it reaches 2.5 volts. Negative trigger means it will trigger on a falling part.

    It still triggers at 2.5 volts. But the waveform is falling.
    Last edited: Jan 14, 2009
  4. Jan 14, 2009 #3
    Assuming Vo is equal to Vpeak then dividing the trigger voltage by Vpeak and finding its sine [ sin(Vt/Vo) ] will give you the phase angle.
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