How to find the phase of a sinusoid?

Discussion in 'Homework Help' started by lam58, Jan 19, 2014.

  1. lam58

    Thread Starter Member

    Jan 3, 2014
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    Hello, I'm a bit confused on this, the question asks to find the phase of a sinusoid at t = 0.008333s and the voltage at 0s.

    Additional info states the wave has a minimum voltage of -3 at 0.008333s and a max of +5v at 0.018333s.

    I found the amplitude to be 4V, the frequency is 50Hz and angular freq is 100∏ rad/s. But I am lost as of how to find the phase at 0.008333 seconds and the voltage at 0s.

    Any help would be much appreciated.
     
  2. studiot

    AAC Fanatic!

    Nov 9, 2007
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    You are correct about the amplitude and frequency.

    How did you find them if not by writing an equation and substituting given information?

    Can you write an equation of the form

    Voltage = A + B sin (ωt + ψ) where ψ is the phase angle by substituting your values.

    You can then find your answers by substituting for t at 0 and .008333 seconds
     
  3. lam58

    Thread Starter Member

    Jan 3, 2014
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    Ok but would it not be A + B cos(ωt + ψ)

    Also I too am in somerset. Bath to be precise. :)

    EDIT: infact would it have to be an ac signal because it's a wave of voltage?
     
    Last edited: Jan 19, 2014
  4. WBahn

    Moderator

    Mar 31, 2012
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    It can be either a cosine or a sine. If the problem doesn't state which, and if there isn't a default function that you normally use, then you can pick either one as long as you make it clear which you chose.

    As for whether it would "have" to be an AC signal depends on whose definition for AC you use. This can be a hot-button topic that takes on the flavor of a religious war, sometimes.

    Given the data you have presented, the waveform is probably not best described as "a sinusoid", but rather "a sinusoid riding on a DC offset" or something similar. But the description you used, coupled with the data, is sufficient for most people to come to the same conclusion that studiot did, namely that it is a sine function with a constant offset (and a cosine function with a constant offset is just as valid).
     
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  5. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Well I'd call it alternating, since it has both positive and negative parts.

    The point of alternating is some of it is positive, some is negative, which is another way to say it reverses (alternates) direction.

    A wiggly graph that is all on one side of the zero or the other is unidirectional, the signal simple getting stronger and weaker.

    Since I started with sine waves, I will keep to them, but yes you could also have a cosine wave.

    Now you have a +5 and a minus 3 so the halfway point is at +1 and the peak values are then symmetrical about V=+1, ie a straight line parallel to the t angle axis. That gives us A = +1.

    From +5 to +1 and from +1 to -3 the curve is antisymmetric, which gives us the amplitude of the sine component, B = 4.

    Now the distance between a peak and trough is ∏ so

    (ωt2+ψ) - (ωt1+ψ) = ∏

    ω(t2-t2) edit = ∏ so ω = ∏/0.01

    gives us the value of ω if we substitute t2=0.018333 and t1=0.008333

    The value of ψ can then be found by comparing the angle when the given sine wave crosses zero going positive and with the angle when a standard sine wave crosses zero (θ=0) in the same way.

    You can also do this with your cosine wave, in which case you will get a different ψ (do you know how much different?)

    Finally now you have all the constants in your cos or sin equation you can substitute for any value of t to find V.

    Our brains are not allways waterlogged or ciderlogged down here on the levels.

    :)

    go well
     
    Last edited: Jan 20, 2014
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  6. lam58

    Thread Starter Member

    Jan 3, 2014
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    I meant though because it can be expressed by the exp function exp(jωt) which can by euler's formula if I remember correctly can always be expressed by cos(ωt). Or something along those lines.

    But I understand what you are saying thanks. :)
     
    Last edited: Jan 20, 2014
  7. lam58

    Thread Starter Member

    Jan 3, 2014
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    Thanks, I actually found it in the end. TBH I know how to do this stuff but it always seems to slip my mind when I need it the most.
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    You can express either sine or cosine as a sum of complex exponentials or you can use the real part of a complex exponential to get cosine or you can use the imaginary part to get sine. Same difference.
     
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