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Difference in signs OF PHASES of -7cos(5x-3) and -cot(3x-4)?

  1. Feb 3, 2012 #1
    1. The problem statement, all variables and given/known data

    The phase of -7cos(5x-3) is -3/5

    The phase of -cot(3x-4) is 4/3


    2. Relevant equations
    3. The attempt at a solution

    Why the difference in the signs of phases? I think the phase of -cot(3x-4) should be -4/3. Its an odd function: -cot(3x-4)=cot(-3x+4) and even then its phase looks to be -4/3.

    Also, would -7cos(5x-3) equal -7cos(-5x+3) because it's an even function?

    Thanks.
     
  2. jcsd
  3. Feb 4, 2012 #2
    Try it yourself. Plug some values into a cosine function and then change the function to negative with the same values, see what you get.
     
  4. Feb 4, 2012 #3
    I did. cos(x)=cos(-x) holds true.

    What about phases though?
     
  5. Jun 3, 2012 #4
    Please, help me determine phase difference

    Hi, All.

    1. The problem statement, all variables and given/known data

    1. f(x)= -7cos(5x-3)= -7cos5(x-3/5)

    cosx is an even function. f(x)=f(-x)

    phase difference here is -3/5

    2. g(x)=- cot(3x-4)= cot(-3x+4)=cot3(-x+4/3)

    cotx is an odd function g(-x)=-g(x)

    phase difference is 4/3


    2. Relevant equations
    3. The attempt at a solution

    I have couple questions:

    1. Why did we derive the phase difference from f(x) and not from f(-x) in the first example ?
    2. Why did we derive the phase difference from f(-x) and not from -f(x) in the second example ?
    3. Is f(x) is preferable to work with to f(-x) and f(-x) to -f(x)?

    Thanks.
     
    Last edited by a moderator: Jun 4, 2012
  6. Jun 4, 2012 #5

    rude man

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    Homework Helper
    Gold Member

    Re: Please, help me determine phase difference

    It just depends on definition of phase "difference."

    If I say A-B is the difference between A and B I would be correct.
    And if I say B-A is the difference between A and B I would still be correct. "Difference" by itself has no polarity.

    BUT: sin(x-ψ) represents a waveform with lagging phase ψ.
    If x = ωt and t is time, then you can see that sin(x-ψ) lags sin(x) by phase angle ψ or equivalently by time τ = ψ/ω.
     
  7. Jun 4, 2012 #6

    berkeman

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    Staff: Mentor

    (Moderator note -- multiple threads merged)
     
  8. Jun 6, 2012 #7
    solve,
    No, it is -7cos(5(x-(3/5)), which means that -7cos(5x-3) has a phase lag of 3/5 radians when compared to -7cos(5x).

    Yes, it is -cot(3(x-(4/3)), which means that -cot(3x-4) has a phase lag of 4/3 radians when compared to -cot(3x). Note that this phase lag is greater than the period of -cot(3x), which is ∏/3.

    What difference? They both show positive values, which means they both lag in phase.

    That would be wrong. -cot(3(x-(-4/3)) is not the equation given.

    Its phase is 4/3 radians and it is lagging.

    Phase is zero for either representation.

    Which equation with respect to what equation?

    Which equation with respect to what equation?

    You could have done so, the values would not change.

    You could have done so, the values would not change.

    It makes no difference, the values will still be the same.

    Ratch
     
  9. Jun 7, 2012 #8
    Hi, Ratch.

    The logic here: -7cos(5x-3)= -7cos5(x-3/5), so the phase is -3/5.

    The logic: -cot(3x-4)= cot(-3x+4)= cot3(-x+4), so the phase looks to be 4/3, but the phase of -cot(3x-4) looks like -4/3, except I am not sure how negative cot influences the phase.


    Not equations, but just functions:

    f(x)= -7cos(5x-3)= f(-x)=-7cos5(x-3/5)

    -g(x)=- cot(3x-4)= g(-x)=cot3(-x+4/3)



    I don't know what's up with me, but asking that bunch of questions is just stupid since, for example, f(x)=f(-x) in relation to an example given above, so "it makes no difference, the values will still be the same." Thanks for that. Except, again, -cot3(x-4/3) seems to have -4/3 as its phase.
     
    Last edited: Jun 7, 2012
  10. Jun 7, 2012 #9
    solve,

    The convention is that a positive phase means the given term is lagging. If you say the phase is minus, then it is leading. That is not correct. See plot below.

    cot(-3x+4)= cot(-3(x-(4/3)) , so yes the phase is 4/3 radians. A negative wave has the same phase as a positive one does.

    Which are defined by equations.

    What are those equations supposed to signify?

    Put in this form, -cot3(x-4/3)==cot(3(x-(4/3)) shows it has a positive or lagging phase. See plot.

    Ratch
     

    Attached Files:

  11. Jun 7, 2012 #10
    I dont think I understand you correctly. For instance, y=sin(x+ pi/4) leads y=sinx by a phase of pi/4. y=sin(x- pi/4) lags behind y=sinx with a phase difference of pi/4. Would that be correct?

    They have to do with the questions about the preference of use between -f(x) and f(-x) i.e -cot(3x-4) or cot(-3x+4) to find the phases. You already answered that question.

    Thanks.
     
  12. Jun 7, 2012 #11
    solve,

    Yes, that is correct. y=sin(x-(-π/4)), shows a -π/4 (minus), which means leading. y=sin(x-(π/4)) shows a π/4 (positive), which is lagging. Isn't that what I averred? See link below.

    http://www.regentsprep.org/Regents/math/algtrig/ATT7/phaseshift.htm

    Ratch
     
  13. Jun 9, 2012 #12
    Also, is it true that cos(pi/2- x)=cos(x- pi/2)= sinx ?

    Thanks, Ratch.
     
  14. Jun 9, 2012 #13
    solve,

    Yes, those equations are well known trig identities.

    Ratch
     
  15. Jun 12, 2012 #14
    Thanks, Ratch.
     
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