Phasors - Converting from sine to cosine

Discussion in 'Homework Help' started by testing12, Mar 27, 2011.

  1. testing12

    Thread Starter Member

    Jan 30, 2011
    80
    2
    Hello Everyone,

    Im having some trouble with phasors.
    My objective is to convert expressions such as:
    - 8 sin(10t rad+70 degrees) and
    120 sin (10t rad -50 degrees)
    -60cos (30t rad +10 degrees)

    to an expression with cosine and the positive amplitude. I cannot seem to get them correct! either i add 90 degrees when im supposed to subtract or subtract when adding is required.

    Can some one please help me, my text does a very brief job of explaining this, and rightfully so since i should know this already!

    Please help.
    Regards.
     
  2. testing12

    Thread Starter Member

    Jan 30, 2011
    80
    2
    For example here is one where I went wrong, but i just dont understand it. Im using the method from the text book. They say to use the vertical axis as sin (wt) and horizontal axis as cos (wt) but the vertical axis is inverted, that is the top is - and bottom is +.

    [​IMG]
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I would keep two relationships in mind

    1. sinθ=cos(90-θ) - for the right angled triangle
    2. cos(α)=cos(-α)
    So take the case of

    v(t)=120sin(10t-50°)

    let θ=[10t-50]

    v(t)=120cos(90°-[10t-50°])=120cos(90°-10t+50°)

    or

    v(t)=120cos(140°-10t)

    let
    α=140°-10t=-(10t-140°)
    &
    -α=(10t-140°)


    so from (2) above
    cos(α)=cos(-α)
    cos(140°-10t)=cos(10t-140°)

    Hence

    v(t)=120cos(10t-140°)
     
  4. testing12

    Thread Starter Member

    Jan 30, 2011
    80
    2
    thank you sir, i was confused. There is so much to learn and feeling the pressure i will review again in the morning.
     
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