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Adding sinusoidal functions using phasors

  1. Mar 21, 2014 #1
    1. The problem statement, all variables and given/known data

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    2. Relevant equations



    3. The attempt at a solution
    $$3cos(20t+10°)-5cos(20t-30°)\\ =3\angle 10°-5\angle -30°\\ =-1.376+3.0209j\\ =3.32\angle -65.51°$$ In the last step, the textbook actually got ##3.32\angle 114.49°##.

    I checked both answers and it seems that the textbook's answer is correct. All I did to get -65.51° is using the arctan function on my calculator. It looks like their angle is 180° greater than that. How did they choose the correct angle? I'm rusty on my trigonometry.
     
  2. jcsd
  3. Mar 21, 2014 #2

    psparky

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    Gold Member

    Don't phasors need to be put in RMS before you work with them in this case?
     
    Last edited: Mar 21, 2014
  4. Mar 21, 2014 #3

    gneill

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    Staff: Mentor

    The arctan() function doesn't discriminate between the cases where the minus sign is to be associated with the numerator or the denominator of the argument; it only sees the overall sign of the single number it is given. To check what quadrant the correct angle should lie in, do a quick sketch of the x (real) and y (imaginary) values on the complex plane. Adjust the result by 180° if required.

    If your calculator has an atan2() function, use that instead as it takes two arguments, one for the x component and one for the y, and sorts out the correct angle automatically. Failing that, your calculator might have rectangular to polar conversion built-in, which will also give the result in the correct quadrant automatically.
     
  5. Mar 21, 2014 #4

    SteamKing

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    Draw a picture of your combined phasor. See the quadrant where the components locate the resultant?

    Generally, the inverse trig functions on your calculator only return the principal angles, which for the tangent is in the range -π/2 ≤ θ ≤ π/2. When doing these types of calculations, you must give your result a separate check to make sure you obtain the correct angle.
     
  6. Mar 21, 2014 #5
    Thanks for help guys. I completely forgot this part of trigonometry and I get it now. It's actually kind of scary that I forgot this because it's not like I stopped doing Math. Too much calculus has made me a bit rusty on my algebra (or at least trigonometry).
     
  7. Mar 21, 2014 #6

    psparky

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    -1.376 + 3.02J never equals -65 degrees.....check your math there...it always equals 114.49 degrees.
     
  8. Mar 21, 2014 #7
    Weird. My calculator is always giving me a principal value of -65.
     
  9. Mar 21, 2014 #8

    gneill

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    Staff: Mentor

    Yes. Because you are calculating:
    $$\phi = arctan\left( \frac{3.02}{-1.376} \right)$$
    Which, after resolving the argument to -2.195, the arctan() function places in the fourth quadrant. It cannot distinguish ##\frac{3.02}{-1.376}## from ##\frac{-3.02}{1.376}## or ##-2.195##, and can only return a corresponding angle that lies within its principal range of +/- 90° .
     
  10. Mar 22, 2014 #9

    psparky

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    Trig is certainly one way, another is simply using the rectangular to polar function on your calculator.
    Or do them both. If they both agree, better yet.
     
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