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Help with graphical technique of adding 2 sinusoids

  1. Dec 20, 2011 #1
    1. The problem statement, all variables and given/known data
    I have a problem understanding how equation 9.11 (see attached) works with Fig. 9.4 (a). Any help would be appreciated.


    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Dec 20, 2011 #2

    Simon Bridge

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    The two equations (9.11) are being represented as a 2D vector (fig 9.4 a).

    The authors have used the phasor representation of the sinusoids and the diagrams represent the vector sums of the phasors. I'd have explained it a bit differently.

    Basically, cosine and sine functions have a relative phase of π/2 - so you can use them as x and y axes. The amount each contributes to the result is it's amplitude, which is what is being plotted on the axis. This does not work if the sinusoids have other than nπ/2: n=0,1,2,...

    In general - use the phasor representation for arbitrary sinusoids and add the phasor arrows head-to-tail just like vectors.
     
  4. Dec 20, 2011 #3
    Try expanding the right hand side, you'll see that it works out :)

    Edit: Oops beaten, cutthroat business this physics stuff :P
     
  5. Dec 20, 2011 #4
    Thanks for that. Makes sense now that I realize they are vectors. Makes you wonder why they didn't make it clearer. Phasors is the subject of the next chapter.
     
  6. Dec 20, 2011 #5

    Simon Bridge

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    Ah well, they are trying to be kind to you by building up to them.
    I always just launch right into it. By now you've had vectors drilled into you so you see them in your sleep so you can handle rotating vectors no problem.

    Never mind - skip ahead to phasors and you'll see it makes better sense.
     
  7. Dec 20, 2011 #6

    NascentOxygen

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    Remember, wherever you see A.cos(wt) you can replace it by A.sin(wt+Pi/2) because this is an exact equivalence.

    By similar reasoning, -cos(wt) = cos(wt+Pi) = cos(wt-Pi)
     
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