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Sum of 2 complex exponentials

  1. Oct 5, 2015 #1
    I appreciate the opportunity afforded by this forum to submit a question.

    I have struggled with the derivation shown in the attached picture. I am certainly unfamiliar with the concept used to include the arctan function in the encircled step.

    Would be highly appreciative of a prompt.

    Best regards,
    wirefree
     

    Attached Files:

  2. jcsd
  3. Oct 5, 2015 #2

    BvU

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    Hello there,

    I then assume you are happy with (4) and wonder how the factor in [ ] brackets is converted to the polar form in the red oval.
    For simplicity I propose to introduce $$ v = A\cos\alpha+B\cos\beta \\ w = A\sin\alpha+B\sin\beta$$ so that we have to convert ## v + jw## to polar form, i.e. to a form ##R\,e^{j\phi}##

    (which is the reverse of what was done to write out (4), where, for example, they wrote$$A\,e^{j\alpha}= A\cos\alpha + j A\sin\alpha$$)

    If we want to solve ## v + jw = R\,e^{j\phi}## for ##R## and ##\phi## and compare with the above, we see that ##\phi## follows from ##{\sin\phi\over\cos \phi} = {w\over v}##, meaning ##\tan\phi = w/v## or ##\phi = \arctan {w\over v}##

    R follows from ##R^2 = (v+jw)(v-jw) = v^2 - j^2 w^2 = v^2 + w^2 ##

    What helps a lot in this is to draw a picture of the complex plane (j axis vertical, real axis horizontal). And a unit circle or a circle with radius R to convert ##R \, e^{j\phi} ## to real and imaginary parts.

    --
     
  4. Nov 14, 2015 #3
    Thank you, BvU.
     
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