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Superposition of Two Travelling Waves, Different Amplitudes

  1. Aug 5, 2015 #1
    1. The problem statement, all variables and given/known data
    I'm looking at an E&M textbook - "Time-Harmonic Electromagnetic Fields". They state:
    "A more general ##x ## polarized field is one consisting of waves traveling in opposite directions with unequal amplitudes - i.e :
    (1) $$ E_{x} = Ae^{-jkz} +Ce^{jkz}$$
    Let ## A ## and ##C## be real. We express the field in terms of an amplitude and phase. This gives:
    (2) $$E_{x} = \sqrt{A^{2}+C^{2}+2ACcos(2kz)} \ e^{-jtan^{-1} \bigg( \frac{A-C}{A+C} tan(kz) \bigg) } $$" ​

    But I can't see where this comes from.

    2. Relevant equations


    3. The attempt at a solution
    It would have though that amplitude is coming out of a cosine rule - that there is a vector triangle with ##\vec{A}## and ##\vec{C}## as two sides. In the argand/phasor diagram, the angle between ##Ae^{-jz} ## and ##Ce^{jz}## is going to be fixed (for any given time) as ##2kz## (or ## 2\pi -2kz ##) So by applying the cosine rule, I'd expect:
    $$ |E_{x}| = |A|^{2}+|C|^{2} -2|A||C|cos(2kz) $$
    But in the book there is a plus. Is this a typo, or have I messed up?

    and I can get the phase from just expanding out ##E_{x} ## into components - from (1):
    $$E_{x}= (A+C) cos(kz) + j(C-A)sin(kz)$$
    So if we want to write ## E_{x} ## as: $$E_{x} = |E_{x}| e^{j\phi} = |E_{x}| (cos\phi + jsin\phi)$$ Then:
    $$ tan(\phi) = \frac{C-A}{A+C} tan(kz) \implies tan(-\phi) = \frac{A-C}{A+C} tan(kz), \therefore \phi =-tan^{-1}\bigg(\frac{A-C}{A+C} tan(kz) \bigg) $$

    If that second bit is right - is the reason that we can't equate amplitude of that equation with the book amplitude because we have ignored time dependence? Just started doing this, so any help is appreciated.

    Thanks :)
     
  2. jcsd
  3. Aug 5, 2015 #2

    blue_leaf77

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    Homework Helper

    Try to calculate the modulus of
    $$E_{x}= (A+C) cos(kz) + j(C-A)sin(kz)$$.
     
  4. Aug 5, 2015 #3
    Ah, that's embarrassing. Out of interest - why is the cosine rule idea not valid? Please see picture.
    **EDIT**: Oops, in the diagram ##\vec{A}## and ##\vec{C}## have the opposite phase to as written in the question.
     

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