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I Sinusoids as Phasors, Complex Exp, I&Q and Polar form

  1. May 10, 2018 #1
    Hi,

    I am going around in circles, excuse the pun, with phasors, complex exponentials, I&Q and polar form...

    1. A cos (ωt+Φ) = Acos(Φ) cos(ωt) - Asin(Φ)sin(ωt)
    Right hand side is polar form .... left hand side is in cartesian (rectangular) form via a trignometric identity?

    2. But then sometimes I read...
    A cos (ωt+Φ) in polar form has corresponding cartesian form of Bcos(ωt)+Csin(ωt), which is fine to understand because this cartesian form gives X and Y coordinates on a cartesian coordinate axes of a vector in that axes.

    3. But point 1 and 2 are different, how can Acos (ωt+Φ) in polar represent Bcos(ωt)+Csin(ωt) in cartesian but also be equal to Acos(Φ) cos(ωt) - Asin(Φ)sin(ωt) via a trignometric identity ---> Is it because Acos(Φ) and Asin(Φ) are constants and therefore also B and C? Might be obvious but I need to ask for my own sanity of seeing so much different ways its written.

    What about if B and C are not constants due to the phase changing with time Φ(t)?

    I am further purplexed by notation used for complex sinusoids.

    3. Acos (ωt+Φ) can be represented as the real part of Aei(ωt+Φ)= Acos(ωt+Φ) + iAsin(ωt+Φ)
    but from point 1, the right hand side of this equation can be then re-written with the trigometric identity in point 1, expanding it into 4 terms which removes the phase from the argument and giving constants, like in point 3. So why cannot it not be written without the Φ in the argument on the right hand side and use different constants
    Aei(ωt+Φ)=Bcos(ωt)+iCsin(ωt)
     
  2. jcsd
  3. May 10, 2018 #2

    kuruman

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    Start with ##e^{\pm i\omega t}=\cos\omega t\pm i\sin\omega t##,
    Now,
    $$\begin{align}
    e^{i (\omega t+\phi)}& =e^{i\omega t}e^{i\phi}=(\cos\omega t+i\sin\omega t)(\cos\phi +i\sin\phi)\nonumber \\
    & =(\cos\omega t~cos\phi-\sin\omega t~sin\phi)+i(\sin\omega t~\cos\phi+\cos\omega t~sin\phi)\nonumber
    \end{align}$$Equating real and imaginary parts, gives you two trig identities
    $$\cos(\omega t+\phi)=\cos\omega t~cos\phi-\sin\omega t~sin\phi~;~~~\sin(\omega t+\phi)=\sin\omega t~\cos\phi+\cos\omega t~sin\phi$$
    You can easily show that $$\cos\omega t= \frac{e^{i\omega t}+e^{-i\omega t}}{2}~;~~~\sin\omega t= \frac{e^{i\omega t}-e^{-i\omega t}}{2i}=-i\frac{e^{i\omega t}-e^{-i\omega t}}{2}.$$
    Then,
    $$
    \begin{align}
    A \cos\omega t +B\sin \omega t &=A\frac{e^{i\omega t}+e^{-i\omega t}}{2}-iB\frac{e^{i\omega t}-e^{-i\omega t}}{2}\nonumber \\
    & = \frac{(A-iB)e^{i\omega t}}{2} +\frac{(A+iB)e^{-i\omega t}}{2}\nonumber
    \end{align}$$
    Note that the last expression is real because it is the sum of a number plus its complex conjugate. You can write the complex coefficient in polar form, ##A-iB = re^{i\phi},~r=\sqrt{A^2+B^2}##. In that case,
    $$
    \begin{align}
    A \cos\omega t +B\sin \omega t &= \frac{re^{i(\omega t +\phi)}}{2} +\frac{re^{-i(\omega t +\phi)}}{2}\nonumber\\
    & =\sqrt{A^2+B^2}\cos(\omega t+\phi)\nonumber
    \end{align}.$$
    Note: In the above expression ##\phi = \arctan(B/A)##.
    I hope this clarifies things for you and convinces you that the complex exponential plays a pivotal role in trigonometric transformations.
     
  4. May 11, 2018 #3

    sophiecentaur

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    Can that be right? I would have thought that the magnitude of the left hand side is constant but the magnitude of the right hand side is not (it's an ellipse) unless B=C.
     
  5. May 11, 2018 #4

    kuruman

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    I don't think that ##B=C## will do it, but ##A=\sqrt{B^2+C^2}## will.
    Suppose ##z = Ae^{i \omega t}=B\cos(\omega t)+iC\sin(\omega t)##, where ##A##, ##B## and ##C## are assumed real.
    Then
    ##|z|^2=(Ae^{i \omega t})(Ae^{-i \omega t})=A^2##
    But also
    ##|z|^2=[B\cos(\omega t)+iC\sin(\omega t)][B\cos(\omega t)-iC\sin(\omega t)]=B^2+C^2##
    The two expressions will match if ##A^2=B^2+C^2## as also shown in post #2.
     
  6. May 12, 2018 #5

    sophiecentaur

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    But the phase has been lost. How can varying B and C coefficients achieve an arbitrary phase for the resultant?
     
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