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Proof of Phasor Method

  1. Jul 16, 2014 #1
    Hello,

    Can someone show me how the inverse Phasor transform of the sum of individual Phasors of sinusoidal functions of the same frequency is the sum of the sinusoids? I could not find any rigorous proof and help appreciated.

    Thanks.
     
  2. jcsd
  3. Jul 17, 2014 #2

    jedishrfu

    Staff: Mentor

  4. Jul 17, 2014 #3
    Somehow I have a feeling the proof is in the Laplace transform...I will keep looking
     
  5. Jul 18, 2014 #4

    olivermsun

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    Science Advisor

    Can you start by writing out an equation for what you say in words in Post #1?
     
  6. Jul 18, 2014 #5
    Here are the equations:

    [tex]
    \sigma = A_{0}\cos(wt + \phi_{0}) + A_{1}\cos(wt + \phi_{1}) + ... + A_{n}\cos(wt + \phi_{n}) = \mathbf{B}\cos(wt + \Phi_{0})
    [/tex]
    [tex]
    \sigma = \mathbf{Re}\{A_{0}e^{\phi_{0}i}e^{wti}\} + \mathbf{Re}\{A_{1}e^{\phi_{1}i}e^{wti}\} + ... + \mathbf{Re}\{A_{n}e^{\phi_{n}i}e^{wti}\}
    [/tex]
    [tex]
    \sigma = \mathbf{Re}\{\mathbf{P}_{0}e^{wti}\} + \mathbf{Re}\{\mathbf{P}_{1}e^{wti}\} + ... + \mathbf{Re}\{\mathbf{P}_{n}e^{wti}\}
    [/tex]

    My question why...???

    [tex]
    \sum\limits_{i = 0}^{n} \mathbf{P}_{i} = \mathbf{P}\{\mathbf{B}\cos(wt + \Phi)\}
    [/tex]

    Thanks.
     
  7. Jul 18, 2014 #6

    jbunniii

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    Gold Member

    What do you mean by ##\mathbf{P}\{\mathbf{B}\cos(wt + \Phi)\}##? Does this equal ##\mathbf{B}\exp(i\Phi)##?

    Also, is it significant that you used boldface for ##\mathbf{B}##? Isn't it just some real number? I want to make sure I understand what you are asking.
     
  8. Jul 18, 2014 #7
    Yes. It's just the Phasor transform.

    Yes, B is a real number.
     
  9. Jul 19, 2014 #8

    jbunniii

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    OK,
    $$\begin{align}
    \sum_{k=0}^{n}A_k\cos(\omega t + \phi_k) &= \sum_{k=0}^{n}A_k[\cos(\omega t)\cos(\phi_k) - \sin(\omega t)\sin(\phi_k)] \\
    &= \left(\sum_{k=0}^{n} A_k \cos(\phi_k)\right) \cos(\omega t) - \left(\sum_{k=0}^{n}A_k\sin(\phi_k)\right)\sin(\omega t) \\
    &= A\cos(\omega t + \Phi) \\
    \end{align}$$
    To compute ##A## and ##\Phi##, we again use the trig identity
    $$A\cos(\omega t + \Phi) = A\cos(\Phi)\cos(\omega t) - A\sin(\Phi)\sin(\omega t)$$
    and compare with what we have above to conclude that
    $$A \cos(\Phi) = \sum_{k=0}^{n} A_k \cos(\phi_k)$$
    and
    $$A\sin(\Phi) = \sum_{k=0}^{n}A_k\sin(\phi_k)$$
    Therefore,
    $$\begin{align}
    \sum_{k=0}^{n}A_k \exp(i\phi_k) &=
    \sum_{k=0}^{n}A_k \cos(\phi_k) + i\sum_{k=0}^{n}A_k\sin(\phi_k)\\
    &= A[\cos(\Phi) + i\sin(\Phi)] = A\exp(i \Phi)
    \end{align}$$
    which gives us what we want.

    If desired, we can calculate ##A## and ##\Phi## explicitly as follows:
    $$A = \sqrt{\left(\sum_{k=0}^{n}A_k \cos(\phi_k)\right)^2 + \left(\sum_{k=0}^{n} A_k \sin(\phi_k)\right)^2}$$
    and
    $$\Phi = \arctan\left(\frac{\sum_{k=0}^{n} A_k \sin(\phi_k)}{\sum_{k=0}^{n}A_k \cos(\phi_k)}\right)$$
     
    Last edited: Jul 19, 2014
  10. Jul 19, 2014 #9

    olivermsun

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    jbunniii has a nice and very thorough demonstration above.

    One of the interesting things you notice is the close relationship between addition of complex exponentials and angle addition in the trig identities. If you keep it in complex exponential form, then you can also show what you say (in words) above by:
    $$
    \begin{align}
    \mathcal{P}^{-1} \Biggl\{ \sum\limits_{j = 0}^{n} \mathbf{P}_j \Biggr\}
    &= \mathcal{P}^{-1} \Biggl\{ \sum\limits_{j = 0}^{n} A_{j}e^{\phi_{j}i} \Biggr\} \\
    &= \mathbf{Re} \Biggl\{ \Biggl[ \sum\limits_{j = 0}^{n} A_{j}e^{\phi_{j}i} \Biggr] e^{wti} \Biggr\} \\
    &= \mathbf{Re} \Biggl\{ \sum\limits_{j = 0}^{n} A_{j}e^{(wt+\phi_{j})i} \Biggr\} \\
    &= \sum\limits_{j = 0}^{n} A_{j}\cos(wt+\phi_{j}) .
    \end{align}
    $$
     
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