How is the Inverse Phasor Transform of Sum of Individual Phasors Proven?

[seminum]

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.

 

[jedishrfu]

While I don't have an answer to your question, I did find this interesting site:

http://music.columbia.edu/cmc/MusicAndComputers/

which covers a discussion of phasors in Chapter 3 which might give you some insight on how to prove it.

Also as an aside, I couldn't help but post this recent news article too:

http://www.dailymail.co.uk/sciencet...tml?ITO=1490&ns_mchannel=rss&ns_campaign=1490

 

[seminum]

Somehow I have a feeling the proof is in the Laplace transform...I will keep looking

 

[olivermsun]

Can you start by writing out an equation for what you say in words in Post #1?

 

[seminum]

Here are the equations:

<br /> \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})<br />
<br /> \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}\}<br />
<br /> \sigma = \mathbf{Re}\{\mathbf{P}_{0}e^{wti}\} + \mathbf{Re}\{\mathbf{P}_{1}e^{wti}\} + ... + \mathbf{Re}\{\mathbf{P}_{n}e^{wti}\}<br />

My question why...?

<br /> \sum\limits_{i = 0}^{n} \mathbf{P}_{i} = \mathbf{P}\{\mathbf{B}\cos(wt + \Phi)\}<br />

Thanks.

 

[jbunniii]

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.

 

[seminum]

Yes. It's just the Phasor transform.

Yes, B is a real number.

 

[jbunniii]

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
$$\Phi = \arctan\left(\frac{\sum_{k=0}^{n} A_k \sin(\phi_k)}{\sum_{k=0}^{n}A_k \cos(\phi_k)}\right)$$

 

[olivermsun]

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?

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:
$$
\begin{align}
\mathcal{P}^{-1} \Biggl\{ \sum\limits_{j = 0}^{n} \mathbf{P}_j \Biggr\}
&= \mathcal{P}^{-1} \Biggl\{ \sum\limits_{j = 0}^{n} A_{j}e^{\phi_{j}i} \Biggr\} \\
&= \mathbf{Re} \Biggl\{ \Biggl[ \sum\limits_{j = 0}^{n} A_{j}e^{\phi_{j}i} \Biggr] e^{wti} \Biggr\} \\
&= \mathbf{Re} \Biggl\{ \sum\limits_{j = 0}^{n} A_{j}e^{(wt+\phi_{j})i} \Biggr\} \\
&= \sum\limits_{j = 0}^{n} A_{j}\cos(wt+\phi_{j}) .
\end{align}
$$