# Proof of Phasor Method

1. Jul 16, 2014

### 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.

2. Jul 17, 2014

### Staff: Mentor

3. Jul 17, 2014

### seminum

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

4. Jul 18, 2014

### olivermsun

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

5. Jul 18, 2014

### seminum

Here are the equations:

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

My question why...???

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

Thanks.

6. Jul 18, 2014

### 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.

7. Jul 18, 2014

### seminum

Yes. It's just the Phasor transform.

Yes, B is a real number.

8. Jul 19, 2014

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

Last edited: Jul 19, 2014
9. Jul 19, 2014

### olivermsun

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}