Phasor Intro

The idea is that since every state in a linear system will have the same frequency (see Sinusoidal Steady-state Response), we don’t want to bother carrying around the $e^{j\omega t}$ term.

Let’s first understand it without phasor notation – solve for $V_{in}=V_0{e}^{j\omega t}$.

We have:

$$\begin{array}{rl}{V}_{out}& =V_{in}\cdot \left(\frac{j\omega RC}{j\omega RC+1}\right)\\ & =V_0{e}^{j\omega t}\cdot \left(\frac{j\omega RC}{j\omega RC+1}\right)\end{array}$$

We could plug in everything ($R,C,\omega$). But $V_{in},V_{out}$ and the current $I$ through the capacitor all contain an $e^{j\omega t}$ (since the system is linear).

Phasor Notation

So, we can drop all $e^{j\omega t}$ terms since it shows up everywhere anyways. This leaves us with a magnitude and phase.

Originally, we started off with $V_{in}=V_0{e}^{j{\omega }_{0}t}$ at a particular frequency ${\omega }_0$. Now, we write it as:

$${\tilde{V}}_{in}=V_0\angle 0°[\text{V}]$$

With a phase $V_{in}=V_0{e}^{j(\omega t+\varphi )}=V_0{e}^{j\varphi }{e}^{j\omega t}$, we would have:

$${\tilde{V}}_{in}=V_0\angle \varphi [\text{V}]$$

We can also write this as Cartesian coordinates.