Phasors Review

Introduction

This is a very sparse review of the motivation for and the use of phasors. It is not necessarily ready for prime time, and it may potentially have errors, typographical and otherwise. Also, there will be logical fallacies that, over time, will be corrected. But it is a start, and hopefully a worthwhile one!

Background

For circuits consisting of independent sources, dependent sources, reactive elements, and resistive elements, as long as certain conditions are met, if the independent sources consist of signals made up of one or more single-frequency sinusoids, the steady-state currents through and voltages across each element will also consist of signals made up of the same frequencies. "Steady-state" here refers to the time when whatever the "initial conditions" of the system are no longer relevant and the system is responding purely to the independent sources. "Certain conditions" include having resistive elements such that any perturbations from steady-state dissipate over time, avoiding doing anything "illegal" like trying to have an instantaneous change in voltage across a capacitor or instantaneous change in current through an inductor, and avoiding doing anything "weird" like having a voltage source in parallel with an inductor or a current source in series with a capacitor. Assuming those certain conditions are met and the circuit has been in place long enough to reach the AC steady state (ACSS), you can examine the circuit one frequency at a time (because, linearity). And for that one frequency at a time, you can assume that every voltage and every current can be written as

\(x(t)=X\,\cos(\omega t+\phi_x)\)

where $$X$$ is the amplitude of the signal and $$\phi_x$$ is the phase of the signal. For a linear, time-invariant circuit with all the input signals at the same frequency $$\omega$$, all AC steady-state signals will also be at that frequency $$\omega$$. If one source has more than one frequency, if if there are several sources with a variety of different frequencies, as long as the system is LTI, you can solve at each frequency individually and then add the results together to get the outputs collectively.

Euler Notation

While $$x(t)=X\,\cos(\omega t+\phi_x)$$ is lovely, taking derivatives of it is not. The $$n^{th}$$ derivative will definitely have an $$\omega^n$$ in front, but then there is the question of whether the result is a cos or sin and whether the result needs a + or - in front of it. One way around that is to use complex exponentials to do the math, then just take the real part at the end to get the signal. Specifically, recall that:

\( Ae^{j(\omega t+\phi)}=A\,\cos(\omega t+\phi)+j\,A\,\sin(\omega t+\phi) \)

such that

\( x(t)=X\,\cos(\omega_xt+\phi_x)=\mathfrak{R}\left\{Xe^{j(\omega_x t+\phi_x)}\right\}=\mathfrak{R}\left\{Xe^{j\omega_x t}e^{\phi_x}\right\}=\mathfrak{R}\left\{Xe^{\phi_x}e^{j\omega_x t}\right\}\)

Given that all the signals in the system at ACSS will be oscillating at the same frequency, all the signals will have the $$e^{j\omega_x t}$$ part and thus the only unique portions will be the amplitude $$X$$ and the phase $$\phi_x$$. Thus, we only need to keep track of those items, and thus, the phasor is born! The phasor $$\mathbb{X}$$ is a complex number specifically storing the amplitude $$X$$ and the phase $$\phi_x$$ for a single frequency sinusoid. There are several ways to represent these two pieces of information; the two most common are:

\( \mathbb{X}=Xe^{j\phi_x}=X\angle \phi_x \)

For a phasor, you need to know the frequency of the system and then you can translate the magnitude and phase of the phasor to the amplitude and phase of a single-frequency sinusoid at that frequency.

Also, taking the $$n^{th}$$ derivative of $$Xe^{\phi_x}e^{j\omega_x t}$$ yields $$\left(j\omega\right)^nXe^{\phi_x}e^{j\omega_x t}$$ meaning we can replace $$\frac{d}{dt}$$ in the time domain with $$j\omega$$ in the frequency domain.

Impedance

Given that every voltage and every current in a system at ACSS will be at the same frequency, the magnitude and phase information can be stored in a phasor. For passive elements, the ratio of the voltage phasor to the current phasor (which can also be described as the magnitude ratio between the voltage and the current and the phase difference between the voltage and the current) is given by a complex number $$\mathbb{Z}$$. That is,

\( \mathbb{Z}=\frac{\mathbb{V}}{\mathbb{I}}=\frac{Ve^{j\phi_v}}{Ie^{j\phi_i}}=\frac{V}{I}e^{j(\phi_v-\phi_i)}=\frac{V}{I}\angle(\phi_v-\phi_i) \)

Reactive Element Impedances

Recall that every current and voltage in ACSS under the influence of same-frequency sinusoids will be $$x(t)=X\,\cos(\omega t+\phi_x)$$. That means the voltage can be represented as $$v(t)=V\,\cos(\omega t+\phi_v)$$ and the current can be represented as $$i(t)=I\,\cos(\omega t+\phi_i)$$. As phasors, $$\mathbb{V}=V\angle \phi_v$$ and $$\mathbb{I}=I\angle\phi_i$$.

Resistors

\( \begin{align*} v&=Ri\\ V\,\cos(\omega t+\phi_v)&=RI\,\cos(\omega t+\phi_i)\\ V&=IR, \phi_v=\phi_i\\ \mathbb{V}&=V\angle \phi_v=IR\angle\phi_i\\ \mathbb{Z}&=\frac{\mathbb{V}}{\mathbb{I}}=\frac{V\angle \phi_v}{I\angle \phi_i}=\frac{IR\angle\phi_i}{I\angle\phi_i}=R \end{align*} \)

Inductors

Note: the following trig mixes radians and degrees. We, the humans, can understand that...just don't put this in a calculator directly

\( \begin{align*} v&=L\frac{di}{dt}\\ V\,\cos(\omega t+\phi_v)&=-L\omega I\,\sin(\omega t+\phi_i)=L\omega I\,\cos(\omega t+\phi_i+90^{\circ})\\ V&=\omega LI, \phi_v=\phi_i+90^{\circ}\\ \mathbb{V}&=V\angle \phi_v=\omega L I\angle(\phi_i+90^{\circ})\\ \mathbb{Z}&=\frac{\mathbb{V}}{\mathbb{I}}=\frac{V\angle \phi_v}{I\angle \phi_i}=\frac{\omega L I\angle(\phi_i+90^{\circ})}{I\angle\phi_i}=\omega L\angle 90^{\circ}=j\omega L \end{align*} \)

Capacitors

Same trig warning...

\( \begin{align*} i&=C\frac{dv}{dt}\\ I\,\cos(\omega t+\phi_i)&=-C\omega V\,\sin(\omega t+\phi_v)=C\omega V\,\cos(\omega t+\phi_v+90^{\circ})\\ I&=\omega CV, \phi_i=\phi_v+90^{\circ}\\ \mathbb{I}&=I\angle \phi_i=\omega C V\angle(\phi_v+90^{\circ})\\ \mathbb{Z}&=\frac{\mathbb{V}}{\mathbb{I}}=\frac{V\angle \phi_v}{I\angle \phi_i}=\frac{V\angle \phi_v}{\omega C V\angle(\phi_v+90^{\circ})}=\frac{1}{\omega C}\angle -90^{\circ}=\frac{1}{j\omega C} \end{align*} \)

Solving Circuits

At this point, if we assume that all the independent sources in a circuit are single-frequency sinusoids, and if there are resistors in the right places to make sure that perturbations get dissipated, then we can re-write and solve the circuit in the "frequency" domain by replacing all the source and measurement labels with phasors and replacing all the reactive elements with their impedances. The advantage here is that all the techniques that you can use with purely resistive networks (including using equivalent resistances, current division, voltage division, and superposition) will now work with the impedances instead of the resistances. All the same caveats apply as well - for instance, with superposition, you are still not allowed to take out any dependent sources. You will then be able to get output phasors as functions of reactive elements. These may end up as functions of frequency $$\omega$$ if you are solving symbolically or they may end up as a complex number if you are using numbers. The magnitude of the complex number will be the amplitude of the resulting sinusoid and the phase angle of the complex number will be the phase of the resulting sinusoid.

Translating

To solve ACSS, you will end up translating the information of a single-frequency sinusoid into a phasor and then later the information in a phasor to a single-frequency sinusoid.

Example

Imagine that the input for a particular circuit is an independent voltage source with $$v_s(t)=3\,\cos(2t+45^{\circ})$$. The frequency for every ACSS signal for this source will be $$\omega=2$$ rad/s and then the numerical version of the phasor would be $$\mathbb{V}_s=3\angle 45^{\circ}$$. Note that if you are presented the signal with sin instead of cos, you should convert it to cos by subtracting 90$$^{\circ}$$ from the angle. That is, if $$v(t)=6\,\sin(5t+20^{\circ})$$, you would first need to write $$v(t)=6\,\cos(5t-70^{\circ})$$ and then $$\mathbb{V}=6\angle -70^{\circ}$$.

Similarly, once you have a phasor result for an output, you will use its magnitude and angle as the amplitude and phase angle for a cosine at the frequency of the original independent source. For example, if you had some $$\mathbb{Y}_o=1.5\angle 25^{\circ}$$ for some voltage $$y(t)$$ based on the source $$v_s(t)$$ above, then $$y_o(t)=1.5\,\cos(2t+25^{\circ})$$.

Example

For a complete example, see https://cad.eecs.umich.edu/ and specifically Example 7.6. My version of that is available Box [[Category:Controls]