Phasor analysis

Phasors are short-hand polar-form complex numbers used in AC circuit analysis. Recall from our understanding of first- and second-order transient circuits that ODEs are fully featured but difficult to impossible to evaluate for combinations of many inductors and capacitors.

Related ideas to know:

• Impedance
• Admittance

Phasor analysis only works with sinusoidal steady-state signals, not general signals. This motivates our exploration of the Laplace transform.

Solving circuits

For a capacitor, the phasor element equations are:

$${\text{I}}_C=j\omega C{\text{V}}_C$$ $${\text{V}}_C=\frac{1}{j\omega C}{\text{I}}_C$$

For an inductor, the element equations are:

$$V_L=j\omega L{\text{I}}_L$$

Kirchhoff’s laws, dividers, superposition still hold in the phasor domain (maybe with extra constraints). Reference directions still (technically) hold in the phasor domain.

Definition

From the page on sinusoids, we have:

$$v(t)=V_A\cos (\omega t+\phi )$$

where $V_A$ is the amplitude, $\omega$ is the angular frequency, and $\phi$ is the phase change. From Euler’s formula, we can rewrite as the real part of a complex number:

$$v(t)=\text{Re}(V_A{e}^{j\phi }{e}^{j\omega t})$$

Then we define the phasor associated with the above sinusoid as:

$$\text{V}=V_A{e}^{j\phi }=V_A\angle \phi$$

Useful representations

Digitally, we express phasors with a bold. On paper, we write a bar above or below it (haven’t found a consistent explanation). Recall that since $\sin$ lags $\cos$ by 90 degrees:

$$A\sin (\omega t\pm \theta )=A\angle \pm \theta -90°$$

In standard form, the phase should be $\phi \in [-\pi ,\pi ]$, so we can shift accordingly to meet those restrictions:

$$V_A\cos (\omega t+\phi )=V_A\cos (\omega t+\phi \pm 180°)$$

Importantly, some elements may have different phase shifts. If $\theta ={\theta }_1$, then they are in phase. If $\theta >{\theta }_2$, then ${\theta }_2$ lags $\theta$ by $\theta -{\theta }_2$. And so on. A phasor may also lead another phasor.

Properties

Phasors are additive, so the sum of sinusoids is the sum of the phasors:

$$\text{V}=\sum {\text{V}}_i$$

Another key property is that multiplying a phasor by $j\omega$ is equivalent to differentiating the corresponding sinusoid. If $v(t)=\text{V}$, then:

$$v^{\prime }(t)=j\omega \text{V}$$