3 - doing algebra with rotating arrows#2 - your introduction to ac circuit analysis

introduction

we learned the basics of phasor algebra in the last post. now, we’re going to take it a step further and learn how to take integrals and derivatives of phasors. this is where things get really interesting, because it opens up a whole new world of possibilities for analyzing ac circuits.

derivative of a sinusoid in phasor form

let’s start with the derivative of a sinusoid in phasor form. if we have a sinusoid like this:

$v(t) = \mathbb{R}\{V_m e^{j(\omega t + \phi)}\} = V_m \cos(\omega t + \phi)$

$\vec{V} = V_me^{j\phi} = V_m\angle\phi$

then the derivative of the sinusoid is:

$j\omega \vec{V}$

proof

$\frac{d}{dt}[v(t)] = -\omega V_m \sin(\omega t + \phi) = \omega V_m \cos(\omega t + \phi + 90^{\circ})$

$\frac{d}{dt}[v(t)] = \mathbb{R}\{\omega V_me^{j(\omega t + \phi + 90^{\circ})}\}$

$\frac{d}{dt}[v(t)] = \mathbb{R}\{\omega V_me^{j\omega t}e^{j\phi}e^{j90^{\circ}}\}$

and we know that $\vec{V} = V_me^{j\phi}$ is the phasor representation of the sinusoid, so we can write:

$\frac{d}{dt}[v(t)] = \mathbb{R}\{\omega e^{j\omega t}e^{j90^{\circ}}\vec{V}\}$

$\frac{d}{dt}[v(t)] = \mathbb{R}\{j\omega \vec{V}e^{j\omega t}\}$

and if we convert this equation to phasor domain, we get:

$\frac{d}{dt}[v(t)] \Leftrightarrow j\omega \vec{V}$

integral of a sinusoid in phasor form

again, if we have a sinusoid like this:

$v(t) = \mathbb{R}\{V_m e^{j(\omega t + \phi)}\} = V_m \cos(\omega t + \phi)$

$\vec{V} = V_me^{j\phi} = V_m\angle\phi$

then the integral of the sinusoid is:

$\frac{\vec{V}}{j\omega}$

proof

$\int v(t) dt = V_m \int \cos(\omega t + \phi) dt$

$\int v(t) dt = \frac{V_m}{\omega} \sin(\omega t + \phi)$

$\int v(t) dt = \frac{V_m}{\omega} \cos(\omega t + \phi - 90^{\circ})$

$\int v(t) dt = \mathbb{R}\{\frac{V_m}{\omega}e^{j(\omega t + \phi - 90^{\circ})}\}$

$\int v(t) dt = \mathbb{R}\{\frac{V_m}{\omega}e^{j\omega}e^{j\phi}e^{-j90^{\circ}}\}$

and we know that $\vec{V} = V_me^{j\phi}$ is the phasor representation of the sinusoid, so we can write:

$\int v(t) dt = \mathbb{R}\{\frac{1}{\omega}\vec{V}e^{j\omega}e^{-j90^{\circ}}\}$

$\int v(t) dt = \mathbb{R}\{\frac{1}{j\omega}\vec{V}e^{j\omega}\}$

and if we convert this equation to phasor domain, we get:

$\int v(t) dt \Leftrightarrow \frac{1}{j\omega}\vec{V}$

summary

the derivative of a sinusoid in phasor form means multiplying the phasor by $j\omega$.

$\frac{d}{dt}[v(t)] \Leftrightarrow j\omega \vec{V}$

the integral of a sinusoid in phasor form means dividing the phasor by $j\omega$.

$\int v(t) dt \Leftrightarrow \frac{1}{j\omega}\vec{V}$

this is a powerful tool for analyzing ac circuits, and it makes it much easier to work with sinusoids in phasor form.