# Electronics/Phasors

## Phasors

Figure One: Rectangular Notation
Figure Two: Polar Notation

Phasors provide a simple means of analyzing linear circuits. At the heart of phasor analysis lies Euler's formula:

${\displaystyle Ae^{j({\omega }t+\theta )}=A\cos({\omega }t+\theta )+jA\sin({\omega }t+\theta )\,}$[1]

A complex exponential can also be expressed as

${\displaystyle Ae^{j({\omega }t+\theta )}=Ae^{j\theta }\cdot e^{j\omega t}}$

${\displaystyle {\tilde {A}}=Ae^{j\theta }}$ is called a phasor. It contains information about the magnitude and phase of a sinusoidal signal, but not the frequency or time. This simplifies use in circuit analysis, since most of the time, all quantities in the circuit will have the same frequency. (For circuits with sources at different frequencies, the principle of superposition must be used.)

A shorthand phasor notation is: ${\displaystyle {\tilde {A}}=A\angle \theta }$

Note that this is simply a polar form, and can be converted to rectangular notation by (see figure one):

${\displaystyle {\tilde {A}}=Ae^{j\theta }=A\angle \theta =A\cos \theta +jA\sin \theta }$

and back again by (see figure two):

${\displaystyle {\tilde {A}}=X+jY={\sqrt {X^{2}+Y^{2}}}\angle {\tan ^{-1}{\frac {Y}{X}}}={\sqrt {X^{2}+Y^{2}}}e^{j\tan ^{-1}{\frac {Y}{X}}}}$. Let us understand how this process works, even though a phasor causes circuits to heat up steady flow in current will explain on a basis that disallows overheating. Since according to the formula the outcome will be there is a time in duration which the current usage stays the same.

## Sinusoidal Signals

To begin, we must first understand what sinusoidal signals are. Sinusoidal signals can be represented as

${\displaystyle A\cos({\omega }t+\theta )\!}$

where A is the amplitude, ${\displaystyle \omega }$ is the frequency in radians per second, and ${\displaystyle \theta }$ is the phase angle in degrees(phase shift). We can return to the sinusoidal signal by taking the real part of Euler's formula:

${\displaystyle A\cos({\omega }t+\theta )={\mathfrak {Re}}(Ae^{j({\omega }t+\theta )})}$

For the moment, consider single-frequency circuits. Every steady state current and voltage will have the same basic form:

${\displaystyle {\tilde {A_{i}}}e^{j{\omega }t}}$ where ${\displaystyle {\tilde {A_{i}}}}$ is a phasor. So we can "divide through" by ${\displaystyle e^{j{\omega }t}\!}$ to get phasor circuit equations. We can solve these equations for some phasor circuit quantity ${\displaystyle {\tilde {Y}}}$, multiply by ${\displaystyle e^{j{\omega }t}\!}$, and convert back to the sinusoidal form to find the time-domain sinusoidal steady-state solution.

## Example

We have three sinusoidal signals with the same frequency added together:

${\displaystyle v(t)=5\cos(200t+30^{\circ })+2\cos(200t+45^{\circ })+\cos(200t-60^{\circ })\!}$

In phasor notation, this is:

${\displaystyle {\tilde {V}}=5e^{j30^{\circ }}+2e^{j45^{\circ }}+e^{-j60^{\circ }}=5\angle 30+2\angle 45+\angle -60}$

We can combine these terms to get one phasor notation. This is done first by separating the real and imaginary components:

${\displaystyle {\text{X: }}5\cos(30^{\circ })+2\cos(45^{\circ })+\cos(-60^{\circ })\approx 6.2443\!}$
${\displaystyle {\text{Y: }}5\sin(30^{\circ })+2\sin(45^{\circ })+\sin(-60^{\circ })\approx 3.04819\!}$

The phasor notation can be written as:

${\displaystyle {\tilde {V}}=6.2443+j3.04819={\sqrt {6.2443^{2}+3.04819^{2}}}\angle {\tan ^{-1}{\frac {3.04819}{6.2443}}}}$
${\displaystyle {\tilde {V}}=6.94861\angle {26.0194}}$

Back to the time domain, we get the answer:

${\displaystyle v(t)=6.94861\cos(200t+26.0194^{\circ })}$

## Footnotes

• j is the Imaginary unit (${\displaystyle j^{2}=-1}$).
• In electrical engineering, the imaginary unit is symbolized by j rather than the symbol i because i is used to denote current in electrical engineering.
• The frequency of the wave, in Hz, is given by ${\displaystyle \omega /2\pi }$.