# Phasors Demystified

Phasors Demystified Swapnil Sunil Jain Aug 7, 2006

Phasors Demystified

Suppose the following integro-differential equation is given in the time-domain11This is not the most general integro-differential equation but it has all the basic elements required for this discussion and hence the reader can easily extend this discussion for the more generalized case.:

 $\displaystyle C_{1}\frac{d}{dt}y(t)+C_{2}\int_{-\infty}^{t}y(t)dt+C_{3}y(t)=x(t)$ (1)

where $y(t)$ and $x(t)$ are sinusoidal waveforms of the same frequency. Now, since $y(t)$ is a sinusoidal function it can be represented as $A_{y}\cos(\omega t+\phi_{y})$ and, similarly, x(t) can be represented as $A_{x}\cos(\omega t+\phi_{x})$. Furthermore, using the properties of complex numbers we can write

 $\displaystyle y(t)=A_{y}\cos(\omega t+\phi_{y})=\Re(A_{y}e^{j\phi_{y}}e^{j% \omega t})$ $\displaystyle x(t)=A_{x}\cos(\omega t+\phi_{x})=\Re(A_{x}e^{j\phi_{x}}e^{j% \omega t})$

Now if we define the quantities $\tilde{Y}$ as $A_{y}e^{j\phi_{y}}$ and $\tilde{X}$ as $A_{x}e^{j\phi_{x}}$ (where $\tilde{Y}$ and $\tilde{X}$ are called phasors), then we can write the above expression in a more compact form as

 $\displaystyle y(t)=\Re(\tilde{Y}e^{j\omega t})$ $\displaystyle x(t)=\Re(\tilde{X}e^{j\omega t})$

Now, using the above expression for $y(t)$ and $x(t)$ we can rewrite our original integro-differential equation as

 $\displaystyle C_{1}\frac{d}{dt}\Re[\tilde{Y}e^{j\omega t}]+C_{2}\int_{-\infty}% ^{t}\Re[\tilde{Y}e^{j\omega t}]dt+C_{3}\Re[\tilde{Y}e^{j\omega t}]=\Re[\tilde{% X}e^{j\omega t}]$

Moving the derivative and the integral inside the $\Re$ operator we get

 $\displaystyle C_{1}\Re\Big{[}\frac{d}{dt}\tilde{Y}e^{j\omega t}\Big{]}+C_{2}% \Re\Bigg{[}\int_{-\infty}^{t}\tilde{Y}e^{j\omega t}dt\Bigg{]}+C_{3}\Re[\tilde{% Y}e^{j\omega t}]=\Re[\tilde{X}e^{j\omega t}]$ $\displaystyle\Rightarrow C_{1}\Re[\tilde{Y}j\omega e^{j\omega t}]+C_{2}\Re[% \tilde{Y}\frac{e^{j\omega t}}{j\omega}]+C_{3}\Re[\tilde{Y}e^{j\omega t}]=\Re[% \tilde{X}e^{j\omega t}]$ $\displaystyle\Rightarrow\Re[\tilde{Y}j\omega C_{1}e^{j\omega t}]+\Re[\frac{% \tilde{Y}C_{2}}{j\omega}e^{j\omega t}]+\Re[\tilde{Y}C_{3}e^{j\omega t}]-\Re[% \tilde{X}e^{j\omega t}]=0$
 $\displaystyle\Rightarrow\Re\Big{[}\tilde{Y}j\omega C_{1}e^{j\omega t}+\frac{% \tilde{Y}C_{2}}{j\omega}e^{j\omega t}+\tilde{Y}C_{3}e^{j\omega t}-\tilde{X}e^{% j\omega t}\Big{]}=0$ $\displaystyle\Rightarrow\Re\Big{[}e^{j\omega t}\Big{(}\tilde{Y}j\omega C_{1}+% \frac{\tilde{Y}C_{2}}{j\omega}+\tilde{Y}C_{3}-\tilde{X}\Big{)}\Big{]}=\Re+j% \Im$
 $\displaystyle\Rightarrow e^{j\omega t}\Big{(}\tilde{Y}j\omega C_{1}+\frac{% \tilde{Y}C_{2}}{j\omega}+\tilde{Y}C_{3}-\tilde{X}\Big{)}=0$
 $\displaystyle\Rightarrow\tilde{Y}j\omega C_{1}+\frac{\tilde{Y}C_{2}}{j\omega}+% \tilde{Y}C_{3}-\tilde{X}=0\quad\mbox{(for t\neq-\infty)}$ (2)

Hence, we have now arrive at the phasor domain expression for (1). You can see from the analysis above that we aren’t adding or losing any information when we transform equation (1) into the ”phasor domain” and arrive at equation (2). One can easily get to (2) by using simple algebraic properties of real and complex numbers   . Furthermore, since (2) can be derived readily from (1), in practice we don’t even bother to do all the intermediate steps and just skip straight to (2) calling this ”skipping of steps” as ”transforming the equation into the phasor domain.”

We can now continue the analysis even further and solve for $y(t)$ which is the whole motivation behind the use of phasors. Solving for $\tilde{Y}$ in (2) we get

 $\displaystyle\Rightarrow\tilde{Y}=\frac{\tilde{X}}{\Big{(}j\omega C_{1}+\frac{% C_{2}}{j\omega}+C_{3}\Big{)}}$

Now, since

 $\displaystyle y(t)=\Re[\tilde{Y}e^{j\omega t}]$

we have

 $\displaystyle y(t)=\Re\Bigg{[}\frac{\tilde{X}e^{j\omega t}}{\Big{(}j\omega C_{% 1}+\frac{C_{2}}{j\omega}+C_{3}\Big{)}}\Bigg{]}$ $\displaystyle\Rightarrow y(t)=\Re\Bigg{[}\frac{A_{x}e^{j\phi_{x}}e^{j\omega t}% }{\Big{(}j\omega C_{1}+\frac{C_{2}}{j\omega}+C_{3}\Big{)}}\Bigg{]}$

The above equation makes sense because you can see that the output y(t) is given completely in terms of the variables $A_{x}$ and $\phi_{x}$ (which depend only on the input sinusoid $x(t)$) and the constants $C_{1}$, $C_{2}$ and $C_{3}$—as we expected! So by converting the integro-differential equation (1) into the phasor domain (2), all the complicated integration and differentiation  operations become simple manipulation of complex variables—which is why phasors are so useful!

Title Phasors Demystified PhasorsDemystified1 2013-03-11 19:26:47 2013-03-11 19:26:47 swapnizzle (13346) (0) 1 swapnizzle (0) Definition