# Circuit Theory/Complex Frequency

### Complex Frequency Why?

Phasor concepts can be extended to cover the turning off and on of circuits using exponentials. For example:

${\displaystyle V_{s}(t)=(1-e^{-{\frac {t}{\tau }}})\cdot f(t)}$ turn on
${\displaystyle V_{s}(t)=(e^{-{\frac {t}{\tau }}})\cdot f(t)}$ turn off

The value of ${\displaystyle \tau }$ (or the time constant) determines how quickly the circuit turns on and off. Adding some constants can determine when the circuit turns on and off.

Why not Laplace transforms? Laplace transforms can model infinite slopes for turning off and on circuits. We are not going to do that in this course. The real world can be modeled with ${\displaystyle \tau }$. Laplace transforms are simple, beautiful and truly amazing in that they can handle infinities and infinitesimals. Scientists and mathematicians are always interested in tools that help them work with the road blocks of infinities and infinitesimals. So far this course has been dealing with ideal inductors, resistors, wires, sources, etc. And if we were to deal with ideal turning off and on, we could not use phasors. We would have to use Laplace. So at this point an exception has been made.

Or one could just consider "Complex Frequencies" as exploring a more complex function in a way similar to phasors.

### Complex Frequency Derivation

Before ${\displaystyle V(t)=V_{m}cos(\omega t+\phi )}$ modeled steady state circuits (never turned off and on).

Now let's consider ${\displaystyle V(t)=V_{m}e^{\sigma t}cos(\omega t+\phi )}$. Sigma or ${\displaystyle \sigma }$ is usually negative, otherwise the circuit will explode. Splitting the charging problem into two problems and adding the results in the end is called "superposition" which usually works. So we can reduce both situations down to exploring the phasor math of this equation.

Substituting ${\displaystyle cos(\omega t+\phi )=\operatorname {Re} (e^{j(\omega t+\phi )})}$ from Euler's equation, we can start a derivation that shows almost nothing changes in the phasor domain!

expression notes
${\displaystyle V(t)=\operatorname {Re} (V_{m}e^{\sigma t}e^{j(\omega t+\phi }))}$ substitution
${\displaystyle V(t)=\operatorname {Re} (V_{m}e^{j\phi }e^{(\sigma +j\omega )t})}$ reorganizing the exponents
${\displaystyle s=\sigma +j\omega }$ defining the complex frequency ${\displaystyle s}$
${\displaystyle V(t)=\operatorname {Re} (V_{m}e^{j\phi }e^{st})}$ after substitution there is no difference
${\displaystyle \mathbb {V} =V_{m}e^{j\phi }}$ Defining the phasor ... nothing different .. ${\displaystyle s}$ instead of ${\displaystyle j\omega }$ from derivatives and integrals
${\displaystyle V(t)=\operatorname {Re} (\mathbb {V} e^{st})}$ Can we assume ${\displaystyle \omega }$ (angular frequency) and ${\displaystyle s}$ (complex frequency) can be treated the same? Yes ... almost.

### time domain to complex frequency domain

The above process shows that converting v(t) and i(t) to phasors is the exact same process. No need to think about ${\displaystyle s}$ at all.

Terminal Equation Phasor Domain Complex Frequency domain
${\displaystyle V=RI}$ ${\displaystyle \mathbb {V} =R\mathbb {I} }$ ${\displaystyle \mathbb {V} =R\mathbb {I} }$
${\displaystyle V=L{\frac {d}{dt}}I}$ ${\displaystyle \mathbb {V} =jwL\mathbb {I} }$ ${\displaystyle \mathbb {V} =sL\mathbb {I} }$
${\displaystyle I=C{\frac {d}{dt}}V}$ ${\displaystyle \mathbb {I} =jwC\mathbb {V} }$ ${\displaystyle \mathbb {I} =sC\mathbb {V} }$

What is interesting is that the complex fequency operators are the same as the Laplace transform operators. This is why they both use the term "complex frequency" to describe s.

What makes the Laplace transform operators so complicated is the different way the transform the functions!

However, as we will see later (in transfer functions), when turning a circuit into a black box, hooking a driving source up to it and then describing the output, the driving function transform doesn't matter. Just the derivative and integral operator transforms matter.

### math in the Complex Frequency Domain

Math in the Complex Frequency Domain, with symbols, does not involve keeping track of the ${\displaystyle {\sqrt {-1}}}$. The symbol ${\displaystyle s}$ contains a complex number. The math programs all work internally in rectangular form. Substituting numerically for ${\displaystyle s}$ with a complex number is simple:

${\displaystyle s=\sigma +j\omega }$.

### complex frequency back to time domain

The unknown function(s) is/are going to look something like this: ${\displaystyle V(t)=\operatorname {Re} (\mathbb {V} e^{st})}$ or ${\displaystyle I(t)=\operatorname {Re} (\mathbb {I} e^{st})}$. Substituting for s in the time domain means working with exponents a little. Getting back into the time domain will not be as simple as putting the unknown function in polar form.

### Summary

Find ${\displaystyle s}$ during the creation of phasors instead of ${\displaystyle \omega }$. Substitue ${\displaystyle s}$ for ${\displaystyle j\omega }$ when replacing the derivative in the terminal relations. Use ${\displaystyle s=\sigma +j\omega }$ when moving back into the time domain.

The goal in the examples is to explore the similarities and differences of the symbol ${\displaystyle s}$ as used in Complex Frequency analysis and in the Laplace transform. The Laplace transform also uses the symbol ${\displaystyle s}$ and also calls it the "complex frequency." But are they the same thing? No. They both use the same complex frequency techniques for transforming the differential and integral operators. But the use different techniques for transforming the functions. Complex frequency transforms functions like phasors and operators like Laplace transforms.