# 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:

$$V_s(t)= (1-e^{-\frac{t}{\tau}})\cdot f(t)$$ turn on

$$V_s(t)= ( e^{-\frac{t}{\tau}})\cdot f(t)$$ turn off The value of
$\tau$ (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 $\tau$. 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 $V(t)=V_m cos(\omega t + \phi)$ modeled steady state circuits
(never turned off and on).

Now let\'s consider $V(t) = V_m e^{\sigma t} cos(\omega t + \phi)$.
Sigma or $\sigma$ 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
$cos(\omega t + \phi) = \operatorname{Re}(e^{j(\omega t + \phi)})$ from
Euler\'s equation, we can start a derivation that shows almost nothing
changes in the phasor domain!

  expression                                                           notes
  -------------------------------------------------------------------- ------------------------------------------------------------------------------------------------------------------
  $V(t) = \operatorname{Re}(V_m e^{\sigma t}e^{j(\omega t + \phi}))$   substitution
  $V(t) = \operatorname{Re}(V_m e^{j\phi} e^{(\sigma+j\omega)t})$      reorganizing the exponents
  $s=\sigma + j\omega$                                                 defining the complex frequency $s$
  $V(t) = \operatorname{Re}(V_m e^{j\phi} e^{st})$                     after substitution there is no difference
  $\mathbb{V}=V_m e^{j\phi}$                                           Defining the phasor \... nothing different .. $s$ instead of $j\omega$ from derivatives and integrals
  $V(t) = \operatorname{Re}(\mathbb{V} e^{st})$                        Can we assume $\omega$ (angular frequency) and $s$ (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 $s$ at all.

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

What is interesting is that the complex frequency 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 $\sqrt{-1}$. The symbol $s$ contains a complex
number. The math programs all work internally in rectangular form.
Substituting numerically for $s$ with a complex number is simple:

$$s = \sigma + j\omega$$.

### complex frequency back to time domain

The unknown function(s) is/are going to look something like this:
$V(t) = \operatorname{Re}(\mathbb{V} e^{st})$ or
$I(t) = \operatorname{Re}(\mathbb{I} e^{st})$. 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 $s$ during the creation of phasors instead of $\omega$. Substitue
$s$ for $j\omega$ when replacing the derivative in the terminal
relations. Use $s=\sigma + j\omega$ when moving back into the time
domain.

The goal in the examples is to explore the similarities and differences
of the symbol $s$ as used in Complex Frequency analysis and in the
Laplace transform. The Laplace transform also uses the symbol $s$ 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 they use different
techniques for transforming the functions. Complex frequency transforms
functions like phasors and operators like Laplace transforms.

