# Circuit Theory/TF Objectives

Thevenin/Norton impedance completely characterizes a circuit from a
load\'s point of view. Differential equations completely describe a
circuit. The impulse response of a circuit is all that is needed to use
the convolution integral. Phasors and complex frequencies completely
describe a circuit. And even while computing the transient response,
complex frequencies associated with the steady state sinusoidal response
appeared. Are they all related? Yes.

The goal is to understand all the above as a transfer function. Why?
Most engineering uses this concept. Here are some electrical
applications:

-   Analog Computers \... (modeling
    before digital computers)
-   Signal Processing "wikilink") \...
    (wireless, cable communication)
-   Control theory \... (motors don\'t
    die and don\'t spin out of control when the load changes)
-   Linear time-invariant system
    theory \... (general math model
    which assumes the values of resistors, inductors and capacitors
    don\'t vary over time, temperature, bias .. but provides a good
    starting point for establishing expectations and experimenting)

The goal is to introduce transfer functions in the context of
electronics.

## Transfer Function Definition

A **Transfer Function** is the ratio of the output of a system to the
input of a system. If we have an input function of *X(s)*, and an output
function *Y(s)*, we define the transfer function *H(s)* to be:

$$H(s) = {Y(s) \over X(s)}$$

## Four RLC Transfer Functions

In the context of RLC circuits, transfer functions are in a
phasor/complex frequency/laplace domain concept. They are not a time
domain concept. They are independent of the forcing function/source.
There are four possible RLC transfer patterns:

  $\frac{output}{input}$           Example Circuit                                                    Transfer Function
  -------------------------------- ------------------------------------------------------------------ --------------------------------------
  $\frac{V_{out}(s)}{I_{in}(s)}$   !impedance                                $H(s) = R + Ls$
  $\frac{I_{out}(s)}{V_{in}(s)}$   !admittance   $H(s) = \frac{1}{R} + Cs$
  $\frac{V_{out}(s)}{V_{in}(s)}$   !voltage divider                    $H(s) = \frac{Ls}{R + Ls}$
  $\frac{I_{out}(s)}{I_{in}(s)}$   !current divider                    $H(s) = \frac{Cs}{\frac{1}{R} + Cs}$

## Applications

The transfer function has many applications as we will soon see. But
it\'s immediate benefit is eliminating the need to find the differential
equation. Reviewing previous examples:

+----------------------+----------------------+----------------------+
| Example              | Differential         | Transfer Function    |
|                      | Equation             |                      |
+======================+======================+======================+
| convolution         | $$v(t) = L{d i(t)    | admittance$$ | ittance "wikilink"): |
| p                    |                      |                      |
| roblem = \frac{\mathbb{I} |
| ral#Example_Impulse_ |                      | (s)}{\mathbb{V}(s)}  |
| Response "wikilink") |                      | = \frac{1}{sL + R}$$ |
|                      |                      |                      |
| <File:Exa            |                      |                      |
| mple40a.png%7CGiven> |                      |                      |
| voltage source, find |                      |                      |
| total current        |                      |                      |
+----------------------+----------------------+----------------------+
| Second Order,       | $I_s = {d^2 i_o \    | current divider:     |
| Source               | over dt^2} + 2 {d i_ |                      |
| Excitem              | o \over dt} + 2 i_o$ | $$H(s) = \fra        |
| ent}{\ |
| 2Source_Excitement/E |                      | mathbb{I}_s(s)} = \f |
| xample46 "wikilink") |                      | rac{\frac{1}{R + Ls} |
|                      |                      | }{\frac{1}{R + Ls} + |
| <File:E              |                      |  \frac{1}{R} + Cs}$$ |
| xample46.png%7CFind> |                      |                      |
| i~o~ giving Is       |                      | $$H(s) = \frac{      |
|                      |                      | \mathbb{I}_o(s)}{\ma |
|                      |                      | thbb{I}_s(s)} = \fra |
|                      |                      | c{1}{s^2 + 2s + 2}$$ |
|                      |                      |                      |
|                      |                      |                      |
|                      |                      | R :=1; L :=1; C :=1; |
|                      |                      |     simplify(        |
|                      |                      | (1/(R + L*s))/(1/(R  |
|                      |                      | + L*s) + 1/R + C*s)) |
+----------------------+----------------------+----------------------+
| Second Order,       | $V_                  | Voltage Divider      |
| Source               | s(t) = L{d (\frac{V_ | $H(s) = \frac{\math  |
| Excitement \over dt}  | bb{V}_{cr}(s)}{\math |
| Theory/2Source_Excit | + LC {d^2 (V_{CR} \o | bb{V}_s(s)} = \frac{ |
| ement/Example45#Homo | ver dt^2}) + V_{CR}$ | \frac{1}{\frac{1}{R} |
| geneous.2FTransient_ |                      |  + Cs}}{\frac{1}{\fr |
| Solution "wikilink") |                      | ac{1}{R} + Cs}+Ls}$\ |
|                      |                      | $H(                  |
| <File:Exa            |                      | s) = \frac{1}{CLs^2  |
| mple45B.png%7CGiven> |                      | + \frac{Ls}{R} + 1}$ |
| V_s, find V_cr       |                      |                      |
+----------------------+----------------------+----------------------+

## Interpreting Complex Roots

After finding the transfer function, the next step is finding the time
constant or the complex frequency roots and from these guess the form of
the solution.

The homogeneous solution is found by setting the denominator of the
transfer function to zero. Factoring the denominator and finding the
complex frequency roots is called finding \"zeros.\"

The numerator (which was one in all the above examples) may have a
complex frequency polynomial (in complicated circuits) that can also be
set equal to zero. The roots are called \"poles.\" Poles influence the
circuit\'s response to different frequencies rather than the time domain
response.

## Summarizing Time

This ends the normal circuit theory course. Time domain analysis could
continue and this is hinted at in the following example. Before WWII it
was important to continue in order to model physical systems with analog
computers. But today, digital computers can model much more accurately.
So stopping time analysis here leaves further study for a specialized
class. Let\'s summarize what we have learned about circuits in the time
domain.

-   Superposition allows replacing multiple sources with one source at a
    time (zeroing the others) and then adding the results of all
    sources.
-   Impedance, admittance, voltage divider and current divider formulas
    lead directly to the differential equations \... which is the same
    math for phasor/complex frequency and the Laplace domain.
-   The Convolution integral eliminates the need to transform driving
    functions (particularly sinusoidal) into the phasor/complex
    frequency/Laplace domain. Instead the source is replaced with 1 volt
    or 1 amp, and the particular solution turns into a DC final
    condition (t = ∞) placed on the homogeneous solution and appears
    with in the differential equation constant.

