ECE 311: Circuits II

Circuits I Recap
Example of Mesh Analysis
$V_{I1}-V_1+R1I{R1}=0$
$V_{I1}=V_1-R1I{R1}$
$V_{I1}=-2-4*(-1)=2V$
$I_{R1}+I1=0$
$I{R1}=-I_1=-1A$

Phaser Analysis

Capacitor DC equ is an open circuit
Inductor DC equ is a closed circuit

Calculate $v{out}(t)$
$-v{in}+R*i+L\frac{di}{dt}+v_c=0$
$i=i_c=C\frac{dvc}{dt}=C\frac{dv{out}}{dt}$

$v{out}(t)$ can be calculated solving the differential equation:
$LC \frac{d^2 v{out}}{dt^2} +RC \frac{dv{out}}{dt}+v{out}=v_{in}$

In circuits II, we will learn how to solve for $v_out$ without having to solve the differential equation.

Capacitors and inductors are store like memory of previous voltages
$v_c(t)=1/C\int{v_c(t)}dt$
$v_L(t)=1/L\int{v_L(t)}dt$

These equations naturally lead to second-order differential equations.

Resistors

$v(t) = Ri(t)$

If the voltage is sinusoidal, the current will be sinusoidal in the same phase and frequency.

The current will follow the same shape as the voltage.

Inductors

$v(t) = L\frac{di(t)}{dt}$

Current lags voltage by 90 degrees. This is because it is based on the derivative of the voltage so when the voltage is at its extremes, the derivative is 0, so the current will be 0 at that time.

The current will always be some kind of sinusoidal function.

$sin(\omega t)=cos(\omega t-\frac{\pi}{2}$

Inductors react to sudden changes in current.

Capacitors

The voltage of a capacity is dependent on the derivative of the voltage over time.

$i(t) = C \frac{dv(t)}{dt}$

The current will therefore lead the voltage by 90 degrees.

Capacitors react to sudden changes in voltage.

Euler Formula

You can use a Taylor Series to describe a function based on one data point.

For example: $e^{jx}$ at $x_0=0$. What is $e^{j3}$?
$e$ is special because it is the only function where it is its own derivative.

$e^{jx} = 1 + jx + \frac{(jx)^2}{2!} + \frac{(jx)^3}{3!} + ...$

If we expand the Taylor Series of sin and cos, we get:
$cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ...$
$sin(x) = 1 - \frac{x^3}{3!} + \frac{x^5}{5!} - ...$

Therefore $e^{jx} = cos(x) + jsin(x)$