Day 3 - Phase Vectors aka Phasors

Systems that consist of only resistors, capacitors, inductors, and wires can be described using linear differential equations.

In these problems we will assume that there were no initial conditions or we have waited a long time such that the transients die out.

The response of a linear system to the sum of scaled sinusoidal inputs is a sum of the same frequencies as the inputs (magnitudes and phases may not be the same).

Therefore the output is represented as scaled, time-shifted versions of the same sinusoids as the ones that comprise the input.

The steady-state output is:

Where A is the maximum amplitude and B is the phase angle.

Now, if we have the source voltage Vm*cos(wt  + theta), we can compute the phasor:

If we are dealing with sin, then we need to subtract 90 (or pi/2) since sin(x) = cos(x - pi/2).

Inverse Phasor Transform
Going form a phasor to a time domain function

When solving an equation with the source voltage we just need to deal with the real parts.

Day 2 - Sinusoidal Signals (Part 1)

Quick review
A*cos(wt + Θ  - Θ)
+ Θ [shifts to the left]
- Θ [shifts to the right]


w = 2πf
w is in radians/sec
f is in Hertz (cycles/sec)   --- f = 1/T where T is the period in seconds


cos(x - π/2) = sin(x)
sin(x + π/2) = cos(x)




Root Mean Square
For a sinusoidal source, the root mean square is the value of a DC source that has the same average power as the sinusoid. The RMS is computed by

1. Take the square of the signal
2. Take the average over one period
3. Take the square root

For sinusoidal source:

Circuit Response to a Source
Supposing we had a circuit like the one shown below where the voltage source followed the equation: Vs = Vm*cos(wt + Θ)

When computing the current, there are two parts that need to be taken into account

• Initial Conditions on the circuit
• Includes the initial voltages across capacitors and initial current in the inductors 
• Called the "homogeneous", "no inputs", or "natural response"
• If it is dissipating energy, it can die out. If it reaches 0, then it is a transient signal
• Input to the system (voltage source, Vs)
• Called the "non-homogeneous", "input driven", "particular", or "forced response"
• If energy is dissipating, the current will remain finite as long as Vs is finite

We now need to solve Kirchoff's Voltage Law:

i(t) is comprised of the two parts mentioned above:

We can solve for:

   where C is an arbitrary constant


The sinusoid for the particular solution (i_p), has the same frequency, but possibly different phase and magnitude

After the transients die off, the steady state output is just the particular solution (which is due to the inputs).

So, the current is

Now, we need to solve for C

So out total solution is:

Day 1 - Imaginary Number

Today we focused on imaginary numbers.

If we break up a complex number into its two components we can make a new plot. Let the imaginary part be the y-axis and let the real part be the y-axis.

We can write the complex number z in terms of a (real part), b (imaginary part), r (the hypotenuse from the origin to Z), and theta (radians from the x-axis to r):

• a+jb Rectangular Form
• re^(j*theta) Polar Form

Next we discussed Euler's Equality: e^(j*theta) = cos(theta) + j*sin(theta)  use to convert from polar form to rectangular form

Therefore when we set theta to pi (cosine is -1 and sine is 0), e^j*pi = -1