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: