6/2/16 - Frequency Filters

For our last class meeting, we start off with the requirements of the course. One of the concepts is the Bode plot. However, the Bode plot was useful particularly in the absence of mathematical computation software and devices such as MATLAB. It takes a complex curve and simplifies it into mainly linear and constant regions.

Next we dive into RC filters. We can solve the for the bandwidth of a RC filter by a quadratic equation with the roots at the first two equations. The bandwidth is defined as the difference between the two frequencies and if the function only has one root, then we have a bandwidth of 0.

We also have the quality factor, which is defined as the cutoff frequency over the bandwidth and we can find it multiple ways.

Here we derive the solutions for the quadratic equation.

Now given a example circuit, we are told to find the two bandwidth frequencies, the bandwidth, and the quality factor. We then solved for the time functions for current at the two frequencies, we see they have the same magnitude but differ by a 90 degree phase shift.

We also included an inspirational note of: "use your jimagination~..."

A close up of the magnitude. Reasons are unknown for this close up.

Here are all the equations we would use for solving RC and RL filters for series or parallel circuits.

Lastly, we are told to design a simple band pass filter to tune into an AM radio station of 640kHz. We are limited by a bandwidth of 5kHz so we don't pick up the signals of other stations.

Given a 8kOhm resistor, we solve for the inductance and capacitance values to allow for us to tune into that frequency and stay within our bandwidth.

Lastly, we are given another circuit to find the potential across the capacitor. We solve for the transfer function using nodal analysis and conclude that this is a low pass filter with the peak value at 0Hz. We can see as the function goes to infinity, the gain reduces to 0.

Plotting the function with a log plot, we can see that it has its peak just around 0 and as the axes go to infinity, the gain drops to 0.

5/26/16 - Transfer Functions

Today we learn about transfer functions, which are functions dependent on frequency.

It can be expressed as gain, impedance, or admittance. With the units of potential over potential or current or current magnitude for gain, or potential over current or visa versa for impedance or admittance.

We do a simple current divider equation to find the ratio of the current input versus output.

Replacing the jomgas (j * omega) with s, we can find the transfer function and then solve for poles and zeros.

Another problem were we use nodal analysis to solve for the gain as a transfer function. We find there is a zero at 0 and a pole at -1000.

Plugging this into a log plot, we can see clearly see where the pole is, as well as the zero at 0.

Another circuit, this time the components are in series, we use a voltage divider to find the transfer function and solve for zeros and poles.

Since the poles are at -100 and -10^6, two separate log plots were used to see the poles and zeros.

For our lab exercise, we hook up a resistor in series with a capacitor and resistor in parallel and are going to measure the voltage across the the resistor in parallel.

Here the circuit set up.

First we fed it the sum of two different sinusoidal signals and observe its output. The blue is the input signal and the yellow is the measured signal. We can see that the yellow is slightly out of phase due the jimaginary component of the capacitor and the gain is roughly 1/4.

The scale of the graphs to show the gain of 1/4.

Our calculations for the output.

Next we feed different frequencies to the circuit.

V = 20[sin(1000πt) + sin(2000πt) + sin(20,000πt)]

Here we calculate our expected voltage from the output.

For 1000Hz.

For 10000Hz.

Lastly, we feed a sweep function into the circuit. A sweep function is simply a sinusoidal signal that increases in frequency over time and repeats itself.

We can see the input signal versus the output signal and we see that as the frequency increases, the less impedance the capacitor has and more potential is across the capacitor than the the resistor,

Our scales for the above graph.

5/24/16 - AC Power

Today we start with the discussion of average power. We can say we find the average power using an integral divided by the period of that integral. We say this method useless to use, however, as the average value of a sinusoidal function is zero.

We solve for the function given to us on a graph. We know the average value is V_p from the middle of the sine function.

Now we're introduced to the concept of the root mean squared, or RMS value. We define the RMS value to be just the amplitude divided by root 2.

Looking back at the previous class meeting and our definition for power of an AC circuit, we can see that the power is simply the product of the RMS values of voltage and current.

Now we're into AC power, since the voltage and current are usually not in phase, the simple method of multiplying voltage times current won't work for most scenarios.

We do define the product of the amplitudes at apparent power, however, with units of Volt Amps (VA).

We can use this apparent power to find the real power, given in units of Watts, by using a factor called the power factor. The power factor is defined as the cosine of the voltage phase angle minus the current phase angle. We can also find it the power factor if we're given the apparent power and real power.

Professor Mason gives us a basis on why the units are Volt Amps and not simply Watts. He explain it is to prevent the units of apparent power from being confused with units of apparent, or complex, power. He uses the units of torque for example, insisting that the units of torque are meter Newtons, instead of Newton meters like work.

Next, we are given a simple phrase to help us remember which circuit element does what to voltage and current.

ELI the ICE man.

Voltage (EMF) Inductance (L) Current (I) and Current (I) Capacitance (C) Voltage (EMF)

Next, we are given a problem to use our new knowledge of complex power. Given a circuit and it's input signal, we calculate the apparent power and the power factor to find the real power.

As one can tell, our units of real power are simply Watts.. times meters over metes times Joules over Joules times seconds over secoonds times Henry Newtons over Newton Henries times fish over fish times gamma times gamma inverse.

There is also a face in there made out of my traumatized gammas.

Next, we discuss complex power in more detail. We are told that the components of apparent power is the real power plus the reactive power. And we can calculate each component given the phase angles of voltage and current and their amplitudes (or RMS).

Next, we have an example problem. Given the voltage and current functions, we can calculate the phasor of complex power. We then use this phasor to calculate the apparent, real, and reactive power.

We then calculated the power factor and the impedance from the voltage over current.

Javier's peech.. I think. 

Next, we are the voltage RMS and frequncy and a power consumption of 4kW dryer motor. We are told that the current power factor is 0.8 and we want to raise it to 0.95.

We calculate apparent power and reactive powers of the two power factors and use it to solve for the difference in reactive power. We then use a formula to find a value of capacitance to meet our requirement.

Next up for lab, we have two 1k resistors and a 1mH inductor hooked up in series. We will be feeding it different frequencies of signals and observing its current.

First we do 5 kHz and solve for the power dissipated, as well as the voltage and current functions.

We can see that there is about a Pi / 3 phase shift as predicted in the calculations.

And here we have the measured RMS values of the current and voltage, which are about what we caluclated. The top value is supposed to be mA/

The same calculation with a much higher frequency.

Again, our values are close and the top value is supposed to be in mA.

Last one, with a much lower frequency.

Again, our values are close, but not on point as expected.

5/19/26 - Oscillators with AC Op Amps

Today we begin by discussion what happens if we use an AC source on an inverting amplifier. We use nodal analysis to solve for the output voltage as a phasor.

We had a short discussion about frequencies and phase shift, it moved to discussing the wavelengths of a microwave and we were asked to calculate for the wavelength of a 60Hz wave.

 We were solving for something...I forgot exactly what it was since I lost the picture of the original problem.

Next, we have the Relaxing Amplifier. Given an DC power supply, the circuit uses no input single to generate a square wave.

Given the last three non-zero digits of my student ID, we take that to be the frequency to solve our values for. We're given that the period of the relaxing amplifier is some shmoo, and the beta is given as other stuff. We take our R1 and R2 values to be the same so we can simply plug in the natural log of 3 into the equation. Moving everything over the one side yields roughly about 500 ohms of resistance.

Since resistors don't come in 500 ohm values, we just hooked up two 1k resistors in parallel.

We set up our circuit in every circuit to observe it's outcome. Given a bit of noise, that's all the circuit needs to jump start its alternating signal.

Hooked up to the analog discovery, we measured the output voltage as a square wave and the period to be approximately 1.3ms, which translates to 769Hz. Given the uncertainty of our components, we accept this result as a success.

Next, we move into AC Power. Here we begin with the equation for the product of two cosine functions, and since we know the voltage and current to be given as a sinusoidal function, we can relate this to find AC power.

Lastly, we are given this circuit to solve for the currents. Using mesh analysis, we find the currents and then use the current to find the voltage. Since only real part of the impedance dissipates power, we use that to find the total power dissipated in the circuit.