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.