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.
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.
No comments:
Post a Comment