05-10-2016

Lecture: Introduction to phase shifting, phasor representation in frequency domain

I. Introduction to Phase shifting:
For today lecture (no lab today), we will learn about how to modify the orginal graph using the idea of pre-cal, which is known as vertical shift and horizontal shift, but for engineering, horizontal shift would be phase shift.

We have to parametrized the two graphs. The graph y(t) is really close to the orignal which only had the vertical shift;therefore we chose to do it first; whereas the green one also had the horizontal shift

II. Phasor representation:

The idea is genius, it is really hard to add the trig identities together. We assumed that voltage and current had the same frequency, therefore, we could use the phasor with polar and imaginary part to simplify our calculation

Given the original time domain function, we could convert it to the polar coordination to multiply and division operation, or to imaginary representation to do the add or subtracting equations

In trig, to find the phase angle, simply take the acrtan of y.x

We practice some operations between polar and imaginary part of phasor representation

For adding, imaginary part will be a better choice

Some more practice for phasors, recall that to to convert from sin to cos , we subtract the angle by 90.

Here given the time domain voltages, we have to combine them into one new voltage. We performed many steps of phasors to get the answer

We find the current with the given impedance of the inductor

III. Conclusion:
Today we spent most of our time to understand the phasor representation of the trig in the frequency domain. For phasors,  we have to use polar to divide or multiply, and imaginary to subtract and add. The capacitor and inductor act like a complex resistors in the AC circuits.

05-05-2016

Lab: Step response of LRC in parallel and series circuit
I. Introduction:
After many exposures to natural response of LRC circuits in parallel and series on Tuesday, today Professor Mason lectured us about step response, the step response is actually natural response except that we add one independent in the circuit, which would add another term when we solve for second order differential equation

Given some constraint conditions, we were asked to find out the circuit here was under damped, critically damped, or over damped

Here are some constraint condition to solve for the problem, since we have parallel LRC, we have di/dt=V/L

Here we have a critically damped condition, the second term has an extra compared to the third term.

II. LRC step response lab:

1. Pre-lab: 

Given the setup above, we were asked to find the change( shoot) of the voltage and current across the capacitor and the inductor. We focused mostly on the right part of the circuit which has LRC in series.

Here is the setup of our circuit, we used the channel 1 and 2 to measure the voltage and current across the capacitor and the inductor

Here is the pictures of our graphs, the orange is the voltage of the input, the blue one is the voltage across the capacitor. 

Here is the zoomed-in graph of voltage across the resistor

Here is our setup of every circuit, there was slightly different numbers between our lab and the theoretical yields, this could due to the fact that we have +-5% in our lab components.

III. Conclusion:
For this lab, we spent much our understanding the behavior of step response LRC circuit, which just adds one more term to the natural reponse LRC circuit, the rest are the same. We also spent some time understood the step response LRC by doing the lab, the lab was almost the same we did on Tuesday except we just needed to add one more voltage indendent source.

05-03-2016

Lab: Series RLC circuit step response

I. Introduction:

Today, we spent some more time to review the inductor and capacitor behave in circuit. Recall that the voltage of the capacitor in DC circuit, in the infinite time, will be the same as the voltage source, thus open circuit. For the inductor, in the infinite amount of time, it will behave like a wire, thus short the circuit

We did some review about the RL and RC circuits

Solving for differential equation, given the RLC circuit

s represents the variable associate with time. Here we solved for s using the quadratic and many algebraic steps

After many derivation, we let the alpha to be damping frequency and w_o to be our natural response frequency, which equivalent to the spring and gravity lab we did back then in 4A

given the RLC circuit, we have to find the damping frequency and natural response frequency, since the delta (b^2-4ac >0) we say that the circuit is overdamped

Some more practice to solve for s, in this case have underdamped, which gave us the imaginary solution.

III. Series RLC circuit Step Response:
This lab will emphasize modeling and testing of a series RLC second order circuit. The circuit will consist of RLC in series, we will measure the voltage across the capacitor

Here is the osilatiion in RLC circuit, this happened because of the bouncing between the capacitor and inductor (the inductor letting more current go through while the capacitor prevents over the time)

Here is our setup of the circuit

Here is our analysis for the pre-lab. There are a total of 3 cases, underdamped, damped, and overdamped

Here is our derivation to find out the two solutions of s

After we found s, we could go back and obtain two more equations to solve for A1 and A2

III. ConclusionFor today lab, we spent most of our time to study the differential equations of RLC circuit. There are a total of three test cases, when the circuit is underdamped (imaginary solution), damped (one real solution), and overdamped (two real solutions), we did the RLC circuit step response to test a series of RLC second order circuit.

04-26-2016

04-26-2016

Lab: Introduction AC circuit, Inverting 

After the second exam, we are now back to inductor and capacitor with the introduction of frequency in voltage and current, which is known as the alternating current (and voltage).

Given the derivative graph, we have to integral the given ones to bring them back to the original shape. The square waves graph is the most intuitive one.

II. Lab:
For this lab, given a capacitor and a resistor with such values, we will apply sinusoids of various frequencies to the circuit and compare the output with our expectation based on analysis

Here is our derivation to find the V_o

Here is our setup. The oscilloscope wires had been taken out for brevity

The voltage of V_out(blue) is lagging the the current source, which was the correct, recalled ELI the ICE man.

When we changed to high frequency, the output displayed some strange peaks, we belived that due to such high frequency, small errors could greatly affect the graphs

With low frequency, the graphs were smoothed out

Here is our derivation for the pre-lab, each times we ran, the amplitude changed and also the frequency

II. Introduction to parametrized function:
Given the current sinusoidal voltage and current, we could used another two equations, with delta linked to u and u linked to r. We can use u and delta function to describe r function

Here we wrote the two definitions of u, delta, and r and their relationships

Given the function in terms of r and u, we can take derivative to bring them back to delta and u and plot them

derivation of voltage using u and r representation

Given circuit, we have to find the current across the capacitor over time.

use the nodal analysis to find the current across the inductor.

III. Conclusion:
Today, we were briefly introduced to the idea of AC circuit and their nearly representation of delta and u. We did the inverting differentiator lab, which gave us the idea of how output voltage with frequency varies. Last but not least, we reviewed some nodal and mesh analysis techniques to solve some voltage and current in the circuit.

04-14-2016

04-14-2016: Passive RC and RL Circuit Natural Response

I. Introduction:

Today, we spent most of our time to discuss the RC and RL. We also reviewed the first differential equation to find out the relationship between current and voltage across the inductor.

Given this simple circuit, our goal is to find out the equivalent inductor. For the inductor, a rule of thumb here is to follow the same as the equivalent resistors. For my notation, + indicated they are in series, not adding.

II. Formula Derivation:

Here is our derivation of the voltage across the capacitor over the course of time. Notice that when the switch is closed and DC current, the capacitor acted like an open circuit. If the switched is open, the capacitor will be fully charged and has its own voltage, ideally a straight line, but in real world, the voltage will be dropped over the course of time

Here is our derivation to find out what time the voltage across the capacitor would be left with 1% of the original voltage, which means it would take 5tau. tau=R*C

After we got the voltage, the current would be voltage divided by the resistor, the power is the multiplication of both voltage and current

Given a simple circuit, our goal is to find out the V(t) and i(t). First of all, we have V(0)=10 V already. We then had to find tau and got our V equation, I=V/R

Here we need to find the smallest tau and the largest tau, since tau is directly proportional to RC, we picked the lowest values for RC and largest values for RC respectively

II. Prelab:

Given a simple circuit,we had to tau, we first find the Rth in the circuit, then use that to find out tau

II. Voltage and current in capacitor and inductor:
1. The capacitor:
For this lab, we were trying to measure the current and voltage of capacitor and inductor over the course of time. We change the voltage by two ways: mechanically, which means that we had to basically take in take out the voltage supply power; or we can use the square wave voltage built in feature of the analog discovery

Here is how the graph looks like when we did it manually, the region we concerned about is the going downslope of the graph

Here is what it should look like if we the built-in feature of square wave voltage instead of doing it mechanically. 

Here we used triggering and single acquisition to find out the slope of the curve

Here is our data. For part A (mechanically), we found the percent to be 35.88%, which was not really favorable in terms of acceptable range. It could due to that the value we were trying measure to small therefore a small change of number could cause a great impact to our percent error

III. Some More derivation

Here we derived the formula to find the current across the inductor

After we got the formula of the current. We know that V=l*di//dt. We took the integral to find the formula for the voltage

Here is our setup for this lab, this setup applied for both the capacitor and the inductor 

The percent error is about 35% for the manual experiment

Here is the portion we need for the inductor. The percent error is about 28% in this case

here is the setup for the inductor circuit

III. Analysis:
Since we don't have an ideally inductor or capacitor, the current or voltage are not perfectly trapped in the circuit, over the course of time, those values would be dissipated into the resistors, causing a great percent error in our lab. Moreover, since we dealt with small value of time division, it could make our measurement really sensitive to the percent error
IV. Conclusion:
For this lab, we spent our time discussing the behaviors of voltage and current in capacitor and inductor circuit. We derived many equations using derivative and integral to the review of what we have learned in our 4b class. In the lab, we verified our knowledge about natural response of the passive RC and RL circuit. The percent error is great due to the time is too small to measure. For the rule of thumb, the capacitor acts like the open circuit with DC, and the inductor acts like short-circuit.