Passive RL Filter Lab

Objective:
The topic was focused on the characteristics of resonant RLC circuit for series and parallel circuits. Most importantly, we covered the concept of filters. Filters are used radio and TV. A filter is a circuit that designated to pass signals with a frequency and reject others. A passive filter consists of Resistor, Inductor, and Capacitor. There are also 4 types of filters which are lowpass, high pass, band pass, and bandstop.

Group Practice:
1. The problem below shows a parallel RLC circuit and are told to find the resonance frequency, quality factor, bandwidth, and the half power frequency. When calculating the quality factor, we notice that Q≥ 10, which means that we can use the equation ω1⋍ω0-B/2 and ω2⋍ωo+B/2 to find the half power frequencies as shown in figure 1.

Figure 1 Calculating resonance frequencies for an RLC parallel circuit.

2. Based on the circuit, we find the transfer function and then noticed that it a second order low pass filter. Furthermore, we calculate the cutoff frequency by taking the magnitude of the transfer function and setting it to 1/sqrt2. We can then solve for the ωc as seen below.

Figure 2. Calculating the cut off frequency and the type of filter. 

Passive RL Filter Lab:

1. The lab requires us to measure the response of the voltage across the resistor and the voltage across the inductor to the input voltage. We are required to plot the frequency response( magnitude and phase of both voltages). 

Figure 3. RL circuit with a sinusoidal input voltage. 

2. For the pre lab, we first calculate the frequency response of vL and vR in the circuit shown in figure 3 as a function of R and L. Since both voltage output share the same frequency, then we will only use the inductor value and use a voltage divider as shown in figure 4. We acquire a resonance frequency of 15915Hz

Figure 4.  Calculating the resonance frequency 

3. The circuit was built based on the lab and the frequency input were converter to hertz and measured the voltage output from the resistor and inductor as well as the input voltage.

Figure 6. Actual Circuit. 

Figure 7. Testing different frequencies. 

Figure 8. Vr at 15915Hz

Figure 9. Vr at 159150Hz

Figure 10. VL at 1591 Hz

Figure 11. VL at 15915Hz

Figure 12. VL at 31830Hz

Figure 13. VL at 159150 Hz

Figure 14. Measurements for V out from resistor and V in 

Figure 15. Measurements for V out from the inductor and V in

Bode Plots Lecture

Objective: 

Today's lecture covered the process in creating Bode plots. The first step is to transform a circuit to its transfer function in terms of the given four possible transformation function gains. A transfer function is written in therms of factors that have real and imaginary parts. So we rewrite it by dividing out the zeros and poles. Thus, placing it in standard form. We then find the magnitude and phase by taking the natural log in terms of decibel values. The plot is created by sketching the asymptote lines and adding them or subtracting them to plot the actual lines.  

Group Practice Problems: 

1. The problem below shows a transfer function and are told to create a bode plot. The first step is to change the transfer function to its standard form and then take to the log of it in order to find its magnitude and phase. We then plot them by recording the corner frequencies, sketch the factors and then combine them by adding or subtracting to sketch the actual lines. 

Figure 1. The Bode plot of a transfer function.

2. We now have a new transfer function where one of our factors is squared. Similarly, to figure 1, we change the transfer function to its standard form and take the log and find the phase. Though we must point out that since one of the values is squared the log is multiplied by 40 instead of 20 when finding the gain or Hdb. 

Figure 2. Plotting the transfer function when one of the poles is squared.

3.  Figure 3 shows the calculation of resonance frequency, upper half and lower half frequencies. We also also calculate the bandwidth and quality factor from an RLC circuit. 

Figure 3.  Finding the resonance frequency and other values for an RLC circuit. 

Learning Outcome: 

We learned to create Bode plot for transfer functions by following some steps. The first step is to change the transfer function to standard form by taking out or dividing out the poles and zeros. We then find the magnitude by taking the log and then finding the phase. With this information, we are then able to create a bode plot by noticing the corner frequencies. We sketch the terms in dotted line and then add them together to obtain the over plot. Also, the phase plot in plotted based on the phase. We then consider resonant circuits in which in an RLC, the capacitive and inductive reactance are equal. From this, we can calculate the resonance frequency, quality factor, bandwidth, upper and lower power frequencies. 

Signals with Multiple Frequency Components Lab

Objective:
Today's lesson covered frequency response which in a circuit, is the variation in its behavior with change in signal frequency. We consider the frequency response of simple circuit using their transfer functions. The transfer function H(ω) is the representation of voltage gain, current gain, transfer impedance, or transfer admittance. We note that we can substitute jധ with an S for easier computation. We are able to find the zeros when we set the numerator to 0 and the poles when we set denominator equal to zero. It is not easy to plot the magnitude and phase of the transfer function. Instead, we use Bode plots. Bode plot are based on logarithms and their properties.

Group Practice Problems:
1. The problem below tells us to find the i0/il or the gain in terms of S where s = jധ. We also need to find the zeros and the poles. In order to answer the problem, we applied current division in order to find the current through the capacitor. We then set up the ration in terms of the gain mention earlier. We then set the numerator and denominator equal to zero where the results from the numerator will give us the zeros and the results from the denominator will give us the poles. We can the graph the problem as seen in figure 1. The graph is not complete.

Figure 1. Find the Zeros and poles. 

Signals with Multiple Frequency Components Lab:

1. The lab requires us to find the magnitude response which is the ratio of the amplitude of the output to the input sinusoid at frequencies of 500Hz, 1000Hz, and 10000Hz. The goal of this lab it to see how the circuits magnitude response affects the shape of a signal applied to the circuit. 

Figure 2. Circuit Schematic

2. The calculation for the gain can be seen below. We expect that the higher the frequency, the voltage output should be zero. Thus, the gain or the magnitude response must be zero. The reason for this is because the capacitor acts like a short circuit. When the frequency is low, the circuit in figure 2 acts asa voltage divider and the output voltage amplitude is half the input voltage amplitude. We can see that in figure 3 of our calculations. the Gain at 500 Hz is .5 which is what we expect. However, it seems that we miscalculated the gain at a frequency of 10kHz. It should be close to 0.

Figure 3. Calculations for magnitude response .

3. The actual measurements for the components are as follow: R1 = 665Ω±.5, 672Ω±.5, and C= 1.03uF±.5.

Figure 4. Actual circuit set up

4.  We will use a waveform generator and apply a custom waveform of 20(sin(1000πt)+sin(2000πt)+sin(20000πt). Instead of doing one frequency at a time we are creating three of them and seeing them in a single sinusoidal wave. The wave produced can be seen below in figure 5.

Figure 5. Application of a custom waveform to the circuit. 

5. Our results show that for the 500Hz, the voltage out is cut in half which will result in a magnitude response of .5 based on the equation ധ=Vout/Vin. As we look at the sinusoidal wave section of 1000Hz, its seem that the voltage output amplitude is in between the voltage input amplitude which means that the values is is less than .5. At a frequency of 10kHz, the output voltage amplitude is larger than the input voltage amplitude which mean that the value is very close to zero.

Figure 6. Vin and Vout measurements at 500Hz,1000Hz, and 10kHz

Figure 7. Vin and Vout measurements closer look.

Lab Learning Outcome:

The purpose of the lab is to apply a custom sinusoidal wave of 20(sin(1000πt)+sin(2000πt)+ sin(20000πt) and see the output voltage from figure 2. Our theoretical calculations show that as the frequency increases, the gain or the magnitude response decreases to zero. The results calculated is due to the idea that at very low frequencies, the capacitor has infinite impedance which mean that we are able to apply a voltage divider which gave us a result of .5. When the frequency was increased to 10kHz, the value goes zero. The reason being that the capacitor acts a like a short circuit which means that there is no current flow to the resistor where the voltage output is represented in figure 2. We created the circuit and measured the components which are:  R1 = 665Ω±.5, 672Ω±.5, and C= 1.03uF±.5. Our experimental results shows that our theoretical calculations do in fact fit with accordingly. We see that the amplitude output voltage is indeed half the amplitude of the input voltage which would give us a gain or magnitude response of .5 The higher the frequency, the magnitude response gets close to zero considering the equations of Vout/Vin. 

Op Amp Relaxation Oscillator Lab

Objective:
The purpose of today's class meeting was to focus on Op Amp AC circuits. The key to analyzing op amp circuits is to treat them as ideal op amps. This means that no current enter the positive and negative intervals of the op amp. The voltages at the positive and negative inputs are zero. The first step in dealing with op amp ac circuits is to transform the circuit to the frequency domain and then apply the ideal op amp analyzation to find the output voltage. We also covered the idea of producing or converting dc to ac by the process of an oscillator. An oscillator is circuit that produces an ac waveform as output when powered a by a dc input. However, in order for sine wave oscillators to sustain oscillation, they must meet the Barkhausen criteria which is that the overall gain must equal 1 or greater and the overall phase shift must be zero. Finally, we covered instantaneous power and average power.

Group Practice:
1. The problem shows an ac op amp and the first step is to convert its frequency domain. We then apply the idea that the op amp is ideal. With this in mind, we can solve for the voltage output more easily. Though, we must first apply nodal analysis at node 1 and 2 as show in figure 1.

Figure 1. AC op amp

2. The problem below shows how to solve for the instantaneous power which is the power at any instant time. If the voltage and current are in the form of a cosine as seen in figure 2, we can calculate the power by multiplying the voltage and current. We then apply trigonometric identity and express in the form as seen below.

Figure 2. Solving for instantaneous power while applying a trigonometric identity. 

3. Below, we have circuit and are told to find the current flowing through the resistor and capacitor, as well as finding the current flowing through the inductor. In order to find the currents, we must first solve for the overall current flowing through the AC power source. Then we can apply a current divider. Though, we must transform the circuit to its phasor domain. We will then find the Zeq in order to find the overall current.

Figure 3. Solving for the currents in the circuit.

Op Amp Relaxation Oscillator Lab:
1. The lab requires us to create an oscillator in which a dc source is converted to ac. Based on figure 4, we must design an op amp relaxation oscillator having a frequency of Hz. However, we must find a convenient capacitor and β = R1/(R1+R2). We choose β=1/2, so R1 and R2 will be 1KΩ. We will use the equation as seen in figure 5 to find the resistance R. But first we find the Period which is t=1/f. Thus, out theoretical T = .00196. We substitute the period, β, and capacitor into the equation in figure 5 and acquire a theoretical resistance R =896ohms

Figure 4. Op Amp Relaxation Oscillator

Figure 5. Theoretical Calculations.

2. We measure our component as follows, R= 874Ω±.5, R1=.985KΩ±.005, R2= .985KΩ±.005, C=.95 μF±.005. The actual set can be seen below.

Figure 6. Actual circuit set up.

3. We tested our oscillator id did not acquire an expected waveforms. We tried fixing the problem but were unable to find the solution on why our plot looked uneven.

Figure 7. Results for voltage across the capacitor. 

4. The results should be similar to the image as seen in figure 8.

Figure 8. Expected results.

Apparent Power and Power Factor Lab

Objective:
Today's class meeting covered effective or RMS value. Through some calculation we found that the effective value is its root mean square. We can then use the rms values in terms of the average power where P = Vrms*Irms*cos(θ៴-θі). We also discussed about apparent power and power factor. The avg power mentioned ealier is the product of two terms where the product of Vrms and Irms is known as the apparent power and the factor cos(θ៴-θі) is known as the power factor. The power factor is the ratio of the average power to the apparent power. In other words, pf = P/S where S is the S = Vrms*Irms. The power factor in this case is dimensionless. Also, the power factor is the cosine of the phase difference between voltage and current.

Group Practice Problems:

1. The problem below proves the relationship between the effective and RMS values. They are in this case equal to each other as seen below. Therefore, we will not refer to the effective values but as RMS values.

Figure 1. RMS and effective values 

2. The problem below requires us to find the apparent power S and power factor pf which is dimensionless. We have a circuit where a resistor is connected in series with an inductor powered by an AC outlet of 210 and a frequency of 50Hz. The 210V is actually the Vrms. The first step is to convert the frequency to angular frequency and transform the circuit to its phasor domain. We will then find the Irms by using the equations I=Vrms/Z where Z is the equivalent and transformed t polar form for easier manipulation. The power factor is calculated by multiplying the Vrms and Irms. The power factor is the cos(θv-θi).

Figure 2. Calculating the power factor and the apparent factor

3. The problem below shows given voltage across  a load and a current through the element in the direction of the voltage drop. We are told to find the complex power, apparent power, the real power, reactive power, power factor and the load impedance. The results are shown below. 

Figure 3. Find the powers f a circuit given v(t) and i(t).

Lab Procedures and Results:

1. We are building a circuit . Given our theoretical values, we will calculate for the following: Irms, Vrms, apparent power, power factor, average power to load, average power dissipated by Rt, and the ratio between the average power Rt and the load. Our calculation can be seen below and we are required to find them while changing the RL= 10, 47 and 100Ω. 

Figure 4. Theoretical calculations procedures. 

2. Our theoretical calculation can be seen below for different RL in figure 5.

Figure 5. Theoretical and experimental values.

3. Values of Components: RT= 10.8ohms, L = 1mH, RL=10.8ohms ,RL=48.3 ohms, RL=99.1 ohms. The actual set up can be seen below. 

Figure 6. Actual set up of circuit.

4. Our experimental measurements are seen below in figure 7, 8, and 9 and can be seen in comparison with our theoretical values seen in figure 5.

Figure 7. Measurements based on RL = 10 ohms.

Figure 8. Measurements based on RL = 47 ohms.

Figure 9. Measurements based on RL=100 ohms.

5. The next step is to acquire the same data but introducing a 1 mF capacitor  parallel to the load. Measurement can be seen in figure 10, 11, and 12. The real value for the capacitor in the actual circuit is .949mF.

Figure 10. 1mF capacitor in parallel to the load where the RL = 10ohms.

Figure 11. 1mF capacitor in parallel to the load where the RL= 47ohms

Figure 12. 1mF capacitor in parallel to the load where the RL= 100ohms.

Lab Learning Outcome:

The objective of the lab is to compare our theoretical results with our experimental results for Irms and Vrms for different load resistor of  RL=10.8ohms±.1 ,RL=48.3 ohms±.1, and RL=99.1 ohms±.1. The load is composed of the inductor which is L = 1mH in series with the RL. There is also and Rt in series with the load which is represented as the transmission line which has a resistance of RT= 10.8ohms±.05. We calculated our theoretical Irms and Vrms based on our formulas seen in figure 4 where are results are seen in figure 5. When comparing them to our experimental results, we acquired a percent error for the Irms of 3.16% when RL=10ohms, 2.39% when RL=47ohms, and 1.46% when RL = 100ohms. The Vrms percent error are .16% for RL=10ohms, 2.47% for RL=47ohms, and 2.05% for RL=100ohms. By having a small percent error, we can conclude that our theoretical calculation serves its purpose in solving for the Irms and Vrms, as well as calculate the apparent power, power factor, and the phase shift.

Group Projects and Practice Problems using the Phasor Domain

Objective:
The new method of transforming the circuit to the phasor domain but now applying all the methods we covered this semester such as nodal analysis, mesh analysis, etc. During calculations, we will be required to create a matrix of complex numbers in order to solve for unknown current or voltages. Though, we must used rref in order to solve for the unknowns when using MATLAB. Superposition are important to solve for a circuit where there are two sources with different angular frequencies being applied. We covered source transformations using the phasor domain. We reflect back to source transformation which is, when a resistor is in parallel with a current source, we can replace it with a voltage source. Also, we can replace a voltage sources which is in series with a resistor with a current source in parallel with the resistor used to calculate the current source.

Group Practice:
 1. The problem below shows a circuit in which there is a current source and are told to find the current through the .1F capacitor. The first step is to find the phasor domains for the inductor and capacitor as shown in figure 1. We can then apply nodal analysis at node one solve simplify into a linear equation. We then apply nodal analysis to the second node and simplify in terms of a linear equation. Since, we are able to solve for voltage 2 in terms of voltage 1. We can use substitution in this case or place it into a matrix and solve in MATLAB. After the voltage is solved we can change it to polar form and solve for the current.

Figure 1. Finding the current through the capacitor.

2. The following problem tells us to find the currents using KVL using 3 loops. However, we notice that one loop is given to us. We apply KVL and place it into system of linear equation and place it into a matrix for MATLAB to solve. It is important to first transform the circuit to its phasor domain before starting to work on it.

Figure 2. Solving for the current using 

3.  The problem below requires us to find the thevenin equivalent from terminals a and b. The first step is to transform the circuit into the phasor domain. We will apply KCL at node 1 and then KVL to the right loop as seen in figure 3. 

Figure 3.

Learning Outcome:
There was no Lab today but we practice using KVL,KCL, mesh analysis, nodal, analysis, thevenin equivalent in the form of a phasor domain. Same process is implemented when dealing with the phasor domain. We also covered the grading rubric and what is expected for the final project presentation. We got together with out partners and created a plan for the upcoming days. There is a pert chart that we must complete and include in our presentation.

RLC Circuit Response Lab

Objective: 
We obtained the step response for RLC circuits when in series and in parallel. Also, we introduced and dealt with second order op amp circuits. In dealing with these circuits, it is important to KVL or KCL and rearranging the terms so that there is a second order differential equations. However, we keep in mind that there is a steady state response which is the final values of what is being found v(t) or i(t). The complete solutions change for overdamped, underdamped, and critically damped. The completed solution consists of the transient response and the steady state response. There is no source free do deal with which means we need to keep in mind of the voltage or current source.

Groupe Practice:
1. The problem below tells us to find i(t) for t>0. First we find i(0) and v(0) before it is opened which we get i(0-)= 0 since inductors acts like a short circuit and the capacitor has a voltage of v(0-) = 20V  when fully charged. Yet, when closed for t>0, the voltage source is disconnected and only the LC circuit remains. We have a source free series LC circuit. We then find the alpha and omega and check whether it is overdamped, underdamped, or critically damped.

Figure 1. Finding the i(t) for t>0

2. The problem below will solve for the natural response and will write the completed response in the end. Also, the coefficients will be solved. However, we first find the initial conditions of the circuit.

Figure 2. Find the initial conditions and solving for the natural response and complete solution. 

RLC Circuit Response Lab Procedures and Results: 

1. This lab requires us to find the maximum overshoot, rise time, and DC gain based on the circuit show in figure 3. We will estimate theses values and compare them to the experimental measurements. 

Figure 3. Circuit schematic for RLC circuit. 

2. For the pre lab, we first write the differential equations of the system which is seen in figure 4. the equations is seen in orange where α(neper frequency) = 1/CR2 + 1/L and ω(undamped natural frequency) = R1/R2LC + 1/LC

Figure 4. Differential equations relating to voltage output and input

3. The actual circuit setup can be seen in figure 5. The measure values for the circuit are: R1= 47Ω, R2= 2.2 ohms L = 1mH, C = 1nF. We apply a 2V step input at low frequency so that the circuit reaches steady state between pulses. Based on the graphs recorded, we will measure the input and output voltage.

Figure 5. Actual RLC circuit.

 4. We acquired an overshoot of overshoot=51.188%, voltage in = 2.0844V, voltage out =

Figure 6. Voltage input and voltage output 

Figure 7. voltage input and voltage output

Learning Outcome:
Based on the RLC circuit schematic, we can calculate the initial values, i(inductor) = Vin/R1+R2 = .04065A and the voltage at the capacitor is Vc=