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.