Understanding Filters in Electronic Circuits: RC, RL, and RLC Circuits, Lab Reports of Engineering

Download Understanding Filters in Electronic Circuits: RC, RL, and RLC Circuits and more Lab Reports Engineering in PDF only on Docsity! Electronic Instrumentation ENGR-4300 Fall 2006 Experiment 2 K.A. Connor, Susan Bonner, Paul Schoch Revised: 9/5/2006 Rensselaer Polytechnic Institute Troy, New York, USA 1 Experiment 2 Complex Impedance, Steady State Analysis, and Filters Purpose: The objective of this exercise is to learn about steady state analysis. You will also be introduced to filters. Equipment Required • DMM (Digit Multimeter) • Rensselaer IOBoard Rev D (with Mobile Studio Desktop) • Oscilloscope (Rensselaer IOBoard) • Function Generator (Rensselaer IOBoard) • DC Power Supply (Rensselaer IOBoard) • Impedance Bridge (located on center table) • Components: 1k resistor, 1μF capacitor, 100mH inductor, 0.068μF (683) capacitor • Protoboard Helpful links for this experiment can be found on the links page for this course: http://hibp.ecse.rpi.edu/~connor/education/EILinks.html#Exp2 Part A – RC circuits, RL circuits, and AC Sweeps Background complex polar coordinates: Complex numbers allow you to express a single number in terms of its real and imaginary parts: z = x + jy. j (i in mathematics) is used to represent the square root of -1. You can also represent a number in the complex plane in terms of polar quantities, as shown in figure A-1. The length of the line segment between a point and the origin and the angle between this segment and the positive x axis form a new complex number: z = Acosθ + jAsinθ where ).(tan 122 x yandyxA −=+= θ Figure A-1 impedance and basic circuit components: Each basic circuit component has an effect on a circuit. We call this effect impedance. You should remember that impedance causes a circuit to change in two ways. It changes the magnitude (amplitude) and the phase (starting position along the time axis). When a resistor is Electronic Instrumentation ENGR-4300 Fall 2006 Experiment 2 K.A. Connor, Susan Bonner, Paul Schoch Revised: 9/5/2006 Rensselaer Polytechnic Institute Troy, New York, USA 2 placed in a circuit, it affects only the amplitude of the voltages. When capacitors and inductors are placed in a circuit, they influence both. We can use complex polar coordinates to represent this influence. A capacitor will change the amplitude by 1/ωC and shift the phase by -90 degrees. An inductor will change the amplitude by ωL and shift the phase by +90 degrees. (Recall that ω is called angular frequency and is equal to 2πf.) We can represent these changes easily in the complex polar plane. We represent impedance by the letter Z. Therefore, LjZ CjC jZRZ LCR ωωω ==−== 1 Experiment The influence of a capacitor In this section, we will examine how a capacitor influences a circuit using Capture/PSpice. • Create the simple RC circuit shown in figure A-2 in Capture o Locate the VSIN source in the SOURCE library. Set the amplitude to 200mV, frequency to 1K and offset to 0V. o Locate the resistor (R) and capacitor (C) in the ANALOG library. Leave the resistor at 1k, but change the value of the capacitor to 1uF. [In PSpice, u represents micro, μ (10-6). o Choose the 0 ground from the ground SOURCE library and put the wires into the circuit. V C1 1u V V1 FREQ = 1K VAMPL = 100mV VOFF = 0 0 R1 1k Figure A-2 • Set up a simulation for this circuit. Choose a Transient Analysis. Run to time 4ms. Choose a step size of 4us. • Place two voltage markers on the circuit: one at the input and one at the corner. The leftmost marker in the circuit shows you the input to the circuit. The rightmost marker shows you how the circuit is influencing this input. • Run the simulation. You should get an output with two sinusoids on it. You should see that the circuit has influenced both the amplitude and the phase of the input. How close is the phase shift to -90 degrees? Note: Use one of the later cycles to determine this. Print this plot and include it in your report. • PSpice has another type of analysis that lets you look at the behavior of a circuit over a whole range of frequencies. It is called an AC sweep. We know that the influence of the capacitor depends on ω and this is related to the frequency of the input signal. Should the behavior of the circuit change at different frequencies? Let’s set up an AC sweep and find out. o Edit your simulation by pressing on the edit simulation button. o Choose AC Sweep/Noise from the drop down list box. Electronic Instrumentation ENGR-4300 Fall 2006 Experiment 2 K.A. Connor, Susan Bonner, Paul Schoch Revised: 9/5/2006 Rensselaer Polytechnic Institute Troy, New York, USA 5 impedances in the complex polar domain in just the same way that we combine resistances in the time domain. Now we can also easily define our transfer function using the voltage divider rule, as illustrated in figure B-1. 21 2 21 2 ZZ ZHVin ZZ ZVout + = + = Figure B-1 Filters: Most electrical signals are made up of many frequencies. In the first experiment, you learned that sound waves within human hearing range cover a range of frequencies from very low to very high. Sometimes we may not want to include all these frequencies in our signal. We may want to filter out the very high frequencies that sound like noise, for instance, so that we only hear the part of the sound that we want to. In electronics a circuit that filters out certain frequencies while allowing others to remain unchanged is called a filter. Figure B-2 The two basic types of filters we will consider in this part are low pass filters (LPF) and high pass filters (HPF). An idealized representation of these two types of filters is shown in figure B-2. Low pass filters filter out high frequencies while allowing low frequencies to pass through unchanged. High pass filters block out low frequencies while allowing high frequencies to pass through unchanged. In an ideal world, the transfer function would be 1 for the frequencies that you want to remain unchanged (a signal multiplied by 1 is the same) and 0 for the frequencies you want to filter out (a signal multiplied by 0 is 0). In an ideal filter, the transition between 1 and 0 is instantaneous. In a real filter, this transition is less exact. Using Transfer Functions to Determine Filter Behavior: We can determine what kind of filter the simple RC circuit pictured in figure B-3 is, by finding the transfer function and examining its behavior at low and high frequencies. First do the complex algebra to determine the capacitor voltage. You will notice that once again we have a voltage divider circuit, except that one of the impedances is imaginary and one is real. For an operating frequency ω π= 2 f , the impedance of the capacitor C1 is equal to Electronic Instrumentation ENGR-4300 Fall 2006 Experiment 2 K.A. Connor, Susan Bonner, Paul Schoch Revised: 9/5/2006 Rensselaer Polytechnic Institute Troy, New York, USA 6 Z j CC1 1 1 = ω , while the impedance of the resistor R1 is Z RR1 1= . Figure B-3 Applying the usual voltage divider relation for this series combination of two impedances gives us V Z Z Z V j C j C R VB C C R A A= + = + 1 1 1 1 1 1 1 1 ω ω Note that the relationship between the input voltage VA and the output voltage VB is now complex. It is useful to be able to set up these expressions and then simplify them for very low and very high frequencies. We will be able to use the resulting expressions to see if our PSpice plots make any sense. We will first look at very low frequencies. We cannot set frequency equal to zero, since parts of our formula will blow up. Rather, we will assume that the frequency is very small, but not zero. Then the capacitive impedance ZC1 will be very much larger than the resistive impedance ZR1 and we can neglect the latter term. We have then, at low frequencies, that V V Z Z V V Vout B C C A A in= ≈ = = 1 1 or, more simply, that V V out in = 1. Note that the input and output voltages are essentially identical. At high frequencies, the capacitive impedance will be the small term, so we can neglect it in the denominator. We cannot neglect it in the numerator, since it is the only term there. Thus, at high frequencies, V V j R C V j R C Vout B A in= ≈ = 1 1 1 1 1 1ω ω or that V V j R C out in = 1 1 1ω . This ratio clearly goes to zero as the frequency goes to infinity. It is also negative imaginary and, thus, has a phase of –90o. Note about “Low” and “High”: While it may seem that terms like low and high, when applied to something like frequency, are fuzzy, ill-defined terms. That is definitely not the case. Here, we mean something quite specific by the expression low frequency. A frequency is only low when the capacitive impedance is so much larger than the resistive impedance that we can neglect the resistive term. We usually need to define the required accuracy to make a statement like this. For example, we can say that a frequency is low as long as the approximate relationship we found between the input and output voltages is within 5% of the full expression. Most of the parts we use in circuits are no more accurate than this, so there is no particular need to do our calculations with better accuracy. In other applications we might need better accuracy. Corner Frequency: When we design a high or low pass filter, we only need to know three things: how it behaves at low frequencies (0 or 1), how it behaves at high frequencies (0 or 1), and at what frequency it Electronic Instrumentation ENGR-4300 Fall 2006 Experiment 2 K.A. Connor, Susan Bonner, Paul Schoch Revised: 9/5/2006 Rensselaer Polytechnic Institute Troy, New York, USA 7 switches (0 to 1 or 1 to 0). For a simple RC or RL filter, the frequency at which it switches is called the corner frequency. It is defined as the point at which the magnitude of the transfer function is equal to 1/√2. Experiment The Transfer Function Equation In this section we will explore the impedance of the capacitor circuit. If we want to apply the voltage divider equation to form the transfer function, then the voltage drop over the two components must add up to the input voltage at all frequencies. • In Capture/PSpice, load the capacitor circuit (figure A-2) or recreate it by replacing the inductor (figure A-3) with a 1uF capacitor. Remove the markers from the circuit. To test the validity of applying a voltage divider to complex impedances, we want to show that, just like with the voltage divider made with resistors, the source voltage (V1) is equal to the sum of the voltages across the resistor and the capacitor (VR + VC). To determine VR and VC, use we will use voltage differential markers. o Chose a differential marker. Place the positive voltage marker (marked with a plus) at the end of the capacitor closest to the source. This follows the convention that voltage is positive at the end where the current enters. Place the negative marker on the end of the capacitor closest to ground. o Place a second differential marker on either side of the resistor. o Place a regular voltage marker at the input (next to the source). • Rerun the transient analysis from part A (4 ms with a step size of 4us). • When you run your simulation, you will obtain three voltages traces. To see the sum of VR and VC, we must add a trace in PSpice. o Find the symbolic name for the VR trace by looking at the bottom of your plot. Each trace on the plot has a name that corresponds to its color. It should look something like V(R1:1,R1:2). If your resistor has a different name or a different polarity relative to ground, the expression will not be exactly the same. o Go to the Trace menu in the PSpice program and choose Add Trace. o Type in the name of the VR trace from the bottom of the plot. o Type in + (plus) or click on the + symbol in the functions menu. o Find the signal corresponding to VC on the plot. Type in the name of the VC trace. o The expression in the bottom box should look something like: V(R1:1,R1:2)+V(C1:2,0). (Since the names and polarities of your components may vary, your expression may not be identical.) o Click on OK. • When have added the trace correctly, you should see that the sum will equal the source voltage. Print out a plot of your results, and include it in your report. • Run a transient simulation for frequencies of 10 Hz (run time = 400ms and step size=400us) and 10 kHz (run time = 0.4ms and step size=0.4us). Place the trace of VR and VC on each plot. Verify that the voltages add as expected. Print these plots and include them in your report. • Since we have a series circuit, we know that the current, I, is the same through all the components. Therefore, V=IZ for each component and for the entire circuit. This means that the transfer function, H=Vout/Vin, becomes independent of current, and simplifies to an expression in terms of the component values only. CR C CR C ZZ Z IZIZ IZ Vin VoutH + = + == Electronic Instrumentation ENGR-4300 Fall 2006 Experiment 2 K.A. Connor, Susan Bonner, Paul Schoch Revised: 9/5/2006 Rensselaer Polytechnic Institute Troy, New York, USA 10 way (i.e. filter it, amplify it, etc.), then we label the input voltage as Vin and the output voltage as Vout. Since it is easiest to reference the phase of the input to zero, we can write a sinusoidal input voltage as Vin = Vo cos(ωt) and the output voltage as Vout = V1 cos(ωt + φo) where ω = 2πf. When working exclusively with AC steady state conditions, it is generally easier to analyze circuits using what is called phasor notation. Euler's identity tells us that exponentials with imaginary arguments can be related to sines and cosines by ej(ωt + φ) = cos(ωt + φ) + jsin(ωt + φ) that permits us to write the output voltage, for example, in the equivalent form Vout = Re{V1 ej(ωt + φ)} = Re{Vout ejωt} where Vout = V1ejφ is a complex number and is called the phasor form of Vout. Much of the time we find it convenient to consider the real and imaginary parts of this complex representation separately Vout = VRe + jVIm where VRe = V1cos(φ) and VIm = V1 sin(φ) Thus, we will also be keeping track of the real and imaginary parts of voltages and currents. When we run PSpice simulations, we will be able to add voltage markers that will do this for us. One can access these markers by starting at the PSpice menu, then going to the Markers menu and, finally, the Advanced menu. If we apply the phasor form to the equations that characterize voltage and current for capacitors and inductors, we obtain the following VL = Re{Vo ejωt } = L dIL/dt = L (d/dt) Re{Io ejωt } = jωL Re{Io ejωt } = jωL IL or more simply, VL = jωL IL and, following the same kind of analysis, IC = jωC VC Thus, instead of a differential equation, we obtain the same kind of algebraic relationship we had for resistors, Ohms Law, except that the impedances (a generalization of resistance) of inductors and capacitors are given by ZL = jωL and ZC = 1/(jωC). For resistors, ZR = R, while in general Z = R + jX. RLC Circuits: Circuits with resistors, inductors and capacitors are called resonant circuits. They find wide application in electronics because there are many circumstances in which we wish to produce or block a single frequency. There are two common types of resonant circuits, parallel and series combinations of a resistor, an inductor and a capacitor. This figure shows series RLC circuits with 4 possible input-output choices. Figure C-1 Oscillating systems usually have such a natural resonance. This frequency, where the inductive and capacitive impedances cancel, is called the resonant frequency. The expression for the resonant frequency is LC 1 0 =ω . More complex filters: Simple RLC circuits can be used to create more types of filters than simple RC or RL circuits. You can create any of the four circuit types shown in figure C-2. Electronic Instrumentation ENGR-4300 Fall 2006 Experiment 2 K.A. Connor, Susan Bonner, Paul Schoch Revised: 9/5/2006 Rensselaer Polytechnic Institute Troy, New York, USA 11 Figure C-2 Each of the above circuits has a low frequency and a high frequency approximation, found in the same manner as we have seen for RC and RL circuits. Looking at the limiting cases of H(jω) for frequencies near zero and for very large frequencies, we can determine which type of filters they are. For example, note that at both high and low frequencies, the transfer function for case C-1(d) is equal to one, and thus the input voltage appears unchanged at the output. [If you want to verify this, redraw the circuit at low and high frequencies replacing the capacitor and the inductor by the appropriate approximation (short or open). You should be able to see that in each case, the output point, C, is at the same voltage as the source. This means that the input signal is appearing at the output. The circuit is “passing” very low and very high frequencies.] Also, since the impedance of a capacitor is negative imaginary and that of an inductor is positive imaginary, there will be a frequency ωo where the net impedance across the output terminals C and D will be zero. Near this frequency, the output voltage will be zero or at least very small. These frequencies are “rejected” by this filter. It must be a band-reject filter. Modeling Electrical Components: When we build circuits, we use a function generator to produce the input voltage, a resistor, an inductor and a capacitor. However, we cannot use just a voltage source and three other components when we model them in PSpice, since real devices usually cannot be modeled by a single parameter. Let us go through each of the four components of circuits and see what is necessary for realistic analysis or simulation. Function Generator – We have already seen several times that the function generator has an internal resistance of 50 ohms. Thus, in a realistic circuit, we must use an ideal sinusoidal voltage source and a resistor to represent the function generator. Resistor – Except at very high frequencies, resistors behave in an essentially ideal manner. Thus it is almost always sufficient to represent the resistor in a circuit as a single resistor. Inductor – Since inductors are made with a long piece of wire, they usually have a significant resistance, in addition to their inductance. Thus, we must include an additional resistor to model the resistance of a real inductor. Capacitor – A capacitor typically consists of two large metal plates separated by a thin insulator. If the insulator is very good, almost no current will flow between the plates. Then, like the resistor, the capacitor will behave in an essentially ideal manner and we don’t need to add any extra components in a circuit to represent a capacitor. No components we can make are really ideal. Resistors also have inductance and, sometimes, capacitance. Inductors have capacitance. Capacitors have resistance and inductance. Fortunately, we have figured out how to make these devices so that they behave in a nearly ideal manner for quite a broad range of frequencies. However, when we push the limits of a circuit, we have to remember that it may behave as if it consists of more components that we can see when we build it. When the RLC circuit is built, we will only see the three components R1, L1 and C1. Experiment Simulation of an RLC Circuit In this section, we will use PSpice to simulate an RLC circuit and plot the magnitude and phase of its transfer function. Electronic Instrumentation ENGR-4300 Fall 2006 Experiment 2 K.A. Connor, Susan Bonner, Paul Schoch Revised: 9/5/2006 Rensselaer Polytechnic Institute Troy, New York, USA 12 • Set up the circuit as shown I figure C-3 below. R2 90 C1 .068n Vin VoutL1 100uH V1 R1 1k Figure C-3 Assume that the amplitude of V1 is 200mV, (400mVp-p.) Note that R3 represents the impedance of the function generator and R2 represents the resistance of the 100mH inductor. • Add a voltage marker between R3 and R1. This is the input voltage. Add a second marker at the upper right hand corner between R2 and C1. This is the output voltage. Perform an AC sweep from 10 Hz to 100kHz. Since we have now done several AC sweeps, can you explain why we do Decade sweeps rather than Linear sweeps? • Create a double plot of the magnitude and phase of the transfer function. o Use the Add Plot Window menu under the Plot menu in Probe. o Add a trace of the absolute value of the ratio of Vout to Vin, to the top plot. You can use the abs() function for this purpose. [The trace expression should look something like abs(V(C1:1)/V(R1:1)). To find the actual names of your input and output voltage traces, look at the bottom left hand corner of your plot.] This is the magnitude of the transfer function. o On the bottom plot display the trace of the phase of the transfer function Vout/Vin. Add a trace of the phase of the ratio of Vout to Vin. You can use the phase function (p) for this purpose. [The trace expression should look something like p(V(C1:1)/V(R1:1)). To find the actual names of your input and output voltage traces, look at the bottom left hand corner of your plot.] What is the phase shift between the output and input at low and high frequencies? o Print out this plot. Label the resonant frequency (extreme point) with the cursors o You should also calculate the value for the theoretical resonant frequency using the equation and mark it on the plot. LC f π2 1 0 = o You will also mark the actual resonant frequency of the circuit you build in the next section on this plot. o Note that the calculated value is close to, but not exactly at, the resonant frequency. The equation gives you a simple mathematical way to get close to the resonant frequency of a circuit. Sometimes the expression will give the exact value, it depends upon the configuration of the circuit. The exact value of the resonant frequency can be found by examining when denominator of the transfer function goes to zero. We do not go into these details in this class. For us the approximation is sufficient. • What type of filter is this? Build the circuit Now you can build the circuit in figure C-3 on your protoboard, hook it to the function generator and compare its behavior to the simulation. • Before you build the circuit, you should measure the exact values of the components. o Measure the DC resistance of the inductor and the resistor using the DMM. Electronic Instrumentation ENGR-4300 Fall 2006 Experiment 2 K.A. Connor, Susan Bonner, Paul Schoch Revised: 9/5/2006 Rensselaer Polytechnic Institute Troy, New York, USA 15 • You should see that the source oscillates as it did before, although at a much higher rate and that the voltage across R2 and the capacitor seems to not change with time at all. Actually, the latter voltage is still oscillating, but at such a small amplitude that you cannot see it. Delete the trace that shows the source voltage. o Click on the node label at the bottom of the PROBE window that corresponds to the side of resistor R1 connected to the source. o Hit the Delete button on your computer keyboard. • Using the cursors, determine the amplitude of the sine wave oscillation across R2 and the capacitor. The voltage will also have a DC level, but we only want to determine the sine wave amplitude. The peak-to-peak amplitude can be determined by subtracting the voltage at a minimum from the voltage at a maximum. The actual amplitude of the sine wave will be half the peak-to-peak value. Write down the amplitude you determined here. • From this amplitude and your knowledge of how voltage dividers work, determine the magnitude of the equivalent impedance of the R2/C1 combination, that we are calling Z. [Hint: Figure D-2 shows the voltage divider. You can use the voltage divider equation: Vz =V1*(Z)/(R1+Z)] 0 V1 R1 50 Equivalent Impedance Z Figure D-2 • Check your answer by replacing the capacitor/resistor combination with a single resistor with the value of Z you calculated. Rerun the simulation. Does the output have the same amplitude as the combination? • Now we can check if the transfer function gives us the same amplitude. Use the parallel rules for impedance to combine 1/jωC and R. Then set up the transfer function for the circuit. Determine its magnitude at 1Meg Hertz. (Don’t forget that ω=2πf). Multiply the value of H at this frequency by the input voltage amplitude (200mV). Is the output amplitude of your plot comparable to the one you calculated? Summary In this section you explored what happens when you combine two impedances in parallel. You have learned that it is possible to find transfer functions for and analyze circuits with components in parallel. You simply need to combine the impedance, Z, of the parallel components into an equivalent impedance using rules similar to those used to combine resistors. Report and Conclusions The following should be included in your report. Everything should be labeled and easy to find. Partial credit will be deducted for poor labeling or unclear presentation. Electronic Instrumentation ENGR-4300 Fall 2006 Experiment 2 K.A. Connor, Susan Bonner, Paul Schoch Revised: 9/5/2006 Rensselaer Polytechnic Institute Troy, New York, USA 16 Part A (16 points) Include the following plots: 1. PSpice transient plot of RC circuit (in figure A-2). (1 pt) 2. PSpice AC sweep plot of the RC circuit voltage. (1 pt) 3. PSpice AC sweep plot of the RC circuit phase. (1 pt) 4. PSpice transient plot of RL circuit (in figure A-3). (1 pt) 5. PSpice AC sweep plot of the RL circuit. (1 pt) 6. PSpice AC sweep plot of RL circuit phase (1 pt) Answer the following questions: 1. What is the amplitude and phase of the output of the RC circuit at 1K hertz? (1 pt) 2. In what frequency range is the amplitude of the output of the RC circuit about equal to the input amplitude? In what frequency range is the amplitude of the output about zero? (2 pt) 3. In what frequency range is the phase of the output of the RC circuit about equal to the input phase? In what frequency range is the phase of the output about -90 degrees? (2 pt) 4. What is the amplitude and phase of the output of the RL circuit at 1K hertz? (1 pt) 5. In what frequency range is the amplitude of the output of the RL circuit about equal to the input amplitude? In what frequency range is the amplitude of the output about zero? (2 pt) 6. In what frequency range is the phase of the output of the RL circuit about equal to the input phase? In what frequency range is the phase of the output about +90 degrees? (2 pt) Part B (24 points) Include the following plots: 1. PSpice plot of capacitor and resistor sum adding to the input voltage at 1k Hz. (1 pt) 2. PSpice plot of capacitor and resistor sum adding to the input voltage at 10 Hz. (1 pt) 3. PSpice plot of capacitor and resistor sum adding to the input voltage at 10k Hz. (1 pt) 4. PSpice plot of transfer function of RC circuit (figure A-2) with corner frequency marked. (2 pt) 5. PSpice plot of transfer function of RL circuit (figure A-3) with corner frequency marked. (2 pt) 6. Mobile Studio picture of RC circuit at 1K hertz. (1 pt) 7. Mobile Studio picture of RC circuit at corner frequency. (1 pt) Answer the following questions: 1. Write out the mathematical expressions for the output voltage of the capacitor for the first RC circuit case you considered (plot 1). Write it in the form V(t)=A Sin(ωt+φ). (1 pt) 2. What kind of filter is the RC circuit? (1 pt) 3. At what frequency did you find the corner on the PSpice plot of the transfer function of the RC circuit? What frequency did you calculate using f=1/(2πRC)? How do the two compare? (3 pt) 4. Find the transfer function of the RC circuit in terms of R, C and jω. Take the limit of this function at very low and very high frequencies. Show that these results are consistent with the PSpice plot of the transfer function. (3 pt) 5. What kind of filter is the RL circuit (figure A-3)? (1 pt) 6. Derive the equation for the corner frequency of the RL circuit. (2 pt) 7. At what frequency did you find the corner on the PSpice plot of the transfer function of the RL circuit? What frequency did you calculate using the equation you derived in the previous question? How do the two compare? (3 pt) 8. At what frequencies did the output of the RC circuit you built look roughly the same as the input? At what frequencies did the output disappear into the noise? (1 pt) Part C (16 points) Include following plots: 1. PSpice plot of the transfer function of the RLC circuit in figure C-3 (magnitude and phase), with three resonant frequency values marked [PSpice value, calculated value, experimental value] (original component values). (3 pt) Electronic Instrumentation ENGR-4300 Fall 2006 Experiment 2 K.A. Connor, Susan Bonner, Paul Schoch Revised: 9/5/2006 Rensselaer Polytechnic Institute Troy, New York, USA 17 2. PSpice plot of the transfer function of the RLC circuit in figure C-3 (magnitude and phase),with the resonant frequency marked (real component values). Also put the 5 experimental points on this plot. (3 pt) Answer following questions: 1. Why is it necessary to plot the phase and the magnitude of the transfer function separately, rather than on the same plot? (1 pt) 2. For the three RLC circuits labeled (a), (b) and (c) in figure C-1 indicate what type of filter the circuit is (high pass, low pass, band reject, or band pass filter) and explain why each is the filter it is. [Filter (d) is discussed in the Background section for part C.] Recall that a capacitor can be modeled as an open circuit al low frequencies and a short at high frequencies. Also recall that an inductor can be modeled as a short at low frequencies and an open circuit at high frequencies. Redraw the circuits at low and high frequencies and consider the value of the output between C and D for each case. You can check your answers in PSpice if you want. (3 pt) 3. What is the phase shift between the output and input of plot 1 above at low and high frequencies? Does the phase shift change when you adjust the values of the components to create plot 2? Why or why not? (2 pt) 4. Determine the resonant frequency of the RLC circuit you analyzed with PSpice. (This occurs at the extreme point.) Calculate the resonant frequency with the equation f = 1/[2π√(LC)]. How similar are they? What factors do you think account for the discrepancy? (3 pt) 5. Why do you suppose it is that, in practice, we generally use filters designed with capacitors and not inductors? (1 pt) Part D ( 12 points) Include the following plots: 1. PSpice AC Sweep plot of circuit in figure D-1 with 5% point marked with cursor (2 pt) Answer the following questions: 1. When the capacitor is added, for what range of frequencies does the capacitor change the voltage across R2 by less than 5%? (1 pt) 2. What is the amplitude of the output of the parallel RC circuit? What is the value you calculated for the equivalent impedance (Z) of the parallel combination at 1Meg Hz? (3 pt) 3. Find the transfer function for the parallel RC circuit. Determine the magnitude of the transfer function at 1Meg Hz. Calculate the amplitude of the output voltage for this circuit at 1Meg Hz? How well does it agree with the output amplitude you found using PSpice? (6 pt) Summary (12 points) 1. Summarize key points (8 pts) 2. Discuss mistakes and problems (2 pts) 3. List member responsibilities (2 pts) Total: 80 points for write up + 20 for attendance = 100 points Attendance: 3 classes (20 points) 2 classes (10 points) 1 class (0 points) out of 20 possible points Minus 5 points for each late. No attendance at all = No grade for experiment.