Applied Network Analysis

ETE 310

Course Description:

        Analysis of circuits in the time and frequency domains employing Laplace transforms methods. Ideal op-amps and applications. Second order passive and active circuits, circuit responses to a variety of input signals, stability analysis of closed loop systems. Computer methods utilized.

First Order Systems        2

OBJECTIVES        2

EQUIPMENT REQUIRED        2

Intro        3

LABORATORY        3

Part 1: RC Filter        3

Part 2: RL Filter A:        7

Part 3: RL Filter B:        9

Conclusion        10

Integrator and First Order Active LPF        11

OBJECTIVES        11

EQUIPMENT REQUIRED        11

Intro        11

LABORATORY        12

Part 1: Integrator        12

Part 2: Active First Order LPF:        14

Conclusion        16

Differentiator and First Order Active HPF        18

OBJECTIVES        18

EQUIPMENT REQUIRED        18

BACKGROUND        18

INTRO        19

LABORATORY        20

Part 1: Differentiator        20

Part 2: Active First Order HPF:        22

Conclusion        24

Second Order RC LPF        25

OBJECTIVES        25

EQUIPMENT REQUIRED        25

BACKGROUND        25

INTRO        27

LABORATORY        27

Part 1: Second Order LPF – Direct cascade of first order LPFs        27

Part 2: Cascaded first order LPFs with opamp buffer        33

Conclusion        36

2nd Order Passive RLC Systems        37

OBJECTIVES        37

EQUIPMENT REQUIRED        37

INTRODUCTION        37

LABORATORY        38

Part 1:        38

Part 2:        42

Part 3:        44

Part 4:        46

Conclusion        47

Active Second Order Sallen-Key Bandpass Filter        48

OBJECTIVES        48

EQUIPMENT REQUIRED        48

INTRODUCTION        48

LABORATORY        49

Part 1:        49

CONCLUSION        55

First Order Systems

 OBJECTIVES

EQUIPMENT REQUIRED

Intro

The purpose of this lab is to examine several filters in the time and frequency domain. Several items will have to be found for each filter, the transfer function and cutoff frequencies. Simulation in MATLAB and PSPICE will be used to show the output in both time and frequency domains.

LABORATORY

Part 1: RC Filter

C:\Users\Mangoner\Pictures\rc_hpf.png

Figure 2: RC Filter

  1. Find the transfer function  for the circuit of Figure 2.

  1. Using MATLAB or hand calculation, find the cutoff frequency for the circuit in both rad/s and Hz.

  1. Enter your transfer function into MATLAB and create a Bode plot for phase and magnitude in Hz. Find the -3 dB frequency on the plot and compare it to your result from 2).

H:\ete 310\lab 2\part 1 bode 3 .png

  1. Perform AC analysis for the circuit in PSpice. Verify that the cutoff frequency is the same between 2), 3), and 4). Provide this plot in your report as well as your schematic. Make sure you show the cursor values in screenshot.

  1. Based on the Bode plot and AC analysis in PSPICE, identify what type of filter this circuit is.

This is a high pass filter based on the Bode plot and AC analysis in Pspice.

  1. Using MATLAB, find an expression for the time domain response of the circuit to a unit step input. Hint: enter your frequency domain transfer function and multiply it by the Laplace transform of your input signal. Finally, take the inverse Laplace transform to find the time domain expression for  

H:\ete 310\lab 2\part 1 bode 6 .png

  1. Find the time constant  of the circuit.

  1. Using PSPICE, perform time domain analysis and obtain a plot  and . Choose the input voltage source “Pulse”, and the on-time (PW or pulse width) should be at least 5 time constants. The period (PER) should be twice the PW. V1 should be 0V and V2 should be 1V. TR and TD represent rise and fall time of the signal. We can choose these to be small, say 1n. Use cursors to find the voltage at

  1. Build the circuit. Apply a sine wave of 1Vpp, 0V DC offset to the circuit. Measure the output voltage for the following frequencies:

0.1fc, 0.2fc, 0.3fc, 0.4fc, 0.5fc, 0.6fc, 0.7fc, 0.8fc, 0.9fc, 1.0fc, 1.1fc, 1.2fc, 1.4fc, 1.7fc, 2.0fc, 2.3fc, 2.6fc, 3.0fc, 4.0fc, 5.0fc, 6.0fc, 7.0fc.

Take a scope capture for the waveform at 0.3fc, 1.0fc, and 7fc. Make sure your scope captures show the measurements on them.

fc

159

V(u)

0.1

15.9

350u

0.2

31.8

184

0.3

47.7

280

0.4

63.6

354

0.5

79.5

430

0.6

95.4

490

0.7

111.3

550

0.8

127.2

600

0.9

143.1

647

1

159

690

1.1

174.9

730

1.2

190.8

748

1.4

222.6

796

1.7

270.3

844

2

318

880

2.3

365.7

901

2.6

413.4

921

3

477

933

4

636

970

5

795

980

6

954

990

7

1113

1

  1. Change your input signal to be a 0 to 1V square wave (don’t forget the offset). Set the frequency to be the same as part 8). Use the oscilloscope cursors to measure the voltage at  Take one screen capture showing the cursor at .

H:\ete 310\lab 2\part 1_10_square 0.png

Part 2: RL Filter A:

C:\Users\Mangoner\Pictures\rl_lpf.png

Figure 3: RL Filter A

  1. Find the transfer function H(s) for the circuit of Figure 2.

  1. Using MATLAB or hand calculation, find the cutoff frequency for the circuit in both rad/s and Hz.

  1. Enter your transfer function into MATLAB and create a Bode plot for phase and magnitude in Hz. Find the -3 dB frequency on the plot and compare it to your result from 2).

  1. Perform AC analysis for the circuit in PSpice. Verify that the cutoff frequency is the same between 2), 3), and 4). Provide this plot in your report as well as your schematic. Make sure you show the cursor values in screenshot.

  1. Based on the Bode plot and AC analysis in PSPICE, identify what type of filter this circuit is.

        LPF

  1. Using MATLAB, find an expression for the time domain response of the circuit to a unit step input. Hint: enter your frequency domain transfer function and multiply it by the Laplace transform of your input signal. Finally, take the inverse Laplace transform to find the time domain expression for Vo
  2. Find the time constant Tao of the circuit.
  3. Using PSPICE, perform time domain analysis and obtain a plot Votand Vint. Choose the input voltage source “Pulse”, and the on-time (PW or pulse width) should be at least 5 time constants. The period (PER) should be twice the PW. V1 should be 0V and V2 should be 1V. TR and TD represent rise and fall time of the signal. We can choose these to be small, say 1n. Use cursors to find the voltage at 

Part 3: RL Filter B:

C:\Users\Mangoner\Pictures\rl_hpf.png

Figure 4: RL Filter B

Find the transfer function H(s) for the circuit of Figure 2.

  1. Using MATLAB or hand calculation, find the cutoff frequency for the circuit in both rad/s and Hz.

  1. Enter your transfer function into MATLAB and create a Bode plot for phase and magnitude in Hz. Find the -3 dB frequency on the plot and compare it to your result from 2).

  1. Perform AC analysis for the circuit in PSpice. Verify that the cutoff frequency is the same between 2), 3), and 4). Provide this plot in your report as well as your schematic. Make sure you show the cursor values in screenshot.
  2. Based on the Bode plot and AC analysis in PSPICE, identify what type of filter this circuit is.
  3. Using MATLAB, find an expression for the time domain response of the circuit to a unit step input. Hint: enter your frequency domain transfer function and multiply it by the Laplace transform of your input signal. Finally, take the inverse Laplace transform to find the time domain expression for Vo
  4. Find the time constant Tao of the circuit.
  5. Using PSPICE, perform time domain analysis and obtain a plot Votand Vint. Choose the input voltage source “Pulse”, and the on-time (PW or pulse width) should be at least 5 time constants. The period (PER) should be twice the PW. V1 should be 0V and V2 should be 1V. TR and TD represent rise and fall time of the signal. We can choose these to be small, say 1n. Use cursors to find the voltage at

Conclusion

We took a look at RC filters such as the LPF, low pass filter, and High Pass Filter, HPF within the frequency domain. The first was a LPF RC circuit, we found the transfer function which we than plotted the bode plot and phase graph in matlab, also simulated it in pspice as well measuring the actual circuit with the scope, we did the same for RL HPF, LPF.


Integrator and First Order Active LPF

 OBJECTIVES

EQUIPMENT REQUIRED

Intro

The purpose of this lab is to examine two first order active circuits in the frequency and time domian. The transfer function, stability, and the output signal will be looked at in the first circuit, the integrator. The second circuit is similar to the first but it adds a resistor in parallel to the feedback capacitor. By adding the resistor there it creates a low pass filter in the feedback.

LABORATORY

Part 1: Integrator

C:\Users\Mangoner\Pictures\integrator.png

Figure 1: Integrator

  1. Find the transfer function  for the circuit of Figure 2.

  1. Find the poles and zeros of the system. Plot them.

H:\ete 310\lab 4\part 1_tf poles.png

  1. Build the circuit of Figure 1. Apply a 1Vpp Square Wave at 500 Hz to the input. Show the Output

H:\ete 310\lab 4\Part 1 3.png

Figure 2: Integrator Output

  1. Reduce your R value by approximately ½ and repeat part 3. What effect does this have on the output? The output signal became steeper since the capacitor is charging at a faster rate.

        Figure 3: Integrator Output with R halvedH:\ete 310\lab 4\Part 1 4.png

Part 2: Active First Order LPF:

C:\Users\Mangoner\Pictures\act-first-ord-lpf.png

Figure 4: Active First Order Low Pass Filter

  1. Find the transfer function  for the circuit of Figure 3.

  1. Plot the poles and zeros of the system. Figure 5

H:\ete 310\lab 4\Part 2 TF Pole.png

  1. Find the cutoff frequency for the system in Hz.

  1. Build the circuit of Figure 2. Apply a 1Vpp Sine Wave to the input. Adjust the frequency until the output is 0.707 of the input. Record the frequency obtained as your measured cutoff frequency. Figure 6

H:\ete 310\lab 4\part 2 4.png

  1. In PSPICE simulate the circuit using an AC sweep and locate the cutoff frequency. Figure 7

H:\ete 310\lab 4\part 2 5.PNG

  1. Enter your transfer function in MATLAB and perform a Bode plot. Again, locate the cutoff frequency of the system and compare. Figure 8

H:\ete 310\lab 4\Part 2 bode.png

  1. In PSPICE, reduce the R1 value by a factor of 2, and describe what effect that has on the system. Hint: look at the transfer function you obtained and think about the overall gain of the system. Figure 9 H:\ete 310\lab 4\part 2 7.PNG

        The effect of changing Rin to half, changes the output by double.

  1. What is the benefit of using this active first order LPF compared to a passive one?

This is beneficial because it adds gain to our filtered source

  1. In order to make our output positive again, we would need to add another inverting amplifier to the output of the present circuit. Why can’t the input of this circuit be connected to the “+” terminal and then the left side of R1 grounded, like in a non-inverting configuration?

Using the positive terminal will make the circuit non-inverting and the transfer function would always be at least 1. This can be problematic if a transfer function less than 1 is desired since it will always be at least 1.

Conclusion

The operational amplifier integrator is an electronic integration circuit. Based on the operational amplifier (op-amp), it performs the mathematical operation of integration with respect to time; that is, its output voltage is proportional to the input voltage integrated over time. The input current is offset by a negative feedback current flowing in the capacitor, which is generated by an increase in output voltage of the amplifier. The output voltage is therefore dependent on the value of input current it has to offset and the inverse of the value of the feedback capacitor. The greater the capacitor value, the less output voltage has to be generated to produce a particular feedback current flow.The input impedance of the circuit is almost zero because of the Miller effect. Hence all the stray capacitances (the cable capacitance, the amplifier input capacitance, etc.) are virtually grounded and they have no influence on the output signal.

With passive filter circuits containing multiple stages, this loss in signal amplitude called “Attenuation” can become quite severe. One way of restoring or controlling this loss of signal is by using amplification through the use of Active Filters. Filter amplification can also be used to either shape or alter the frequency response of the filter circuit by producing a more selective output response, making the output bandwidth of the filter more narrower or even wider. Then the main difference between a “passive filter” and an “active filter” is amplification. The advantage of this configuration is that the op-amps high input impedance prevents excessive loading on the filters output while its low output impedance prevents the filter's cutoff frequency point from being affected by changes in the impedance of the load.


Differentiator and First Order Active HPF

 OBJECTIVES

EQUIPMENT REQUIRED

BACKGROUND

This laboratory will examine two types of first order, active systems. An active system is one which uses op amps or some other semiconductor device that can provide gain. This is contrasted with passive systems which use only passive components (R’s, L’s, and C’s). Recall that first order systems have only one effective frequency dependent component. Also, remember that in the frequency domain, integration is represented by a transfer function of  and differentiation is represented by a transfer function of .

Nodal analysis can be used to derive the transfer function of opamp circuits by keeping in mind the ideal op amp rules:

  1. No current flows into the inverting or non-inverting inputs of the opamp.
  2. The non-inverting and inverting inputs are at the same voltage.

C:\Users\Mangoner\Pictures\non-inverting-amp.png

Figure 1: Non-inverting Amp

For example, the node equations for the non-inverting amplifier of Figure 1 can be written by realizing that the current through  is equal to the current through  (since no current flows into either input). The voltage in between these components must be , by the ideal op amp rules. The node equations can be written as follows:

Solving for  gives (try it):

These results were obtained for resistors, which are independent of frequency. These results could be generalized to impedance where there are frequency dependent components as well. In that case the gain would change as the impedance of different parts of the circuit change with frequency.

INTRO

For this lab we will explore first order active systems to see what changes in an active high pass with several resistor placements. The first system looked at was the differentiator so the output would be the derivative of the input square signal. Since it is an active system the gain can be changed by changing the resistance. Using MATLAB and Pspice for the second system, we can see the the cutoff frequency and the output. The simulation should show that the output will be filtered to allow high frequencies and the op amp will create a gain.

LABORATORY

Part 1: Differentiator

C:\Users\Jason\Pictures\differentiator.png

Figure 2: Differentiator

  1. Find the transfer function  for the circuit of Figure 2.

  1. Build the circuit of Figure 2. Apply a 1Vpp Square Wave at 500 Hz to the input. Set your scope to measure the output on channel 2. Obtain a scope capture for this. The output should resemble the negative of the derivative of the input.H:\ete 310\Lab 5\part 1 #2.png
  2. Reduce your R value by approximately ½ and repeat part 3. What effect does this have on the output? Hint: think of RC time constants, i.e. the time it takes for a capacitor to charge to a certain voltage.

H:\ete 310\Lab 5\Part 1 #3.png

  1. Conceptually, predict if the gain of this circuit increases, decreases, or remains constant when frequency is increased. Think about the standard inverting amplifier gain of –(Zf/Zi). Explain.

We predict that the gain will go up because the capacitor is inversely proportional to frequency which would decrease the impedance as the frequency increases but because we have a feedback resistance it would make the gain increase.

  1. Try to increase the frequency of your circuit and see if your prediction was correct. Take a screenshot at a higher frequency for your report.

H:\ete 310\Lab 5\Part 1 #5.png

Part 2: Active First Order HPF:

C:\Users\Jason\Pictures\act-first-ord-hpf.png

Figure 3: Active First Order High Pass Filter

  1. Find the transfer function  for the circuit of Figure 3.

  1. Plot the poles and zeros of the system.

H:\ete 310\Lab 5\part 2 #2.png

  1. Find the cutoff frequency for the system in Hz.

  1. Build the circuit of Figure 2. Apply a 1Vpp Sine Wave to the input. Adjust the frequency until the output is 0.707 of the input. Record the frequency obtained as your measured cutoff frequency.

H:\ete 310\Lab 5\part 2 #4.png

  1. In PSPICE simulate the circuit using an AC sweep and locate the cutoff frequency.

H:\ete 310\Lab 5\part 2 #5.PNG

  1. Enter your transfer function in MATLAB and perform a Bode plot. H:\ete 310\Lab 5\part 2 #6.png
  2. In PSPICE, reduce the R1 value by a factor of 2, and describe what effect that has on the system. Hint: look at the transfer function you obtained and think about the overall gain of the system.

H:\ete 310\Lab 5\part 2 #7.png

  1. What is the benefit of using this active first order HPF compared to a passive one?

This is beneficial because it adds gain to our filtered source.

  1. Conceptually, what is the DC gain and high frequency gain of the circuit? Think about the impedance of the capacitor at DC and the impedance when frequency is infinity.

DC gain is when frequency goes to 0 which causes the capacitor to act as a short due to low impedance. High frequency gain is when the frequency is really high which causes the capacitor to have low impedance which acts as an open.

Conclusion

The basic Op-amp Differentiator circuit is the exact opposite to that of the Integrator Amplifier circuit that we looked at in the previous lab. This circuit produces a voltage output which is directly proportional to the input voltage’s rate-of-change with respect to time. So as we input a square wave we end up with a triangle because the rate of change with respect to time of our square wave is constant which then results in a slope. The figure to the right shows other responses based on input signal. op-amp differentiator output voltage

The input signal to the differentiator is applied to the capacitor. The capacitor blocks any DC content so there is no current flow to the amplifier summing point, resulting in zero output voltage. The capacitor only allows AC input voltage changes to pass through and whose frequency is dependant on the rate of change of the input signal.

At low frequencies the reactance of the capacitor is “High” resulting in a low gain ( Rƒ/Xc ) and low output voltage from the op-amp. At higher frequencies the reactance of the capacitor is much lower resulting in a higher gain and higher output voltage from the differentiator amplifier. Which we prove with the Pspice simulations.

Second Order RC LPF

 OBJECTIVES

EQUIPMENT REQUIRED

BACKGROUND

This laboratory will explore second order RC low pass filters. All previous circuits have contained only one inductor or capacitor, now we will examine circuits with two frequency dependent components. A second order filter should perform better at blocking the signal at the desired frequencies. How well the filter blocks a signal is known as its roll-off. You should be familiar with the approximating Bode plots using the +/- 20dB/decade response at each zero or pole. A first order filter lowpass filter should provide a roll-off of -20dB/decade, a second order filter should provide a roll-off of -40dB/decade, and so on. We can construct a simple second order RC LPF by simply cascading two first order LPFs. One might be tempted to think the resulting transfer function would be just the product of the individual transfer functions, but this assumption is incorrect. Recall that the first order RC LPF provided a transfer function of:

So, assume that the R and C values are identical for each first order filter, then let us multiply the two transfer functions together and see what we obtain for a new transfer function :

Now that there is a result to compare to, let's analyze the circuit using nodal analysis and see if our assumption was correct.

C:\Users\Mangoner\Pictures\2nd-order-passive-rc-lpf.png

Figure 1: Second Order Passive LPF

Writing the node equation for , and assuming the current directions shown in the figure:

Eq 1:

Combine terms and multiply both sides by R:

Eq 2, using voltage divider:

Vo

Plug Eq 2 into Eq 1:

Subtract  from both sides and factor:

Now, comparing  and , a discrepancy is found. The coefficient of s in the denominator is different between the two approaches. Nodal analysis provides the correct result. The second circuit interacts with the first circuit and “loads” the first circuit. Increasing the impedance of the second circuit will help a bit, but inserting an opamp buffer is the best at isolating the two filter stages. Though, using a buffer for each filter stage is costly and wasteful. In Part 2, we will examine the opamp buffer for this purpose.

INTRO

This lab will explore second order systems in the time and frequency domain. Doing the needed calculations in both domains will take some time, so for this lab more experience in using MATLAb will be gained. MATLAB was used to calculate the transfer function,and the cutoff frequency.  With the transfer function the poles, bode plot, and the step response can be found.

LABORATORY

Part 1: Second Order LPF – Direct cascade of first order LPFs

  1. Using just the boxed node equations in the “Background” section, use MATLAB to solve these equations for , thus allowing you to find the transfer function . Your answer in MATLAB should be a symbolic answer in terms of s, R, and C. Assume R1 = R2 = R and C1 = C2 = C. No numeric values are used at this point. Make sure your transfer function matches the results found in that section.

  1. Derive an expression for magnitude of the transfer function, and then use it to find the -3 dB cutoff frequency in rad/s for this system in terms of R and C.

  1. Using MATLAB, substitute your component values of R and C into your expression from step 1 using the command “subs(Vo,[R C],[10e3 100e-9] )”, to obtain a numerical transfer function.

  1. Enter your expression from step 3 into a “tf” variable/object. Use MATLAB to plot the poles and zeros of the system. Is the system stable?

H:\ete 310\lab 6\part1 #4.png

  1. Use MATLAB to find the time domain response of the system to a unit step input. Use the function step(sys), where sys is a transfer function variable. Using plot cursors, find the time it takes for the waveform to reach 63.2% of its final value (1V).

H:\ete 310\lab 6\part1 #5.png

  1. Using MATLABs bode plotting function, find the -3 dB cutoff frequency for the system. Also, find the -6 dB frequency (this is where the output voltage is half of the input voltage).H:\ete 310\lab 6\part1 #6.png
  2. From your MATLAB bode plot, find the “roll-off” of the filter from 500 Hz to 5kHz (one decade in frequency). Record the gain in dB at 500 Hz, then again at 5kHz, and then find the difference. Compare this to the bode plot of a first order low pass filter we did in a prior lab with R=10k and C = 100nF.

H:\ete 310\lab 6\part1 #7 2.pngH:\ete 310\lab 6\part1 #7 1.png

  1. Build the circuit of Figure 2. Apply a 10Vpp Sine Wave at 10 Hz to the input. Increase frequency until output voltage drops to -3 dB (7.07 V) and -6 dB (5.0 V). Obtain screenshots for each of these measurements.

H:\ete 310\lab 6\part 1 #8.png

part 1 8 revis.png

  1. Apply a 0 to 5 V Square Wave (5Vpp, +2.5VDC offset) with a frequency slow enough for the output to charge to near maximum in the positive cycle of the waveform. Using oscilloscope cursors, find the time it takes for the waveform to reach 63.2% of its final value (5V). Compare with results from step 5.

H:\ete 310\lab 6\part 1 9 redo.png

  1. Use PSPICE to repeat the measurements of step 6. Compare results. 

Part 2: Cascaded first order LPFs with opamp buffer

C:\Users\Mangoner\Pictures\buffered-2nd-order-lpf.png

Figure 2: Buffered Second Order LPF

  1. Using nodal analysis or voltage divider, write the node equations for the circuit of Figure 2. Use MATLAB symbolic toolbox to solve these equations to find the transfer function of the system.

        

  1. Derive an expression for magnitude of the transfer function, and then use it to find the -3 dB cutoff frequency in rad/s for this system in terms of R and C.

        

  1. Using MATLAB, substitute your component values of R and C into the equation and obtain a numerical transfer function.

        

  1. Enter your transfer function expression into MATLAB as a transfer function variable (object), then plot the poles and zeros of the system.

C:\Users\cachapman\Desktop\part 2 #4 .png

  1. Using MATLABs bode plotting function, find the -3 dB cutoff frequency for the system. Also, find the -6 dB frequency (this is where the output voltage is half of the input voltage).

        C:\Users\cachapman\Desktop\part 2 #5 .png

  1. Build the circuit of Figure 2. Apply a 10Vpp Sine Wave at 10 Hz to the input. Increase frequency until output voltage drops to -3 dB (7.07 V) and -6 dB (5.0 V). Obtain screenshots for each of these measurements. H:\ete 310\lab 6\part 2 6 1.png

H:\ete 310\lab 6\part 2 6.png

  1. Use PSPICE to repeat the measurements of step 5 and 6.

par 2 # 7.PNG

Conclusion

We worked with cascaded RC Low-Pass Filters, learning that things in practice do not work as the theory says it should. Using two first order RC LPFs in cascade we make our roll-off slope become steeper from a single RC LPF of -20db/decade to -40db/decade as we see to the right figure. Though when identical RC filter stages are cascaded together as we did, the output gain at the required cut-off frequency ( ƒc ) is reduced because they load each other; to reduce the loading effect we can make the impedance of each following stage 10x the previous stage, so R2 = 10 x R1 and C2 = 1/10th C1. second order low pass filter response curve

In addition we can fix this issue by adding buffer between the RC LPFs as we did and this results due to the unity op-amp allow voltage and not current to pass and being a high impedance it does not allow the cascade RC LPF to load each other.


2nd Order Passive RLC Systems

 OBJECTIVES

EQUIPMENT REQUIRED

INTRODUCTION

In this lab a second order passive RLC system will show that by rearranging the circuit so that the output reading can be taken from different parts will provide several types of filters. This lab will also show how the values of a resistor will determine the type of system the circuit will be.

LABORATORY

Part 1:

  1. Find the transfer function of the series RLC low-pass configuration (output across C) in terms of R, L, and C.

  1. From your transfer function, find the natural frequency , damping ratio , and quality factor Q for the circuit, in terms of R, L, and C

  1. Using your symbolic transfer function from step 2, plug in the values for the components from the circuit of figure 1 and find a numerical representation of the transfer function,  (in both Hz and rad/s), , and .

  1. From step 2, determine what values of R will make the system underdamped, overdamped, or critically damped (given L = 33mH and C = 100nF).

        Since  is greater than 1 our system is overdamped.

  1. Plot the poles and zeros of the system.

H:\ete 310\lab 8\part 1 # 5.png

  1. Create a bode plot for the system, and find the resonant (natural) frequency

H:\ete 310\lab 8\part 1 # 6.png

  1. Using MATLAB, find the step response of the circuit. Does the circuit oscillate about the steady state value? If so, what is the observed frequency of oscillation? (use cursors to measure the time difference between peaks).

Due to the circuit transfer function, we see that our poles are two real which indicates the circuit is overdamped so it doesn't oscillate.

  1. Change the R value to be 100 . Repeat steps 5-7. Compare results.

H:\ete 310\lab 8\part 1 # 8 bode.pngH:\ete 310\lab 8\part 1 # 8 pz.png

  1. Build the circuit of Figure 1. Use R = 100. Apply a 5Vpp Square Wave to the input. You should observe oscillation. Set the frequency to be slow enough so you can see the oscillation decay to zero. Using oscilloscope cursors, measure the time difference between peaks to estimate the frequency of oscillation. Record frequency and peak value. Repeat this for R = 10k. You should not see any oscillation. Obtain screenshots for each these measurements.

H:\ete 310\lab 8\part 1 #9.png

  1. Set your input to be a sine wave. Start at a low frequency of 10 Hz and then start increasing the frequency while observing the output. Convince yourself that the circuit behaves as a low-pass filter.
  2. Adjust the frequency until near the calculated resonant frequency. Slowly adjust the frequency up and down until you find the maximum output voltage for your actual circuit. What is gain of the circuit at resonance? Swap out the resistor for the 10k and observe the effect on the gain at this frequency. Remember that the resonant frequency only depends on L and C.

H:\ete 310\lab 8\part 1 # 11.png

  1. Use PSPICE to generate a frequency response with R=100 and R = 10k. Compare with results obtained in the lab.

Part 2:

  1. Rearrange the circuit so that the output is taken across the both the inductor and capacitor.

  1. Start at 10 Hz and increase frequency to 50 kHz while observing the output. Take a screenshot for the system response at the resonant frequency.

                

H:\ete 310\lab 8\part 2 # 2.png

  1. Find the transfer function of the circuit.

Part 3:

  1. Rearrange the circuit so that the output is taken across the resistor alone.

  1. Start at 10 Hz and increase frequency to 50 kHz while observing the output. Take a screenshot for the system response at the resonant frequency.

H:\ete 310\lab 8\PART 3 #2.png

  1. Find the transfer function of the circuit.

Part 4:

  1. Rearrange the circuit so that the output is taken across the inductor alone.

  1. Start at 10 Hz and increase frequency to 50 kHz while observing the output. H:\ete 310\lab 8\part 4 #2.png
  2. Find the transfer function of the circuit.

        

Conclusion

In this lab we took a look at the classic RLC circuit and took a look at the signal across each component. We see that depending where we look at the signal from a simple RLC circuit we can get a LPF, HPF, BPF. Rearrange the circuit so that the output is taken across the both the inductor and capacitor you get a notch filter. Rearrange the circuit so that the output is taken across the resistor alone we get a BPF. Rearrange the circuit so that the output is taken across the inductor alone we end up with a HPF.


Active Second Order Sallen-Key Bandpass Filter

 OBJECTIVES

EQUIPMENT REQUIRED

INTRODUCTION

For this lab the Sallen Key second order active system will be looked at to see if this circuit is capable of changing the quality factor and what impact the quality factor will have on the system.  The properties of the op amp will also be looked at to see if it benefits the system in any way.

C:\Users\Mangoner\Pictures\sk-bpf.png

Figure 1: Sallen-Key 2nd Order Active BPF

LABORATORY

Part 1:

  1. Using nodal analysis, write the node equations for the system of figure 1 (three equations). Assume  and .  and  can be treated as providing a gain . Use the gain  to simplify the analysis instead of using  and  in your node equations, as described in the background section.

>> pretty (eq1)

       Va - Vi   Va - Vo

0 == - ------- - ------- - C s (Va - Vb)

          R1        R1

>> pretty (eq2)

     Vb (C R1 s + 1)

0 == --------------- + C s (Va - Vb)

            R1

>> pretty (eq3)

            Vo

Vb == ---

             K

  1. Using the equations from step 1, find the transfer function for the system (solve the node equations for ), using the method of your choice. It should be in terms of , , and .

H(s) =

  1. From your transfer function, find the natural frequency , damping ratio , and quality factor Q for the circuit, in terms or , , and  You may find that some of these variables cancel.

  1. From your results of step 3, determine  when  is equal to the following values: 1, 2, 3, 4, 4.5. What resistor values should be used for  to obtain these gains for ? Can the gain  be less than one for this circuit?

= .35

= .47

= .707

=1.414

=2.828

No the gain cannot be less than 1 because the op-amp is non-inverting thus the gain has to be at least 1.

  1. Using your symbolic transfer function from step 2, plug in the values for the components from the circuit of figure 1 and find a numerical representation of the transfer function,  (in both Hz and rad/s), , and .

sys =

 

        2500 s

  -------------------

  s^2 + 2500 s + 2e06

  1. Plot the poles and zeros of the system.

  1. Create a bode plot for the system, and find the resonant (natural) frequency, and the bandwidth of the circuit. The bandwidth is the difference of the two -3dB cutoff frequencies.

H:\ete 310\lab 9\part 1 # 7.png

  1. Build the circuit of Figure 1. Apply a 5Vpp Sine Wave at 50 Hz to the input. Increase frequency until output voltage seems to peak (reach a maximum value). Record frequency and peak value. Obtain screenshots for each these measurements.

H:\ete 310\lab 9\lab 9 part 8.png

  1. We can approximate the bandwidth of the circuit by measuring the frequencies where the output drops by 0.707 from its maximum value. Reduce the frequency until output voltage is 0.707 of what it was in step 7. Record this as the lower cutoff frequency. Now repeat the procedure by increasing frequency until you observe this point again. The difference of the two frequencies is an approximate bandwidth for the system.H:\ete 310\lab 9\lab 9 part 9 lower.pngH:\ete 310\lab 9\lab 9 part 9 upper.png

The bandwidth is

  1. Change the feedback resistance RA of the circuit to 33K. What is the value of Q and K for this new value of RA? Repeat the measurements of step 8. With a higher Q value, did the peak to peak voltage increase or decrease? Did the bandwidth increase or decrease? Make a new bode plot and compare to the previous one. Also, make a new pole zero plot for the changed system.

Peak to Peak voltage increased with a higher Q.

Bandwidth Decreased.

H:\ete 310\lab 9\part 10.png

H:\ete 310\lab 9\part 10 upper.pngH:\ete 310\lab 9\part 10 lower.png

The bandwidth is

  1. Use PSPICE to repeat the measurements of step 7-9. Compare results.

CONCLUSION

This lab we made a Second Order Sallen-Key Bandpass Filter we learned how this type of filter was different in that in this arrangement the gain is used to change the system parameters, like the Q factor which is the quality factor of the circuit. With higher Q values it allows us to make shaper filters with the bandwidth to be smaller or larger. Because we want to allow or block a particular frequency and not typically a range of frequencies.