ELE - In_Lab_&_Post_Lab 2

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Feb 20, 2024

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The Step Response for Second-Order Circuits (In-Lab & Post-Lab) By Justin Pope Course: ELE 302 Section: 17 Instructor: Dr. Venkatesh TA Name: Rene Gilliland-Rocque Date Submitted: Saturday, October 7, 2023

5.0 IN-LAB IMPEMENTATION & MEASUREMENTS (5 marks in total): Part I: The Step Response of a Second-Order Bandpass Circuit Figure 5.0: Second-Order Bandpass Filter (a) Step 1: Connect Channel (1) of the oscilloscope to display the open-circuit voltage v s (t) of the function generator. Set the following: Trigger: source ± Channel (1), and slope ± rising. Adjust the controls of the function generator to provide a square-wave signal v s (t) with a peak-to-peak value of 10V and DC offset of 5V at a frequency of 20Hz. On FG: Waveform → Square, Frequency → 20 Hz, Amplitude → 5 Vpp* , Offset → 2.5 V* *Due to FG matching impedance Next, set the following: Channel (2): Vertical-position ± one division above the bottom of the screen, coupling ± dc, and V/div ± 1V. Time: Time/div ± 50μs when v s (t) is rising. Blank off Channel (1). (b) Step 2: Construct the circuit shown in Figure 5.0 . Set R=3.4kΩ. Connect Channel (2) to display the step response v o (t). Plot v o (t) on Graph 1.0. (c) Step 3: Use the cursors on the oscilloscope to measure points on the v o (t) display to calculate the parameters (σ and ω) that characterize the step response as: ࠵? ࠵? (t) = A ࠵? −࠵?࠵? sin(ωt) (d) Step 4: Record your results in Table 1.0. (e) Step 5: Demonstrate the correct operation of your setup to your TA. (1 mark) (f) Step 6: Set R=1.4kΩ, and repeat as in Step 2 and Step 3. Record your results in Table 1.0. (g) Step 7: Set R=500Ω, and repeat as in Step 2.

3.4kΩ 1.4kΩ 500Ω

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Table 1.0 (0.5 marks) R (KΩ) T v o (T/4) v o (3T/4) σ ω 3.4 (Red) 124μs 4.03 V -1.20V 19539.43 50670.85 rad/s 1.4 (Yellow ) 115μs 6.2975 V -82.50 mV 75393.21 54636.39 rad/s Part II: The Step Response of a Second-Order Lowpass Circuit Figure 6.0: Second-Order Lowpass Filter (h) Step 8: Connect the circuit shown in Figure 6.0. Set R=10kΩ. Use Channel (2) to display the step response v o (t). Plot v o (t) on Graph 2.0. (i) Step 9: Use the cursors on the oscilloscope to measure points on the v o (t) display to calculate (and record in Table 2.0 ) the parameters (σ and ω) that characterize the step response as: ࠵? ࠵? (t) = B +A ࠵? −࠵?࠵? cos(ωt + θ) (j) Step 10: Set R=1.4kΩ, and repeat as in Step 8.

(k) Step 11: Set R=1kΩ, and repeat as in Step 8. 10kΩ 1.4kΩ

1kΩ

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Table 2.0 (0.5 marks) R (KΩ) T v o (T/4) v o (3T/4) σ ω 10 (Red) 124μs 16.315 V 5.395V 17848.56 50670.85 Part III: The Step Response of a Second-Order Highpass Circuit (l) Step 12: Connect the circuit shown in Figure 7.0. Connect Channel (2) to display the step response v o (t). Plot v o (t) on Graph 3.0. (m) Step 13: Use the cursors on the oscilloscope to measure points on v o (t) display to calculate (and record in Table 3.0 ) the parameters (σ and ω) that characterize the step response as: ࠵? ࠵? (t) = A ࠵? −࠵?࠵? cos(ωt + θ) (n) Step 14: Demonstrate the correct operation of your setup to your TA. (1 mark)

8kΩ Table 3.0 (0.5 marks) R (KΩ) T v o (T/4) v o (3T/4) σ ω 8 119μs 9.555 V -6.431 V 6654.36 52799.88

6.0POST-LAB QUESTIONS (2 marks in total, 2/3 marks for each question): (1) By examining your plots on Graph 1.0, answer the following: a) What are the effects of varying the value of R on the step response of a second-order bandpass circuit? As resistance increases, the period (T) increases for the graph. As a result, it takes more time for the circuit to reach steady state. I also noticed the amplitude of the graph to increase with increasing resistance, which also contributes to the reason it takes more time for the circuit to reach steady state. (2) By examining your plots on Graph 2.0, answer the following: a) What are the effects of varying the value of R on the step response of a second-order lowpass circuit? As resistance increases, we notice more oscillations in the graphs, hence, taking more time for the circuit to reach steady state. By observing the 1kΩ resistor graph, the circuit went straight to steady state, whereas, the 10kΩ circuit took longer to stabilize and reach steady state. (3) Suppose that the 8kΩ-resistor is removed from the circuit in Fig (2.6), what effects will this have on the step response? If we were to remove the 8kΩ resistor, the graph would reach steady state at a faster rate. Using our analysis and observations from the previous questions, we can conclude that as resistance increases, the time it takes for a circuit to reach steady state takes longer. In comparison, we are removing a resistor, therefore, the circuit will reach steady state much faster. The graph will also be observed to have a larger period and the amplitude decreases.

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