EE 242 Lab 5 Pre-Lab

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California Polytechnic State University, San Luis Obispo *

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242

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Electrical Engineering

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Apr 3, 2024

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Stephanie Lee Professor McDonald EE 242-04 15 February 2024 EE 242 Experiment 5 Pre-Lab 1. For the low-pass RC filter, calculate the phase and magnitude of the transfer function at frequencies of 20, 50, 100, 200, 500, 1000, and 1500 Hz. Tabulate the result and then sketch the magnitude and the phase of the response as a function of the frequency (use log scale for frequency). Use R=50 k, C=0.02 µF Calculations: H(f) = 1/(1+j(2πf)(RC) H(f) = 1/(1+j0.006283*f) H(20) = 1/(1+ 0.006283j(20)) H(20) = 1/(1.007864756 7.162455807) => 0.992196615 7.162455807 The calculations for the other frequencies are the same: f (Hz) 20 50 100 200 500 1000 1500 |H(f)| 0.992196615 0.954028 0.846733 0.622677 0.303314 0.157177 0.105511 phase -7.16245581 -17.4406 -32.1419 -51.4881 -72.3432 -80.9569 -83.9434
2. Determine the half power frequency fc defined as the frequency at which the magnitude of the transfer function is at 1/√2 of its maximum value. Assuming C is fixed at 0.02 µF, calculate the half power frequencies corresponding to R=20 k, 50 k, and 100 k. 1/√2 = 1/√(1 2 +(2πfRC) 2 ) 2 = 1+(2πfRC) 2 1 = 2πfRC f c = 1/(2πRC) At 20k: 1/(2π(20x10 3 )(0.02x10 -6 )) => f c at 20k = 397.89 Hz At 50k: 1/(2π(50x10 3 )(0.02x10 -6 )) => f c at 50k = 159.15 Hz At 100k: 1/(2π(100x10 3 )(0.02x10 -6 )) => f c at 100k = 79.58 Hz 3. Repeat steps above for the high-pass filter. a. For the high-pass RC filter, calculate the phase and magnitude of the transfer function at frequencies of 20, 50, 100, 200, 500, 1000, and 1500 Hz. Tabulate the result and then sketch the magnitude and the phase of the response as a function of the frequency (use log scale for frequency). R=50 k, C=0.02 µF H(f) = j(2πf)RC/(1+j2πfRC) H(f) = j(0.00628)f/(1+0.00628fj) H(20) = j(0.00628)(20)/(1+0.00628(20)) => 0 + 0.125j/(1+0.125j) H(20) = 0.125 90/(1.007864756 7.162455807) => 0.124683 82.83754 The calculations for the other frequencies are the same: f 20 50 100 200 500 1000 1500 |H(f)| 0.124683 0.299717 0.532018 0.782479 0.952891 0.98757 0.994418 phase 82.83754 72.55941 57.85809 38.51189 17.65679 9.043061 6.056611
b. Determine the half power frequency fc defined as the frequency at which the magnitude of the transfer function is at 1/√2 of its maximum value. Assuming C is fixed at 0.02 µF, calculate the half power frequencies corresponding to R=20 k, 50 k, and 100 k. f c = 1/(2πRC) At 20k: 1/(2π(20x10 3 )(0.02x10 -6 )) => f c at 20k = 397.89 Hz At 50k: 1/(2π(50x10 3 )(0.02x10 -6 )) => f c at 50k = 159.15 Hz At 100k: 1/(2π(100x10 3 )(0.02x10 -6 )) => f c at 100k = 79.58 Hz
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