301 Lab06 EC.pdf

# 301 Lab06 EC.pdf - ECEN 301 Student_Jonathan Bernal Sec_3...

• Lab Report
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Page 1 of 14 ECEN 301 Student _Jonathan Bernal______ Sec __3 Electrical Engineering for Engineers Brigham Young University Lab 6: R,L,C Circuits Step and Frequency Response TA Pass-off (10 pts) Objectives Observe the AC characteristics of capacitors & inductors. Understand the behavior of low-pass and high-pass filter circuits. Calculate components necessary to build filter circuits with specific characteristics. Learn how to measure the RC time constant ( τ ) of a filter circuit. Pre-lab Please do the pre-lab work before coming to the lab . You will then be able to begin the lab work immediately and get meaningful help from the instructors. 1. Review Low-Pass Filters and High-Pass Filters in your textbook (Rizzoni). 2. Review section on “Transient Response of First-Order Circuits" of your textbook (Rizzoni). 3. The circuits shown below are often called filter circuits. One is low pass, meaning that low frequencies and DC from the source will go through to the output but at high frequencies the output will be attenuated (reduced). The other is high-pass, meaning that high frequencies from the source will go through to the output but at low frequencies the output will be attenuated.
Page 2 of 14 (2 pts) Low Pass (2 pts) High Pass Why did you select the filter type for each circuit above? (High-pass/Low-pass) (2 pts each) Circuit 1 – At high frequencies. The output drops away. Vin is parallel to capacitor Circuit 2 – At low frequencies the resistance goes to infinity. High frequencies are going to pass through. Vin is in series with Capacitor. 4. The primary characteristic of an RC filter is its corner frequency . The corner frequency is the frequency at which the output is attenuated (cut) by 3 dB (or by a factor of -1 ) The corner frequency of an RC filter is given by f c = . a. Using the formula above, design a high-pass RC filter with a corner frequency of 1.6 kHz ± 10% and a resistance of R=10 k W ± 10%. Show your calculations (3 pts) If Fc = 1.6 kHz and R = 10 kOhm, C = 1/(2*pi*R*Fc) = 1/( 2*pi*10kOhm*1.6kHz) So C = 9.95 nF + or – 10 % b.

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• Fall '17
• Dr.Nordin
• RC circuit, High-pass filter, Low-pass filter, kHz, corner frequency, rc filter

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