University:
Kentucky State University
Course:
PHY 212 | General Physics II
Academic year:

2021
Views:

27

Pages:

61
Author:

Quintin Ware

**Chapter #16: Frequency Response:** **Low-pass Filter:** [Image of a frequency response graph for a low-pass filter] * **wc:** cutoff frequency **Resonance:** [Image of a circuit diagram with a capacitor and an inductor] * A circuit having at least one capacitor and one inductor. * I need at least 1 (C) & 1 (L) to reach resonance! **@ resonance:** * I & V are in phase (pure resistive) * |Zin| = R **@ resonance:** * The amplitude of the response is either max or min (|Vout| is max or min) **Parallel resonant CKT:** [Image of a parallel RLC circuit] ω = 0, s = jω When |Ic| = |IL| They cancel each other Zin = R **Resonant CKT** Where Ic & IL ≠ 0 but they cancel each other. Y(jω) = 1/Zin Y(jω) = 1/R + jωC + 1/jωL It's easier to deal with admittance since it's a parallel RLC CKT. Y(jω) = 1/R + j(ωC - 1/ωL) **@ resonance** Y(jω) = 1/R Since Zin = R @ resonance, maximum impedance @ resonance. **1st condition of resonance** ω₀C - 1/ω₀L = 0 From @, 1/ω₀C = ω₀L **Q₀, α, ζ & Complex frequency:** [Image of a complex plane with axes labeled ω and α] When α = 0: ωd = ω₀ Also: R = √(L/C) **Example:** A parallel resonant CKT is composed of elements: R = 8kΩ, L = 50mH, C = 80nF Find: 1. ω₀ = 1/√(LC) = 15.811 k rad/sec 2. Q₀ = ω₀RC = 10.119 (very high Q₀) 3. ωd = ω₀√(1-(1/2Q₀)²) = ω₀√(1-ζ²) = (15.791) k rad/sec 4. α = ω₀/2Q₀ = 781 rad/sec 5. ζ = α/ω₀ = 1/(2Q₀) = 0.0494