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The center frequency is fo At the center (resonant) frequency, XL Q 1- RC/L 2T√LC 1 — (8.0 N)²(150 pF)/5.0 μH 2π√√(5.0 μH) (150 pF) 2π foL = 2π(5.79 MHz) (5.0 μH) = 183 XL 183 Rw 8.0 Ω Z₁ = Rw (Q² + 1) = 8.0 N(22.8² + 1) = 4.17 kN (purely resistive) Vout(min) = 22.8 Next, use the voltage-divider formula to find the minimum output voltage magnitude. 560 Ω 4.73 ΚΩ RL R₁ + Z₂ 5.81 MHz Vin 10 V1.18 V

The center frequency is fo At the center (resonant) frequency, XL Q 1- RC/L 2T√LC 1 — (8.0 N)²(150 pF)/5.0 μH 2π√√(5.0 μH) (150 pF) 2π foL = 2π(5.79 MHz) (5.0 μH) = 183 XL 183 Rw 8.0 Ω Z₁ = Rw (Q² + 1) = 8.0 N(22.8² + 1) = 4.17 kN (purely resistive) Vout(min) = 22.8 Next, use the voltage-divider formula to find the minimum output voltage magnitude. 560 Ω 4.73 ΚΩ RL R₁ + Z₂ 5.81 MHz Vin 10 V1.18 V

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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I need to calculate the Vout min in the equations below, but with the RW at 1 K ohm instead of 8 Ohm. My calculators aren't taking the center frequency equation at the beginning so I can't work out the problem with a 1 k Ohm winding resistance.

The center frequency is \[ f_0 = \frac{\sqrt{1 - R_W^2 C/L}}{2\pi \sqrt{LC}} = \frac{\sqrt{1 - (8.0 \, \Omega)^2 (150 \, \text{pF}) / 5.0 \, \mu \text{H}}}{2\pi \sqrt{(5.0 \, \mu \text{H})(150 \, \text{pF})}} = 5.81 \, \text{MHz} \] At the center (resonant) frequency, \[ X_L = 2\pi f_0 L = 2\pi (5.79 \, \text{MHz})(5.0 \, \mu \text{H}) = 183 \, \Omega \] \[ Q = \frac{X_L}{R_W} = \frac{183 \, \Omega}{8.0 \, \Omega} = 22.8 \] \[ Z_r = R_W (Q^2 + 1) = 8.0 \, \Omega (22.8^2 + 1) = 4.17 \, \text{k} \Omega \quad \text{(purely resistive)} \] Next, use the voltage-divider formula to find the minimum output voltage magnitude. \[ V_{\text{out(min)}} = \left( \frac{R_L}{R_L + Z_r} \right) V_{\text{in}} = \left( \frac{560 \, \Omega}{4.73 \, \text{k} \Omega} \right) 10 \, V = 1.18 \, V \]
Transcribed Image Text:The center frequency is \[ f_0 = \frac{\sqrt{1 - R_W^2 C/L}}{2\pi \sqrt{LC}} = \frac{\sqrt{1 - (8.0 \, \Omega)^2 (150 \, \text{pF}) / 5.0 \, \mu \text{H}}}{2\pi \sqrt{(5.0 \, \mu \text{H})(150 \, \text{pF})}} = 5.81 \, \text{MHz} \] At the center (resonant) frequency, \[ X_L = 2\pi f_0 L = 2\pi (5.79 \, \text{MHz})(5.0 \, \mu \text{H}) = 183 \, \Omega \] \[ Q = \frac{X_L}{R_W} = \frac{183 \, \Omega}{8.0 \, \Omega} = 22.8 \] \[ Z_r = R_W (Q^2 + 1) = 8.0 \, \Omega (22.8^2 + 1) = 4.17 \, \text{k} \Omega \quad \text{(purely resistive)} \] Next, use the voltage-divider formula to find the minimum output voltage magnitude. \[ V_{\text{out(min)}} = \left( \frac{R_L}{R_L + Z_r} \right) V_{\text{in}} = \left( \frac{560 \, \Omega}{4.73 \, \text{k} \Omega} \right) 10 \, V = 1.18 \, V \]
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