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Problem 4. The switch of the Circuit shown is closed at time t = 0, and at time t = 0, i (0) = 3A & v(0) = 2V. = (t) = 8e-3t t = 0 452 -ww- 0.5F 2H {/i(1) 192 1Ω + v(t) (a) Derive the second order differential equation of i(t) over t for t≥ 0. (b) Find the natural response of i(t) for t ≥ 0. (c) What is the basic reason why the natural response goes to zero, as time goes to the Brief catalog of the Operational Amplifier circuits

Problem 4. The switch of the Circuit shown is closed at time t = 0, and at time t = 0, i (0) = 3A & v(0) = 2V. = (t) = 8e-3t t = 0 452 -ww- 0.5F 2H {/i(1) 192 1Ω + v(t) (a) Derive the second order differential equation of i(t) over t for t≥ 0. (b) Find the natural response of i(t) for t ≥ 0. (c) What is the basic reason why the natural response goes to zero, as time goes to the Brief catalog of the Operational Amplifier circuits

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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• Problem 4. The switch of the Circuit shown is closed at time t = 0, and at time t = 0, i(0) = 3A & v(0) = 2V. is(t) = 8e-3t ■ Vin m R₁ U10- (a) Derive the second order differential equation of i(t) over t for t≥ 0. (b) Find the natural response of i(t) for t≥ 0. (c) What is the basic reason why the natural response goes to zero, as time goes to the Brief catalog of the Operational Amplifier circuits (Based on the ideal operational amplifier) R₁ U₂0-W (a) Inverting amplifier Vout= R₂ ww R₁ Rn Xo t = 0 ww R₁ R₁ R₁ 492 -ww- U₂+ R₁ out=R₁ in Ovout 0.5F R₁ Rn (d) Summing amplifier Vin - 2H {\i(t) 10 R₁2 R₁ Vout = (1 + R ₁₂) (b) Noninverting amplifier R₂K₁ U10-W + ART Juin R₂K₂ U20-W RJK3 V30-W R₂/(1-(K₁ + K₂ + K3)): R(K4-1) v(t) Vin Rp. (c) Voltage follower (buffer amplifier) Ovout Vin Ovout K4(K₁v1 + K₂v2 + K303) (e) Noninverting summing amplifier
Transcribed Image Text:• Problem 4. The switch of the Circuit shown is closed at time t = 0, and at time t = 0, i(0) = 3A & v(0) = 2V. is(t) = 8e-3t ■ Vin m R₁ U10- (a) Derive the second order differential equation of i(t) over t for t≥ 0. (b) Find the natural response of i(t) for t≥ 0. (c) What is the basic reason why the natural response goes to zero, as time goes to the Brief catalog of the Operational Amplifier circuits (Based on the ideal operational amplifier) R₁ U₂0-W (a) Inverting amplifier Vout= R₂ ww R₁ Rn Xo t = 0 ww R₁ R₁ R₁ 492 -ww- U₂+ R₁ out=R₁ in Ovout 0.5F R₁ Rn (d) Summing amplifier Vin - 2H {\i(t) 10 R₁2 R₁ Vout = (1 + R ₁₂) (b) Noninverting amplifier R₂K₁ U10-W + ART Juin R₂K₂ U20-W RJK3 V30-W R₂/(1-(K₁ + K₂ + K3)): R(K4-1) v(t) Vin Rp. (c) Voltage follower (buffer amplifier) Ovout Vin Ovout K4(K₁v1 + K₂v2 + K303) (e) Noninverting summing amplifier
The complete response is the sum of the natural response and the forced response X = Xn+Xf Natural response of a first-order circuit Natural response of a second-order circuit CASE Overdamped Critically damped Underdamped az S1 = FORCING FUNCTION K Kt K₁² K sin cot Ke at Forced response of a first-order, or a second-order circuit + A₁ NATURAL FREQUENCIES $1,52 = -α± √√² - 0² $1,$₂=-α S1, S2 = -x±j√√/0²-a² -α±jood dx dt d²x dt² Solution of the Second-Order Differential Equation d²x x(t) = xn(t) + xf (t) a2 dt² + a。x = f(t) + A1 dx dt "xn(t)" = Aest Xn(t) = Ke-t/t + a。x = 0 -a₁ + √²-4a₂a0 242 - Xn(t) = A₁e³₁ª + A₂e³₂t ASSUMED RESPONSE A At + B At² +Bt+C A sin cot + B cos cot Ae-at NATURAL RESPONSE, Xn Ale+Azer (A₁+A₂t)e-at (A, cos coat+A₂ sin coat)e xn(t) = ? (a₂s² + a₁s+ao) = 0 -α₁ - √²-4a₂a0 242 NATURAL RESPONSE, Xn A₁ est + A₂e21 -α1 (A₁+A₂t)e (A₁ cos wat+A₂ sin wat)ext
Transcribed Image Text:The complete response is the sum of the natural response and the forced response X = Xn+Xf Natural response of a first-order circuit Natural response of a second-order circuit CASE Overdamped Critically damped Underdamped az S1 = FORCING FUNCTION K Kt K₁² K sin cot Ke at Forced response of a first-order, or a second-order circuit + A₁ NATURAL FREQUENCIES $1,52 = -α± √√² - 0² $1,$₂=-α S1, S2 = -x±j√√/0²-a² -α±jood dx dt d²x dt² Solution of the Second-Order Differential Equation d²x x(t) = xn(t) + xf (t) a2 dt² + a。x = f(t) + A1 dx dt "xn(t)" = Aest Xn(t) = Ke-t/t + a。x = 0 -a₁ + √²-4a₂a0 242 - Xn(t) = A₁e³₁ª + A₂e³₂t ASSUMED RESPONSE A At + B At² +Bt+C A sin cot + B cos cot Ae-at NATURAL RESPONSE, Xn Ale+Azer (A₁+A₂t)e-at (A, cos coat+A₂ sin coat)e xn(t) = ? (a₂s² + a₁s+ao) = 0 -α₁ - √²-4a₂a0 242 NATURAL RESPONSE, Xn A₁ est + A₂e21 -α1 (A₁+A₂t)e (A₁ cos wat+A₂ sin wat)ext
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