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Alexander-Sadiku Fundamentals of Electric Circuits Chapter 8 Second-Order Circuits Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 Second-Order Circuits Chapter 8 8.1 8.2 8.3 8.4 8.5 Examples of 2nd order RCL circuit The source-free series RLC circuit The source-free parallel RLC circuit Step response of a series RLC circuit Step response of a parallel RLC 2 8.1 Examples of Second Order RLC circuits (1) What is a 2nd order circuit? A second-order circuit is characterized by a secondorder differential equation. It consists of resistors and the equivalent of two energy storage elements. RLC Series RLC Parallel RL T-config RC Pi-config 3 8.2 Source-Free Series RLC Circuits (1) • The solution of the source-free series RLC circuit is called as the natural response of the circuit. • The circuit is excited by the energy initially stored in the capacitor and inductor. 4 8.2 Source-Free Series RLC Circuits (2) To eliminate the integral, we differentiate with respect to t and rearranging the term The 2nd order of expression d 2 i R di i 0 2 L dt LC dt 5 8.2 Source-Free Series RLC Circuits (3) There are three possible solutions for the following 2nd order differential equation: d 2 i R di i 0 2 L dt LC dt => d 2i di 2 2 a w 0i 0 2 dt dt where a R 2L and w0 1 LC General 2nd order Form The types of solutions for i(t) depend on the relative values of a and w. 6 8.2 Source-Free Series RLC Circuits (4) There are three possible solutions for the following 2nd order differential equation: d 2i di 2 2 a w 0i 0 2 dt dt 1. If a > wo, over-damped case i(t ) A1e s1t A2e s2t 2 where s1, 2 a a w0 2 2. If a = wo, critical damped case i(t ) ( A2 A1t )eat where s1, 2 a 3. If a < wo, under-damped case i (t ) e at ( B1 cos w d t B2 sin w d t ) where wd w02 a 2 7 8.2 Source-Free Series RLC Circuits (5) Example 1 If R = 10 Ω, L = 5 H, and C = 2 mF in 8.8, find α, ω0, s1 and s2. What type of natural response will the circuit have? • Please refer to lecture or textbook for more detail elaboration. Answer: underdamped 8 8.2 Source-Free Series RLC Circuits (6) Example 2 The circuit shown below has reached steady state at t = 0-. If the make-before-break switch moves to position b at t = 0, calculate i(t) for t > 0. • Please refer to lecture or textbook for more detail elaboration. Answer: i(t) = e–2.5t[5cos1.6583t – 7.538sin1.6583t] A 9 8.3 Source-Free Parallel RLC Circuits (1) 0 Let 1 i (0) I 0 v(t )dt L v(0) = V0 Apply KCL to the top node: t v 1 dv vdt C 0 R L dt Taking the derivative with respect to t and dividing by C The 2nd order of expression d 2 v 1 dv 1 v0 2 RC dt LC dt 10 8.3 Source-Free Parallel RLC Circuits (2) There are three possible solutions for the following 2nd order differential equation: d 2v dv 2a w02v 0 2 dt dt where a 1 2RC and w0 1 LC 1. If a > wo, over-damped case v(t ) A1 e s1t A2 e s2t where s1, 2 a a 2 w0 2 2. If a = wo, critical damped case v(t ) ( A2 A1t ) e at where s1, 2 a 3. If a < wo, under-damped case v(t ) e at ( B1 cos wd t B2 sin wd t ) where wd w02 a 2 11 8.3 Source-Free Parallel RLC Circuits (3) Example 3 Refer to the circuit shown below. Find v(t) for t > 0. • Please refer to lecture or textbook for more detail elaboration. Answer: v(t) = 66.67(e–10t – e–2.5t) V 12 8.4 Step-Response Series RLC Circuits (1) • The step response is obtained by the sudden application of a dc source. The 2nd order of expression vs d 2 v R dv v 2 L dt LC LC dt The above equation has the same form as the equation for source-free series RLC circuit. • The same coefficients (important in determining the frequency parameters). • Different circuit variable in the equation. 13 8.4 Step-Response Series RLC Circuits (2) The solution of the equation should have two components: the transient response vt(t) & the steady-state response vss(t): v(t ) vt (t ) vss (t ) The transient response vt is the same as that for source-free case vt (t ) A1e s1t A2 e s2t (over-damped) vt (t ) ( A1 A2t )e at (critically damped) vt (t ) e at ( A1 cos wd t A2 sin wd t ) (under-damped) The steady-state response is the final value of v(t). vss(t) = v(∞) The values of A1 and A2 are obtained from the initial conditions: v(0) and dv(0)/dt. 14 8.4 Step-Response Series RLC Circuits (3) p.p. 8.7 Having been in position for a long time, the switch in the circuit below is moved to position b at t = 0. Find v(t) and vR(t) for t > 0. • Please refer to lecture or textbook for more detail elaboration. Answer: v(t) = {10 + [(–2cos3.464t – 1.1547sin3.464t)e–2t]} V vR(t)= [2.31sin3.464t]e–2t V 15 8.5 Step-Response Parallel RLC Circuits (1) • The step response is obtained by the sudden application of a dc source. The 2nd order of expression d 2i 1 di i Is 2 dt RC dt LC LC It has the same form as the equation for source-free parallel RLC circuit. • The same coefficients (important in determining the frequency parameters). • Different circuit variable in the equation. 16 8.5 Step-Response Parallel RLC Circuits (2) The solution of the equation should have two components: the transient response vt(t) & the steady-state response vss(t): i(t ) it (t ) iss (t ) The transient response it is the same as that for source-free case it (t ) A1e s1t A2 e s2t (over-damped) it (t ) ( A1 A2t )e at (critical damped) it (t ) e at ( A1 cos wd t A2 sin wd t ) (under-damped) The steady-state response is the final value of i(t). iss(t) = i(∞) = Is The values of A1 and A2 are obtained from the initial conditions: i(0) and di(0)/dt. 17 8.5 Step-Response Parallel RLC Circuits (3) Example 5 Find i(t) and v(t) for t > 0 in the circuit shown in circuit shown below: • Please refer to lecture or textbook for more detail elaboration. Answer: v(t) = Ldi/dt = 5x20sint = 100sint V 18