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EKT101 Electric Circuit Theory Chapter 5 First-Order and Second Circuits 1 First-Order and Second Circuits Chapter 5 5.1 Natural response of RL and RC Circuit 5.2 Force response of RL and RC Circuit 5.3 Solution of natural response and force response in RL and RC Circuit 5.4 Natural and force response in series RLC Circuit 5.5 Natural and force response in parallel RLC Circuit 2 5.1 Natural response of RL and RC circuit (1) • A first-order circuit is characterized by a firstorder differential equation. By KCL iR iC 0 Ohms law v dv C 0 R dt Capacitor law • Apply Kirchhoff’s laws to purely resistive circuit results in algebraic equations. • Apply the laws to RC and RL circuits produces differential equations. 3 5.1 Natural response of RL and RC circuit (2) • The natural response of a circuit refers to the behavior (in terms of voltages and currents) of the circuit itself, with no external sources of excitation. Time constant RC Decays more slowly Decays faster • The time constant of a circuit is the time required for the response to decay by a factor of 1/e or 36.8% of its initial value. • v decays faster for small and slower for large . 4 5.1 Natural response of RL and RC circuit (3) DC source disconnected The key to working with a source-free RC circuit is finding: v(t ) V0 e t / where RC 1. The initial voltage v(0) = V0 across the capacitor. 2. The time constant = RC. 5 5.1 Natural response of RL and RC circuit (4) Example 1 Refer to the circuit below, determine vC, vx, and io for t ≥ 0. Assume that vC(0) = 30 V. • Please refer to lecture or textbook for more detail elaboration. Answer: vC = 30e–0.25t V ; vx = 10e–0.25t ; io = –2.5e–0.25t A 6 Solution 1 vC(0) = 30 V. = vC = 30e–0.25t V vx = 10e–0.25t io = –2.5e–0.25t A 7 5.1 Natural response of RL and RC circuit (5) Example 2 The switch in circuit below is opened at t = 0, find v(t) for t ≥ 0. • Please refer to lecture or textbook for more detail elaboration. Answer: V(t) = 8e–2t V 8 Solution 2 1/6 F V(t) = 8e–2t V 9 5.1 Natural response of RL and RC circuit (6) • A first-order RL circuit consists of a inductor L (or its equivalent) and a resistor (or its equivalent) By KVL vL vR 0 di L iR 0 dt Inductors law Ohms law di R dt i L i (t ) I 0 e Rt / L 10 5.1 Natural response of RL and RC circuit (7) A general form representing a RL i (t ) I 0 e where • • • t / L R The time constant of a circuit is the time required for the response to decay by a factor of 1/e or 36.8% of its initial value. i(t) decays faster for small and slower for large . The general form is very similar to a RC source-free circuit. 11 5.1 Natural response of RL and RC circuit(8) Comparison between a RL and RC circuit A RL source-free circuit i (t ) I 0 e t / where L R A RC source-free circuit v(t ) V0 e t / where RC 12 5.1 Natural response of RL and RC circuit(9) The key to working with a source-free circuit is finding: i (t ) I 0 e t / where RL L R 1. The initial voltage i(0) = I0 through the inductor. 2. The time constant = L/R. 13 5.1 Natural response of RL and RC circuit(10) Example 3 Find i and vx in the circuit. Assume that i(0) = 5 A. Answer: i(t) = 5e–53t A 14 3 Solution 3 find rth 1 5 15 5.1 Natural response of RL and RC circuit(11) Example 4 For the circuit, find i(t) for t > 0. • Please refer to lecture or textbook for more detail elaboration. Answer: i(t) = 2e–2t A 16 17 Unit-Step Function (1) • The unit step function u(t) is 0 for negative values of t and 1 for positive values of t. 0, u (t ) 1, t0 t0 0, u (t to ) 1, t to 0, u (t to ) 1, t to t to t to 18 Unit-Step Function (2) Represent an abrupt change for: 1. voltage source. 2. for current source: 19 5.2 Force response of RL and RC Circuit (1) • The step response of a circuit is its behavior when the excitation is the step function, which may be a voltage or a current source. • Initial condition: v(0-) = v(0+) = V0 • Applying KCL, dv v Vs u (t ) c 0 dt R or v Vs dv u (t ) dt RC • Where u(t) is the unit-step function 20 5.3 Solution of natural and force response in RL and RC Circuit • Integrating both sides and considering the initial conditions, the solution of the equation is: t0 V0 v(t ) t / V ( V V ) e 0 s s Final value at t -> ∞ Complete Response = = Natural response (stored energy) V0 e–t/τ Initial value at t = 0 + + t 0 Source-free Response Forced Response (independent source) Vs (1–e–t/τ) 21 5.2 Force response of RC Circuit(3) Three steps to find out the step response of an RC circuit: 1. The initial capacitor voltage v(0). 2. The final capacitor voltage v() — DC voltage across C. 3. The time constant . v (t ) v () [v (0) v ()] e t / Note: The above method is a short-cut method. You may also determine the solution by setting up the circuit formula directly using KCL, KVL , ohms law, capacitor and inductor VI laws. 22 5.2 Force response of RC Circuit(4) Example 5 Find v(t) for t > 0 in the circuit in below. Assume the switch has been open for a long time and is closed at t = 0. Calculate v(t) at t = 0.5. • Please refer to lecture or textbook for more detail elaboration. Answer: v(t ) 15e 2t 5 and v(0.5) = 0.5182V 23 5.2 Force response of RL Circuit(1) • The step response of a circuit is its behavior when the excitation is the step function, which may be a voltage or a current source. • Initial current i(0-) = i(0+) = Io • Final inductor current i(∞) = Vs/R • Time constant = L/R t Vs Vs i (t ) ( I o )e u (t ) R R 24 5.2 Force response of RL Circuit(2) Three steps to find out the step response of an RL circuit: 1. The initial inductor current i(0) at t = 0+. 2. The final inductor current i(). 3. The time constant . i (t ) i () [i (0) i ()] e t / Note: The above method is a short-cut method. You may also determine the solution by setting up the circuit formula directly using KCL, KVL , ohms law, capacitor and inductor VI laws. 25 5.2 Force response of RL Circuit(3) Example 6 The switch in the circuit shown below has been closed for a long time. It opens at t = 0. Find i(t) for t > 0. • Please refer to lecture or textbook for more detail elaboration. Answer: i(t ) 2 e10t 26 Example 6 (solution) Apply source transformation 30V i ( 0) 3 A 30 2A i(t ) 2 e10t 27 Second order circuit 5.4 NATURAL AND FORCE RESPONSE IN SERIES RLC CIRCUIT 28 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 29 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. The 2nd order of expression d 2i R di i 0 2 dt L dt LC How to derive and how to solve? 30 Source-Free Series RLC Circuits (2) Method will be illustrated during the lecture 31 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. 32 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 33 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 34 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 35 t>0 under-damped case 37 5.4 NATURAL AND FORCE RESPONSE IN PARALLEL RLC CIRCUIT 38 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 39 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 40 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 41 42 : v(t) = 66.67(e–10t – e–2.5t) V 43 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. 44 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. 45 Step-Response Series RLC Circuits (3) Example 4 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 46 47 48 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. 49 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. 50 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 51 v(t) = Ldi/dt = 5x20sint = 100sint V 52