1. Chapter 5 First-Order and Second Circuits 1

2. First-Order and Second Circuits Chapter 5 5.1Natural response of RL and RC Circuit 5.2Force response of RL and RC Circuit 5.3Solution of natural response and force response in RL and RC Circuit 5.4 Natural and force response in series RLC Circuit 5.5Natural and force response in parallel RLC Circuit 2

3. 5.1 Natural response of RC circuit (1) A first-order circuit is characterized by a first-order differential equation. 3 Apply Kirchhoff’s laws to purely resistive circuit results in algebraic equations. Apply the laws to RC and RL circuits produces differential equations. Ohms lawCapacitor law By KCL

4. 5.1 Natural response of 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. 4 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 . Time constant Decays more slowly Decays faster

5. 5.1 Natural response of RC circuit (3) The key to working with a source-free RC circuit is finding: 5 1.The initial voltage v(0) = V 0 across the capacitor. 2.The time constant  = RC. where DC source disconnected

6. 5.1 Natural response of RC circuit (4) Example 1 Refer to the circuit below, determine v C, v x, and i o for t ≥ 0. Assume that v C (0) = 30 V. 6 Please refer to lecture or textbook for more detail elaboration. Answer: v C = 30e –0.25t V ; v x = 10e –0.25t ; i o = –2.5e –0.25t A

7. Solution 1 7 v C (0) = 30 V. = v C = 30e –0.25t V v x = 10e –0.25t i o = –1.25e –0.25t A

8. 5.1 Natural response of RC circuit (5) Example 2 The switch in circuit below is opened at t = 0, find v(t) for t ≥ 0. 8 Please refer to lecture or textbook for more detail elaboration. Answer: V(t) = 8e –2t V

9. 9 Solution 2 1/6 F V(t) = 8e –2t V

10. 5.1 Natural response of RL circuit (6) A first-order RL circuit consists of a inductor L (or its equivalent) and a resistor (or its equivalent) 10 By KVL Inductors lawOhms law

11. 5.1 Natural response of RL circuit (7) 11 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. A general form representing a RL where

12. 5.1 Natural response of RL circuit(8) 12 A RL source-free circuit where A RC source-free circuit where Comparison between a RL and RC circuit

13. 5.1 Natural response of RL circuit(9) The key to working with a source-free RL circuit is finding: 13 1.The initial voltage i(0) = I 0 through the inductor. 2.The time constant  = L/R. where

14. 5.1 Natural response of RL circuit(10) Example 3 Find i and v x in the circuit. Assume that i(0) = 5 A. 14 Answer: i(t) = 5e –53t A

15. Solution 3 find r th 15 1 5 3 ? ? ?

16. 5.1 Natural response of RL circuit(11) Example 4 For the circuit, find i(t) for t > 0. 16 Please refer to lecture or textbook for more detail elaboration. Answer: i(t) = 2e –2t A

17. 17 t=0, SC i(0) = current division ??

18. Unit-Step Function (1) The unit step function u(t) is 0 for negative values of t and 1 for positive values of t. 18

19. Unit-Step Function (2) 1. voltage source. 2. for current source: 19 Represent an abrupt change for :

20. 5.2 Force response of 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. 20 Initial condition: v(0-) = v(0+) = V 0 Applying KCL, or Where u(t) is the unit-step function

21. 5.2 Force response of RC Circuit(2) Three steps to find out the step response of an RC circuit: 21 1.The initial capacitor voltage v(0). 2.The final capacitor voltage v() — DC voltage across C. 3.The time constant . 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(3) 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. 22 Please refer to lecture or textbook for more detail elaboration. Answer: and v(0.5) = 0.5182V

23. Solution Ex 5 1 - The initial capacitor voltage v(0). 2- final capacitor voltage v(  ) Use KCL at nod to get v(∞) 3- time constant . Use R th  =RC 23

24. 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. 24 Initial current i(0-) = i(0+) = I o Final inductor current i(∞) = Vs/R Time constant  = L/R

25. 5.2 Force response of RL Circuit(2) Three steps to find out the step response of an RL circuit: 25 1.The initial inductor current i(0) at t = 0+. 2.The final inductor current i(). 3.The time constant . 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.

26. 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. 26 Please refer to lecture or textbook for more detail elaboration. Answer:

27. Example 6 (solution) 27 30V Apply source transformation 30 2A

28. 5.3Solution of natural and force response in RL and RC Circuit Integrating both sides and considering the initial conditions, the solution of the equation is: 28 Final value at t -> ∞ Initial value at t = 0 Source-free Response Complete Response = Natural response + Forced Response (stored energy) (independent source) = V 0 e –t/τ + V s (1–e –t/τ )

29. Second order circuit 29

30. Examples of Second Order RLC circuits (1) What is a 2nd order circuit? 30 A second-order circuit is characterized by a second- order differential equation. It consists of resistors and the equivalent of two energy storage elements. RLC SeriesRLC Parallel RL T-configRC Pi-config

31. Source-Free Series RLC Circuits (1) 31 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 How to derive and how to solve?

32. Source-Free Series RLC Circuits (2) At t=0, So, Eliminate integral, differentiate to t, rearrange  32

33. Source-Free Series RLC Circuits (3) 33 There are three possible solutions for the following 2nd order differential equation: The types of solutions for i(t) depend on the relative values of  and  => General 2nd order Form where

34. 34 There are three possible solutions for the following 2nd order differential equation: 1. If  >  o, over-damped case where 2. If  =  o, critical damped case where 3. If  <  o, under-damped case where Source-Free Series RLC Circuits (4)

35. 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? 35 Please refer to lecture or textbook for more detail elaboration. Answer: underdamped

36. 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. 36 Please refer to lecture or textbook for more detail elaboration. Answer: i(t) = e –2.5t [5cos1.6583t – 7.538sin1.6583t] A

37. t>0 under-damped case

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40. Source-Free Parallel RLC Circuits (1) 40 The 2nd order of expression Let v(0) = V 0 Apply KCL to the top node: Taking the derivative with respect to t and dividing by C

41. Source-Free Parallel RLC Circuits (2) 41 There are three possible solutions for the following 2nd order differential equation: 1. If  >  o, over-damped case where 2. If  =  o, critical damped case where 3. If  <  o, under-damped case where

42. Source-Free Parallel RLC Circuits (3) Example 3 Refer to the circuit shown below. Find v(t) for t > 0. 42 Please refer to lecture or textbook for more detail elaboration. Answer: v(t) = 66.67(e –10t – e –2.5t ) V

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44. 44 : v(t) = 66.67(e –10t – e –2.5t ) V

45. Step-Response Series RLC Circuits (1) 45 The step response is obtained by the sudden application of a dc source. The 2nd order of expression 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.

46. Step-Response Series RLC Circuits (2) 46 The solution of the equation should have two components: the transient response v t (t) & the steady-state response v ss (t):  The transient response v t is the same as that for source-free case  The steady-state response is the final value of v(t).  v ss (t) = v(∞)  The values of A 1 and A 2 are obtained from the initial conditions:  v(0) and dv(0)/dt. (over-damped) (critically damped) (under-damped)

47. 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 v R (t) for t > 0. 47 Please refer to lecture or textbook for more detail elaboration. Answer: v(t) = {10 + [(–2cos3.464t – 1.1547sin3.464t)e–2t]} V v R (t)= [2.31sin3.464t]e –2t V

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50. Step-Response Parallel RLC Circuits (1) 50 The step response is obtained by the sudden application of a dc source. The 2nd order of expression 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.

51. Step-Response Parallel RLC Circuits (2) 51 The solution of the equation should have two components: the transient response v t (t) & the steady-state response v ss (t):  The transient response i t is the same as that for source-free case  The steady-state response is the final value of i(t).  i ss (t) = i(∞) = I s  The values of A 1 and A 2 are obtained from the initial conditions:  i(0) and di(0)/dt. (over-damped) (critical damped) (under-damped)

52. 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: 52 Please refer to lecture or textbook for more detail elaboration. Answer: v(t) = Ldi/dt = 5x20sint = 100sint V

53. 53 v(t) = Ldi/dt = 5x20sint = 100sint V