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九十七學年度第二學期 電路學（二）授課綱要

九十七學年度第二學期 電路學（二）授課綱要. 課本： Fundamentals of Electric Circuits, 3rd edition. by Charles K. Alexander and Matthew N. O. Sadiku. 計分方式： 平時： 40 % （ 出席： 2 0% ，小考、演習課： 20%） 期中考： 30 % 期末考： 30 %. 內容： Second-Order Circuits Sinusoids and Phasors Sinusoidal Steady-State Analysis 期中考

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九十七學年度第二學期 電路學（二）授課綱要

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  1. 九十七學年度第二學期 電路學（二）授課綱要 課本： Fundamentals of Electric Circuits, 3rd edition. by Charles K. Alexander and Matthew N. O. Sadiku 計分方式： 平時：40 % （出席：20%，小考、演習課：20%） 期中考：30 % 期末考：30 %

  2. 內容： • Second-Order Circuits • Sinusoids and Phasors • Sinusoidal Steady-State Analysis期中考 • AC Power Analysis • Three-Phase Circuits • Magnetically Coupled Circuits • Frequency Response  • 期末考

  3. 電路學(二) Chapter 8 Second-Order Circuits

  4. Second-Order CircuitsChapter 8 8.1 Examples of 2nd order RCL circuit 8.2 Finding Initial and Final Values 8.3 The source-free series RLC circuit 8.4 The source-free parallel RLC circuit 8.5 Step response of a series RLC circuit 8.6 Step response of a parallel RLC 8.7 General Second-Order Circuits

  5. RL T-config RC Pi-config RLC Series RLC Parallel 8.1 Examples of Second Order RLC circuits (1) What is a 2nd order circuit? A second-order circuit is characterized by a second-order differential equation. It consists of resistors and the equivalent of two energy storage elements.

  6. 8.2 Finding Initial and Final Values (1) v(0), i(0), dv(0)/dt, di(0)/dt, i(∞), v(∞) The voltage of a capacitor is always continued The current of a inductor is always continued

  7. 8.2 Finding Initial and Final Values (2) Example 1The switch has been closed for a long time. Find (a) v(0+), i(0+), (b) dv(0+)/dt, di(0+)/dt, (c) i(∞), v(∞)

  8. 8.2 Finding Initial and Final Values (3) Example 2In the circuit, calculate (a) iL(0+), vC(0+), vR(0+), (b) diL(0+)/dt, dvC(0+)/dt, dvR(0+)/dt, (c) iL(∞), vC(∞), vR(∞)

  9. The 2nd order of expression 8.3 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. How to derive and how to solve?

  10. where General 2nd order Form 8.3 Source-Free Series RLC Circuits (2) 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 a and w.

  11. 1. If a > wo, over-damped case where 2. If a = wo, critical damped case where 3.If a < wo, under-damped case where 8.3 Source-Free Series RLC Circuits (3) There are three possible solutions for the following 2nd order differential equation:

  12. 8.3 Source-Free Series RLC Circuits (4) Example 3 If R = 10 Ω, L = 5 H, and C = 2 mF in the circuit, find α, ω0, s1 and s2. What type of natural response will the circuit have?

  13. 8.3 Source-Free Series RLC Circuits (5) Example 4 (p. 324) Find i(t) in the circuit. Assume the circuit has reached steady state at t = 0-.

  14. 8.3 Source-Free Series RLC Circuits (6) Example 5 (p.326) The circuit shown below has reached steady state at t = 0-. If the make-before-break switch moves to position b att = 0, calculate i(t) for t > 0.

  15. The 2nd order of expression 8.4 Source-Free Parallel RLC Circuits (1) Let v(0) = V0 Apply KCL to the top node: Taking the derivative with respect to t and dividing by C

  16. 1. If a > wo, over-damped case where 2. If a = wo, critical damped case where 3.If a < wo, under-damped case where 8.4 Source-Free Parallel RLC Circuits (2) There are three possible solutions for the following 2nd order differential equation:

  17. 8.4 Source-Free Parallel RLC Circuits (3) Example 6 (p.328) In the parallel circuit, find v(t) for t > 0,assuming v(0) = 5 V, i(0) = 0, L = 1 H, and C = 10 mF. Consider these cases: R = 1.932 , R = 5 , and R = 6.25 .

  18. 8.4 Source-Free Parallel RLC Circuits (4) Example 7 (p.330) Find v(t) for t > 0 in the RLC circuit.

  19. 8.4 Source-Free Parallel RLC Circuits (5) Example 8 (p.331) 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

  20. 1 H 1  What type of response is exhibited by the circuit ? 1 F

  21. What type of response is exhibited by the circuit ? 1  1 H 1 F

  22. The 2nd order of expression 8.5 Step-Response Series RLC Circuits (1) • The step response is obtained by the sudden application of a dc source. • 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.

  23. (over-damped) (critically damped) (under-damped) 8.5 Step-Response Series RLC Circuits (2) The solution of the equation should have two components: the transient responsevt(t) & the steady-state responsevss(t): • The transient response vt is the same as that for source-free case • 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.

  24. 8.5 Step-Response Series RLC Circuits (3) Example 9 (p.333) For the circuit, find v(t) for t > 0. Consider these cases: R = 5 , R = 4 , and R = 1 .

  25. 8.5 Step-Response Series RLC Circuits (4) Example 10 (p.336) 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

  26. 8.6 Step-Response Parallel RLC Circuits (1) • 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.

  27. (over-damped) (critical damped) (under-damped) 8.6 Step-Response Parallel RLC Circuits (2) The solution of the equation should have two components: the transient response it(t) & the steady-state response iss(t): • The transient response it is the same as that for source-free case • 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.

  28. 8.6 Step-Response Parallel RLC Circuits (3) Example 11 (p.337) In the circuit, find i(t) and iR(t) for t > 0.

  29. 8.6 Step-Response Parallel RLC Circuits (4) Example 12 (p.339) 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

  30. 8.7 General Second-Order Circuits (1) Given a 2nd-order circuit, we determine its step response x(t) by following steps: • Determine the initial conditions x(0) and dx(0)/dt. • Turn off the independent sources and find the form of the transient response xt(t). • Obtain the steady-state response xss(t) = x(∞). • The total response x(t) = xt(t) + xss(t)Determine the constants associated with xt(t) by imposing the initial conditions.

  31. 8.7 General Second-Order Circuits (2) Example 13 (p.339) Find complete response v and then i in the circuit.

  32. 8.7 General Second-Order Circuits (3) Example 13 (p.341) Find vo(t)for t > 0 in the circuit.

  33. 8.8 Second-Order Op Amp Circuits (1) Because inductors are bulky and heavy, we only consider RC second-order op amp circuit. Example 14 (p.344) In the op amp circuit, find vo(t)for t > 0 when vs = 10u(t) mV. Let R1 = R2 = 10 k, C1 = 20 μF, and C2= 100 μF

  34. 8.8 Duality (1) Two circuits are said to be duals of one another if they are described by the same charactering equations with dual quantities interchanged. • Construct the dual circuit of a given circuit • Place a node at the center of each mesh. Place the reference node outside the given circuit. • Draw lines between these nodes such that each line crosses an element. Replace the element by its dual. • Determine the polarity of voltage sources and direction of current sources. A voltage source produce clockwise mesh current has a dual current source entering to the corresponded node.A current source is in the same direction as the mesh current has a dual voltage source which positive terminal connected to the corresponded node.

  35. 8.8 Duality (2) Example 15 (p.351) Construct the dual of the circuit.

  36. 8.8 Duality (3) Example 16 (p.351) Obtain the dual of the circuit.

  37. 8.9 Applications (1) Example 17 (p.353) Assuming that the switch in the figure is closed prior to t = 0-, find the inductor voltage vL(t)for t > 0.

  38. 8.9 Applications (2) Example 18 (p.353) The output of a D/A converter is shown in the figure (a). If the RLC circuit in figure (b) is used as smoothing circuit, determine the output voltage vo(t).

  39. 3, 4, 5, 10, 12, 14, 15, 19, 25, 29, 33, 37, 41, 43, 46, 49, 53, 55, 58, 62, 63, 65, 67, 75, 76, 78

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