2. NO.11 Chapter 6 First-Order Circuit 一阶电路

3. 1.RC and RL Circuits 2.Initial Conditions 3.First-order Circuit Zero-input Response 4.First-order Circuit Zero-state Response 5.First-order Circuit Complete Response 6.Three quantities in finding the response 7.Applications Items:

4. Introduction 换路：即电路变化 C t = 0 UsUs R K + – uCuC RC Circuits t = 0 + – uRuR UsUs R K RLRL R Circuits + – uRuR UsUs R RLRL

5. Before K is switched on i = 0, u C = 0 i = 0, u C = U s (I) Dynamic circuit i + – uCuC UsUs R C Introduction to dynamic circuits Steady state analysis K + – uCuC UsUs R C i t = 0 K is switched on long time

6. K + – uCuC UsUs R C i Initial state Transient state Steady state t1t1 USUS ucuc t 0 ? i Response of a Circuit Transient response of an RL or RC circuit is – Behavior when voltage or current source are suddenly applied to or removed from the circuit due to switching. – Temporary behavior Dynamic Circuit ： circuit which containing energy-store elements (L, C). 过渡过程

7. (II) Why the circuit produce transient process 1. The circuit containing L and C elements Inductor current cannot change instantaneously Capacitor voltage cannot change instantaneously 2. voltage or current source are suddenly applied to or removed from the circuit switching

8. 动态电路及过渡过程 动态电路 dynamic circuit 特点：当改变原 来的工作状态时会有过渡过程。（瞬态） 原稳态 开关动作 接入 ( 去掉 ) 电源 换路 过渡过程 新稳态 Transient state

9. 研究动态电路基本规律的意义： 认识、掌握过渡过程 ( 暂态过程 transient state) 的现象和规律。 动态电路的基本规律反映出一般动态 系统（机械、自动控制系统等）的普遍规 律。

10. FORMULATING RC AND RL CIRCUIT EQUATIONS A first-order circuit is characterized by a first- order differential equation. kk

11. 6-1 Initial conditions of the circuits Initial conditions ： The values of voltages and currents and the derivatives at t = 0+: 1. Initial conditions ： The values of voltages and currents and the derivatives at t = 0+: 0 t 确定微分常数时须利用电路初始条件 f(0 + ) just prior to switchingjust after switching Switching at t=0

12. 2. Switching rule 换路规则： At the instant of switching, if the current flowing through the capacitance is finite, the charges remain unchanged; if the voltage across the inductance is finite, the current flowing through the inductance remains unchanged. u C (0 + )=u C (0 - ) i L (0 + )=i L (0 - ) Note ： Only the voltage on a capacitor and the current through an inductor cannot chang abruptly 。

13. 3. 电路中其他初始电压、电流的一般求解方法： 具体求法是： 画出 t=0 + 电路：在该电路中若 u C (0 + )= u C (0 - )= U S ， 电容用一个电压源 U S 代替，若 u C (0 + )= 0 则电容用 a short circuit 代替。若 i L (0 + )= i L (0 - )=I S ，电感用一 个电流源 I S 代替，若 i L (0 + )= 0 则电感作 an open circuit 。 由 t=0 + 电路来求得 电阻电路

14. S(t=0) + - + - + - t 0- 0+ 5A10V 0 5A 0010V 5A10V 0 0 -10A -10V 15A Example 1

15. Assume Close the switch S when t=0. Find i 1 (0 + ) 、 i 2 (0 + ) 、 i 3 (0 + ) 、 u c (0 + ) 、 u L (0 + ). example2 = = 0 t=0 + circuit

16. 求初始值的一般步骤： (1) 根据 t=0 - 时的电路，求出 u C (0 - ) 及 i L (0 - ) 。 (2) 由换路定则 : u C (0 + ) = u C (0 - ), i L (0 + ) = i L (0 - ) 。 (3) 作出 t=0 + 时的等效电路，并在图上标出各待 求量。 (4) 由 t=0 + 等效电路，求出各待求量的初始值。

17. §6-2 Zero-input response of first-order RC circuits 一阶 RC 电路的零输入响应 Assume that at time t=0, shut the key K, and u c (0 - )=U 0 Find ： u c (t) ， i c (t) （ t≥0 ） (t=0) Capacitor Discharging 即：放电过程 discharge Solution: Zero input response ： The behavior (in terms of the voltages and currents) of the circuit itself, with no external sources of excitation.

18. Solution: -u R +u c =0 By KVL: ∵ u R =i R, Eq.(7-1) is a homogeneous equation because the right side is zero. Eq.(7-1) A solution in the form of an exponential u c =Ke st t≥0 Eq.(7-2) where K and s are constants to be determined.

19. characteristic equation Substituting the trial solution into Eq.(7-1) yields OR Eq.(7-3) R T Cs+1=0 a single root of the characteristic equation zero -input response of the RC circuit: ∵ u c (0 + )=U 0

20. The time constant 时间常数 τ=RC the zero-input response: τ 小：过渡过程短 τ 大：过渡过程长 3  ～ 5  It is customary to assume that it takes 3  ～ 5  for the circuit to reach the new steady state. 0.368U 0 k (t=0) (t=0) u c (t) U0U0

21. t≥0 i (t=0)

22. Note: The key to working with a source-free RC circuit is finding: (1)The initial voltage u(0)=U 0 across the capacitor; (2)The time constant τ;

23. Example 7.1 Let u c (0-)=15V. Find u C,u x, and i x for t>0. uCuC uxux

24. R0R0 R L U0U0 uLuL 1 2 S(t=0) i §6-3 Zero-input response of first-order RL circuits Assume : S 2 Find : i L (t), t≥0

25. From the initial conditions ： Let i = Ae pt Then (Lp+R)e pt =0Lp+R=0 Hence ： =

26. Then ： R0R0 R L U0U0 uLuL 12 S(t=0) i

27. Curve ： t O i L, u L -RI 0 I0I0 i L (t) u L (t)

28. §6-2 and §6-3 Zero-input Response (t=0) Capacitor Discharging u c (0 - )=U 0 K(t=0) i L uLuL R τ=RC

29. 为什么ＲＣ的 τ 与Ｒ成正比，ＬＣ的却成反比？ 因为 RC 的 i 0 =U 0 /R, 此时Ｒ的耗能为 U 2 0 ／ Ｒ, 与Ｒ成反比，即说明Ｒ越大，过渡时 间持续越长， τ 与Ｒ成正比； 而 LC 的 i 0 = Ｉ 0, 此时Ｒ的耗能为Ｉ 2 0 Ｒ, 与 Ｒ成正比，即说明Ｒ越大，过渡时间持 续越短， τ 与Ｒ成反比． K(t=0) i L uLuL R

30. R0R0 R L U0U0 uLuL 1 2 S(t=0) i Zero-input response of first-order RL circuits Assume : S 2 Find : i L (t), u R (t), t≥0 Multisim

31. 第９周周一 期中考试 范围 : 1-6 章 方式 : 闭卷 ( 英文试题 )

32. 例 7.3.2 下图是一台 300kw 汽轮发电机的励磁回路。已知 R=0.189Ω, L=0.398H, U=35V ，电压表量程为 50V ，内阻 R V =5kΩ 。 t=0 时开关 S 打开（设 S 打开前电路已稳定）. 求： 1 、 i(0 + ) 2 、 τ （ t≥0 ） 3 、 i 和 u v （ t≥0 ） 4 、 S 刚断开时的 u v (0 + ) Multisim τ=L/R=79.6μs, D _ + i(0-)=i(0+)=185A, 解:解: t=0

33. §6-4 First-order Circuit Complete Response 全响应

34. UsUs For t ≥0  If u c (0)=0, it is Zero-State Response. 零状态响应 u c (t) t=0 1. Complete Response of a RC Circuit

35. How to find the complete response of the RC circuit ?  If u c (0) ≠ 0, it is Complete Response. 全响应 UsUs For t ≥0 u c (t) t=0

36. divide solution v(t) into two components: The homogeneous solution (natural response) is the general solution of Eq.1 when the input is set to zero. total particular homogeneous

37. The particular solution ( forced response) : seek a particular solution of the equation t ≥0 Now combining the forced and natural responses, we obtain Setting U F (t)=U S meets this condition.

38. UsUs  Zero-State Response : u c (0)=0 u c (t) t=0 ∴ K= -U S using the initial condition:

39. O USUS t u C (t) i(t) u C (t), i(t) 即：充电过程 charge

40.  不论 R 、 C 如何，电源充 电能量的一半被 R 吸收， 一半转换为电容的电场能 量，充电效率为 50 ％。

41. UsUs u(t) t=0 ∴ K= U 0 -U S  Complete Response : u c (0)=U 0 The complete response of the RC circuit: u c (0)=U 0

42. Complete Response : Step response of first-order RC circuit U0U0 USUS

43. Note: 全响应 = 零输入响应 + 零状态响应 Zero-input responseZero-State Response The superposition principle u c (t) USUS t=0 u c (0)=U 0

44. The RL circuit is the dual of the RC circuit: L ISIS The complete response of the RL circuit is 2. Complete Response of a RL Circuit

45. +_+_ u c (t) 30Ω 20Ω t=0 K I S =1A 0.5F Example1 : Example1 : The switch is opened at t=0. u c (0) =5V, Find u C (t) ． (t≥0) icic 方法一：（经典法）求解微分方程

46. Example1 : Example1 : The switch is opened at t=0. u c (0) =5V, Find u C (t) ． (t≥0) t≥0 时的电路 方法二： 先利用戴维南等效， 再套全响应公式 icic

47. Solution: Zero-input: Zero-state: Complete response:

48. 3 、 Step function 阶跃函数 1 The unit step function 单位阶跃激励 : ε(t) t 0 The unit step function ε (t) is 0 for negative values of t and 1 for positive values of t.

49.  延迟单位阶跃信号

50. We use the step function to represent an abrupt change in voltage or current, for example, the voltage May be expressed : U A ε (t-t 0 ) abab Equivalent circuit: abab UAUA = t=t 0

51. 30Ω +_+_ u c (t) 20Ω t=0 K I S 1A 0.5F Example 1: Example 1: The switch is opened at t=0. u c (0) =5V, Find u C (t), (t≥0)

52. Solution: Zero-input: Zero-state: Complete response:

53. usus Example 2: Find u C (t), for t≥0. Solution : 方法一 : 分段分析 t o usus 10V 1S u c (0) =0V,

54. usus t o usus 10V 1S t≥1s U C (1) u c (0) =0V,

55. 1) 用分段函数表示

56. 方法二 : 用阶跃函数表示 usus t o usus 10V 1S t o t o u ´ u "

57. t o u 10V 1S1S usus According to the superposition principle :

59. s(t=0) Determine the expressions for u C (t) and i(t). (t ≥0 ) Example ： i(t) 0.1F

60. 50/7V 10/7Ω uCuC t≥0t≥0 i 0.1F s(t=0)

61. The question is changed to ‘Determine i(t)’. (t ≥0 ) Example ： i(t) 0.1F Is there another way?

62. The complete response of a first-order circuits depends on three quantities: 1.The initial value of state variable(U 0+ or I 0+ ) 2.The final value of state variable(U S or I S ) 3.The time constant  (RC or GL) f(0 + ) f( ∞ ) 三要素法 Note:

63. (1) Get f(0+) --- use 0+ equivalent circuit (2) Get f(  )---use  equivalent circuit (3) Get τ---calculate the equivalent resistance R, τ=RC or L/ R Then, Note: method of “three quantities” can be applied in step response on any branch of First-order circuit. 6. Method of “three quantities” (method 3)

64. Solution ： Using the method of “three quantities” (1) u c (0+), i(0+) s(t=0) Determine the expressions for u C (t) and i(t). (t ≥0 ) Example ： i(t) 0.1F ? ×

65. Find i(0+): s(t=0) i 0.1F i(0+): Equivalent circuit at t=0+ 5V

66. s(t=0) (2) u c (∞), i(∞) 10V 2Ω 5Ω u C (∞) i(∞)

67. s(t=0) (3) τ (t ≥0) R=2//5=10/7Ω R 2Ω 5Ω

68. 4、4、 s(t=0) i (t≥0)

69. s(t=0) i (t≥0)

70.  How to get initial value f(0+) ? 1.the capacitor voltage and inductor current are always continuous in some conditions. ( 换路定则 ) Vc(0 + )=Vc(0 - ); I L (0 + )=I L (0 - ) 2. ---use 0+ equivalent circuit. C: substituted by voltage source; L: substituted by current source. 3. Find f(0+) in the above DC circuit. Note:

71.  How to get final value f(∞) ?  How to get time constant τ? The key point is to get the equivalent resistance R. R is the Thévenin equivalent resistance “seen” by the inductor (or the capacitor) Use ∞ equivalent circuit(stead state) to get f(∞). C: open circuit; L: short circuit.

72. Example 2: Example 2: The switch is closed at t=0. i L (0) =2A, Find i L (t), u L (t), i (t) (t≥0) Solution ： “Three quantities”

73. (1) Find f(0 + ):

74. When t→ ∞ (2) Find f(∞ ):

75. (3)Find ： R a b 外加电源法

76. (4)According to:

77. §6-5 Applications Differential Circuit and Integral Circuit

78. 1 、 Differential Circuit o t uiui /V 10 t1t1 t2t2 tPtP o t - 10 uCuC u O /V 10  <<t P 微分电路的条件 (1)  << t P ； (2) u R as output R C uiui uOuO uCuC i Example:

79. 2. Integral Circuit o t uiui /V 10 t1t1 t2t2 t o u O /V tPtP  >> t P C R uiui uOuO uRuR i 积分电路的条件 (1)  >> t P ； (2) u C as output

80. §6-6 一阶电路的冲击响应 （不讲）

81. 作业 11 ： 《电路》 : P150 6-1 《 Fundamentals of Electric Circuits 》 : P284 7.5 P286 7.20

82. 作业 12 ： 《 Fundamentals of Electric Circuits 》 : P288-289 7.36 7.43 《电路》 P155 6-23

83. 作业 13: 《电路》 P152-155 6-8 6-21 《 Fundamentals of Electric Circuits 》 : P291 7.55