2. 2 、 Zero-input Response, Zero-state Response,Complete Response Outline ： 4 、 Basic Signals ： Unit-step Function and Unit-impulse Function 3 、 Steady-state Response Transient Response 1 、 Switching Theorem Three-factor Method Chapter 6 First-Order Circuit (transient state)

3. L i u i u C R i u u=Ri For the direct current case, l is short-circuited For the direct current case,c is open-circuited

4. before the movement of K i = 0, u C = 0 i = 0, u C = U s 一、 transition process of circuit K + – uCuC UsUs R C i t = 0 Long after K is closed Transition process ： the process be undergone by the circuit transienting from one s teady state to another steady state. switching ： the circuit (structure or parameter) changes. i + – uCuC UsUs R C K is closed Summary

5. 2. the structure or parameter of the circuit are changed 三、 D ifferences between steay-state analysis and transient analysis steady state transient 1. long after switching ； just after switching: i L 、 u C time-varying 3.Circuit discribed with algebraic equations ； Circuit discribed with differential equations 2. I L 、 U C keep unchanged 二、 Reasons for transition process 1.The circuit contains energy-storage elements L 、 M 、 C

6. 一、 about t = 0 - and t = 0 + switching operation occurs at t=0 ， － ∞ 0 - 0 0 + +∞ K + – uCuC UsUs R C i t = 0 initial condition is ： (1)the value of u ， i,when t = 0 +, (2) the value at t = t 0+, when closing the switch at the instant of t= t 0 original steady state original steady - state final value the instant of switching transition process new steady state initial value after switching §6-1 Equation of Dynamic Circuit and Initial Conditions

7. if i (  ) is finit value i ucuc C + - calculate u c (0 + ) at t = 0+ 1.The Capacitor 二、 Switching Theorem u c (0 + ) = u c (0 - ) Conclusion ： At the instant of switching,if the current through capacitor is finite value, voltage across capacitor (charge) keeps unchanged before and after switching.

8. iLiL u L + - find the value of i L (0 + ) at t = 0 + 2.The Inductor i L (0 + )= i L (0 - ) if u(  ) is finite value Conclusion ： at the instant of switching ， if the inductor voltage is finit value, the inductor currentremains unchanged before and after switching. Switching Theorem ： capacitor voltage and inductor current cannot change abrubtly at the instant of switching. bear in mind u c (0 + ) = u c (0 - )i L (0 + )= i L (0 - )

9. solution ： (1) find u C (0 - ) according to 0 - circuit example1 +-+- 10V i iCiC +uC-+uC- k 10kΩ 40kΩ Open k at t = 0,find i C (0 + ) ？ i C (0 -- ) =0 i C (0 + )= i C (0 -- )=0 +-+- 10V +uC-+uC- 10kΩ 40kΩ iCiC u C (0 + ) = u C (0 - )=8V (2)according to switching theorem +-+- 10V i iCiC + 8V - 10kΩ t=0 + equivalent circuit (3)find i C (0 + ) according to equivalent circuit at t=0 + 三、 Determination for initial conditions

10. i L (0 + )= i L (0 - ) =10/(1+4)=2A +uL-+uL- 10V 11 44 0 + circuit 2A Solutoin ： ∴ u L (0 + )=u L (0 - )=0 iLiL +uL-+uL- L 10V K 11 44 Close k at t = 0, compute u L (0 + )? Additional example +uL-+uL- 44 2A

11. 3. Draw equivalent circuit diagram at t=0 + 。 Replace capacitor (inductor ) by voltage source （ current source) 。 Get the value at t=0 + ， the direction is identical to assumed direction of capacitor voltage and inductor current. 4. Find the value of desired variables at t=0 + within 0 + circuit diagram. 1.Find u C (0 - ) or i L (0 - ) with the aid of the circuit(steady state) before switching 。 2. Determine u C (0 + )or i L (0 + ) with application of switching theorem 。 Steps of finding initial conditions: Link

12. i L (0 + ) = i L (0 - ) = I S u C (0 + ) = u C (0 - ) = RI S u L (0 + )= - RI S 0 + circuit diagram uLuL + – iCiC R ISIS R I S + – Compute i C (0 + ), u L (0 + ) ？ K(t=0) + – uLuL iLiL C + – uCuC L R ISIS iCiC Solution ：（ 1 ） Initiial value （ 2 ） At t=0+ Additional Example ：

13. 一、 Zero-input response of RC Given u C (0 - )=U 0, compute u C and i 。 solution: i K(t=0) + – uRuR C + – uCuC R u C -u R =u C -Ri=0 eigenfunction root RCp+1=0 eigenequation hence §6-2 Zero-input Response of First-Order Circuit Excitation (source)is switched off ， the response is caused only by initial energy- storage of capacitor and inductor assume Review ： first-order diffential equation contain only one edergy- storage component

14. initial value is u C (0 + )=u C (0 - )=U 0 A=U0A=U0 let  =RC,  is called time constant of one-order circuit t U0U0 uCuC 0 I0I0 t i 0 ∴ i K(t=0) + – uRuR C + – uCuC R

15. The value of time constant  reflects the length of time spent on a transient. the largerer the ,the longer the transient duration Small  accords with short transient duration If initial value of voltage is fixed ： R derease （ C remain unchanged ） i=u/R smaller discharging current longer discharging time U0U0 t ucuc 0 small   C increase （ R remain unchanged ） W=0.5Cu 2 bigger energy-storage  = R C Link big

16. In engineering perspective,a transient lasts 3  - 5  U 0 0.368 U 0 0.135 U 0 0.05 U 0 0.007 U 0 t 0  2  3  5  U 0 U 0 e -1 U 0 e -2 U 0 e -3 U 0 e -5 If U c decays at original velocity plot the tangent across starting point  ： the lenghth of time that is spent by capacitor voltage on decaying to 36.8% of the original voltage 。 U0U0 t ucuc 0  0.368U 0

17. eigenequation Lp+R=0 from initial value i (0 + )= I 0, derive, A= i(0 + )= I 0 i (0 + ) = i (0 - ) = i K(t=0) USUS L + – uLuL RR1R1 二. Zero-input Response of RL Circuit Assume let  = L/R, time constant of first- order RL t I0I0 i 0  -RI 0 uLuL t

18.  L big （ R remains unchanged ） big initial energy R small （ L remains unchanged ） samll energy cosumed in the process of discharging bigger , slower discharging The initial value is fixed ：  = L/R Large  ---- slow discharging process t I0I0 i 0 The response is determined by initial conditions 、 paticualr solution(steady-state value ）、 time constant. steady-state value initial value time constant Three-factor Method ： summarize the law for first-order circuit and get,

19. 1. Zero-input response of first-order circuit is caused by the initial value of energy-storage components. 2. The speed of decay lie on time constant  for RC circuit,  = RC, for RL circuit  = L/R 3. All the responses within a circuit bear the same time constant 。 Summary and Review:

20. Zero state response ： The response caused by the excitation of energy-storage component without initial energy. write equations ： the form of solution ： complementary solution of homogeneous equations paticular solution of homogeneous equations 一、 Zero state response of RC i K(t=0) USUS +– uRuR C + – uCuC R u C (0 - )=0 u c (0 - )=0 Calculat capacitor voltage u c (t) and current i(t )? Given §6-3 Zero State Response of First-Order Circuit

21. The changing rule is determined by circuit structure and parameter complete solution: u C (0 + )=A+U S = 0  A= - U S from initial condition u C (0 + )=0 fixed integral constant A homogeneous equation, complementary solution: ： paticular solution （ Forced response 、 steady-state response ） ： complementary solution （ free response ， transient response ） 因为它由输入激励决定，称为强制分量；它也是 电路的稳态解，也称为稳态分量

22. foced response (steady state) free response(transient state) t ucuc -U S uC'uC' uC"uC" USUS t i 0 

23. One half of the energy provided by source supply is dissipated on resistors ， the other half is converted into electric field energy stored in capcitors. stored by the capacitor ： provided by source supply ： dissipated by the resistor Energy Relationship ：

24. 二. Zero State Response of RL Circuit iLiL K(t=0) USUS +– uRuR L + – uLuL R i L (0 - )=0 Calculate inductor current i L ( t ) and voltage u L ( t )? Given uLuL USUS t 0 t iLiL 0 R U S  A 

25.  = L / R eq = L / (R 1 // R 2 ) + - R1R1 R2R2 L example 4 example 5 R eq C  = R eq C R1R1 R2R2 L R eq Method of computing time constant 

26. i L (0 + ) = i L (0 - ) = 1 A u V (0 + )= - 10000V iLiL K(t=0) + – uVuV L=4H R=10  V R V 10k  10V open switch K at t=0, find that the voltmeter is damaged,why ？ Full scale of voltmeter is ： 50V Analysis ： ∴ result in is demaged 。 V Additional examples

27. 一、 Complete Response of First-order Circuit Complete response ： energy-storage components have initial energy at the instant of switching,and the curcuit contains excitation after switching operation. i K(t=0) USUS +– uRuR C + – uCuC R Compute: u c (t)? u c (0 - )=U 0 ， Given Substitute u c (0 + )=U 0 i in the equation obtain ： A=U 0 - U S 1.Differential Method Complete response=Zero state response+Zero input response =Steady state response+Transient response §6-4 Complete Response of first- order Circuit and The General Equation

28. Solution ： Find Thevenin's equivalent 2A 2i12i1 + - i1i1 4Ω 2 0.1H uLuL + - iLiL place the switch at 2 circuit u oc 2 uLuL 0.1H + - iLiL R eq + - Example5 2A 2i12i1 + - i1i1 4Ω + - 8V 0.1H 2Ω uLuL + - iLiL 1 2 S The circuit has reached steady state when the switch is placed at 1 。 change switch from 1 to 2 at t=0; Find voltage u L for t≥0 +

29. 2i12i1 + - i1i1 4Ω 2 u - + 12V 2 uLuL 0.1H + - iLiL 10Ω + -

30. 1.Definition.delay of unit-step function t  ( t ) 0 1 一.Unit-step function t  ( t-t 0 ) t0t0 0 1 3. Unit-step function constitutes complicated signals example 1 A t0t0 t f(t)f(t) 0 A  ( t ) t f(t)f(t) A 0 t0t0 -A  ( t-t 0 ) Supplement: Unit-Step Function and Unit-Impulse Function

31. 1t 1 f(t) 0 t 0 t f(t)  ( t-t 0 ) 0 t0t0 t  ( t-t 0 ) t0t0 0 1 example4 Descript switchingoperation with  ( t ) K + – uCuC UsUs R C i t = t 0 + – uCuC U s  ( t - t 0 ) R C i

32. 1. Unit-step Function p(t) area （ intensity ）： 2.unit-impulse function  (t)  1/  t p(t)p(t) 0  / 2 1/  t p(t)p(t) -  / 2 let ： 二、 The Unit-impulse Function

33. Definition: t (t)(t) (1) 0 t  (t-t 0 ) t0t0 0 （1）（1） ≥0 + ） ≤0- ） The area value stays 1 when amplitude tends to be ∞ 3. The delay of unit-impulse function,  ( t-t 0 )

34. similarly ： (2) Screening t ≤0 - t ≥0 + t (t)(t) (1) 0 f(t)f(t) f(0) t0t0 because when t≠0 ，  (t)=0 ， hence, f(t)  (t)= f(0)  (t) 4 、 Properties of  Function （ 1 ） relationship with ε(t)

35. additional example The integral of multiply f(t) by  (t) take out the function value at t=0,namely f(0); The integral of multiply f(t) by  (t-t 0 ) take out the function value at t= t 0, namely,f(t 0 ).

36. u C (0 - )=0 i C + – uCuC R unit-step response ： zero-state response due to unit-step function t ucuc 1 t 0 i §6-5 Unit -Step Response of First-order Circuit

37. 10kΩ 10 kΩ +-+- icic 100  F u C (0 - )=0 10kΩ +-+- icic 100  F u C (0 - )=0 solution ： 10kΩ usus +-+- icic 100  F u C (0 - )=0 0.5 10 t(s) u s (V) 0 + example ： in the diagram below,Uc(0-)=0 ， find Ic(t) due to Us ？

38. 10kΩ +-+- icic 100  F u C (0 - )=0 10kΩ +-+- icic 100  F u C (0 - )=0 +-+- icic 100  F u C (0 - )=0 5kΩ Thevenin's equivalence +-+- icic 100  F u C (0 - )=0 5kΩ Thevenin's equivalence

39. Piecewise form ： t(s) i C (mA) 0 1 -0.632 0.5 0.368

40. =0 §6-6 Unit-impulse Response of First -Order Circuit 一.Unit-impulse response of RC circuit iCiC iRiR C + – uCuC R t (t)(t) (1) 0 Is it possible that u C is an unit-impulse function? 1.zero-state response demonstration: assume KCL equation is invalid; can't be unit-impulse function unit-impulse response ： the zero-state response due to unit- impulse function

41. iCiC iRiR C + – uCuC R δ(t) t 0 2. After t≥0+,δ(t)=0, it therefore can be regarded as zero-input response of Uc(0+)=1/C

42. iLiL δ(t) + – uRuR L + – uLuL R +– =0 ψ=Li=1, 1.zero-state responce 2.zero-input responce δ(t) t 0 二、 The RL Circuit

43. For any linear circuit that not change with time excitation zero-state responce unit-step responce unit-impulse respomce 三、 Relationship Between Unit-Step Response And Unit Impulse Response

44. Circuits zero-state responce unit-step responce s(t) unit-impulse responce h(t) iLiL iS iS L R C + – uCuC R usus + - iLiL uS uS L R + - iCiC iRiR C + – uCuC R isis

45. in deduction cource,take care of iLiL 4δ(t)V L R1R1 + - + - uL uL R2R2 solution ：  after t≥0 +, δ(t)=0 。 deduct without considering  t≥0 pay attention to the state at t=0 example6-7 Given circuit diagram ， i L (0-)=0 ， R 1 =6Ω ， R 2 =4Ω ， L=100mH find unit-impulse responce i L and u L 。 p150

46. 1 Conditions shouled be met when applying three-factor method ： dc or first-order circuit stimulated by sine wave 2 Conditions shouled be met when employing switching theorem ： capacitor voltage and inductor current are finit values CL 、 KVL must be satisfied regardless of wheather the capacitor voltage and inductor current change abrubtly or not. Homework ： 6-1 、 3 、 4 、 12 、 13 、 18 、 21 、 23 、 26 、 27 Summary and Review ：

47. Homework Problems ： Notice that in Example 6-26 、 6-27 the value at t=0 cannot be discarded,namely unit-step function  （ t) Lecture ： 6-23