課程名稱 : 電路學 (2) 授課教師 : 楊武智 期 :96 學年度第 2 學期

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1 課程名稱 : 電路學 (2) 授課教師 : 楊武智 學 期 :96 學年度第 2 學期 1

2 基礎電路學 主要在探討 電路分析技巧 教材分二部份 : 電路時域分析及電路頻域分析 時域分析 ( 電路學 (1)): 電路變數及元件 電路簡化技巧 電路分析技巧 RL 及 RC 電路的自然及步進響應及 RLC 電路的自然及步進響應 頻域分析 ( 電路學 (2)): 正弦性信號穩態分析 拉普拉斯轉換簡介 拉普拉斯轉換的電路分析及 PSpice 的電路分析 2

3 本教材內容, 主要為依據書本 Introduction Circuits for Electrical and Computer Engineering, by Nilsson/Riedel, 2002 所編撰講義, 僅供選課學生參考 3

4 網路講義檔案兩種 :.pdf 一般講義檔只有文字說明.pcm 有聲動態講義檔有文字及語音說明對象 1. 課本閱讀困難 2. 上課無法充分理解 3. 自習用 可由 PowerCam 軟體讀取 PowerCam 可由學校義守大學 > 電算中心 > 電 算中心首頁 > 校內授權軟體 安裝 或由 安裝 4

5 如何使用 PowerCam 請參考下步驟 Step 1 Step 2 5

6 Step 1 6

7 Step 2 7

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9 Chapter Eight Introduction to the Laplace transform 拉普拉斯轉換簡介

10 Why another analytical technique is needed First, we wish to consider the transient behavior of circuits whose describing equations consists of more than a single node-voltage or mesh current differential equations. 10

11 Second, we wish to determine the transient response of circuits whose signal source vary in ways more complicated than the simple dc level jumps (in Chapter 5 and 6). 11

12 Third, we can use the Laplace transform to introduce the concept of the transfer function as a tool for analyzing the steady-state sinusoidal response of a circuit when the frequency of the sinusoidal source is varied. 12

13 Finally, we wish to relate, in a systematic fashion, the time-domain behavior of a circuit to its frequency-domain behavior. 13

14 8.1 Definition of the Laplace transform The Laplace transform of a function is given by the expression st L { f( t)} = f( t) e dt (8.1) 0 where the symbol L { f ( t )} is read "the Laplace transform of f(t)." 14

15 The Laplace transform of f(t) is also denoted as F(s); F(s) = L {f(t)} This notation emphasizes that when the integral in Eq. (8.1) has been evaluated, the resulting expression is a function of s. 15

16 In our application, "t" represents the time domain, and, because the exponent of "e" in the integral must be dimensionless, s must have the dimension of reciprocal time, or frequency. So, after transformation it is in frequency domain or s-domain. 16

17 The Lapalce transform transforms the problem from the time domain to the frequency domain. After obtaining the frequency-domain expression for the unknown, we can inverse-transform it back to the time-domain. 17

18 The transform is similar to another logarithmic transform. It can change a multiplication or division problem, such as A = BC, into a simple addition or subtraction problem: log A = log BC = log B + log C. Antilogs are used to carry out the inverse process. 18

19 The phasor (Chapter 7) is another transform, it converts a sinusoidal signal into a complex number for easier, algebraic computation of circuit values. After determining the phasor value of a signal, we transform it back to its time-domain expression. 19

20 In our engineering analysis, we assume that the integral always converges. To emphasize the lower limit is zero, this equation is called one-sided, or unilateral, Laplace transform. In the two-sided, or bilateral case, the lower limit is. We don't use the two-sided form here. An example is shown in Fig

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22 In Fig. 8.1(b), to avoid ambiguity of discontinuity at t = 0, we will use 0 as the lower limit and introduce an impulse function at t = 0. (We use the notation 0 and 0 + to denote values of t just to the left and right of the origin, respectively. ) 22

23 The one - sided Laplace transform ignores f ( t ) for t < 0. What happens prior to 0 is accounted for by the initial conditions. 23

24 As an example, for parallel RLC circuit with signal source f(t), the node equation can be expressed as v 1 t dv + vdτ + I + C = f () t 0 0 R L dt Its Laplace transform can be written as v 1 t dv L + vdτ + I C = L{f(t) } R L dt In the left - hand side, we can express it as v 1 t dv + vdτ + { I 0 } + C R L 0 dt L L L L 24

25 Thus we divide the Laplace transform into two types: functional transform and operational ( 運算的 ) transform. A functional transform is the Laplace transform of a specific function, such as step function, impulse ( 脈衝 ) function, sin ωt, t, e -at, and so on. An operational transform defines a general property of the Laplace transform, such as finding the transform of the derivative or integral of a specific function. 25

26 8.2 Step function In Laplace transform approach, we use a step function to represent an abrupt change. It can lead to the concept of the impulse function. 26

27 Fig. 8.2 illustrates the step function. It is zero for t < 0. The symbol for the step function is Ku(t). 27

28 28

29 The mathematical definition of the step function is, Ku(t) = 0, t < 0, Ku(t) = K, t > 0. If K = 1, the function is called the unit step. 29

30 The step function is not defined at t = 0. In situations where we need to define the transition between 0 and 0 +, we assume that it is linear and that Ku(0) = 0.5K Fig. 8.3 illustrates the linear transition from 0 to

31 31

32 A step function that occurs at t = a can be expressed as, Ku(t a) = 0, t < a, Ku(t a) = K, t > a. Note that this step function is 0 when the argument t a is negative, and it is K when the argument is positive. Fig. 8.4 illustrates this function. 32

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34 A step function equals to K for t < a can be written as Ku(a t). Fig. 8.5 illustrates this function. Thus, Ku(a t) = K, t < a, Ku(a t) = 0, t > a. 34

35 35

36 Example 8.1 (p.399) Use the step functions to write an expression for the function illustrated in Fig

37 37

38 The expression for f(t) is f () t = 2[ tut () ut ( 1)] + ( 2t+ 4)[ ut ( 1) ut ( 3)] + (2t 8)[ u( t 3) u( t 4)] 38

39 8.3 The impulse function The concept of impulse function (also known as Dirac delta function) enables us to define the derivative at a discontinuity, and thus to define the Laplace transform of that derivative. 39

40 An impulse is a signal of infinite amplitude and zero duration. It is an artificial signal and a mathematical model. Impulse doesn t exist in natural, but some circuit signals come very close to approximating its definition. Impulsive voltages and currents occur in circuit analysis either because of a switching operation or because the circuits is excited by an impulsive source. 40

41 To define the derivative of a function at a discontinuity, we first assume that the function varies linearly across the discontinuity, as shown in Fig. 8.8, where we observer that as ε 0, an abrupt discontinuity occurs at the origin. 41

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43 When we differentiate the function, the derivative between ε and + ε exists. It is a constant of value 1/2 ε. -a(t- ε ) For t > ε, the derivative is ae. Fig. 8.9 shows these observations graphically. 43

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45 As ε approaches zero, the value of f ± ε approaches infinity. '( t ) between At the same time, the duration of this large value is approaching zero. Furthermore, the area under f remains constant as ε 0. '( t ) between ± ε In this example, the area is unity. 45

46 As ε approaches zero, we say that the function f '( t) between ± ε approaches a unit impluse function, denoteed as δ(t). Thus the derivative of f(t) at the origin approaches a unity impulse function as ε approaches zero, or f(0) δ(t) as ε 0. 46

47 If the area under the impluse function curve is other than unity, the impulse function is denoted as K δ(t), where K is the area, or referred to as the strength of the impulse function. 47

48 To summarize, an impulse function is created from a variable parameter function whose parameter approaches zero. The variable-parameter function must exhibit the following three characteristics as the parameter approaches zero: 1. The amplitude approaches infinity. 2. The duration of the function approaches zero. 3. The area under the variable-parameter function is constant as the parameter changes. 48

49 Many different variable-parameter functions have the above mentioned characteristics. In Fig. 8.8, we use a linear fucntion f(t) = 0.5t/ ε for ε t ε Its derivative f'(t) is a variable-parameter function. and the area under it is 1. Another example, as shown in Fig. 8.10, is a function of K f(t)= e 2ε t / ε 49

50 Another example of variable-parameter function: 50

51 0 K t/ ε K t/ ε Area = e dt + e dt 2ε 0 2ε K K = + = K 2 2 which tells us that the area under the curve is constant and equal to K units. Therefore, as ε 0, f(t) Kδ(). t 51

52 Mathematically, the impulse function is defined as - K δ(t)dt = K, δ () t = 0, t 0. An impluse that occurs at t K δ(t a). = a is denoted as The graphic symbol for the impulse function is an arrow. The strength of the impulse is given parenthetically next to the head of the arrow. 52

53 Fig shows impulse Kδ(t) and Kδ(t a). 53

54 An important property of the impulse function is the sifting property, which is expressed as - f(t) δ(t a)dt = f(a). where the function f(t) is assumed to be continuous at t = a; that is, at the location of impulse. It shows that the impulse function sifts out everything except the value of f(t) at t = a. 54

55 By using the shifting property, we can find the Laplace transform of δ(t): L -st { δ(t)} δ(t)e dt δ(t)dt 0 0 = = = 1. 55

56 We can also define the derivatives of the impulse function and the Laplace transform of these derivatives. We discuss the first derivative, along with its transform and then state the result for the higher-order derivatives. 56

57 The function illustrated in Fig. 8.12(a) generates an impulse function as ε 0. Fig. 8.12(b) shows the derivative of this impulse generated function, which is defined as the derivative of the impulse [δ (t)] as ε 0. The derivative of impulse function sometimes is referred to as a moment function, or unit doublet. 57

58 58

59 To find the Laplace transform of δ (t), we do the integration of the function, and then let ε 0. L { δ ( )} = lim + = = = = s lim ε 0 lim ε 0 lim ε 0 ( ) 0 ε 1 st 1 st t 2 e dt 2 e dt ε 0 ε ε + 0 ε e sε se sε 2 sε se sε + e 2 sε s se 2ε s 2 + se 2s ε 2 sε In deriving the result, we had to use L'Hopital's rule twice to evaluate the indeterminate form 0/0. 59

60 Higher-order derivatives may be generated in a manner similar to that used to generate the first derivative. For the n th derivative of the impulse function, we find L (n) n { (t)} s. δ = 60

61 An impulse function can also be thought of as a derivative of a step function. δ () t = du() t dt Fig presents the graphic interpretation of this derivative. Fig. 8.13(a) approaches a unit step function as ε 0. The function shown in Fig. 8.13(b) is the derivative of this step function, and appraoches a unit impulse as ε 0. 61

62 f(t) is a unit step function as ε 0 62

63 8.4 Functional transforms A functional transform is simply the Laplace transform of a specified function of t. Because we are limiting to the unilateral, or one-sided, Laplace transform, we define all functions to be zero for t < 0. In the previous section we show the Laplace transform of the unit impulse function. It is equal to 1. 63

64 Next illustration is the Laplace transform of unit step function: L 0 st 0 st {()} ut = f() te dt= u(0) edt+ 1e dt st st e = 0+ 1e dt = = + 0 s s + 64

65 The Laplace transform of the decaying exponential function shown in Fig is at at st ( a+ s) t L { e } = e e dt = e dt = + s a 65

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67 The Laplace transform of sinusoidal function shown in Fig is st L {sin ωt} = (sin ωt) e dt jωt jωt e e = e 0 2 j e e = 0 2 j 0 st ( s jω) t ( s+ jω) t dt = 2 j s jω s+ jω ω = 2 2 s + ω dt 67

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69 Table 8.1 gives an abbreviated list of Laplace transform pairs. 69

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71 8.5 Operational transform Operational transforms indicate how mathematical operations performed on either f(t) or F(s) are converted into the opposite domain. 71

72 The operations of primary interest are Multiplication by a constant Addition (subtraction) Differentiation Integration Translation in the time domain Translation in the frequency domain Scale change 72

73 Multiplication by a constant If L { f( t)} = F( s), then L { Kf( t)} = KF( s). 73

74 Addition (Subtraction) If L { f ( t)} = F( s), 1 1 L { f ( t)} = F ( s), 2 2 L { f ( t)} = F ( s), 3 3 then L { f () t + f () t f ()} t = F() s + F () s F () s

75 Differentiation Differentiation in the time domain corresponds to multiplying F(s) by s and then subtracting the initial value of f(t) that is, f(0 ) from this product: L df () t = sf() s f (0) dt 75

76 Which can be obtained from the definition of the Laplace transform, or L df(t) df(t) e st = dt dt 0 dt We can evaluate the integral by integrating by parts. That is, For f(x) = u(x)v (x) then u(x)v (x)dx = u(x)v(x) u (x)v(x)dx. 76

77 -st Letting u = e and dv = [df(t)/dt]dt yields df(t) st st L = e f() t f()( t se dt). dt 0 0 Because we are assuming that f(t) is Laplace -st transformable, the evaluation of e f(t) at t = is zero. Therefore the right-hand side reduces to - -st f(0 ) + s f(t)e dt = sf() s f (0) 0 - Thus, df(t) = sf () s f L (0) dt 77

78 Laplace transform of higher-order derivatives For example to find the second derivative of f(t), we let df () t g(t) = dt and then Gs () = sfs () f(0) But because dg(t) = dt () 2 d f t dt 2 78

79 We write 2 dg(t) d f () t L = L sg() s g(0) 2 = dt dt 2 = sfs () sf(0) df (0 ) dt Which gives 2 df () t 2 df (0) L sfs () sf(0) 2 = dt dt 79

80 By successively applying the preceding process, we can find Laplace transform of the nth derivative, L n df () t n n 1 n 2 df (0 ) sfs () s f(0) s n = dt dt df (0 ) n df (0 ) dt n dt 2 n 3 s 2 80

81 Integration Integration in the time domain corresponds to dividing by s in the s domain. t L { f( x) dx} 0 = F() s s 81

82 This can be obtained by integration-by-parts formula. t -st Let u = f ( x) dx, dv = e dt st e Then du = f () t dt, v =. s This yields -st -st t e t e L { f ( x) dx} = f ( x) dx + f ( t) dt - s s st Fs ( ) = 0 + f( t) e dt = s 0 s 82

83 Translation in the time domain If we start with any function f(t)u(t), we can represent the same function, translated in time by the constant a, as f(t a)u(t a). Translation in the time domain corresponds to multiplication by an exponential in the frequency domain. as L { f( t a) u( t a)} = e F( s) a> 0 83

84 Proof: L -st -st {f(t a)u(t a)} = u(t a)f(t a)e dt = f(t a)e dt - 0 a Now we change the variable of integration. We let x = t a. Then x = 0 when t = a, x = when t =, and dx = dt. Thus L s( x+ a) {f(t a)u(t a)} = f ( x) e dx 0 = e sa sx as 0 f( x) e dx= e F( s) 84

85 For example, when we know that Ltut { ( )} = 1 s 2 we can write the Laplace transform ( t a) u( t a) directly as L{( t a) u( t a )} = e s as 2 85

86 Translation in the frequency domain Translation in the frequency domain corresponds to multiplication by an exponential in the time domain. 86

87 at From L { e f ( t)} = F( s + a), thus, knowing that L {cosωt} = s 2 s 2 + ω the relation can be used to derive -at s+ a L {e cos ωt} = 2 2 ( s+ a) + ω 87

88 Scale change The scale-change property gives the relationship between f(t) and F(s) when the time variable is multiplied by a positive constant. 88

89 L { f( ta)} 1 s = F a a Thus, knowing that s L { cos t} =, 2 s + 1 we can use to deduce that 1 s/ ω s L {cosωt} = = ω ( s/ ω) + 1 s + ω 89

90 90

91 8.6 Applying the Laplace transform We now illustrate how to use the Laplace transform to solve the ordinary integrodifferential equations that describe the behavior of lumped-parameter circuits. 91

92 Consider the circuit shown in Fig Assume no initial energy is stored in the circuit at the instant when the switch, which is shorting the dc current source, is opened. The problem is to find the time-domain expression for v(t) when t 0. 92

93 93

94 vt () 1 t dvt () + vxdx ( ) + C = I ut ( ) R L 0 dt dc By using Laplace transform, we obtain V(s) 1 V ( s) CsV [ ( s) v(0 )] = Idc R L s s The initial voltage on the capacitor v(0 That gives 1 1 Idc V() s + + sc =. R sl s - ) is zero. 94

95 Then Idc / C V( s) =. (8.40) 2 s + (1/ RC) s+ (1/ LC) To find v(t) we must inverse-transform (next section) the expression for V(s). v(t) = L -1 { V ( s)} 95

96 Simplifying the notation by dropping t and s: L 1 {v} = V or v = L { V} L 1 {i} = I or i = L {} I L 1 {f} = F or f = L { F} 96

97 8.7 Inverse transform A rational function of s is the function that can be expressed in the form of a ratio of two polynomials the s such that no nonintegral powers of s appear in the polynomials. For linear, lumped-parameter circuits whose component values are constant, the s-domain expression for the unknown voltages and currents are always rational functions of s. If we can inverse-transform rational functions of s, we can solve for the time-domain expressions for voltage and currents. lumped: 集總 97

98 In general, we need to find the inverse transform of a function that has the form n n 1 as n + an 1s + + as 1 + a0 m m 1 m + m N(s) F(s) = = D(s) b s b s b s b (8.42) The coefficients a and b are real constants, and the exponents m and n are positive integers. The ratio N(s)/D(s) is called a proper rational function if m > n, and an improper rational function if m n. 98

99 Partial fraction expansion: Proper rational functions - Distinct real roots of D(s) - Distinct complex roots of D(s) - Repeated real roots of D(s) - Repeated complex roots of D(s) Partial fraction expansion: Improper rational functions 99

100 Partial fraction expansion: Proper rational functions An example: s+ 6 K1 K2 K3 K s(s + 3)(s + 1) s s + 3 ( s + 1) s + 1 The evaluation of K's is traightforward. By substituing different values for s in the left-hand and right-hand sides of the above equation. We can find various K's. From Table 8.1 L s+ 6 = ( K + K e + K te + K e ) u( t) s(s + 3)(s + 1) -1 3t t t

101 Distinct real roots of D(s): For example, Fs () 96( s+ 5)( s+ 12) K K K3 ss ( + 8)( s+ 6) s s+ 8 s = ( s+ 5)( s+ 12) Ks Ks ( s+ 8)( s+ 6) s+ 8 s+ 6 s= = K1 + + s= 0 s= 0 or 96(5)(12) 8(6) = K = 120, Similarly, we can find K = 72, K =

102 We proceed to find the inverse transform : L 96(s+5)(s+12) = ( e 72 e ) u( t) ss ( + 8)( s+ 6) 1 6t 8t 102

103 Distinct complex roots of D(s): For example, Fs () 100( s + 3) K K K3 ( s + 6)( s + 6s+ 25) s+ 6 s+ 3 j4 s+ 3+ j4 1 2 =

104 L 1 100( s + 3) ( s 6)( s 6s 25) 6 3 [ 12 t t e 20e cos(4t )] u( t) (8.60) = + 104

105 In distinct complex roots: First, in physically realizable circuits, complex roots are always appear in conjugate pairs. Second, the coefficients associated with these conjugate pairs are themselves conjugates. Thus for complex conjugate roots, you actually need to calculate only one of the two coefficients. 105

106 * s+ a K K + 2α + α + β + α β + α + β 1 2 Assume : = + s 2 s 2 2 s j s j ( s+ a)( s+ α jβ) K ( s+ α jβ) = K + + α + α + β + α + β 2 2 s 2 s s j ( s+ a) K2( s+ α jβ) = K1 + s+ α + jβ s+ α + jβ Let s = α + j β, we have K 1 Similarly, we can find K 2 β ja ( α) = 2β β + ja ( α) = 2β 106

107 * * K K + s+ α jβ s+ α + jβ = = θ jθ K K e K = = θ * jθ K K e K L K K + = Ke e + Ke e s+ α jβ s+ α + jβ * 1 jθ ( α jβ) t jθ ( α+ jβ) t ( )( ) ( ) ( j( βt+ θ) j( βt+ θ) e + e ) αt j( βt+ θ) j( βt+ θ) αt 2 = K e e + e = K e = ( αt ) 2K e cos( βt+ θ) 2 107

108 Repeated real roots of D(s): For example, Fs () 100( s + 25) K1 K2 K3 K4 = ss ( + 5) s ( s+ 5) ( s+ 5) s

109 Repeated complex roots of D(s): For example, Fs () = 768 ( s + 6s+ 25) 2 2 * * K1 K2 K1 K2 = ( s+ 3 j4) s + 3 j4 ( s + 3+ j4) s + 3+ j4 109

110 P

111 Partial fraction expansion: Improper rational functions For example: Fs () = s + 9s s s s s s 4s 10 2 = s s + 9s+ 20 s 4s s+ 4 s+ 5 2 =

112 Then we have inverse-transform 2 d δ() t dδ() t f(t) = + + t e e u t 2 dt dt 4t 5t 4 10 δ ( ) (20 50 ) ( ) 112

113 8.8 Poles and zeros of F(s) F(s) n n 1 as n + an 1s + + as 1 + a0 m m 1 m + m N(s) = = D(s) b s b s b s b (8.42) The rational function can also be expressed as F(s) = K(s+ z )(s+ z ) (s+ z ) 1 2 n (s + p )(s + p ) (s + p ) 1 2 m where K is the constant a n / b. m 113

114 The roots of the denominator polynomial, that is, p, p,, p, are called the poles of F(s); they are 2 3 m the values of s at which F(s) becomes infinitely large. p 1, The roots of the numerator polynomial, that is, z z, z,, z, are called the zeros of F(s); they are 2 3 n the values of s at which F(s) becomes zero. 1, 114

115 For example, function F(s) has the expression Fs () = 2 8s + 120s s + s + s + s+ = 4( s+ 5)( s+ 10) ( s+ 1)( s+ 2)( s+ 3)( s+ 4) The poles are 1, 2, 3, and 4. The zeros are 5 and

116 The poles and zeros plotted in the s plane: An example F( s) 10( s + 5)( s = s( s + 10)( s j4)( s j8)( s j4) j8) 116

117 F() s can also have either an rth-order pole or an rth-order zero at infinity. For example, the function F( s) = 4/ 2 s will reduce to zero when s =. This function has second-order zero at infinity. Similarly, the function F() s = s will have a pole at infinity. 117

118 118

119 In practice, because R, L, and C are positive constants, the poles must lie in the left half of the s plane. This comes from the fact that the response of a linear lumped-parameter circuit must be bounded. 119

120 8.9 Initial- and final-value theorems The initial - and final - value theorem can be used to determine the behavior of f(t) at 0 and from F(s). The initial lim t 0 + f ( t) = - value theorem states that lim sf( s), s and the lim f t ( t) final - value threorem states that = lim sf( s). s 0 120

121 Proof of initial-value theorem From operational transform of differentiation, df df st L = sf() s f (0) = e dt. dt 0 dt We take the limit df st lim[ sf( s) f (0 )] = lim e dt. s s 0 dt The righthand side may be written as lim s df e dt st df d st t+ e dt + 0 dt 121

122 -st As s, (df/dt)e 0; hence the second integral vanishes + in the limit. The first integral reduces to f(0 ) f (0 ), which is independent of s. Thus right-hand side becomes df st lim e dt f + (0 ) f = (0 ) s 0 dt The left-hand side may be written as lim[ sf( s) f (0 )] = lim[ sf( s)] f (0 ). s s So we have the proof + lim sf( s) = f (0 ) = lim f ( t) s t

123 Proof of final-value theorem From operational transform of differentiation, df df st L = sf() s f (0) = e dt. 0 dt dt Here we take the limit as s 0: df st lim[ sf( s) f (0 )] = lim e dt. s 0 s 0 0 dt 123

124 The right-hand side can be reduced to lim df lim t st df df e dt dt dy. s 0 0 = = dt 0 dt t 0 dy Carrying out the integration process yields lim[ f( t) f(0 )] = lim[ f( t)] f(0 ). t t 124

125 Hence we have lim[ sf( s) f (0 )] = lim[ sf( s)] f (0 ) s 0 s 0 = lim[ f( t)] f(0 ). t That reduces to the final-value theorem. lim[ sf( s)] = lim[ f ( t)] s 0 t The final-value theorem is useful only if f( ) exists. This condition is true only if all the poles of F(s), except for a simple pole at the origin, lie in the left half of the s plane. 125

126 The application of initial- and final-value theorem Consider the transform pair given by Eq. (8.60). 126

127 L 1 100( s + 3) ( s 6)( s 6s 25) = [ 12 t t e 20e cos(4t )] u( t) (8.60) 127

128 The initial-value theorem gives lim sf( s) = 0 s lim f( t) = 0 t 0 + The final-value thereom gives lim sf( s) = 0 s 0 lim f( t) = 0 t 128

129 The real value of the initial- and final-value theorems lies in being able to test the s- domain expressions before working out the inverse transform. For example, consider the expression for V(s) given by Eq. (8.40). 129

130 Idc / C V() s =. (8.40) 2 s + (1/ RC) s+ (1/ LC) Applying the initial-value thereom yields si ( dc / C) lim sv ( s) = lim = 0. s s 2 s + (1/ RC) s+ (1/ LC) Applying the final-value thereom gives lim sv ( s) = lim s 0 s 0 si ( / C) dc 2 s RC s L + (1/ ) + (1/ = 0. C) The derived expression for V(s) correctly predicts the initial and final values of v(t). 130

131 End of Chapter Eight 131