Dorf, R.C., Wan, Z., Johnson, D.E. “Laplace Transform” The Electrical Engineering Handbook Ed. Richard C. Dorf Boca Raton: CRC Press LLC, 2000 6 Laplace Transform 6.1 Definitions and Properties Laplace Transform Integral?Region of Absolute Convergence?Properties of Laplace Transform?Time-Convolution Property?Time-Correlation Property?Inverse Laplace Transform 6.2 Applications Differentiation Theorems?Applications to Integrodifferential Equations?Applications to Electric Circuits?The Transformed Circuit?Thévenin’s and Norton’s Theorems?Network Functions?Step and Impulse Responses?Stability 6.1 Definitions and Properties Richard C. Dorf and Zhen Wan The Laplace transform is a useful analytical tool for converting time-domain signal descriptions into functions of a complex variable. This complex domain description of a signal provides new insight into the analysis of signals and systems. In addition, the Laplace transform method often simplifies the calculations involved in obtaining system response signals. Laplace Transform Integral The Laplace transform completely characterizes the exponential response of a time-invariant linear function. This transformation is formally generated through the process of multiplying the linear characteristic signal x(t) by the signal e –st and then integrating that product over the time interval (–￥, +￥). This systematic procedure is more generally known as taking the Laplace transform of the signal x(t). Definition: The Laplace transform of the continuous-time signal x(t) is The variable s that appears in this integrand exponential is generally complex valued and is therefore often expressed in terms of its rectangular coordinates s = s + j w where s = Re(s) and w = Im(s) are referred to as the real and imaginary components of s, respectively. The signal x(t) and its associated Laplace transform X(s) are said to form a Laplace transform pair. This reflects a form of equivalency between the two apparently different entities x(t) and X(s). We may symbolize this interrelationship in the following suggestive manner: Xs xtedt st () ()= - -￥ +￥ ò Richard C. Dorf University of California, Davis Zhen Wan University of California, Davis David E. Johnson Birmingham-Southern College ? 2000 by CRC Press LLC X(s) = +[x(t)] where the operator notation + means to multiply the signal x(t) being operated upon by the complex expo- nential e –st and then to integrate that product over the time interval (–￥, +￥). Region of Absolute Convergence In evaluating the Laplace transform integral that corresponds to a given signal, it is generally found that this integral will exist (that is, the integral has finite magnitude) for only a restricted set of s values. The definition of region of absolute convergence is as follows. The set of complex numbers s for which the magnitude of the Laplace transform integral is finite is said to constitute the region of absolute convergence for that integral transform. This region of convergence is always expressible as s + < Re(s) < s – where s + and s – denote real parameters that are related to the causal and anticausal components, respectively, of the signal whose Laplace transform is being sought. Laplace Transform Pair Tables It is convenient to display the Laplace transforms of standard signals in one table. Table 6.1 displays the time signal x(t) and its corresponding Laplace transform and region of absolute convergence and is sufficient for our needs. Example. To find the Laplace transform of the first-order causal exponential signal x 1 (t) = e –at u(t) where the constant a can in general be a complex number. The Laplace transform of this general exponential signal is determined upon evaluating the associated Laplace transform integral (6.1) In order for X 1 (s) to exist, it must follow that the real part of the exponential argument be positive, that is, Re(s + a) = Re(s) + Re(a) > 0 If this were not the case, the evaluation of expression (6.1) at the upper limit t = +￥ would either be unbounded if Re(s) + Re(a) < 0 or undefined when Re(s) + Re(a) = 0. On the other hand, the upper limit evaluation is zero when Re(s) + Re(a) > 0, as is already apparent. The lower limit evaluation at t = 0 is equal to 1/(s + a) for all choices of the variable s. The Laplace transform of exponential signal e –at u(t) has therefore been found and is given by Xs e ute dt e dt e sa at st sat sat 1 0 0 () () () () () == = -+ -- -+ +￥ -￥ +￥ -+ +￥ òò L[ ()] Re() Re()eut sa sa at- = + >- 1 for ? 2000 by CRC Press LLC Properties of Laplace Transform Linearity Let us obtain the Laplace transform of a signal, x(t), that is composed of a linear combination of two other signals, x(t) = a 1 x 1 (t) + a 2 x 2 (t) where a 1 and a 2 are constants. The linearity property indicates that + [a 1 x 1 (t) + a 2 x 2 (t)] = a 1 X 1 (s) + a 2 X 2 (s) and the region of absolute convergence is at least as large as that given by the expression TABLE 6.1 Laplace Transform Pairs Time Signal Laplace Transform Region of x(t) X(s) Absolute Convergence 1. e –at u(t)Re(s) > –Re(a) 2. t k e –at u(–t s) > –Re(a) 3. –e –at u(–t)Re(s) < –Re(a) 4. (–t) k e –at u(–t s) < –Re(a) 5. u(t)Re(s) > 0 6. d(t) 1 all s 7. s k all s 8. t k u(t) Re(s) > 0 9. Re(s) = 0 10. sin w 0 t u(t)Rs) > 0 11. cos w 0 t u(t e(s) > 0 12. e –at sin w 0 t u(t)Rs) > –Re(a) 13. e –at cos w 0 t u(t e(s) > –Re(a) Source: J.A. Cadzow and H.F. Van Landingham, Signals, Systems, and Transforms, Englewood Cliffs, N.J.: Prentice-Hall, 1985, p. 133. With permission. 1 sa+ k sa k ! ()+ +1 1 ()sa+ k sa k ! ()+ +1 1 s dt dt k k d() k s k ! +1 sgn t t t = 3 < ì í ? 10 10 , –, 2 s w w 0 2 0 2 s + s s 2 0 2 +w w w()sa++ 2 0 2 sa sa + ++() 2 0 2 w ? 2000 by CRC Press LLC where the pairs (s 1 + ; s + 2 )< Re(s) < min(s – 1 ; s – 2 ) identify the regions of convergence for the Laplace transforms X 1 (s) and X 2 (s), respectively. Time-Domain Differentiation The operation of time-domain differentiation has then been found to correspond to a multiplication by s in the Laplace variable s domain. The Laplace transform of differentiated signal dx(t)/dt is Furthermore, it is clear that the region of absolute convergence of dx(t)/dt is at least as large as that of x(t). This property may be envisioned as shown in Fig. 6.1. Time Shift The signal x(t – t 0 ) is said to be a version of the signal x(t) right shifted (or delayed) by t 0 seconds. Right shifting (delaying) a signal by a t 0 second duration in the time domain is seen to correspond to a multiplication by e –st0 in the Laplace transform domain. The desired Laplace transform relationship is where X(s) denotes the Laplace transform of the unshifted signal x(t). As a general rule, any time a term of the form e –st0 appears in X(s), this implies some form of time shift in the time domain. This most important property is depicted in Fig. 6.2. It should be further noted that the regions of absolute convergence for the signals x(t) and x(t – t 0 ) are identical. FIGURE 6.1Equivalent operations in the (a) time-domain operation and (b) Laplace transform-domain operation. (Source: J.A. Cadzow and H.F. Van Landingham, Signals, Systems, and Transforms, Englewood Cliffs, N.J.: Prentice-Hall, 1985, p. 138. With permission.) FIGURE 6.2Equivalent operations in (a) the time domain and (b) the Laplace transform domain. (Source: J.A. Cadzow and H.F. Van Landingham, Signals, Systems, and Transforms, Englewood Cliffs, N.J.: Prentice-Hall, 1985, p. 140. With permission.) max( ; ) Re() min( ; )ss ss ++ -- << 12 12 s + dxt dt sXs () () é ? ê ù ? ú = +[( )] ()xt t eXs st -= - 0 0 ? 2000 by CRC Press LLC Time-Convolution Property The convolution integral signal y(t) can be expressed as where x(t) denotes the input signal, the h(t) characteristic signal identifying the operation process. The Laplace transform of the response signal is simply given by where H(s) = + [h(t)] and X(s) = + [x(t)]. Thus, the convolution of two time-domain signals is seen to correspond to the multiplication of their respective Laplace transforms in the s-domain. This property may be envisioned as shown in Fig. 6.3. Time-Correlation Property The operation of correlating two signals x(t) and y(t) is formally defined by the integral relationship The Laplace transform property of the correlation function f xy (t) is in which the region of absolute convergence is given by FIGURE 6.3Representation of a time-invariant linear operator in (a) the time domain and (b) the s-domain. (Source: J. A. Cadzow and H. F. Van Landingham, Signals, Systems, and Transforms, Englewood Cliffs, N.J.: Prentice-Hall, 1985, p. 144. With permission.) yt hxt d() ()( )=- -￥ ￥ ò ttt Ys HsXs() ()()= ft t xy xtyt dt() ()( )=+ -￥ ￥ ò F xy sXsYs() ()()=- max( , ) Re() min( , )-<<--+ +-ss ss xxyy s ? 2000 by CRC Press LLC Autocorrelation Function The autocorrelation function of the signal x(t) is formally defined by The Laplace transform of the autocorrelation function is and the corresponding region of absolute convergence is Other Properties A number of properties that characterize the Laplace transform are listed in Table 6.2. Application of these properties often enables one to efficiently determine the Laplace transform of seemingly complex time functions. TABLE 6.2Laplace Transform Properties Signal x(t) Laplace Transform Region of Convergence of X(s) Property Time Domain X(s) s Domain s + < Re(s) < s – Linearity a 1 x 1 (t) + a 2 x 2 (t) a 1 X 1 (s) + a 2 X 2 (s) At least the intersection of the region of convergence of X 1 (s) and X 2 (s) Time differentiation sX(s) At least s + < Re(s) and X 2 (s) Time shift x(t – t 0 )e –st0 X(s) s + < Re(s) < s – Time convolution H(s)X(s) At least the intersection of the region of convergence of H(s) and X(s) Time scaling x(at) Frequency shift e –at x(t) X(s + a) s + – Re(a) < Re(s) < s – – Re(a) Multiplication (frequency convolution) x 1 (t)x 2 (t) Time integration At least s + < Re(s) < s – Frequency differentiation (–t) k x(t) At least s + < Re(s) < s – Time correlation X(–s)Y(s max(–s x– , s y+ ) < Re(s) < min(–s x+ , s y– ) Autocorrelation function X(–s)X(s max(–s x– , s x+ ) < Re(s) < min(–s x+ , s x– ) Source: J.A. Cadzow and H.F. Van Landingham, Signals, Systems, and Transforms, Englewood Cliffs, N.J.: Prentice-Hall, 1985. With permission. ft t xx xtxt dt() ()( )=+ -￥ ￥ ò F xx sXsXs() ()()=- max( , ) Re() min( , )-<<-- ++-ss ss xy x s y dxt dt () hxt dt()( )tt- -￥ ￥ ò 1 **a X s a ? è ? ? ? ÷ ss +- < ? è ? ? ? ÷ <Re s a 1 2 12 pj XuXsud cj cj ()( ) -￥ +￥ ò - ss ss ss ss ++ -- ++ -- +<<+ +<<+ () () () () () () () () Re() 12 12 12 12 s c xd t ()tt -￥ ò 1 0 s Xs X() ()for = dXs ds k k () xtytzdt()( )+ -￥ +￥ ò xtxtzdt()( )+ -￥ +￥ ò ? 2000 by CRC Press LLC Inverse Laplace Transform Given a transform function X(s) and its region of convergence, the procedure for finding the signal x(t) that generated that transform is called finding the inverse Laplace transform and is symbolically denoted as The signal x(t) can be recovered by means of the relationship In this integral, the real number c is to be selected so that the complex number c + jw lies entirely within the region of convergence of X(s) for all values of the imaginary component w. For the important class of rational Laplace transform functions, there exists an effective alternate procedure that does not necessitate directly evaluating this integral. This procedure is generally known as the partial-fraction expansion method. Partial Fraction Expansion Method As just indicated, the partial fraction expansion method provides a convenient technique for reacquiring the signal that generates a given rational Laplace transform. Recall that a transform function is said to be rational if it is expressible as a ratio of polynomial in s, that is, The partial fraction expansion method is based on the appealing notion of equivalently expressing this rational transform as a sum of n elementary transforms whose corresponding inverse Laplace transforms (i.e., generating signals) are readily found in standard Laplace transform pair tables. This method entails the simple five-step process as outlined in Table 6.3. A description of each of these steps and their implementation is now given. I. Proper Form for Rational Transform.This division process yields an expression in the proper form as given by TABLE 6.3Partial Fraction Expansion Method for Determining the Inverse Laplace Transform I. Put rational transform into proper form whereby the degree of the numerator polynomial is less than or equal to that of the denominator polynomial. II. Factor the denominator polynomial. III. Perform a partial fraction expansion. IV. Separate partial fraction expansion terms into causal and anticausal components using the associated region of absolute convergence for this purpose. V. Using a Laplace transform pair table, obtain the inverse Laplace transform. Source: J.A. Cadzow and H.F. Van Landingham, Signals, Systems, and Transforms, Englewood Cliffs, N.J.: Prentice-Hall, 1985, p. 153. With permission. xt Xs() [()]=+ ±1 xt j Xseds st cj cj () ()= -￥ +￥ ò 1 2p Xs Bs As bs bs bsb sas asa m m m m n n n () () () == + +×××+ + + +×××+ + - - - - 1 1 10 1 1 10 Xs Bs As Qs Rs As () () () () () () = =+ ? 2000 by CRC Press LLC in which Q(s) and R(s) are the quotient and remainder polynomials, respectively, with the division made so that the degree of R(s) is less than or equal to that of A(s). II. Factorization of Denominator Polynomial. The next step of the partial fraction expansion method entails the factorizing of the nth-order denominator polynomial A(s) into a product of n first-order factors. This factorization is always possible and results in the equivalent representation of A(s) as given by The terms p 1 , p 2 , . . ., p n constituting this factorization are called the roots of polynomial A(s), or the poles of X(s). III. Partial Fraction Expansion. With this factorization of the denominator polynomial accomplished, the rational Laplace transform X(s) can be expressed as (6.2) We shall now equivalently represent this transform function as a linear combination of elementary transform functions. Case 1: A(s) Has Distinct Roots. where the a k are constants that identify the expansion and must be properly chosen for a valid representation. and a 0 = b n The expression for parameter a 0 is obtained by letting s become unbounded (i.e., s = +￥) in expansion (6.2). Case 2: A(s) Has Multiple Roots. The appropriate partial fraction expansion of this rational function is then given by As spsp sp n ( ) ( )( )...( )=- - - 12 Xs Bs As bs b s b spsp sp n n n n n () () () ( )( ) ( ) == + + ×××+ - - ××× - - - 1 1 0 12 Xs sp sp sp n n ()=+ - + - + ×××+ - a aa a 0 1 1 2 2 a kk spXs k n sp k =- = = ( ) ( ) , ,...,for 1 2 Xs Bs As Bs spAs q () () () () ()() == - 11 Xs sp sp nq As q q () () () ()=+ - + ×××+ - +- ( ) a a a 0 1 1 1 1 other elementary terms due to the roots of 1 ? 2000 by CRC Press LLC The coefficient a 0 may be expediently evaluated by letting s approach infinity, whereby each term on the right side goes to zero except a 0 . Thus, The a q coefficient is given by the convenient expression (6.3) The remaining coefficients a 1 , a 2 , … , a q–1 associated with the multiple root p 1 may be evaluated by solving Eq. (6.3) by setting s to a specific value. IV. Causal and Anticausal Components.In a partial fraction expansion of a rational Laplace transform X(s) whose region of absolute convergence is given by it is possible to decompose the expansion’s elementary transform functions into causal and anticausal functions (and possibly impulse-generated terms). Any elementary function is interpreted as being (1) causal if the real component of its pole is less than or equal to s + and (2) anticausal if the real component of its pole is greater than or equal to s – . The poles of the rational transform that lie to the left (right) of the associated region of absolute convergence correspond to the causal (anticausal) component of that transform. Figure 6.4 shows the location of causal and anticausal poles of rational transform. V. Table Look-Up of Inverse Laplace Transform.To complete the inverse Laplace transform procedure, one need simply refer to a standard Laplace transform function table to determine the time signals that generate each of the elementary transform functions. The required time signal is then equal to the same linear combi- nation of the inverse Laplace transforms of these elementary transform functions. FIGURE 6.4Location of causal and anticausal poles of a rational transform. (Source: J.A. Cadzow and H.F. Van Landing- ham, Signals, Systems, and Transforms, Englewood Cliffs, N.J.: Prentice-Hall, 1985, p. 161. With permission.) a 0 0== ?+￥ lim () s Xs a q q sp spXs Bp Ap =- = = ()() () () 1 1 11 1 ss +- <<Re()s ? 2000 by CRC Press LLC Defining Terms Laplace transform: A transformation of a function f(t) from the time domain into the complex frequency domain yielding F(s). where s = s + jw. Region of absolute convergence:The set of complex numbers s for which the magnitude of the Laplace transform integral is finite. The region can be expressed as where s + and s – denote real parameters that are related to the causal and anticausal components, respectively, of the signal whose Laplace transform is being sought. Related Topic 4.1 Introduction References J.A. Cadzow and H.F. Van Landingham, Signals, Systems, and Transforms, Englewood Cliffs, N.J.: Prentice-Hall, 1985. E. Kamen, Introduction to Signals and Systems, 2nd Ed., Englewood Cliffs, N.J.: Prentice-Hall, 1990. B.P. Lathi, Signals and Systems, Carmichael, Calif.: Berkeley-Cambridge Press, 1987. 6.2 Applications 1 David E. Johnson In applications such as electric circuits, we start counting time at t = 0, so that a typical function f(t) has the property f(t) = 0, t < 0. Its transform is given therefore by which is sometimes called the one-sided Laplace transform. Since f(t) is like x(t)u(t) we may still use Table 6.1 of the previous section to look up the transforms, but for simplicity we will omit the factor u(t), which is understood to be present. Differentiation Theorems Time-Domain Differentiation If we replace f(t) in the one-sided transform by its derivative f￠(t) and integrate by parts, we have the transform of the derivative, 1 Based on D.E. Johnson, J.R. Johnson, and J.L. Hilburn, Electric Circuit Analysis, 2nd ed., Englewood Cliffs, N.J.: Prentice- Hall, 1992, chapters 19 and 20. With permission. Fs ftedt st () ()= - -￥ ￥ ò ss +- <<Re()s Fs ftedt st () ()= - ￥ ò 0 ? 2000 by CRC Press LLC (6.4) We may formally replace f by f ￠ to obtain or by (6.4), (6.5) We may replace f by f ￠ again in (6.5) to obtain + [f -(t)], and so forth, obtaining the general result, (6.6) where f (n) is the nth derivative. The functions f, f ￠, … , f (n–1) are assumed to be continuous on (0,￥), and f (n) is continuous except possibly for a finite number of finite discontinuities. Example 6.2.1 As an example, let f(t) = t n , for n a nonnegative integer. Then f (n) (t) = n! and f(0) = f ￠(0) = … = f (n–1) (0) = 0. Therefore, we have or (6.7) Example 6.2.2 As another example, let us invert the transform which has the partial fraction expansion where and +[()] () ()￠ =-ft sFs f0 ++[ ( )] [ ( )] ( )￠￠ = ￠ - ￠ft s ft f0 +[ ()] () () ()￠￠ =--￠ft sFs sf f 2 00 +[ ()] () () () () () ( ) ft sFs sf sf f nnn n n =- -￠ -- -- -12 1 00 0L ++[!] [ ]nst nn = ++[] [!] ! ; , , ,t s n n s n n nn ===? + 1 012 1 Fs ss () () = + 8 2 3 Fs A s B s C s D s ()=+++ + 32 2 AsFs s == = 3 0 4() Ds Fs s =+ =- =- ()()21 2 ? 2000 by CRC Press LLC To obtain B and C, we clear F(s) of fractions, resulting in Equating coefficients of s 3 yields C = 1, and equating those of s 2 yields B = –2. The transform is therefore so that Frequency-Domain Differentiation Frequency-domain differentiation formulas may be obtained by differentiating the Laplace transform with respect to s. That is, if F(s) = + [ f(t)], Assuming that the operations of differentiation and integration may be interchanged, we have From the last integral it follows by definition of the transform that (6.8) Example 6.2.3 As an example, if f(t) = cos kt, then F(s) = s/(s 2 + k 2 ), and we have We may repeatedly differentiate the transform to obtain the general case from which we conclude that 84 2 2 2 23 =++ ++ +-()() ()sBs Cs s Fs ss ss () !! =-+- + 2 2 2 11 1 2 32 ft t t e t ()=-+- - 221 dF s ds d ds fte dt st () ()= - ￥ ò 0 dF s ds d ds fte dt tf t e dt st st () [() ] [()] = =- - ￥ ￥ - ò ò 0 0 +[()] () tf t dF s ds =- +[ cos ] () tkt d ds s sk sk sk =- + ? è ? ? ? ÷ = - + 22 22 222 dFs ds tftedt n n ns () [( ) ( )]=- ￥ - ò 0 ? 2000 by CRC Press LLC (6.9) Properties of the Laplace transform obtained in this and the previous section are listed in Table 6.4. Applications to Integrodifferential Equations If we transform both members of a linear differential equation with constant coefficients, the result will be an algebraic equation in the transform of the unknown variable. This follows from Eq. (6.6), which also shows that the initial conditions are automatically taken into account. The transformed equation may then be solved for the transform of the unknown and inverted to obtain the time-domain answer. Thus, if the differential equation is the transformed equation is The transform X(s) may then be found and inverted to give x(t). TABLE 6.4One-Sided Laplace Transform Properties f(t) F(s) 1. cf(t) cF(s) 2. f 1 (t) + f 2 (t) F 1 (s) + F 2 (s) 3. sF(s) – f(0) 4. 5. 6. e –at f(t) F(s + a) 7. f(t – t)u(t – t) e –st F(s) 8. F(s)G(s) 9. f(ct), c > 0 10. t n f(t), n = 0,1,2, ... (–1) n F (n) (s) dft dt () dft dt n n () sFs sf sf sf f nn n n n () () () () () --￠ - ￠￠ -- -- - -() 12 1 00 00 1 L fd t ()tt 0 ò Fs s () fg fgtd t *= - ò ()( )ttt 0 1 c F s c ? è ? ? ? ÷ +[()]() () ; ,,,tft dFs ds n nn n n =- = ?1012 ax ax ax ft n n n n() - -() + +?+ = () 1 1 0 asXs sx x asXssx x aXs Fs n nn n n nn n () - () -?- () [] + () - () -?- () +?+ () = () -() - -- -() 11 1 12 2 0 00 00 ? 2000 by CRC Press LLC Example 6.2.4 As an example, let us find the solution x(t), for t > 0, of the system of equations Transforming, we have from which The partial fraction expansion is from which Integration Property Certain integrodifferential equations may be trans- formed directly without first differentiating to remove the integrals. We need only transform the integrals by means of Example 6.2.5 As an example, the current i(t) in Fig. 6.5, with no initial stored energy, satisfies the system of equations, Transforming yields ￠ + ￠+= = ￠ = - xxxe xx t 43 0102 2 () , () sXs s sXs Xs s 2 24 13 1 2 () [() ] ()--+ -+ = + Xs ss sss () ()()() = ++ +++ 2 813 123 Xs sss ()= + - + - + 3 1 1 2 1 3 xt e e e ttt ()=-- --- 3 23 FIGURE 6.5An RLC circuit. + fd Fs s t () () tt 0 ò é ? ê ù ? ú = di dt iidtut i t ++ = = ò 25 00 0 () () sIs Is s Is s () () ()++=2 51 ? 2000 by CRC Press LLC or Therefore the current is Applications to Electric Circuits As the foregoing example shows, the Laplace transform method is an elegant procedure than can be used for solving electric circuits by transforming their describing integrodifferential equations into algebraic equations and applying the rules of algebra. If there is more than one loop or nodal equation, their transformed equations are solved simultaneously for the desired circuit current or voltage transforms, which are then inverted to obtain the time-domain answers. Superposition is not necessary because the various source functions appearing in the equations are simply transformed into algebraic quantities. The Transformed Circuit Instead of writing the describing circuit equations, transforming the results, and solving for the transform of the circuit current or voltage, we may go directly to a transformed circuit, which is the original circuit with the currents, voltages, sources, and passive elements replaced by transformed equivalents. The current or voltage transforms are then found using ordinary circuit theory and the results inverted to the time-domain answers. Voltage Law Transformation First, let us note that if we transform Kirchhoff’s voltage law, we have where V i (s) is the transform of v i (t).The transformed voltages thus satisfy Kirchhoff’s voltage law. A similar procedure will show that transformed currents satisfy Kirchhoff’s current law, as well. Next, let us consider the passive elements. For a resistance R, with current i R and voltage v R , for which v R = Ri R the transformed equation is (6.10) This result may be represented by the transformed resistor element of Fig. 6.6(a). Inductor Transformation For an inductance L, the voltage is v L = L di L /dt Is ss s () () = ++ = ++ é ? ê ù ? ú 1 25 1 2 2 14 22 it e t t () . sin= - 05 2 A vtvt vt n12 0() () ()+ +×××+ = Vs Vs Vs n12 0() () ()+ +×××+ = Vs RIs RR () ()= ? 2000 by CRC Press LLC Transforming, we have (6.11) which may be represented by an inductor with impedance sL in series with a source, Li L (0), with the proper polarity, as shown in Fig. 6.6(b). The included voltage source takes into account the initial condition i L (0). Capacitor Transformation In the case of a capacitance C we have which transforms to (6.12) This is represented in Fig. 6.6(c) as a capacitor with impedance 1/sC in series with a source, v C (0)/s, accounting for the initial condition. We may solve Eqs. (6.10), (6.11), and (6.12) for the transformed currents and use the results to obtain alternate transformed elements useful for nodal analysis, as opposed to those of Fig. 6.6, which are ideal for loop analysis. The alternate elements are shown in Fig. 6.7. Source Transformation Independent sources are simply labeled with their transforms in the transformed circuit. Dependent sources are transformed in the same way as passive elements. For example, a controlled voltage source defined by FIGURE 6.6Transformed circuit elements. FIGURE 6.7Transformed elements useful for nodal analysis. Vs sLIs Li LL () () ()=-0 v C idtv CCC t =+ ò 1 0 0 () Vs sC Is s v CCC () () ()=+ 11 0 ? 2000 by CRC Press LLC v 1 (t) = Kv 2 (t) transforms to V 1 (s) = KV 2 (s) which in the transformed circuit is the transformed source controlled by a transformed variable. Since Kirch- hoff’s laws hold and the rules for impedance hold, the transformed circuit may be analyzed exactly as we would an ordinary resistive circuit. Example 6.2.6 To illustrate, let us find i(t) in Fig. 6.8(a), given that i(0) = 4 A and v(0) = 8 V. The transformed circuit is shown in Fig. 6.8(b), from which we have This may be written so that Thévenin’s and Norton’s Theorems Since the procedure using transformed circuits is identical to that using the phasor equivalent circuits in the ac steady-state case, we may obtain transformed Thévenin and Norton equivalent circuits exactly as in the phasor case. That is, the Thévenin impedance will be Z th (s) seen at the terminals of the transformed circuit with the sources made zero, and the open-circuit voltage and the short-circuit current will be V oc (s) and I sc (s), respectively, at the circuit terminals. The procedure is exactly like that for resistive circuits, except that in the transformed circuit the quantities involved are functions of s. Also, as in the resistor and phasor cases, the open- circuit voltage and short-circuit current are related by (6.13) FIGURE 6.8(a) A circuit and (b) its transformed counterpart. Is ss ss () [( )] () () = ++- ++ 2348 32 // / Is sss () =- + + + - + 13 1 20 2 3 3 it e e e ttt () =- + - --- 13 20 3 23 A VsZsIs oc th sc () ()()= ? 2000 by CRC Press LLC Example 6.2.7 As an example, let us consider the circuit of Fig. 6.9(a) with the transformed circuit shown in Fig. 6.9(b). The initial conditions are i(0) = 1 A and v(0) = 4 V. Let us find v(t) for t > 0 by replacing everything to the right of the 4-W resistor in Fig. 6.9(b) by its Thévenin equivalent circuit. We may find Z th (s) directly from Fig. 6.9(b) as the impedance to the right of the resistor with the two current sources made zero (open circuited). For illustrative purposes we choose, however, to find the open-circuit voltage and short-circuit current shown in Figs. 6.10(a) and (b), respectively, and use Eq. (6.13) to get the Thévenin impedance. The nodal equation in Fig. 6.10(a) is from which we have From Fig. 6.10(b) The Thévenin impedance is therefore FIGURE 6.9(a) An RLC parallel circuit and (b) its transformed circuit. FIGURE 6.10Circuit for obtaining (a) V oc (s) and (b) I sc (s). Vs ss s Vs oc oc () () 3 1 24 1 6 ++ = Vs s s oc () () = - + 46 8 2 Is s s sc ()= -6 6 Zs Vs Is s s s s s s th oc sc () () () () == - + é ? ê ê ù ? ú ú - é ? ê ê ù ? ú ú = + 46 8 6 6 24 8 2 2 ? 2000 by CRC Press LLC and the Thévenin equivalent circuit, with the 4 W connected, is shown in Fig. 6.11. From this circuit we find the transform from which Network Functions A network function or transfer function is the ratio H(s) of the Laplace transform of the output function, say v o (t), to the Laplace transform of the input, say v i (t), assuming that there is only one input. (If there are multiple inputs, the transfer function is based on one of them with the others made zero.) Suppose that in the general case the input and output are related by the differential equation and that the initial conditions are all zero; that is, Then, transforming the differential equation results in from which the network function, or transfer function, is given by (6.14) FIGURE 6.11Thévenin equivalent circuit terminated in a resistor. Vs s ss ss () () ()() = - ++ = - + + + 46 24 16 2 20 4 vt e e tt ()=- + -- 16 20 24 V a dv dt a dv dt a dv dt av b dv dt b dv dt b dv dt bv n n o n n n o n o oo m m i m m m i m i oi ++++ =+ +++ - - - - - - 1 1 1 1 1 1 1 1 L L v dv dt dv dt v dv dt dv dt o o n o n i i m i m () () () () () () 0 00 0 00 0 1 1 1 1 === ==== = - - - - LL ()( )( as as as aVs bs bs bsbVs n n n n o m m m m i ++++ =++++ - - - - 1 1 10 1 1 10 L L Hs Vs Vs bs bs bsb as as as a o i m m m m n n n n () () () == + +×××+ + + +×××+ + - - - - 1 1 10 1 1 10 ? 2000 by CRC Press LLC Example 6.2.8 As an example, let us find the transfer function for the transformed circuit of Fig. 6.12, where the transfer function is V o (s)/V i (s). By voltage division we have (6.15) Step and Impulse Responses In general, if Y(s) and X(s) are the transformed output and input, respectively, then the network function is H(s) = Y(s)/X(s) and the output is (6.16) The step response r(t) is the output of a circuit when the input is the unit step function u(t), with transform 1/s. Therefore, the transform of the step response R(s) is given by (6.17) The impulse response h(t) is the output when the input is the unit impulse d(t). Since + [d(t)] = 1, we have from Eq. (6.16), (6.18) Example 6.2.9 As an example, for the circuit of Fig. 6.12, H(s), given in Eq. (6.15), has the partial fraction expansion, so that If we know the impulse response, we can find the transfer function, from which we can find the response to any input. In the case of the step and impulse responses, it is understood that there are no other inputs except the step or the impulse. Otherwise, the transfer function would not be defined. FIGURE 6.12An RLC circuit. Hs Vs Vs s s s ss o i () () () () ( )( ) == ++ = ++ 4 43 4 13/ Ys HsXs() ()()= Rs Hss() ()= / ht Hs Hs() [()] [()]== -- ++ 11 1/ Hs ss ()= - + + + 2 1 6 3 ht e e tt () =- + -- 26 3 V Hs ht() [()]=+ ? 2000 by CRC Press LLC Stability An important concern in circuit theory is whether the output signal remains bounded or increases indefinitely following the application of an input signal. An unbounded output could damage or even destroy the circuit, and thus it is important to know before applying the input if the circuit can accommodate the expected output. This question can be answered by determining the stability of the circuit. A circuit is defined to have bounded input–bounded output (BIBO) stability if any bounded input results in a bounded output. The circuit in this case is said to be absolutely stable or unconditionally stable. BIBO stability can be determined by examining the poles of the network function (6.14). If the denominator of H(s) in Eq. (6.14) contains a factor (s – p) n , then p is said to be a pole of H(s) of order n. The output V o (s) would also contain this factor, and its partial fraction expansion would contain the term K/(s – p) n . Thus, the inverse transform v o (t) is of the form (6.19) where v 1 (t) results from other poles of V o (s). If p is a real positive number or a complex number with a positive real part, v o (t) is unbounded because e pt is a growing exponential. Therefore, for absolute stability there can be no pole of V o (s) that is positive or has a positive real part. This is equivalent to saying that V o (s) has no poles in the right half of the s-plane. Since v i (t) is bounded, V i (s) has no poles in the right half-plane. Therefore, since the only poles of V o (s) are those of H(s) and V i (s), no pole of H(s) for an absolutely stable circuit can be in the right-half of the s-plane. From Eq. (6.19) we see that v i (t) is bounded, as far as pole p is concerned, if p is a simple pole (of order 1) and is purely imaginary. That is, p = jw, for which which has a bounded magnitude. Unless V i (s) contributes an identical pole jw, v o (t) is bounded. Thus, v o (t) is bounded on the condition that any jw pole of H(s) is simple. In summary, a network is absolutely stable if its network function H(s) has only left half-plane poles. It is conditionally stable if H(s) has only simple jw-axis poles and possibly left half-plane poles. It is unstable otherwise (right half-plane or multiple jw-axis poles). Example 6.2.10 As an example, the circuit of Fig. 6.12 is absolutely stable, since from Eq. (6.15) the only poles of its transfer function are s = –1, –3, which are both in the left half-plane. There are countless examples of conditionally stable circuits that are extremely useful, for example, a network consisting of a single capacitor with C = 1 F with input current I(s) and output voltage V(s). The transfer function is H(s) = Z(s) = 1/Cs = 1/s, which has the simple pole s = 0 on the jw-axis. Figure 6.13 illustrates a circuit which is unstable. The transfer function is which has the right half-plane pole s = 2. FIGURE 6.13Unstable circuit. vt Ate Ate Ae vt on npt n npt pt () ()= + +×××+ + - - -1 1 2 11 etjt pt =+cos sinww Hs IsVs s i () ()() ( )==-//12 ? 2000 by CRC Press LLC Defining Terms Absolute stability: When the network function H(s) has only left half-plane poles. Bounded input–bounded output stability: When any bounded input results in a bounded output. Conditional stability: When the network function H(s) has only simple jw-axis poles and possibly left half- plane poles. Impulse response, h(t): The output when the input is the unit impulse d(t). Network or transfer function: The ratio H(s) of the Laplace transform of the output function to the Laplace transform of the input function. Step response, r(t): The output of a circuit when the input is the unit step function u(t), with transform 1/s. Transformed circuit: An original circuit with the currents, voltages, sources, and passive elements replaced by transformed equivalents. Related Topics 3.1 Voltage and Current Laws ? 3.3 Network Theorems ? 12.1 Introduction References R.C. Dorf, Introduction to Electric Circuits, 2nd ed., New York: John Wiley, 1993. J.D. Irwin, Basic Engineering Circuit Analysis, 3rd ed., New York: Macmillan, 1989. D.E. Johnson, J.R. Johnson, J.L. Hilburn, and P.D. Scott, Electric Circuit Analysis, 3rd ed., Englewood Cliffs, N.J.: Prentice-Hall, 1997. J.W. Nilsson, Electric Circuits, 5th ed., Reading, Mass.: Addison-Wesley, 1996. ? 2000 by CRC Press LLC