5.1 Introduction. 5.2 Definition of Laplace transorm

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1 5.1 Introduction In thi chapter, we will introduce Laplace tranform. Thi i an extremely important technique. For a given et of initial condition, it will give the total repone of the circuit compriing of both natural and forced repone in one operation. The idea of Laplace tranform i analogou to any familiar tranform. For example, Logarithm are ued to change a multiplication or diviion problem into a impler addition or ubtraction problem and Antilog are ued to carry out the invere proce. Thi example point out the eential feature of a tranform: They are deigned to create a new domain to make mathematical manipulation eaier. After evaluating the unknown in the new domain, we ue invere tranform to get the evaluated unknown in the original domain. The Laplace tranform enable the circuit analyt to convert the et of integrodifferential equation decribing a circuit to the complex frequency domain, where thay become a et of linear algebraic equation. Then uing algebraic manipulation, one may olve for the variable of interet. Finally, one ue the invere tranform to get the variable of interet in time domain. Alo, in thi chapter, we expre the impedance in domain or complex frequency domain. Hence, we may analyze a circuit uing one of the reduction technique uch a Thevenin theorem or ource tranformation dicued in earlier chapter. 5. Definition of Laplace tranorm A tranform i a change in the mathematical decription of a phyical variable to facilitate computation. Keeping thi definition in mind, Laplace tranform of a function f(t) i defined a Lff(t)g F () Z 1 f(t)e t dt (5.1) Here the complex frequency i + j!. Since the argument of the exponent e in equation (5.1) mut be dimenionle, it follow that ha the dimenion of frequency and unit of invere econd (ec 1 ).

2 33 j Network Theory The notation implie that once the integral ha been evaluated, f(t), a time domain function i tranformed to F (), a frequency domain function. If the lower limit of integration in equation (5.1) i 1, then it i called the bilateral Laplace tranform. However for circuit application, the lower limit i taken a zero and accordingly the tranform i unilateral in nature. The lower limit of integration i ometime choen to be to permit f(t) to include (t) or it derivative. Thu we hould note immediately that the integration from to + i zero except when an impule function or it derivative are preent at the origin. Region of convergence The Laplace tranform of a ignal f(t) a een from equation (5.1) i an integral operation. It exit if f(t)e t i abolutely integrable. That i Z 1 f(t)e t dt < 1. Cleary, only typical choice of will make the integral converge. The range of that enure the exitence of X() define the region of convergence (ROC) of the Laplace tranform. A an example, let u take x(t) e 3t, t. Then X() Z 1 Z 1 x(t)e (+j!)t dt e ( +3)t e j!t dt The above integral converge if and only if +3< or >3. Thu, >3 define the ROC of X(). Since, we hall deal only with caual ignal(t ) we avoid explicit mention of ROC. Due to the convergence factor, e t, a number of important function have Laplace tranform, even though Fourier tranform for thee function do not exit. But thi doe not mean that every mathematical function ha Laplace tranform. The reader hould be aware that, for example, a function of the form e t doe not have Laplace tranform. The invere Laplace tranform i defined by the relationhip: L 1 ff ()g f(t) 1 j +j1 Z j1 F ()e t d (5.) where i real. The evaluation of integral in equation (5.) i baed on complex variable theory, and hence we will avoid it ue by developing a et of Laplace tranform pair.

3 Laplace Tranform j Three important ingularity function The three important ingularity function employed in circuit analyi are: (i) unit tep function, u(t) (ii) delta function, (t) (iii) ramp function, r(t). They are called ingularity function becaue they are either not finite or they do not poe finite derivative everywhere. The mathematical definition of unit tep function i u(t) ; t < 1; t > (5.3) The tep function i not defined at t. Thu, the unit tep function u(t) i for negative value of t, and 1 for poitive value of t. Often it i advantageou to define the unit tep function a follow: 1; t + u(t) ; t A dicontinuity may occur at time other than t ; for example, in equential witching, the unit tep function that occur at t a i expreed a u(t a). Figure 5.1 The unit tep function Figure 5. The tep function occuring at t a Figure 5.3 The tep function occuring at t a ; t a<or t<a Thu; u(t a) 1; t a>or t>a Similarly, the unit tep function that occur at t a i expreed a u(t + a). ; t+ a<or t< a Thu; u(t + a) 1; t+ a>or t> a

4 33 j Network Theory We ue tep function to repreent an abrupt change in voltage or current, like the change that occur in the circuit of control engineering and digital ytem. For example, the voltage ; t<a v(t) K; t>a may be expreed in term of the unit tep function a v(t) Ku(t a) (5.4) The derivative of the unit tep function u(t) i the unit impule function (t). 8 That i; (t) d < ; t < dt u(t) undefined; t : ; t > (5.5) The unit impule function alo known a dirac delta fucntion i hown in Fig The unit impule may be viualized a very hort duration pule of unit area. Thi may be expreed mathematically a: Z + (t)dt 1 (5.6) where t denote the time jut before t and t + denote the time jut after t. Since the area under the unit impule i unity, it i a practice to write 1 beide the arrow that i ued to ymbolize the unit impule function a hown in Fig When the impule ha a trength other than unity, the area of the impule function i equal to it trength. For example, an impule function 5(t) ha an area of 5 unit. Figure 5.5 how impule function, (t +), 5(t) and (t 3). Figure 5.4 The circuit impule function Figure 5.5 Three impule function

5 Laplace Tranform j 333 An important property of the unit impule function i what i often called the ifting property; which i exhibited by the following integral: Z t f(t ); t f(t)(t t )dt 1 <t <t ; t 1 >t >t t 1 for a fintie t and any f(t) continuou at t. Integrating the unit tep function reult in the unit ramp function r(t). or Figure 5.6 how the ramp function. r(t) r(t) Z t 1 u()d tu(t) (5.7) ; t t; t Figure 5.6 The unit ramp function In general, a ramp i a function that change at a contant rate. Figure 5.7 The unit ramp function delayed by t Figure 5.8 The unit ramp function advanced by t A delayed ramp function i hown in Fig Mathematically, it i decribed a follow: ; t t r(t t ) t t ; t t

6 334 j Network Theory An advanced ramp function i hown in Fig Mathematically, it i decribed a follow: ; t t r(t + t ) t + t ; t t It i very important to note that the three igularity function are related by differentiation a or by integration a (t) du(t) ; u(t) dr(t) dt dt u(t) Z t 1 (t)dt; r(t) Z t 1 u()d 5.4 Functional tranform A functional tranform i imply the Laplace tranform of a pecified function of t. Here we make an aumption that f(t) i zero for t< Decaying exponential function f(t) e at u(t), where a> and u(t) i the unit tep function. L e at u(t) F () Z 1 Z 1 f(t)dt e at e t dt e (+a)t ( + a) 1 + a 1 t 5.4. Unit tep function f(t)u(t) Z 1 Lfu(t)g F () e t dt 1

7 Laplace Tranform j Impule function f(t)(t) Z 1 Lf(t)g F () (t)e t dt e t t 1 Pleae note that we have ued the ifting property of an impule function Sinuoidal function f(t) in!t; t Since in!t 1 e j!t e j!t j and Lfe at g 1 + a we have Lfin!tg F () 1 Z 1 e j!t e j!t e t dt j 1 1 j j! 1 + j!! +! Table 5.1 give a lit of important Laplace tranform pair. It include the function of mot interet in an introductory coure on circuit application. Table 5.1 Important tranform pair f(t)(t ) F () (t) 1 u(t) 1 t 1 e at 1 + a! in!t +! co!t +!

8 336 j Network Theory f(t)(t ) F () t n n! n+1 te at 1 ( + a) e at in!t e at co!t! ( + a) +! + a ( + a) +! All function in the above table are repreented without multiplied by u(t), ince we have explicity declared that t. 5.5 Operational tranform (propertie of Laplace tranform) Operational tranform indicate how mathematical operation performed on either f(t) or F () are converted into the oppoite domain. Following operation are of primary interet. Note: The ymbol mean by the definition Linearity If then Proof: Lff 1 (t)g F 1 () and Lff (t)g F () Lfa 1 f 1 (t)+a f (t)g a 1 F 1 ()+a F () Z 1 Lfa 1 f 1 (t)+a f (t)g [a 1 f 1 (t)+a f (t)]e t dt Z 1 Z 1 a 1 f 1 (t)e t dt + a f (t)e t dt a 1 F 1 ()+a F () EXAMPLE 5.1 Find the Laplace tranform of f(t) (A + Be bt u(t)). We have the tranform pair Lfu(t)g 1 and Lfe bt u(t)g 1 + b

9 Laplace Tranform j 337 Thu, uing linearity property, 5.5. Time hifting Lff(t)g F () LfAu(t)g + LfBe bt u(t)g A + B + b (A + B) + Ab ( + b) If Lfx(t)g X(), then for any real number t, Lfx(t t )u(t t )g e t X() Proof: Z 1 Lfx(t t )u(t t )g x(t t )u(t t )e t dt 1; t t > or t>t Since; u(t t ) ; t t < or t<t we get; Lfx(t t )u(t t )g Uing the tranformation of variable, we get; Lfx(t t )u(t t )g Z 1 t x(t t )e t + t EXAMPLE 5. Find the Laplace tranform of x(t), hown in Fig Z 1 t dt x()e (+t ) d Z 1 e t x()e d e t X() Figure 5.9

10 338 j Network Theory Figure 5.1(a) Uing Fig. 5.1(a) and 5.1(b), we can write Figure 5.1(b) x(t) x 1 (t)+x (t) u(t ) u(t 4) We know that, Lfu(t)g 1 and uing time hifting property, we have Lfx(t)g X() 1 e 1 e 4 ) X() 1 (e e 4 ) Shifting in domain (Frequency-domain hifting) If Lfx(t)g X(), then Proof: Lfe t x(t)g X( ) Z1 Lfe t x(t)g Z 1 e t x(t)e t dt x(t)e ( )t dt X( )

11 Laplace Tranform j 339 EXAMPLE 5.3 Find the Laplace tranform of x(t) Ae at co(! t + )u(t). Given x(t)ae at co(! t + )u(t) Ae at [co! t co in! t in ]u(t) A co e at co! tu(t) A in e at in! tu(t) We know the tranform pair, Lfco! tu(t)g +! and Lfin! tu(t)g! +! Applying frequency hifting property, we get L e at co! tu(t) +! + a ( + a) +! and L e at in! tu(t)! +! Finally, applying linearity property, we get!!+a!+a ( + a) +! LfAe at co(! t + )u(t)g A co Lfe at co! tu(t)g Ain Lfe at in! tu(t)g A co ( + a)! ( + a) +! A in ( + a) +! A[( + a) co θ ω in θ] ( + a) + ω Time caling If Lfx(t)g X(), then Lfx(at)g 1 a X a

12 34 j Network Theory Proof: put Z 1 Lfx(at)g x(at)e t dt at ) adt d Hence Lfx(at)g Z 1 x()e τ a Z 1 1 a d 1 a x()e a d 1 a X a EXAMPLE 5.4 Find the Laplace tranform of x(t) in(! t)u(t). We know the tranform pair, Applying caling property, Lfin! tu(t)g! +! Lfin! tu(t)g 1 6! ! Time differentiation ω +4ω If Lfx(t)g X(), then Proof: Let Then dx(t) L X() x() dt y(t) dx(t) dt Lfy(t)g Y () Z 1 Z 1 dx(t) e t dt dt y(t)e t dt

13 Laplace Tranform j 341 Integrating by part yield Hence; L Y ()e t x(t) Z1 1 x(t)( e t )dt lim t!1 [e t x(t)] x() + x() + X() dx(t) Y () X() x() dt Z 1 x(t)e t dt Therefore, differentiation in time domain i equivalent to multiplication by in the domain. Whenever x(t) i dicontinuou at t (like a tep function), then x() hould be read a x( ). The differentiation property can be extended to yield L d n x(t) dt n n X() n 1 x() x n 1 () When x(t) i dicontinuou at the origin, the argument on the right ide of the above equation hould be read a. Accordingly for a dicontinuou function x(t) at the origin, we get d n x(t) L dt n n X() n 1 x( ) x n 1 ( ) EXAMPLE 5.5 Find the Laplace tranform of x(t) in! tu(t). We find that, x() We know that, Lfin! tu(t)g! +! dx(t)! in! t co! tu(t) dt! in! tu(t) (5.8)

14 34 j Network Theory Applying time caling property, 3 Lfin! tu(t)g 1 6! !! +(! ) Taking Laplace tranform on both the ide of equation (5.8), we get dx(t) L! Lfin! tu(t)g dt ) X() x() ) X() EXAMPLE 5.6 Solve the econd order linear differential equation! +(! ) ω [ +(ω ) ] y (t)+5y (t)+6y(t) x(t) with the initial condition, y(), y ()1and x(t) e t u(t). Taking Laplace tranform on both the ide of the given differential equation, we get Y () y() y () +5jY () y()j +6Y ()X() where X() Lfe t u(t)g 1 +1 Subtituting the initial condition, we get ( +5 +6)Y () ) Y () ( + 1)( + )( +3) Uing partial fraction expanion, we get Y () Taking invere Laplace tranform, we get y(t) 1 e t +6e t 9 e 3t, t

15 Laplace Tranform j Integration in time domain For a caual ignal x(t), If then Proof: Dividing both ide by yield y(t) Z t x()d; Lfy(t)g Y () X() Lfx(t)g X() 1Z x(t)e t dt X() Z 1 x(t) e t dt Integrating the right-hand ide by part, we get X() X() e t y(t) y(t) e t 1 t 1 t + Z 1 y(t) e t ( )dt Z 1 y(t)e t dt The firt term on the right-hand ide evaluate to zero at both limit, becaue Hence; e 1 and y() Z Y() X() x()d Thu, integration in time domain i equivalent to diviion by in the domain. EXAMPLE 5.7 Conider the RC circuit hown in Fig The input i the rectangular pule hown in Fig Find i(t) by auming circuit i initially relaxed.

16 344 j Network Theory Figure 5.11 Figure 5.1 Applying KVL to the circuit repreented by Fig. 5.11, we get Ri(t)+ 1 C ) Ri(t)+ 1 C Z t Z t i()d v(t) Taking Laplace tranform on both the ide, we get We know the tranform pair, i()d V o [u(t a) u(t b)] RI()+ 1 C I()V o (e a e b ) ) I() V o R + 1 (e a e b ) RC Lfe at u(t)g 1 + a and then uing the time-hift property, we can find invere of I(). That i; i(t) V o R e t RC u(t) v o t!t a R e t RC u(t) t!t b ) i(t) V i o (t a) (t b) he RC u(t a) e RC u(t b) R Differentiation in the domain For a ignal x(t), t,wehave Lf tx(t)g dx() d

17 Laplace Tranform j 345 Proof: For a caual ignal, x(t), the Laplace tranform i given by Z 1 Lfx(t)g X() x(t)e t dt Differentiating both the ide with repect to, weget Z dx() x(t)( te t )dt d ) Z dx() [ tx(t)]e t dt d Hence; Lf tx(t)g dx() d or In general; Lft n x(t)g ( 1) n dn X() d n EXAMPLE 5.8 Find the Laplace tranform of x 1 (t) te 3t u(t). We know that, Lfe at u(t)g 1 + a Hence Lfe 3t u(t)g 1 +3 Uing the differentiation in domain property, Convolution Lfx 1 (t)g X 1 () d 1 ( +3) d Lftx(t)g dx() d 1 +3 If and then Lfx(t)g X() Lfh(t)g H() Lfx(t) h(t)g X()H() where indicate the convolution operator.

18 346 j Network Theory Proof: x(t) h(t) Z 1 1 x()h(t )d Since x(t) and h(t) are caual ignal, the convolution in thi cae reduce to x(t) h(t) Hence; Lfx(t) h(t)g Z 1 Z 1 Interchanging the order of integral, we get Lfx(t) h(t)g Z 1 x()h(t )d 4 Z 1 x()h(t )d5 e t dt x() 4 3 Z 1 h(t )e t dt5 d Uing the change of variable t in the inner integral, we get Z 1 Z 1 3 Lfx(t) h(t)g x()e 4 h()e d5 d X()H() Pleae note that thi theorem reduce the complexity of evaluating the convolution integral to a imple multiplication. EXAMPLE 5.9 Find the convolution of h(t) e t and f(t) e t. 3 h(t) f(t)l 1 fh 1 ()F ()g L L e t e t, t

19 Laplace Tranform j 347 EXAMPLE 5.1 Find the convolution of two indentical rectangular pule. Each rectangular pule ha unit amplitude and duration equal to T econd. Alo, the pule i centered at t T. Let the pule be a hown in Fig From the Fig. 5.13, we can write x(t) u(t) u(t T ) Taking Laplace tranform, we get X() 1 1 e T T Figure 5.13 T Let Then; 1 (1 e T ) y(t) x(t) x(t) Y()X () 1 e T ) Y () 1 e T + 1 e 4T Taking invere Laplace tranform, we get y(t)tu(t) (t T )u(t T )+(t 4T)u(t 4T) r(t) r(t T )+r(t 4T) y(t) x(t) x(t) Initial-value theorem Figure 5.14 The initial-value theorem allow u to find the initial value x() directly from it Laplace tranform X(). If x(t) i a caual ignal, then; x() lim X()!1 (5.9) Proof: To prove thi theorem, we ue the time differentiation property. dx(t) L dt X() x() Z 1 dx The problem with are better undertood after the invere Laplace tranform are tudied. dt e t dt (5.1)

20 348 j Network Theory If we let!1, then the integral on the right ide of equation (5.1) vanihe due to damping factor, e t. Thu; EXAMPLE 5.11 Find the initial value of We know the tranform pair: lim!1 [X() x()] ) x() lim!1 X() F () +1 ( +1) f() lim X() lim!1!1 ( +1) +3 lim!1 + lim!1 ( +1) Lfe bt + b co atg ( + b) + a Hence, invere Laplace tranform of F () yield f(t) e t co 3t At t,wegetf()1. Thi verifie the theorem Final-value theorem The final-value theorem allow u to find the final value x(1) directly from it Laplace tranform X(). If x(t) i a caual ignal, then lim x(t) lim X() t!1! Proof: The Laplace tranform of dx(t) i given by dt X() x() Z 1 dx(t) e t dt dt

21 Laplace Tranform j 349 Taking the limit! on both the ide, we get lim[x() x()] lim!! Z 1 dx(t) e t dt dt Z 1 dx(t) dt Z 1 dx(t) dt dt h i lim! e t dt Since; we get; Hence; x(t)j 1 x(1) x() lim[x() x()] lim[x()] x()!! x(1) x() lim[x() x()]! x(1) lim! [X()] Thi prove the final value theorem. The final value theorem may be applied if, and only if, all the pole of X() have a real part that i negative. The final value theorem i very ueful ince we can find x(1) from X(). However, one mut be careful in uing final value theorem ince the function x(t) may not have a final value a a t!1. For example, conider x(t) in at having X(). Now we know + a lim in at doe not exit. However, if we uncarefully ue the final value theorem in thi cae, we t!1 would obtain: lim X() lim a!! + a Note that the actual function x(t) doe not have a limiting value a t!1. The final value theorem ha failed becaue the pole of X() lie on the j! axi. Therefore, we conclude that for final value theorem to give a valid reult, pole of X() hould not lie to right ide of the -plane or on the j! axi. EXAMPLE 5.1 Find the final value of X() 1 ( +1) +1 Conider a function, X() P (). The root of the denomoniator polymial, Q() are called pole () and the Q() root of the numerator polynomial, P () are called zero (O).

22 35 j Network Theory lim x(t)x(1) t!1 We know the Laplace tranform pair lim! [X()] lim! 1 ( +1) +1 Hence; Thu; L e at in b bt ( + a) + b x(t)l 1 fx()g L 1 1 ( +1) +1 e t in 1t x(1) Thi verifie the reult obtained from final-value theorem Time periodicity Let u conider a function x(t) that i periodic a hown in Fig The function x(t) can be repreented a the um of time-hifted function a hown in Fig Figure 5.15 A periodic function Figure 5.16 Decompoition of periodic function Hence; x(t)x 1 (t)+x (t)+x 3 (t)+ x 1 (t)+x 1 (t T )u(t T )+x 1 (t T )u(t T )+ (5.11) where x 1 (t) i the waveform decribed over the firt period of x(t). That i, x 1 (t) i the ame a the function x(t) gated over the interval <t<t. gating mean the function x(t) i multiplied by 1 over the interval t T and elewhere by.

23 Laplace Tranform j 351 Taking the Laplace tranform on both ide of equation (5.11) with the time-hift property applied, we get X()X 1 ()+X 1 ()e T + X 1 ()e T + ) X()X 1 ()(1 + e T + e T + ) But 1 + a + a a ; jaj < 1 1 Hence, we get X()X 1 () 1 e T (5.1) In equation (5.1), X 1 () i the Laplace tranform of x(t) defined over firt period only. Hence, we have hown that the Laplace tranform of a periodic function i the Laplace tranform evaluated over it firt period divided by 1 e T. EXAMPLE 5.13 Find the Laplace tranform of the periodic ignal x(t) hown in Fig Figure 5.17 From Fig. 5.17, we find that T Second. The ignal x(t) conidered over one period i donoted a x 1 (t) and hown in Fig. 5.18(a). Figure 5.18(a) Figure 5.18(b) Figure 5.18(c)

24 35 j Network Theory The ignal x 1 (t) may be viewed a the multiplication of x A (t) and g(t). That i; x 1 (t)x A (t)g(t) [ t + 1][u(t) u(t 1)] ) x 1 (t) tu(t)+tu(t 1) + u(t) u(t 1) tu(t)+(t 1+1)u(t 1) + u(t) u(t 1) tu(t)+(t 1)u(t 1) + u(t 1) + u(t) u(t 1) u(t) tu(t)+(t 1)u(t 1) u(t) r(t)+r(t 1) Taking Laplace Tranform, we get X 1 () e 1+e Hence; X() X 1() 1 e ( 1+e ) T (1 e ) 5.6 Invere Laplace tranform The invere Laplace tranform of X() i defined by an integral operation with repect to variable a follow: x(t) 1 +j1 Z j1 X()e t d (5.13) Since i complex, the olution requrie a knowledge of complex variable. In otherword, the evaluation of integral in equation (5.13) require the ue of contour integration in the complex plane, which i very difficult. Hence, we will avoid uing equation (5.13) to compute invere Laplace tranform. In many ituation, the Laplace tranform can be expreed in the form X() P () Q() (5.14) where P ()b m m + b m 1 m b 1 + b Q()a n n + a n 1 n a 1 + a ; a n 6 The function X() a defined by equation (5.14) i aid to be rational function of, ince it i a ratio of two polynomial. The denominator Q() can be factored into linear factor.

25 Laplace Tranform j 353 A partial fraction expanion allow a trictly proper rational function P () to be expreed a a Q() factor of term whoe numerator are contant and whoe denominator correpond to linear or a combination of linear and repeated factor. Thi in turn allow u to relate uch term to their correponding invere tranform. For performing partial fraction technique on X(), the function X() ha to meet the following condition: (i) X() mut be a proper fraction. That i, m<n. When X() i improper, we can ue long diviion to reduce it to proper fraction. (ii) Q() hould be in the factored form. EXAMPLE 5.14 Find the invere Laplace tranform of X() where; +4 X() ( +) ( + 1)( +3) K K +3 K 1 ( +1)X()j 1 ( +) ( +3) 1 1 K ( +3)X()j 3 ( +) ( +1) 1 3 Hence; X() We know that: Lfe t u(t)g 1 + Therefore; x(t) [e t + e 3t ]u(t) EXAMPLE 5.15 Find the invere Laplace tranform of X() + +5 ( + 3)( +5)

26 354 j Network Theory Let X() K K +5 + K 3 ( +5) where K 1 ( +3)X()j ( +5) 3 K 1 d 1! d [( +5) X()] 5 d + +5 d ( +3) K 3 ( +5) X() ( +3) 1 5 Then X() ( +5) Taking invere Laplace tranform, we get x(t)e 3t e 5t 1te 5t ; t or x(t) (e 3t e 5t 1te 5t )u(t) Reinforcement problem R.P 5.1 Find the Laplace tranform of: (a) coh(at) (b) inh(at) (a) coh(at) 1 [eat + e at ] We know the Laplace tranform pair: Lfe at g 1 + a and Lfe at g 1 a

27 Laplace Tranform j 355 Applying linearity property, we get, Lfcoh(at)g 1 Lfeat g + 1 Lfe at g 1 1 a a a (b) inh at 1 [eat e at ] Applying linearity property, Lfinh(at)g 1 1 a 1 + a a a R.P 5. Find the Laplace tranform of f(t) co(!t + ). Given f(t) co(!t + ) co co!t in in!t Applying linearity property, we get, Lff(t)g F () co Lfco!tg in Lfin!tg co +! in! +! co θ ω in θ + ω R.P 5.3 Find the Laplace tranform of each of the following function: (a) x(t) t co(t +3 )u(t) (b) x(t) tu(t) 4 d dt (t) (c) x(t) 5u t 3 (d) x(t) 5e t u(t)

28 356 j Network Theory (a) Let u firt find the Laplace tranform of co(t +3 )u(t) L fco(t +3 )u(t)g L fco 3 co tu(t) in 3 in tu(t)g co 3 Lfco tu(t)g in 3 Lfin tu(t)g co 3 in co 3 in 3 +4 The Laplace tranform of x(t) i now found by uing differentiation in domain property. L t co(t +3 ) d Lfco(t +3 d )u(t)g d co 3 in 3 d +4 p 3 3 d d p 3 3 d d d d " p! d d 3 d d # " p d 3 +4! 1 d p p p ( ) ( +4) 1 ( +4) ( +4) + 8 1p p 3 ( +4) 3 p! # p! ( +4) 3 (b) x(t) tu(t) 4 d dt (t) Lfx(t)g X() Lftu(t)g 4L d dt (t)

29 Laplace Tranform j 357 d We know that whenever a function f(t) i dicontinuou at the origin, we have L dt f(t) F () f( ). Applying thi relation to the econd term on the right ide of the above equation, we get X() 1 4[ 1 ( )] 4[ ] t (c) x(t) 5u 3 Uing caling property, 4 Lff(at)g 1 a F a we get; Lfx(t)g X() Lfu(t)g! (d) x(t) 5e t u(t) We know the Laplace tranform pair: Hence; 5 Lfe at u(t)g 1 + a n o Lfx(t)g X() 5L e 1 t u(t) R.P 5.4 Find the Laplace tranform of the following function: (a) x(t) t co at (b) x(t) 1 in at inh(at) a (c) x(t) in!t t!

30 358 j Network Theory (a) x(t) t co at We know that Let Hence Lftf(t)g d d F () f(t) co at ) F () + a Lft co atg Lftf(t)g d d a ( + a ) (b) x(t) 1 in at inh at a 1 a 1 eat in at 1 e at in at 1 4a e at in at e at in at We know the hifting in domain property: + a Lfe t f(t)g F ()j!( ) Applying thi property along with linearity property, we get (c) x(t) 1 t in!t Lfx(t)g X() 1 Lfe at 4a in atg Lfe at in atg 1 4a 1 4a a a + a! a + a!+a a ( a) + a a ( + a) + a [( a) + a ][( + a) + a ] We know that Lff(t)g F () Z 1 f(t)e t dt

31 Laplace Tranform j 359 Hence; Z 1 Z 1 Z 1 F ()d f(t) e t ddt Z 1 Z 1 L e t f(t) t f(t) e t dt t f(t) t 1 dt In the preent cae; f(t) in!t 1 j ej!t 1 j e j!t ej!t +e j!t 4 Hence; F() j! 4 + j! Hence, 1 X()L t in!t 1 L t f(t) Z 1 Z x F ()d lim f(x)dx x!1 ln(x j!) ln( j!) lnx +ln + ln(x + j!) ln( + j!) lim x!1 4 1 x 4 ln +4! ln +4! x 1 4 ln +4ω x!1 R.P 5.5 Conider the pule hown in Fig. R.P. 5.5, where f(t) e t for <t<t. Find F () for the pule.

32 36 j Network Theory Figure R.P. 5.5 The dicrete pule f(t) could be imagined a the product of ignal x(t) and g(t) a hown in Fig. R.P. 5.5(a) and (b) repectively. That i; f(t)x(t)g(t) e t [u(t) u(t T )] e t u(t) e t u(t T ) e t u(t) e (t T +T ) u(t T ) e t u(t) e T e (t T ) u(t T ) Hence; Lff(t)g F () 1 et e T 1 e T ( ) T ( 1 e ) ( ) Figure R.P. 5.5(a) Alternate method: Lff(t)g F () Z T Z 1 e t e t dt 1 e T ( ) ( ) f(t)e t dt

33 Laplace Tranform j 361 R.P 5.6 Find the Laplace tranform of f(t) hown in Fig. R.P Figure R.P. 5.6 f(t) i a dicrete pule and can be expreed mathematically a: f(t)x(t)g(t) in t[u(t) u(t 1)] in tu(t) in tu(t 1) in tu(t) in[(t 1 + 1)]u(t 1) ) f(t) in tu(t) in[(t 1)] co (t 1) co[(t 1)] in u(t 1) in tu(t) + in[(t 1)]u(t 1) Hence; F()Lff(t)g + + e 1 + π + π [1 + e ] Figure R.P. 5.6(a) R.P 5.7 Determine the Laplace tranform of f(t) hown in Fig. R.P Figure R.P. 5.7

34 36 j Network Theory We can write, f(t)x(t)g(t) 5 t [u(t) u(t )] 5 tu(t) 5 tu(t ) 5 tu(t) 5 (t +)u(t ) 5 tu(t) 5 (t )u(t ) 5u(t ) Hence; Lff(t)g F () e 5 e 5 [1 e e ] Figure R.P. 5.7(a) R.P 5.8 Find the Laplace tranform of f(t) hown in Fig. R.P Figure R.P. 5.8 The equation of a traight line i y mx + c, where m lope of the line and c intercept on y-axi. Hence, f(t) 5 3 t +5 When f(t), let u find t.

35 Laplace Tranform j 363 That i; 5 3 t +5 ) t 4: Second Mathematically, f(t)x(t)g(t) 5 3 t +5 [u(t) u(t 4:)] 5 3 tu(t)+5 tu(t 4:)+5u(t) 5u(t 4:) tu(t)+5 (t 4:+4:)u(t 4:) 3 +5u(t) 5u(t 4:) 5 3 tu(t)+5 (t 4:)u(t 4:)+7u(t 4:) 3 +5u(t) 5u(t 4:) 5 3 tu(t)+5 (t 4:)u(t 4:)+u(t 4:)+5u(t) 3 Hence; F()Lff(t)g e 4: + e 4: e 4: +6e 4: Figure R.P. 5.8(a) R.P 5.9 If f( ) 3 and 15u(t) 4(t) 8f(t) +6f (t), find f(t) (hint: by taking the Laplace tranform of the differential equation, olving for F () and by inverting, find f(t)). Given, 15u(t) 4(t) 8f(t)+6f (t) Taking Laplace tranform on both the ide, we get 15 48F ()+6[F () f( )] 15 ) 48F ()+6F () Therefore; F()(6 +8) 18 + ) F () (6 +8) (6 +8) K 1 + K

36 364 j Network Theory The contant K 1 and K are found uing the theory of partial fraction. +15 K : K 6 5: Hence; F() 1:875 5: i Taking the invere, we get f(t) h e 4 3 t u(t) R.P 5.1 Find the invere Laplace tranform of the following function: +1 (a) F () e (b) F () (a) F () ( +) +9 ( +) 1 ( +) +9 + ( +) +3 1 ( +) +3 + ( +) ( +) +3 The determination of the Laplace invere make ue of the following two Laplace tranform pair: Lfe bt a in atg ( + b) + a Hence; Lfe bt + b co atg ( + b) + a f(t)l 1 ff ()g e t co 3t 3e (b) F () e t in 3t

37 Laplace Tranform j 365 Let where F ()e X() 3 X() ( +1) ( +1) +4 ) x(t) 3 4 e t in 4t Since F ()e X() we get; f(t)x(t 1) Therefore; f(t) 3 4 e (t 1) in[4(t 1)]; t > 1 f(t) 3 4 e (t 1) in[4(t 1)]u(t 1) Laplace tranform method for olving a et of differential equation: 1. Identify the circuit variable uch a inductor current and capacitor voltage.. Obtain the differential equation decribing the circuit and keep a watch on the initial condition of the circuit variable. 3. Obtain the Laplace tranform of the variou differential equation. 4. Uing Cramer rule or a imilar technique, olve for one or more of the unknown variable, obtaining the olution in domain. 5. Find the invere tranform of the unknown variable and thu obtain the olution in the time domain. R.P 5.11 Referring to the RL circuit of Fig. R.P. 5.11, (a) write a differential equation for the inductor current i L (t). (b) Find I L (), the Laplace tranform of i L (t). (c) Solve for i L (t). Figure R.P. 5.11

38 366 j Network Theory (a) Applying KVL clockwie,weget 1i L (t)+5 di L 5u(t ) dt (b) Taking Laplace tranform of the above equation, we get 1I L ()+5[I L () i L ( )] 5 e 5 ) I L () e +5i L ( ) 5 +1 e ( +) e K1 + K + where K K 1 1 Hence; I L () 1 e ( +) ( +) (c) Taking Invere Laplace tranform, we get i L (t) 1 u(t) e t u(t) t!t e t u(t) 1 u(t ) e t u(t ) e t u(t) R.P 5.1 Obtain a ingle integrodifferential equation in term of i C for the circuit of Fig. R.P Take the Laplace tranform, olve for I C (), and then find i C (t) by making ue of invere tranform. Figure R.P. 5.1

39 Laplace Tranform j 367 Applying KVL clockwie to the right-meh, we get 4u(t)+i C +1 Taking Laplace tranform, we get Z 1 i C dt +4[i C :5(t)] I C()+ 1I C() +4I C () ) I C () Taking invere Laplace tranform, we get i C (t).4δ(t) 1.6e t u(t) Amp. R.P 5.13 Refer the circuit hown in Fig. R.P Find i() and i(1) uing initial and final value theorem. Figure R.P Applying KVL we get i + di dt 1 Taking Laplace tranform, on both the ide, we get I()+[I() i( )] 1 ) I()+[I() 1] 1 ) I()[1+] 1 +

40 368 j Network Theory 1 ) I() (1+) (1+) According to initial value theorem, i() lim!1 I() ( +5) lim! !1 lim We know from fundamental for an inductor, i( + )i( )i(). Hence, i() found uing initial value theorem verifie the initial value of i(t) given in the problem. From final value theorem, i(1) lim! I() ( +5) lim! A R.P 5.14 Find i(t) and v C (t) for the circuit hown in Fig. R.P when v C () 1Vandi() A.The input ource i v i 15u(t) V. Chooe R o that the root of the characteritic equation are real. Figure R.P. 5.14

41 Laplace Tranform j 369 Applying KVL clockwie,weget L di dt + v C + R i v i (t) (5.15) The differential equation decribing the variable v C i C dv C dt i (5.16) The Laplace tranform of equation (5.15) i L[I() i() + V C ()+RI() V i ()] (5.17) The Laplace tranform of equation (5.16) i C[V C () v C () I()] (5.18) Noting that i(), ubtituting for C and L and rearranging equation (5.17) and (5.18), we get, [R + ]I()+V C ()V i () 15 (5.19) I()+ 1 V C()5 (5.) Putting equation (5.19) and (5.) in matrix form, we get 3 R + 1 " # 15 I() V C () (5.1) Solving for I() uing Cramer rule, we get I() 5 + R + The invere Laplace tranform of I() will depend on the value of R. The equation + R + i defined a the characteritic equation. For the root of thi equation to be real, it i eential that b 4ac. Thi mean that; R 4 1 ) R p The condition b 4ac i with repect to algebraic equaion ax + bx + c.

42 37 j Network Theory Let u chooe the value of R a 3 5 Then I() ( + 1)( +) ) I() K K + where K K Hence; I() Taking invere Laplace tranform, we get i(t) 5e t u(t) 5e t u(t) Pleae note that t give i()and t 1 give i(1). Solving the matrix equation (5.1) for V C (), uing Cramer rule, we get Subtituting the value of R,weget V C () 1 +1R +3 ( + R +) V C () 1( +3 +3) ( + 1)( +) Uing partial fraction expanion, we can write, V C () K 1 + K +1 + K 3 + where, K 1 15, K 1, K 3 5 Hence; V C () Taking invere Laplace tranform, we get v C (t) 15u(t) 1e t u(t)+5e t u(t) Verification: Putting t,weget Thi check the validity of reult obtained. v C () V v C (1)15 +15V

43 Laplace Tranform j 371 R.P 5.15 For the circuit hown in Fig. R.P. 5.15, the teady tate i reached with the 1 V ource. At t, witch K i opened. What i the current through the inductor at t 1 econd? Figure R.P At t, the circuit i a hown in Fig. 5.15(a). i ( + )i ( ):5 A Figure R.P. 5.15(a) Figure R.P. 5.15(b) For t +, the circuit diagram i a hown in Fig. 5.15(b). Applying KVL clockwie to the circuit, we get 8i(t)+4 di dt

44 37 j Network Theory Taking Laplace tranform, we get 8I()+4[I() i( )] ) 8I()+4I()4:5 ) [ + ]I():5 I() :5 + Taking invere Laplace tranform, we get, i(t) :5e t At t :5 ec, we get R.P 5.16 Refer the circuit hown in Fig. R.P Find: i(.5).5e A (a) v o (t) for t (b) i o (t) for t (c) Doe your olution for i o (t) make ene when t?explain. Figure R.P Figure R.P. 5.16(a)

45 Laplace Tranform j 373 KCL at node A: (for t ) Taking Laplace tranform, Hence; I dc 1 L I dc Z t V o() L v o dt + v o R + C dv o dt + V o() R + CV o() I dc V o () C RC LC Subtituting the value of I dc, R, L, and C, we find that V o () Uing partial fraction, we get where K 1, and K 1; +1; ; ( + )( + 8) V o () K K + 8 Hence; V o () Taking invere Laplace tranform, we get v o (t) e t u(t) e 8t u(t) (b) i o (t) C dv o dt Hence I o () C [V o () v o ()] For t, ince the witch wa in cloed tate, the circuit wa not activated by the ource. Thi mean that v o () v o ( )v o ( + )and i L ( + )i L ( ): Then; I o ()CV o () ; +1; ( + )( + 8) K K + 8

46 374 j Network Theory We find that, K 1 1 3, and K Therefore; I o () Taking invere Laplace tranform, we get i o (t) 4e 8t u(t) e t u(t)ma (c) i o ( + )4 1 3mA Ye. The initial inductor current i zero by hypothei i L ( + )I L ( ). Alo, the initial reitor current i zero becaue v o ( + ) v o ( ). Thu at t +, the ource current appear in the capacitor. R.P 5.17 Refer the circuit hown in Fig. R.P The circuit parameter are R 1kΩ, L 8 mh and C 1nF, if V dc 7V, find: (a) v o (t) for t (b) i o (t) for t (c) Ue initial and final value theorem to check the inital and final value of current and voltage. Figure R.P At t, witch i open and at t +, the witch i cloed. Since at t, the circuit i not energized by dc ource, i o ( )and v o ( ). Then by the hypothei, that the current in an inductor and voltage acro a capacitor cannot change intantaneouly, i o ( + )i o ( )and v o ( + )v o ( )

47 Laplace Tranform j 375 The KCL equation when the witch i cloed (for t + )igivenby C dv o dt + v o R + 1 Z t (v o V dc )d L ) C dv o dt + v o R + 1 L ) C dv o dt + v o R + 1 L Z t Z t v o d 1 L Z t v o d 1 L V dct V dc d Laplace tranform of the above equation give C [V o () v o ()] + V o() R + 1 V o () 1 V dc L L Since v o () i ame a v o ( ),weget CV o ()+ V o() R + 1 V o () V dc L L V dc ) V o () LC + 1 RC + 1 LC Subtituting the value of V dc, R, L, and C, weget V o () [ ] ( 1 )( ) where 1 ; 5 p Hence; Uing partial fraction, we get 5 j V o () ( + 5) j35)( j35) V o () K 1 + K + 5 j35 + K j We find that K K ( 5 + j35)(j7)

48 376 j Network Theory 5 p 5 /171:87 V o () 7 + 5p 5 /171: j p 5 / 171: j35 Taking invere Laplace tranform, we get v o (t) h7+5 p 5 /171:87 e (5 j35)t +5 p i 5 / 171:87 e (5+j35)t u(t) The invere of V o () can be expreed in a better form by following the technique decribed below: Let u conider a tranformed function where F () C + jd + a j! + m / + a j! + C jd + a + j! m / + a + j! m p c + d and tan 1 d c The invere tranform of F () i given by f(t) me at co(!t + )u(t) (For the proof ee R.P. 5.19) In the preent context, m 5 p 5; 171:87! 35 and a 5 ( Thi mean that, L 1 5 p 5 /171: j p ) 5 / 171: j35 5 p 5e 5t co (35t + 171:87 ) 1 p 5e 5t co (35t + 171:87 ) Hence; v o (t) 7+1 p 5e 5t co (35t ) u(t)

49 Laplace Tranform j 377 (b) i o (t) v o R + C dv o dt Taking Laplace tranform on both the ide, we get I o () V o() R + C V o () v o ( ) ) I o () V o() R + CV o() ) I o ()CV o () + 1 RC Vdc L RC + 1 RC + 1 LC Subtituting the value of V dc, R, L, and C, weget We find that, I o () 87:5( + 1) ( + 5 j35)( j35) K 1 + K + 5 j35 + K j35 87:5 1 K mA 87:5(5 + j35) K ( 5 + j35)(j7) 1:5/ 16:6 ma I o () 7 + 1:5/ 16:6 + 5 j35 + 1:5 /16: j35 The invere Laplace tranform yield, i i o (t) h7+1:5/ 16:6 e (5 j35)t +1:5 /16:6 e (5+j35)t u(t) 7+5e 5t co(35t 16.6 ) u(t)ma V dc (c) V o () LC RC LC From Final Value theorem: v o (1) lim v o (t) lim V o () V dc LC 7V t!1! LC The ame reult may be obtained by putting t 1 in the expreion for v o (t).

50 378 j Network Theory From initial value theorem : v o () lim!1 V o () V dc 6 LC!1 6 lim + 1 RC + 1 LC Thi verifie our beginning analyi that v o ( + ) v o ( ). The ame reult may be obtained by putting t in the expreion for v o (t). + 1 RC We know that, I o () V dc L + 1 RC + 1 LC From final value theorem : I o (1) lim! I o () lim! 6 V dc L RC RC + 1 LC V dc RC L 1 LC V dc R mA The ame reult may be obtained by putting t 1 in the expreion for i o (t). From initial value theorem : i o () lim I o ()!1 lim 6 V dc!1 4 L RC + 1 RC LC Thi agree with our initial analyi that the initial current through the inductor i zero. The ame reult can be obtained by putting t in the expreion for i o (t).

51 Laplace Tranform j 379 R.P 5.18 Apply the initial and final value theorem to each of the function given below: (a) F () (b) F () ( +6 +8) Since in F () referred in (a) and (b) are improper 1 fraction, the correponding time domain counterpart, f(t) contain impule. Thu, neither the initial value theorem nor the final value theorem may be applied to thee tranformed function. R.P 5.19 Find the invere Laplace tranform of F () c + jd + a j! + Expreing c + jd and c jd in the exponenetial from, we get, c jd + a + j! where Hence; F () me j + a j! + me j + a + j! m p c + d and tan 1 d c f(t)l 1 ff ()g me j e (a j!)t u(t)+me j e (a+j!)t u(t) me at e j(+!t) u(t)+me at e j(+!t) u(t) " # me at e j(+!t) + e j(+!t) u(t) me at co(θ + ωt)u(t) R.P 5. Find the initial and final value of f(t) when F () Initial value theorem f() lim!1 F () lim! If the degree of the numerator polynomial i greater than or equal to the degree of the denominator polynomial, the fraction i aid to be improper.

52 38 j Network Theory Final value theorem: The pole of F () are given by finding the root of the denominator polynomial. That i, +1 ) ( 1) ) 1; 1 Since both the pole of F () lie to the right of the plane, final value theorem cannot be ued to find f(1). R.P 5.1 Find i(t) for the circuit of Fig.R.P. 5.1, when i 1 (t) 7e 6t A for t and i(). Alo find i(1). Figure R.P. 5.1 Refer Fig. R.P. 5.1(a). KCL at node v 1 : v i 7e 6t Alo; v 1 3i +4 di dt 1 Hence; 5 ) ) 3i +4 di dt 4 di + i 7e 6t 5 dt i 7e 6t di dt +i 35 4 e 6t Taking Laplace tranform of the differential equation, we get Figure R.P. 5.1(a) [I() i()] + I() 35 4 ) I() ( + )( +6)

53 Laplace Tranform j 381 Uing partial fraction expanion, we get I() K K +6 and find that K and K Hence; I() ) i(t) e t Circuit element model and initial condition 1 +6 e 6t u(t) In the analyi of a circuit, the Laplace tranform can be carried one tep further by tranforming the circuit itelf rather than the differential equation. Earlier we have een how to repreent a circuit in time domain by differential equation and then ue Laplace tranform to tranform the differential equation into algebraic equation. In thi ection, we will ee how we can repreent a circuit in domain uing the Laplace tranform and then analyze it uing algebraic equation Reitor The voltage-current relationhip for a reitor R i given by Ohm law: Taking Laplace tranform on both the ide, we get v(t) i(t)r (5.) V () I()R (5.3) Fig (a) how the repreentation of a reitor in time domain and Fig. 5.19(b) in frequency domain uing Laplace tranform. Figure 5.19(a) Reitor repreented in time domain Figure 5.19(b) Reitor repreented in the frequency domain uing Laplace tranform The impedance of an element i defined a Z() V () I()

54 38 j Network Theory provided all initial condition are zero. Pleae note that the impedance i a concept defined only in frequency domain and not in time domain. In the cae of a reitor, there i no initial condition to be et to zero. Compariion of equation (5.) and (5.3) reveal that, reitor R ha ame repreentation in both time and frequency domain Capacitor For a capacitor with capacitance C, the time-domain voltage-current relationhip i Z t v(t) 1 i()d + v() C (5.4a) The domain characterization i obtained by taking the Laplace tranform of the above equation. That i, V () 1 v() I()+ C (5.4b) To find the impedance of a capacitor, et the initial condition v() to zero. Then from equation (5.4b), we get Z() V () I() 1 a the impedance of the capacitor. With the help of equation C (5.4b), we can draw the frequency domain repreentation of a Capacitor and the ame i hown in Fig. 5.(b). Thi equivalent circuit i drawn o that the KVL equation repreented by equation (5.4 b) i atified. Performing ource tranformation on the equivalent domain circuit for a capacitor which i hown in Fig. 5.(b), we get an alternate frequency domain repreentation a hown in Fig. 5.(c). Figure 5.(a) A capacitor repreented in time domain (b) A capacitor repreented in the frequency domain (c) Alternate frequency domain repreentation for a capacitor

55 Laplace Tranform j Inductor For an inductor with inductance L, the time domain voltage-current relation i v(t) L di(t) dt (5.5) The Laplace tranform of equation (5.5) yield, V () LI() Li() (5.6) To find impedance of an inductor, et the initial condition i() to zero. Then from equation (5.6), we get Z() V () I() L (5.7) which repreent the impedance of the inductor. Equation (5.6) i ued to get the frequency domain repreentation of an inductor and the ame i hown in Fig. 5.1(b). The erie connection of element correpond to um of the voltage in equation (5.6). Converting the voltage ource in Fig.5.1(b) into an equivalent current ource, we get an alternate repreentation for the inductor in frequency domain which i a hown in Fig. 5.1(c). To find the frequency domain repreentation of a circuit, we replace the time domain repreentation of each element in the circuit by it frequency domain repreentation. Figure 5.1(a) An inductor repreented in time domain (b) An inductor repreented in the frequency domain (c) An alternate frequency domain repreentation To find the complete repone of a circuit, we firt get it frequency domain repreentation. Next, uing KV L or KCL, we find the variable of interet in doamin. Finally, we ue the invere Laplace tranform to repreent the variable of interet in time domain.

56 384 j Network Theory EXAMPLE 5.16 Determine the voltage v C (t) and the current i C (t) for t for the circuit hown in Fig. 5.. C C Figure 5. We hall analyze thi circuit uing nodal technique. Hence we repreent the capacitor in frequency domain by a parallel circuit ince it i eaier to account for current ource than voltage ource while handling nodal equation. The ymbol for witch indicate that at t it i cloed and at t +, it i open. The circuit at t i hown in Fig. 5.3(a). Let u aume that at t, the circuit i in teady tate. Under teady tate condition, capacitor act a on open circuit a hown in Fig. 5.3(a). i 1 ( ) v C ( ) V Hence; v C () v C ( + )v C ( )4V Fig. 5.3(b) repreent the frequency domain repreentation of the circuit hown in Fig. 5.. KCL at top node: Figure 5.3(a) V C () + 3 V C()+ ) V C () Figure 5.3(b)

57 Laplace Tranform j 385 Invere Laplace tranform yield Alo; v C (t) I C () V C() h6 e 3 t i u(t)v + 3 ) i C (t) 3 e 3 t u(t)a EXAMPLE 5.17 Determine the current i L (t) for t for the circuit hown in Fig Figure 5.4 At t, witch i cloed and at t +, it i open. Let u aume that at t, the circuit i in teady tate. In teady tate, capacitor i open and inductor i hort. The equivalent circuit at t i a hown in Fig. 5.5(a). i L ( ) A v C ( )188V Therefore; i L () i L ( + )i L ( )1A v C () v C ( + )v C ( )8V Figure 5.5(a) For t +, the circuit in frequency domain i a hown in Fig. 5.5(b). We will ue KVL to find i L (t). Hence, we ue erie circuit to repreent both the capacitor and inductor in the frequency domain. Thee erie circuit contain voltage ource rather than current ource. It i eaier to account for voltage ource than current ource when writing meh equation. Thi jutifie the election of erie repreentation for both the capacitor and inductor.

58 386 j Network Theory Applying KVL clockwie to the right meh, we get 8 + I L()+4I L () 4+8I L () ) I L () + ( +1)+1 ) I L () + +5 ( +1) ) I L () ( +1) ( +1) Figure 5.5(b) We know the Laplace tranform pair: L e at + a co bt ( + a) + b and L e at b in bt ( + a) + b Hence; i L (t) e t co t + 1 e t in t u(t)a EXAMPLE 5.18 Find v o (t) of the circuit hown in Fig Figure 5.6 The unit tep function u(t) i defined a follow: 1; t + u(t) ; t Figure 5.7(a)

59 Laplace Tranform j 387 Since the circuit ha two independent ource with u(t) aociated with them, the circuit i not energized for t. Hence the initial current through the inductor i zero. That i, i L ( ). Since current through an inductor cannot change intantaneouly, i L () i L ( + )i L ( ) Alo; v C () v C ( + )v C ( ) The equivalent circuit for t + in frequency domain i a hown in Fig. 5.7(b). Figure 5.7(b) KCL at upernode: V 1 () ) V 1 () ) V 1 () +1 + V () V () + V () + V () The contraint equation: Applying KVL to the path compriing of current ource! voltage ource! inductor, we get, V 1 () V () V () V 1 () 1 + ) V 1 () V () 1 +

60 388 j Network Theory Putting the above two equation in matrix form, we get V 1() V () 1 + Solving for V () and then applying the principle of voltage diviion, we get V o () 1 V ( ) () ( + )(3 +3 +) ) V o () ( + )( +:5 j:646)( +:5+j:646) Uing partial fraction, we can write V o () K K +:5 j:646 + K +:5+j:646 We find that K 1 :5 K :316 / 37:76 Hence; V o () :5 :316 / 37:76 :316 /37: :5 j:646 +:5+j:646 We know that, L a e at u(t) L 1 m / + a j! + m / + a + j! me at co(!t + )u(t) Hence, v o (t).5e t u(t)+.63e :5t co [.646t ] u(t) EXAMPLE 5.19 For the network hown in Fig. 5.8, find v o (t), t>, uing meh equation. Figure 5.8

61 Laplace Tranform j 389 The tep function u(t) i defined a follow. 1; t + u(t) ; t Since the circuit i not energized for t, there are no initial condition in the circuit. For t +, the frequency domain equivalent circuit i hown in Fig. 5.9(b). Figure 5.9(a) Figure 5.9(b) By inpection, we find that I 1 () KVL clockwie for meh : 4 +1[I () I 1 ()] + I ()+1[I () I 3 ()] 4 ) I 1()+I ()[1++1] I 3 () Subtituting the value of I 1 (),weget 4 +4I () I 3 () ) 4I () I 3 () 6 KVL clockwie for meh 3: 1[I 3 () I ()] + I 3 ()+1I 3 () ) I ()+I 3 ()[ +] Putting the KVL equation for meh and meh 3 in matrix form, we get I 6 3 () I 3 ()

62 39 j Network Theory Solving for I 3 (), uing Cramer rule, we get 1:5 I 3 () ) V o ()I 3 () 1 1: Uing partial fraction, we can write V o () K 1 + K We find that, K ; and K Hence; V o () ) v o (t) 6 i h1 e 7 4 t u(t) 7 EXAMPLE 5. Ue meh analyi to find v o (t), t> in the network hown in Fig Figure 5.3 The circuit i not energized for t becaue the independent current ource i aociated with u(t). Thi mean that there are no initial condition in the circuit. The frequency domain circuit for t + i hown in Fig By inpection we find that: I 1 () 4 ; I () I x() I x ()I 3 () 4 ) I () I 3 () 4 ) I () 1 I 3 () 4

63 Laplace Tranform j 391 KVL for meh 3give [I 3 () I ()] + 1 I 3 () 4 +1 I 3 () ) I 3 () 1 I 3 () 4 + I 3 () 4 + I 3 () 4( ) ) I 3 () ( +4) and V o () 1[I 3 ()] 4( ) ( +4) Figure 5.31 By partial fraction, we can write V o () K 1 + K +4 We find that K 1 ; K 6 Hence; V o () 6 +4 Taking invere Laplace tranform, we get v o (t) u(t) 6e 4t u(t) EXAMPLE 5.1 Uing the principle of uperpoition, find v o (t) for t>. Refer the circuit hown in Fig Figure 5.3 Since both the independent ource are aociated with u(t), which i zero for t, the circuit will not have any initial condition. The frequency domain circuit for t + i hown in Fig. 5.33(a).

64 39 j Network Theory Figure 5.33(a) A a firt tep, let u find the contribution to V o () due to voltage ource alone. Thi need the deactivation of the current ource. Referring to Fig. 5.33(b), we find that 4 I() ) V o1 ()I()[1] Next let u find the contribution to the output due to current ource alone. Refer to Fig. 5.33(c). Uing the principle of current diviion, I 1 () ) V o ()1[I 1 ()] + + Finally adding the two contribution, we get Figure 5.33(b) Figure 5.33(c) We find that, Hence; V o ()V o1 ()+V o () K 1 +1 j1 + K1 +1+j1 K 1 p / 45 p p / 45 /+45 V o () +1 j j

65 We know that: L 1 m / + a jb + m / me at co(bt + )u(t) + a + jb Laplace Tranform j 393 Hence; v o (t) p e t co(t +45 )u(t) EXAMPLE 5. (a) Convert the circuit in Fig to an appropriate domain repreentation. (b) Find the Thevein equivalent een by 1Ω reitor. (c) Analyze the implified circuit to find an expreion for i(t). Figure 5.34 (a) Since the independent current ource ha u(t) in it, the circuit i not activated for t.in otherword, all the initial condition are zero. Fig.5.35 (a) how the domain equivalent circuit for t +. Figure 5.35(a) (b) Sine we are intereted in the current in 1Ω uing the Thevenin theorem, remove the 1Ω reitor from the circuit hown in Fig. 5.35(a). The reulting circuit thu obtained i hown in Fig. 5.35(b).

66 394 j Network Theory Z t () i found by deactivating the independent current ource. Z t ()(5+:1)jj :5 : Ω Referring to Fig (b), V t () 3 [Z t()] 7: ( ) Volt The Thevenin equivalent circuit along with 1Ω reitor i hown in Fig (c). I() V t() Z t ()+1 7: ( ) 7: ( )( + 614) Figure 5.35(b) Figure 5.35(c) Uing partial fraction, we get Taking invere Laplace tranform, we get I() :5 + : : i(t).5+.8e 4886t.58e 614t u(t)a Check: and i():5+:8 :58 i(1):5: Thee could be verified by evaluating i(t) at t and t 1 uing the concept explained in Chapter 4. EXAMPLE 5.3 Refer the RLC circuit hown in Fig Find the complete repone for v(t) if t +. Take v()v.

67 Laplace Tranform j 395 Figure 5.36 Since we wih to analyze the circuit given in Fig uing KVL, we hall repreent L and C in frequency domain uing erie circuit to accomodate the initial condition. Accordingly, we get the frequency domain circuit hown in Fig (a). Applying KVL clockwie to the circuit hown in Fig (a), we get I() ) I() ( )( + 16) 9 Hence; V()I() ( +3) ( + 16) Uing partial fraction, we get V () + K1 + K +3 + K 3 ( +3) + K 4 j4 + K 4 + j4 Figure 5.36(a) Solving for K 1, K, K 3, and K 4,weget 88 K 1 ( +3) ( + 16) K d 88 d ( : + 16) 3 K 3 88 ( + 16) 3: K 4 ( +3) ( + j4) :36 / 16: j4