110_1_ee110_mtr_ex

110_1_ee110_mtr_ex - 514 The Laplace Transform in Circuit...

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Unformatted text preview: 514 The Laplace Transform in Circuit Analysis figure 13.13 The step response of a parallet RLC circuit. Figure £3.14 The s~domain equivaient circuit for the circnit shown in Fig. 13.13. "the Step Response of a Parallel. Eircuit Next we analyze the parallei RLC circuit, shown in Fig. 13.13, that we first 2 analyzed in Example 8,7. The problem is to find the expression for iL alter the constant current source is switched across the parallel elements. The ' initiai energy stored in the circuit is zero. As before, we begin by constructing the odomain equivalent circuit ? shown in Fig. 13.14. Note how easiiy an independent source can be trans '4 formed from the time domain to the frequency domain. We transform the source to the 3 domain simply by determining the Laplace transform of its time-domain function. Here, opening the switch results in a step change in the current applied to the circuit. Therefore the s-domain current source is 33“ 6011(1)}, or [ck/s. To find I L, we first soive for V and then use e 1 (1319) 5L ' ' IL to establish the s-doinain expression for IL. Summing the currents away from the top node generates the expression V V [dc CV + + we 2 W. 13.20 5 R sL s ( ) Solving Eq. 13.20 for V gives 1 C C V : —?—ww~——L/—————. (13.21) S" in (1/RC)s + (1/LC) Substitutingfiq. 13.21 into Eq. 13.19 gives 1 LC [L m ----- (13.22) stsz + (1./RC)s + (1/LC)] Substituting the numerical vaiues of R, L, C, and [dc into Eq. 13.22 yield ' 384 x 105 5(52 + 64,000: + 16 x 108)' IL — Before expanding Eq. 13.23 into a sum of partiai fractions, we facto quadratic term in the denominator: i 384 x 105 " s(s + 32,000 — j24,000)(s + 32,000 + j24,000)‘ IL (13. Now, we can test the s-domain expression for IL by checking to 3 Whether the final~vaiue theorem predicts the correct value for Q t = 00. All the poles of IL, except for the first-order poie at the origin, in the left half of the s piane, so the theorem is applicable. We know fro the behavior of the circuit that after the switch has been open for a to time, the inductor will short-circuit the current sonrce. Therefore, the fin value of L; must be 24 inA. The limit of 51L as s we 0 is ' 384 X 105 liars]; : w 2 24 mA. {13.2 H) “it x 108 (Currents in the 5 domain carry the dimension of ampere—seconds, the dimension 01” ML will be amperes.) Thus our s—domain expressio checks out. «Mn24mmmmwe«aswwawmvwwmm‘wakfimmwmwwwwmmm' ‘ 7;; 133 Applications 515 . We now proceed with the partial fraction. expansion of Eq. 13.24: K1 K2 I — +- m ~ L s s + 32,000 M j24,000 4: + S + 32,000 + 124,000 e partial fraction coefficients are __ 2.619% 105 “K I 1 16X :08 m 24 x 30”“3, 384 x 105 (“32,000 + j24,00{})(j48,000) : 20 x 10 ““ 35126.87". 40 cos (24,000t + 126.87”) E —24 cos 24,000? — 32 sin 24,000t. f we weren’t using a previous solution as a check, We would test 313.29 to make sure that iL(0) satisfied the given initiai conditions and 00) satisfied the known behavior of the circuit. _"bj'e'ctiiiie' 2——.—Know how-to analyze a circuit in the s domaih and-be able totransfur'mans, domain saluti'on to 'the‘ - madam-ai'n' . - - - . . . .- . ‘3 4 eoergy storedin the circuit shown isfzero I Answefi (a) I : 4fiifsz+ 1.235: 5 j ' at the time whe'nithe switch'is closed.- ' = _. - I (b) I. : (soégo'az Sin 0_8i:)u(z3): a) Findithe s-domain expression for I. (C) V a EGGS/(32+ 1125* 31).? ~. . ' (div '2 [20050-63 cas--(0.8i.’+=;3637mm) 3V, .- b)_.F_in_d_.the time~domainexpression.foriwhen _ . . _ _ _ __-- L‘_ 0.3.: 4.80" - -‘4_H..' I Is?) i the s-domain expression for V. ' '_d)*F1nd the timedomain expression for *0 when __ - . ' 30TH." (E‘hapterfroblems .1315. and 13.16.- _ I ...
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110_1_ee110_mtr_ex - 514 The Laplace Transform in Circuit...

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