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iC (0+ ) = C K1 ( 500) e0 + K2 e0 + K2 ( 500) (0) e0 5 = C [K1 ( 500) + K2 ] K2 = 5 = 25 106 C
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Transient Analysis
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Finally, the complete solution is: v(t) = 25 106 te 500t V A plot of the voltage response of this critically damped circuit is shown in Figure 549
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2 15 V 10 9 1 Critically damped circuit voltage response
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05 0 0 0002 0004 0006 0008 001 0012 0014 0016 0018 Time (s) 002
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Critically damped circuit current response 25 2 A 10 6 15 1 05 0 05 0 0002 0004 0006 0008 001 0012 0014 0016 0018 Time (s) 002
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Figure 549 Focus on Computer-Aided Tools: The Matlab m- le containing the numerical analysis
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and plotting commands for this example may be found in the CD that accompanies this book
EXAMPLE 512 Complete Response of Underdamped Second-Order Circuit
Problem
t=0 L C + 12 V iL(t) R v load _
Determine the complete response of the circuit of Figure 550
Solution
Known Quantities: Source voltage, resistor, capacitor, inductor values Find: The load voltage as a function of time
Schematics, Diagrams, Circuits, and Given Data: R = 10
; C = 10 F; L = 5 mH
Assumptions: No energy is stored in the capacitor and inductor before the switch closes;
ie, vC (0 ) = 0 V; iL (0 ) = 0 A
Analysis: Since the load voltage is given by the expression vload = RiL (t), we shall solve for the inductor current
Part I
Circuits
1 Apply KVL to determine the circuit differential equation: VB vL (t) vC (t) vR (t) = 0 VB L 1 diL dt C
t
iL dt iL R = 0
d 2 iL 1 R diL 1 dVB + iL = =0 t >0 + 2 dt L dt LC L dt 2 We determine the characteristic polynomial by substituting s for d/dt: s2 + 1 R s+ =0 L LC 1 L 2R 2 R L
s1,2 =
4 LC
= 1,000 j 4359 These are complex conjugate roots, therefore we have an underdamped circuit with natural response given by equation 565: iLN (t) = K1 e( 1,000+j 4,359)t + K2 e( 1,000 j 4,359)t 3 The forced response is zero, as stated earlier, because of the behavior of the capacitor as t : F = 0 4 The complete solution is therefore equal to the natural response: iL (t) = iLN (t) = K1 e( 1,000+j 4,359)t + K2 e( 1,000 j 4,359)t 5 The initial conditions for the energy storage elements are vC (0+ ) = 0 V; iL (0+ ) = 0 A 6 To evaluate the coef cients K1 and K2 , we consider the initial conditions iL (0+ ) and diL (0+ )/dt The rst of these is given by iL (0+ ) = 0, as stated above Thus, i(0+ ) = 0 = K1 e0 + K2 e0 K1 + K 2 = 0 K1 = K2 To use the second initial condition, we observe that 1 diL + (0 ) = vL (0+ ) dt L and we note that the inductor voltage can change instantaneously; ie, vL (0 ) = vL (0+ ) To determine vL (0+ ) we need to apply KVL once again at t = 0+ : VS vC (0+ ) vR (0+ ) vL (0+ ) = 0 vR (0+ ) = iL (0+ )R = 0; Therefore vL (0+ ) = VS 0 0 = 12 and vL (0+ ) diL + (0 ) = = 2, 400 dt L Now we can obtain a second equation in K1 and K2 , di + (0 ) = ( 1, 000 + j 4, 359)K1 e0 ( 1, 000 j 4, 359)K2 e0 dt V vC (0+ ) = 0
5
Transient Analysis
and since K1 = K2 2,400 = K1 [( 1,000 + j 4,359) ( 1,000 j 4,359)] K1 = 2,400 = j 02753 j 8,718
K2 = K1 = j 02753 Note that K1 and K2 are complex conjugates Finally, the complete solution is: vLoad (t) = RiL (t) = 10 j 02753e( 1,000+j 4,359)t + j 02753e( 1,000 j 4,359)t = 2753e 1,000t j ej 4,359t + j e j 4,359t = 5506e 1,000t sin(4,359t) V The output voltage of the circuit is plotted in Figure 551
4 3 2 Volts 1 0 1 2 0 0002 0004 0006 Time (s) 0008 001
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