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= 2,500 s1 = 2087;
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(5,0002 4 106
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s2 = 4,7913
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These are real, distinct roots, therefore we have an overdamped circuit with natural response given by equation 563: iN (t) = K1 e 2087t + K2 e 4,7913t 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: i(t) = iN (t) = K1 e 2087t + K2 e 4,7913t 5 The initial conditions for the energy storage elements are vC (0+ ) = 5 V; iL (0+ ) = 0 A
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Circuits
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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 diL + 1 di + (0 ) = (0 ) = vL (0+ ) dt 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+ ) = i(0+ )R = 0 vC (0+ ) = 5 Therefore vL (0+ ) = VS vC (0+ ) vR (0+ ) = 25 5 0 = 20 V and we conclude that 1 di + (0 ) = vL (0+ ) = 20 dt L Now we can obtain a second equation in K1 and K2 , di + (0 ) = 20 = 2087K1 e0 47913K2 e0 dt and since K1 = K2 20 = 2087K2 47913K2 K1 = 436 103 K2 = 436 10 3 Finally, the complete solution is: i(t) = 436 10 3 e 2087t 436 10 3 e 4,7913t A To compute the desired quantity, that is, vC (t), we can now simply integrate the result above, remembering that the capacitor initial voltage was equal to 5 V: vC (t) = 1 C
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i(t)dt + vC (0)
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i(t)dt = 106
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436 10 3 e 2087t 436 10 3 e 4,7913t dt
106 436 10 3 2087t e 1 ( 2087) 106 436 10 3 4,7913t e 1 ( 4,7913) t >0
2087t 2087t 4,7913t
= 209e 2087t + 209 + 09e 4,7913t 09 = 20 209e vC (t) = 25 209e + 09e + 09e 4,7913t V
The capacitor voltage is plotted in Figure 547
5
Transient Analysis
30 25 20 Volts 15 10 5 0 0 0005 001 0015 002 Time (s) 0025 003
Figure 547 Overdamped circuit capacitor voltage response
Focus on Computer-Aided Tools: The Matlab m- le containing the numerical analysis and plotting commands for this example may be found in the CD that accompanies this book An EWB circuit simulation is also included in the CD
EXAMPLE 511 Complete Response of Critically Damped Second-Order Circuit
Problem
Determine the complete response of v(t) in the circuit of Figure 548
t=0 + v(t) _ C iC(t) R iR(t) L iL(t) IS
Solution
Known Quantities: Resistor, capacitor, inductor values
L=2H R = 500
C = 2 F IS = 5 A
Find: The capacitor voltage as a function of time
L = 2 H t = 0+
Schematics, Diagrams, Circuits, and Given Data: IS = 5 A; R = 500
; C = 2 F;
Assumptions: The capacitor voltage and inductor current are equal to zero at
Analysis:
1 Apply KCL to determine the circuit differential equation: IS iL (t) iR (t) iC (t) = 0 IS
t
v(t)dt
dv(t) v(t) C =0 R dt
d v 1 1 dIS 1 dv + v= =0 t >0 + dt 2 RC dt LC C dt We note that the DC forcing function is zero, because the inductor acts as a short circuit in the steady state, and the voltage across the inductor (and therefore across the parallel circuit) will be zero as t
Part I
Circuits
2 We determine the characteristic polynomial by substituting s for d/dt: s2 + 1 1 s+ =0 RC LC 1 1 2RC 2 1 RC
s1,2 =
4 LC
= 500 s1 = 500
1 (1, 000)2 106 2 s2 = 500
These are real, repeated roots, therefore we have a critically damped circuit with natural response given by equation 564: vN (t) = K1 e 500t + K2 te 500t 3 The forced response is zero, as stated earlier, because of the behavior of the inductor as t : F = 0 4 The complete solution is therefore equal to the natural response: v(t) = vN (t) = K1 e 500t + K2 te 500t 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 vC (0+ ) and dvC (0+ )/dt The rst of these is given by vC (0+ ) = 0, as stated above Thus, v(0+ ) = 0 = K1 e0 + K2 0e0 K1 = 0 To use the second initial condition, we observe that 1 dv + dvC (0+ ) (0 ) = = iC (0+ ) dt dt C and we note that the capacitor current can change instantaneously; ie, iC (0 ) = iC (0+ ) To determine iC (0+ ) we need to apply KCL once again at t = 0+ : IS iL (0+ ) iR (0+ ) iC (0+ ) = 0 iL (0+ ) = 0; Therefore iC (0+ ) = IS 0 0 0 = 5 A and we conclude that 1 dv + (0 ) = iC (0+ ) = 5 dt C Now we can obtain a second equation in K1 and K2 , iC (t) = C dv = C K1 ( 500) e 500t + K2 e 500t + K2 ( 500) te 500t dt iR (0+ ) =
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