barcode reader using vb net source code Figure 545 Response of underdamped second-order circuit in Software

Creation QR Code JIS X 0510 in Software Figure 545 Response of underdamped second-order circuit

Figure 545 Response of underdamped second-order circuit
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As a nal note, we return to the special situation in Case III when R is identically zero We de ned this earlier as the resonance condition The resonant solution is not characterized by an exponential decay (damping), and gives rise to a pure sinusoidal waveform, oscillating at the natural frequency, Resonant circuits nd application in lters, which are presented in 6 We shall not discuss this case any longer in the present chapter Forced and Complete Response of Second-Order Circuits Once we obtain the natural response using the techniques described in the preceding section, we may nd the forced response using the same method employed for rstorder circuits Once again, we shall limit our analysis to a switched DC forcing
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function, for the sake of simplicity (the form of the forced response when the forcing function is a switched sinusoid is explored in the homework problems) The form of the forced differential equation is a d 2 x(t) dx(t) +b + cx(t) = F dt 2 dt (573)
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where F is a constant Therefore we assume a solution of the form xF (t) = XF = constant, and we substitute in the forced equation to nd that XF = F c (574)
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Finally, in order to compute the complete solution, we sum the natural and forced responses, to obtain x(t) = xN (t) + xF (t) = K1 es1 t + K2 es2 t + F c (575)
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For a second-order differential equation, we need two initial conditions to solve for the constants K1 and K2 These are the values of x(t) at t = 0 and of the derivative of x(t), dx/dt, at t = 0 To complete the solution, we therefore need to solve the two equations x(t = 0) = x0 = K1 + K2 + and dx (t = 0) = x0 = s1 K1 + s2 K2 dt (577) F c (576)
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To summarize, we must follow the steps in the accompanying methodology box to obtain the complete solution of a second-order circuit excited by a switched DC source
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Solution of Second-Order Circuits 1 Write the differential equation for the circuit 2 Find the roots of the characteristic polynomial, and determine the natural response 3 Find the forced response 4 Write the complete solution as the sum of natural and forced responses 5 Determine the initial conditions for inductor currents and capacitor voltages 6 Apply the initial conditions to the complete solution to determine the constants K1 and K2 Although these steps are straightforward, the successful application of this technique will require some practice, especially the determination of the initial conditions and the computation of the constants There is no substitute for practice in gaining familiarity with these techniques! The following examples should be of help in illustrating the methods just described
5
Transient Analysis
EXAMPLE 510 Complete Response of Overdamped Second-Order Circuit
Problem
t=0 + vC(t) _ C
Determine the complete response of the circuit of Figure 546
+ vR(t) _ +
Solution
i(t)
vL(t) _
Known Quantities: Resistor, capacitor, inductor values; source voltage Find: The capacitor voltage as a function of time
R = 5000 L = 1 H C = 1 F VS = 25 V
Vs = 25 V
Schematics, Diagrams, Circuits, and Given Data: R = 5 k ; C = 1 F; L = 1 H;
Assumptions: The capacitor has been charged (through a separate circuit, not shown) prior to the switch closing, such that vC (0) = 5 V Analysis:
1 Apply KVL to determine the circuit differential equation: VS vC (t) vR (t) vL (t) = 0 VS
t
idt iR L
di =0 dt
R di 1 1 dVS d i + + i= =0 t >0 dt 2 L dt LC L dt We note that the above equations that we have chosen the series (inductor) current as the variable in the differential equation; we also observe that the DC forcing function is zero, because the capacitor acts as an open circuit in the steady state, and the current will therefore be zero as t 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 =
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