Question 9.19: The switch in Figure 9–12 has been open for a long time. At ......
The switch in Figure 9–12 has been open for a long time. At t = 0 the switch is closed. Find i(t) for t ≥ 0.
The governing equation for the second-order circuit in Figure 9–12 is found by combining the element equations and a KVL equation around the loop with the switch closed:
KVL: υ_{R}(t) + υ_{L}(t) + υ_{C}(t) = 0
Resistor: υ_{R}(t) = Ri(t)
Inductor: υ_{L}(t) = L\frac{di(t)}{dt}
Capacitor: υ_{C}(t) = \frac{1}{C} \int_{0}^{t}{i(τ)dτ + υ_{C}(0)}
Substituting the element equations into the KVL equation yields
L\frac{di(t)}{dt} + Ri(t) + \frac{1}{C} \int_{0}^{t}{i(τ)dτ + υ_{C}(0)} = 0
Using the linearity property, the differentiation property, and the integration property, we transform this second-order integrodifferential equation into the s domain as
L[sI(s) − i_{L}(0)] + RI(s) + \frac{1 I(s)}{C s} + υ_{C}(0) \frac{1}{s} = 0
Solving for I(s) results in
I(s) = \frac{si_{L}(0) − υ_{C}(0)/L}{s^{2} + \frac{R}{L} s + \frac{1}{LC}} A-s
Prior to t = 0, the circuit was in a dc steady-state condition with the switch open. In dc steady state, the inductor acts like a short circuit and the capacitor like an open circuit, so the initial conditions are iL(0−) = 0 A and υ_{C}(0−) = V_{A} = 10 V. Inserting the initial conditions and the numerical values of the circuit parameters into the equation for I(s) gives
I(s) = -\frac{10}{s^{2} + 400s + 2 × 10^{5}}A-s
The denominator quadratic can be factored as (s + 200)² + 400² and I(s) written in the following form:
I(s) = – \frac{10}{400}\left[\frac{400}{(s + 200)^{2} + (400)^{2}}\right]A-s
Comparing the quantity inside the brackets with the entries in the F(s) column of Table 9–2, we find that I(s) is a damped sine with α = 200 and β = 400. By linearity, the quantity outside the brackets is the amplitude of the damped sine. The inverse transform is
i(t) = [−0.025e^{−200t} \sin 400t] u(t) A
Substituting this result back into the circuit integrodifferential equation yields the following term-by-term tabulation:
υ_{L}(t) = L\frac{di(t)}{dt} = +5e^{−200t} \sin 400t − 10e^{−200t} \cos 400t V
υ_{R}(t) = Ri(t) = -10e^{−200t} \sin 400t V
υ_{C}(t) = \frac{1}{C} \int_{0}^{t}{i(τ)dτ = + 5e^{−200t} \sin 400t + 10e^{−200t} \cos 400t − 10 V}
υ_{C}(0) = + 10 V
The sum of the far right-hand sides of these equations is zero. This result shows that the waveform i(t) found using Laplace transforms does indeed satisfy the circuit integrodifferential equation and the initial conditions.
| T A B L E 9–2 BASIC LAPLACE TRANSFORM PAIRS | ||
| S_{IGNAL} | W_{AVEFORM} f(t) | T_{RANSFORM} F (s) |
| Impulse | δ(t) | 1 |
| Step function | u(t) | \frac{1}{s} |
| Ramp | tu(t) | \frac{1}{s^{2}} |
| Exponential | [e^{- αt}] u(t) | \frac{1}{s + α} |
| Damped ramp | [te^{- αt}] u(t) | \frac{1}{(s + α)^{2}} |
| Sine | [\sin βt] u(t) | \frac{β}{s^{2} + β^{2}} |
| Cosine | [\cos βt] u(t) | \frac{s}{s^{2} + β^{2}} |
| Damped sine | [e^{−αt}\sin βt] u(t) | \frac{β}{(s + α)^{2} + β^{2}} |
| Damped cosine | [e^{−αt}\cos βt] u(t) | \frac{(s + α)}{(s + α)^{2} + β^{2}} |