Question 8.8: The input voltage is the unit-step function, u(t).Assume tha...

The excitation and the impedance, in the Laplace transform domain, are
V(s)=\frac{1}{s} and Z(s) = R +L_{s}
Therefore,
I_{L}(s)=\frac{V(s)}{Z(s)}=\frac{1}{s(R + Ls)}=\frac{1}{Ls(s + R/L)} =\frac{1}{R}\left(\frac{1}{s}-\frac{1}{(s + R/L)} \right)
Applying initial and final value theorems, we get
i(0^{+} )=\underset{s\rightarrow \infty }{\lim } s\frac{1}{s(R + Ls)}=0

\underset{t\rightarrow \infty }{\lim }i(t)=\underset{s\rightarrow 0 }{\lim } s\frac{1}{s(R + Ls)}=\frac{1}{R}
For the example, R = 2 and\underset{t\rightarrow \infty }{\lim }i(t)=0.5These values can be used to check the complete response derived. Taking the inverse Laplace transform, we get
i_{L} (t) = \frac{1}{R} (1 −e^{-\frac{R}{L}t })u(t).
WithZ_{L}(s)=Ls,
V_{L}(s)=I_{L}(s)Z_{L}(s)=\frac{1}{(s + R/L)} and v_{L}(t)=e^{-\frac{R}{L}t }u(t).
The unit-step responses are
v_{L}(t)=e^{-\frac{R}{L}t }u(t) and i_{L} (t) = \frac{1}{R} (1 −e^{-\frac{R}{L}t })u(t).
Since the impulse is the derivative of the unit-step function, as the circuit is linear, the impulse response is the derivative of the response to the unit-step excitation. The derivative of these expressions, the unit-impulse responses are
v_{L}(t)=-\frac{R}{L}e^{-\frac{R}{L}t }u(t) and i_{L}(t) =\frac{1}{L}e^{-\frac{R}{L}t }u(t).
Figure 8.12a, b shows, respectively, the voltage across the inductor and the current through for the unit-step source voltage for the example. The voltage is 1 at t = 0 and asymptotically approaches zero as t → ∞. For larger values of inductance, the rate of fall is low. The current is 0 at t = 0 and reaches the limit 1/2 = 0.5. The inductor influences the nature of the response only during the initial period, called the transient interval. In steady state, the impedance of the inductor is zero, as frequency of excitation is zero. That is, the excitation is constant. Figure 8.13a, b shows, respectively, the voltage across the inductor and the current through for the unit-impulse source voltage. These responses are the derivative of the responses for the unit-step excitation. The voltage across the resistor and that induced in the inductor are of the same magnitude with opposite polarities, since the excitation is zero for t > 0 and KVL has to be satisfied.With the source connected, the inductor stores energy 0.5Li²(t)and the resistor dissipates energy at the rate off Ri²(t).When the source disconnected, the stored energy is eventually dissipated by the resistor. The rate of energy dissipation, which is controlled by the ratio L/R, decreases with the time and finally both the energy and the current become zero. The larger the value of L/R, the smaller will be the rate of decay. The value of the exponential signal e^{-kt} u(t) is 1 at t = 0. At t = 1/k, the value gets reduced to e^{-1} = 1/e = 0.3679. At t = 2/k, the value gets reduced to e^{-2}=e^{-1}e^{-1}=0.1353The value 1/k is called time constant of the corresponding circuit. The time constant is useful for comparisof different responses. The time constant for the RL circuit is L/R. The transient response is zero, only when t → ∞. For practical purposes, the duration of four time constants is considered as the duration of the transient response, as the response increases to 98% of the final value for a growing exponential and the response reduces to below 2% for a decaying exponential.AC or DC and transient or steady state, the constraint is that KVL and KCL are to be satisfied at any part of a linear circuit at any instant. The usual difference of impedances replacing resistors has to be taken into account. The DC excitation is replaced by an AC excitation.