Question 8.9: The input voltage is v(t) = sin(t)u(t). Assume that the seri...
The input voltage is v(t) = sin(t)u(t). Assume that the series circuit consists of only the resistor with value R = 2Ω and the inductor with value L = 3 H, shown in Fig. 8.11b. Let the initial current through the inductor be zero. Find the current through the circuit and the voltage across the inductor after the excitation is applied.
The excitation and the impedance, in the Laplace transform domain, are
V(s)=\frac{1}{s^{2}+1 } and Z(s) = R +Ls
Therefore,
I_{L} (s) =\frac{V (s)}{Z(s)} =\frac{1}{(s^{2} + 1)(R + Ls) }=\frac{1}{L(s^{2} + 1)(s + R/L)}
=\frac{1}{3}\left(\frac{−9(0.5 + j/3)/13}{s − j}+\frac{−9(0.5 − j/3)/13}{s + j}+\frac{(0.6923)}{(s + 2/3)} \right)
=\frac{1}{3}\left(\frac{−0.3462 − j0.2308}{s − j}+\frac{−0.3462 + j0.2308}{s + j}+\frac{(0.6923)}{(s + 2/3)} \right)
For this example, applying the initial value theorem, we get i(0^{+} )=0. The final value theorem is not applicable, as the response is oscillatory. Taking the inverse Laplace transform, we get
i_{L}(t)=(0.8320cos(t −2.5536)+0.6923e^{-\frac{2}{3}t } )/3u(t) = (0.2773 cos(t −2.5536)+0.2308e^{-\frac{2}{3} t} )u(t)
Since v_{L} (t) = L(di_{L}(t)/dt), by differentiating, we get
v_{L} (t) = (0.8320 sin(t − 2.5536) − 0.4615e^{-\frac{2}{3}t })u(t) = (0.8320 cos(t − 0.9828) − 0.4615e^{-\frac{2}{3}t })u(t)
component while the exponential component is the transient. The magnitude of the steady-state current component is
\left|\frac{-j}{2+j3} \right|=0.2774.
The transient response eventually dies down. For cos(t)u(t) input, since the cosine function is the derivative of the sine function, we have to differentiate the expressions.