Question 6.14: A battery with zero internal impedance has an open circuit v...
A battery with zero internal impedance has an open circuit voltage of 100 volts. At a time t = 0, this battery is switched into a 50 Ω air-dielectric coaxial cable via a 150 Ω resistor. The cable is 300 m long and is terminated in a load of 33.3Ω.
(a) Sketch a bounce diagram for the first 4 µs after the switch is closed.
(b) Draw a graph of the voltage that appears across the load impedance as a function of time.
(c) Find the asymptotic values of VL and IL t \rightarrow \infty .
(a) The two reflection coefficients and the incident voltage step that propagates on the line are given by
\begin{gathered} \mathcal{R}_{B}=\frac{Z_{B}-Z_{C}}{Z_{B}+Z_{C}}=\frac{150-50}{150+50}=\frac{1}{2} \\\\ \mathcal{R}_{L}=\frac{Z_{L}-Z_{C}}{Z_{L}+Z_{C}}=\frac{33.3-50}{33.3+50}=-\frac{1}{5} \end{gathered}
V_{1}=\frac{Z_{C}}{Z_{B}+Z_{C}} V_{B}=\frac{50}{150+50} 100=25 V
In the (b) Since the coaxial cable is filled with air, the velocity of propagation equals 3 \times 10^{8} m / s . The voltage signal takes 1 µs s to travel from one end to the other.
(c) The asymptotic voltages and currents as computed from (6.59) and (6.60) are for V _{ t \rightarrow \infty}=18.2 V and I _{ t \rightarrow \infty}=0.55 A.
V =\frac{Z_{ L }}{Z_{B}+Z_{L}} V_{B} (6.59)
I =\frac{ V }{Z_{ L }}=\frac{1}{Z_{B}+Z_{ L }} V _{ B } (6.60)
