Question 1 of 8: Step Response of a Transmission Line Feeding a Resistive Load

Question 1: Step Response of a Transmission Line Feeding a Resistive Load (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Step EMF $V_0=12$ V, generator (source) resistance $R_g=50\ \Omega$; line length $l=10$ km, characteristic impedance $Z_0=50\ \Omega$, propagation velocity $v=2\times10^8$ m/s; the load is two semi-infinite lines of the same $Z_0=50\ \Omega$ in parallel, so it presents a pure resistance with no reflections of its own.

Find. The generator terminal current $i_g(t)$ for $0\le t\le150\ \mu\text{s}$.

Approach. This is a Bergeron (bounce-diagram) transmission-line transient problem: find the load's effective resistance, the round-trip delay, the reflection coefficients at load and source, and superpose the launched and returning waves at the generator end.

1. Reduce the load to a single resistance. Two identical infinitely long $50\ \Omega$ lines in parallel present a pure resistance (an infinite line never returns a reflection), so $$Z_L=\frac{Z_0\cdot Z_0}{Z_0+Z_0}=\frac{50\times50}{100}=25\ \Omega .$$
2. One-way and round-trip delay. $$T=\frac{l}{v}=\frac{10\times10^3}{2\times10^8}=50\ \mu\text{s},\qquad 2T=100\ \mu\text{s}.$$ Both fall inside the requested 0–150 μs window.
3. Reflection coefficients. At the load, $$\Gamma_L=\frac{Z_L-Z_0}{Z_L+Z_0}=\frac{25-50}{25+50}=-\frac{1}{3}.$$ At the generator, because $R_g=Z_0=50\ \Omega$ the source is matched to the line, so $$\Gamma_g=\frac{R_g-Z_0}{R_g+Z_0}=0 .$$ A matched source absorbs any wave arriving from the load completely and launches no further wave — this is a single-bounce problem.
4. Launched wave at $t=0$. The step EMF divides between $R_g$ and the line's input impedance $Z_0$ (a semi-infinite line looks purely resistive to the source before any reflection returns): $$V_1^{+}=V_0\,\frac{Z_0}{R_g+Z_0}=12\times\frac{50}{100}=6\ \text{V},\qquad I_1^{+}=\frac{V_1^{+}}{Z_0}=\frac{6}{50}=0.12\ \text{A}.$$ This current flows at the generator terminals for $0\le t<2T$.
5. Wave reflected at the load and its return. The reflected voltage wave launched from the load back toward the source is $$V_1^{-}=\Gamma_L\,V_1^{+}=\left(-\tfrac13\right)(6)=-2\ \text{V}.$$ It arrives back at the generator terminals at $t=2T=100\ \mu\text{s}$. A backward-travelling wave's current contribution is $-V^-/Z_0$ (current and voltage waves have opposite-sign reflection coefficients), so it adds $$\Delta I=-\frac{V_1^{-}}{Z_0}=-\frac{-2}{50}=0.04\ \text{A}$$ to the terminal current. Because $\Gamma_g=0$, this wave is fully absorbed at the source and nothing further is launched, so the terminal current is constant from $t=2T$ onward.
6. Terminal current after the bounce. $$I_g(t\ge2T)=I_1^{+}+\Delta I=0.12+0.04=\boxed{0.16\ \text{A}}.$$ Consistency check — a matched source that has fully absorbed all reflections is electrically indistinguishable from a plain series DC circuit, so the final current must equal the simple steady-state value: $$I_{ss}=\frac{V_0}{R_g+Z_L}=\frac{12}{50+25}=0.16\ \text{A}\ \checkmark$$