Question 3 of 8: Pulse Echoes from Two Shunt Resistors on a Long Line

Question 3: Pulse Echoes from Two Shunt Resistors on a Long Line (20 marks)

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. A matched generator launches a short pulse down a semi-infinite line that carries two identical shunt resistors near its far end.

Find. (i) the energy in each of the first two pulses that come back to the generator terminals, and (ii) the time between their arrivals.

Approach. Treat each shunt resistor as a junction whose impedance is $R$ in parallel with the continuing line, obtain one voltage reflection and one transmission coefficient, multiply the coefficients along each return path, and square to convert voltage ratios into energy ratios.

1. Size the pulse against the geometry. The pulse occupies $v_p\tau = (3\times 10^{8})(1\times 10^{-6}) = 300\ \text{m}$ of line — exactly the resistor spacing. The 10 GHz carrier only guarantees that the 1 µs burst contains $10^{4}$ cycles, so the burst behaves as a well-defined wave packet and ordinary steady-state coefficients apply to it.
2. Reduce the first junction. Looking to the right at $R_1$, the wave sees the resistor in parallel with the continuing infinite line: $$Z_{\text{eq}} = \frac{R Z_0}{R + Z_0} = \frac{377 \times 377}{754} = 188.5\ \Omega .$$ Hence $$\boxed{\ \Gamma = \frac{Z_{\text{eq}} - Z_0}{Z_{\text{eq}} + Z_0} = -\frac{1}{3}, \qquad \tau_v = 1 + \Gamma = \frac{2}{3}\ }$$ An energy audit confirms the split: $\Gamma^2 = 1/9$ returns, $\tau_v^2 = 4/9$ continues down the line and $4/9$ is burned in the resistor, summing to unity. By symmetry the second junction has the same coefficients, and so does the first junction when a wave arrives at it from the far side.
3. First return pulse. The earliest echo is the direct reflection at $R_1$: $$E_1 = \Gamma^2 E_0 = \frac{1}{9}(1000\ \text{J}) = 111.1\ \text{J} \quad (11.1\% \text{ of the launched energy}).$$
4. Second return pulse. The next echo crosses $R_1$, reflects from $R_2$ and crosses $R_1$ again, so its voltage carries three factors: $$\frac{V_2}{V_{\text{inc}}} = \tau_v\,\Gamma\,\tau_v = \left(\frac{2}{3}\right)\left(-\frac{1}{3}\right) \left(\frac{2}{3}\right) = -\frac{4}{27},$$ $$E_2 = \left(\frac{4}{27}\right)^{2} E_0 = \frac{16}{729}(1000\ \text{J}) = 21.95\ \text{J} \quad (2.19\%).$$ The first echo is $81/16 = 5.06$ times the energy of the second.
5. Time interval between the arrivals. The second echo travels the extra 300 m twice: $$\Delta t = \frac{2s}{v_p} = \frac{600}{3\times 10^{8}} = 2.00\ \mu\text{s}.$$ Absolute arrival times follow from the 10 km run: $t_1 = 2L/v_p = 66.7\ \mu\text{s}$ and $t_2 = 68.7\ \mu\text{s}$.
6. Confirm the echoes are separable, and that they stop there. Each echo is itself 1 µs long while the arrivals are 2 µs apart, so the two pulses are cleanly resolved with a 1 µs gap. Because $Z_g = Z_0$ the generator absorbs both without re-reflecting them, so no further echoes are generated at the source end; later, weaker echoes (energy $16/729 \times 1/9$ and below) arrive at 2 µs intervals from the multiple bounces trapped between the two resistors.

The physical picture is worth stating plainly. Neither resistor is a mismatch in the sense of a bad termination — each is a perfectly good 377 Ω element — yet each still reflects, because bridging a resistor across a line that continues beyond it halves the impedance seen at that plane. That is why a tap or a monitoring bridge on a long feeder is never “invisible”, and why time-domain reflectometry can locate such taps from the size and timing of their echoes.