Question 1 of 8: Pulse Echoes from Two Shunt Taps on an Infinite Line

Question 1: Pulse Echoes from Two Shunt Taps on an Infinite 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.

Find. The amplitude and arrival time of the first three pulses observed at the generator terminals, plotted against time, starting with the outgoing pulse itself.

Approach. Find the launched amplitude from the source divider, treat each shunt resistor as a junction whose reflection coefficient follows from the resistor in parallel with the line that continues beyond it, then time each returning echo by its round trip.

1. Amplitude launched onto the line. The generator sees the line's characteristic impedance as a pure resistance, so the pulse divides between $R_g$ and $Z_0$:$$V^{+} = E\,\frac{Z_0}{R_g + Z_0} = 1000\left(\frac{377}{754}\right) = 500\ \text{V}.$$Exactly half the EMF appears on the line, which is the signature of a matched source.
2. What the first 754 Ω resistor actually does. The tap is not a termination: the line continues past it and, being infinite, presents its own $Z_0$ to the right of the tap. The junction impedance is therefore the resistor in parallel with that continuing line,$$Z_J = \frac{R_1 Z_0}{R_1 + Z_0} = \frac{754(377)}{1131} = 251.33\ \Omega,$$which is well below $Z_0$ even though $R_1 = 2Z_0$.
3. Reflection and transmission at a tap. With that junction impedance,$$\Gamma_J = \frac{Z_J - Z_0}{Z_J + Z_0} = \frac{251.33 - 377}{251.33 + 377} = -0.200,\qquad \tau_J = 1 + \Gamma_J = 0.800.$$Both taps are identical, and the same pair of numbers applies to a wave arriving from either side.
4. The source end takes no part. Because $R_g = Z_0 = 377\ \Omega$, the generator reflection coefficient is$$\Gamma_g = \frac{R_g - Z_0}{R_g + Z_0} = 0.$$Every echo that reaches the generator is absorbed there, so no pulse is ever launched a second time and the arrivals at the terminals are simply the individual echoes, one per scattering path.
5. Transit times. At $3\times10^{8}$ m/s the line covers 300 m per microsecond, so$$T_1 = \frac{10\ \text{km}}{v_p} = 33.33\ \mu\text{s},\qquad T_{12} = \frac{1\ \text{km}}{v_p} = 3.33\ \mu\text{s}.$$The echo from tap 1 needs one round trip, $2T_1 = 66.67\ \mu\text{s}$; the echo from tap 2 needs $2(T_1 + T_{12}) = 73.33\ \mu\text{s}$.
6. Echo from the first tap. The incident 500 V pulse is reflected once:$$V_2 = \Gamma_J V^{+} = (-0.200)(500) = -100\ \text{V}\quad\text{at } 66.67\ \mu\text{s}.$$It is inverted because the shunt resistance drags the junction impedance below $Z_0$.
7. Echo from the second tap. This path crosses tap 1 going out, reflects at tap 2, and crosses tap 1 again coming back, so it carries the transmission factor twice:$$V_3 = \tau_J\,\Gamma_J\,\tau_J\,V^{+} = (0.800)(-0.200)(0.800)(500),$$$$\boxed{V_1 = +500\ \text{V at } 0,\quad V_2 = -100\ \text{V at } 66.67\ \mu\text{s},\quad V_3 = -64\ \text{V at } 73.33\ \mu\text{s}.}$$

The three pulses are 1 μs wide and are separated by tens of microseconds, so they stand cleanly apart on the time axis with no overlap to resolve. Plotting them gives the record an operator would see on a time-domain reflectometer looking into this line: one large outgoing pulse, then two smaller inverted returns whose spacing, 6.67 μs, immediately reveals that the two discontinuities are 1 km apart.

It is worth recording what continues down the line as well, because it shows where the missing energy went: 400 V passes tap 1, 320 V passes tap 2, and that 320 V pulse travels off to infinity and never returns.