Pulse reflections from two matched shunt taps on an infinite line: 22-Elec-A7 Electromagnetics, May 2017, Q1

Question 1: Pulse reflections from two matched 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.

Given. A matched pulse generator feeds a semi-infinite line that carries two identical shunt resistors near its far end, and launches a single pulse of known energy.

Given data
Quantity	Symbol	Value
Generator internal impedance	$R_g$	$377\ \Omega$
Line characteristic impedance	$Z_0$	$377\ \Omega$
Propagation velocity	$v_p$	$3\times 10^{8}$ m/s
Distance generator to first tap	$d_1$	$\approx 10$ km
Tap spacing	$\Delta d$	$500$ m
Shunt resistors (two)	$R_1 = R_2$	$377\ \Omega$
Energy in the launched pulse	$W^{+}$	$1$ J

Find. (I) the largest pulse duration for which the two echoes arrive at the generator without overlapping, and (II) the power level and arrival time of each of those first two echoes.

Figure 1.1 — the 377 Ω line continues past both taps, so each tap resistor works in parallel with the line's own 377 Ω.

Approach. Find the reflection coefficient a tap presents (it is not a match), note that the generator is matched so nothing re-reflects there, then let the round-trip delays set the timing and the two-way transmission factors set the amplitudes.

Impedance seen at a tap. A length of lossless line that has not yet heard from its termination always presents $Z_0$, and the line continues beyond each tap for ever, so the resistor is in parallel with $Z_0$: $$Z_J = \frac{R\,Z_0}{R + Z_0} = \frac{377 \times 377}{754} = 188.5\ \Omega .$$
Reflection and transmission at a tap. With $Z_J = Z_0/2$, $$\Gamma_J = \frac{Z_J - Z_0}{Z_J + Z_0} = \frac{188.5 - 377}{188.5 + 377} = \boxed{-\tfrac{1}{3}} , \qquad \tau_J = 1 + \Gamma_J = \tfrac{2}{3}.$$ A 377 Ω resistor bridged across a 377 Ω line is therefore not a matched termination; two thirds of the incident wave continues past it and one third comes back inverted.
The generator end absorbs everything. Because $R_g = Z_0$, $\Gamma_g = (Z_0-Z_0)/(Z_0+Z_0) = 0$: every echo that reaches the generator is absorbed there and is never re-launched. Each echo therefore appears once, and the observed train is short.
Round-trip delays. The echo from a discontinuity at distance $d$ returns after $t = 2d/v_p$: $$t_1 = \frac{2 \times 10\,000}{3\times 10^{8}} = 66.67\ \mu\text{s}, \qquad t_2 = \frac{2 \times 10\,500}{3\times 10^{8}} = 70.00\ \mu\text{s}.$$
(I) Non-overlap condition. The two echoes are separated at the generator by the extra round trip over the 500 m gap, that is by 1000 m of travel: $$\Delta t = t_2 - t_1 = \frac{2\,\Delta d}{v_p} = \frac{1000}{3\times 10^{8}} = 3.333\ \mu\text{s},$$ so the pulses stay clear of one another provided the pulse duration $T$ satisfies $T \le \Delta t$, giving $$\boxed{T_{\max} = \frac{2\,\Delta d}{v_p} = 3.33\ \mu\text{s}}$$ (equivalently a pulse 1 km long on the line). Note that the 10 km distance sets only when the echoes arrive, not how closely they crowd — which is why the question can say “about 10 km”.
Pulse power and amplitude. Packing the 1 J into that longest permitted width gives the incident power and, from $P = |V^{+}|^{2}/Z_0$, the wave amplitude: $$P^{+} = \frac{W^{+}}{T_{\max}} = \frac{1}{3.333\times 10^{-6}} = 300\ \text{kW}, \qquad V^{+} = \sqrt{P^{+} Z_0} = \sqrt{300\,000 \times 377} = 10.63\ \text{kV}.$$
(II) First echo. The pulse simply reflects off tap 1: $$P_{r1} = \Gamma_J^{2}\,P^{+} = \tfrac{1}{9}\times 300 = \boxed{33.3\ \text{kW}} \quad\text{arriving at } 66.67\ \mu\text{s},$$ carrying $W_{r1} = \tfrac{1}{9} = 0.111$ J.
(II) Second echo. This wave must pass tap 1 going out, reflect from tap 2, and pass tap 1 coming back, so its amplitude factor is the product $\tau_J \Gamma_J \tau_J$: $$\frac{V_{r2}}{V^{+}} = \tfrac{2}{3}\left(-\tfrac{1}{3}\right)\tfrac{2}{3} = -\tfrac{4}{27} = -0.1481, \qquad P_{r2} = \left(\tfrac{4}{27}\right)^{2} P^{+} = \boxed{6.58\ \text{kW}}$$ arriving at $70.00\ \mu\text{s}$ and carrying $0.0220$ J. Using $V^{+}$ instead of the transmitted wave here is the classic way to overstate this echo by more than a factor of two.
The plot. Both echoes are rectangular, $3.33\ \mu\text{s}$ wide, with the second one exactly abutting the first. The next echo (two extra bounces between the taps, factor $\tau_J\Gamma_J^{3}\tau_J = -4/243$, i.e. $0.0813$ kW) arrives at $73.33\ \mu\text{s}$ and is about $81$ times weaker than the second echo ($410$ times weaker than the first), so the trace is effectively the two pulses drawn below.

Figure 1.2 — powers of the first two reflected pulses at the generator terminals; each is 3.33 μs wide, and the generator is matched so neither is re-launched.

Final results
Quantity asked	Result
Reflection coefficient at each tap	$\Gamma_J = -1/3$ (with $Z_J = 188.5\ \Omega$)
(I) Longest non-overlapping pulse width	$T_{\max} = 3.33\ \mu$s (1 km of line)
Incident pulse power at that width	$300$ kW ($V^{+} = 10.63$ kV)
First echo	$33.3$ kW for $3.33\ \mu$s, arriving at $66.67\ \mu$s ($0.111$ J)
Second echo	$6.58$ kW for $3.33\ \mu$s, arriving at $70.00\ \mu$s ($0.0220$ J)
Third echo (for scale)	$0.081$ kW at $73.33\ \mu$s

Topic. Transient reflection from a shunt discontinuity on a matched-source line.

Key relations. $\Gamma = (Z-Z_0)/(Z+Z_0)$ and $\tau = 1+\Gamma$ at any junction; a shunt $R$ across a continuing line gives $Z_J = R\,\|\,Z_0$; round-trip delay $t = 2d/v_p$; pulse power $P = |V|^{2}/Z_0$ and energy $W = PT$.

Why this works. On a lossless line the wave meets each discontinuity with no knowledge of what lies beyond the next one, so amplitudes simply multiply along the path a given echo takes. Because the source is matched, the lattice (bounce) diagram truncates: every returning wave dies at the generator, so the echo train is just one term per distinct path through the taps. The energy bookkeeping closes too — the 1 J splits into $1/9$ J returned by tap 1, $16/729$ J returned by tap 2, and the remainder carried away down the infinite line.

Common pitfall. Believing that $R = Z_0$ tapped across a line is a match: it is not, because the line itself continues and contributes a second $Z_0$ in parallel. The other frequent slip is timing the echoes with the one-way delay $d/v_p$ instead of the round trip $2d/v_p$.

Source. Pozar, Microwave Engineering 4th ed. §2.1–2.3; Ulaby, Fundamentals of Applied Electromagnetics ch. 2 (transient response, bounce diagrams).

Question 1 of 8: Pulse reflections from two matched shunt taps on an infinite line

Question 1: Pulse reflections from two matched shunt taps on an infinite line (20 marks)