Question 1 of 8: Pulse echoes from two shunt taps on an endless line
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2018, 16-Elec-A7 Electromagnetics. Three hours, closed book, one approved Casio or Sharp calculator. Eight questions, all of equal value; the rubric says any five constitute a complete paper, so a candidate answers five — but all eight are worked here, because the set is a study resource. Aids printed on the paper: $\varepsilon_0 = 8.85\times10^{-12}$ F/m, $\mu_0 = 4\pi\times10^{-7}$ H/m, $e = 1.6\times10^{-19}$ C.
Reference texts. M. N. O. Sadiku, Elements of Electromagnetics, 7th ed.; W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed.; F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; D. M. Pozar, Microwave Engineering, 4th ed.; C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed.
Constants used throughout. Question 6 hands us $\omega = 2\pi\times10^{10}$ rad/s together with $k = (2\pi/3)\ \text{cm}^{-1}$, so the setter’s phase velocity is $\omega/k = 3.000\times10^{8}$ m/s exactly; Question 3’s cut-off arithmetic only closes on round numbers under the same choice. Accordingly $c = 3\times10^{8}$ m/s and $\eta_0 = 120\pi = 376.99\ \Omega$ are used here. The paper’s own aid list would give $c = 2.99863\times10^{8}$ m/s and $\eta_0 = 376.83\ \Omega$ — a 0.05 % shift that changes no answer except at one deliberate boundary case, flagged in Question 3.
Question 1: Pulse echoes from two shunt taps on an endless line (20 marks)
Given. A pulse generator whose internal resistance equals the line impedance feeds an endless feeder that is bridged by two equal resistors, one a kilometre beyond the other.
Given data
Quantity
Symbol
Value
Generator internal impedance
$R_g$
377 Ω
Line characteristic impedance
$Z_0$
377 Ω
Propagation velocity
$v_p$
$3\times10^{8}$ m/s
Distance to the first tap
$d_1$
10 km
Distance to the second tap
$d_2$
11 km
Each tap resistor
$R$
377 Ω
Energy in the launched pulse
$W^{+}$
1 J
Find. The longest pulse that still yields separated echoes at the generator, the energy carried by the first of those echoes, and whether the train stops after two.
The bench as specified: a matched pulse generator, an endless 377 Ω feeder, and two 377 Ω resistors bridged across it 10 km and 11 km out.
Approach. Treat each tap as its resistor in parallel with the impedance the onward line still presents, read the reflection and transmission coefficients from that junction impedance, and then time the echoes by round trips.
The tap is not a match. A length of lossless line that has not yet heard from anything downstream presents its own $Z_0$, so the wave arriving at the first tap sees the bridging resistor in parallel with the continuing line: $$Z_J = \frac{R\,Z_0}{R + Z_0} = \frac{377 \times 377}{754} = 188.5\ \Omega .$$ This is the step most candidates skip — a resistor equal to $Z_0$ tapped across a through line is emphatically not a termination.
Reflection and transmission at a tap. With that junction impedance, $$\Gamma_J = \frac{Z_J - Z_0}{Z_J + Z_0} = \frac{188.5 - 377}{188.5 + 377} = \boxed{-\tfrac{1}{3}}, \qquad \tau = 1 + \Gamma_J = \tfrac{2}{3}.$$ So one third of the incident amplitude turns back, two thirds carries on, and the tap resistor absorbs the balance.
Time the two echoes. Each echo is a full round trip to its own tap: $$t_1 = \frac{2 d_1}{v_p} = \frac{2 \times 10^{4}}{3\times10^{8}} = 66.67\ \mu\text{s}, \qquad t_2 = \frac{2 d_2}{v_p} = \frac{2 \times 1.1\times10^{4}}{3\times10^{8}} = 73.33\ \mu\text{s}.$$
The pulse-length limit follows from their spacing. Two echoes are distinct only while the pulse is shorter than the gap between their arrivals, and that gap is the round trip over the tap spacing alone: $$t_2 - t_1 = \frac{2(d_2 - d_1)}{v_p} = \frac{2 \times 10^{3}}{3\times10^{8}} = \boxed{\tau_p \lt 6.67\ \mu\text{s}} .$$ Note what does not enter: the 10 km run to the first tap only sets when the train arrives, never how finely it is resolved.
Energy in the first echo. Every pulse on this feeder rides the same $Z_0$, so energy scales as amplitude squared: $$W_1 = \Gamma_J^{2}\,W^{+} = \left(\tfrac{1}{3}\right)^{2}\times 1\ \text{J} = \boxed{0.111\ \text{J}} .$$ The first transit therefore splits the joule three ways — $1/9$ J back towards the generator, $4/9$ J onward down the line, and the remaining $4/9$ J burned in the first tap resistor.
Count the rest of the train. Because $R_g = Z_0$ the generator is matched and $\Gamma_g = 0$, so nothing that returns is ever sent out again. The two taps, however, ring against each other. The second echo has amplitude $\tau\,\Gamma_J\,\tau = -4/27$ and energy $W_2 = (4/27)^{2} = 0.0220$ J; a wave that bounces once more between the taps returns as $\tau\,\Gamma_J^{3}\,\tau = -4/243$, i.e. $W_3 = 2.71\times10^{-4}$ J at $t = 80\ \mu\text{s}$. Each further echo is $\Gamma_J^{2} = 1/9$ of its predecessor in amplitude, so the answer is $\boxed{\text{yes — an unending, geometrically decaying train}}$, spaced $6.67\ \mu\text{s}$ apart, though the third is already 400 times weaker in energy than the first and would be lost in noise.
The echoes arriving back at the generator terminals. All are negative because a shunt tap lowers the impedance seen by the wave; successive echoes sit one tap-spacing round trip (6.67 μs) apart.
Final results
Quantity asked
Result
Junction impedance at either tap
$Z_J = 188.5\ \Omega$
Reflection coefficient at a tap
$\Gamma_J = -1/3$
Upper bound on the pulse length
$\tau_p \lt 2\,\Delta d / v_p = 6.67\ \mu\text{s}$
First echo: arrival and energy
$66.67\ \mu\text{s}$, $\;W_1 = 0.111$ J
Second echo: arrival and energy
$73.33\ \mu\text{s}$, $\;W_2 = 0.0220$ J
More than two echoes?
Yes — an infinite train, each $1/81$ of the previous in energy
