Question 1 of 8: Energy in the Reflected Pulse, and the Highest Non-Overlapping Repetition Rate
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2018 — 16-Elec-A7, Electromagnetics. Closed book, three hours, approved Casio or Sharp calculator only. Eight questions, all of equal value; the paper states that any five constitute a complete paper, so each question carries 20 of the 100 marks that are actually graded. All eight are solved here, because this set is a study resource rather than an examination script.
Aids given on the paper. $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$, $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$, $e = 1.6\times10^{-19}\ \text{C}$. Those two constants fix $c = 1/\sqrt{\mu_0\varepsilon_0} = 2.9986\times10^{8}\ \text{m/s}$ and $\eta_0 = \sqrt{\mu_0/\varepsilon_0} = 376.82\ \Omega$ — which is why Question 1 quotes a 377 Ω line: it is free space itself, rendered as a cable.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission-line transients, stubs, waveguide modes); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. (Biot–Savart, motional EMF, plane waves); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed.; F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short-element fields and the reactive near-field boundary).
Constants used throughout: the paper's own aid list is taken as authoritative, so $c = 2.9986\times10^{8}$ m/s and $\eta_0 = 376.82\ \Omega$. Question 1 supplies its own propagation velocity ($3\times10^{8}$ m/s) and that stated value is used there. Question 5 is the one place where the difference between $2.9986\times10^{8}$ and $3.00\times10^{8}$ changes the wording of the answer, and both readings are worked through explicitly.
Question 1: Energy in the Reflected Pulse, and the Highest Non-Overlapping Repetition Rate (20 marks)
Given. A matched pulse generator drives an endless 377 Ω line that carries one shunt tap 5 km from its terminals.
Given data
Quantity
Symbol
Value
Generator internal impedance
$R_g$
377 Ω
Pulse duration
$\tau_p$
1 µs
Energy delivered to a matched 377 Ω load
$W$
2 J
Line characteristic impedance
$Z_0$
377 Ω
Propagation velocity
$v_p$
$3\times10^{8}$ m/s
Distance to the tap
$d$
5 km
Shunt resistor at the tap
$R$
377 Ω
Line beyond the tap
—
continues to infinity
Find. (I) the energy carried by one reflected pulse, and (II) the highest pulse repetition frequency at which an outgoing pulse and a returning echo never overlap at the generator terminals, together with the power-versus-time plot of the first two echoes.
The line as the question describes it: matched generator, one shunt resistor 5 km out, and line continuing beyond the tap without end.
Approach. Convert the stated 2 J into the amplitude of the wave actually launched, form the junction reflection coefficient from the tap resistor in parallel with the onward line, scale the energy by $\Gamma^2$, and then set the repetition period from the round-trip delay plus the pulse width.
Turn the stated energy into a generator EMF. If the pulse were delivered to a 377 Ω resistor sitting directly on the generator, the divider would put half the EMF across it, so $P = W/\tau_p = 2\ \text{J}/1\ \mu\text{s} = 2\ \text{MW}$, $V_L = \sqrt{P R_g} = \sqrt{(2\times10^{6})(377)} = 27.46\ \text{kV}$ and $E = 2V_L = 54.92\ \text{kV}$.
Find the wave the line actually carries. A line of characteristic impedance $Z_0$ presents exactly $Z_0$ to the generator until something reflects back, so the launched wave is another simple divider: $$V^{+} = E\,\frac{Z_0}{R_g + Z_0} = \frac{E}{2} = 27.46\ \text{kV},\qquad P^{+} = \frac{(V^{+})^{2}}{Z_0} = 2\ \text{MW}.$$ The launched pulse therefore carries $\boxed{W^{+} = P^{+}\tau_p = 2\ \text{J}}$ — the same 2 J the question quotes, which is the point of the way it is worded.
Build the junction impedance from the tap and the onward line. This is the step the question is really testing. The 377 Ω resistor is not a termination: the line carries on past it, and that continuing line looks like its own $Z_0$. The arriving wave therefore sees $$Z_J = R \parallel Z_0 = \frac{(377)(377)}{377+377} = 188.5\ \Omega .$$
Reflect off that junction. $$\Gamma_J = \frac{Z_J - Z_0}{Z_J + Z_0} = \frac{188.5 - 377}{188.5 + 377} = -\frac{1}{3}.$$ The echo is inverted and one third of the incident amplitude, so its voltage is $-9.15$ kV.
Scale the energy (answer I). Energy in a travelling pulse goes as the square of its amplitude at fixed $Z_0$, so $$W_{\text{ref}} = \Gamma_J^{2}\,W^{+} = \tfrac{1}{9}(2\ \text{J}) \;\Rightarrow\; \boxed{W_{\text{ref}} = 0.222\ \text{J}} ,$$ carried as a 1 µs pulse of $P_{\text{ref}} = W_{\text{ref}}/\tau_p = 222\ \text{kW}$.
Audit the energy before going on. The wave continuing past the tap has amplitude $(1+\Gamma_J)V^{+} = \tfrac{2}{3}V^{+} = 18.31$ kV, so it carries $\tfrac{4}{9}(2) = 0.889$ J, and the tap resistor — which sees that same voltage — dissipates $18.31^2/377 \times 1\ \mu\text{s} = 0.889$ J. Adding up, $0.222 + 0.889 + 0.889 = 2.000$ J. Nothing is missing, which confirms both $\Gamma_J$ and the 2 J starting point.
Time the echo (answer II). The pulse must travel out and back: $$T_{rt} = \frac{2d}{v_p} = \frac{2(5000)}{3\times10^{8}} = 33.33\ \mu\text{s}.$$ The echo therefore occupies the window from 33.33 µs to 34.33 µs after its own pulse left. For the next outgoing pulse not to be sitting on top of it, the repetition period must be at least that long: $$T \ge T_{rt} + \tau_p = 34.33\ \mu\text{s} \;\Rightarrow\; \boxed{f_{\text{PRF,max}} = 29.1\ \text{kHz}} .$$
Establish how many echoes there are. The generator impedance equals $Z_0$, so $\Gamma_g = 0$: the returning echo is completely absorbed at the generator terminals and never starts a second round trip. Each outgoing pulse therefore produces exactly one echo, and the “first two reflected pulses” are the echoes of the first two outgoing pulses, not two bounces of one pulse.
Generator-terminal waveform over one repetition period at the maximum rate: the launched pulse, then the single inverted echo arriving one round trip later, ending exactly where the next pulse begins.
Because the incident pulses are identical and the line is linear, both plotted echoes are identical rectangles: 222 kW high and 1 µs wide, the first centred at $t = 33.8\ \mu\text{s}$ and the second one repetition period later at $t = 68.2\ \mu\text{s}$. The reflected power is positive in both, even though the reflected voltage is negative — power does not carry the sign of $\Gamma$.
Engineering note, not required for the marks: strictly, higher repetition rates also avoid overlap whenever the echo happens to land in a gap between outgoing pulses (any period $T$ with $T \ge 2\ \mu\text{s}$ that divides 34.33 µs exactly will do, for instance $T = 17.17\ \mu\text{s}$). Those interleaved solutions are fragile — a few per cent of cable-length or velocity error destroys them — so the robust design answer is the 29.1 kHz above, and that is what a marker expects.
