Question 1 of 8: Steady-state terminal-voltage pattern of a pulsed line with a shunt tap
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2017 — 16-Elec-A7 Electromagnetics. Three hours, closed book (one Casio or Sharp approved calculator). Eight questions of equal value; the rubric states that any five questions constitute a complete paper and that only the first five answered will be marked. All eight are solved here, because the set is a study resource rather than a sitting. Aids printed on the paper: $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}$ and $\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}$.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission-line transients, stub networks, waveguides); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (plane waves, polarisation, Biot–Savart, inductance); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. (magnetostatics, guided waves); F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed. (wave polarisation, radiation); C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short-element fields).
Constants used throughout. The aid sheet's $\varepsilon_0$ and $\mu_0$ give $c = 2.9986 \times 10^{8}\ \text{m/s}$ and $\eta_0 = 376.82\ \Omega$, while every round figure the paper quotes (10 km at $3 \times 10^{8}\ \text{m/s}$, 300 and 600 MHz, 10 GHz giving a 3 cm wavelength) is built on $c = 3.00 \times 10^{8}\ \text{m/s}$ and $\eta_0 = 120\pi = 376.99\ \Omega$. Those are the values used below; the two sets differ by less than 0.05 % and no answer or conclusion changes.
Question 1: Steady-state terminal-voltage pattern of a pulsed line with a shunt tap
Given. A matched pulse generator drives a semi-infinite line that carries one shunt resistor 10 km out.
Given data
Quantity
Symbol
Value
Open-circuit EMF amplitude
$E$
$38.8\ \text{kV}$
Pulse width
$t_p$
$1\ \mu\text{s}$
Pulse repetition frequency
$\mathrm{PRF}$
$10\ \text{kHz}$
Generator internal impedance
$R_g$
$377\ \Omega$
Line characteristic impedance
$Z_0$
$377\ \Omega$
Propagation velocity
$v_p$
$3 \times 10^{8}\ \text{m/s}$
Shunt resistor across the line
$R = Z_0/2$
$188.5\ \Omega$
Distance generator to resistor
$d$
$10\ \text{km}$
Find. The repeating pattern of the voltage at the generator terminals, with every voltage level and every time interval labelled.
Figure 1.1 — the pulser, the 10 km line, the shunt tap, and the line continuing to infinity. Only one echo exists because the source is matched.
Approach. Launch a single wave through the source divider, reflect it once at the shunt tap (which is in parallel with the onward line), and note that a matched generator absorbs the return — so the pattern is one pulse pair repeated at the pulse repetition rate.
Launch the incident wave. Before any reflection has had time to return, the generator sees only the line's own characteristic impedance, so the launched amplitude is a simple resistive divider:$$V^{+} = E\,\frac{Z_0}{R_g + Z_0} = 38.8\ \text{kV} \times \frac{377}{377 + 377} = 19.4\ \text{kV}$$
Find what the tap actually presents. The line does not stop at the resistor — it carries on to infinity, and a semi-infinite lossless line always looks like $Z_0$. The junction impedance is therefore the resistor in parallel with that onward line:$$Z_J = R \parallel Z_0 = \frac{(Z_0/2)\,Z_0}{Z_0/2 + Z_0} = \frac{Z_0}{3} = 125.7\ \Omega$$A resistor equal to half of $Z_0$ is emphatically not a matched termination.
Reflect at the junction. With the junction impedance in hand,$$\Gamma_J = \frac{Z_J - Z_0}{Z_J + Z_0} = \frac{Z_0/3 - Z_0}{Z_0/3 + Z_0} = \boxed{-\tfrac{1}{2}}$$so the wave sent back toward the generator has amplitude $\Gamma_J V^{+} = -9.7\ \text{kV}$, and the wave that continues past the tap and never returns has amplitude $(1 + \Gamma_J)V^{+} = +9.7\ \text{kV}$.
Check the generator end. Because $R_g = Z_0$, the source reflection coefficient is $\Gamma_g = (R_g - Z_0)/(R_g + Z_0) = 0$: the returning pulse is absorbed entirely in the internal impedance. There is exactly one echo per launched pulse — no staircase, no decaying train.
Put in the timing. The one-way transit is $d/v_p = 10\,000/(3 \times 10^{8}) = 33.33\ \mu\text{s}$, so the echo reappears at the generator after the round trip$$t_{rt} = \frac{2d}{v_p} = 66.67\ \mu\text{s}$$The repetition period is $T = 1/\mathrm{PRF} = 100\ \mu\text{s}$, and since $66.67 + 1 \lt 100$ each echo lands inside its own period: successive pulses never overlap and the steady-state pattern is simply this pair repeated.
Assemble the waveform. Taking $t = 0$ at the start of any launched pulse, the terminal voltage is $+19.4\ \text{kV}$ for $0 \lt t \lt 1\ \mu\text{s}$, zero from $1\ \mu\text{s}$ to $66.67\ \mu\text{s}$, $-9.7\ \text{kV}$ for $66.67\ \mu\text{s} \lt t \lt 67.67\ \mu\text{s}$, and zero for the remaining $32.33\ \mu\text{s}$ before the next pulse.
Cross-check against the DC limit. With $\Gamma_g = 0$ the first transit is already final, so a continuous drive would settle at the plain resistive divider across the junction: $E\,Z_J/(R_g + Z_J) = 38.8 \times 125.7/502.7 = 9.7\ \text{kV}$, exactly the amplitude computed for the onward wave. The two routes agree, which confirms $\Gamma_J$. For completeness the pulse powers are $P^{+} = (V^{+})^{2}/Z_0 = 998\ \text{kW}$ incident and $\Gamma_J^{2}P^{+} = 250\ \text{kW}$ returned.
Figure 1.2 — the steady-state generator terminal voltage over one 100 μs period: a +19.4 kV launched pulse, a −9.7 kV echo 66.67 μs later, then 32.33 μs of quiet.
