Question 1 of 8: Pulse energy delivered to two parallel semi-infinite lines
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2015 — 07-Elec-A7 Electromagnetics. Three hours, closed book; one of two approved calculators permitted. Eight questions, all of equal value (20 marks each). The rubric states that any five questions constitute a complete paper and that only the first five presented are marked — all eight are worked here, because this set is a study resource rather than a sat examination. Aids printed on the paper: $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$ and $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$, from which $\eta_0=\sqrt{\mu_0/\varepsilon_0} = 376.8\ \Omega$ and $c = 1/\sqrt{\mu_0\varepsilon_0} = 3.00\times10^{8}\ \text{m/s}$.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission-line transients, matching stubs, waveguide modes); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed.; W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (plane waves, normal incidence); F. T. Ulaby, Fundamentals of Applied Electromagnetics, 8th ed. (polarisation, interference, radiation); C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short current elements).
Question 1: Pulse energy delivered to two parallel semi-infinite lines (20 marks)
Given. A single rectangular pulse is launched down a long lossless line and eventually meets a junction where two further lines of identical construction continue away to infinity.
Given data
Quantity
Symbol
Value
Generator internal impedance
$R_g$
$377\ \Omega$
Generator EMF pulse
$E,\ \tau$
$12\ \text{V}$ for $2\ \mu\text{s}$
Driving line length
$d$
$10\ \text{km}$
Characteristic impedance
$Z_0$
$377\ \Omega$
Propagation velocity
$v_p$
$3\times10^{8}\ \text{m/s}$
Termination
—
two semi-infinite lines, each $377\ \Omega$, in parallel
Find. The energy carried by the pulse that travels away on each of the two infinitely long lines (and hence on the pair together).
Figure Q1. The 2 μs pulse launched by the matched generator, and its division at the junction J into a reflected pulse and one transmitted pulse on each semi-infinite line.
Approach. Find the wave the generator actually launches, replace each semi-infinite line by its characteristic impedance to get the junction load, apply the reflection and transmission coefficients once, and convert the transmitted amplitude into energy through the line's own power law.
Launch the wave from the generator. A line that has not yet heard from its termination presents exactly $Z_0$ at its input, so the source sees a simple divider: $$V^{+} = E\,\frac{Z_0}{R_g+Z_0} = 12\times\frac{377}{377+377} = \boxed{6.00\ \text{V}}$$ The generator resistance equals $Z_0$, so the generator end is matched and any wave that returns there is absorbed without further reflection — the problem has exactly one round of reflection.
Confirm the pulse and the line do not overlap in time. The one-way transit time is $T_d = d/v_p = 10\,000/(3\times10^{8}) = 33.3\ \mu\text{s}$, and the pulse occupies only $\tau = 2\ \mu\text{s}$, i.e. $v_p\tau = 600\ \text{m}$ of line. The pulse therefore arrives at the junction complete and isolated, and the transient can be treated as a single event rather than a staircase.
Replace the two semi-infinite lines by one resistance. An infinitely long lossless line never returns a reflection, so at its input it is indistinguishable from a resistor of $Z_0$. Two such lines joined in parallel present $$Z_J = \frac{Z_0}{2} = \frac{377}{2} = 188.5\ \Omega$$ This is the single most important step: the load is not $377\ \Omega$ even though every line in sight is a $377\ \Omega$ line.
Reflect and transmit at the junction. With the junction resistance in hand, $$\Gamma_J = \frac{Z_J-Z_0}{Z_J+Z_0} = \frac{188.5-377}{188.5+377} = -\frac{1}{3}, \qquad 1+\Gamma_J = \frac{2}{3}$$ The reflected pulse is $\Gamma_J V^{+} = -2.00\ \text{V}$ and the voltage that appears at the junction node is $$V_t = (1+\Gamma_J)V^{+} = \tfrac{2}{3}\times 6.00 = \boxed{4.00\ \text{V}}$$ Because the two continuing lines share that node, each of them launches a 4.00 V wave — the node voltage is common, only the current divides.
Convert the transmitted amplitude to energy. A travelling wave on a lossless line carries $P = V^{2}/Z_0$, so each infinite line carries $$P = \frac{(4.00)^2}{377} = 42.44\ \text{mW}, \qquad W = P\tau = 42.44\ \text{mW}\times 2\ \mu\text{s} = \boxed{84.9\ \text{nJ}}$$ and the two lines together remove $2\times 84.9 = 169.8\ \text{nJ}$ from the junction.
Audit the energy. The incident pulse carried $W_i = (V^{+})^2\tau/Z_0 = 191.0\ \text{nJ}$ and the reflected pulse carries $W_r = (\Gamma_J V^{+})^2\tau/Z_0 = 21.2\ \text{nJ}$, exactly $|\Gamma_J|^2 = 1/9$ of the incident energy. The balance $191.0 - 21.2 = 169.8\ \text{nJ}$ reproduces the transmitted total, which is the cheapest possible check that the transmission coefficient was applied to the right quantity.
The reflected 2.00 V pulse travels back for a further 33.3 μs and is then absorbed in the matched generator resistance, so nothing further happens on any of the three lines.
