Question 1 of 8: Pulse energy split at a junction of two semi-infinite lines
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2014 — 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 the 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 $\eta_0=\sqrt{\mu_0/\varepsilon_0}\approx 377\ \Omega$ and $c = 3\times10^{8}\ \text{m/s}$.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission lines, waveguides); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed.; W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (statics, Faraday's law, plane waves); F. T. Ulaby, Fundamentals of Applied Electromagnetics, 8th ed. (polarisation, radiation); C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short elements, image theory).
Question 1: Pulse energy split at a junction of two semi-infinite lines (20 marks)
Given. A single rectangular pulse is launched down a lossless line and meets a junction at which two further lines of identical construction continue away to infinity.
Given data
Quantity
Symbol
Value
Pulse energy
$W_i$
$1\ \text{J}$
Pulse duration
$T$
$1\ \mu\text{s}$
Characteristic impedance (all three lines)
$Z_0$
$377\ \Omega$
Phase velocity (all three lines)
$v_p$
$3\times10^{8}\ \text{m/s}$
Load
$Z_J$
two semi-infinite $Z_0$ lines in parallel
Find. The energy carried by the reflected pulse, and the energy carried by the pulse that travels away on each of the two lines forming the load.
One-wire schematic of the junction. Each semi-infinite line presents a pure resistance $Z_0$, so the pair in parallel loads the driving line with $Z_0/2$.
Approach. Replace each infinitely long line by the pure resistance $Z_0$ it presents at its input, compute the junction reflection coefficient, and split the pulse energy using $|\Gamma|^2$ and $1-|\Gamma|^2$ — the two load lines then share the transmitted energy equally by symmetry.
Represent each load line by its characteristic impedance. A line of infinite length never returns an echo, so its input impedance is exactly its characteristic impedance, $Z_{in}=Z_0=377\ \Omega$. Two such inputs in parallel give $$Z_J=\frac{Z_0}{2}=\frac{377}{2}=188.5\ \Omega .$$
Reflection and transmission coefficients at the junction. $$\Gamma_J=\frac{Z_J-Z_0}{Z_J+Z_0}=\frac{188.5-377}{188.5+377}=-\frac{1}{3},\qquad \tau_J=1+\Gamma_J=\frac{2}{3}.$$ The negative sign says the junction is a step down in impedance, so the reflected voltage is inverted.
Energy in the reflected pulse. The reflected wave travels back on the same $Z_0$, so its power is $|\Gamma_J V^{+}|^{2}/Z_0$ and its duration is unchanged. The energy therefore scales as the square of the voltage ratio: $$W_r=|\Gamma_J|^{2}W_i=\tfrac{1}{9}(1\ \text{J})\;\Rightarrow\;\boxed{W_r = 0.111\ \text{J}} .$$
Energy delivered into the load pair. What is not reflected is transmitted, $W_t=\left(1-|\Gamma_J|^{2}\right)W_i=\tfrac{8}{9}=0.889\ \text{J}$.
Split the transmitted energy between the two lines. The two load lines are identical and see the same junction voltage $\tau_J V^{+}$, so each carries $\left(\tau_J V^{+}\right)^{2}/Z_0$ — exactly half of $W_t$: $$W_A=W_B=\frac{W_t}{2}=\frac{4}{9}\ \text{J}\;\Rightarrow\;\boxed{W_A=W_B=0.444\ \text{J}} .$$
Independent check through the actual voltages. The incident power is $P_i=W_i/T=1\ \text{MW}$, so $V^{+}=\sqrt{P_iZ_0}=\sqrt{(10^{6})(377)}=19.42\ \text{kV}$. The junction voltage is $\tau_JV^{+}=12.94\ \text{kV}$, giving $(12.94\ \text{kV})^{2}/377\ \Omega = 444\ \text{kW}$ on each load line, i.e. $0.444\ \text{J}$ in $1\ \mu\text{s}$. Energy balances: $0.111+2(0.444)=1.000\ \text{J}$.
The result is worth reading physically. Although the junction is a mismatch, only one ninth of the energy comes back; the load lines being infinite means the pulse leaves the junction once and never returns, so there is no bounce sequence to sum. The pulse itself occupies $v_pT = 300\ \text{m}$ of line, but that length never enters the arithmetic precisely because no second reflection ever arrives. Note also the numerical coincidence that each load line receives $\tau_J^{2}=4/9$ of the incident energy: that is a genuine identity here, because each load line carries the junction voltage across the same $Z_0$ as the driving line.
