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22-Elec-A7 Electromagnetics · May 2016

Question 1 of 8: Step response of a line feeding two semi-infinite lines

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, May 2016 — 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.

Constants. The paper prints $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$ and $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$ as aids. Its own numbers imply the classroom values $c = 3.00\times10^{8}\ \text{m/s}$ and $\eta_0 = 120\pi = 377\ \Omega$: the offset in Question 3 is exactly $\lambda/8$ only if $\lambda = 3.00\ \text{cm}$ at 10 GHz, and the 2.25 cm guide of Question 4 places a cutoff exactly on 20 GHz only for the same value. Those are the constants used throughout; the more precise values shift every field and guide wavelength here by less than 0.3 %, and the one place where the difference matters is flagged in Question 4.

Reference texts. D. M. Pozar, Microwave Engineering, 4th ed., §2.1–2.7 (transmission-line theory, standing waves, impedance transformation). M. N. O. Sadiku, Elements of Electromagnetics, 7th ed., ch. 11 (transmission lines) and ch. 12 (waveguides). W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed., ch. 10–11 (transmission lines, uniform plane waves). F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed., ch. 2 and ch. 7 (line transients, wave polarisation). C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed., ch. 4 (the infinitesimal/short current element).

Question 1: Step response of a line feeding two semi-infinite lines (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A matched-source step generator drives a long lossless feeder whose far end splits into two identical lines that continue away for ever, so nothing is ever reflected from beyond the junction.

Given data
QuantitySymbolValue
Generator EMF (step)$E$$12\ \text{V}$
Generator internal resistance$R_g$$50\ \Omega$
Feeder characteristic impedance$Z_0$$50\ \Omega$
Feeder length$d$$10\ \text{km}$
Propagation velocity$v_p$$2\times10^{8}\ \text{m/s}$
Termination$Z_J$two semi-infinite $50\ \Omega$ lines in parallel

Find. The generator-terminal current $i(t)$ over $0 \le t \le 150\ \mu\text{s}$, drawn to scale.

E = 12 V stepR(g) = 50 Ωd = 10 km, vp = 2×10⁸ m/sZ₀ = 50 Ωjunction J∞Z₀ = 50 Ω∞Z₀ = 50 Ω2 semi-infinite linesLoad seen at J: Z(J) = Z₀/2 = 25 Ω Γ(J) = (Z(J) − Z₀)/(Z(J) + Z₀) = −1/3Round-trip delay 2d/v(p) = 100 μs -- until then the terminals cannot know the load.R(g) = Z₀, so the echo is absorbed: ONE step, no further reflections.
Figure 1.1 — the feeder and its junction. A semi-infinite line of characteristic impedance $Z_0$ is indistinguishable from a resistor of $Z_0$, so two of them in parallel present $Z_0/2$.

Approach. Work forward in time: the feeder presents $Z_0$ until news of the junction can return, so the initial current follows from a simple resistive divider; then add the single reflected wave that arrives one round trip later, and confirm the result against the DC circuit.

  1. Fix the round-trip delay. The one-way transit is $T = d/v_p = 10\times10^{3}/(2\times10^{8}) = 50\ \mu\text{s}$, so the earliest the generator can learn anything about the far end is $2T = 100\ \mu\text{s}$.
  2. Launch the incident wave. During the first round trip the feeder behaves as a pure resistance $Z_0$, so $$V^{+} = E\,\frac{Z_0}{R_g + Z_0} = 12\cdot\frac{50}{100} = 6.00\ \text{V}, \qquad i(0^{+}) = \frac{V^{+}}{Z_0} = \boxed{0.120\ \text{A}}$$
  3. Evaluate the junction. A line of characteristic impedance $Z_0$ that never ends absorbs everything sent into it and so is electrically a resistor of $Z_0$. Two in parallel give $Z_J = Z_0/2 = 25\ \Omega$, whence $$\Gamma_J = \frac{Z_J - Z_0}{Z_J + Z_0} = \frac{25 - 50}{25 + 50} = -\tfrac{1}{3}$$
  4. Return the reflected wave. The wave $V^{-} = \Gamma_J V^{+} = -2.00\ \text{V}$ travels back and reaches the generator at $t = 2T$. A backward voltage wave carries current $-V^{-}/Z_0 = +0.040\ \text{A}$, so the terminal current steps up to $$i(t \\gt 2T) = 0.120 + 0.040 = \boxed{0.160\ \text{A}}$$
  5. Stop the bouncing. The generator resistance equals $Z_0$, so $\Gamma_g = 0$: the returning wave is absorbed and there is no second echo. The waveform is therefore a single step, and it must already be the steady state.
  6. Check against the DC circuit. A lossless line is just wire at DC, so the final current is the plain divider $i(\infty) = E/(R_g + Z_J) = 12/75 = 0.160\ \text{A}$, which matches step 4 exactly.

The two results bracket the whole waveform, so the plot is a single riser: a flat $0.120\ \text{A}$ from the instant the step is applied until $t = 100\ \mu\text{s}$, then a jump to $0.160\ \text{A}$ held for the rest of the window. Nothing further happens before $150\ \mu\text{s}$, and indeed nothing further happens ever.

i (t) [A]0.120 A0.160 A02550751001251502d/v(p) = 100 μst (μs)launched wave onlysteady state (DC divider)Generator-terminal current, Q10.120 A while only the launched wave is present, then 0.160 A.
Figure 1.2 — the required plot. The step at $100\ \mu\text{s}$ is the round-trip echo from the junction; because the source is matched, the staircase has exactly one riser.

It is worth reading the steady state physically. The load node settles at $V_J = (1 + \Gamma_J)V^{+} = 4.00\ \text{V}$, and the $0.160\ \text{A}$ divides equally between the two continuing lines, each carrying $0.080\ \text{A}$ away from the junction and never returning it. The generator delivers $E i = 1.92\ \text{W}$, of which $i^{2}R_g = 1.28\ \text{W}$ heats the source resistance and $0.64\ \text{W}$ streams away down the two infinite lines.

Final results
QuantityValue
One-way transit time $T = d/v_p$$50\ \mu\text{s}$
Round-trip delay $2T$$100\ \mu\text{s}$
Launched wave $V^{+}$$6.00\ \text{V}$
Junction impedance $Z_J$$25\ \Omega$
Junction reflection coefficient $\Gamma_J$$-1/3$
Generator current, $0 \lt t \lt 100\ \mu\text{s}$$\mathbf{0.120\ \text{A}}$
Generator current, $t \gt 100\ \mu\text{s}$$\mathbf{0.160\ \text{A}}$
Current into each semi-infinite line$0.080\ \text{A}$
Steady-state junction voltage$4.00\ \text{V}$
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