22-Elec-B9 Electromagnetic Field, Transmission Lines, Antennas, and Radiation · May 2016

Question 2 of 8: Pulse Echoes from Two Shunt Resistors on a Matched Line

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

Notes on this paper

Paper format. National Exams, May 2016 — 07-Elec-B9, Electromagnetic Field, Transmission Lines, Antennas, and Radiation. Three hours; closed book; any non-communicating calculator permitted. Eight questions, each of equal value (20 marks); the paper states that five questions constitute a complete exam paper and that the first five appearing in the answer book are the ones marked. The only constants printed on the cover page are the free-space permittivity and permeability, and a Smith chart is bound in as page 4 for use with Question 3. All eight questions are worked here, because this set is a study resource rather than a three-hour sitting.

Reference texts. D. M. Pozar, Microwave Engineering (transmission lines, the Smith chart, waveguides, resonators); M. N. O. Sadiku, Elements of Electromagnetics (Maxwell's equations, plane waves, guided waves); W. H. Hayt and J. A. Buck, Engineering Electromagnetics (field theory and wave propagation); C. A. Balanis, Antenna Theory: Analysis and Design (short elements, arrays, radiation resistance); F. T. Ulaby, Fundamentals of Applied Electromagnetics (radar, link budgets, applied propagation).

Units and constants. The paper works in SI throughout. Where a question supplies a rounded propagation velocity (for example 3×10⁸ m/s) that value is used as given; where it does not, the exact free-space constants are used, giving $c = 2.998\times10^{8}\ \text{m/s}$ and $\eta_0 = 376.7\ \Omega$.

Question 2: Pulse Echoes from Two Shunt Resistors on a Matched Line (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.

Given data — Question 2
Quantity	Symbol	Value
Generator internal resistance	$R_g$	$377\ \Omega$
Line characteristic impedance	$Z_0$	$377\ \Omega$
Propagation velocity	$v$	$3\times10^{8}\ \text{m/s}$
Pulse duration	$\tau_p$	$1\ \mu\text{s}$
Pulse power on the line	$P$	$100\ \text{kW}$
First shunt resistance, at 10 km	$R_1$	$754\ \Omega$
Second shunt resistance, 450 m further	$R_2$	$754\ \Omega$
Line length beyond the second resistor	—	Infinite (no end reflection)

Find. The voltage appearing at the generator terminals over 0 to 120 µs: the amplitude and the arrival time of the launched pulse and of every echo that returns within that window.

The line as given: a matched generator, two 754 Ω resistors shunted across a line that continues past both of them to infinity.

Approach. Convert the stated pulse power into a launched wave amplitude using the matched-generator condition, find the reflection and transmission coefficients of a shunt resistor sitting on a line that continues past it, then track every path that starts at the generator and returns to it inside 120 µs.

Fix the amplitude of the launched wave. Because $R_g = Z_0$, the generator is matched to the line: the line presents $Z_0$ to the source, the terminal voltage is half the EMF, and the power delivered is the source's maximum available power. Writing the pulse power in terms of the forward wave $V^{+}$ on a line of impedance $Z_0$, $$P = \frac{(V^{+})^{2}}{Z_0} \;\Longrightarrow\; V^{+} = \sqrt{P Z_0} = \sqrt{(100\times10^{3})(377)}$$ $$\boxed{V^{+} = 6140\ \text{V}}$$ The corresponding EMF is $E = 2V^{+} = 12\,280\ \text{V}$, and the check $E^{2}/(4R_g) = 100\ \text{kW}$ confirms it.
Find what a shunt resistor does to a line that continues past it. This is the step that decides the whole answer. The wave arriving at the first resistor sees $R_1$ in parallel with the characteristic impedance of the onward line, not $R_1$ alone: $$Z_j = R_1 \parallel Z_0 = \frac{(754)(377)}{754 + 377} = 251.33\ \Omega.$$ The reflection and transmission coefficients follow at once: $$\Gamma = \frac{Z_j - Z_0}{Z_j + Z_0} = \frac{251.33 - 377}{251.33 + 377} = -0.200, \qquad \tau = 1 + \Gamma = 0.800.$$ A resistor equal to twice $Z_0$ therefore returns a fifth of the incident amplitude, inverted, and passes four fifths onward. The power audit $\Gamma^{2} + \tau^{2} + \tau^{2}(Z_0/R_1) = 0.04 + 0.64 + 0.32 = 1$ accounts for the reflected wave, the transmitted wave, and the fraction burned in the resistor itself, and is the cheapest way to be sure the coefficients are right.
Compute the transit times. Both resistors are identical, so the geometry alone sets the schedule: $$t_1 = \frac{10\,000}{3\times10^{8}} = 33.333\ \mu\text{s}, \qquad t_2 = \frac{10\,450}{3\times10^{8}} = 34.833\ \mu\text{s}.$$ The 450 m section between the two resistors takes $1.5\ \mu\text{s}$ to traverse one way.
Establish that no echo is ever re-launched. At the generator, $\Gamma_g = (R_g - Z_0)/(R_g + Z_0) = 0$ because $R_g = Z_0$ exactly. Every returning wave is therefore absorbed in the source resistance on arrival, and the terminal voltage at that instant is simply the amplitude of the arriving wave. This is what keeps the answer to a short, finite list instead of an endless bounce diagram — and it is also why the far end being infinite matters only for the onward-travelling energy.
Track the first echo. The launched pulse reaches $R_1$, reflects with $\Gamma = -0.2$, and returns: $$V_1 = \Gamma V^{+} = (-0.200)(6140) = -1228\ \text{V} \quad\text{at}\quad t = 2t_1 = 66.667\ \mu\text{s}.$$
Track the echo from the second resistor. This wave passes through the first junction going out, reflects off the second, and passes back through the first junction returning, so it carries two transmission factors: $$V_2 = \tau\,\Gamma\,\tau\,V^{+} = (0.800)(-0.200)(0.800)(6140) = -785.9\ \text{V}$$ arriving at $t = 2t_2 = 69.667\ \mu\text{s}$. Note that this is not $\Gamma^{2}V^{+}$; forgetting the pair of transmission factors is the standard slip here.
Track the waves that rattle in the 450 m section. A wave returning from $R_2$ meets $R_1$ from the far side and is partly reflected back down again, so the section between the two resistors acts as a weak resonator. Each extra round trip inside it multiplies the amplitude by $\Gamma^{2} = 0.04$ and delays the arrival by $2\times1.5 = 3\ \mu\text{s}$: $$V_3 = \tau^{2}\Gamma^{3}V^{+} = -31.4\ \text{V at } 72.667\ \mu\text{s}, \qquad V_4 = \tau^{2}\Gamma^{5}V^{+} = -1.26\ \text{V at } 75.667\ \mu\text{s}.$$ The next term is $-0.05\ \text{V}$ at $78.667\ \mu\text{s}$, which is below any reasonable plotting resolution, so the record is effectively complete.
Assemble the plot. Each entry is a rectangular pulse of the same $1\ \mu\text{s}$ width as the original, since nothing in the system disperses it. The terminal voltage is zero everywhere else in the window; in particular nothing at all happens between $1$ and $66.667\ \mu\text{s}$, and nothing after about $77\ \mu\text{s}$.

Generator terminal voltage, 0 to 120 µs. The launched pulse, then four inverted echoes clustered between 66.7 and 75.7 µs; the vertical scales for the positive and negative excursions differ so the small echoes remain visible.

Check: interpretation of “a one microsecond pulse of 100 kW power”. An EMF by itself carries no power, so the 100 kW is read here as the power the pulse delivers into the line during the microsecond it lasts. Because the generator is matched, this is also its maximum available power, and the two readings coincide: $P = (V^{+})^{2}/Z_0 = E^{2}/(4R_g) = 100\ \text{kW}$ gives the same $V^{+} = 6140\ \text{V}$ either way. If instead a marker intended 100 kW to be the total drawn from the EMF (half of it burned in $R_g$), every voltage in the answer scales by $1/\sqrt{2}$ and the timing is unchanged.

Question 2 — generator terminal voltage, 0 to 120 µs
Event	Arrival time	Amplitude	Path
Launched pulse	$0 - 1\ \mu\text{s}$	$+6140\ \text{V}$	—
Echo from $R_1$	$66.667\ \mu\text{s}$	$-1228\ \text{V}$	$\Gamma$
Echo from $R_2$	$69.667\ \mu\text{s}$	$-785.9\ \text{V}$	$\tau^{2}\Gamma$
One extra bounce in the 450 m section	$72.667\ \mu\text{s}$	$-31.4\ \text{V}$	$\tau^{2}\Gamma^{3}$
Two extra bounces	$75.667\ \mu\text{s}$	$-1.26\ \text{V}$	$\tau^{2}\Gamma^{5}$
Junction coefficients	$\Gamma = -0.200$, $\tau = 0.800$ at each 754 Ω shunt; $\Gamma_g = 0$ at the generator

Topic. Time-domain reflectometry on a line with shunt discontinuities and a matched source.

Key relations. $V^{+} = \sqrt{PZ_0}$; a shunt $R$ on a continuing line presents $Z_j = R \parallel Z_0$, giving $\Gamma = (Z_j - Z_0)/(Z_j + Z_0)$ and $\tau = 1 + \Gamma$; round-trip delay $= 2\ell/v$; power audit $\Gamma^{2} + \tau^{2} + \tau^{2}(Z_0/R) = 1$.

Why this works. A pulse short compared with every transit time behaves like a probe: it visits each discontinuity once per path and returns a scaled copy, so the terminal record is a catalogue of paths rather than a superposition of overlapping waveforms. That is precisely how a cable fault locator works — the delay gives the distance and the sign of the echo says whether the discontinuity is a series break (positive) or a shunt load (negative). The matched source is what makes the record readable: with $\Gamma_g = 0$ each echo appears exactly once, whereas a mismatched generator would re-launch every echo and fill the trace with second-generation replicas that are easily mistaken for further faults.

Common pitfall. Treating the shunt resistor as the line's termination and using $\Gamma = (R - Z_0)/(R + Z_0)$, which for $R = 754\ \Omega$ gives $+0.333$ — wrong in both magnitude and sign, because the line does not stop at the resistor. The companion error is computing the second echo as $\Gamma^{2}$ instead of $\tau^{2}\Gamma$, which understates it by a factor of 3.2 and gives it the wrong sign.

Source. Pozar, Microwave Engineering, §2.3 (the terminated lossless line); Ulaby, Fundamentals of Applied Electromagnetics, Ch. 2 (transients on transmission lines and bounce diagrams).

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