Pulse Train on a Mismatched Transmission Line: 22-Elec-A7 Electromagnetics, December 2016, Q1

Question 1: Pulse Train on a Mismatched Transmission 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. A pulse generator whose internal resistance equals the line impedance drives a long, lossless, air-spaced line into a load of half that impedance, and each launched pulse carries a stated energy.

Given data
Quantity	Symbol	Value
Generator internal resistance	R_g	377 Ω
Line characteristic impedance	Z₀	377 Ω
Propagation velocity	v_p	3 × 10⁸ m/s
Line length	d	10 km
Load resistance	R_L	377/2 = 188.5 Ω
Pulse width	τ	1 µs
Pulse repetition frequency	PRF	10 kHz
Energy per outgoing pulse	W	1 J

Find. The complete voltage waveform at the generator terminals — every distinct amplitude, the instant at which it appears, how long it lasts, and how the pattern repeats.

Figure 1.1 — The matched-source line and the resulting terminal voltage: one launched pulse and one inverted echo in every 100 µs period.

Approach. Because the generator resistance equals the line impedance the source end is matched, so the reflection coefficient there is zero and exactly one echo can ever return; the pulse energy fixes the launched amplitude, the load mismatch fixes the echo amplitude, and the round-trip delay fixes when the echo arrives relative to the pulse repetition period.

Convert the stated pulse energy into a launched wave amplitude. The energy quoted is that of the pulse travelling away from the generator, so the instantaneous power carried by the forward wave during the 1 µs window is $$P_{\text{pulse}}=\frac{W}{\tau}=\frac{1\ \text{J}}{1\times10^{-6}\ \text{s}}=1.00\times10^{6}\ \text{W}$$ A forward-travelling wave on a lossless line carries $P=(V^{+})^{2}/Z_{0}$, so $$V^{+}=\sqrt{P_{\text{pulse}}Z_{0}}=\sqrt{(1.00\times10^{6})(377)}=\boxed{19.42\ \text{kV}}$$
Back out the generator EMF for completeness. At the instant a pulse is launched the line has not yet heard from its termination and therefore presents exactly $Z_{0}$ to the source. The generator, its internal resistance and that $Z_{0}$ form a simple divider: $$V^{+}=E\,\frac{Z_{0}}{R_{g}+Z_{0}}=\frac{E}{2}\qquad\Longrightarrow\qquad E=2V^{+}=38.83\ \text{kV}$$ The generator terminal voltage during the launch is therefore only half the open-circuit EMF, which is the value the question asks us to describe.
Compute the transit times that organise the waveform. The one-way delay and the round trip are $$T_{d}=\frac{d}{v_{p}}=\frac{10\,000}{3\times10^{8}}=33.33\ \mu\text{s},\qquad 2T_{d}=66.67\ \mu\text{s}$$ while the pulses are launched every $$T_{\text{PRF}}=\frac{1}{\text{PRF}}=\frac{1}{10\times10^{3}}=100\ \mu\text{s}$$ Since $2T_{d}=66.67\ \mu\text{s}$ is comfortably less than 100 µs, each echo returns and dies away before the next pulse leaves. The waveform is therefore a clean, non-overlapping repeating pattern rather than a superposition — a point worth stating explicitly, because it is the whole reason the answer is simple.
Evaluate the two reflection coefficients. At the load, $$\Gamma_{L}=\frac{R_{L}-Z_{0}}{R_{L}+Z_{0}}=\frac{188.5-377}{188.5+377}=-\frac{1}{3}$$ and at the generator, $$\Gamma_{g}=\frac{R_{g}-Z_{0}}{R_{g}+Z_{0}}=\frac{377-377}{377+377}=0$$ The vanishing source reflection coefficient is the decisive feature: whatever comes back from the load is absorbed entirely in $R_{g}$, so the bounce sequence stops after a single return.
Find the echo amplitude seen at the generator terminals. The wave reflected from the load has amplitude $$V^{-}=\Gamma_{L}V^{+}=-\tfrac{1}{3}\left(19.42\ \text{kV}\right)=\boxed{-6.472\ \text{kV}}$$ By the time it arrives the original pulse has long since passed, so the terminal voltage during that 1 µs window is just $V^{-}$ itself — a negative pulse, because the load is lighter than the line and reflects with inverted sign.
Assemble the terminal waveform. Within each 100 µs repetition period the generator terminals show, in order: a positive 19.42 kV rectangular pulse of 1 µs duration starting at $t=0$; nothing at all for the next 65.67 µs; a negative 6.472 kV pulse of 1 µs duration starting at $t=2T_{d}=66.67\ \mu\text{s}$; and then nothing until the next launch at $t=100\ \mu\text{s}$. The pattern repeats indefinitely at the 10 kHz rate.
Audit the energy, which is the cheapest check on the whole answer. The fraction of energy returned is $|\Gamma_{L}|^{2}$, so $$W_{\text{reflected}}=|\Gamma_{L}|^{2}W=\tfrac{1}{9}(1\ \text{J})=0.111\ \text{J},\qquad W_{\text{load}}=1-0.111=0.889\ \text{J}$$ Averaged over the repetition period the generator launches $1\ \text{J}\times10\ \text{kHz}=10.0\ \text{kW}$, of which 8.89 kW reaches the load and 1.11 kW is dissipated back in the generator’s own internal resistance. Recomputing $W_{\text{reflected}}$ directly from the echo waveform, $(V^{-})^{2}\tau/Z_{0}=0.111\ \text{J}$, reproduces the same figure and confirms the amplitude.

For reference, the voltage that actually appears across the load at $t=T_{d}=33.33\ \mu\text{s}$ is $(1+\Gamma_{L})V^{+}=12.94$ kV — useful if the same paper is reworked from the load’s point of view, though it is not what this question asks for.

Question 1 — generator terminal voltage
Event	Terminal voltage	Starts at	Duration
Launched pulse (forward wave V⁺)	+19.42 kV	t = 0	1 µs
Quiescent interval	0 V	t = 1 µs	65.67 µs
Echo from the load (Γ_L = −1/3)	−6.472 kV	t = 2d/v_p = 66.67 µs	1 µs
Quiescent interval	0 V	t = 67.67 µs	32.33 µs
Pattern repeats	—	t = 100 µs	every 100 µs
Peak generator EMF	38.83 kV	—	—
Energy split per pulse	0.889 J to load, 0.111 J back into R_g	—	—

Topic. Transient (bounce-diagram) analysis of a pulse-driven transmission line.

Key relations. Forward-wave power $P=(V^{+})^{2}/Z_{0}$; source divider $V^{+}=EZ_{0}/(R_{g}+Z_{0})$; reflection coefficient $\Gamma=(Z-Z_{0})/(Z+Z_{0})$; one-way delay $T_{d}=d/v_{p}$; returned energy fraction $|\Gamma_{L}|^{2}$.

Why this works. A pulse shorter than the transit time never overlaps its own reflection, so the line behaves as a pure delay element rather than as a resonant structure and the terminal voltage is simply the sum of whichever waves happen to be present at the terminals at that instant. During launch only the forward wave is present, and the line looks like a resistor of $Z_{0}$ because no information about the far end has arrived yet. When the echo returns, the source-end reflection coefficient decides whether the story continues: a matched source ends it immediately, and any other source impedance would produce a geometrically decaying staircase of further echoes at 66.67 µs intervals.

Common pitfall. Placing the echo at the one-way delay $T_{d}$ instead of the round trip $2T_{d}$, and computing the returned energy from $V^{+}$ rather than from the reflected amplitude $\Gamma_{L}V^{+}$. A third trap is to treat the 377 Ω source resistance as merely “a number in the divider” and miss that it makes $\Gamma_{g}=0$, which is what limits the answer to two pulses per period.

Source. Pozar, Microwave Engineering, 4th ed., §2.1–2.3 (lossless-line waves and reflection); Hayt & Buck, Engineering Electromagnetics, 9th ed., ch. 10 (transmission-line transients and bounce diagrams).

Question 1 of 8: Pulse Train on a Mismatched Transmission Line

Question 1: Pulse Train on a Mismatched Transmission Line (20 marks)