Question 1 of 8: Pulsed Transmission Line — Generator Terminal Voltage

Question 1: Pulsed Transmission Line — Generator Terminal Voltage (equal value)

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 source of internal (Thévenin) resistance $R_g=377\ \Omega$ emits rectangular pulses of amplitude $V_0=38.8$ kV, width $\tau=1\ \mu\text{s}$, repeating at $f_{PRF}=10$ kHz; the line has $Z_0=377\ \Omega$, $v=3\times10^8$ m/s; a resistive load $Z_L=Z_0/2=188.5\ \Omega$ sits $d=10$ km from the generator.

Find. The repeating (steady-state) waveform of the voltage at the generator terminals, with every voltage level and time interval labelled.

Approach. Because $R_g=Z_0$, the source is matched to the line: it launches one clean pulse per EMF pulse and absorbs anything that returns with no re-reflection, so this is a single-bounce bounce-diagram problem repeated once per period.

1. Pulse launched onto the line. A semi-infinite matched line looks purely resistive ($Z_0$) to the source, so the EMF divides as a resistive divider between $R_g$ and $Z_0$: $$V_1=V_0\,\frac{Z_0}{R_g+Z_0}=38.8\times\frac{377}{377+377}=\boxed{19.4\ \text{kV}}.$$ This 1 μs-wide pulse appears at the generator terminals for $0\le t<1\ \mu\text{s}$, then the terminals sit at 0 V (no source, no wave present) until a reflection returns.
2. Round-trip delay. $$t_d=\frac{d}{v}=\frac{10\times10^3}{3\times10^8}=33.33\ \mu\text{s},\qquad 2t_d=66.67\ \mu\text{s}.$$
3. Reflection at the load. $$\Gamma_L=\frac{Z_L-Z_0}{Z_L+Z_0}=\frac{188.5-377}{188.5+377}=-\frac{1}{3}.$$ The reflected pulse has amplitude $V_2=\Gamma_L V_1=-\tfrac13\times19.4=\boxed{-6.47\ \text{kV}}$ and the same 1 μs width (a lossless line preserves pulse shape).
4. Return to the generator. Because $R_g=Z_0$, the source's own reflection coefficient is $$\Gamma_g=\frac{R_g-Z_0}{R_g+Z_0}=0,$$ so when the $-6.47$ kV pulse arrives back at $t=2t_d=66.67\ \mu\text{s}$ it is absorbed completely in $R_g$ — it appears at the terminals for one pulse width ($66.67$ to $67.67\ \mu\text{s}$) and launches nothing further. There is exactly one round trip per EMF pulse; the pattern is already in steady state from the very first pulse.
5. Period check. $T_{PRF}=1/f_{PRF}=100\ \mu\text{s}$, and the round trip ($66.67\ \mu\text{s}$) is shorter than $T_{PRF}$, so the reflected pulse always returns and dies out well before the next EMF pulse fires — the four-part pattern below repeats exactly every $100\ \mu\text{s}$ with no overlap or accumulation.