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

Question 1: Pulse Train on a Doubly Mismatched 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 rectangular pulse generator drives a resistive load through a lossless line that is mismatched at both ends, so every pulse launched on the line rings back and forth.

Find. The shortest line length whose steady-state load waveform is again a single train of $1\ \mu$s pulses at 20 kHz, and the amplitude of those pulses.

Approach. Find the reflection coefficients at both ends, follow the launched pulse round the bounce diagram to get the train of echoes the load actually sees, then choose the delay that makes successive echoes land exactly on the pulses of later generator cycles, and sum the resulting geometric series.

1. Reflection coefficients at the two ends. The mismatch at each termination is measured by $\Gamma = (Z-Z_0)/(Z+Z_0)$, so$$\Gamma_L = \frac{25-50}{25+50} = -\tfrac{1}{3},\qquad \Gamma_g = \frac{100-50}{100+50} = +\tfrac{1}{3}.$$Neither end absorbs its incident wave, so the pulse survives many round trips.
2. Amplitude launched onto the line. At the instant the pulse starts, the generator sees the line as a pure $50\ \Omega$ resistance, so the incident wave is the simple divider$$V^{+} = E\,\frac{Z_0}{R_g+Z_0} = 15\times\frac{50}{150} = 5.00\ \text{V}.$$
3. The echo train the load sees. Writing $T_d = d/v_p$ for the one-way delay, the first pulse reaches the load at $t = T_d$ with amplitude $(1+\Gamma_L)V^{+}$; it then returns to the generator, re-reflects, and comes back, so successive arrivals are spaced by the round trip $2T_d$ and are scaled by $\Gamma_L\Gamma_g$ each time:$$v_L\big(T_d + 2nT_d\big) = (1+\Gamma_L)V^{+}\,(\Gamma_L\Gamma_g)^{n} = 3.333\left(-\tfrac{1}{9}\right)^{n}\ \text{V}.$$
4. Timing condition for an unaltered waveform. Left alone, those echoes would appear as extra pulses between the wanted ones and the load PRF would no longer be 20 kHz. The waveform keeps its shape only if every echo falls exactly on top of a pulse belonging to a later generator cycle, which requires the round-trip delay to be a whole number of repetition periods:$$2T_d = n\,T_{\text{PRF}},\qquad n = 1,2,3,\dots$$The shortest line is the case $n = 1$, giving $\boxed{T_d = 25\ \mu\text{s}}$.
5. Line length. Converting the delay to a physical length with the stated propagation velocity,$$d = v_p T_d = (2\times10^{8})(25\times10^{-6}) = 5\,000\ \text{m},$$$$\boxed{d_{\min} = 5.0\ \text{km}}$$
6. Steady-state pulse amplitude. With every echo superposed on a later pulse, the height of each pulse in the steady state is the sum of the whole geometric series of arrivals:$$V_L = \frac{(1+\Gamma_L)V^{+}}{1-\Gamma_L\Gamma_g} = \frac{3.3333}{1-(-1/9)} = \frac{3.3333}{10/9},$$$$\boxed{V_L = 3.00\ \text{V (1 }\mu\text{s pulses at 20 kHz)}}$$
7. Independent check. When all the echoes coincide the line no longer stores any net energy between pulses, so during a pulse the circuit behaves as if the line were a plain piece of wire. The DC divider then gives$$V_L = E\,\frac{R_L}{R_g+R_L} = 15\times\frac{25}{125} = 3.00\ \text{V},$$which reproduces the series sum exactly and confirms both the reflection algebra and the timing argument.