Question 3 of 10: Step transient on a doubly mismatched 75 Ω line

Question 3: Step transient on a doubly mismatched 75 Ω line

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 20 V step is applied at $t = 0$ to a 3 m lossless 75 Ω line whose one-way transit time is 20 ns, fed from 50 Ω and terminated in 675 Ω (Figure 3).

Find. The phase velocity on the line, and the voltage waveform at the mid-span probe over the first 100 ns, with every step time and level identified.

[Figure not reproduced: Figure 3 (redrawn). A 20 V step drives a 75 Ω line through a 50 Ω source into a 675 Ω load; the one-way transit time is T = 20 ns. See the official exam paper.]

Approach. Get $v_p$ from length over transit time, then run a bounce diagram: the probe sits at mid-span, so it sees each forward wave at $T/2$ into that transit and each backward wave at $3T/2$, giving one step every 20 ns.

1. (a) Phase velocity from the transit time. The “transient time through the line” is the one-way delay, so $$\boxed{v_p = \frac{\ell}{T} = \frac{3\ \text{m}}{20\times10^{-9}\ \text{s}} = 1.5\times10^{8}\ \text{m/s}}$$ which is exactly $c/2$; the implied effective permittivity is $\epsilon_{r,\text{eff}} = (c/v_p)^2 = 4$, a plausible solid polyethylene-loaded cable.
2. Set up the reflection coefficients. At the resistive terminations $$\Gamma_L = \frac{R_L - Z_0}{R_L + Z_0} = \frac{675 - 75}{750} = 0.80, \qquad \Gamma_g = \frac{R_g - Z_0}{R_g + Z_0} = \frac{50 - 75}{125} = -0.20$$ Both are non-zero, so the line is doubly mismatched and the transient is an infinite — but rapidly decaying — staircase of ratio $\Gamma_L\Gamma_g = -0.16$ per round trip.
3. Launch the first wave. At $t = 0^+$ the source sees the line as a plain $Z_0$ resistor, because no reflection has had time to return: $$V_1^+ = v_g\,\frac{Z_0}{R_g + Z_0} = 20\left(\frac{75}{125}\right) = 12.0\ \text{V}$$
4. Time the arrivals at the mid-span probe. The probe is $1.5$ m from the source, i.e. half a transit, so the $k$-th forward wave reaches it at $t = T/2 + 2kT = 10 + 40k$ ns and the $k$-th backward wave (which has travelled $2\ell - z = 4.5$ m) at $t = 3T/2 + 2kT = 30 + 40k$ ns. Within the first 100 ns that gives discontinuities at $10,\ 30,\ 50,\ 70$ and $90$ ns — one every 20 ns, which is the signature of a probe exactly at mid-span.
5. Accumulate the wave amplitudes. Successive waves are $V_1^+ = 12.000$ V, $V_1^- = \Gamma_L V_1^+ = 9.600$ V, $V_2^+ = \Gamma_g V_1^- = -1.920$ V, $V_2^- = \Gamma_L V_2^+ = -1.536$ V and $V_3^+ = \Gamma_g V_2^- = 0.3072$ V. The probe voltage is the running sum, so $$\boxed{\begin{aligned} v(1.5\ \text{m}, t) = &\ 12.000\ \text{V} \ (10 \le t \lt 30\ \text{ns}) \\ &\ 21.600\ \text{V}\ (30 \le t \lt 50) \\ &\ 19.680\ \text{V}\ (50 \le t \lt 70) \\ &\ 18.144\ \text{V}\ (70 \le t \lt 90) \\ &\ 18.451\ \text{V}\ (90 \le t \lt 130) \end{aligned}}$$ Note the waveform overshoots to 21.6 V before settling: the large positive $\Gamma_L$ nearly doubles the first arrival, and the negative $\Gamma_g$ then pulls it back.
6. Check the final value against the d.c. circuit. A lossless line is simply a wire at d.c., so the steady state must be the resistive divider $$v(\infty) = v_g\,\frac{R_L}{R_g + R_L} = 20\left(\frac{675}{725}\right) = 18.621\ \text{V}$$ Summing the bounce series to 4 µs reproduces 18.621 V, so the staircase is converging on the right value and both reflection coefficients are correct. The remaining error at 90 ns is under 1 %.