Question 3 of 7: Step transient on a doubly mismatched line

Question 3: Step transient 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 10 V step is applied at $t = 0$ to a 2 m lossless line through a 50 Ω generator resistance; the line is 100 Ω and is terminated in 25 Ω.

Find. The one-way transit time, and labelled snapshots of $v(z)$ at $t = 4$ ns and $t = 14$ ns.

[Figure not reproduced: Figure 3 — the transient circuit, redrawn from the printed figure. z runs from the generator terminals (z = 0) to the load (z = l = 2 m). See the official exam paper.]

Approach. Get the transit time, then the initial launched step from the resistive divider the generator sees against $Z_0$, then the two reflection coefficients, and finally place the travelling edge at $v_p t$ for each instant.

1. One-way transit time. The step travels the line at the stated speed, so $$T = \frac{l}{v_p} = \frac{2}{2\times 10^{8}} = \boxed{10\ \text{ns}}.$$ This single number governs the whole answer: $t = 4$ ns is before the wave reaches the load, and $t = 14$ ns is 4 ns after it did.
2. Size the launched wave. At $t = 0^{+}$ the generator cannot see the load — a length of lossless line that has not yet heard from its far end presents exactly $Z_0$. The launched step is therefore a plain resistive divider: $$V_1^{+} = V_g\,\frac{Z_0}{R_g + Z_0} = 10\,\frac{100}{150} = 6.67\ \text{V}.$$
3. Compute both reflection coefficients. $$\Gamma_L = \frac{R_L - Z_0}{R_L + Z_0} = \frac{25-100}{25+100} = -0.600, \qquad \Gamma_g = \frac{R_g - Z_0}{R_g + Z_0} = \frac{50-100}{50+100} = -0.333.$$ Both ends are mismatched, so the transient is a decaying staircase rather than a single step; only the first two terms are needed for the times asked.
4. Snapshot at $t = 4$ ns. Since $4\ \text{ns} \lt T$, only the forward wave exists. It has advanced $$z_{\text{edge}} = v_p t = (2\times 10^{8})(4\times 10^{-9}) = 0.80\ \text{m},$$ so $$\boxed{v(z,\,4\ \text{ns}) = 6.67\ \text{V for } 0 \le z \le 0.80\ \text{m},\quad 0\ \text{V beyond}}$$ with the leading edge at $z = 0.80$ m travelling towards the load.
5. Find the reflected wave. The forward step reaches the load at $t = T = 10$ ns and immediately launches $$V_1^{-} = \Gamma_L V_1^{+} = (-0.600)(6.67) = -4.00\ \text{V}$$ back towards the generator. The load voltage at that instant jumps to $(1+\Gamma_L)V_1^{+} = 2.67\ \text{V}$.
6. Snapshot at $t = 14$ ns. In the 4 ns since reflection the return wave has run $(2\times 10^{8})(4\times 10^{-9}) = 0.80$ m back from the load, so its falling edge sits at $$z_{\text{edge}} = l - 0.80 = 1.20\ \text{m}.$$ Behind the edge (nearer the generator) only the forward wave has arrived; ahead of it the two waves superpose: $$\boxed{v(z,\,14\ \text{ns}) = 6.67\ \text{V for } 0 \le z \lt 1.20\ \text{m};\ \ 2.67\ \text{V for } 1.20 \lt z \le 2\ \text{m}}$$ with the falling edge at $z = 1.20$ m travelling towards the source.
7. Cross-check the eventual steady state. A lossless line is just wire at DC, so the final load voltage must be the plain divider $V_\infty = 10\,(25)/(50+25) = 3.33\ \text{V}$. Summing the whole bounce series independently, $$V_\infty = \frac{V_1^{+}(1+\Gamma_L)}{1 - \Gamma_L\Gamma_g} = \frac{6.67(0.400)}{1 - 0.200} = 3.33\ \text{V},$$ which agrees exactly and confirms both reflection coefficients.