Step Response on a Mismatched Transmission Line: 17-Phys-A3 Electromagnetics, Undated paper, Q3

Question 3: Step Response 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.

Given data
Quantity	Symbol	Value
Source resistance	$R_g$	50 Ω
Characteristic impedance	$Z_0$	75 Ω
Load resistance	$Z_L$	675 Ω
Line length	$l$	3 m
One-way transit time	$T$	20 ns
Source step	$v_g(t)$	$20\,u(t)$ V
Observation point	$z$	-1.5 m (line midpoint)

Find. $v=l/T$; the voltage waveform $v(-1.5\text{m},t)$ for $0\le t\le100$ ns, with every discontinuity time and level.

Approach. Compute the generator and load reflection coefficients; build the bounce sequence of forward/backward wave amplitudes launched every $T$ seconds; since $z=-1.5$ m is the exact midpoint, every wave front reaches it $T/2$ after leaving either end, so add up arrivals in time order up to $t=100$ ns.

Part (a) — Speed of propagation. $$v=\frac{l}{T}=\frac{3}{20\times10^{-9}}=\boxed{1.5\times10^8\ \text{m/s}}.$$
Reflection coefficients. $$\Gamma_g=\frac{R_g-Z_0}{R_g+Z_0}=\frac{50-75}{125}=-0.2,\qquad \Gamma_L=\frac{Z_L-Z_0}{Z_L+Z_0}=\frac{600}{750}=0.8.$$
Bounce sequence. The first wave launched at $t=0$ has amplitude $V_1^+=v_g\dfrac{Z_0}{R_g+Z_0}=20\times\dfrac{75}{125}=12$ V. Each subsequent wave is the previous one times the reflection coefficient at the end it just reached: $$\begin{aligned}V_1^+&=12, & V_1^-&=\Gamma_LV_1^+=9.6,\\ V_2^+&=\Gamma_gV_1^-=-1.92, & V_2^-&=\Gamma_LV_2^+=-1.536,\\ V_3^+&=\Gamma_gV_2^-=0.3072,&&\ldots\end{aligned}$$
Part (b) — Waveform at the midpoint. Every forward wave $V_n^+$ (launched from the source at $t=2(n-1)T$) reaches $z=-1.5$ m at $t=2(n-1)T+T/2$; every backward wave $V_n^-$ (launched from the load at $t=(2n-1)T$) reaches it at $t=(2n-1)T+T/2$. Summing arrivals in order up to $t=100$ ns: $$\boxed{v(-1.5\text{m},t)=\begin{cases}0, & 0\le t<10\text{ ns}\\ 12.00\ \text{V}, & 10\le t<30\text{ ns}\\ 21.60\ \text{V}, & 30\le t<50\text{ ns}\\ 19.68\ \text{V}, & 50\le t<70\text{ ns}\\ 18.144\ \text{V}, & 70\le t<90\text{ ns}\\ 18.451\ \text{V}, & 90\le t\le100\text{ ns}\end{cases}}$$ The sequence is oscillating in towards the DC steady state $v_g Z_L/(R_g+Z_L)=20(675)/725=18.62$ V, consistent with the values above straddling 18.62 V with shrinking amplitude.

Voltage at the line midpoint $z=-1.5$ m versus time. Each step corresponds to a wavefront (launched every $T=20$ ns from one end) arriving at the midpoint $T/2=10$ ns later.

Final results
Quantity	Value
Speed of propagation, $v$	$1.5\times10^8$ m/s
$v(-1.5\text{m},t)$, $t\in[0,10)$ ns	0 V
$t\in[10,30)$ ns	12.00 V
$t\in[30,50)$ ns	21.60 V
$t\in[50,70)$ ns	19.68 V
$t\in[70,90)$ ns	18.144 V
$t\in[90,100]$ ns	18.451 V

Topic: Bounce-diagram (multiple-reflection) analysis of a step response on a lossless line.

Key relations: $\Gamma=(Z-Z_0)/(Z+Z_0)$; $V_1^+=v_gZ_0/(R_g+Z_0)$; each bounce multiplies the previous wave by the reflection coefficient at the end just reached; steady state $=v_gZ_L/(R_g+Z_L)$.

Why this works: A lossless line only delays and (at each end) partially reflects a step; because $z=-1.5$m is the exact midpoint, EVERY wavefront (whichever direction) arrives there exactly $T/2$ after leaving its end, which is what makes tracking a single time-ordered list of arrivals straightforward instead of needing a full bounce diagram grid.

Common pitfall: Forgetting that $\Gamma_g$ uses the SOURCE resistance $R_g$ (not $Z_0$) in the numerator, and using the wrong transit time for an off-midpoint observation point (here it happens to be $T/2$ from both ends, which is special to this position).

Source: Hayt & Buck, Engineering Electromagnetics, Ch. 11 (transmission-line transients, bounce diagrams).

Question 3 of 10: Step Response on a Mismatched Transmission Line

Question 3: Step Response on a Mismatched Transmission Line (20 marks)