Question 3 of 7: Step Response on a Mismatched Transmission Line
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
17-Phys-A3, Electromagnetics — National Exam, December 2019. 3-hour closed-book exam (Casio or Sharp approved calculators only); any FIVE of the seven questions constitute a complete paper and only the first five as they appear in a candidate's answer book are marked, each of equal value, with full justification required for marks. All seven printed questions are solved below as a complete study resource.
Reference texts: Sadiku, Elements of Electromagnetics (7th ed.) — plane waves and boundaries, transmission lines, waveguides, Gauss's and Faraday's laws; Hayt & Buck, Engineering Electromagnetics (9th ed.) — transmission-line transients and bounce diagrams; Pozar, Microwave Engineering (4th ed.) — transmission-line theory, rectangular waveguide cutoff; Balanis, Antenna Theory (4th ed.) — wave polarization and radiation fundamentals.
Question 3: Step Response on a Mismatched Transmission Line (20 marks)
Given. Line length $L=2$ m, $v=2\times10^8$ m/s, $Z_0=100\ \Omega$; source resistance $R_g=50\ \Omega$ in series with $v_g(t)=10\,u(t)$ V; load resistance $R_L=25\ \Omega$; $z=0$ at the generator, $z=L$ at the load.
Given data
Quantity
Symbol
Value
Line length
$L$
2 m
Propagation velocity
$v$
$2\times10^8$ m/s
Characteristic impedance
$Z_0$
100 Ω
Source resistance / step
$R_g/v_g$
50 Ω / 10 V step
Load resistance
$R_L$
25 Ω
Find. The one-way transit time $T$, and the voltage-vs-position snapshot at $t=4$ ns and $t=14$ ns, with the wavefront position and direction of travel marked.
Approach. This is a single-step bounce-diagram problem: compute the launched step amplitude via the resistive divider at the source, find $T$, then track the leading edge (and, after the first reflection, the reflected edge) as it propagates at speed $v$.
Part (a) — One-way transit time.
$$T=\frac{L}{v}=\frac{2}{2\times10^8}=\boxed{10\ \text{ns}}.$$
Part (b) — Launched step amplitude. At $t=0^+$ the line looks purely resistive ($Z_0$) to the source, so
$$V_1=v_g\,\frac{Z_0}{R_g+Z_0}=10\times\frac{100}{150}=\boxed{6.67\ \text{V}}.$$
This step propagates toward the load at speed $v$, starting at $z=0$.
Snapshot at $t=4$ ns (before the wave reaches the load, since $t\ <\ T$). The leading edge has travelled
$$z_{front}=vt=(2\times10^8)(4\times10^{-9})=0.8\ \text{m}.$$
So $V(z)=6.67$ V for $0\le z<0.8$ m (behind the front, step already arrived) and $V(z)=0$ for $z$ between $0.8$ m and $2$ m (undisturbed line ahead of the front). The edge is travelling toward the load.
Reflection at the load ($t=T=10$ ns).
$$\Gamma_L=\frac{R_L-Z_0}{R_L+Z_0}=\frac{25-100}{125}=-0.6,$$
so a reflected step of amplitude $V_2=\Gamma_L V_1=-0.6\times6.67=\boxed{-4.00\ \text{V}}$ launches from the load back toward the source at $t=10$ ns.
Snapshot at $t=14$ ns (4 ns after the load reflection, since $t-T=4$ ns). The reflected edge has travelled $v(t-T)=0.8$ m back from the load, i.e. it now sits at
$$z_{refl}=L-v(t-T)=2-0.8=\boxed{1.2\ \text{m}}\ \text{(from the generator)}.$$
Behind this edge (toward the load, $z$ between $1.2$ m and $2$ m) both the incident and reflected steps are present:
$$V(z)=V_1+V_2=6.67-4.00=\boxed{2.67\ \text{V}}.$$
Ahead of the reflected edge (toward the source, $z<1.2$ m) only the original incident step has arrived: $V(z)=6.67$ V. The reflected edge is travelling toward the source; it has not yet reached $z=0$ (that happens at $t=20$ ns), so no further reflection has occurred by $t=14$ ns.
Snapshot at $t=4$ ns: incident step (6.67 V) has advanced to $z=0.8$ m, travelling toward the load.
Snapshot at $t=14$ ns: the load-reflected step ($-4.00$ V) has travelled back to $z=1.2$ m, giving 2.67 V behind it and 6.67 V still ahead of it (toward the source).
