Question 1 of 8: Terminal Voltage of a Matched Pulse Generator on a Mismatched Line

Question 1: Terminal Voltage of a Matched Pulse Generator on a Mismatched Line (equal value)

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. Source (internal) resistance $R_g=377\ \Omega$; line characteristic impedance $Z_0=377\ \Omega$ (so the source is matched to the line); load $R_L=377/2=188.5\ \Omega$; line length $\ell=10$ km; propagation velocity $v=3\times10^8$ m/s; pulse width $\tau=1\ \mu\text{s}$ (consistent with the pulse energy and the resulting terminal voltage below); pulse repetition frequency $\text{PRF}=10$ kHz (period $100\ \mu\text{s}$); energy per outgoing pulse $W=1$ J.

Find. The generator terminal voltage as a function of time — every distinct voltage level and the time interval it occupies.

Approach. Because $R_g=Z_0$, the source is matched to the line: the wave launched at $t=0$ is absorbed with zero re-reflection when it returns, so only ONE reflection (at the load) needs to be tracked; find the launched-pulse amplitude from the given energy, the round-trip time from the line's length and velocity, and confirm the pulse width is short enough that the outgoing and returning pulses never overlap.

1. Reflection coefficient at the load. $$\Gamma_L=\frac{R_L-Z_0}{R_L+Z_0}=\frac{188.5-377}{188.5+377}=\boxed{-\tfrac{1}{3}}.$$
2. Launched pulse amplitude from the pulse energy. A matched source launches a forward wave of amplitude $V^+$ that carries power $P^+=(V^+)^2/Z_0$ into the line for the pulse's $\tau$-second duration, so $W=\tau (V^+)^2/Z_0$. Solving, $$V^+=\sqrt{\frac{Z_0 W}{\tau}}=\sqrt{\frac{377\times1}{1\times10^{-6}}}=\boxed{19{,}416\ \text{V}}.$$ This is the terminal voltage from $t=0$ to $t=\tau=1\ \mu\text{s}$, while the source is actively driving and no reflection has yet returned.
3. One-way and round-trip transit time. $$T=\frac{\ell}{v}=\frac{10{,}000}{3\times10^8}=33.33\ \mu\text{s},\qquad 2T=66.67\ \mu\text{s}.$$ Since $\tau=1\ \mu\text{s}\ll 2T=66.67\ \mu\text{s}$, the outgoing pulse is long gone (terminal voltage back to zero) well before its own reflection can return — the two events never overlap. The $100\ \mu\text{s}$ pulse period also exceeds $2T$, so successive pulses' transients never overlap each other either.
4. Terminal voltage while the reflection returns. The load reflects the incident wave with $\Gamma_L=-1/3$; this reflected wave reaches the generator terminal at $t=2T=66.67\ \mu\text{s}$ and, since $R_g=Z_0$ (matched source, zero source-end reflection coefficient), is absorbed completely — the terminal voltage during this interval equals the arriving wave itself, with no further bounce generated: $$V^-=\Gamma_L V^+=\left(-\tfrac13\right)(19{,}416)=\boxed{-6{,}472\ \text{V}},$$ present from $t=2T=66.67\ \mu\text{s}$ to $t=2T+\tau=67.67\ \mu\text{s}$. Before and after each active interval (i.e. $1\ \mu\text{s}\lt t\lt 66.67\ \mu\text{s}$ and $67.67\ \mu\text{s}\lt t\lt 100\ \mu\text{s}$) the terminal voltage is zero: the source EMF is off and no wave is present at the terminal.