Question 1 of 7: Terminated Transmission Line — Standing-Wave Measurement

Question 1: Terminated Transmission Line — Standing-Wave Measurement (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 lossless line of characteristic impedance $Z_0=50\ \Omega$ and length $l=3.75$ m is driven at $f=50$ MHz and terminated in an unknown load $Z_L$. The measured standing-wave envelope $|V(z)|$ (z measured from the load) read off Figure 1(b) has $V_{max}=1.50$ V at $z=0$, $1.5$ m and $3.0$ m, and $V_{min}=0.50$ V at $z=0.75$ m and $2.25$ m; the trace ends at $z=l=3.75$ m on a minimum.

[Figure not reproduced: Standing-wave envelope |V(z)| as printed in Figure 1(b). Maxima (1.50 V) at z = 0, 1.5, 3.0 m and minima (0.50 V) at z = 0.75, 2.25 m, so λ/2 = 1.5 m and λ = 3 m; the voltage is a MAXIMUM at the load and a minimum at the source end z = l = 3.75 m. See the official exam paper.]

Find. The propagation velocity $v$, the standing-wave ratio $S$, the load impedance $Z_L$, and the input impedance $Z_{in}$ seen by the source at $z=l$.

Approach. Read the wavelength off the maxima/minima spacing to get $v$; take the max/min ratio for $S$; use the location of the first minimum (equivalently, the maximum sitting at the load) to fix the phase of $\Gamma_L$ and $S$ to fix its magnitude, then back out $Z_L$; finally transform $Z_L$ along the line by length $l$ for $Z_{in}$.

1. Part (a) — Speed of light along the line. Successive minima (0.75 m, 2.25 m) — and successive maxima (0, 1.5, 3.0 m) — are separated by $\lambda/2$, so $$\lambda = 2\times(2.25-0.75) = 3\ \text{m}, \qquad v = f\lambda = (50\times10^6)(3) = \boxed{1.5\times10^8\ \text{m/s}}.$$ (This is $c/2$, consistent with a line dielectric of $\varepsilon_r=(c/v)^2=4$, though that value isn't asked for here.)
2. Part (b) — Standing wave ratio. $$S = \frac{V_{max}}{V_{min}} = \frac{1.50}{0.50} = \boxed{3.0}.$$
3. Part (c) — Reflection coefficient magnitude and phase. From $S$, $$|\Gamma_L| = \frac{S-1}{S+1} = \frac{2}{4} = 0.5.$$ With $\beta = 2\pi/\lambda = 2\pi/3\ \text{rad/m}$, the first voltage minimum from the load occurs where the total round-trip phase is $-\pi$: $\theta_L - 2\beta d_{min} = -\pi$, so $$\theta_L = 2\beta d_{min} - \pi = 2\left(\frac{2\pi}{3}\right)(0.75) - \pi = \pi-\pi = 0.$$ The graph confirms this directly: $|V|$ is a maximum (1.50 V) right at the load, $z=0$, which happens only when $\Gamma_L$ is real and positive. So $\Gamma_L = 0.5\angle0^\circ = +0.5$.
4. Load impedance. $$Z_L = Z_0\,\frac{1+\Gamma_L}{1-\Gamma_L} = 50\times\frac{1.5}{0.5} = \boxed{150\ \Omega\ (=150+j0\ \Omega)}.$$ Check: a voltage maximum at the load means $Z_L$ is purely resistive and larger than $Z_0$, with $Z_L = S\,Z_0 = 3\times50 = 150\ \Omega$. The “unknown complex load” therefore turns out to have zero reactance — the measured pattern, not the wording, decides this.
5. Part (d) — Input impedance at the source, $z=l=3.75$ m. $$\beta l = \frac{2\pi}{3}(3.75) = 2.5\pi\ \text{rad} \equiv 90^\circ \pmod{360^\circ},$$ i.e. $l = 1.25\lambda$, an odd number of quarter-wavelengths. Here $\tan\beta l\to\infty$, so the general formula $Z_{in}=Z_0\,(Z_L+jZ_0\tan\beta l)/(Z_0+jZ_L\tan\beta l)$ reduces to the quarter-wave-transformer result: $$Z_{in}=\frac{Z_0^2}{Z_L}=\frac{50^2}{150} = \boxed{16.7\ \Omega\ (\text{purely resistive})}.$$ Check against Figure 1(b): the trace ends on a voltage MINIMUM at $z=l$, where the impedance must be real and equal to $Z_0/S = 50/3 = 16.7\ \Omega$ — the same value.