Question 1 of 7: Standing-wave measurement on a mismatched line
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 16-Elec-A7 Electromagnetics, December 2019 — 3 hours, closed book, approved calculator only. Seven questions of equal value (20 marks each); the paper states that five completed questions constitute a complete paper. All seven are solved here, because the set is intended as a study resource.
Reference texts (22-Elec-A7 Electromagnetics). F. T. Ulaby & U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; M. N. O. Sadiku, Elements of Electromagnetics, 7th ed.; W. H. Hayt & J. A. Buck, Engineering Electromagnetics, 9th ed.; D. M. Pozar, Microwave Engineering, 4th ed.; C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed.
Constants used throughout. $c = 3.00\times 10^{8}\ \text{m/s}$, $\varepsilon_0 = 8.854\times 10^{-12}\ \text{F/m}$, $\mu_0 = 4\pi\times 10^{-7}\ \text{H/m}$, $\eta_0 = 376.7\ \Omega$. Where a boundary case could turn on the choice between $c = 3\times 10^{8}$ and $c = 2.998\times 10^{8}$ m/s, both readings are stated.
Figure reading. In Figure 1(b) the maxima sit at z = 0, 1.5 and 3.0 m and the minima at z = 0.75 and 2.25 m, so the wavelength is 3.00 m. Three independent checks confirm that reading — the quarter-wave maximum-to-minimum spacing, the fact that the stated 3.75 m line is then exactly 1.25 wavelengths, and the fact that z = l then lands on a voltage minimum where the quarter-wave transformer result must give $Z_{\text{in}} = Z_0/S$.
Question 1: Standing-wave measurement on a mismatched line (20 marks)
Given. A lossless 50 Ω line of length 3.75 m is driven at 50 MHz through a 50 Ω source resistance, and the measured voltage envelope of Figure 1(b) is read from the printed figure as follows.
Given data (z is measured from the load)
Quantity
Symbol
Value
Characteristic impedance
$Z_0$
$50\ \Omega$
Source frequency
$f$
$50\ \text{MHz}$
Line length
$l$
$3.75\ \text{m}$
Envelope maximum
$V_{\max}$
$1.50\ \text{V}$ at $z = 0,\ 1.5,\ 3.0\ \text{m}$
Envelope minimum
$V_{\min}$
$0.50\ \text{V}$ at $z = 0.75,\ 2.25,\ 3.75\ \text{m}$
Find. The phase velocity on the line, the standing-wave ratio, the load impedance, and the impedance the generator sees at the input plane $z = l$.
[Figure not reproduced: Figure 1(a) — the terminated line, redrawn from the printed figure. The coordinate z is measured back from the load. See the official exam paper.]
[Figure not reproduced: Figure 1(b) — the measured voltage envelope, redrawn from the printed figure. Successive maxima are half a wavelength apart and the load plane z = 0 sits on a maximum. See the official exam paper.]
Approach. Read the wavelength from the spatial period of the envelope, convert it to a phase velocity, get $|\Gamma|$ from the ratio of the envelope extremes, fix the phase of $\Gamma$ from the position of the maximum relative to the load, and transform the resulting $Z_L$ through the line.
Extract the wavelength from the envelope. On a lossless line the envelope repeats every half wavelength, and a maximum and its neighbouring minimum are a quarter wavelength apart. The printed figure gives a maximum at $z = 0$ and the first minimum at $z = 0.75\ \text{m}$, so $$\lambda = 4\,(0.75 - 0) = 3.00\ \text{m}.$$ The maximum-to-maximum spacing confirms it independently: $1.50\ \text{m} = \lambda/2$.
Convert the wavelength to a phase velocity. The signal speed on the line follows from $v_p = f\lambda$, with $f = 50\ \text{MHz}$: $$v_p = (50\times 10^{6})(3.00) = \boxed{1.50\times 10^{8}\ \text{m/s}}$$ which is exactly $c/2$, so the line's effective relative permittivity is $\varepsilon_{r,\text{eff}} = (c/v_p)^2 = 4.0$ — entirely reasonable for a solid-dielectric feeder.
Read the standing-wave ratio straight off the envelope. By definition $S$ is the ratio of the envelope extremes: $$S = \frac{V_{\max}}{V_{\min}} = \frac{1.50}{0.50} = \boxed{3.00}$$ (dimensionless, and $S \ge 1$ always).
Convert $S$ to a reflection-coefficient magnitude. Inverting $S = (1+|\Gamma|)/(1-|\Gamma|)$ gives $$|\Gamma_L| = \frac{S-1}{S+1} = \frac{3-1}{3+1} = 0.500 .$$ The incident wave amplitude then follows as $|V^{+}| = V_{\max}/(1+|\Gamma_L|) = 1.50/1.5 = 1.00\ \text{V}$, and this reproduces $V_{\min} = |V^{+}|(1-|\Gamma_L|) = 0.50\ \text{V}$, so the two envelope readings are mutually consistent.
Fix the phase of $\Gamma_L$ from the position of the maximum. A voltage maximum occurs where the incident and reflected waves add in phase, i.e. where $\theta_\Gamma - 2\beta z = 0$. The printed figure puts a maximum at the load plane itself ($z = 0$), so $$\theta_\Gamma = 0 \quad\Longrightarrow\quad \Gamma_L = +0.500 \angle 0^{\circ}.$$ A purely real, positive $\Gamma_L$ means the load is purely resistive and larger than $Z_0$ — despite the question calling it “an unknown complex load”, the measurement itself rules out any reactive part.
Invert the reflection coefficient to get the load. With $\Gamma_L = (Z_L - Z_0)/(Z_L + Z_0)$, $$Z_L = Z_0\,\frac{1+\Gamma_L}{1-\Gamma_L} = 50\,\frac{1.500}{0.500} = \boxed{150\ \Omega\ (\text{purely resistive})}$$ which is the familiar shortcut $Z_L = S Z_0$ that applies only when the load sits at a voltage maximum.
Measure the line length in wavelengths. With $\lambda = 3.00\ \text{m}$, $$\frac{l}{\lambda} = \frac{3.75}{3.00} = 1.25 \quad\Longrightarrow\quad \beta l = 2\pi(1.25) = 2.5\pi\ \text{rad} \equiv 90^{\circ}.$$ Because impedance repeats every half wavelength, the extra full wavelength does nothing: electrically the line is a quarter-wave transformer.
Transform the load to the input plane. At $\beta l = 90^{\circ}$ the general expression $Z_{\text{in}} = Z_0(Z_L + jZ_0\tan\beta l)/(Z_0 + jZ_L\tan\beta l)$ collapses to the quarter-wave result $$Z_{\text{in}} = \frac{Z_0^{2}}{Z_L} = \frac{50^{2}}{150} = \boxed{16.67\ \Omega\ (\text{purely resistive})}.$$ Two independent checks agree: $z = l = 3.75\ \text{m}$ is one of the listed voltage minima, and at a minimum the impedance must be $Z_0/S = 50/3 = 16.67\ \Omega$; and evaluating the full formula numerically just short of the singular point returns $16.667 - j0.000\ \Omega$.
