22-Elec-B9 Electromagnetic Field, Transmission Lines, Antennas, and Radiation · May 2016

Question 3 of 8: Load Impedance and Generator EMF from the Standing-Wave Pattern

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, May 2016 — 07-Elec-B9, Electromagnetic Field, Transmission Lines, Antennas, and Radiation. Three hours; closed book; any non-communicating calculator permitted. Eight questions, each of equal value (20 marks); the paper states that five questions constitute a complete exam paper and that the first five appearing in the answer book are the ones marked. The only constants printed on the cover page are the free-space permittivity and permeability, and a Smith chart is bound in as page 4 for use with Question 3. All eight questions are worked here, because this set is a study resource rather than a three-hour sitting.

Reference texts. D. M. Pozar, Microwave Engineering (transmission lines, the Smith chart, waveguides, resonators); M. N. O. Sadiku, Elements of Electromagnetics (Maxwell's equations, plane waves, guided waves); W. H. Hayt and J. A. Buck, Engineering Electromagnetics (field theory and wave propagation); C. A. Balanis, Antenna Theory: Analysis and Design (short elements, arrays, radiation resistance); F. T. Ulaby, Fundamentals of Applied Electromagnetics (radar, link budgets, applied propagation).

Units and constants. The paper works in SI throughout. Where a question supplies a rounded propagation velocity (for example 3×10⁸ m/s) that value is used as given; where it does not, the exact free-space constants are used, giving $c = 2.998\times10^{8}\ \text{m/s}$ and $\eta_0 = 376.7\ \Omega$.

Question 3: Load Impedance and Generator EMF from the Standing-Wave Pattern (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 500 MHz source of internal resistance $R_g = 50\ \Omega$ feeds an unknown load through $\ell = 25\ \text{cm}$ of lossless $Z_0 = 50\ \Omega$ line on which $v = 2\times10^{8}\ \text{m/s}$. The measured standing-wave ratio is $S = 2$, the first voltage minimum lies $d_{\min} = 15\ \text{cm}$ back from the load terminals, and the load absorbs $P_L = 10\ \text{W}$.

Find. The load impedance $Z_L$, and the magnitude of the generator EMF that produces 10 W in it.

Standing-wave envelope on the 25 cm line. S = 2 sets the ratio of the dashed bounds; the minimum 15 cm from the load fixes the phase of Γ, and the generator end (d = 25 cm) lands on a maximum, where Z is real and equal to S·Z0 = 100 Ω.

Approach. The standing-wave ratio fixes the magnitude of the reflection coefficient and the position of the minimum fixes its phase, which together give $Z_L$; rotating the same reflection coefficient through the line length gives $Z_{\text{in}}$, and because the line is lossless the 10 W must also cross the input plane, which sizes the EMF.

Convert the wavelength first — every angle depends on it. $$\lambda = \frac{v}{f} = \frac{2\times10^{8}}{500\times10^{6}} = 0.400\ \text{m} = 40\ \text{cm}.$$ Two ratios are worth writing down immediately, because they are what the setter has engineered: $d_{\min}/\lambda = 15/40 = 0.375$ and $\ell/\lambda = 25/40 = 0.625$. Both are exact eighths of a wavelength, which is the tell that the arithmetic will close cleanly.
Get the magnitude of the reflection coefficient from the SWR. $$|\Gamma| = \frac{S - 1}{S + 1} = \frac{2 - 1}{2 + 1} = \frac{1}{3} = 0.3333.$$
Get its phase from the position of the minimum. At a voltage minimum the incident and reflected waves are exactly out of phase, so the local reflection coefficient is real and negative: $\Gamma(d_{\min}) = -|\Gamma|$, an angle of $180^\circ$. Rotating from that plane back to the load (towards the load, so the phase advances) through $d_{\min}$, $$\Gamma_L = \Gamma(d_{\min})\,e^{+j2\beta d_{\min}}, \qquad 2\beta d_{\min} = 2\left(\frac{2\pi}{0.40}\right)(0.15) = 4.712\ \text{rad} = 270^\circ.$$ Hence the angle of $\Gamma_L$ is $180^\circ + 270^\circ = 450^\circ \equiv 90^\circ$, and $$\Gamma_L = \tfrac{1}{3}\,e^{\,j90^\circ} = j\,0.3333.$$ On the Smith chart this is the same operation: enter at the $S = 2$ circle where it crosses the real axis on the left (the voltage-minimum point, $z = 0.5$), then move $0.375\lambda$ on the "wavelengths towards load" scale.
Convert to impedance. $$z_L = \frac{1 + \Gamma_L}{1 - \Gamma_L} = \frac{1 + j0.3333}{1 - j0.3333} = 0.800 + j0.600,$$ $$\boxed{Z_L = Z_0\,z_L = 40 + j30\ \Omega}$$ The load is inductive, with $|Z_L| = 50\ \Omega$ — it happens to sit on the $|z| = 1$ circle, which is a coincidence of this data set and not a general result. As a check, a load of $40 + j30$ does give $|\Gamma| = |{-10 + j30}|/|90 + j30| = 31.62/94.87 = 1/3$, as required.
Rotate the same reflection coefficient to the generator end. Moving $0.625\lambda$ towards the generator retards the phase by $2\beta\ell = 450^\circ \equiv 90^\circ$, so $$\Gamma_{\text{in}} = \Gamma_L\,e^{-j2\beta\ell} = \tfrac{1}{3}e^{\,j90^\circ}e^{-j90^\circ} = +\tfrac{1}{3},$$ which is real and positive. The input therefore sits at the voltage maximum of the pattern: $$\boxed{Z_{\text{in}} = Z_0\,\frac{1 + \tfrac{1}{3}}{1 - \tfrac{1}{3}} = 2 Z_0 = 100\ \Omega \ \text{(purely resistive)}}$$ Equivalently $Z_{\text{in}} = S\,Z_0$, which is exactly what a voltage maximum must give.
Size the EMF from the power. The line is lossless, so the 10 W absorbed by the load also crosses the input plane. With $Z_{\text{in}}$ purely resistive, the rms input current follows directly, and the EMF is that current through the series combination of source resistance and input impedance: $$I_{\text{in}} = \sqrt{\frac{P}{Z_{\text{in}}}} = \sqrt{\frac{10}{100}} = 0.3162\ \text{A (rms)},$$ $$|E| = I_{\text{in}}\left(R_g + Z_{\text{in}}\right) = (0.3162)(50 + 100)$$ $$\boxed{|E| = 47.43\ \text{V rms} = 67.08\ \text{V peak}}$$ The input voltage itself is $V_{\text{in}} = I_{\text{in}}Z_{\text{in}} = 31.62\ \text{V rms}$, and $I_{\text{in}}^{2}Z_{\text{in}} = 10\ \text{W}$ closes the loop.
Note what the source is not doing. With $Z_{\text{in}} = 100\ \Omega$ against $R_g = 50\ \Omega$ the generator is not conjugately matched, so it is not delivering its maximum available power: that would be $|E|^{2}/(4R_g) = 11.25\ \text{W}$ into a $50\ \Omega$ input. The 10 W actually delivered is 89 % of it, which is the price of the standing wave.

Question 3 — results
Quantity	Symbol	Value
Wavelength on the line	$\lambda$	$40\ \text{cm}$ ($\ell = 0.625\lambda$, $d_{\min} = 0.375\lambda$)
Reflection coefficient at the load	$\Gamma_L$	$0.3333\,\angle\,90^\circ$
Load impedance	$Z_L$	$\mathbf{40 + j30\ \Omega}$
Input impedance at the generator	$Z_{\text{in}}$	$100\ \Omega$ (real; a voltage maximum)
Input voltage / current	$V_{\text{in}}$, $I_{\text{in}}$	$31.62\ \text{V rms}$, $0.3162\ \text{A rms}$
Generator EMF	$\|E\|$	$\mathbf{47.43\ \text{V rms}}$ ($67.08\ \text{V peak}$)

Topic. Recovering an unknown load from slotted-line measurements, and sizing the source from a power constraint.

Key relations. $|\Gamma| = (S-1)/(S+1)$; $\Gamma$ is real and negative at a voltage minimum; $\Gamma_L = \Gamma(d)e^{+j2\beta d}$; $Z = Z_0(1+\Gamma)/(1-\Gamma)$; $Z_{\text{in}} = S Z_0$ at a maximum and $Z_0/S$ at a minimum; $P = I^{2}\operatorname{Re}(Z)$ for rms quantities.

Why this works. The standing-wave pattern carries exactly two independent numbers — a ratio and a position — which is precisely what a complex $\Gamma_L$ needs. The minimum is used rather than the maximum because it is sharper and therefore locates more precisely in a real slotted-line measurement, and because its phase is unambiguous. Once $\Gamma_L$ is known the line is fully characterised, and since a lossless line cannot absorb power the load power can be imposed at any convenient plane; choosing the input plane is easiest here because the rotation happened to land on a purely resistive point.

Common pitfall. Rotating the wrong way. Moving from the measured minimum to the load advances the phase of $\Gamma$ ($+2\beta d$), while moving from the load towards the generator retards it ($-2\beta\ell$); interchanging the two returns $Z_L = 40 - j30\ \Omega$, a conjugate error that no SWR check can catch. A second trap is forgetting that a stated power is rms-based, and mixing a peak voltage into $P = V^{2}/(2Z)$ or $V^{2}/Z$ inconsistently — a factor-of-two error in power and $\sqrt2$ in the EMF.

Source. Pozar, Microwave Engineering, §2.4 (the Smith chart) and §2.3 (standing waves and the slotted line); Sadiku, Elements of Electromagnetics, Ch. 11.

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