22-Elec-A7 Electromagnetics · May 2013

Question 6 of 8: Power Capability of a Voltage-Limited Line

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

Notes on this paper

Paper format. National Examinations, May 2013 — 07-Elec-A7, Electromagnetics. Three hours, closed book; one of two approved calculators (Casio or Sharp) permitted. Eight questions, all of equal value; any five constitute a complete paper and only the first five presented are marked. All eight are solved here, since the set is a study resource. Aids printed on the cover page: ε₀ = 8.85 × 10⁻¹² F/m and μ₀ = 4π × 10⁻⁷ H/m; Question 2 additionally supplies the quarter-wave relation.

Reference texts (22-Elec-A7 Electromagnetics).

D. M. Pozar, Microwave Engineering, 4th ed. — transmission lines, matching, waveguides.
M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. — plane waves, guided waves, radiation.
W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. — field theory and transients on lines.
C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. — short elements, image theory.
F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed. — oblique incidence, surface currents.
S. J. Chapman, Electric Machinery Fundamentals, 5th ed. — rotating-loop induction and torque (Question 3).

Conventions used throughout. The paper itself quotes propagation velocities as $3\times10^{8}\ \text{m/s}$, so free space is taken as $c=3.00\times10^{8}\ \text{m/s}$ and the intrinsic impedance as $\eta_0=\mu_0 c=120\pi=376.99\ \Omega$. Lines are treated as lossless, and "rms" is stated explicitly wherever the exam asks for it.

Question 6: Power Capability of a Voltage-Limited Line (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 50 Ω line whose insulation limits the voltage anywhere on it to 2000 V peak carries a 300 MHz signal into a load that produces a standing-wave ratio of 1.5.

Given data
Quantity	Symbol	Value
Characteristic impedance	$Z_0$	50 Ω
Propagation velocity	$v_p$	$3\times10^{8}$ m/s
Voltage constraint (envelope peak)	$V_{max}$	2000 V peak
Operating frequency	$f$	300 MHz ($\lambda=1.00$ m)
Standing-wave ratio produced by the load	$s$	1.50

Find. (i) the greatest power the line can pass to such a load without exceeding 2000 V anywhere, and (ii) the greatest line length for which that voltage limit need not bind.

Figure 6.1 — Standing-wave envelope for $s=1.5$: maxima of 2000 V alternate with minima of 1333 V every half wavelength, so a line no longer than $\lambda/2$ can be positioned between two maxima.

Approach. Convert the standing-wave ratio to a reflection coefficient, cap the forward-wave amplitude using the envelope maximum, form the net delivered power from the difference of the forward and reverse powers, and then use the half-wavelength periodicity of the maxima to answer part (ii).

Convert SWR to reflection coefficient. $$|\Gamma|=\frac{s-1}{s+1}=\frac{1.50-1}{1.50+1}=0.200$$
Cap the forward wave using the voltage limit. The envelope maximum occurs where the incident and reflected waves add in phase: $$|V|_{max}=|V^{+}|\left(1+|\Gamma|\right)\le2000\ \text{V}\quad\Rightarrow\quad |V^{+}|\le\frac{2000}{1.200}=\boxed{1666.7\ \text{V peak}}$$ The corresponding minimum is $|V|_{min}=|V^{+}|(1-|\Gamma|)=1333$ V, consistent with $s=V_{max}/V_{min}=1.5$.
Form the net delivered power. The load receives the forward power less the reflected power. For peak (not rms) phasor amplitudes, $$P=\frac{|V^{+}|^{2}}{2Z_0}\left(1-|\Gamma|^{2}\right) =\frac{1666.7^{2}}{2\times50}\left(1-0.200^{2}\right) =27\,778\times0.960=\boxed{26.7\ \text{kW}}$$
Cross-check with the maximum–minimum form. An equivalent and quicker expression is $P=V_{max}V_{min}/(2Z_0)$: $$P=\frac{2000\times1333}{2\times50}=26.7\ \text{kW}$$ which agrees, confirming the algebra. For reference, a matched load ($s=1$) on the same line would carry $2000^{2}/(2\times50)=40.0$ kW, so the 1.5 SWR costs a third of the line's capability. This is the sense in which 26.7 kW is the lowest upper bound: it is the greatest power guaranteed deliverable for any load of that SWR, and no such load can do better without breaching the voltage limit.
Locate the voltage maxima for part (ii). The envelope repeats every half wavelength, because the relative phase of the two travelling waves advances by $2\beta z$: $$\lambda=\frac{v_p}{f}=\frac{3\times10^{8}}{300\times10^{6}}=1.00\ \text{m},\qquad \frac{\lambda}{2}=0.500\ \text{m}$$
Deduce the longest permissible length. If the line is shorter than the spacing between successive maxima, then for a favourable load phase angle the whole line can sit between two maxima, so the 2000 V envelope peak never physically occurs on it and the constraint stops binding. The limiting case has the two maxima falling exactly at the two terminals: $$\ell_{max}=\frac{\lambda}{2}=\boxed{0.500\ \text{m}=50\ \text{cm}}$$ Any longer line must contain at least one interior maximum, whatever the load phase, and the 2000 V restriction then applies in full.

Check: the relaxation in part (ii) assumes the load phase angle can be chosen (or trimmed with a short length of line) so that the standing-wave pattern sits favourably relative to the terminals. The 50 cm figure is therefore the limiting length under the most favourable phase, not a guarantee for an arbitrary load of the same SWR magnitude.

Final results — Question 6
Quantity	Value
Reflection coefficient magnitude	0.200
Largest permissible forward-wave amplitude	1667 V peak
Envelope minimum on the line	1333 V peak
(i) Lowest upper bound on delivered power	26.7 kW
Comparison: matched line on the same voltage limit	40.0 kW
Wavelength at 300 MHz	1.00 m
(ii) Longest length for which the limit may be relaxed	0.500 m (50 cm) = $\lambda/2$

Topic. Voltage-limited power capability of a mismatched transmission line.

Key relations. $|\Gamma|=(s-1)/(s+1)$; $|V|_{max}=|V^{+}|(1+|\Gamma|)$ and $|V|_{min}=|V^{+}|(1-|\Gamma|)$; net power $P=|V^{+}|^{2}(1-|\Gamma|^{2})/(2Z_0)=V_{max}V_{min}/(2Z_0)$; the envelope pattern repeats every $\lambda/2$.

Why this works. On a mismatched line the two travelling waves alternately reinforce and oppose one another, so the voltage envelope has fixed maxima and minima whose ratio is the SWR. Dielectric breakdown is governed by the largest voltage that appears anywhere on the line, which is the envelope maximum — so that is what the rating must be applied to, and the forward wave must be scaled down accordingly. The net power, on the other hand, is the difference between the forward and reverse power flows, because the reflected wave carries energy back toward the generator. Combining the two gives the familiar result that a mismatch always reduces the power a voltage-rated line can deliver. The half-wavelength periodicity follows from the round-trip phase $2\beta z$: the relative phase of the two waves advances at twice the rate of either wave alone.

Common pitfall. Computing power as $V_{max}^{2}/(2Z_0)$, which treats the envelope maximum as a travelling-wave amplitude and overstates the answer by 50 % here. The second pitfall is putting the maxima $\lambda$ apart instead of $\lambda/2$, which doubles the answer to part (ii).

Source. Pozar, Microwave Engineering, 4th ed., §2.3 (standing waves, SWR and power); Hayt & Buck, Engineering Electromagnetics, 9th ed., Ch. 10.

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