Question 5 of 8: Power limit set by the standing-wave voltage maximum
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2015 — 07-Elec-A7 Electromagnetics. Three hours, closed book; one of two approved calculators permitted. Eight questions, all of equal value (20 marks each). The rubric states that any five questions constitute a complete paper and that only the first five presented are marked — all eight are worked here, because this set is a study resource rather than a sat examination. Aids printed on the paper: $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$ and $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$, from which $\eta_0=\sqrt{\mu_0/\varepsilon_0} = 376.8\ \Omega$ and $c = 1/\sqrt{\mu_0\varepsilon_0} = 3.00\times10^{8}\ \text{m/s}$.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission-line transients, matching stubs, waveguide modes); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed.; W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (plane waves, normal incidence); F. T. Ulaby, Fundamentals of Applied Electromagnetics, 8th ed. (polarisation, interference, radiation); C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short current elements).
Question 5: Power limit set by the standing-wave voltage maximum (20 marks)
Given. A mismatched but low-VSWR feeder whose insulation sets a hard ceiling on the voltage anywhere along it.
Given data
Quantity
Symbol
Value
Standing wave ratio
$S$
$1.5$
Characteristic impedance
$Z_0$
$50\ \Omega$
Propagation velocity
$v_p$
$2\times10^{8}\ \text{m/s}$
Maximum allowed line voltage
$V_{\max}$
$10\ \text{kV}$ rms
Find. The largest power that may be transmitted without the voltage anywhere on the line exceeding the stated limit.
Figure Q5. Standing-wave envelope. The insulation limit is a ceiling on the crests, not on the mean, so the deliverable power is set by the product of the crest and the trough.
Approach. Convert the SWR into a reflection coefficient, use the ceiling to cap the forward wave rather than the net voltage, and then compute the net power as forward minus reflected.
Turn the SWR into a reflection coefficient. By definition $S = (1+|\Gamma|)/(1-|\Gamma|)$, so $$|\Gamma| = \frac{S-1}{S+1} = \frac{1.5-1}{1.5+1} = 0.200$$ Only the magnitude matters here; the phase merely slides the pattern along the line and cannot change the size of the crests.
Locate the ceiling on the forward wave. Along the line the incident and reflected waves alternately add and subtract, giving $$V_{\max} = |V^{+}|(1+|\Gamma|), \qquad V_{\min} = |V^{+}|(1-|\Gamma|)$$ The insulation limit applies at the crests, so it caps the forward wave first: $$|V^{+}| = \frac{V_{\max}}{1+|\Gamma|} = \frac{10.0}{1.200} = 8.333\ \text{kV rms}$$ and the corresponding trough is $V_{\min} = V_{\max}/S = 6.667\ \text{kV rms}$.
Form the net power. Forward and reflected waves carry power independently on a lossless line, so the power actually delivered is the difference $$P = \frac{|V^{+}|^{2}}{Z_0}\left(1-|\Gamma|^{2}\right) = \frac{(8333)^{2}}{50}\left(1-0.04\right)$$ using rms amplitudes throughout, which is why the denominator is $Z_0$ and not $2Z_0$.
Evaluate, and note the compact form. $$P = \frac{V_{\max}V_{\min}}{Z_0} = \frac{(10.0\times10^{3})(6.667\times10^{3})}{50} = \boxed{1.333\ \text{MW}}$$ The two routes agree identically because $(1+|\Gamma|)(1-|\Gamma|) = 1-|\Gamma|^2$; the product-of-extremes form is the one worth remembering.
Compare with the matched case. If the line were perfectly matched the same insulation would allow $V_{\max}^{2}/Z_0 = 2.00\ \text{MW}$. A standing wave ratio of only 1.5 therefore costs a third of the power rating, $$\frac{P}{P_{\text{matched}}} = \frac{1}{S} = 0.667$$ a de-rating that is entirely due to the crests being higher for the same average flow, not to any loss.
Note where the crests sit. Voltage maxima repeat every half wavelength, and at, say, 1 MHz on this line $\lambda = v_p/f = 200\ \text{m}$, so successive crests are 100 m apart. Any section shorter than $\lambda/2$ can be guaranteed to miss a crest, which is the standard argument for relaxing the restriction on a short jumper.
The result is a reminder that voltage rating and power rating are different things on a mismatched line. The generator here is quite capable of driving 2 MW into a matched load with the same insulation; it is the standing wave, not the source, that imposes the 1.333 MW ceiling.
