Question 4 of 8: Power limit set by the voltage ceiling on a mismatched line

Question 4: Power limit set by the voltage ceiling on a mismatched 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 mismatched but lossless feeder whose insulation sets a ceiling on the peak voltage that may appear anywhere on it.

Find. The largest power deliverable to the load without exceeding the voltage ceiling for a line longer than half a wavelength; and whether a shorter line could permit more.

Approach. Convert the standing-wave ratio into $|\Gamma|$, cap the forward wave so that the envelope maximum just reaches the ceiling, and compute the net power as forward minus reflected. Then ask whether a short line can avoid containing a maximum at all.

1. Reflection coefficient magnitude. $$|\Gamma|=\frac{S-1}{S+1}=\frac{0.5}{2.5}=0.200 .$$
2. Where the ceiling binds. The envelope runs between $V_{max}=V^{+}(1+|\Gamma|)$ and $V_{min}=V^{+}(1-|\Gamma|)$, and successive maxima are $\lambda/2$ apart. A line longer than $\lambda/2$ therefore contains at least one maximum wherever the load happens to place the pattern, so the ceiling must be applied to the maximum.
3. Cap the forward wave. $$V^{+}\le\frac{V_{max}}{1+|\Gamma|}=\frac{5000}{1.2}=4167\ \text{V peak},\qquad V_{min}=0.8V^{+}=3333\ \text{V peak}.$$
4. Net power delivered. Forward minus reflected power, for peak-referenced phasors, $$P=\frac{|V^{+}|^{2}}{2Z_0}\left(1-|\Gamma|^{2}\right)=\frac{(4167)^{2}(0.96)}{2(50)}\;\Rightarrow\;\boxed{P_{max}=167\ \text{kW}} .$$ Two equivalent shortcuts give the same figure: $P=V_{max}V_{min}/(2Z_0)$ and $P=V_{max}^{2}/(2Z_0S)$.
5. Contrast with the matched-line rating. If the same line were matched, $V_{max}^{2}/(2Z_0)=250\ \text{kW}$ would be permissible. The mismatch costs a factor $1/S$, i.e. one third of the rating, even though only 4 % of the incident power is actually reflected.
6. Part 2: a line shorter than $\lambda/2$. Yes, the answer can be different, and specifically larger. Because maxima are $\lambda/2$ apart, a line shorter than that need not contain one; whether it does depends on the phase of $\Gamma_L$ and on where the line ends. In the most favourable case the line occupies only the neighbourhood of a voltage minimum, so the largest voltage present approaches $V_{min}$, and the cap becomes $V^{+}\le 5000/(1-|\Gamma|)=6250\ \text{V}$, giving $$P\to\frac{(6250)^{2}(0.96)}{2(50)}=375\ \text{kW}=S^{2}\times 167\ \text{kW}.$$

For any specific short line the honest procedure is to evaluate $|V(z)|=|V^{+}||1+\Gamma_Le^{-2j\beta z}|$ over the actual extent of the line and cap that maximum at 5 kV, which yields something between 167 kW and 375 kW. The limiting factor $S^{2}$ shows how much rating is being sacrificed purely to the possibility of a maximum appearing: at $\mathrm{SWR}=1.5$ a designer who can guarantee the line is short and positioned near a minimum recovers more than twice the power. In practice one does not rely on that, since the load impedance drifts with frequency and temperature and moves the pattern along the line.