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

Question 8 of 8: Elevation Angle and Frequency Dependence for a Short Vertical Element

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 8: Elevation Angle and Frequency Dependence for a Short Vertical Element (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 vertical current element of length $l = 1\ \text{m}$ radiating $P = 10\ \text{W}$ into free space at $f = 10\ \text{MHz}$; the observation point lies on a sphere of radius $R = 10\ \text{km}$ where the power density is $S = 9\times10^{-9}\ \text{W/m}^{2}$; and the directivity is supplied as the maximum-to-average ratio, 1.5.

Find. (i) the elevation angle of the observation point; (ii) the power density there if the generator frequency is halved to 5 MHz.

Elevation cut of the short vertical element's sin-squared pattern. The maximum lies along the horizon and the null straight up, so the observation point sits 29.74° above the horizon.

Approach. Convert the radiated power into the peak power density on the sphere using the supplied directivity, then invert the element's $\sin^{2}$ pattern to find the angle; for part (ii), identify which quantities actually depend on frequency.

Confirm the element is electrically short, so the pattern is the standard one. $$\lambda = \frac{c}{f} = \frac{3\times10^{8}}{10\times10^{6}} = 30\ \text{m}, \qquad \frac{l}{\lambda} = \frac{1}{30} = 0.0333.$$ At one thirtieth of a wavelength the current is essentially uniform and the pattern is the ideal $\sin^{2}\theta$ doughnut, with $\theta$ measured from the element axis.
Find the average power density on the sphere. All the radiated power crosses the sphere, so $$S_{\text{avg}} = \frac{P}{4\pi R^{2}} = \frac{10}{4\pi(10^{4})^{2}} = 7.958\times10^{-9}\ \text{W/m}^{2}.$$
Scale up to the peak using the supplied ratio. The aid states that the maximum-to-average ratio is 1.5 — this is the directivity of a short element, $D = 1.5$ (1.76 dBi): $$S_{\max} = 1.5\,S_{\text{avg}} = 1.194\times10^{-8}\ \text{W/m}^{2}.$$ The measured $9\times10^{-9}\ \text{W/m}^{2}$ is below this, as it must be, so the point lies off the pattern maximum — which is what makes the question answerable.
Invert the pattern to get the angle. The element is vertical, so its maximum lies in the horizontal plane and the pattern in terms of elevation angle $\alpha$ above the horizon is $$S(\alpha) = S_{\max}\sin^{2}\theta = S_{\max}\cos^{2}\alpha, \qquad \theta = 90^\circ - \alpha.$$ Solving, $$\cos^{2}\alpha = \frac{S}{S_{\max}} = \frac{9\times10^{-9}}{1.194\times10^{-8}} = 0.7540, \qquad \cos\alpha = 0.8683,$$ $$\boxed{\alpha = 29.74^\circ \text{ above the horizon}}$$ By symmetry the same density occurs at $29.74^\circ$ below the horizon and at every azimuth, so the locus is a pair of cones; the question asks for the elevation angle, which is $29.74^\circ$.
Part (ii): identify what the frequency actually controls. Two quantities in this problem depend on frequency and two do not. The pattern shape $\sin^{2}\theta$ and the directivity $D = 1.5$ are properties of a short element and contain no frequency at all. What does depend on frequency is the radiation resistance, $$R_r = 80\pi^{2}\left(\frac{l}{\lambda}\right)^{2} \propto f^{2},$$ which falls from $0.8773\ \Omega$ at 10 MHz to $0.2193\ \Omega$ at 5 MHz — a factor of four.
Apply the question's own premise. The generator is stated to deliver 10 W to the element, and only the frequency is changed. With the radiated power held at 10 W and the pattern unchanged, every term in $S(\alpha) = 1.5P\cos^{2}\alpha/(4\pi R^{2})$ is unchanged, so $$\boxed{S = 9\times10^{-9}\ \text{W/m}^{2}\ \text{ at }5\ \text{MHz} \ \text{ — unchanged}}$$ What changes is what the generator must do to keep delivering that power: the drive current rises from $I = \sqrt{P/R_r} = 3.376\ \text{A rms}$ to $6.752\ \text{A rms}$, exactly double, and the element becomes correspondingly harder to feed — a sub-ohm, highly reactive load. Halving the frequency has made the antenna a much worse radiator without making its radiation any different, which is the real lesson of the sub-part.
Note the check that keeps the reasoning honest. Both parts close on the same identity: $S(\alpha)\,4\pi R^{2}/(1.5\cos^{2}\alpha)$ returns 10 W at either frequency. Any answer to (ii) other than "unchanged" must be accompanied by a statement of what was held constant instead of the radiated power.

Check: the alternative reading of part (ii). If a marker intends instead that the element current stays fixed while the frequency is halved — that is, the generator is not re-matched and simply drives the same 3.376 A into a load whose radiation resistance has dropped four-fold — then the radiated power falls to $I^{2}R_r = 2.50\ \text{W}$ and the density at the point falls in proportion to $2.25\times10^{-9}\ \text{W/m}^{2}$. That reading contradicts the question's own words ("delivers 10 W power"), so the unchanged $9\times10^{-9}\ \text{W/m}^{2}$ is given as the answer; the quarter-value figure is recorded here because the two differ by a factor of four and the distinction is worth stating explicitly in an answer book.

Question 8 — results
Quantity	Symbol	Value
Wavelength / electrical length at 10 MHz	$\lambda$, $l/\lambda$	$30\ \text{m}$, $0.0333$ (electrically short)
Average power density on the sphere	$S_{\text{avg}}$	$7.958\times10^{-9}\ \text{W/m}^{2}$
Maximum power density (D = 1.5)	$S_{\max}$	$1.194\times10^{-8}\ \text{W/m}^{2}$
(i) Elevation angle of the point	$\alpha$	$\mathbf{29.74^\circ}$ above the horizon
(ii) Power density at 5 MHz	$S$	$\mathbf{9\times10^{-9}\ \text{W/m}^{2}}$ — unchanged
Radiation resistance, 10 MHz → 5 MHz	$R_r$	$0.8773\ \Omega \rightarrow 0.2193\ \Omega$
Drive current required, 10 MHz → 5 MHz	$I$	$3.376 \rightarrow 6.752\ \text{A rms}$

Topic. Pattern, directivity and radiation resistance of the short (Hertzian) current element.

Key relations. $S(\theta) = S_{\max}\sin^{2}\theta$ with $\theta$ from the element axis, so $S(\alpha) = S_{\max}\cos^{2}\alpha$ in elevation; $S_{\text{avg}} = P/(4\pi R^{2})$; $S_{\max} = D\,S_{\text{avg}}$ with $D = 1.5$; $R_r = 80\pi^{2}(l/\lambda)^{2}$ for a uniform current element; $I = \sqrt{P/R_r}$.

Why this works. Directivity and radiation resistance answer two different questions, and this problem separates them cleanly. Directivity is a pure pattern property — how the radiated power is distributed — and for any short element it is 1.5 no matter how short, because the $\sin^{2}\theta$ shape does not change. Radiation resistance is a feed-point property — how hard it is to get power into the antenna — and it collapses as $(l/\lambda)^{2}$. That is why very short antennas are perfectly good radiators of whatever power you manage to force into them, and simultaneously very difficult to feed: the useful resistance sinks towards zero while the loss resistance of the conductors and the matching network does not, so efficiency, not pattern, is what fails at low frequency.

Common pitfall. Measuring the angle from the wrong reference: using $\sin^{2}\alpha$ rather than $\cos^{2}\alpha$ returns $60.3^\circ$, the complement of the right answer, and looks entirely plausible. In part (ii), the classic error is to assume that halving the frequency must change the radiated field, and to divide by four without noticing that the question fixes the delivered power; the discipline is to ask what is held constant before scaling anything.

Source. Balanis, Antenna Theory, §4.1–4.2 (infinitesimal dipole: fields, directivity, radiation resistance); Ulaby, Fundamentals of Applied Electromagnetics, Ch. 9.

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