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

Question 5 of 8: Width Between Nulls of a Three-Element Broadside Array

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 5: Width Between Nulls of a Three-Element Broadside Array (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. Three identical vertical current elements, equally spaced $d = 20\ \text{m}$ along a north–south line, all carrying equal currents in phase, at $f = 30\ \text{MHz}$. The observation is in the horizontal plane, where a vertical element radiates uniformly.

Find. The angular width, in the horizontal plane, between the two nulls that bound the main beam pointing east.

Left: the array geometry, three in-phase vertical elements on a north-south line radiating broadside (east). Right: the array factor in the horizontal plane, showing the 19.19° main beam and the grating lobes forced by the 2λ spacing.

Approach. Because the elements are vertical, their own pattern is constant in the horizontal plane and the shape there is the array factor alone; find the first zero of the three-element uniform array factor either side of broadside and double the offset.

Establish the wavelength and the electrical spacing. $$\lambda = \frac{c}{f} = \frac{3\times10^{8}}{30\times10^{6}} = 10\ \text{m}, \qquad \frac{d}{\lambda} = \frac{20}{10} = 2.00.$$ The elements are two full wavelengths apart, which is a very wide spacing — a point returned to below.
Confirm that east is indeed the main beam. With all currents in phase the array is broadside: the maximum lies perpendicular to the array axis. A north–south axis therefore puts principal maxima due east and due west, as the question states.
Write the array factor. Measuring $\phi$ from the broadside (east) direction in the horizontal plane, the progressive phase between adjacent elements is $\psi = \beta d\sin\phi$, and for $N$ equal in-phase currents $$|AF| = \left|\frac{\sin(N\psi/2)}{\sin(\psi/2)}\right|, \qquad \psi = \frac{2\pi}{\lambda}d\sin\phi = 4\pi\sin\phi \ \text{ rad}.$$ The element pattern contributes nothing in this plane, so $|AF|$ is the whole horizontal pattern.
Locate the first null. The numerator vanishes when $N\psi/2 = m\pi$; the first zero that is not cancelled by the denominator is $m = 1$, giving $\psi = 2\pi/N$. Hence $$\beta d \sin\phi_{\text{null}} = \frac{2\pi}{3} \;\Longrightarrow\; \sin\phi_{\text{null}} = \frac{\lambda}{Nd} = \frac{10}{(3)(20)} = \frac{1}{6},$$ $$\phi_{\text{null}} = \arcsin(0.16667) = 9.594^\circ.$$
Double it for the width between nulls. The pattern is symmetric about broadside, so the two nulls bounding the main beam sit at $\pm 9.594^\circ$ and $$\boxed{\text{FNBW} = 2\arcsin\!\left(\frac{\lambda}{Nd}\right) = 19.19^\circ}$$ That is a beam roughly $19^\circ$ wide between nulls, or about $8.9^\circ$ between half-power points — a usefully sharp beam for a 30 MHz installation, obtained from an aperture $Nd = 60\ \text{m}$ across.
State the consequence of the wide spacing. The main beam is not the only full-amplitude beam. Principal maxima recur wherever $\psi$ is a multiple of $2\pi$, that is $\sin\phi = m/2$, giving grating lobes at $$\phi = \pm30^\circ \ \text{ and } \ \pm90^\circ$$ in addition to the intended $\phi = 0$. The array therefore radiates just as strongly $30^\circ$ either side of east, and endfire along the array axis (north and south) as well. The grating lobes have the same null-to-null width as the main beam when measured in $\sin\phi$, but they broaden in angle away from broadside — about $22.3^\circ$ for the lobes at $\pm30^\circ$ and about $67^\circ$ for the endfire lobes — and with $N = 3$ there is one subsidiary minor lobe between adjacent principal maxima. Suppressing them would require $d < \lambda$, i.e. spacing under 10 m — but the question asks only for the width of the main beam, which is unaffected.

Check: which plane, and which pattern. The answer is the width in the horizontal plane, where the vertical elements are omnidirectional, so it is a pure array-factor result. In the vertical plane containing east, the element's own $\sin^{2}$ pattern (Question 8) applies instead and the beam is bounded by the ground rather than by array nulls. The question's wording — "the main beam pointing east" — identifies the horizontal cut.

Question 5 — results
Quantity	Symbol	Value
Wavelength	$\lambda$	$10\ \text{m}$
Element spacing	$d/\lambda$	$2.00$
Array type	—	Uniform, in phase ⇒ broadside (east–west)
First null off broadside	$\phi_{\text{null}}$	$9.594^\circ$
Width between nulls	FNBW	$\mathbf{19.19^\circ}$
Grating lobes (consequence of $d = 2\lambda$)	—	$\pm30^\circ$ and $\pm90^\circ$

Topic. Beamwidth of a uniform linear broadside array, and grating lobes from wide element spacing.

Key relations. $\psi = \beta d \sin\phi + \alpha$ (here the progressive phase $\alpha = 0$); $|AF| = |\sin(N\psi/2)/\sin(\psi/2)|$; nulls at $\sin\phi = m\lambda/(Nd)$; $\text{FNBW} = 2\arcsin[\lambda/(Nd)]$; grating lobes wherever $\sin\phi = m\lambda/d$; pattern multiplication: total pattern = element pattern × array factor.

Why this works. The array factor is a geometric series whose sum is the Dirichlet kernel, and its first zero is reached when the phase spread across the whole aperture is one full cycle — which is why the beamwidth depends on the total aperture $Nd$ and not on $N$ and $d$ separately. That is the fundamental trade of array design: beamwidth buys aperture, but filling that aperture with too few elements leaves gaps in the sampled wavefront and the pattern repeats itself as grating lobes. Here the aperture is $6\lambda$, which is why the beam is sharp, and the sampling is only one element every $2\lambda$, which is why the repeats appear.

Common pitfall. Measuring $\phi$ from the array axis instead of from broadside, which turns the $\sin\phi$ into a $\cos\phi$ and returns the complement of the right answer. A second error is quoting $\lambda/(Nd)$ in radians as the full width rather than the half width, halving the answer; and a third is using the small-angle form $\text{FNBW} \approx 2\lambda/(Nd) = 19.10^\circ$ without noting it is an approximation — here it happens to be within half a percent, but it degrades quickly for shorter arrays.

Source. Balanis, Antenna Theory, §6.2–6.3 (N-element linear array, uniform amplitude and spacing); Ulaby, Fundamentals of Applied Electromagnetics, Ch. 9.

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