22-Elec-B9 Electromagnetic Field, Transmission Lines, Antennas, and Radiation · May 2013
Question 5 of 8: Beam Direction of a Five-Element Phased Array
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2013 — 07-Elec-B9,
Electromagnetic Field, Transmission Lines, Antennas, and Radiation.
Three hours, closed book; an approved Casio or Sharp calculator is permitted.
Eight questions are printed and five constitute a complete paper,
each of equal value (20 marks); the first five appearing in the answer book are
the ones marked. The cover page supplies only two aids,
$\varepsilon_0 = 8.85\times 10^{-12}\ \text{F/m}$ and
$\mu_0 = 4\pi\times 10^{-7}\ \text{H/m}$, plus a Smith chart on page 4.
Candidates are urged to state any interpretation assumptions with their answers.
All eight questions are solved below, because this set is a study
resource rather than a three-hour sitting.
Reference texts. D. M. Pozar, Microwave Engineering,
4th ed. (transmission lines, the Smith chart, waveguides and cavities);
M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. and
W. H. Hayt & J. A. Buck, Engineering Electromagnetics, 9th ed.
(plane waves, oblique incidence, radiation); C. A. Balanis,
Antenna Theory: Analysis and Design, 4th ed. (short elements, arrays,
apertures); F. T. Ulaby & U. Ravaioli, Fundamentals of Applied
Electromagnetics, 8th ed. (radar equation, ground-plane imaging). SI units
throughout; the paper's own $c = 3\times 10^{8}\ \text{m/s}$ and
$Z_0 = 377\ \Omega$ are used so that the printed aids reproduce exactly.
Check: reading of the printed data. Some symbols are printed loosely. Question 3 prints
“10GH” and “1KWs total energy”; these are read as a
10 GHz carrier and a total pulse energy of
$1\ \text{kW}\cdot\text{s} = 1000\ \text{J}$. Question 6 prints
“15 MH” (15 MHz) and the aid
$R_{\text{rad}} = (2\pi/3)Z_0 (l/\lambda)^2$. Question 7 prints
“10GH” (10 GHz) and the aid
$\theta_0 = 1.2\lambda/d$. Question 8 prints
“an 600 angle” ($60^\circ$).
Question 5: Beam Direction of a Five-Element Phased Array
(20 marks)
Given. A uniform linear array of five short vertical elements
lying along a horizontal north–south line.
Given data
Quantity
Symbol
Value
Number of elements
$N$
5
Element spacing
$d$
0.40 m
Frequency
$f$
1 GHz
Wavelength
$\lambda$
0.30 m
Progressive phase, going south
$\alpha$
$-120^\circ$ per element
Element type
—
short vertical current element (omnidirectional in azimuth)
Find. The direction(s) in which the array radiates its
maximum.
Figure 5.1 — Plan view. Blue
dots are the five elements on the north–south line; red arrows are the
horizontal-plane directions in which the array factor reaches its maximum.
Approach. Form the array factor for a uniform linear array,
set its total inter-element phase to zero (modulo $2\pi$), and solve for the
cone angle measured from the array axis; then check how many solutions the
spacing allows.
Compute the electrical spacing.
$\lambda = c/f = 0.30\ \text{m}$, so
$$\frac{d}{\lambda} = \frac{0.40}{0.30} = 1.333, \qquad
\beta d = 2\pi\frac{d}{\lambda} = 480^\circ .$$
The elements are more than half a wavelength apart, which warns immediately that
grating lobes are possible.
Write the array factor.
Numbering the elements from north to south so that element $n$ carries phase
$n\alpha$ with $\alpha = -120^\circ$,
$$\text{AF}(\theta) = \sum_{n=0}^{4} e^{jn\psi}, \qquad
\psi = \beta d\cos\theta + \alpha ,$$
where $\theta$ is measured from the array axis pointing south.
Impose the maximum condition.
All five contributions add in phase when $\psi = 2\pi m$, i.e.
$$480^\circ\cos\theta - 120^\circ = 360^\circ m
\quad\Longrightarrow\quad
\cos\theta = \frac{1 + 3m}{4}.$$
Enumerate the visible solutions.
Only $|\cos\theta| \le 1$ is physical, which admits three values of $m$:
$$\boxed{\ \theta = 75.5^\circ,\quad \theta = 0^\circ,\quad
\theta = 120^\circ \ \text{(all measured from due south)}\ }$$
corresponding to $m = 0$, $m = +1$ and $m = -1$ respectively. Each satisfies
$|\text{AF}| = N = 5$, so all three are full-strength maxima.
Turn the cone angles into compass directions.
Because a short vertical element radiates equally in all azimuths, each
$\theta$ defines a cone about the north–south axis; the element
pattern is strongest in the horizontal plane, so the observable maxima lie where
each cone meets that plane:
due south; $75.5^\circ$ either side of south (azimuths
$104.5^\circ$ and $255.5^\circ$); and $120^\circ$ from south, that is
$60^\circ$ either side of north (azimuths $60^\circ$ and $300^\circ$).
Identify the intended main beam.
The $m = 0$ solution is the beam the phasing was designed to produce, so the
main beam points $75.5^\circ$ away from south on the cone about the array axis
— toward azimuth $104.5^\circ$ (S $75.5^\circ$ E) and its mirror image
$255.5^\circ$ (S $75.5^\circ$ W). The other two are grating lobes of equal
amplitude, a direct consequence of $d \gt \lambda/2$.
It is worth being explicit that this array does not have a single beam. A
progressive lag of $120^\circ$ over a spacing of $1.333\lambda$ is satisfied
three times as the observation direction swings from the south axis round to the
north axis, and a uniform array gives every one of those directions the same
peak. If a single beam were required the spacing would have to be reduced below
$\lambda/2$ — at 1 GHz that means under 15 cm — or the phase
increment reduced so that only one solution stays inside the visible region.
