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IN THE EC SYLLABUS

Antennas and antenna arrays

An antenna converts a guided wave into a radiated one and back, and every standard result in this topic grows from one idea: the far field of a short current element. The Hertzian dipole leads to the half-wave dipole and the quarter-wave monopole over a ground plane, and adding identical elements with chosen phases gives linear arrays. Solving questions here means doing a small set of calculations cleanly: radiation resistance and radiated power, directivity or gain from a radiation-intensity expression, effective aperture and the Friis link budget, and the array factor with its broadside or end-fire phasing, nulls, grating lobes and pattern multiplication.

2 previous-year questions · 2 solved · 5 practice questions

Conventions, field regions and radiated power

Conventions used throughout: SI units, free space unless stated, with F/m, H/m, m/s and intrinsic impedance . Fields are phasors with time dependence , so an outgoing spherical wave carries the factor , with . Spherical coordinates measure from the axis, with unit vectors . Currents and fields are peak (amplitude) phasors, so every time-average power carries a factor . The reflection coefficient at the antenna terminals is , with the antenna impedance and the feed-line impedance.

Around any antenna the field falls into regions. Close in, the reactive near field (terms in and ) stores energy that flows out and back every cycle and carries no net power. Beyond it, in the radiating near-field (Fresnel) region, the angular shape of the field still changes with distance. Far out, in the far-field (Fraunhofer) region, and are transverse to , perpendicular to each other and in phase, fall off as , and satisfy . Their angular shape no longer depends on . That shape is the radiation pattern, and every parameter in this topic (radiation intensity, directivity, beamwidth, gain, array factor) is defined in the far field.

FAR-FIELD DISTANCE

= largest dimension of the antenna (a length here, not the directivity). At the path-length difference between the centre and the edge of the antenna is (phase error ). For an antenna large compared with the wavelength, the reactive near field extends to about and the radiating near field from there to . For an electrically small antenna () the binding condition is ; the radiation and induction terms of a Hertzian dipole are equal in magnitude at .

TIME-AVERAGE POWER DENSITY AND TOTAL RADIATED POWER

The middle form holds in the far field only, where and are in phase and . is a peak phasor. In a lossless medium is the same through every closed surface that encloses the antenna.

EXAM TIP

Peak or rms? These notes use peak values, so and . If a question gives rms values, use and . The radiation resistance itself is the same either way; only the factor moves.

Hertzian dipole: fields and radiation resistance

The Hertzian (infinitesimal) dipole is a -directed current element of length at the origin, carrying the same current along its whole length. A common working limit is ; if the current really is uniform, the results below stay within 1% for up to about . It is the building block of wire antennas: any wire is a superposition of such elements. Its magnetic vector potential is , and followed by Maxwell's curl equation for gives the fields below.

HERTZIAN DIPOLE: COMPLETE FIELDS

All other components (, , ) are zero. Valid for and . The terms are the radiation field, the terms the induction field and the terms (one in , one in ) the quasi-static (electric dipole) field. Multiplying out shows that only the product of the two terms is real, so the induction and quasi-static terms carry no net power.

Close to the dipole () the highest powers of dominate: has the shape of an electrostatic dipole's field, that of the magnetostatic field of a current element, and and are out of phase, so the power there is mostly reactive. Far away () only the terms survive, becomes negligible, and and form an outgoing spherical wave.

HERTZIAN DIPOLE: FAR FIELD

Valid for . In free space V/m. The pattern is : maximum broadside (), nulls along the axis (), and the same for every (omnidirectional in the plane ).

HERTZIAN DIPOLE: RADIATED POWER AND RADIATION RESISTANCE

The numerical forms assume free space (); is the peak current. grows as , which is why electrically short antennas are poor radiators: a small competes with the ohmic loss resistance.

Radiation resistance is the resistance which, carrying the antenna's reference current, would dissipate exactly the power the antenna radiates: . It is part of the real part of the input impedance, and it represents power leaving as radiation, not heat. The ohmic loss resistance , referred to the same current, accounts for the heat.

SHORT DIPOLE (CURRENT FALLING LINEARLY TO ZERO AT THE ENDS)

Centre-fed wire of total length (roughly ) with a triangular current: at the feed, zero at the ends. Its average current is , so its far field is half that of a Hertzian dipole of the same length with current : power and are one quarter. The pattern (), directivity (1.5) and beamwidth () are unchanged.

EXAM TIP

Read the current distribution before choosing the formula. 'Hertzian', 'infinitesimal' or 'elementary' dipole, a current element , or a short wire with uniform current (for example a top-loaded one): . A short centre-fed dipole whose current falls to zero at the tips: . The directivity is 1.5 in both cases, because the pattern is in both.

Radiation intensity, directivity, gain and effective aperture

Radiation intensity is the power radiated per unit solid angle. Because far-field power density falls as , depends on direction only. An isotropic radiator with the same total power would have in every direction. Directivity compares the actual antenna with that isotropic reference, and gain does the same after charging the antenna for its losses.

RADIATION INTENSITY AND TOTAL RADIATED POWER

Units W/sr. The element of solid angle is ; dropping the is an easy error to make.

DIRECTIVITY

A pure ratio set by the shape of the pattern alone; losses do not enter. 'Directivity' with no direction stated means the maximum, . dBi means decibels relative to an isotropic radiator; values relative to a half-wave dipole (dBd) are 2.15 dB lower.

BEAM SOLID ANGLE

is the normalized power pattern (). is the solid angle through which all the power would flow if stayed at inside it and were zero outside. The approximation uses the half-power beamwidths and in the two principal planes, in radians, and holds only for a single narrow main lobe with small side lobes.

DIRECTIVITY FROM BEAMWIDTHS (PENCIL BEAMS ONLY)

. It ignores the power in side lobes and so tends to overestimate . Do not use it for broad patterns such as a dipole's; integrate instead.

Half-power beamwidth (HPBW) is the angle, in a stated plane, between the two directions on either side of the maximum where the radiation intensity falls to half its maximum (field of maximum, dB). First-null beamwidth (FNBW) is the angle between the first nulls either side of the main lobe. For the Hertzian dipole, at and , so its HPBW is in any plane containing the dipole axis. Its beam solid angle is , giving .

RADIATION EFFICIENCY AND GAIN

= power accepted at the antenna terminals; is referred to the same current as . , with equality only for a lossless antenna. Gain as defined here excludes impedance mismatch. If the feed line of impedance is mismatched to the antenna impedance , the realized gain is with .

EFFECTIVE APERTURE

For a receiving antenna matched to its load and polarization-matched to an incident wave of power density arriving from the direction in which the gain is . For a lossless antenna , the maximum effective aperture . Hertzian or short dipole: ; half-wave dipole: . For a large aperture antenna (horn, reflector) of physical area , , with equality only for uniform, in-phase illumination of a lossless aperture; the ratio is the aperture efficiency.

EXAM TIP

Directivity by integration, fast. If for and zero below, then and : gives 4, gives 6. If over the full sphere, and . First check whether the expression you are given is a field pattern or a power pattern: goes as the square of the field.

WORKED EXAMPLE

A Hertzian dipole (uniform current) of length 1.5 m radiates in free space at 10 MHz with a peak current of 2 A. Its ohmic loss resistance is . Find (a) the radiation resistance, (b) the radiated power, (c) the radiation efficiency, (d) the gain in dBi, and (e) the far-field amplitude at 3 km in the broadside direction.

Solution
  1. m, so . The current is stated to be uniform, so the Hertzian-dipole formulas apply; at this length they are accurate to better than 0.2%.

  2. (a) .

  3. (b) W.

  4. (c) .

  5. (d) , so dBi.

  6. (e) 3 km is 100 wavelengths, so this is the far field. V/m.

  7. Check (e) through power: W/sr, so W/m², and V/m. The two agree.

Answer: (a) , (b) 3.95 W, (c) 0.798, (d) about 0.78 dBi, (e) about 6.28 mV/m.

Half-wave dipole and quarter-wave monopole

A thin centre-fed dipole of total length along the -axis carries a standing-wave current , which is zero at both ends; is the standing-wave amplitude. For this becomes for , with its maximum at the feed. Summing Hertzian-dipole fields along the wire gives the results below.

CENTRE-FED DIPOLE OF ANY LENGTH: FAR FIELD

Thin wire with the sinusoidal current above, far field. gives the half-wave result below. For the pattern is : (3.82 dBi), HPBW , and referred to . The broadside directivity peaks near (); longer dipoles break into more lobes and the broadside lobe shrinks.

HALF-WAVE DIPOLE: FAR FIELD

Thin wire, free space for the second form, = peak current at the feed. Nulls along the wire axis, maximum at , omnidirectional in ; the beam is slightly narrower than the Hertzian dipole's .

HALF-WAVE DIPOLE: RADIATION PARAMETERS

In free space , where , and . W. The feed sits at a current maximum, so is also the input resistance. The half-power points are at and . The reactance is tuned out by making the dipole slightly shorter than (typically about to , shorter for thicker wire); the resistance at resonance is then slightly below 73 Ω.

A monopole is a vertical wire of height fed against a large conducting ground plane. By image theory, an infinite perfectly conducting plane can be replaced by the mirror image of the current: a vertical current element's image is a vertical element carrying current in the same direction, while a horizontal element's image carries current in the opposite direction. So a monopole plus its image is a dipole of total length , and in the upper half-space () the fields are exactly that dipole's. Below the plane there is no field. For the same feed current the monopole therefore radiates only the upper half of the dipole's power, into half the solid angle.

MONOPOLE FROM ITS EQUIVALENT DIPOLE

Monopole of height over an infinite, perfectly conducting ground plane, compared with a dipole of total length with the same current distribution. Same , half the power, hence twice the directivity. A finite or lossy ground changes these values.

QUARTER-WAVE MONOPOLE

over an infinite perfect ground plane: half the half-wave dipole's impedance, twice its directivity. The pattern above the ground is the half-wave dipole's, with its maximum along the ground plane () and half power at , i.e. about above the ground plane.

SHORT MONOPOLE

Monopole of height over an infinite perfect ground plane. From the image rule: the equivalent dipole of length has or , and the monopole has half of that. The directivity is 3 in both cases (twice 1.5).

Antenna

Current

HPBW (E-plane)

Hertzian dipole,

Uniform,

1.5 (1.76 dBi)

Short dipole,

Triangular, at the feed

1.5 (1.76 dBi)

Half-wave dipole

,

1.64 (2.15 dBi)

Quarter-wave monopole, infinite perfect ground

,

3.28 (5.16 dBi)

Maximum along the ground; half power about above it

EXAM TIP

Radiation resistance is always quoted for a stated reference current. The general-length formulas refer it to the standing-wave amplitude , while the input resistance refers to the feed current . For a thin lossless dipole, . The two are equal only when is an odd multiple of , which is why 73 Ω and 36.5 Ω are also the input resistances of the half-wave dipole and quarter-wave monopole. The relation fails near , where the feed sits at a current minimum and the input resistance is very large.

EXAM TIP

Monopole trap: with the same feed CURRENT as a dipole, the monopole has the same and half the power. With the same input POWER, the monopole's is double the dipole's, because the power is squeezed into half the space. Either way the directivity doubles.

EXAM TIP

Folded half-wave dipole: a second half-wave conductor of the same radius runs parallel to the first, close to it, and is joined to it at both ends; one conductor is fed at its centre. The two conductors carry equal currents, so for the same radiated power the feed current is halved and the input resistance is about . The pattern and directivity are those of the ordinary half-wave dipole.

Friis transmission equation

A transmitting antenna of gain is fed with power , and a receiving antenna of gain is a distance away. In the direction of the receiver the power density is ; the product is the effective isotropic radiated power (EIRP). The receiver collects , where . Multiplying gives the Friis equation. By reciprocity (linear, passive, isotropic media), an antenna has the same pattern, gain and impedance whether it transmits or receives.

FRIIS TRANSMISSION EQUATION

Conditions: each antenna in the other's far field, free space with no obstacles or multipath, both antennas matched to their lines and to each other in polarization, and , taken in the direction of the other antenna. Use gains, not directivities, when the antennas have loss. is the free-space path loss.

FRIIS EQUATION IN DECIBELS

. The constant 32.44 dB uses m/s (the exact speed of light gives 32.45 dB). With fixed gains, rises by 6 dB for every doubling of distance or of frequency. With fixed effective apertures instead, the received power rises with frequency, because then sits in the denominator.

FRIIS EQUATION WITH MISMATCH AND POLARIZATION LOSS

, = reflection coefficients at the two antenna terminals. PLF = polarization loss factor: for two linearly polarized antennas whose polarization directions are at angle (so 0 for orthogonal linear polarizations), and 0.5 between a linearly polarized and a circularly polarized antenna.

EXAM TIP

Friis traps. (1) Convert dBi to a ratio before multiplying (), or stay in dB and add; never mix the two. (2) is the free-space wavelength at the operating frequency. (3) Power falls as but field strength as : doubling the distance halves the field and quarters the power ( dB).

WORKED EXAMPLE

A 3 GHz point-to-point link in free space uses a transmitter power of 5 W, a transmitting antenna of gain 20 dBi and a receiving antenna of gain 15 dBi, 2 km apart. Both antennas are matched, aligned boresight to boresight and polarization-matched. Take m/s. Find the received power in dBm and in watts, and the effective aperture of the receiving antenna.

Solution
  1. m.

  2. Free-space path loss: dB.

  3. dBm.

  4. dBm, i.e. mW W W.

  5. Receiving aperture: , so m².

  6. Check by power density: W/m², and W. The two agree.

Answer: dBm W; m².

Linear arrays and pattern multiplication

An array feeds several identical elements with chosen amplitudes and phases to shape a beam no single element could. In the far field every element sees the observer in the same direction , and the elements' distances differ only in the phase term. Element at is nearer the observer by , so its field arrives with extra phase ; the difference in the amplitude is negligible. The sum of these phasors, with each element's own feed phase added, is the array factor.

ARRAY FACTOR OF A UNIFORM LINEAR ARRAY

identical, identically oriented elements on the -axis at spacing , equal amplitudes, element fed with ( = progressive phase, positive when each element leads the one before it). is measured from the array axis; for an array along another axis use the cosine of the angle from that axis (for the -axis, ). has its maximum value at , and again at if those values are reached.

TWO-ELEMENT ARRAY

Two equal-amplitude elements, the one at leading the one at by . Nulls where , provided that value lies in the visible range (for example, with has no null at all). With unequal amplitudes, , , whose minimum (reached where is visible) is : there is no complete null unless .

Phase of element at

Maxima

Nulls

(in phase)

(broadside)

(antiphase)

(end-fire, both ways)

(lags by )

only (cardioid toward )

(in phase)

MAIN-BEAM DIRECTION AND PHASING

Set in the wanted direction; a real beam direction exists only if . With the convention the beam leans toward the end whose elements lag: for end-fire along , each element lags the one before it by , exactly the phase a wave travelling along picks up between elements. Varying electronically steers the beam (a phased array).

NULLS AND SIDE LOBES OF A UNIFORM ARRAY

Solve for and keep only solutions with . Small arrays have higher side lobes (about dB for ). With uniform amplitudes, changing or only moves the side lobes in angle (or out of the visible range); lowering their level needs a tapered amplitude distribution.

BROADSIDE ARRAY: NULLS AND BEAMWIDTHS

Uniform array with ; with and not a multiple of . The approximate forms (in radians unless marked) need . The beamwidth depends on the array length , not on alone.

ORDINARY END-FIRE ARRAY: NULLS AND BEAMWIDTHS

Uniform array with , beam toward ; with and not a multiple of . The square-root approximations (radians) need . For the same length, the end-fire beam is much wider than the broadside beam in the plane containing the axis, but it is a pencil beam around the axis instead of a fan all round it.

HANSEN-WOODYARD (INCREASED-DIRECTIVITY) END-FIRE ARRAY

Uniform amplitudes, beam toward , large . The extra phase puts just outside the visible range, which narrows the main beam. The spacing makes at , which keeps the back radiation small. The price is higher side lobes. The factor 1.8 is approximate (about 1.79 to 1.81 for from 10 to 50).

CONDITION FOR NO GRATING LOBES

A grating lobe is a second full maximum (), where . As runs from to , covers the visible range ; there is no grating lobe if this range contains no multiple of other than the main beam's . At exactly (broadside) grating lobes appear along the axis; at exactly (end-fire) a second beam appears at .

DIRECTIVITY OF UNIFORM ARRAYS OF ISOTROPIC ELEMENTS (APPROXIMATE)

Valid for and no grating lobes. For a broadside array with , exactly, for any (see the exact formula below).

EXACT DIRECTIVITY OF A UNIFORM ARRAY OF ISOTROPIC ELEMENTS

Equal amplitudes, progressive phase , valid when the main beam lies in the visible range (). When is a multiple of (for example ), every sine vanishes and for any ; this includes the two-beam end-fire array at . The same happens for end-fire at with , because .

PATTERN MULTIPLICATION

For identical, identically oriented elements; the array factor carries the feed amplitudes and phases, and mutual coupling is ignored. In dB the two patterns add. A null of either factor is a null of the total, and the total maximum lies where the product peaks, which need not be at the maximum of either factor.

Pattern multiplication also works for arrays of arrays: a set of identical sub-arrays spaced along an axis is itself an array whose 'element' is the sub-array. The binomial array is the standard case. Its amplitudes are binomial coefficients (1:2:1, 1:3:3:1, ...), because expands into exactly those coefficients, so an -element binomial array is two-element arrays multiplied together. At broadside, : no side lobes at all, but a wider main beam than a uniform array of the same length.

Analysing an array pattern

  1. Place the array on an axis, pick element 0 as the reference, and write , where is the angle from the array axis and is the progressive phase (positive = each element leads the one before it).

  2. Find the visible range: as goes from to , runs from down to .

  3. Main beam(s): , plus if they fall in the visible range (grating lobes).

  4. Nulls: with not a multiple of (for two elements, ) inside the visible range; convert each to .

  5. Multiply by the element pattern and add the element's own nulls (pattern multiplication).

  6. Directivity and beamwidth from the formulas above, or by integrating when the elements are not isotropic.

EXAM TIP

Watch the angle and the sign. The array factor depends on the angle from the ARRAY AXIS: for an array along it is a function of , which in the -plane is . And these formulas take as the phase by which each element LEADS the one before it: if the element at larger lags by , then and the beam leans toward .

EXAM TIP

An end-fire array at is a trap: with the visible range is , so there is a second full beam at and the array radiates equally both ways along its axis. Single-direction end-fire needs , and a two-element cardioid needs with .

WORKED EXAMPLE

A uniform linear array of 10 isotropic elements lies along the -axis. (a) With , find the progressive phase for a broadside beam, the first-null directions, the FNBW, the approximate HPBW and the directivity. (b) The same array is re-phased for ordinary end-fire toward . Is acceptable? (c) With , find , the FNBW, the approximate HPBW and the directivity of the end-fire array.

Solution
  1. (a) Broadside: . and .

  2. First nulls: , so and . FNBW (the approximation rad is close).

  3. HPBW (solving exactly gives ).

  4. Directivity: broadside with gives exactly, i.e. 10 dB.

  5. (b) End-fire toward needs . The visible range is then , and occurs at : a grating lobe as strong as the main beam. So is not acceptable for a single end-fire beam.

  6. (c) , so and . The visible range contains no , so there is no grating lobe.

  7. First null: , so and FNBW .

  8. HPBW (an exact solution gives ).

  9. Directivity: . The exact formula gives exactly 10 here, because every term vanishes: (10 dB).

Answer: (a) ; nulls at and ; FNBW ; HPBW ; . (b) No: a full-strength grating lobe appears at . (c) ; FNBW ; HPBW ; .

PREVIOUS-YEAR QUESTIONS

2 questions from the EC papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE ECE 2026 · GATE ECE 2020

GATE 2026 EC Q.21

GATE 2026 · ECQ.21 · 1 mark · MCQ

Consider the Friis' transmission equation , where and are the received and the transmitted powers, respectively. and are the gain of transmitting and receiving antennas, respectively, is the distance between the transmitting and receiving antennas, and is the wavelength in free space.

Given: , and .

Choose the distance (), in km, from the following options at which the received power, ?

  1. (A)

  2. (B)

  3. (C)

  4. (D)

Answer and solution
See the question in the paper
GATE 2026 EC Q.21: answer and solution

GATE 2020 EC Q.43

GATE 2020 · ECQ.43 · 2 marks · MCQ

For an infinitesimally small dipole in free space, the electric field in the far field is proportional to , where . A vertical infinitesimally small electric dipole () is placed at a distance () above an infinite ideal conducting plane, as shown in the figure. The minimum value of , for which one of the maxima in the far field radiation pattern occurs at , is

GATE 2020 EC question 43: A horizontal hatched line is the infinite conducting plane, labelled 'Infinite conducting plane' beneath it;Tap to open full size
Figure from the GATE 2020 EC paper. It is part of the question.
Figure description

A horizontal hatched line is the infinite conducting plane, labelled 'Infinite conducting plane' beneath it; the origin 0 is marked on it and the horizontal axis to the right is labelled y. A dotted vertical line rises from the origin and is labelled Z at the top, with an arrowhead. On this z-axis, at height h above the plane, sits a short upward arrow labelled I delta l (the vertical current element, pointing in +z). A double-headed vertical arrow to the left, running from the plane up to the dipole, is labelled h. A thin straight line leaves the origin up and to the right into the first quadrant (the observation direction), and the angle between it and the vertical z-axis is marked theta with a small arc.

  1. (A)

  2. (B)

  3. (C)

  4. (D)

Answer and solution
GATE 2020 EC Q.43: answer and solution

PRACTICE

5 questions to test yourself

PRACTICE 11 mark · NAT

A Hertzian dipole of length 2 cm carries a uniform current and radiates in free space at 300 MHz. Take m/s. Its radiation resistance, in ohms (rounded off to three decimal places), is ______.

Answer and solution

Answer: (accepted range )

  1. m, so .

  2. .

PRACTICE 22 marks · MCQ

Two identical Hertzian dipoles, both oriented along the -axis, are placed on the -axis at and and fed with equal, in-phase currents. In which directions is the total far field zero?

  1. (A)

    Along only

  2. (B)

    Along and

  3. (C)

    Along and

  4. (D)

    Along only

Answer and solution

Answer: (B)

  1. Element pattern: an -directed dipole radiates as the sine of the angle from its own axis, . It is zero along and equal to 1 along and .

  2. Array factor (origin at the centre, , , array along ): . It is zero at () and equal to 2 everywhere in the -plane.

  3. By pattern multiplication the total is zero along (element null) and along (array null). Along both factors are at their maximum, so the total is maximum there.

PRACTICE 31 mark · MCQ

A quarter-wave monopole stands on an infinite, perfectly conducting ground plane. Compared with a half-wave dipole in free space carrying the same feed current, the monopole has

  1. (A)

    the same radiation resistance and the same directivity

  2. (B)

    half the radiation resistance and twice the directivity

  3. (C)

    twice the radiation resistance and half the directivity

  4. (D)

    half the radiation resistance and the same directivity

Answer and solution

Answer: (B)

  1. By image theory the monopole and its image form a half-wave dipole, so above the ground plane the fields, and hence , equal the dipole's.

  2. There is no field below the plane, so for the same current the monopole radiates half the power: .

  3. Same with half the power gives doubled: .

PRACTICE 42 marks · NAT

A 10 GHz link operates in free space. The transmitter delivers 100 W to a matched antenna of gain 20 dBi, and a receiving antenna of gain 30 dBi is 5 km away. The two antennas face each other along their boresights, and the receiving antenna is matched to its load and polarization-matched. Take m/s. The power delivered to the receiver load, in microwatts (rounded off to two decimal places), is ______.

Answer and solution

Answer: (accepted range )

  1. m. , .

  2. Receiving aperture: m².

  3. Power density at the receiver: W/m².

  4. W W.

  5. Check in dB: dB, so dBm W.

PRACTICE 52 marks · MSQ

A uniform linear array of isotropic elements lies along the -axis with spacing and progressive phase (each element leads the previous one by ), with measured from the axis. Which of the following statements is/are correct?

  1. (A)

    With , the first nulls on either side of the main beam are at and .

  2. (B)

    With , the array has a single main beam, directed along .

  3. (C)

    The normalized array factor can be written as with .

  4. (D)

    With , the directivity is exactly 4.

Answer and solution

Answer: (A), (C), (D)

  1. , so and . C is correct.

  2. A: with the first nulls are at , i.e. , , . Correct.

  3. B: with , runs from 0 at to at . At , again, so there is a second full beam at . Incorrect.

  4. D: a broadside array of isotropic elements with has exactly, here 4. Correct.

OFFICIAL GATE 2027 SYLLABUS WORDING

Copied from the GATE 2027 syllabus; see all 2027 syllabi.

GATE EC

dipole and monopole antennas, linear antenna arrays

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