L15 - Aperture Distributions and Efficiency#

ECE 444 · Fall 2026

Aperture Distributions and Efficiency

Change the distribution and you change the pattern.

Lesson 15 · Antennas, Phased Arrays, and Radar Systems · Dr. Neil Rogers

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Learning Objectives

  1. I can describe aperture distributions and calculate aperture efficiency for a given illumination.

Why Return to Apertures?

  • Lessons 6, 13, and 14 found the beamwidth and sidelobes of a wire, a horn, a dish.

  • Each depends only on the aperture field \(E_a\), which an array sets element by element.

Amplitude shapes the beam; phase steers it.

In Lesson 6 we derived the pattern of a line source, and in Lessons 13 and 14 we found the beamwidth, sidelobe level, and aperture efficiency of a horn and a reflector. Each of those results was worked out for a particular antenna, but none of them depends on the antenna. The far field depends only on the field across the radiating opening, the aperture distribution \(E_a\), so the same relationships hold for any aperture. This lesson collects them in a form that applies to every aperture: the shape of \(E_a\) sets the sidelobe level and the efficiency, and the aperture’s length in wavelengths sets the beamwidth.

That generalization is what we need for the rest of Module 3. A horn or a reflector fixes its aperture distribution with metal. A phased array sets it with the amplitude and phase of each element, so we can choose the distribution we want. Lesson 15 tells us which distribution produces which pattern, and Lessons 16 through 22 show how the PHASER’s element weights produce that distribution.

Part 1: The Aperture as Source

  • An aperture is the opening a wave leaves through: a horn mouth, a reflector face, an array.

  • The far field depends only on \(E_a\), the tangential field across the opening.

  • Anything behind it matters only through the \(E_a\) it produces.

Field across an aperture on the left and the far-field pattern it produces on the right

An aperture is the opening a wave leaves through: the mouth of a horn, the projected face of a reflector, the radiating surface of a patch, or the row of elements on the PHASER board. Whatever is behind it, the far field depends on only one thing — the tangential electric field across that opening, which is the aperture distribution \(E_a\). Change the distribution and you change the pattern. Leave the distribution alone and no change behind the aperture can alter the far field.

The Transform Relationship

Lesson 6 established the relationship. For a source confined to a region, the far field is the Fourier transform of the source distribution, and Module 1 carried that out for the line source. Take the one-dimensional aperture of length \(L\) lying along \(x\), radiating into the half-space in front of it, and measure the angle \(\theta\) from broadside (the aperture normal). The space factor is

\[S(\theta) = \int_{-L/2}^{L/2} E_a(x)\ e^{\ jkx\sin\theta}\ dx\]

with \(k = 2\pi/\lambda\). The exponent is the extra path from the element at \(x\) relative to the aperture center, converted to phase. Toward a far-field point at angle \(\theta\), the rays from the center and from \(x\) are parallel, and they differ in length by \(x\sin\theta\), a phase of \(kx\sin\theta\).

A one-dimensional aperture with its center and a point a distance x from it. Parallel rays leave both toward a far-field direction at angle theta from broadside, and the two paths differ by x sine theta, a phase of k x sine theta

That integral is a Fourier transform: the aperture coordinate \(x\) is the variable, and the transform variable is the space frequency

\[u = \frac{L}{\lambda}\sin\theta\]

  • \(S(\theta)\) is the Fourier transform of \(E_a(x)\) in the space frequency \(u\).

  • The shape of \(\vert S\vert\) against \(u\) depends only on the shape of \(E_a\).

  • \(\theta\) runs from broadside, so \(u\) carries \(\sin\theta\) and nulls fall at integer \(u\).

Working in \(u\) instead of \(\theta\) is what makes the results in this lesson reusable. The shape of \(\vert S\vert\) against \(u\) depends only on the shape of the illumination. The aperture’s size in wavelengths, \(L/\lambda\), sets how many degrees correspond to each unit of \(u\), and nothing else.

Note

Module 1 wrote the line-source pattern in the polar angle measured from the wire axis. Module 3 measures \(\theta\) from broadside instead, which is what every phased-array plot and every PHASER readout uses, so the space frequency here carries \(\sin\theta\) rather than \(\cos\theta\). Lesson 16 makes that substitution explicit once; after that, broadside is \(\theta = 0\) everywhere in the module.

The symbol \(u\) also changes scale. Lesson 6 wrote the uniform line source as \(\sin u/u\) with \(u = (kL/2)\cos\theta\), so its nulls sat at \(u = \pi\) and its first sidelobe at \(u = 4.493\). The \(u\) here is Lesson 6’s \(u\) divided by \(\pi\), with \(\sin\theta\) in place of \(\cos\theta\), which puts the nulls at the integers and the first sidelobe at \(u = 1.430\). Lesson 14’s circular aperture keeps the \(\pi\), \(u = (\pi D/\lambda)\sin\theta\).

Key Point: Shape and Size

The shape of the illumination sets the sidelobe level and the beamwidth constant. The size in wavelengths scales the beamwidth and leaves the sidelobes alone.

Top: the uniform aperture pattern against space frequency u is one curve for every length, and apertures 2, 5, and 20 wavelengths long see it out to u = 2, 5, and 20. Bottom: the same three apertures against angle have beams 25.6, 10.2, and 2.5 degrees wide, and every first sidelobe sits at minus 13.3 dB

The far field is the Fourier transform of the aperture field, and every aperture and array design in this course sets both its shape and its size.

The figure shows both halves of that statement for a uniform aperture. Against \(u\) there is one pattern, and the aperture length only decides how much of it is visible: real angles reach \(\vert u\vert = L/\lambda\), so a \(2\lambda\) aperture sees out to \(u = 2\) and a \(20\lambda\) aperture out to \(u = 20\). Against \(\theta\) the same three apertures have beams 25.6°, 10.2°, and 2.5° wide, and every one of them puts its first sidelobe at −13.3 dB.

Part 2: The Uniform Aperture

Start with the simplest illumination, a constant field \(E_a(x) = E_0\) across the whole opening. The integral is elementary:

\[\begin{split}\begin{aligned} S(\theta) &= E_0\int_{-L/2}^{L/2} e^{\ jkx\sin\theta}\ dx \\ &= E_0\ \frac{e^{\ jkx\sin\theta}}{jk\sin\theta}\Bigg|_{-L/2}^{L/2} \\ &= E_0 L\ \frac{\sin\left(\tfrac{1}{2}kL\sin\theta\right)}{\tfrac{1}{2}kL\sin\theta} \end{aligned}\end{split}\]

With \(k = 2\pi/\lambda\) the argument is \(\tfrac{1}{2}kL\sin\theta = \pi u\), so the normalized field pattern of a uniform aperture is a sinc function of the space frequency:

\[\vert F(u)\vert = \left\vert \frac{\sin \pi u}{\pi u}\right\vert, \quad u = \frac{L}{\lambda}\sin\theta .\]

That expression gives three numbers we will use throughout the module: the null positions, the beamwidth, and the first sidelobe level.

Nulls and Beamwidth

\[\begin{split}\begin{aligned} \sin\pi u &= 0 \\ \sin\theta_\text{null} &= \pm\frac{\lambda}{L},\ \pm\frac{2\lambda}{L},\ \dots \end{aligned}\end{split}\]

  • An aperture shorter than \(\lambda\) has no null in real space, so it radiates broadly.

\(\sin\pi u\) vanishes at \(u = \pm 1, \pm 2, \pm 3, \dots\), so the first null sits at \(\sin\theta = \lambda/L\). An aperture shorter than one wavelength has no null anywhere in real space, which is why a small aperture radiates broadly however we feed it.

The beamwidth comes from the half-power points. Solving

\[\frac{\sin \pi u}{\pi u} = \frac{1}{\sqrt{2}}\]

gives \(\pi u = 1.3916\), so the half-power points sit at \(u = \pm 0.4429\) and

\[\theta_\text{HP} \approx 0.886\ \frac{\lambda}{L}\ \text{rad} = 50.8^\circ\ \frac{\lambda}{L}.\]

  • The half-power points sit at \(u = \pm 0.4429\).

  • Memorize 0.886: Lesson 20’s array beamwidth and every PHASER beamwidth prediction use it.

The half-power points are at \(\sin\theta = \pm 0.4429\ \lambda/L\). The beamwidth spans both points, and for a beam a few degrees wide the sine is the angle in radians:

\[\begin{split}\begin{aligned} \theta_\text{HP} &= 2\arcsin\left(0.4429\ \frac{\lambda}{L}\right) \\ &\approx 0.886\ \frac{\lambda}{L}\ \text{rad} \end{aligned}\end{split}\]

The small-angle step is good to within 1% for \(L \ge 2\lambda\). Lesson 6’s \(2\lambda\) line source gave 25.6° exactly against 25.4° from the approximation.

The 0.886 is worth memorizing. It is the constant behind the array beamwidth formula in Lesson 20 and behind every beamwidth prediction you will make on the PHASER.

First Sidelobe

  • The first sidelobe peaks at \(u = 1.430\), where \(\vert F\vert = 0.217\), or −13.3 dB.

  • The level does not depend on \(L\). A longer aperture narrows the beam and leaves the sidelobe 13.3 dB down.

Uniform aperture pattern in decibels with half-power width, first null and first sidelobe marked

The first sidelobe peaks where the slope of \(\sin\pi u/\pi u\) is zero, which happens where \(\tan\pi u = \pi u\). The first root past the main lobe is \(u = 1.430\), where \(\vert F\vert = 0.217\), or −13.3 dB. The first sidelobe level does not depend on \(L\). Making a uniform aperture longer narrows the beam and raises the gain, and it leaves the first sidelobe exactly 13.3 dB below the peak. The shape of the illumination sets the sidelobes, and the uniform shape fixes them at −13.3 dB.

Worked Example: Beamwidth of a 10λ Aperture

  • At \(L = 10\lambda\): \(\theta_\text{HP} = 5.08^\circ\), first nulls at ±5.74°, first sidelobe at −13.3 dB.

  • At 10 GHz that aperture is 0.30 m. At 3 GHz the same 0.30 m is \(3\lambda\), and the beam opens to 16.9°.

Worked example — beamwidth of a 10-wavelength aperture

A uniformly illuminated aperture is \(L = 10\lambda\) long. Its half-power beamwidth is

\[\theta_\text{HP} = 50.8^\circ \times \frac{\lambda}{10\lambda} = 5.08^\circ,\]

its first nulls are at \(\sin\theta = 0.1\), or \(\theta = \pm 5.74^\circ\), and its first sidelobe is 13.3 dB down. At 10 GHz that aperture is 0.30 m long. At 3 GHz the same 0.30 m is only \(3\lambda\), and the beam opens to 16.9°.

Rectangular and Circular Apertures

A rectangular aperture is no harder to analyze when the illumination separates, \(E_a(x,y) = E_x(x)E_y(y)\). The double integral factors, and the pattern in each principal plane is the line-source result for that dimension:

\[\begin{split}\begin{aligned} \theta_{\text{HP},x} &= 0.886\ \frac{\lambda}{L_x} \\ \theta_{\text{HP},y} &= 0.886\ \frac{\lambda}{L_y} \end{aligned}\end{split}\]

  • A tall narrow aperture makes a wide flat beam, and a wide short aperture makes a narrow tall one.

  • A uniform circle of diameter \(D\): \(\theta_\text{HP} \approx 1.029\ \lambda/D = 59^\circ\ \lambda/D\), first sidelobe −17.6 dB.

Left: the X-band aperture to scale, 0.682 m wide and 0.152 m tall, cosine illumination across and uniform up. Right: the half-power contour of its beam, 3.0 degrees wide in azimuth and 10.0 degrees tall in elevation. The long dimension makes the narrow beam

The figure draws the aperture that the X-band worked example below arrives at, 0.682 m by 0.152 m at 10 GHz. The 0.682 m dimension makes the 3° azimuth beam and the 0.152 m dimension the 10° elevation beam, so the beam’s footprint is the aperture turned 90°.

The aperture efficiency of Part 3 factors the same way. Both integrals in its ratio split into an \(x\) integral times a \(y\) integral, so

\[\eta_\text{ap} = \eta_x\ \eta_y\]

where each factor is the line-source ratio for its own dimension.

A circular aperture of diameter \(D\) does not separate. Lesson 14 integrated Lesson 6’s radiation vector over a uniformly lit disk and found a Bessel function in place of the sinc:

\[\begin{split}\begin{aligned} \vert\mathbf{N}(\theta)\vert &\propto \left\vert\frac{2J_1(u)}{u}\right\vert \\ u &= \frac{\pi D}{\lambda}\sin\theta \end{aligned}\end{split}\]

Its half-power beamwidth and first sidelobe are

\[\begin{split}\begin{gathered} \theta_\text{HP} \approx 1.029\ \frac{\lambda}{D}\ \text{rad} = 59^\circ\ \frac{\lambda}{D} \\ \text{first sidelobe} = -17.6\ \text{dB} \end{gathered}\end{split}\]
Uniform square and uniform circular apertures of the same width against sine of angle times width in wavelengths: the circle's half-power beam is 1.029 against 0.886 and its first sidelobe minus 17.6 dB against minus 13.3 dB; an inset shows that the circle, projected onto one cut, is already tapered toward its edges

The half-power point is at \(u = 1.616\), and the same small-angle step as the line source gives

\[\theta_\text{HP} \approx 2\left(\frac{1.616}{\pi}\right)\frac{\lambda}{D} = 1.029\ \frac{\lambda}{D}.\]

The circle is a little wider in beam than a square of the same width and a little better in sidelobes, and for the same reason: the edges of a circular aperture carry less of the total area than the edges of a rectangle, so the aperture is already mildly tapered as seen along any cut.

Part 3: Aperture Efficiency

Lesson 2 defined the effective aperture and the aperture efficiency \(\eta_\text{ap}\), the fraction of the physical area \(A\) that the antenna uses:

\[A_e = \frac{G\lambda^2}{4\pi} = \eta_\text{ap} A\]

Lesson 13 derived \(\eta_\text{ap}\) from the aperture field for the horn, and Lesson 14 used the same ratio for the reflector. This part restates that result and evaluates it for the illuminations the rest of the module uses.

Lesson 13 derived the directivity of an aperture from its field:

\[\begin{split}\begin{aligned} \eta_\text{ap} &= \frac{\left\vert\int_A E_a\ dS'\right\vert^2}{A\int_A \vert E_a\vert^2\ dS'} \\ D &= \eta_\text{ap}\ \frac{4\pi A}{\lambda^2} \end{aligned}\end{split}\]

The ratio is coherent gain over available gain. By the Cauchy-Schwarz inequality it never exceeds one, and it equals one only for constant amplitude and constant phase.

Read that ratio as coherent gain over available gain. The numerator grows with field that adds in phase; the denominator is the power the aperture had to radiate. By the Cauchy-Schwarz inequality the ratio never exceeds one, and it equals one only when \(E_a\) has constant amplitude and constant phase over the whole aperture. Uniform illumination is the most efficient illumination there is, and every departure from it — a taper, a phase error, a piece of aperture with nothing on it — lowers the efficiency.

The derivation compares two quantities. At \(\theta = 0\) every phase factor in the radiation integral is 1, so the boresight field is the coherent sum of the aperture field, \(\int_A E_a\ dS'\). The power leaving is the plane-wave power flowing through the opening. In Lesson 13’s notation,

\[\begin{split}\begin{aligned} U_{\max} &= \frac{k^2}{8\pi^2\eta_0}\left\vert\int_A E_a\ dS'\right\vert^2 \\ P_{\text{rad}} &= \frac{1}{2\eta_0}\int_A \vert E_a\vert^2\ dS' \end{aligned}\end{split}\]

and Lesson 2’s directivity is their ratio:

\[\begin{split}\begin{aligned} D &= \frac{4\pi U_{\max}}{P_{\text{rad}}} \\ &= \frac{4\pi}{\lambda^2}\ \frac{\left\vert\int_A E_a\ dS'\right\vert^2}{\int_A \vert E_a\vert^2\ dS'} \\ &= \eta_\text{ap}\ \frac{4\pi A}{\lambda^2} \end{aligned}\end{split}\]

The chain gives directivity. Gain is \(G = \eta_\text{rad} D\) (Lesson 2), and the walls of a horn and the surface of a reflector dissipate so little that \(\eta_\text{rad} \approx 1\) (Lesson 13). For those antennas

\[G \approx \eta_\text{ap}\ \frac{4\pi A}{\lambda^2}\]

Evaluated with the amplitude of \(E_a\) alone, the ratio is what Lesson 14 called the taper efficiency \(\eta_t\) (Lesson 13 wrote \(\eta_\text{taper}\)). The worked example and the taper table below give \(\eta_t\); an aperture with no other loss has \(\eta_\text{ap} = \eta_t\).

Worked Example: Efficiency of a Cosine Illumination

\[\begin{split}\begin{aligned} \eta_t &= \frac{\left(\int_{-1/2}^{1/2}\cos(\pi\xi)\ d\xi\right)^2}{1 \times \int_{-1/2}^{1/2}\cos^2(\pi\xi)\ d\xi} \\ &= \frac{(2/\pi)^2}{1 \times (1/2)} \\ &= \frac{8}{\pi^2} = 0.811 \end{aligned}\end{split}\]

  • A horn’s cosine illumination delivers 81% of the gain its area could support, a 0.9 dB loss.

Left: the cosine illumination across the aperture with its mean, 0.637, which is the coherent sum. Right: its square with mean 0.500, the available power. The taper efficiency is 0.637 squared over 0.500, or 0.811

Worked example — efficiency of a cosine illumination

Take \(E_a(x) = \cos(\pi x/L)\) over \(-L/2 \le x \le L/2\), which is the illumination inside the mouth of a pyramidal horn in its broad dimension. Work in \(\xi = x/L\) so the aperture runs from \(-1/2\) to \(1/2\) and its length is 1:

\[\begin{split}\begin{aligned} \int_{-1/2}^{1/2}\cos(\pi\xi)\ d\xi &= \frac{2}{\pi} \\ \int_{-1/2}^{1/2}\cos^2(\pi\xi)\ d\xi &= \frac{1}{2} \end{aligned}\end{split}\]
\[\eta_t = \frac{(2/\pi)^2}{1 \times (1/2)} = \frac{8}{\pi^2} = 0.811 .\]

The cosine-illuminated aperture delivers 81% of the gain its area could support, a loss of 0.9 dB. The same arithmetic gives 0.75 for a triangular illumination and \(2/3\) for \(\cos^2\).

Other Aperture-Efficiency Losses

  • \(\eta_t\) is one factor of \(\eta_\text{ap}\). Spillover, phase error, blockage, and cross-polarization multiply in.

  • Lesson 14’s reflector came to 0.66 with \(\eta_t = 0.85\); a horn comes to 0.49.

  • None of these losses is heat, so \(\eta_\text{ap}\) is not \(\eta_\text{rad}\).

Lesson 14's efficiency budget as a waterfall: spillover, taper, blockage, surface error, and the remaining losses take an ideal aperture from 1.00 to 0.66

Two cautions on using \(\eta_\text{ap}\) in practice. First, the taper efficiency \(\eta_t\) is only one of its factors. A real reflector also loses to spillover past the rim (Lesson 14’s \(\eta_s\)), to phase error across the surface, to feed and strut blockage, and to cross-polarization, and \(\eta_\text{ap}\) as measured is the product of all of them. Lesson 14’s efficiency budget took an ordinary reflector from 1.00 to 0.66 that way, with a taper factor of only 0.85. A horn’s cosine taper alone gives 0.81, and Lesson 13 found the rest in the flare’s phase error: \(0.78 \times 0.77 \times 0.81 = 0.49\), which is why horns come in near 0.5. Second, \(\eta_\text{ap}\) is not radiation efficiency. \(\eta_\text{rad}\) accounts for power turned into heat; \(\eta_\text{ap}\) accounts for power that radiates but does not end up on boresight.

Key Point: Gain From Area

\[G = \eta_\text{ap}\ \frac{4\pi A}{\lambda^2}\]

Area and wavelength set the ceiling, and the illumination decides how close the antenna comes to that ceiling. Use \(\eta_\text{ap} \approx 0.5\) for a horn and 0.55 to 0.7 for a good reflector when no better value is available.

For a low-loss aperture, \(\eta_\text{rad} \approx 1\) and gain equals directivity.

Part 4: Tapering and the Pattern

  • Tapering lets the illumination fall off toward the edges.

  • The uniform aperture’s sharp edge produces its −13.3 dB sidelobes; a rounded edge lowers them.

  • The outer aperture then works below full strength, so the beam widens and the gain falls.

Uniform, cosine, and cosine-squared patterns against space frequency on a log scale, each with the dashed envelope its sidelobes follow: the uniform sidelobes fall 6 dB per octave, the cosine's 12, and the cosine-squared's 18

Tapering means letting the illumination fall off toward the edges of the aperture instead of stopping abruptly. The uniform aperture’s −13.3 dB sidelobes come from the sharp edge: the transform of a function with a step in it decays slowly. Round the edge off and the sidelobes fall away much faster. The outer part of the aperture no longer works at full strength, so the aperture behaves as though it were shorter and less complete than it is, the beam widens, and the gain falls.

The far sidelobes show how much the edge matters. Each level of smoothness at the edge, first the illumination itself and then its slope, adds one power of \(u\) to the decay. The uniform illumination steps to zero, and its sidelobes fall as \(1/u\), or 6 dB per octave of \(u\). The cosine reaches zero continuously but with a slope there, and its sidelobes fall as \(1/u^2\), or 12 dB per octave. The \(\cos^2\) illumination arrives with zero slope as well, and its sidelobes fall as \(1/u^3\), or 18 dB per octave.

The Taper Trade

Four illuminations cover most of the ground, and their numbers are the ones this course uses everywhere:

Taper

Sidelobe (\(\text{dB}\))

HPBW (\(\times\ \lambda/L\))

\(\eta_t\)

\(\Delta G\) (\(\text{dB}\))

Uniform

−13.3

0.886

1.00

0

Cosine

−23

1.19

0.81

−0.9

Triangular

−26.5

1.28

0.75

−1.25

Cosine\(^2\)

−31.5

1.44

0.667

−1.8

The table reads as one continuous trade. Going from uniform to \(\cos^2\) lowers the first sidelobe by 18 dB, widens the beam by 63%, and loses 1.8 dB of gain. There is no illumination that lowers sidelobes and narrows the beam at the same time. When a radar system needs low sidelobes to keep clutter and jamming out of the receiver, it accepts a wider beam or a larger aperture to get them.

The four illuminations plotted as first sidelobe against beamwidth constant: uniform at minus 13.3 dB and 0.886 with full gain, cosine at minus 23.0 dB and 1.189 with minus 0.91 dB of gain, triangular at minus 26.5 dB and 1.276 with minus 1.25 dB, and cosine squared at minus 31.5 dB and 1.441 with minus 1.76 dB. Every step to lower sidelobes widens the beam and loses gain

Illumination and Length

The widget below computes the pattern of each illumination directly from the aperture integral. Pick an illumination and watch three things: the sidelobe level and the half-power constant change together, while the aperture length slider moves the pattern in angle without touching either. Set the aperture to \(2\lambda\) and step through the illuminations to see the sidelobes leave visible space entirely, then set it to \(20\lambda\) and confirm that the first sidelobe of the uniform case is still exactly 13.3 dB down.

Four Illuminations and Their Patterns

Four aperture illuminations and the patterns they produce in decibels

  • From uniform to \(\cos^2\), the first sidelobe drops 18 dB, the beam widens 63%, and gain falls 1.8 dB.

  • No illumination lowers sidelobes and narrows the beam at once.

  • A low-sidelobe radar accepts a wider beam or a larger aperture.

Part 5: Designing in Wavelengths

Every result so far depends on \(L/\lambda\) and \(A/\lambda^2\), never on \(L\) or \(A\) alone. That gives two scaling rules that let us size hardware before you know anything else:

  • Beamwidth scales as \(\lambda/L\). Double the aperture in wavelengths and the beam is half as wide.

  • Gain scales as \(A/\lambda^2\). Double both aperture dimensions in wavelengths and the gain rises by a factor of four, or 6 dB.

Both rules apply whether the aperture or the frequency changes. An antenna moved from 10 GHz to 20 GHz doubles its size in wavelengths, halves both beamwidths, and gains 6 dB, provided the feed keeps illuminating it the same way. That proviso matters on real hardware, since a feed horn’s illumination pattern is itself frequency-dependent, but the scaling is the right first estimate.

The figure follows the 0.30 m uniform aperture of the \(10\lambda\) worked example from 1 to 20 GHz. Its beamwidth falls as \(1/f\): the exact \(2\arcsin(0.4429\ \lambda/L)\) gives 17.0° at 3 GHz, where the small-angle \(50.8^\circ\ \lambda/L\) gives 16.9°, and 5.08° at 10 GHz. A uniform 0.30 m square over the same band gains 6 dB per doubling of frequency, from 20.5 dBi at 3 GHz to 31.0 dBi at 10 GHz.

One 0.30 m uniform aperture from 1 to 20 GHz. Top: the half-power beamwidth narrows from 53 degrees at 1 GHz to 17.0 at 3 GHz and 5.08 at 10 GHz. Bottom: the gain of a uniform 0.30 m square rises 6 dB per doubling of frequency, 20.5 dBi at 3 GHz and 31.0 dBi at 10 GHz

Worked Example: Sizing an X-Band Aperture

  • Specification at 10 GHz, \(\lambda = 0.03\ \text{m}\): 3° azimuth, 10° elevation, azimuth sidelobes below −20 dB.

  • Cosine in azimuth: \(L_x = 0.682\ \text{m} = 22.7\lambda\).

  • Uniform in elevation: \(L_y = 0.152\ \text{m} = 5.08\lambda\).

\[G = 0.81\ \frac{4\pi(0.104)}{(0.03)^2} = 30.7\ \text{dBi}\]

Worked example — sizing an X-band aperture

Size a rectangular aperture at 10 GHz for a 3° beam in azimuth, a 10° beam in elevation, and azimuth sidelobes no higher than −20 dB. At 10 GHz, \(\lambda = 3\times10^8/10^{10} = 0.03\ \text{m}\).

Quantity

Work

Result

Azimuth illumination

uniform gives only −13.3 dB; cosine gives −23 dB

cosine, constant 1.19

Azimuth length

\(L_x = 1.19\lambda/\theta_\text{HP} = 1.19(0.03)/0.05236\)

\(0.682\ \text{m} = 22.7\lambda\)

Elevation length

uniform is fine; \(L_y = 0.886(0.03)/0.1745\)

\(0.152\ \text{m} = 5.08\lambda\)

Aperture area

\(A = 0.682 \times 0.152\)

\(0.104\ \text{m}^2\)

Aperture efficiency

\(\eta_x\eta_y = 0.81 \times 1.00\)

0.81

Gain

\(G = 0.81(4\pi)(0.104)/(0.03)^2 = 1176\)

30.7 dBi

Largest dimension

diagonal, \(\sqrt{0.682^2 + 0.152^2}\)

0.699 m

Far-field distance

\(2D^2/\lambda = 2(0.699)^2/0.03\)

33 m

Lesson 5 defined \(D\) in \(2D^2/\lambda\) as the antenna’s largest dimension, which for a rectangle is its diagonal.

Gain as the Design Output

Worked example — sizing an X-band aperture, checking the design

Check the gain against the pencil-beam estimate from Lesson 2: \(41{,}253/(3 \times 10) = 1375\), or 31.4 dBi, is the lossless geometric bound, and the practical constant of \(26{,}000\) to \(32{,}400\) gives 29.4 to 30.3 dBi. The computed 30.7 dBi sits just above that band, which is the right place for an aperture whose only loss is a known amplitude taper. A real antenna with spillover and a feed in front of it would land inside it.

The design has two further consequences. The 3° azimuth requirement is what made this antenna 0.68 m wide, and the sidelobe requirement made it 34% wider than a uniform aperture with the same beamwidth would have been. Also, a 33 m far-field distance means this antenna cannot be pattern-tested in any ordinary room, which is the compact-range and near-field-scanning problem from Lesson 9.

The sidelobe specification chooses the illumination, the illumination fixes the beamwidth constant, and the beamwidth fixes the length. The gain comes out last: it is a design output, not a design input.

The X-band design chain. The azimuth sidelobe limit of minus 20 dB picks the cosine illumination, constant 1.19, and the 3 degree beam then fixes the azimuth length at 0.682 m. Elevation has no sidelobe limit, so it stays uniform, constant 0.886, and the 10 degree beam fixes 0.152 m. Area 0.104 square meters at efficiency 0.81 gives 30.7 dBi, just above the practical band of 29.4 to 30.3 dBi and below the 31.4 dBi pencil-beam bound

Notice how the requirements mapped onto the aperture. The sidelobe specification chose the illumination, the illumination fixed the beamwidth constant, the beamwidth specification then fixed the length, and only after all of that did the gain come out. For an aperture antenna, gain follows from the design; it is not a design input.

Summary: The Aperture-to-Pattern Relationship

Symbol / idea

What it is

Number to remember

\(E_a(x)\)

aperture distribution, the field across the opening

the pattern depends on nothing else

\(u = (L/\lambda)\sin\theta\)

space frequency; pattern \(=\) Fourier transform of \(E_a\)

shape sets \(u\)-pattern, size sets the angle scale

\(\vert F\vert = \vert\sin\pi u/\pi u\vert\)

uniform line source or aperture

nulls at integer \(u\)

Summary: Beamwidth and Sidelobes

Symbol / idea

What it is

Number to remember

\(\theta_\text{HP}\)

half-power beamwidth of a uniform aperture

\(0.886\ \lambda/L\), or \(50.8^\circ\ \lambda/L\)

first sidelobe

uniform aperture, any length

−13.3 dB; circular uniform, −17.6 dB and \(1.029\ \lambda/D\)

Summary: Efficiency, Gain, and the Trade

Symbol / idea

What it is

Number to remember

\(\eta_\text{ap}\), \(\eta_t\)

coherent gain over available gain; \(\eta_t\) is the amplitude-only part

\(\eta_t = 1.00\) / 0.81 / 0.75 / 0.667 for uniform / cos / triangular / cos\(^2\)

\(G = \eta_\text{ap}\ 4\pi A/\lambda^2\)

gain of a low-loss aperture antenna (\(\eta_\text{rad} \approx 1\))

horn \(\approx 0.5\), good reflector 0.55 to 0.7

taper trade

lower sidelobes widen the beam and reduce gain

\(-13.3 \to -31.5\) dB widens the beam 63% and loses 1.8 dB

Looking Ahead to Lesson 16

  • Lesson 16 replaces the integral with a sum over \(N\) elements: the array factor.

  • The 0.886 constant and the taper trade carry over to element weights.

  • Taper efficiency becomes a sum; the PHASER’s Hann preset is the \(\cos^2\) row, sampled.

\[\eta_t = \frac{\left(\sum a_n\right)^2}{N\sum a_n^2}\]

Lesson 16 samples the aperture. Replace the continuous illumination with \(N\) discrete elements spaced \(d\) apart, replace the integral with a sum, and the space factor becomes the array factor — the same Fourier relationship with the same beamwidth constant and the same sidelobe trade, now expressed in element weights you can change electronically. Pattern multiplication follows immediately: the full pattern is the element factor times the array factor. Everything in this lesson survives that step, so we need these numbers in hand before Lesson 16.

Note

An array is an aperture sampled at discrete points, so the same trade appears there with sums in place of integrals. The array version of the taper efficiency is the same \(\eta_t\) with sums,

\[\eta_t = \frac{\left(\sum a_n\right)^2}{N\sum a_n^2},\]

which is the identical ratio of coherent to available gain evaluated over \(N\) element amplitudes. Lesson 22 and its lab use it on the PHASER. The Hann preset is the \(\cos^2\) row above sampled at the element positions, and the Blackman preset tapers harder than any row in the table.

A continuous uniform aperture of length L above, and below it the same length occupied by eight equally spaced elements carrying the same total excitation

The Fourier view carries the rest of the module. Steering the beam in Lesson 18 is a linear phase ramp across the aperture, which shifts the transform. Sidelobe control in Lesson 22 is the taper table applied to element amplitudes. Grating lobes in Lesson 23 are what happens when the sampling is too coarse. Before Lesson 16, be able to state the uniform-aperture beamwidth constant, its first sidelobe level, and the definition of aperture efficiency without looking them up. The midterm pattern-measurement project is due 2 Oct, and the beamwidth and sidelobe numbers you predict for it come from this lesson.