Ferkans — Interactive Telecom Tutor

Why a Physics Chapter Now?

Every capacity expression in Part I of this book — from the Marzetta scaling law to the ZF precoder — rested on a single assumption that we never put under the microscope: the channel between a base station and a user is a superposition of plane waves. A plane wave is a mathematical idealisation. It says that as the wave crosses the base-station array, every antenna element sees the same wavefront, and the only thing that changes from element to element is a linear phase. That is why the steering vector is $\mathbf{a}(\theta) = [1,\, e^{-j\kappa d\sin\theta},\, e^{-j 2\kappa d\sin\theta},\,\ldots]^T$ : the angle $\theta$ is the only parameter that matters.

Now look at a specific base station we might actually deploy for 6G. Consider an aperture $D = 1$ m at carrier frequency $f_0 = 100$ GHz. The wavelength is $\lambda = 3$ mm, so $D/\lambda = 333$ . The far-field boundary — the distance beyond which the plane-wave model is a good approximation — is the Fraunhofer distance $d_F = 2 D^2/\lambda \approx 667$ m. For a user standing $50$ m from this array, the plane-wave model is wrong. The wave reaches the array as a spherical wave, and the phase across the array is a quadratic — not linear — function of the element coordinate. This is the near field.

The point is that near-field effects are not an exotic regime relevant only to antenna measurement chambers. For XL-MIMO and for the frequencies we are moving into, they will be the rule, not the exception. Before we can do any signal processing in the near field, we need to revisit the physics.

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Definition:
Radiation Regions of an Antenna Aperture

For a radiating aperture of largest linear dimension $D$ (the aperture diameter of the array, or its diagonal for planar arrays) at wavelength $\lambda$ , the space around it is divided into three regions by their distance $d$ from the aperture centre:

Reactive near field ( $d < d_R$ ): dominated by stored, non-radiating energy. Roughly $d_R \approx 0.62\sqrt{\frac{D^3}{\lambda}}.$ Field components tangent and normal to the aperture are comparable in magnitude, and there is no well-defined radiation pattern.
Radiating near field / Fresnel region ( $d_R \leq d < d_F$ ): the field is approximately a spherical wave, the angular radiation pattern depends on $d$ , and beam "focus" is range-dependent.
Far field / Fraunhofer region ( $d \geq d_F$ ): the phase fronts are effectively planar across the aperture, the radiation pattern becomes independent of $d$ , and the array response factorises into $e^{-j\kappa d}/d$ times the classical plane-wave steering vector.

The Fraunhofer distance is $d_F \triangleq \frac{2 D^2}{\lambda}.$ This is the range at which the path-length difference between the centre of the aperture and its edge, relative to a target at distance $d$ , equals $\lambda/16$ — a phase error of $\pi/8$ , which is the conventional threshold for the quadratic term in the Taylor expansion to be negligible.

The number $2D^2/\lambda$ is a convention; different authors use $D^2/\lambda$ or $5 D^2/\lambda$ depending on how strict they want the plane-wave approximation to be. The $2 D^2/\lambda$ definition is the classical Balanis value and is what the rest of this chapter uses.

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Example: Three Fraunhofer Distances You Should Memorise

Compute the Fraunhofer distance $d_F = 2 D^2/\lambda$ for three representative deployments and say whether a user at a typical cell distance of $10$ m lies in the near field or the far field.

(a) Sub-6 GHz $4 \times 4$ array: $f_0 = 3.5$ GHz, $16$ elements on a half-wavelength grid.

(b) mmWave $64 \times 1$ ULA: $f_0 = 28$ GHz, $64$ elements on a half-wavelength grid.

(c) Sub-THz XL-MIMO: $f_0 = 100$ GHz, aperture $D = 1$ m (roughly $666$ elements on a half-wavelength grid).

Solution

Convert to wavelengths

$\lambda = c / f_0$ . For $3.5$ GHz, $\lambda \approx 8.57$ cm; for $28$ GHz, $\lambda \approx 1.07$ cm; for $100$ GHz, $\lambda = 3$ mm.

(a) Sub-6 GHz $4\times 4$ array

A $4 \times 4$ planar array with $\lambda/2$ spacing has side $3(\lambda/2) = 12.9$ cm; the diagonal is $D = \sqrt{2}\cdot 12.9 \approx 18.2$ cm. Then $d_F = 2 D^2/\lambda = 2\cdot(0.182)^2/0.0857 \approx 0.77\ \text{m}.$ A user at $10$ m is very comfortably in the far field. This is why the plane-wave model served us well through decades of sub-6 GHz cellular engineering — and through the massive MIMO analysis in Chapters 1–6 of this book.

(b) mmWave $64 \times 1$ ULA

Half-wavelength spaced ULA: $D = 63 \cdot (\lambda/2) = 63 \cdot 0.00535 \approx 0.337$ m. $d_F = 2\cdot (0.337)^2 / 0.0107 \approx 21.2\ \text{m}.$ A user at $10$ m is inside $d_F$ — this is the near field. A $64$ -element mmWave array on a lamp post already has near-field users in its normal operating range.

(c) Sub-THz 1 m XL-MIMO

With $D = 1$ m and $\lambda = 3$ mm, $d_F = 2\cdot 1^2/0.003 \approx 667\ \text{m}.$ A user at $10$ m is very deep in the near field. Practically every user in the cell is in the near field of this array. Cell-scale near-field propagation is one of the defining features of sub-THz communications.

The pattern

$d_F$ scales quadratically with $D$ and linearly with $f_0$ . Doubling the aperture quadruples $d_F$ ; doubling the frequency doubles $d_F$ . The combined effect of pushing $D$ up and $\lambda$ down in future systems is that near-field ranges grow very rapidly.

Fraunhofer Distance vs Aperture and Carrier

Sweep the aperture $D$ and read off $d_F = 2 D^2/\lambda$ for several carrier frequencies. The two horizontal lines at $10$ m and $100$ m mark typical indoor and micro-cell distances — you can see where the curves cross these lines and identify when cell-range near-field effects become unavoidable.

Parameters

Low carrier (GHz)3.5

High carrier (GHz)140

Min aperture

D

(m)0.05

Max aperture

D

(m)2

Theorem: Fresnel Approximation via Second-Order Taylor Expansion

Consider a point source at distance $d \gg D$ from a planar aperture of diameter $D$ , centred at the origin, with a point on the aperture at coordinate $\mathbf{u} = (u_1, u_2, 0)$ of magnitude $\|\mathbf{u}\| \leq D/2$ . The distance from the source at $(d, 0, 0)$ to the aperture point is $r(\mathbf{u}) = \sqrt{(d - 0)^2 + u_1^2 + u_2^2} = d\sqrt{1 + \frac{u_1^2 + u_2^2}{d^2}}.$ Taylor expanding $\sqrt{1 + x} = 1 + x/2 - x^2/8 + \mathcal{O}(x^3)$ and tracking terms gives $r(\mathbf{u}) = d + \frac{u_1^2 + u_2^2}{2 d} - \frac{(u_1^2 + u_2^2)^2}{8 d^3} + \mathcal{O}(d^{-5}).$ The plane-wave model keeps only the zeroth-order term $r \approx d$ , the Fresnel (quadratic-phase) model keeps the first two terms, and the exact model is needed when even the cubic term matters.

The aperture is small compared to $d$ , so the path length is close to $d$ plus a small correction. The first correction is the quadratic spherical-wavefront term $(u_1^2 + u_2^2)/(2d)$ , which describes the curvature of the wavefront as it passes the aperture. Dropping it is what turns a spherical wave into a plane wave.

Show Hint

Write $r = d\sqrt{1 + x}$ with $x = (u_1^2 + u_2^2)/d^2$ .

Apply $\sqrt{1+x} \approx 1 + x/2 - x^2/8$ valid for $|x| \ll 1$ .

The Fraunhofer criterion sets the maximum magnitude of the quadratic residual across the aperture to $\lambda/16$ .

Proof

Factor and expand

$r(\mathbf{u}) = d\sqrt{1 + \frac{u_1^2 + u_2^2}{d^2}} = d\left(1 + \frac{u_1^2+u_2^2}{2 d^2} - \frac{(u_1^2+u_2^2)^2}{8 d^4} + \mathcal{O}(d^{-6})\right).KATEXPLACEHOLDER0ENDr(\mathbf{u}) = d + \frac{u_1^2+u_2^2}{2d} - \frac{(u_1^2+u_2^2)^2}{8 d^3} + \mathcal{O}(d^{-5}).$ $

Turn path length into phase

The phase contribution at the aperture point $\mathbf{u}$ is $e^{-j\kappa r(\mathbf{u})}$ with $\kappa = 2\pi/\lambda$ . Pulling out the common factor $e^{-j\kappa d}$ , $e^{-j\kappa r(\mathbf{u})} = e^{-j\kappa d}\cdot \exp\!\left\{-j\kappa\frac{u_1^2+u_2^2}{2 d}\right\}\cdot \exp\!\left\{+j\kappa\frac{(u_1^2+u_2^2)^2}{8 d^3}\right\}\cdots$ The middle factor is the Fresnel quadratic phase. It depends on $d$ , so the array response is range-dependent — this is the formal reason near-field beams focus to a point, not a direction.

Fraunhofer criterion

The plane-wave approximation drops both correction factors. For it to be accurate, the quadratic-phase term must be small across the aperture: $\max_{\|\mathbf{u}\| \leq D/2} \kappa \frac{\|\mathbf{u}\|^2}{2d} \leq \kappa\frac{(D/2)^2}{2 d} = \frac{\pi D^2}{4 d\,\lambda} \leq \frac{\pi}{8}.$ Rearranging, $d \geq \frac{2 D^2}{\lambda} = d_F. \quad\blacksquare$

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Reactive vs Radiating Near Field

The reactive near-field boundary $d_R \approx 0.62\sqrt{D^3/\lambda}$ marks where evanescent, non-radiating components of the field become negligible. Communication systems effectively never operate in the reactive region — $d_R$ is typically tens of centimetres even for large apertures, and any user that far from a base station is also mechanically implausible. In this chapter, "near field" always means the radiating / Fresnel region $d_R \leq d < d_F$ , where waves are spherical but well-behaved and carry power outward.

Historical Note: Who Was Fraunhofer? And Why Does He Own This Distance?

1787–1826

Joseph von Fraunhofer was a German optician — he spent most of his working life in Benediktbeuern, measuring the dark absorption lines in the solar spectrum that now bear his name. The "Fraunhofer diffraction" regime is named after his work on how light from a small aperture spreads out when observed far from the aperture. In that regime, the observed pattern is (up to a quadratic phase we rarely see) the spatial Fourier transform of the aperture illumination — which is exactly why far-field beamforming is a spatial-frequency problem and the FFT structures so useful.

"Near-field" communications is the regime that Fraunhofer's approximation does not cover. In classical optics this is called the Fresnel regime, after Augustin-Jean Fresnel, who developed the quadratic-phase approximation a few years before Fraunhofer's death. Our hero in the near field is therefore really Fresnel, but the boundary distance — the frontier between the two regimes — keeps Fraunhofer's name.

Common Mistake: Far Field Is Not 'Very Far Away'

Mistake:

A common intuition is that "far field" means the user is, in absolute terms, far from the array — tens or hundreds of metres. This leads engineers to assume that all cellular links must be far-field simply because cells are tens of metres across.

Correction:

"Far field" is a statement about the ratio $d / d_F$ , not the absolute distance $d$ . For a large aperture $D$ , an "obviously far away" user at $30$ m may still be well inside $d_F$ . For the sub-THz 1 m aperture in Example EThree Fraunhofer Distances You Should Memorise, $d_F \approx 667$ m, so every user in a cell of radius $100$ m is in the near field. The only way to know is to compute $d_F$ from $D$ and $\lambda$ .

⚠️Engineering Note

Near-Field Effects in 3GPP and IEEE 802.11

Current 5G and Wi-Fi standards all assume far-field propagation when specifying channel models, codebook designs and beam management procedures. The 3GPP TR 38.901 geometry-based stochastic model maps angular clusters to plane-wave steering vectors; the 5G NR Type-II codebook is a discrete Fourier basis derived from ULA plane-wave steering; and Wi-Fi sounding assumes linear phase progression across antennas.

For 6G at FR3 (7–24 GHz) and FR4 (above 100 GHz), study item TR 38.901-E is evaluating spherical-wave channel models and near-field spatial non-stationarity. Field trials of XL-MIMO prototypes at HHI and Huawei have already shown that users within roughly $d_F/5$ see a measurable breakdown of the plane-wave approximation in terms of beam-alignment losses.

Practical Constraints

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3GPP TR 38.901 plane-wave model fails for $d / d_F < 1$ at sub-THz.
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5G NR Type-II codebook is DFT-based, i.e., purely angular.
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6G study items (Rel-19+) are actively considering near-field extensions.

📋 Ref: 3GPP TR 38.901 v18.0.0

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Key Takeaway

Far or near is about $d/d_F$ , and $d_F$ grows as $2 D^2/\lambda$ . Doubling the aperture quadruples the near-field range; halving the wavelength doubles it. For 6G XL-MIMO at sub-THz, the near field is the normal operating range, not an edge case. Any signal processing that assumes a plane wave will pay for that assumption in beam-alignment loss, capacity loss, or both.

Quick Check

A base station at $f_0 = 140$ GHz has a square array with side $0.5$ m (diagonal $D = 0.5\sqrt{2} \approx 0.71$ m). A user stands at $d = 30$ m. Is the user in the near field?

No — 30 m is well beyond any reasonable near-field distance.

Yes — $d_F \approx 470$ m, so $d / d_F \approx 0.064 \ll 1$ .

Only the edge elements of the array see near-field behaviour; the centre sees a plane wave.

Impossible to say without knowing the user's angle.