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Latest News || 27 October 2026: Dr Soumava Mukherjee's invited talk at Queen's University Belfast joined device-level hardware innovation with channel-level control. || 22 October 2026: I received a Certificate of Appreciation for serving as a reviewer for IEEE IMAS 2026. || 15 October 2026: I presented a poster on beam-domain ML fingerprinting for RIS-aided mmWave localisation at the Centre for Wireless Innovation's annual symposium. || 15 October 2026: CWI's annual symposium brought research and industry perspectives on 6G, ISAC, optical wireless, AI, and future networks to Queen's University Belfast. || 14 October 2026: Erik Nordell of Ericsson visited our Keysight Lab for a live RIS-assisted video streaming demonstration.
Knowledge Hubreconfigurable intelligent surface4. Localization Perspective

reconfigurable intelligent surface learning note

4. Localization Perspective

How RIS helps localization through virtual anchors, controllable paths, near-field wavefronts, and channel-parameter estimation.

At a glance

  • The RIS as a virtual anchorIts position is known, so the reflected path's delay and angle locate the user even when the direct path is blocked.
  • Bandwidth for range, aperture for angleRange error falls as 1/B and angle error shrinks with surface size; both improve with SNR.
  • Near field adds rangeWavefront curvature across a large surface carries distance information, not only direction.

Why Localization Needs RIS

  • Localization: map radio observations to position.
  • Useful observations: time delay, angle, phase, Doppler, and received power.
  • Problem: blockage removes direct paths; uncontrolled reflections are harder to interpret geometrically.

RIS changes this situation because it creates a known controllable path.

References for this section3

Virtual Anchors

A physical anchor is a transmitter or receiver with known position. An RIS can behave like a virtual anchor because its location is known and its reflection pattern is controlled.

If the receiver estimates the delay and angle of the RIS-reflected path, it can infer geometric constraints. With multiple RIS states, multiple RIS panels, or a direct path plus reflected path, the position becomes easier to solve.

The receiver does not directly observe a floor-plan position:

  1. Pilots: reveal channel observations at known signal locations.
  2. Channel estimate: extracts path gain, delay, angle, and possibly Doppler.
  3. Position estimate: maps those parameters to geometry.
  4. RIS role: known surface location and state can make one reflected path easier to identify.

The RIS as a virtual anchor: how precisely can we place the user?

1. Formula and symbols
σd=c2πβ2γ,β=B/12\sigma_d=\frac{c}{2\pi\beta\sqrt{2\gamma}},\quad\beta=B/\sqrt{12}
σθ=6π2cos⁡2θ γ(M2−1)\sigma_\theta=\sqrt{\frac6{\pi^2\cos^2\theta\,\gamma(M^2-1)}}
σp≈σd2+(rσθ)2\sigma_p\approx\sqrt{\sigma_d^2+(r\sigma_\theta)^2}
  • BB: bandwidth
  • β\beta: RMS bandwidth for the stated flat spectrum
  • γ\gamma: linear SNR
  • MM: elements per side
  • θ\theta: departure angle
  • rr: range from anchor
  • σd\sigma_d: Local range-error standard deviation
  • σθ\sigma_\theta: Local angle-error standard deviation
  • σp\sigma_p: Approximate local position uncertainty

Angle error uses radians.

known leg 39.7 mr = 32.6 mRIS (anchor)BSUsererror ellipse enlarged ×10 · drag the user
Bandwidth
RIS elements per side
Carrier
2. Work with the current inputs
σd=c2πβ2 SNR,  β=B12  ⇒  σd=3×108⋅122π⋅400×106⋅2⋅12.5=8.3 cm\sigma_d = \frac{c}{2\pi\beta\sqrt{2\,\mathrm{SNR}}},\ \ \beta = \tfrac{B}{\sqrt{12}} \;\Rightarrow\; \sigma_d = \frac{3\times10^{8}\cdot\sqrt{12}}{2\pi\cdot 400\times10^{6}\cdot\sqrt{2\cdot 12.5}} = 8.3\ \text{cm}
σθ=6π2cos⁡2θ  SNR (M2−1)=6π2⋅0.7372⋅12.5⋅(322−1)=0.5368∘\sigma_\theta = \sqrt{\frac{6}{\pi^{2}\cos^{2}\theta\;\mathrm{SNR}\,(M^{2}-1)}} = \sqrt{\frac{6}{\pi^{2}\cdot 0.737^{2}\cdot 12.5\cdot(32^{2}-1)}} = 0.5368^\circ
σp≈σd2+(r σθ)2=(8.3 cm)2+(32.6 m×9.37×10−3)2=31.6 cm\sigma_p \approx \sqrt{\sigma_d^{2} + (r\,\sigma_\theta)^{2}} = \sqrt{(8.3\ \text{cm})^{2} + (32.6\ \text{m}\times 9.37\times10^{-3})^{2}} = 31.6\ \text{cm}
3. Result
SNR via the RIS11.0 dBdirect path blocked
Range error σ_d8.3 cmrange 32.6 m
Angle error σ_θ0.537°AoD 42.5°
Position error31.6 cmCRB, one snapshot
The BS knows where the RIS is. From the reflected path it measures two things: total delay (giving range from the RIS, once the known BS–RIS leg is removed) and the departure angle across the surface. Range and angle cross at one point. Their accuracy follows the Cramér–Rao bound, the best any unbiased estimator can do: wider bandwidth sharpens range, a larger surface sharpens angle, and both improve with SNR. SNR comes from the same plate link model (50 dBm EIRP, 30 dB per building crossing). Assumes synchronised nodes and a known BS–RIS geometry. Drag the user.
References for this section3

Measurements Used for RIS Localization

  • ToA: Time of arrival gives path length.
  • TDoA: Time difference of arrival gives relative path length.
  • AoA: Angle of arrival gives direction at the receiver.
  • AoD: Angle of departure gives direction from the transmitter or RIS.
  • RSS: Received signal strength can help but is less reliable.
  • Doppler: Motion-induced frequency shift supports tracking.
References for this section3

Near-Field Localization

Large RIS panels and high carrier frequencies create near-field behavior. In the near field, wavefronts are spherical rather than planar.

This is not just a complication. It is also an opportunity. Spherical wavefront curvature contains range information. That means a large RIS may help estimate both angle and distance, not only direction.

Near or far field? Watch the wavefront bend

1. Formula and symbols
Δϕedge=2πλ(r2+(D/2)2−r)\Delta\phi_{\rm edge}=\frac{2\pi}{\lambda}(\sqrt{r^2+(D/2)^2}-r)
rFF≈2D2/λr_{\rm FF}\approx2D^2/\lambda
  • DD: aperture size
  • λ\lambda: wavelength
  • rr: range
  • Δϕedge\Delta\phi_{\rm edge}: edge-to-centre spherical phase difference in radians

The far-field distance is the stated approximate phase-error convention.

0.01°0.1°1°10°100°1000°10⁴°10⁵°10 cm1 m10 m100 m1 kmπ/8 limitnear fieldedge phase error vs a flat waveuser distance (log scale)
Carrier
Elements across the aperture
2. Work with the current inputs
Δϕedge=2πλ(r2+(D2)2−r)≈πD24λrΔϕedge=π8  ⟺  r=2D2λ=2⋅(34.3 cm)21.1 cm=21.93 m\begin{gathered}\Delta\phi_{\text{edge}} = \frac{2\pi}{\lambda}\left(\sqrt{r^{2}+\left(\tfrac{D}{2}\right)^{2}} - r\right) \approx \frac{\pi D^{2}}{4\lambda r} \\[4pt] \Delta\phi_{\text{edge}} = \frac{\pi}{8} \iff r = \frac{2D^{2}}{\lambda} = \frac{2\cdot(34.3\ \text{cm})^{2}}{1.1\ \text{cm}} = 21.93\ \text{m}\end{gathered}
3. Result
Aperture D34.3 cm
Rayleigh distance21.93 m
Edge phase error27.74°limit 22.5°
RegimeNear fieldrange is observable
A user straight in front of the surface sends a spherical wave. The curve shows how much extra phase the edge elements see compared with a flat wave, as the user moves away. Close in, that curvature is large and carries range information; far away the wave looks flat and the surface only learns direction. The usual rule draws the line at π/8 (22.5°), which happens at the Rayleigh distance 2D²/λ (shaded: near field).
References for this section3

Key Challenges

  • RIS phase pattern design for localization, not only data rate.
  • Synchronization between transmitter, receiver, and RIS controller.
  • Hardware impairments and phase quantization.
  • Multipath association: deciding which estimated path corresponds to which reflector.
  • Mobility and real-time tracking.
  • Near-field channel modeling.
References for this section3

Takeaway

  • Why RIS helps: a controlled path can act as a known geometric reference.
  • Where strongest: blocked direct links or large near-field apertures with range-sensitive phases.
References for this section3

Check your understanding

0 of 4 answered

  1. Which two measurements of the RIS path locate the user in the virtual-anchor demo?

  2. At the same SNR, doubling the bandwidth changes the range error how?

  3. Far from the RIS, why does the angle term often dominate the position error?

  4. Inside the Rayleigh distance, what extra information does the surface observe?

Complete references and further reading