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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 surface1. What is a Reconfigurable Intelligent Surface (RIS)?

reconfigurable intelligent surface learning note

1. What is a Reconfigurable Intelligent Surface (RIS)?

A detailed beginner-friendly explanation of RIS, smart radio environments, and why RIS is important for 6G wireless networks.

At a glance

  • A programmable wallHundreds of small unit cells each set the phase of the wave they reflect.
  • Phases that add upAligned at the user, N reflections add coherently, so received power grows with N².
  • Redirects, never amplifiesA passive RIS adds no energy and suffers double path loss, so placement matters.

The Basic Idea

An RIS is a programmable surface that changes how an incoming radio wave leaves it.

  • What it controls: reflection, scattering, focusing, and, in some designs, transmission.

  • Physical idea: use part of the environment to shape a propagation path.

  • What it does not do: a passive RIS does not create new signal energy.

  • Ordinary path: walls and objects reflect, scatter, or block the base-station signal.

  • RIS path: one reflection becomes adjustable, as the figure below shows.

  • Smart radio environment: a wall, ceiling, or facade helps control the radio channel.

  • Uses: coverage, communication, sensing, and localization.

  • Base station: still transmits the original signal.

  • Surface: adjusts its response to redirect that signal toward the user.

  • Outcome: position, orientation, and phase settings determine whether the reflected path helps.

A programmed surface opens a path around the blockage

1. Formula and symbols
Δϕ=2πdλ(sin⁡θout−sin⁡θin)\Delta\phi=\frac{2\pi d}{\lambda}(\sin\theta_{\rm out}-\sin\theta_{\rm in})
  • dd: cell spacing
  • λ\lambda: wavelength
  • Δϕ\Delta \phi: required adjacent phase step

angles are measured from the surface normal in the shown geometry. A plain mirror instead fixes equal incident/reflected angles.

direct path blockedprogrammed reflectionRISBSUser
Surface on the facade
2. Work with the current inputs
θin=55.2∘,θout=34.7∘\theta_{\rm in}=55.2^\circ,\quad\theta_{\rm out}=34.7^\circ
Δϕ=180∘[sin⁡(34.7∘)−sin⁡(55.2∘)]=−45.2∘\Delta\phi=180^\circ[\sin(34.7^\circ)-\sin(55.2^\circ)]=-45.2^\circ

Here d = λ/2. The programmed phase step applies in RIS mode; mirror mode uses equal input/output angles.

3. Result
Signal at the userVia RIS path
Angle in55°
Angle out35°chosen by the phases
A plain wall reflects like a mirror: angle out equals angle in, so the energy misses the user. An RIS adds a phase gradient across its elements (colours) that tilts the reflection toward the user. Angles are measured from the surface normal; the geometry is illustrative.
References for this section3

How RIS Differs from a Relay

  • Relay: receives, processes, and retransmits; it needs active RF and processing hardware.

  • Passive RIS: changes the incident field without decoding or regenerating it.

  • Benefit: lower power use than a typical active relay.

  • Limit: it cannot fully rebuild a weak or corrupted signal.

FeatureRelayPassive RIS
Receives and retransmitsYesNo, mainly reflects/scatters
RF chainsUsually yesUsually no
Power consumptionHigherLower
Noise amplificationPossibleNo active receiver noise, but double path loss exists
Main strengthSignal regenerationEnvironment control

Mirror, RIS or relay: same spot, three answers

1. Formula and symbols
γDF=min⁡(γ1,γ2)\gamma_{\rm DF}=\min(\gamma_1,\gamma_2)
RDF=12Blog⁡2(1+γDF)R_{\rm DF}=\tfrac12B\log_2(1+\gamma_{\rm DF})
Rsurface=Blog⁡2(1+γsurface)R_{\rm surface}=B\log_2(1+\gamma_{\rm surface})
  • γ1\gamma _{1}: Linear SNR of the first decode-forward relay hop
  • γ2\gamma _{2}: Linear SNR of the second decode-forward relay hop
  • BB: bandwidth

Half-duplex relay rate uses half the time

mirror/RIS use their modeled reflected-path SNR. All SNRs inside logarithms are linear.

MirrorBSUser231 Mbit/sSNR 6.0 dBRISBSUser1068 Mbit/sSNR 32.1 dBRelay (DF)DFBSUser583 Mbit/sSNR 35.1 dB
Plate size (elements)
2. Work with the current inputs
MirrorRISRelay
Where the wave goesFixed: angle out = angle inProgrammableAnywhere (re-transmits)
RF power at the deviceNoneNone (only control electronics)Amplifiers and radio chains
Noise addedNoneNone (passive)Receiver noise; amplify-and-forward also amplifies it
DuplexFullFullUsually half: listen, then talk
Extra delayNoneNoneProcessing delay (decode, re-encode)
Path lossDouble (two hops)Double (two hops)Two separate hops, each recovered
SNR here6.0 dB32.1 dB35.1 dB
Rate here231 Mbit/s1068 Mbit/s583 Mbit/s
γmirror=γplate sinc⁡2 ⁣(πDλ(sin⁡θu−sin⁡θs)),γRIS=γplateLRIS,Crelay=12Blog⁡2 ⁣(1+min⁡(γ1,γ2))\gamma_{\text{mirror}} = \gamma_{\text{plate}}\,\operatorname{sinc}^{2}\!\left(\tfrac{\pi D}{\lambda}(\sin\theta_u - \sin\theta_s)\right), \quad \gamma_{\text{RIS}} = \frac{\gamma_{\text{plate}}}{L_{\text{RIS}}}, \quad C_{\text{relay}} = \tfrac{1}{2}B\log_2\!\left(1 + \min(\gamma_1, \gamma_2)\right)
3. Result
Mirror231 Mbit/sSNR 6.0 dB
RIS1068 Mbit/sSNR 32.1 dB
Relay (DF)583 Mbit/sSNR 35.1 dB
Best hereRIS
One geometry, three devices on the wall, all at 28 GHz with 100 MHz. The BS (30 dBm, 20 dBi) lights the wall from 30°. Mirror and RIS are the same 17.1 cm plate (1,024 half-wavelength cells): the mirror reflects only around the mirror angle, while the RIS points its beam at the user, paying 2 dB of reflection and quantisation loss. The relay is a decode-and-forward box with 10 dBi antennas and 23 dBm of its own power: two clean hops, but it must listen, then talk, so it halves the rate (half duplex). Move the user off the mirror angle and watch the mirror collapse.
References for this section3

What is Inside an RIS?

An RIS contains many small unit cells. Each unit cell can impose a phase shift, amplitude change, polarization change, or transmission/reflection coefficient on an incident wave.

Typical hardware components include:

  • Metallic patches or metasurface elements.
  • PIN diodes, varactors, MEMS switches, or tunable materials.
  • A biasing/control network.
  • A controller that receives configuration commands from the network.

The surface works collectively. One unit cell alone does little. Hundreds or thousands of cells together can shape a wavefront.

Inside an RIS: one unit cell, layer by layer

1. Formula and symbols
d≈λ/2=c/(2fc),ϕ∈{0,π}d\approx\lambda/2=c/(2f_c),\qquad\phi\in\{0,\pi\}
  • dd: illustrative unit-cell pitch
  • λ\lambda: wavelength
  • fcf_c: carrier
  • cc: wave speed
  • ϕ\phi: selected ideal 1-bit reflection phase

Physical cell dimensions and achievable phases are design dependent.

the surfacepatchsubstrategroundbias layerdiode OFFincomingreflected (in phase)
Layer
Diode state
2. Work with the current inputs
d≈2997924582×28×109 m≈5.4 mm,ϕ=0∘d\approx\frac{299792458}{2\times28\times10^9}\ \mathrm{m}\approx5.4\ \mathrm{mm},\quad\phi=0^\circ

PIN diode (or varactor). Bridges a small gap in the patch. A PIN diode is either on or off, which changes the patch's electrical size and flips the reflection phase by about 180°: one bit. A varactor instead tunes its capacitance smoothly for a wide, continuous phase range.

3. Result
SelectedPIN diode (or varactor)
Reflection phase0°design target, 1-bit cell
Cell size at 28 GHz≈ 5.4 mmhalf a wavelength
Left: the surface is a grid of identical cells. Right: one cell pulled apart. Click a layer (or use the buttons) to read what it does, and switch the diode to see the reflected wave flip by 180° relative to the incoming one. Layer sizes are typical of 1-bit PIN-diode designs at mmWave and drawn for clarity, not to scale.

The controller: how a new phase pattern reaches the surface

1. Formula and symbols
Tupdate=Tlatency+Nb/Rlink+Nb/(kfclk)T_{\rm update}=T_{\rm latency}+Nb/R_{\rm link}+Nb/(k f_{\rm clk})
  • NN: elements
  • bb: bits per element
  • RlinkR_{\rm link}: link bit rate
  • kk: parallel chains
  • fclkf_{\rm clk}: chain clock
  • TlatencyT_{\rm latency}: fixed link latency

The two serialization times are added in this model.

BS / edgecomputes phasesWired EthernetControllerFPGA / MCUback of the panelshowing 8 of 16 chainsshift registers, then one latch
Control link
Elements
Bits per element
2. Work with the current inputs
Tupdate=Tlatency+NbRlink+Nbk fclk=100.0 μs+40961000×106+409616⋅50×106=109.2 μsT_{\text{update}} = T_{\text{latency}} + \frac{N b}{R_{\text{link}}} + \frac{N b}{k\, f_{\text{clk}}} = 100.0\ \mu\mathrm{s} + \frac{4096}{1000\times10^{6}} + \frac{4096}{16\cdot 50\times10^{6}} = 109.2\ \mu\mathrm{s}
3. Result
Pattern size0.50 kB4,096 bits
Link time104.1 µs
Shift-in time5.1 µs
Total update109.2 µs9,156 patterns/s; coherence at 30 km/h ≈ 543.5 µs
The BS (or an edge server) computes the pattern and sends it over a control link to a small controller, usually an FPGA or microcontroller board mounted on the back of the panel. The controller clocks the bits into chains of shift registers; when every chain is full, one latch pulse updates all elements at once. Total update time = link latency + transfer + shifting. Link rates and latencies here are illustrative.
References for this section3

Why Phase Control Matters

  • Aligned RIS phases at a user: reflected waves reinforce.
  • Opposing phases at an interferer: a designed pattern may reduce unwanted energy there.
  • Constraint: improving one direction can affect other users and sidelobes.

The simplified design goal is:

ϕ⋆=arg⁡max⁡ϕ∣hd+∑n=1Ngn ejϕnfn∣2\boldsymbol{\phi}^{\star} = \arg\max_{\boldsymbol{\phi}} \Big| h_d + \sum_{n=1}^{N} g_n\, e^{j\phi_n} f_n \Big|^{2}
  • hdh_d: direct BS-to-user channel.
  • fnf_n: BS-to-element-nn channel.
  • gng_n: element-nn-to-user channel.

For one user the answer is simple: each element cancels its own path phase, ϕn=arg⁡hd−arg⁡(fngn)\phi_n = \arg h_d - \arg(f_n g_n), so every reflected term lines up with the direct path.

This is why RIS is not just a mirror. A mirror follows the law of reflection. An RIS can implement anomalous reflection, focusing, splitting, beam steering, and in advanced designs simultaneous transmission and reflection.

Aligned reflections add up; random ones cancel

1. Formula and symbols
E=∑n=1Nejϕn,q=∣E∣/NE=\sum_{n=1}^Ne^{j\phi_n},\quad q=|E|/N
Lrelative=20log⁡10qL_{\rm relative}=20\log_{10}q
  • NN: equal-amplitude elements
  • ϕn\phi_n: each reflected phase
  • EE: summed field
  • qq: amplitude fraction of perfect alignment
  • LrelativeL_{\rm relative}: relative received-power change in dB
phases scattered
2. Work with the current inputs
q=(2.651)2+(−0.806)212=0.2309q=\frac{\sqrt{(2.651)^2+(-0.806)^2}}{12}=0.2309
Lrelative=20log⁡10(0.2309)=−12.7 dBL_{\rm relative}=20\log_{10}(0.2309)=-12.7\ \mathrm{dB}

The displayed dB calculation floors the amplitude fraction at 0.001 for numerical readability.

3. Result
Elements12
Sum vs perfect alignment23%amplitude
Received power penalty-12.7 dB
Each small arrow is one element's reflected wave at the user, equal in strength. Placed head to tail, their sum (yellow) is what the user receives. With aligned phases the arrows form a straight line N times as long as one; with scattered phases they curl up and partly cancel.
References for this section3

Why RIS Matters for 6G

RIS is attractive for 6G because future networks are expected to use:

  • Dense deployments.
  • mmWave and sub-THz bands.
  • Massive MIMO and large apertures.
  • Sensing and localization.
  • Energy-efficient infrastructure.
  • AI-driven radio control.

High-frequency signals provide large bandwidth but are easily blocked. RIS can add controllable non-line-of-sight paths and improve coverage without installing a full active base station everywhere.

More elements, more coherent gain

1. Formula and symbols
G=N2,GdB=20log⁡10N,D≈N λ/2G=N^2,\quad G_{\rm dB}=20\log_{10}N,\quad D\approx\sqrt N\,\lambda/2
  • NN: equal-strength, coherently aligned cells
  • GG: power gain over a single cell under the same per-cell model
  • DD: illustrative square-panel side at half-wavelength pitch

It is not an absolute end-to-end link gain.

020406080gain (dB)24 dBN = 1636 dBN = 6448 dBN = 25660 dBN = 1,02472 dBN = 4,096+12 dB for every 4× elements
Elements on the surface (N)
2. Work with the current inputs
GdB=20log⁡10(1024)=60.2 dBG_{\rm dB}=20\log_{10}(1024)=60.2\ \mathrm{dB}
D≈1024×0.535 cm=17.1 cmD\approx\sqrt{1024}\times0.535\ \mathrm{cm}=17.1\ \mathrm{cm}
3. Result
Elements1,02432 × 32 grid
Ideal coherent gain60.2 dB
Panel side at 28 GHz≈ 17.1 cmhalf-wavelength spacing
Ideal gain of N perfectly phased elements over a single element: the received amplitude grows with N, so power grows with N², or 20·log₁₀N dB. A real RIS link also suffers double path loss (BS to RIS, then RIS to user), which is why practical surfaces need hundreds to thousands of elements.
References for this section3

Correct Perspective: RIS is Not Magic

RIS redirects existing energy; a passive surface is not a perfect signal booster. Check these limits:

  • Double path loss: BS-to-RIS and RIS-to-user attenuation both matter.
  • Channel estimation difficulty: Passive surfaces cannot easily sense the channel themselves.
  • Finite phase resolution: Real hardware may have 1-bit, 2-bit, or limited tuning states.
  • Control overhead: Large surfaces require many control parameters.
  • Hardware nonidealities: Mutual coupling, phase-dependent amplitude, and frequency selectivity can reduce ideal gains.

Link budget calculator: direct path vs RIS path

1. Formula and symbols
Pr,RIS=PtGtGr(4π)2χA2d12d22P_{r,\rm RIS}=\frac{P_tG_tG_r}{(4\pi)^2}\frac{\chi A^2}{d_1^2d_2^2}
SNR=Pr/(kTBF)\mathrm{SNR}=P_r/(kTB F)
  • PtP_t: Transmitted power
  • PrP_r: Received power
  • GtG_t: Linear transmit antenna gain
  • GrG_r: Linear receive antenna gain
  • χ\chi: product of the incidence/reflection cosine factors and linear extra-loss factor
  • AA: physical area
  • d1d_{1}: Distance from source to RIS
  • d2d_{2}: Distance from RIS to user
  • kk: Boltzmann constant
  • TT: Noise temperature in kelvin
  • BB: Receiver bandwidth
  • FF: Linear receiver noise factor
building facade52.0 m39.7 m32.6 mnormalRISBSUserdrag the user or the RIS
Carrier frequency
RIS elements (N)
Bandwidth
2. Work with the current inputs
Direct
-75.7 dBm
Via RIS
-70.0 dBm
Both (power sum)
-69.0 dBm
Scale -130 to -30 dBm; the orange tick marks the noise floor.
Link summary
Path

50.0 dBm leaves the BS → -70.0 dBm reaches the user → SINR 17.0 dB → 567 Mbit/s

  1. BS transmits50.0 dBm30 dBm + 20 dBi antenna

    −93.4 dB39.7 m to the RIS

  2. RIS reflects+65.0 dB+70.1 dB aperture −3.2 dB angles −2.0 dB loss

    −91.6 dB32.6 m to the user

  3. User receives-70.0 dBm

    vs -87.0 dBmnoise + interference at the receiver

  4. SINR17.0 dB

    × 100 MHzB · log₂(1 + SINR)

  5. Data rate567 Mbit/sShannon limit
Budget itemDirectVia RIS
Transmit power30.0 dBm30.0 dBm
BS antenna gain (beam on target)+20.0+20.0
User antenna gain0.00.0
Path length52.0 m39.7 + 32.6 m
Free-space loss−95.7−93.4 −91.6
Building penetration−30.0clear
RIS aperture gain (capture + re-radiate)—+70.1
Oblique angles (θᵢ 49°, θᵣ 43°)—−3.2
RIS reflection + quantisation loss—−2.0
Received power-75.7 dBm-70.0 dBm
SINR11.3 dB17.0 dB
Shannon rate385 Mbit/s567 Mbit/s
Step-by-step calculation (updates with the controls)
  1. 1Carrier wavelength

    Every loss and gain below depends on the wavelength, so we start there.

    λ=cf=3×108 m/s28.0×109 Hz=10.71 mm\lambda = \frac{c}{f} = \frac{3\times10^{8}\ \text{m/s}}{28.0\times10^{9}\ \text{Hz}} = 10.71\ \text{mm}
  2. 2Free-space path loss (Friis)

    A wave spreads over a growing sphere, so its power density falls with distance squared. In decibels:

    FSPL(d)=20log⁡10 ⁣(4πdλ) dB\mathrm{FSPL}(d) = 20\log_{10}\!\left(\frac{4\pi d}{\lambda}\right)\ \text{dB}
    FSPL(d0)=20log⁡10 ⁣(4π⋅52.00.01071)=95.7 dB(BS→user)FSPL(d1)=20log⁡10 ⁣(4π⋅39.70.01071)=93.4 dB(BS→RIS)FSPL(d2)=20log⁡10 ⁣(4π⋅32.60.01071)=91.6 dB(RIS→user)\begin{aligned} \mathrm{FSPL}(d_0) &= 20\log_{10}\!\left(\tfrac{4\pi\cdot 52.0}{0.01071}\right) = 95.7\ \text{dB} && \text{(BS} \to \text{user)}\\ \mathrm{FSPL}(d_1) &= 20\log_{10}\!\left(\tfrac{4\pi\cdot 39.7}{0.01071}\right) = 93.4\ \text{dB} && \text{(BS} \to \text{RIS)}\\ \mathrm{FSPL}(d_2) &= 20\log_{10}\!\left(\tfrac{4\pi\cdot 32.6}{0.01071}\right) = 91.6\ \text{dB} && \text{(RIS} \to \text{user)} \end{aligned}
  3. 3Direct path: received power

    In linear units the link is a product of gains and losses; in decibels the product becomes a sum. Start from the transmit power, add the antenna gains, subtract the losses:

    Prdir=Pt Gt GrFSPL(d0) Lpen  ⟹  Prdir[dBm]=Pt+Gt+Gr−FSPL(d0)−LpenP_r^{\text{dir}} = \frac{P_t\,G_t\,G_r}{\mathrm{FSPL}(d_0)\,L_{\text{pen}}} \;\Longrightarrow\; P_r^{\text{dir}}[\text{dBm}] = P_t + G_t + G_r - \mathrm{FSPL}(d_0) - L_{\text{pen}}
    Prdir=30.0+20.0+0.0−95.7−30.0=−75.7 dBmP_r^{\text{dir}} = 30.0 + 20.0 + 0.0 - 95.7 - 30.0 = \mathbf{-75.7\ dBm}

    The direct line crosses the building, so the penetration loss applies.

  4. 4RIS aperture and its gain

    The surface is N elements spaced half a wavelength apart, so its physical area is

    A=N(λ2)2=1024×(5.35 mm)2=293.5 cm2A = N\left(\frac{\lambda}{2}\right)^{2} = 1024\times(5.35\ \text{mm})^{2} = 293.5\ \text{cm}^2

    An aperture of area A has antenna gain 4πA/λ². The RIS uses it twice: once to capture the incoming wave and once to re-radiate it toward the user. So in decibels:

    GRIS=20log⁡10 ⁣(4πAλ2)=20log⁡10 ⁣(4π×0.0293(0.01071)2)=70.1 dBG_{\text{RIS}} = 20\log_{10}\!\left(\frac{4\pi A}{\lambda^{2}}\right) = 20\log_{10}\!\left(\frac{4\pi\times0.0293}{(0.01071)^{2}}\right) = 70.1\ \text{dB}
  5. 5RIS path: received power

    Chain the two hops: free space to the surface, the surface gain, free space to the user. Tilted angles shrink the projected area, which adds a cosine factor for each side:

    PrRIS=PtGtGr(λ4πd1)2(4πAλ2)2(λ4πd2)2cos⁡θicos⁡θrLRIS Lpen=PtGtGr(A4πd1d2)2cos⁡θicos⁡θrLRIS Lpen\begin{aligned} P_r^{\text{RIS}} &= P_t G_t G_r \left(\frac{\lambda}{4\pi d_1}\right)^{2}\left(\frac{4\pi A}{\lambda^{2}}\right)^{2}\left(\frac{\lambda}{4\pi d_2}\right)^{2}\frac{\cos\theta_i\cos\theta_r}{L_{\text{RIS}}\,L_{\text{pen}}}\\ &= P_t G_t G_r\left(\frac{A}{4\pi d_1 d_2}\right)^{2}\frac{\cos\theta_i\cos\theta_r}{L_{\text{RIS}}\,L_{\text{pen}}} \end{aligned}

    Notice the λ terms cancel: for a fixed physical area, the RIS path does not get weaker at higher frequency, unlike the direct path. In decibels:

    PrRIS[dBm]=Pt+Gt+Gr−FSPL(d1)−FSPL(d2)+GRIS+10log⁡10(cos⁡θicos⁡θr)−LRIS−Lpen\begin{aligned} P_r^{\text{RIS}}[\text{dBm}] = {}& P_t + G_t + G_r - \mathrm{FSPL}(d_1) - \mathrm{FSPL}(d_2)\\ &+ G_{\text{RIS}} + 10\log_{10}(\cos\theta_i\cos\theta_r) - L_{\text{RIS}} - L_{\text{pen}} \end{aligned}
    10log⁡10(cos⁡49∘⋅cos⁡43∘)=−3.2 dB10\log_{10}(\cos49^\circ\cdot\cos43^\circ) = -3.2\ \text{dB}
    PrRIS=30.0+20.0+0.0−93.4−91.6+70.1−3.2−2.0+0.0=−70.0 dBmP_r^{\text{RIS}} = 30.0 + 20.0 + 0.0 - 93.4 - 91.6 + 70.1 - 3.2 - 2.0 + 0.0 = \mathbf{-70.0\ dBm}
  6. 6Noise power at the receiver

    Thermal noise is kTB. At room temperature (290 K), kT is −174 dBm per hertz; the receiver adds its noise figure on top:

    N=kTB F  ⟹  N[dBm]=−174+10log⁡10B+NFN = kTB\,F \;\Longrightarrow\; N[\text{dBm}] = -174 + 10\log_{10}B + \mathrm{NF}
    N=−174+10log⁡10(100×106)+7.0=−174+80.0+7.0=−87.0 dBmN = -174 + 10\log_{10}(100\times10^{6}) + 7.0 = -174 + 80.0 + 7.0 = \mathbf{-87.0\ dBm}
  7. 7SINR

    Signal, interference and noise powers add in linear units (milliwatts), not in dBm, so convert first with P[mW] = 10^(P[dBm]/10):

    SINR=SI+N,I=−130.0 dBm=1.00×10−13 mW,N=2.00×10−9 mW\mathrm{SINR} = \frac{S}{I + N}, \qquad I = -130.0\ \text{dBm} = 1.00\times10^{-13}\ \text{mW}, \qquad N = 2.00\times10^{-9}\ \text{mW}
    SINRdir=10−75.7/101.00×10−13+2.00×10−9=2.68×10−82.00×10−9=13.4=11.3 dB\mathrm{SINR}_{\text{dir}} = \frac{10^{-75.7/10}}{1.00\times10^{-13} + 2.00\times10^{-9}} = \frac{2.68\times10^{-8}}{2.00\times10^{-9}} = 13.4 = \mathbf{11.3\ dB}
    SINRRIS=10−70.0/101.00×10−13+2.00×10−9=9.94×10−82.00×10−9=49.8=17.0 dB\mathrm{SINR}_{\text{RIS}} = \frac{10^{-70.0/10}}{1.00\times10^{-13} + 2.00\times10^{-9}} = \frac{9.94\times10^{-8}}{2.00\times10^{-9}} = 49.8 = \mathbf{17.0\ dB}

    With no interference (I ≪ N), SINR equals the SNR.

  8. 8Shannon capacity: the data-rate ceiling

    The highest error-free rate over a bandwidth B with this SINR. Real modulation and coding reach a fraction of it.

    C=Blog⁡2(1+SINR)C = B\log_{2}(1 + \mathrm{SINR})
    Cdir=100 MHz×log⁡2(1+13.4)=385 Mbit/sC_{\text{dir}} = 100\ \text{MHz}\times\log_{2}(1 + 13.4) = \mathbf{385\ Mbit/s}
    CRIS=100 MHz×log⁡2(1+49.8)=567 Mbit/sC_{\text{RIS}} = 100\ \text{MHz}\times\log_{2}(1 + 49.8) = \mathbf{567\ Mbit/s}
  9. 9Both paths together

    If the user receives both paths with unrelated phases, their powers add (an RIS that also phase-aligns with the direct path could do up to 3 dB better when the two are equal):

    Prboth=10log⁡10 ⁣(10−75.7/10+10−70.0/10)=−69.0 dBmSINRboth=18.0 dB,Cboth=601 Mbit/s\begin{aligned} P_r^{\text{both}} &= 10\log_{10}\!\left(10^{-75.7/10} + 10^{-70.0/10}\right) = -69.0\ \text{dBm}\\ \mathrm{SINR}_{\text{both}} &= 18.0\ \text{dB}, \qquad C_{\text{both}} = \mathbf{601\ Mbit/s} \end{aligned}
3. Result
Direct received power-75.7 dBmSINR 11.3 dB
RIS received power-70.0 dBmSINR 17.0 dB
Noise floor-87.0 dBm
RIS advantage+5.7 dBRIS path is stronger
Top view, one grid square = 10 m; drag the user or the RIS. Direct path: Friis free-space loss, plus a penetration loss when the line crosses the building. RIS path: a flat surface of N half-wavelength elements (17.1 cm per side) with ideal phases, Pr = Pt·Gt·Gr·(A/(4π·d₁·d₂))²·cosθᵢ·cosθᵣ, minus a lumped reflection and quantisation loss. The full step-by-step maths is below the table. A rough far-field estimate: ground reflections, diffraction and element patterns are ignored, and icons are not to scale.
References for this section3

Takeaway

  • Core idea: program part of the radio environment rather than relying only on uncontrolled reflections.
  • Where useful: blocked coverage, localization, interference management, or sensing when the RIS path geometry is favorable.
References for this section3

Check your understanding

0 of 4 answered

  1. A plain wall reflects a wave arriving 55° from its normal. Where does the reflection go?

  2. With ideal phases, how does the power received through an RIS grow with the number of elements N?

  3. Why does even a large passive RIS struggle at long range?

  4. Moving from 2-bit to 1-bit phase control costs roughly how much gain?

Complete references and further reading