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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 Hubris assisted isac system3. Sharing the Surface Between Communication and Sensing

ris assisted isac system learning note

3. Sharing the Surface Between Communication and Sensing

Connect user rate, reciprocal echo strength and a fixed-aperture partition baseline before attempting joint optimization.

At a glance

  • User and target can differThe user decodes data; the sensing receiver estimates an echo from a target.
  • Partitioning has a measurable costIn this ideal baseline, communication power scales as α² and sensing echo power as (1 − α)⁴.
  • Compare under equal resourcesJoint designs need the same total aperture, transmit power and overhead as their baselines.

System Model in Words

The BS transmits a known waveform carrying data. A user decodes that data; the BS or a separate sensing receiver processes target echoes. A RIS changes the effective channel for both tasks.

Specify the receiver first. A BS receiver using a reciprocal RIS path has a four-hop budget. A receiver mounted on the RIS does not have that same return route. Neither case should inherit the other’s path loss by assumption.

References for this section3

Communication Path

Let hdh_d denote the complex direct channel and hrh_r the complex RIS-assisted channel. A scalar user model is:

y=(hd+hr)x+n,γc=Pt∣hd+hr∣2I+N.y=(h_d+h_r)x+n,\qquad \gamma_c=\frac{P_t|h_d+h_r|^2}{I+N}.
  • xx: transmitted symbol.
  • yy: received sample.
  • nn: receiver noise sample.
  • PtP_t: transmit power.
  • II: interference power.
  • NN: noise power.
  • γc\gamma_c: signal-to-interference-plus-noise ratio.

Channels add as complex amplitudes: a reflected path can help or hurt depending on its phase.

Ideal spectral efficiency is R=log⁡2(1+γc)R=\log_2(1+\gamma_c) in bit/s/Hz. Actual throughput also depends on bandwidth, coding, scheduling and overhead.

References for this section3

Sensing Path

  • Direct echo: BS → target → BS.
  • Reciprocal RIS echo: BS → RIS → target → RIS → BS.
  • Mixed echo: only the outgoing or return route uses the RIS.

Separate these echoes by their expected delay and spatial response when possible. Unresolved paths can interfere. A communication SINR target does not establish a sensing detection probability.

References for this section3

Surface Partitioning

A simple baseline dedicates a fraction α\alpha of the physical aperture to communication and 1−α1-\alpha to reciprocal sensing. Assume fixed illumination, coherent steering, independent subarrays and no useful cross-beams.

γc=γc,fullα2,PsPs,full=(1−α)4.\gamma_c=\gamma_{c,\mathrm{full}}\alpha^2,\qquad \frac{P_s}{P_{s,\mathrm{full}}}=(1-\alpha)^4.

The squared communication factor comes from coherent one-way aperture gain. The fourth-power sensing factor comes from using the sensing subarray on both passes. At a 50/50 split the factors are 1/4 and 1/16. The lab fixes full-aperture user SNR at 100 (20 dB); it shows a baseline, not the best possible trade-off.

How much surface can each task afford?

1. Formula and symbols
R=log⁡2(1+γ0α2)R=\log_2(1+\gamma_0\alpha^2)
Ps/Ps,full=(1−α)4P_s/P_{s,\rm full}=(1-\alpha)^4
  • α\alpha: surface fraction for communication
  • 1−α1-\alpha: sensing fraction
  • γ0\gamma _{0}: full-surface linear communication SNR
  • RR: bit/s/Hz
  • PsP_s: echo power

The fourth power follows two traversals of the sensing sub-aperture.

C: communication cells (50%)S: sensing cells (50%)C: 4.70 bit/s/HzS: 6.25% echo powerAt 50/50: user power is 1/4; echo power is 1/16.
2. Work with the current inputs
R=log⁡2(1+100α2)=log⁡2(1+100×0.502)=4.70R=\log_2(1+100\alpha^2)=\log_2(1+100\times0.50^2)=4.70
Ps/Ps,full=(1−α)4=(1−0.50)4=0.0625P_{s}/P_{s,\mathrm{full}}=(1-\alpha)^4=(1-0.50)^4=0.0625
3. Result
Communication spectral efficiency4.70 bit/s/Hz
Echo / full sensing aperture6.25%
Sensing echo loss-12.0 dB
Illustrative partition baseline: fixed total aperture and power, 20 dB full-aperture user SNR. Communication power scales as area squared; the reciprocal sensing echo uses its subarray twice and scales as area to the fourth power. Direct paths, interference and cross-beams are ignored; this is not a joint-optimization bound.
References for this section3

Joint Optimization

A joint design can instead choose all element phases and BS beams together. The variables are coupled: improving one beam can move energy away from another.

  • Objective: for example, maximize echo SNR subject to minimum user SINR.
  • Constraints: total BS power, surface losses, available phase states and, for active panels, amplifier power.
  • Baseline: compare against partitioning using equal aperture, power, scan time and receiver assumptions.

A feasible algorithm is not automatically globally optimal. Report both the achieved sensing metric and communication metric, including their assumptions.

References for this section3

Takeaway

Use partitioning to understand the cost of sharing, then evaluate whether a joint design earns its additional complexity. Next, examine geometry and phase control, because a stronger echo is only one part of localisation.

References for this section3

Check your understanding

0 of 4 answered

  1. At a 50/50 area split, what fraction of the full sensing-aperture echo remains?

  2. Full-aperture user SNR is 100 in linear units. At α = 0.5, what is user SNR?

  3. For linear user SNR γ = 15, what is ideal spectral efficiency log₂(1 + γ)?

  4. What sensing area fraction gives 0.8⁴ = 40.96% of the full sensing echo?

Complete references and further reading