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Frequency Reuse Underwater: Capacity of an Acoustic Cellular Network

Frequency Reuse Underwater: Capacity of an Acoustic Cellular Network. Milica Stojanovic millitsa@mit.edu. Background and motivation. B: total system bandwidth N: reuse number R: cell radius D: reuse distance. 2. 1. 7. 3. Hexagonal cells: N=i 2 +j 2 +ij Є {1,3,4,7,9,12,…}

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Frequency Reuse Underwater: Capacity of an Acoustic Cellular Network

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  1. Frequency Reuse Underwater: Capacity of an Acoustic Cellular Network Milica Stojanovic millitsa@mit.edu

  2. Background and motivation B: total system bandwidth N: reuse number R: cell radius D: reuse distance 2 1 7 3 Hexagonal cells: N=i2+j2+ij Є {1,3,4,7,9,12,…} D=(3N)1/2, R=QR Q: reuse factor 1 6 4 2 5 1 7 3 Radio channel: P(x)~1/xn 1 Performance requirement: SIR ≥ SIRo 6 4 1 5 Worst case: cell edge. Six co-channel interferers: SIR=P(R)/6P(D) => (3N)n/2≥ 6SIRo N=7 ensures more than 17 dB of SIR R D 1 1 Spatial frequency reuse in terrestrial radio networks enables large area coverage within limited bandwidth. How does this concept apply to underwater acoustic networks?

  3. Problem statement xn xkax(f) Acoustic propagation: 2D: base stations on the surface (radio-based infrastructure) or on the bottom (cable-based infrastructure) Design: Given the bandwidth B [Hz] and a desired density of users ρ [users/km2], design a cellular system — i.e., find cell radius and reuse number (R,N) — such that the performance requirements are met: (i) co-channel SIR ≥ SIRo (ii) per-user bandwidth W≥Wo Capacity analysis: What is the maximal user density ρmax that can be supported by a cellular architecture within a given bandwidth B?

  4. 10log a(f) I(x) A simplifying approximation? for B0<<fmin : rarely true for an acoustic system not a very good approximation System design: SIR in acoustic channels Bo=B/N for TDMA; (B/N)/U for FDMA A(x,f)=A0xk ax(f) • R does not cancel out. SIR depends on both R,N. • for every N, radius must satisfy R≥Ro(N). SIR also depends on frequency allocation (fmin, B).

  5. p.s.d. of ambient noise adjust transmission power to ensure SNR>>SIR; SINR≈SIR SIR, SNR and SINR Attenuation grows with fmin: for signal and interference. Overall effect: SIR increases with fmin. Bandwidth allocation across N cells (each B/N): (f1- f2), (f2- f3), …, (fN-1- fN,) design for the worst case, fmin=f1, band-edge. SIR used as a figure of merit for system design  interference dominates over noise.

  6. Desired user density: Number of users per cell: no closed form solution except with approximation for I(x): R≥R0(N) (R still doesn’t cancel out.) … Ro(7) Ro(4) Ro(3) SIR requirement and the minimal cell radius User bandwidth requirement and the maximal cell radius At least one user per cell:

  7. fmin=10 kHz . .. … ….. ……… …………… …………………. ………………………………. ……………………………………………………. Cell radius and the reuse number: Admissible region (R,N) Possible solutions (R,N) for the cellular system topology lie between R0(N), R1(N), and the straight line (αρ)-1/2. • R0(N) decays faster than R1(N) • Admissible values of N: • between Nmin, Nmax • For every admissible N, there is • a range of admissible cell radii • Small N: • +easier frequency allocation • +minimal loss with guard band insertion

  8. Sensitivity of solution (R,N) to performance requirements (SIRo,Wo) and system parameters (ρ,B) Stricter performance requirements cause the admissible region to narrow. Increasing SIR0 causes R0 to increase. Increasing W0 causes R1 to decrease. No solution! It is not possible to employ cellular system architecture to meet the required (SIRo,Wo) for the given (ρ,B).

  9. Design constraint:  in order for a valid design to exist, must have For any given N, there are two possibilities: (a) (b) depends on B  corresponding regions in the (ρ,B) space Capacity analysis Def. capacity = maximal user density that can be supported within given bandwidth. Maximal user density that can be supported for a given N (conditional capacity): Maximal user density (capacity):

  10. ρmax ρmax(N) Capacity: example • Capacity increases with bandwidth. • Below the solid curve lies the region (ρ,B) • for which a cellular system can be designed. • Capacity-achieving architectures are • characterized by N that grows with ρmax. x Practical system: small N, ρmax (N). • Capacity depends on • system requirements (SIR0,W0) • (less for stricter) • band-edge frequency fmin • (through R0) ρmax = ρmax (Ni) for BЄ[NiW0,Ni+1W0) ; Ni=3,4,7,9,etc. Ni is the capacity-achieving architecture in this region

  11. Capacity and the band-edge frequency fmin Increasing fmin improves the SIR, allows R0(N) to be reduced. x Moving to higher frequencies enables support of a greater user density within the same bandwidth. Is there a price to be paid?

  12. Power and bandwidth allocation Greater fmin : + greater SIR, capacity - greater attenuation  must increase transmission power System design assumption: SNR>>SIR, i.e. SINR≈SIR  use SIR as a figure of merit. Ex. System design: SIR0=15 dB, W0=1 kHz; ρ~1 user/km2, B=20 kHz; N≤7 determine fmin from the capacity curves: ρmax(7) satisfactory at fmin=20 kHz determine R from the admissible region: fmin R0(7)  R=1 km Is the SNR>>SIR assumption justified? Def. PTn: transmission power for the n-th cell operating in the frequency region [fn,fn+B/N]. fn=fmin+(n-1)B/N, n=1,2,…N. PTn,min: minimal power for which SINRn≈SIRn e.g., deviation of no more than 1 dB.

  13. Determining the transmission power: SIR, SINR • Equal power for all cells: • SINRn curves are shifts of SINRN. • Design assumption must be • justified for the highest band, n=N. • It will automatically hold for all • lower bands, n<N. • Power is wasted. Unequal power allocation: A cell operating in a lower band requires less power to meet the 1 dB deviation rule than one operating in a higher band.

  14. R=1 km R=R0 Minimal transmission power, cell radius, and the band-edge frequency SIR0=15 dB, W0=1 kHz; ρ=1 user/km2, B=20 kHz; N=7 Fixed cell radius R: PTn,min increases with fmin Minimal cell radius R0: PTn,mindecreases with fmin Savings in power that result from transmission over a shorter distance outweigh the expenses required for transmission at a higher frequency.

  15. Summary • Spatial frequency reuse  appealing for underwater acoustic networks? • Acoustic propagation: simple rules of cellular radio system design do not apply. • Reuse number N and cell radius R must satisfy a set of constraints in order to constitute • an admissible solution for a cellular network topology. The range of solutions depends on • performance requirements (SIRo,Wo); band-edge frequency (fmin) • system parameters (ρ,B), • System capacity—maximal user density that can be supported within given bandwidth: • ρmax increases with B, • defines a boundary for the region (ρ,B) in which a cellular system can be designed. • Capacity-achieving architectures : N that grows with ρmax (but that is okay). • Capacity increases as the operational bandwidth is moved to higher frequencies. • Transmission power does not have to be increased if the cell radius is kept minimal. • Efficient power allocation: unequal across cells (more to higher bands). • Relationships between various system parameters are complex, • but design rules are (relatively) simple.

  16. R0 Once N is chosen, R can have any value between and R1. Number of users per Hz of occupied bandwidth To maximize C (s.t.c. SIR≥SIRo, W≥Wo), choose R=R1(N). (Alternatively, to maximize per-user bandwidth W, choose minimal R. ) Depending upon R(N), W,SIR, C and U will have values between some min. and max. W C SIR U ρ=0.25 users/km2, B=50 kHz, SIRo=15 dB, Wo=1 kHz.

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