Frequency Reuse Underwater: Capacity of an Acoustic Cellular Network

Milica Stojanovic

millitsa@mit.edu

Background and motivationB: total system bandwidthN: reuse numberR: cell radiusD: reuse distance

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Hexagonal cells:N=i2+j2+ij Є {1,3,4,7,9,12,…}D=(3N)1/2, R=QRQ: reuse factor

Radio channel: P(x)~1/xn

Performance requirement:SIR ≥ SIRo

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

Spatial frequency reuse in terrestrial radio networks enables large area coverage within limited bandwidth.How does this concept apply to underwater acoustic networks?

DR

Problem statement

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

Acoustic propagation:

2D: base stations on the surface (radio-based infrastructure) or on the bottom (cable-based infrastructure)

xn xkax(f)

Capacity analysis:

What is the maximal user density ρmax that can be supported by a cellular architecture within a given bandwidth B?

System design: SIR in acoustic channels

A(x,f)=A0 xk ax(f)

)QR(I)R(IkQ

61

6P(D)P(R)

SIR

df)f,x(A

1BP

=)x(P0B+minf

minf0

T ∫

∫0B+minf

minf

x- (f)dfa=I(x)

R does not cancel out. SIR depends on both R,N.for every N, radius must satisfy R≥Ro(N).

10log a(f)

I(x)

A simplifying approximation?

-x00min

-x0 aB)f(aB)x(I for B0<<fmin :

rarely true for an acoustic system

not a very good approximation

Bo=B/N for TDMA; (B/N)/U for FDMA

SIR also depends on frequency allocation (fmin, B).

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.

adjust transmission power to ensure

SNR>>SIR; SINR≈SIR

p.s.d. of ambient noiseR1)-Q(

ak

Q6

1SIR 0

SIR requirement and the minimal cell radius

0SIR)QR(I

)R(IkQ61

SIR R≥R0(N)

Ro(3)Ro(4)Ro(7)…

no closed form solution except with approximation for I(x):

0

0

0

alog10)1Q(

4Qlog10kSIRlog10R

R1)-Q(a

kQ

6

1SIR

(R still doesn’t cancel out.)

User bandwidth requirement and the maximal cell radius

Desired user density: Number of users per cell: 2/33α ,RραRρπ 22

ρ

02 WRραN/B

W0

1 NWB

αρ1

)N(RR

At least one user per cell:

αρ1

R1Rρα 2

Cell radius and the reuse number: Admissible region (R,N)

)N(RR}αρ1

),N(Rmax{)N(R 100 Possible solutions (R,N) for the cellular system topologylie between R0(N), R1(N), andthe straight line (αρ)-1/2.

•Small N: +easier frequency allocation +minimal loss with guard band insertion

•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

.

.. … ….. ……… …………… …………………. ………………………………. …………………………………………………….

fmin=10 kHz

αρ1

Sensitivity of solution (R,N) to performance requirements (SIRo,Wo) and system parameters (ρ,B)

Stricter performance requirementscause 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).

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):

NW/B,0

NW/B),N(NRα/)W/B()N(ρ

o

o20o

max

Maximal user density (capacity): )N(ρsupρ maxN

max

)N(RR)N(R 10

in order for a valid design to exist, must have )N(R)N(R 10

Design constraint:

00 NW

Bαρ1

}αρ1

),N(Rmax{

depends on B

corresponding regions in the (ρ,B) space)N(Rα/1ρ 2

0

)N(Rα/1ρ 20

For any given N, thereare two possibilities:

(a)

(b)

Capacity: example

ρmax

ρmax(N)

•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.

ρmax = ρmax (Ni) for BЄ[NiW0,Ni+1W0) ; Ni=3,4,7,9,etc.

Ni is the capacity-achieving architecture in this region

Capacity depends on

•system requirements (SIR0,W0) (less for stricter)

•band-edge frequency fmin

(through R0)

Practical system: small N, ρmax (N).

x

Capacity and the band-edge frequency fmin

Increasing fmin improves the SIR, allows R0(N) to be reduced.

Moving to higher frequencies enables support of a greater user density withinthe same bandwidth.

Is there a price to be paid?

x

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.

N/Bnf

nfdf)f(NnZ

)QR(nIk)QR(nS6nZ

)R(nIkRnS

nSINR

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 oneoperating in a higher band.

Minimal transmission power, cell radius, and the band-edge frequency

R=1 km

R=R0

Fixed cell radius R:PTn,min increases with fmin

Minimal cell radius R0:PTn,min decreases with fmin

SIR0=15 dB, W0=1 kHz; ρ=1 user/km2, B=20 kHz; N=7

Savings in power that result from transmission over a shorter distance outweighthe expenses required for transmission at a higher frequency.

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.

Number of users per Hz of occupied bandwidth

(Alternatively, to maximize per-user bandwidth W, choose minimal R. )

Once N is chosen, R can have any value between and R1. R0

To maximize C (s.t.c. SIR≥SIRo, W≥Wo), choose R=R1(N).

Depending upon R(N), W,SIR, C and U will have values between some min. and max.

N/BRαρ

C2

W

SIR

C

U

ρ=0.25 users/km2, B=50 kHz, SIRo=15 dB, Wo=1 kHz.