transmit beamforming for acoustic ofdm

Transmit Beamforming for Acoustic OFDM

Milica Stojanovic

Northeastern University

October 2020

Milica Stojanovic Transmit Beamforming for Acoustic OFDM

Hey, I’m talking to you!


Transmit beamforming

I Why:- efficient use of power- avoidance of unintended listeners- spatial division of multiple users

I How: Assign a weight to each transmit element so that theirsignals add constructively at the receiver.

I Problem: Weights depend on the channel. Downlink channelmust be inferred from uplink. Estimate is noisy and delayed.

I What has been done: A lot in radio, a little in underwateracoustic systems (adaptive modulation, time reversal).


OFDM

I bandwidth scalability

I ease of FFT processing

I possibility of differentially coherent detection

I proven methods for acoustic Doppler compensation

I ideal platform for broadband beamforming


Basic setup: Multiple transmitters

+

noise

RX

TX1

data in

TX2

TXM

.

.

.

data out

CHANNEL 1

CHANNEL 2

CHANNEL M


Basic setup: Multiple paths

+

noise

RX

TX1

data in

TX2

TXM

.

.

.

data out

CHANNEL 1

CHANNEL 2

CHANNEL M


Basic setup: Multiple frequencies

+

noise

RX

TX1

data in

TX2

TXM

.

.

.

data out

CHANNEL 1

CHANNEL 2

CHANNEL M

K frequencies

P paths


Optimal beamforming (sorry, no pretty pictures)

I Hmk =

∑p h

mp γ

mp (fk)e−j2πfkτ

mp : channel transfer function on

the k-th carrier, m-th element

I wmk : beamformer weight on the k-th carrier, m-th element

I yk = dk∑

m wmk Hm

k + zk = dkwTk Hk + zk : signal received on

the k-th carrier

I Maximum SNR: wk ∼ H∗kI Normalization: 1

K

∑k w

Hk wk = 1

no extra power expenditure on account of beamforming


Channel estimation via feedback (uplink)

noise

RX

TX1

estimatesout

TX2

TXM

.

.

.

pilots in

CHANNEL 1

CHANNEL 2

CHANNEL M

+

+

+


Channel estimation via feedback (uplink)

noise

TX

RX1

estimatesout

RX2

RXM

.

.

.

pilots in

CHANNEL 1

CHANNEL 2

CHANNEL M

+

+

+


Channel estimation

I Downink channel has to be estimated from the uplink pilot.

I If Hdn = Hup, the only problem is noise.

I During the time it takes to close the feedback loop (∼ 2d/c),the system geometry could change (ever so slightly), henceHdn 6= Hup.

I xmk = Hm,upk + zm,upk : received pilot signals

I Hdn = L(X) = Hup + αZup: estimated channel

I α = 0: noiseless estimate

α =√

LK , L = dBTmpe: LS estimation in the IR domain

α = 1: time reversal (TR)


Beamforming with a channel estimate

I yk = dk∑

m wmk Hm,dn

k + zdnk = dkwTk H

dnk + zdnk

I Noises are characterized by σ2dn, σ2

up.

I W ∼ Hdn∗: Beamformer weights are still determinedaccording to the maximum SNR rule, but a channel estimateis used instead of the unknown true value.

I Case studies: perfect channel knowledge, delayed channel(noiseless estimate), TR, IR; no adjustment (wm

k = 1√M

).


How good will the channel estimate be?

I The time to close the feedback loop allows for the followingchanges to occur:(i) small-scale fading coefficients γmp(ii) system geometry (e.g. rx drifts at speed v in direction θR).

I Neither is fully predictable.

I Both can cause a significant change in the channel response.

I Grand question: Is there some feature of the channel thatchanges slowly enough that it can withstand the feedbackdelay, yet be exploited to formulate an efficient beamformer?


Beamforming in the principal path’s direction

TX array

RXprincipal path𝜃0

d0

d0 sin 𝜃0

plane wave


Beamforming in the principal path’s direction

I Principal path is stable (no surface interaction).

I Drifting will cause tx-rx positioning to change by a few metersover a few seconds, but this change is negligible compared toa transmission distance on the order of kilometers. Thechange in the principal path’s angle of arrival is thus expectedto be negligible.

I Note: receiver must still compensate for the Doppler shift(on either side of the link).


Beamforming: Classical approach (array processing)

I Plane wave propagation, equally-spaced array elements (d0),signal coming from direction θ0:τm0 = τ0

0 + m∆τ0, m = 0, . . . ,M − 1∆τ0 = d0

c sin θ0

I If the signal is properly synchronized, τ00 = 0.

I For a narrowband signal of frequency f0, beamformer weights

are wm0 = 1√

Me j2πf0τ

m0 ∼ e j2πmf0

d0c

sin θ0 (or ej2πm

d0λ0

sin θ0)

I Acoustic communication signal is not narrowband, sobeamforming weights are set for each frequency fk ,

wmk = e j2πmfk

d0c

sin θ0


Channel-based vs. angle-based beamforming

I Hmk =

∑p h

mp γ

mp (fk)e−j2πfkτ

mp =∑

p hmp γ

mp (fk)e−j2πfkτ

0p︸︷︷︸

hmp (fk )

e−j2πmfk∆τp

I wmk ∼ Hm∗

k =

hm∗0 (fk)︸︷︷︸h0e

−j2πfkτ00

e j2πmfkd0c

sin θ0︸︷︷︸steering to θ0

+∑

p 6=0 hm∗p (fk)︸︷︷︸

?

e j2πmfkd0c

sin θp︸︷︷︸steering to θp

I Optimal beamforming: match the phase and gain of everypath and array element.

I Beamforming in the principal direction: treat all other paths asnuisance (give up on predicting small-scale fading coefficientson the downlink), and just steer the beam in direction θ0.


Estimating the principal angle of arrival

I Let wmk (θ) = 1√

Me j2πmfk

dc

sin θ

I Beamforming on the k-th carrier of the uplink signal wouldyield

∑m wm

k (θ)xmk = wTk (θ)xk.

I Total power after beamforming is S(θ) =∑

k |wTk (θ)xk|2

I Principal angle is estimated as that angle for which the poweris maximized: θ0 = arg max

θS(θ)

I Beamformer weight are set to wmk = wm

k (θ0).


Notes on narrowband beamforming

I Narrowband beamforming rests on the assumption that thesignal bandwidth is much smaller than the center frequency.

I The weights assigned to different transmit elements are thesame for all signal frequencies.

I If the weights are evaluated for f0, then wmk = wm

0 .

I For angle-based narrowband beamforming,

θ0 = arg maxθ∑

k |wT0 (θ)xk|2, wm

k = wm0 = 1√

Me j2πmf0

dc

sin θ0

I Question: How much is the performance degraded undernarrowband assumption over underwater acoustic channels?


Numerical illustration

I System geometry:d = 1 km, h = 100 m, hR = 20 m, hT = 70 m (top)v = 0.5 m/s, θR = 45◦ ⇒ d increases, hR decreases by0.47 m between uplink/downlink time

I Frequency occupancy: f0 = 10 kHz, B = 5 kHz

I Transmit array: M = 12, d0 = 0.345 m (2.3λmax or 3.45λmin)

I hmp , τmp calculated from system geometry; c=1500 m/s in

water, 1300 m/s in bottom

I Small-scale fading: independent across array elements;σs , σb ∼ λ0, Bδp = 10−4 Hz

I Signal: K = 1024, differential QPSK

I Noise: σ2up = σ2

dn = σ2, SNR= 1σ2 with 1

MK

∑m

∑k

|Hm,upk |2 = 1


Multipath structure

delay [ms]0 0.005 0.01 0.015 0.02 0.025

uplin

k

-1

-0.5

0

0.5

1path gains

fist array elementlast array element

delay [ms]0 0.005 0.01 0.015 0.02 0.025

dow

nlin

k

-1

-0.5

0

0.5

1

fist array element

last array element


Frequency response of the channel: Uplink vs. downlink

frequency [kHz]×10

4

1 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4 1.45 1.5

chan

nel r

espo

nse

mag

nitu

de (

first

arr

ay e

lem

ent)

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

uplinkdownlink


Frequency response of the channel: Across the array

frequency [kHz]×10

4

1 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4 1.45 1.5

chan

nel r

espo

nse

mag

nitu

de (

uplin

k)

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

first arary elementsecond array element


Performance results

SNR [dB]10 15 20 25

MS

E [

dB

]

-35

-30

-25

-20

-15

-10

-5

0

5

10

CB-BB, true channelCB-BB, delayed channelCB-BB, TR estimateCB-BB, IR estimateAB-BB, true angleAB-BB, estimated angleCB-NB, true channelCB-NB, delayed channelAB-NB, true angleAB-NB, estimated anglenone

CB/AB=channel based/angle based, BB/NB=broadband/narrowband

delayed channel = uplink channel, noiseless


Inside angle-based beamforming: S(θ)

SNR=20 dBθ0 = 0.05 rad (0.05-0.0537 up; 0.0504-0.0542 down; 0.0524 est.)

θ [rad]-0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4

S(θ

)

×104

0

1

2

3

4

5

6

7

20000

40000

60000

80000

30

210

60

240

90

270

120

300

150

330

180 0

Re

-1.5 -1 -0.5 0 0.5 1 1.5

Im

-1.5

-1

-0.5

0

0.5

1

1.5


Resolution and ambiguity limits

I λmaxM∆max

< d0 ≤ λmin∆min

∆max ,min = max, min(p,q){sin θp − sin θq}

I LHS: Paths p and q are resolvable.

I RHS: There is no ambiguity as to whether the angle of path pis x degrees above or below the angle of path q.

I Note: include only those paths that are stable and notnegligible in strength.


Conclusion

I When the channel varies over the time it takes to close thefeedback, discrepancy between the uplink and the downlink issignificant enough to rule out channel-based beamforming.

I A possible solution is angle-based beamforming in thedirection of the principal path (that which has no surfaceinteraction and is stable). Rx/tx motion over theuplink-downlink should not cause a significant change in theangle of a typical geometry.

I Next steps:angle trackingbroadband null steeringmultiple userssimultaneous tx/rx beamforming.


transmit beamforming for acoustic ofdm

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