SS5 - Underwater Communication Basicsw3.ualg.pt/~sjesus/aulas/2010/pdeet/ss5.pdf · Outline of SS5...

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Doctoral Program on Electrical Engineering and Communications SS5 - Underwater Communication Basics ergio M. Jesus ([email protected]) Universidade do Algarve, PT-8005-139 Faro, Portugal www.siplab.fct.ualg.pt 3 March 2010 SS5 - Underwater Communication Basics

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Page 1: SS5 - Underwater Communication Basicsw3.ualg.pt/~sjesus/aulas/2010/pdeet/ss5.pdf · Outline of SS5 The Underwater Acoustic Communication Problem (1) UComs: an estimation or a detection

Doctoral Program on Electrical Engineering and Communications

SS5 - Underwater Communication Basics

Sergio M. Jesus([email protected])

Universidade do Algarve,PT-8005-139 Faro, Portugal

www.siplab.fct.ualg.pt

3 March 2010

SS5 - Underwater Communication Basics

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Outline of SS5

• The Underwater Acoustic Communication Problem(1) UComs: an estimation or a detection problem ?(2) Basic structure

• Baseband Digital Data Transmission(1) Binary case: correlator-receiver and detector(2) Antipodal and On-off signalling(3) Pulse Amplitude Modulation(4) Multisignal(5) constellation diagrams

• Carrier Modulation(1) Phase Shift Keying (PSK)(2) Quadrature Amplitude Modulation (QAM)

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The Underwater Acoustic Communications Problem

Let us assume the (usual) data model / discrete version{p(t) = h(t) ∗ s(t)y(t) = p(t) + w(t) or

{p = Hsy = p + w

As an estimation problem: estimate deterministic signal s(t),

sLS/ML = (HtH)−1Hty

As a detection problem: use various si(t) to code information (signalling) →correlator-receiver (or MF)

T (y) = ytpi > γ

where pi = Hsi, but...H is unknown !!

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Basic structure

Generic building block structure

+

+s(t) s(t)^

Coder Modulator

Emitter Channel

Channel

Noise

Demod. Decoder

Receiver

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Baseband Digital Data Transmission

Assume an additive white Gaussian noise channel

y(t) = s(t) + w(t)

with w(t) : N [0, σ2δ(t)].as opposed to...

Band limited white Gaussian noise channel

y(t) = h(t) ∗ s(t) + w(t)

with h(t) non ideal system, H(f) = H0 exp[−jθ(ω)], amplitude and phase distortion.

• H0 not constant with frequency

• θ(ω) not linear

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Binary signalling

Transmit 0’s and 1’s

Code or signal

0→ s0(t)

1→ s1(t) 0 ≤ t ≤ Tb,

Tb is the bit duration in seconds.

Data rate is Rb = 1/Tb bits/s.

Assume that bits are equally probable and are independent

Received signal is

y(t) = si(t) + w(t), i = 0, 1, 0 ≤ t ≤ Tb

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Optimal correlator - receiver (1)

X

0

T()dt

s(t)1

X

0

T()dt

s(t)0

y(t)

t=Tb

b

b

y1

y0

DetecOutput

y0 =∫ Tb

0

y(t)s0(t)dt

y1 =∫ Tb

0

y(t)s1(t)dt

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Optimal correlator - receiver (2)

Assume a 0 is transmitted, so y(t) = s0(t) + w(t),

y0 =∫ Tb

0

y(t)s0(t)dt =∫ Tb

0

s20(t)dt+∫ Tb

0

w(t)s0(t)dt= E + w0

y1 =∫ Tb

0

y(t)s1(t)dt =∫ Tb

0

s0(t)s1(t)dt+∫ Tb

0

w(t)s1(t)dt= w1

where it was assumed that

Orthogonality:

∫ Tb

0

s0(t)s1(t)dt = 0⇒ s0(t) ⊥ s1(t)

Equal energy: E =∫ Tb

0

s20(t)dt =∫ Tb

0

s21(t)dt

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Optimal correlator - receiver (3)

w0 =∫ Tb

0

w(t)s0(t)dt

w1 =∫ Tb

0

w(t)s1(t)dt

E[wi] =∫ Tb

0

si(t)E[w(t)]dt = 0

σ2i =

∫ Tb

0

∫ Tb

0

si(t)s1(t′)E[w(t)w(t′)]dtdt′

= σ2

∫ Tb

0

si(t)si(t′)δ(t− t′)dtdt′

= σ2

∫ Tb

0

s2i (t)dt = Eσ2

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Optimal correlator - receiver (4)

PDFs

p(y0/s0) =1√2πσ

e−(y0−E)2/2σ2

p(y1/s0) =1√2πσ

e−y21/2σ

2

DetectorThreshold γ = E/2

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Antipodal Signalling

Antipodal waveforms are such that s0(t) = −s1(t), so

0→ s0(t) = s(t)

1→ s1(t) = −s(t)

Received signal: y(t) = ±s(t) + w(t) → single correlator - receiver with s(t)

Correlator output

r(t) ={ E + w if s(t)−E + w if -s(t)

with

w =∫ Tb

0

n(t)s(t)dtE[w] = 0σ2w = σ2E

Detector: threshold γ = 0

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On-off Signalling

On-off signalling uses the scheme where

y(t) ={w(t) if 0 is transmitteds(t) + w(t) if 1 is transmitted

so

y ={w if 0 is transmittedE + w if 1 is transmitted

therefore

p(y/0) = 1√2πσ

e−y2/2σ2

p(y/1) = 1√2πσ

e−(y−E)2/2σ2

Detector: threshold γ = E/2.

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Probability of error for binary signalling

Antipodal

Pe = Q(√2E

σ2

)Orthogonal

Pe = Q(√ E

σ2

)On-off

Pe = Q(√ E

2σ2

)

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Pulse Amplitude Modulation (PAM)

Signalling with M = 2k multiple amplitude levels: k = log2M bits every T seconds

sm(t) = Amg(t), 0 ≤ t ≤ T, m = 0, 1, . . . ,M − 1

where Am = (2m−M + 1)d, and

g(t) ={√

1/T 0 ≤ t ≤ T0 otherwise

normalized energy rectangular pulse (pulse shape).

Bit rate Rb = 1/Tb → symbol rate Rs = 1/T = 1/kTb. So Rs = Rb/k.

Detector: pick level m with the shortest Euclidian distance Dm = |y − Am| (sinceunidimensional).

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Probability of error for PAM

Pe =2(M − 1)

MQ(√3(log2M)Eav

(M2 − 1)σ2

)Eav = average signal energy

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Multisignal signalling

Multidimensional signal set such that

Orthogonal:

∫ T

0

si(t)sk(t)dt = Eδik, i, k = 0, 1, . . . ,M − 1

Identical energy: E =∫ T

0

s2i (t)dt, i = 0, 1, . . . ,M − 1

Received signal: y(t) = si(t) + w(t), 0 ≤ t ≤ T, i = 0, . . . ,M − 1

Allows for a multidimensional representation

s0 = (√E , 0, . . . , 0)

s1 = (0,√E , . . . , 0)

... = ...

sM−1 = (0, . . . , 0,√E)

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Multisignal detector

+

12E1-

+

12E1-

+

12E1-

X

0

T()dt

s(t)1

X

0

T()dt

s(t)

X

0

T()dt

s(t)M

2

r(t)

t=T

.

.

.

....

.

PickMax

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Constellation diagrams

Binary constellations

|

0

|

0

|

0−√E

√E

√E√E√E

Multi level (PAM) Multisignal

|0|

0−3d 3d−d d

√E

√E√E

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Carrier modulation

Baseband: PAM signal sm(t) = Amg(t), Sm(f) has same bandwidth as G(f) = F [g(t)]

Modulation

Carrier

Baseband Bandpass

sm(t)

cos(2πfct)

um(t) = sm(t) cos(2πfct)

Spectral content: Um(f) = Am2 [G(f − fc) +G(f + fc)]

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Phase-shift keying (PSK)

Information on signal phase: θm = 2πmM , m = 0, 1, . . . ,M − 1.

um(t) = Ag(t) cos(2πfct+

2πmM

), m = 0, 1, . . . ,M − 1

when g(t) is a square pulse of amplitude√

2T and energy of um(t) is Es

um(t) =

√2ET

cos(2πfct+

2πmM

)=

√2ET

cos(2πmM

)cos 2πfct−

√2ET

sin(2πmM

)sin 2πfct,

m = 0, 1, . . . ,M − 1, 0 ≤ t ≤ T

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Probability of error for PSK

Pe ≈ 2Q(√kEb

σ2sin

π

M

)Eb = bit energy

Detector: θy = arctan Im(y)Re(y)

and pick signal sm with closestphase to θy

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Quadrature Amplitude Modulation (QAM)

Exploit both signal space and amplitude where signals in quadrature (amplitude andphase modulation) with M2 = 2.

umn(t) = Amg(t) cos(2πfct+ θn), m = 1, . . . ,M1, n = 1, . . . ,M2

Rectangularand circular QAMconstellationsM1 = 2k1, M2 = 2k2

k1 + k2 = log2M1M2

Rs = Rb/(k1 + k2)

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Bibliography

1. J.P. Proakis and M. Salehi, “Contemporary Communications Systems”,

Brooks Cole Thomson Learning, 2000.

2. A.B. Carlson, P.B. Crilly e J.C. Rulledge “Communications Systems”,

4 Edition, McGraw-Hill, 2002.

3. J.G. Proakis, “Digital Communications”, 3rd Edition, McGraw-Hill, 1995.

SS5 - Underwater Communication Basics S.M.Jesus