Principles of Direct Sequence & Frequency Hopping …...Principles of Direct Sequence & Frequency Hopping SSS Professor A. Manikas Imperial College London EE303 - Communication Systems

Principles ofDirect Sequence & Frequency Hopping SSS

Professor A. Manikas

Imperial College London

EE303 - Communication SystemsAn Overview of Fundamentals

Prof. A. Manikas (Imperial College) EE303: DS & FH SSS 30 Nov 2011 1 / 77

Table of Contents1 Introduction

Modelling of PN-signals in SSSEquivalent Energy Unitisation E¢ciency (EUE)Comments

2 Classification of Jammers in SSS3 Direct Sequence SSS

Introductory Concepts and Mathematical ModellingPSD(f ) of a Random Pulse SignalPSD(f ) of a PN-Signal b(t) in DS-SSSsPSD(f ) of DS/BPSK Spread Spectrum Tx Signal s(t)

4 DS-BPSK Spread SpectrumOutput SNIRBit Error Probability with Jamming

5 DS-SSS on the (pe , EUE, BUE)-parameter plane6 Anti-Jam Margin7 Some Comments on DS-QPSK-SSS8 Frequency Hopping SSSs9 Appendices

Appendix A. Block Diagram of a Typical Spread Spectrum SystemAppendix B. BPSK/DS/SS Transmitter and ReceiverAppendix C. QPSK/DS/SS Transmitter and ReceiverAppendix D. Proof of SNIRout (Equation 27)


Introduction

Introduction

(a) SSS (jammers): (b) CDMA (K users):


Introduction Modelling of PN-signals in SSS

Modelling of PN-signals in SSSConsider

!

({α[n]}=a sequence of ± 10sc(t) = an energy signal of duration Tc , e.g. c(t) = rect

ntTc

o

We have seen that the PN signal b(t) can be modelled as follows:I DS-SSS (Examples: DS-BPSK, DS-QPSK):

b(t) = ∑n

α[n].c(t ! nTc ) (1)

I FH-SSS (Examples: FH-FSK)

b(t) = ∑nexp {j(2πk[n]F1t + φ[n])} .c(t ! nTc ) (2)

where {k[n]} is a sequence of integers such that

{α[n]} 7! {k[n]} (3)

and with φ[n] = random: pdfφ[n] =12π rect

n12π

o


Introduction Equivalent Energy Unitisation E¢ciency (EUE)

Equivalent Energy Unitisation E¢ciency (EUE)Remember:

F Jamming source, or, simply Jammer = intentional interferenceF Interfering source = unintentional interference

F With area-B = area-A we can find NjF Pj = 2% areaA = 2% areaB = NjBj ) Nj =

PjBj


Introduction Equivalent Energy Unitisation E¢ciency (EUE)

ifBJ = qBss ; 0 < q ' 1 (4)

then

EUEJ =EbNJ

=Ps .BJPJ .rb

=Ps .q.BssPJ .B

= PG% SJRin % q (5)

EUEequ =Eb

N0 +NJ(6)

= PG% SJRin % q| {z }EUEj

%(N0Nj+ 1

)!1(7)

where

SJRin ,PsPJ

(8)


Introduction Comments

CommentsEUEequ (or EUEJ ): very important since bit error probabilities aredefined as function of EUEequ (or of EUEJ )For a specified performance

!*the smaller the SIRin ) the better for the signalthe larger the SIRin ) the better for the jammer

Jammer limits the performance of the communication systemi.e. e§ects of channel noise can be ignoredi.e. Jammer Power( Pn ) EUEequ =

EbN0+NJ

' EbNJ= EUEJ

""

N.B. exception to this:i) non-uniform fading channelsii) multiple access channels

9 ∞ number of possible jamming waveformsThere is no single jamming waveform that is worst for all SSSsThere is no single SSS that is best against all jamming waveforms.


Classification of Jammers in SSS

Classification of Jammers in SSS


Direct Sequence SSS Introductory Concepts and Mathematical Modelling

Direct Sequence SSSIntroductory Concepts & Mathematical Modelling

If BPSK digital modulator is employed then the BPSK-signal can bemodelled as:

BPSK s(t) = Ac · sin(2πFc t +m(t) · π2 ) (9)

where the data waveform can be modelled as follows:

m(t) , ∑na[n] · c1(t ! n · Tcs )| {z }

rect{ t!n·TcsTcs }

; nTcs ' t < (n+ 1) · Tcs (10)

with {a[n]} = sequ. of independent data (message) bits (±1’s)



Equation (9) can be rewritten as follows:

BPSK s(t) = Ac ·m(t) · cos(2πFc t)

) BPSK can be considered as(

PMAM

(11)

remember:



The PSD(f )’s of m(t) and s(t) are shown below



If a DS/BPSK modulator is employed, then the SS-transmitted signalis

DS/BPSK : s(t) = Ac sin

0

@2πFc t +

±1z }| {m(t)b(t)

π

2

1

A (12)

= Acm(t)b(t) cos(2πFc t) (13)

where

8>>>>>>>>>><

>>>>>>>>>>:

m(t) , ∑na[n] · c1(t ! nTcs )| {z }

"rect{ t!nTcsTcs }

; nTcs ' t < (n+ 1)Tcs

b(t) = ∑k

α[k ] · c2(t ! kTcs )| {z }"

rect{ t!kTcTc }

; kTc ' t <

chip-duration

(k + 1)#Tc

3333333333333333

(14)



BPSK DS/CDMA Transmitter & Receiver

TX (see also Appendix-B):

Rx (see also Appendix-B):



The PN-sequence {α[l ]} (whose elements have values ±1)

is M times faster than the data sequence {α[n]} .

i.e.Tcs = M · Tc (15)

i.e.PN-signal Bandwidth ( data-Bandwidth (16)



Systems which have coincident data and SS code clocks

are often said to have a “data privacy feature”I such systems are easy to build and can be combined in single units

Tcs =

5qMTc

Note: chip = Tc = smallest time increment



If the above “data privacy feature” is taken into account

then8>>>><

>>>>:

m(t) , ∑na[n].c1(t ! n · Tcs ) nTcs < t < (n+ 1)Tcs

b(t) = ∑k

α[k ].c2(t ! kTc ) kTc < t < (k + 1)Tc

with Tcs = MTc ;4 kM

5= n;

(17)

where

n · Tcs + k 0 · Tc ' t < n · Tcs + (k 0 + 1) · Tc8k 0 = 0, 1, . . . ,M ! 1 = kmodM



Conclusion

DS/BPSK similar to BPSK"

except that the apparent data rate is M times faster+

signal spectrum is M times wider

ThereforePG = Bss

B = M (18)

Note:1 message cannot be recovered without knowledge of PN-sequence i.e.PRIVACY

2 typical:F PN-chip-rate ! several M bits/secF data rate ! few bits/sec



PSD of DS/BPSK/SS Transmitted Signal

Tx signal

s(t) = m(t) · b(t) · Ac · cos(2πFc t)

+PSDs (f ) = PSDm(f ) 0 PSDb(f ) 0 PSDAc cos(2πFc t)(f )

+PSDs (f ) = PSDm(f ) 0 PSDb(f ) 0 PSDAc cos(2πFc t)(f )

remember

PSDAc cos(2πFc t)(f ) =A2c4· (δ (f ! Fc ) + δ (f + Fc ))


Direct Sequence SSS PSD(f ) of a Random Pulse Signal

PSD(f) of a Random Pulse SignalFor a random pulse signal m(t) (i.e. a sequence of pulses where thereis an invariant average time of separation Tcs between pulses) with allpulses of the same form but with random amplitudes and statisticalindependent random time of occurrence, then:

PSDm(f ) =1Tcs

· En|Fourier Transform of a single pulse|2

o



Example: PSD of a random BINARY signal m(t)Consider a random binary sequence of 0’s and 1’s. This binarysequence is transmitted as random signal m(t) with 1’s and 0’s beingsent using the pulses shown below.

For instance a random binary sequence/waveform could be

If 1’s and 0’s are statistically independent with Pr(0) = Pr(1) = 0.5,the PSD of the transmitted signal can be estimated as follows:



Solution:

PSDm(f ) =1Tcs

· E

8<

:

666666FT

0

@

1

A

666666

29=

;

=1Tcs

· E:a2 · T 2cs · sinc

2(fTcs );

=1Tcs

· T 2cs · sinc2 {fTcs} · E

:a2;

| {z }=(!A2) 12+(+A2)

12

= TcsA2 sinc2 {fTcs}

Rmm(τ) = ? for you!


Direct Sequence SSS PSD(f ) of a PN-Signal b(t) in DS-SSSs

PSD(f) of a PN-Signal b(t) in DS-SSSsAutocorrelation function: Rbb(τ) = 1

NcTcRcc (τ)~ Rαα(τ)

where c(t) =rect tTc =

1 Rcc (τ) = TcΛ( τTc)

2 Rαα(τ) , (Nc + 1)repNcTc {δ(τ)}!repTc {δ(τ)}

3 autocorrelation function Rbb(τ) =1




i.e.

Rbb(τ) =1


=1

NcTc(Nc + 1)repNcTc

*TcΛ

(τ

Tc

)<

!1

NcTcrepTc

*TcΛ

(τ

Tc

)<

=Nc + 1Nc

· repNcTc

*Λ(

τ

Tc

)<!1Nc



Thus, we have

Rbb(τ) =Nc+1Nc

repNcTcn

Λ=

τTc

>o! 1

Nc(19)

By using the FT tables the PSD(f )= FT{Rbb(τ)} of the signal b(t)is:

PSDb(f ) =Nc+1N 2c

comb 1Nc Tc

:sinc2 {f · Tc}

;! 1

Ncδ(f ) (20)


Direct Sequence SSS PSD(f ) of DS/BPSK Spread Spectrum Tx Signal s(t)

PSD(f) of DS/BPSK Spread Spectrum Tx Signal s(t)

s(t) = m(t) · b(t) · Ac · cos(2πFc t)

) PSDs (f ) = PSDm(f ) 0 PSDb(f ) 0A2c4 (δ(f ! Fc ) + δ(f + Fc ))| {z }

term 1

Ignore (for the time being) the e§ects of m(t)

PSDterm1(f ) = A2c4 · PSDb(f ) 0 (δ(f ! Fc ) + δ(f + Fc ))

= A2c4 ·*Nc+1N 2c

· comb 1Nc Tc

:sinc2(f · Tc )

;!1Nc· δ(f )

<

0 (δ(f ! Fc ) + δ(f + Fc ))

= A2c4 ·

Nc+1N 2c

·=comb 1

Nc Tc

:sinc2 {(f ! Fc )Tc}

;

+ comb 1Nc Tc

:sinc2 {(f + Fc )Tc}

;>

!A2c4

1Nc(δ(f ! Fc ) + δ(f + Fc ))



i.e.

PSDterm1(f ) =A2c4·Nc + 1N2c

·=comb 1

Nc Tc

:sinc2 {(f ! Fc )Tc}

;

+ comb 1Nc Tc

:sinc2 {(f + Fc )Tc}

;>

!A2c41Nc(δ(f ! Fc ) + δ(f + Fc )) (21)



Above the e§ects of m(t) have not been taken into account.I If m(t) is used,then each discrete frequency in Equation (21) becomes a sinc2

function.I There are two cases:

CASE-1:

1Tcs< 1

2NcTc

CASE-2:

1Tcs1 1

NcTc

the peaks will merge into acontinuous smooth spectrum



CASE-1:



CASE-2:



N.B.:I If

b(t) = random

thens(t) = Acm(t)b(t) cos(2πFc t)

I If the e§ects of m(t) are ignored then

PSDs (f ) = A2cT2c sinc {fTc} 0

14(δ(f ! Fc ) + δ(f + Fc ))

+

PSDs (f ) =Tc4A2c=sinc2 {(f ! Fc )Tc}+ sinc2 {(f + Fc )Tc}

>

i.e. PSD is similar to ‘CASE-2’ above


DS-BPSK Spread Spectrum Output SNIR

DS-BPSK Spread Spectrum:Output SNIR

Consider the block diagram of a SS-Communication System whichemploys a BPSK digital modulator:



N.B.: Bss = 1Tc; B = 1

Tcs; PG = Bss

B = TcsTc

Then, at point T1 , we have

s(t) = Ac ·m(t) · b(t) · cos(2πFc t) (22)

and at point^T1 :

s(t) + n(t) + j(t) (for k = 1) (23)

At the input of the receiver the Signal-to-Noise-plus-InterferenceRatio (SNIRin) is:

SNIRin =EUE

PG · (1+ JNRin)=EUEequPG

(24)



at point^T :

(s(t) + n(t) + j(t)) · b(t ! τ) · 2 cos(2πFc t + θ) (25)

at point T0 :

From the analysis presented in Appendix-D we have

Punwanted =

8><

>:

N0Tcs+ A2c

Tcs ·Tc · τ2 · cos2(θ) + Pjout for |τ| ' Tc

N0Tcs+ A2c ·Tc

Tcs· cos2(θ) + Pjout for |τ| > Tc

3333333(26)



Therefore

SNIRout =PdesiredPunwanted

=

8>><

>>:

A2c ·Λ2( τTc )·cos

2(θ)

N0Tcs+ A2cTcs ·Tc ·τ

2 ·cos2(θ)+Pjoutfor |τ| ' Tc

0 for |τ| > Tc

33333333(27)

if τ = 0 and θ = 0 (i.e. the system is synchronized) then:

SNIR out-maxmatched filter

=A2c

N0Tcs+ Pj · TcTcs

=2 · Ps

N0Tcs+

Pj ·TcTcs

=2 · EbTcs

N0Tcs+

Pj ·TcTcs

=2 · Eb

N0 + Pj · Tc| {z }=

PjBss=Nj

or, equivalently


= 2EUE equ (28)



However (see Equation 48, Appendix D),

SNIRin =EUE


(29)

Therefore,


= 2EUEequ= 2.PG.SNIRin (30)


DS-BPSK Spread Spectrum Bit Error Probability with Jamming

Bit Error Probability with Jamming

A. CONSTANT POWER BROADBAND JAMMER:

From the “Detection Theory” topic we know that thebit-error-probability pe for a Binary Phase-Shift Key (BPSK)communication system is given by:

pe = T

8<

:p2 · EUE| {z }

SNRout, matched filter

9=

; where EUE =EbN0

(31)

Consider a DS/BPSK SSS which operates in the presence of aconstant amplitude broadband jammer with double sided powerspectral density

PSDj (f ) =Nj2

(32)



Then,pe = T

np2 · EUEequ

o

where EUEequ =Eb

N0+Njwith Nj =

PJBss

If we make the assumption that Nj ( N0 then

pe = T:p2 · EUEJ

;(33)

where EUEJ =EbNj

This is known as the BASELINE PERFORMANCE of a DS/BPSKSSS



B. PULSE JAMMER:

Consider a DS/BPSK SSS which operates in the presence of ajammer which transmits “broadband noise” with large power but onlya fraction of the time.

The double-sided power spectral density of the jammer is given by:

PSDj (f ) =Nj2ρ

(34)

whereI ρ , the fraction of time the jammer is “on”.

I Pj = average jamming power

I Pjρ = actual power during a jamming pulse duration



Let the jammer pulse duration be greater than Tcs (data bit time).

I Then*Pr(jammer = “on” ) = ρPr(jammer = “o§” ) = 1! ρ

and the bit-error-probability is given by:

pe = (1! ρ)T

(s

2EbN0

)

| {z }'0 (very small)

+ ρT

8><

>:

vuut2 Eb

N0 +Njρ

9>=

>;(35)

which can be simplified to

pe = ρ·T{2ρ · EUEJ} where EUEJ =EbNj

(36)



By plotting the above equation for di§erent values of ρ we get:



Note thatI the value of which maximizes pe decreases with increasing values ofEUEj

I there is a value of ρ which maximizes the probability of error pe . Thisvalue can be found by di§erentiating Equation-36 with respect to ρ.That is

dpedρ

= 0) ρ0 =

(0.709EUEj

if EUEj > 0.709

1 if EUEj ' 0.709(37)

I Therefore

pemax = maxρ

nρ · T

nq2ρEUEj

oo(38)

pemax = ρ0 · Tnq

2ρ0EUEjo

=) pemax =

(0.083EUEj

if EUEj > 0.709

T:p

2EUEj;

if EUEj ' 0.709(39)



N.B.:I when jammer pulse length is shorter than a data bit time Tcs then theabove expression is not valid.

I However, Equation-39 represents an UPPER BOUND on thebit-error-probability pe .

The next graph illustratesI the bit-error-prob. plotted against the EUEj for a baseline jammer(i.e. ρ = 1) and

I the worst case jammer (i.e. ρ = ρ0) for a DS/BPSK SSS.

Note the huge di§erence between the two curves.




DS-SSS on the (pe , EUE, BUE)-parameter plane

DS-SSS on the ( pe, EUE, BUE)-parameter plane


Anti-Jam Margin

Anti-Jam Margin

An important parameter of SSS is the ANTIJAM MARGIN which isdefined as follows:

dB(AJM) , dB(EUEequ) ! dB(EUE which corresponds to the pe ,PR )

) dB(AJM) , 10 0 log(EUEequ)! 10 0 log(EUEpe ,PR )

AJM represents a safety margin against jammer (or against jammerplus noise).


Some Comments on DS-QPSK-SSS

Some Comments on DS-QPSK-SSSIn a DS-QPSK-SSS the binary information is represented by twodi§erent data signals mI (t) and mQ (t) and two di§erent PN-signalbI (t) and bQ (t)

data signals and PN-signals:

8>>>>>>>>>><

>>>>>>>>>>:

inphase

8>><

>>:

mI (t) , ∑n(even)

a[n] · rectnt!n·TcsTcs

o

bI (t) , ∑k (even)

α[k ] · rectnt!k ·TcTc

o

quadrature

8>><

>>:

mQ (t) , ∑n(odd )

a[n] · rectnt!n·TcsTcs

o

bQ (t) , ∑k (odd )

α[k ] · rectnt!k ·TcTc

o

(40)

with

Tcs = PGTc ;FkPG

G= n;



Note that mI (t) is formed by the even data bits and mQ (t) by theodd ones. At each transmitter, the two data signals are firstlymultiplied by two di§erent code waveforms bI (t) and bQ (t), and thenby two di§erent carriers

Acp2· cos(2πFc t + φ) and

Acp2· sin(2πFc t + φ) (41)

while the transmitted signal s(t) is formed by the sum of these twocomponents:

s(t) = Acp2·mI (t) · bI (t) · cos(2πFc t + φ)

+ Acp2·mQ (t) · bQ (t) · sin(2πFc t + φ)

(42)

where Ac and Fc are assumed common for all carriers.It is important to point out that now the amplitude of the carriers haschanged from Ac to Acp

2so that the power of the transmitted signal

s(t) still remains equal to A2c2 .



DS-QPSK-SSS implementation No.1: QPSK1

(Note: if = Tc2 delay is inserted then OQPSK1 - o§set QPSK1)

Transmitter:

s(t) =p2Ps · cos(2πFc t +

tan!1(bQ (t)·mQ (t)bI (t)·mI (t)

)

z}|{ψ(t) )



Receiver:



DS-QPSK-SSS implementation No.2: QPSK2

(Note: if = Tc2 delay is inserted then OQPSK2)

Transmitter:

s(t) =p2Ps · cos(2πFc t +

tan!1(bQ (t)·m(t)bI (t)·m(t)

)

z}|{ψ(t) )



Receiver:


Frequency Hopping SSSs


Consider that the following part of the spectrum has been allocatedto a FH/SSS:

Let us partition the above spectrum onto L di§erent frequency slotsof bandwidth F1 (or of bandwidth qF1 where q is a constant). ThenBss = q · F1 · L

L = 2m



Define the following symbols:Tc = hop duration (i.e. hop rate rhop = 1

Tc)

Tcs = message bit duration (i.e. bit rate rb = 1Tcs)

M = number of hops per message bit (i.e. Tcs = M · Tc )

FH/SSS:

8<

:

Fast hop ! rhop > rbslow hop ! rhop < rbbalance hop ! rhop = rb

The frequency slot is constant in each time chip Tc , BUT changesfrom chip-to-chip.This can be represented by the following diagram:



In general, the No. of di§erent frequency slots L over which the signalmay hop, is a power of 2.



F1 is, in general, equal to 1Tc,

i.e. F1 = 1Tc(but this is not a necessary requirement).

several FH signals occupy a common RF channel

FH model of b(t) - (complex representation):

b(t) = ∑nexp {j(2π.k [n].F1t + φn)} · rect

*t ! nTcTc

<

where

I k [n] =f{PN-sequ {α[n]}}I k [n] is an integer that is formed by a codeword which is formed by oneor more m-sequences



Transmitter-Receiver path:



A Di§erent Implementation:



L frequencies are produced by the digital Frequ. Synthesizer,separated by F1

Bs ' q · F1 · L (43)

Note:I if reception= coherent :

F more di¢cult to achieveF places constraints on the transmitted signal and transmitted medium

I if reception= non-coherent :

F PN-gen. can run at a considerably slower rate in this type of systemthan in a DS system



FH: non-coherent ) poorer performance against thermal noise.

Performance:I Coherent FSK (CFSK)

pe ,CFSK = Tnp

EUEo

I Non-Coherent FSK (NFSK)

pe ,NFSK =12exp

(!EUE2

)



A serious problem is the

very strong signals at receiver

swapping out the e§ects

of weaker signal

*z }| {“near-far” problem :

I DS: severe problemI FH: much more susceptible

acquisition: much faster in FH than in DS

PG = BssB = it is not very good criterion for FH


AppendicesAppendix A. Block Diagram of a Typical Spread Spectrum

System

AppendicesAppendix A. Block Diagram of a Typical SSS

(terrestrial & satellite comm. systems)


Appendices Appendix B. BPSK/DS/SS Transmitter and Receiver

Appendix B. BPSK/DS/SS Transmitter and ReceiverBPSK/DS/SS Transmitter


Appendices Appendix B. BPSK/DS/SS Transmitter and Receiver

BPSK/DS/SS Receiver


Appendices Appendix C. QPSK/DS/SS Transmitter and Receiver

Appendix C. QPSK/DS/SS Transmitter and Receiver

Transmitter





Receiver:


Appendices Appendix D. Proof of SNIRout (Equation 27)

Appendix D. Proof of SNIRout (Equation 27)



N.B.: Bss = 1Tc; B = 1

Tcs; PG = Bss

B = TcsTc

Then at point T1 we have

s(t) = Ac ·m(t) · b(t) · cos(2πFc t) (44)

and at point^T1 :

s(t) + n(t) + j(t) (for k = 1) (45)

At the input of the receiver the Signal-to-Noise-plus-InterferenceRatio (SNIRin) is:

SNIRin = PsPn+Pj

= Ps/Pn1+Pj/Pn

=Ps ·Tcs

N0 ·Bss ·Tcs1+JNRin

=EUE

Bss ·Tcs1+JNRin

= EUEPG·(1+JNRin)



i.e.

8><

>:

SNIRin = PsPn+Pj

EUEPG·(1+JNRin)

or

SNIRin =A2c/2

N0 ·Bss+Pj

(46)

However,

EUEequ =Eb

No +Nj=

Eb/N01+ Pj

BssN0

=EUE

1+ PjPn

=EUE

1+ JNRin(47)

Therefore,

SNIRin =EUE


(48)



at point^T :

(s(t) + n(t) + j(t)) · b(t ! τ) · 2 cos(2πFc t + θ) (49)

at point T0 :

i) signal term:

m0(t) =AcTcs

·Z Tcs

0m(t) · b(t) · b(t ! τ) · cos(θ) · dt

= ±AcTcs

· cos(θ) ·Z MTc

0b(t) · b(t ! τ) · dt

) m0(t) = ±Ac · RMbb(τ) · cos(θ) (50)



However,

Pm0 = E:m20(t)

;= A2c · E

nRM

2

bb (τ)o· cos2(θ)

+

Pm0 = A2c ·=Var

nRMbb(τ)

o+ E2

nRMbb(τ)

o>· cos2(θ)

=) Pm0 = A2c · Var

nRMbb(τ)

o· cos2(θ)

"due to Code Noise

+ A2c · E2nRMbb(τ)

o· cos2(θ)

"desired term

(51)



However:

E {m0(t)} =

8<

:

±Ac · E:RMbb(τ)

;· cos(θ) for |τ| ' Tc

0 for |τ| > Tc

333333(52)

i.e.power of

desired signal part= Pdesired = E2 {m0(t)} (53)

=

8><

>:

A2c ·Λ2=

τTc

>· cos2(θ) for |τ| ' Tc

0 for |τ| > Tc

3333333



In addition

Var {m0(t)} = Varn±AcRMbb(τ) cos(θ)

o

= A2cVarnRMbb(τ)

ocos2(θ)

) Pcode-noise = Var {m0(t)}

if 0 ' τ ' Tc then

Var {m0(t)} = A2cVarnRMbb(τ)

ocos2(θ)

= A2cR2cc (τ ! Tc )M · T 2c

cos2(θ)

=A2cTcTcs

τ2

T 2ccos2(θ) (54)

if τ > Tc then

Var {m0(t)} = A2cVarnRMbb(τ)

ocos2(θ)

= A2c1Mcos2(θ)

=A2cTcTcs

cos2(θ) (55)



ii) noise term

nout (t) =1Tcs

Z Tcs

0n(t) · b(t ! τ) · 2 cos(2πFc t + θ) · dt

Pn!out = E:n2out (t)

;= . . . for you . . .

. . . ) Pn!out = N0Tcs

(56)



iii) jammer

If Bj 5 Bssthen Jammer = Narrowband wrt the desired SS-Signal.

In this case the jammer j(t) can be modelled as:

j(t) = Aj (t) · cos(2πFj t + φj (t))

and the jout (t) can be estimated as follows:

jout (t) =1Tcs

Z Tcs

0j(t) · b(t ! τ) · 2 cos(2πF0t + θ) · dt

= for you . . .Prof. A. Manikas (Imperial College) EE303: DS & FH SSS 30 Nov 2011 75 / 77


for you . . .

) Pjout =PjPG (57)

N.B.: If jammer = wideband, i.e. if Bj 6 Bss , then Pjout = ?



From the previous analysis we have

Punwanted =

8><

>:

N0Tcs+ A2c

Tcs ·Tc · τ2 · cos2(θ) + Pjout for |τ| ' Tc

N0Tcs+ A2c ·Tc

Tcs· cos2(θ) + Pjout for |τ| > Tc

3333333(58)

Therefore

SNIRout =PdesiredPunwanted

=

8>><

>>:

A2c ·Λ2( τTc )·cos

2(θ)

N0Tcs+ A2cTcs ·Tc ·τ

2 ·cos2(θ)+Pjoutfor |τ| ' Tc

0 for |τ| > Tc

33333333(59)


Principles of Direct Sequence & Frequency Hopping …...Principles of Direct Sequence & Frequency Hopping SSS Professor A. Manikas Imperial College London EE303 - Communication Systems

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