@Y. C. Jenq1 Digital Communication Theory and Systems Y. C. Jenq Department of Electrical & Computer...

@Y. C. Jenq 1

Digital Communication Theory and Systems

Department of Electrical & Computer Engineering

Portland State University

P. O. Box 751

Portland, OR 97207

[email protected]

@Y. C. Jenq 2

Outlines

Pulse Amplitude Modulation (PAM)– Probability of Error under AWGN– Optimal Receiver & Matched Filter– Geometric Representation

Multi-Dimensional Orthogonal Signals– PSK Systems– QAM Systems– FSK Systems

@Y. C. Jenq 3

Outlines-continued

Optimal Receiver for M-ary Orthogonal Signals with Additive White Gaussian Noise– PAM Systems– PSK & QAM Systems– FSK Systems

Probability of Error for Optimal Detector with Additive White Gaussian Noise– PAM Systems– PSK & QAM Systems– FSK Systems

@Y. C. Jenq 4

Outlines-continued

Digital Transmission through Band-limited Additive White Gaussian Noise Channels – Base-band Channels & Intersymbol Interference (ISI)– Pass-Band Channels

Power Spectrum of Digitally Modulated Signals Signal Design for Band-limited Channels Probability of Error with ISI and AWGN System Design and Channel Equalization Multi-Carrier Modulation and OFDM

@Y. C. Jenq 5

PAM Base-band Systems

Binary PAM System (binary antipodal signaling)1 - represented by a pulse with amplitude A0 - represented by a pulse with amplitude -A

s1(t)

t

A

Tbs2(t)

t

-A

Tb

Tb : bit interval (second)

Rb = 1/Tb : bit rate (bit/sec)

@Y. C. Jenq 6

A simple Receiver – Sample & Threshold Detector

r(t) = sm(t) + n(t) > 0?Am = 1

Am = -1

yes

no

Probability of Error

Pe = Pr{Am=1} Pr{R<0} + Pr{Am=-1} Pr{R>0}

Pe = Q (Am /n) = Q ([S/R]1/2)

R

Gaussian Noise

Probability of Error for Binary PAM Systems with AWGN

@Y. C. Jenq 7

The Optimal Receiver – The Matched Filter

s(t) h(t) = s(T-t) y(t)

T T T 2Tttt

y(t) = 0

ts()h(t-)d = 0

ts()s(T-t+)d

y(T) = 0

Ts()s()d

Probability of Error for Binary PAM Systems with AWGN

@Y. C. Jenq 8

The Matched Filter

The received signal: r(t) = s(t) + n(t)

Matched filter output: y(t)

y(t) = 0

tr()h(t-)d= 0

t[s()+n()]h(t-)d

Sampled at t=T

y(T) = 0

T s()h(T-)d + 0

Tn()h(T-)d

= ys(T) + yn(T)

Output Signal to Noise Ratio: (S/N)o

(S/N)o = ys2

(T)/ E{yn2

(T)}

@Y. C. Jenq 9

The Matched Filter

If the noise is White with the power spectral density No/2 The h(t) that maximizes the output signal to noise ratio, (S/N)o , is the Matched Filter:

h(t) = s(T-t)

And the maximum output signal to noise ratio is

(S/N)o = (2/No)0

Ts()s()d = 2Es/No = Es/(No /2)

= Signal Energy/Noise Power Spectral Density

@Y. C. Jenq 10

Probability of Error for Binary PAM Systems with AWGN and Matched Filter Receiver

The received signal: r(t) = s(t) + n(t) = Amg(t) + n(t)

the Matched Filter output: y(t)

y(t) = 0

tr()g(T-t+)d

= 0

t Amg()g(T-t+)d + 0

tn()g(T-t+)d

Sampled at t=T

y(T) = 0

T Amg()g()d + 0

Tn()g()d

= ys(T) + yn(T)= AmEg+ (Gaussian Random Variable with = EgNo/2)

Probability of Error: Pe = Q ([Am2Eg /(No/2)]1/2)

Pe = Q([ Signal Energy/Noise Power Spectral Density]1/2)

@Y. C. Jenq 11

Probability of Error for Binary PAM Systems with Integrate & Dump Receiver

The received signal: r(t) = s(t) + n(t) = Amg(t) + n(t)

the Intergrated & Dump output: y(t)

y(t) = 0

tr()d

= 0

t Amg()d + 0

tn()d

Sampled at t=T

y(T) = 0

T Amg()d + 0

Tn()d

= ys(T) + yn(T)

= Am 0

Tg()d + (Gaussian Random Variable with = TNo/2)

Probability of Error: Pe = Q (Am 0

Tg()d /(No/2)1/2)

Pe >= Q([ Signal Energy/Noise Power Spectral Density]1/2)

@Y. C. Jenq 12

M-ray PAM Base-band Systems

M-ary PAM System : k bits per Symbol

s1(t)

t

3A

T

s4(t)

t

-3A

T

T : symbol interval (second), T = k Tb (second)

R = 1/T : symbol rate (symbol/sec), Rb = k R (bit/sec)

s2(t)

tA

T

s3(t)

t-A

T

@Y. C. Jenq 13

sm(t) = Am gT(t)m = 1, 2, 3, .. , M, 0 t T

Tt

1gT(t)

Energy of the signal

Em=

s2

m(t) dt = A2m

g2

T(t) dt = A2m Eg

M-ary PAM Signals

@Y. C. Jenq 14

• Orthogonal Functions• Ortho-normal Functions• Gram-Schmidt Orthogonalization Procedure• Basis Functions

1(t) = s1(t)/(E1)1/2, E1 = s1

2(t)dt

c21 = s2(t)1(t)dt, d2(t) = s2(t) - c21 1(t)

2(t) = d2(t)/(E2)1/2 , E2 = d2

2(t)dt

sm(t) = n=1,N sm,n n(t), sm,n= sm(t)n(t)dt,

Geometric Representation of Signals

@Y. C. Jenq 15

Geometric Representation of PAM Signals

For M-ary PAM signals sm(t) = Am gT(t), m = 1, 2, 3, .. , M, 0 t T

Let (t) = (1/Eg)1/2 gT(t) 0 t T

Then sm(t) = sm(t),

where sm = Am (Eg)1/2, m= 1, 2, …, M

(PAM signals are one-dimensional signals)

Signal Energy: Em= sm2 = Eg Am

2 , m= 1, 2, …, M

Euclidean Distance: dmn = (|sm - sn|2 )1/2 = (Eg (Am – An)2 )1/2, 0d

@Y. C. Jenq 16

Two-Dimensional Band-pass Signals:Carrier Phase Modulation

Phase Modulation signals:

um(t) = gT(t) cos(2 fct + 2 m/M), m = 1, 2, 3, .. , M,

0 t T

Let 1(t) = (2/Eg)1/2 gT(t) cos(2 fct ), 0 t T

and2(t) = -(2/Eg)1/2 gT(t) sin(2 fct ), 0 t T

Then

um(t) = Es1/2cos(2 m/M)1(t) + Es

1/2sin(2 m/M)2(t)

Or um = [Es1/2cos(2 m/M) Es

1/2sin(2 m/M)]

@Y. C. Jenq 17

Two-Dimensional Band-pass Signals:Carrier Phase Modulation

PSK (Phase-Shift Keying) signals:

um(t) = (2Es/T)1/2cos(2 fct + 2 m/M), m = 1, 2, 3, .. , M,

0 t T

Let 1(t) = (2/T)1/2 cos(2 fct ), 0 t T

and2(t) = -(2/T)1/2 sin(2 fct ), 0 t T

Then

um(t) = Es1/2cos(2 m/M)1(t) + Es

1/2sin(2 m/M)2(t)

Or um = [Es1/2cos(2 m/M) Es

1/2sin(2 m/M)]

@Y. C. Jenq 18

PSK Signal Point Constellations

Es1/2

Es1/2

Es1/2

Es1/2

M=2

M=8M=8

M=4

000

001011

010

110

111 110

100

0001

11 10

1 0

Gray Encoding

@Y. C. Jenq 19

PSK Signal Point ConstellationsMinimum Distances

Es1/2

Es1/2

Es1/2

Es1/2

M=2

M=8M=8

M=4

000

001011

010

110

111 110

100

0001

11 10

1 0

dmn = 2Es1/2 dmin = (2Es)1/2

dmin = [(2-21/2)Es]1/2

@Y. C. Jenq 20

Two-Dimensional Band-pass Signals:Quadrature Amplitude Modulation

QAM (Quadrature Amplitude Modulation) signals:

um(t) = AmcgT(t)cos(2 fct)+AmsgT(t)sin(2 fct), m = 1, 2, 3, .. , M,

0 t T

Let 1(t) = (2/Eg)1/2gT(t)cos(2 fct ), 0 t T

and2(t) = (2/Eg)1/2gT(t) sin(2 fct ), 0 t T

Then um(t) = Amc(Eg/2)1/21(t) + Ams(Eg/2)1/22(t)

Or um = [Amc(Eg/2)1/2 Ams(Eg/2)1/2]

@Y. C. Jenq 21

QAM Signal Point Constellations

[Eg(Amc2+Ams

2)/2]1/2

M=2

M=4

@Y. C. Jenq 22

QAM-PSK Signal Constellations

@Y. C. Jenq 23

For N-dimensional Waveforms

sm(t) = n=1,N sm,n n(t), sm,n= um(t)n(t)dt,

sm=( sm,1, sm,2, sm3, ……. ,sm,N )

For orthogonal signals

sm(t) = Es1/2

m(t), m = 1, 2, 3, .. , M, 0 t T

sm=( 0, 0, 0, … , Es1/2, …. ,0 )

M-th position

Multi-Dimensional Waveforms

@Y. C. Jenq 24

Frequency-Shift Keying (FSK)

FSK (Frequency-Shift Keying) signals:

um(t) = (2Es/T)1/2cos(2 fct + 2 m f t), m = 0, 2, 3, .. , M-1,

0 t T

Define correlation coefficients mn as

mn= (1/Es) T um(t)un(t)dt,

= sin[2(m-n)f T]/[2(m-n)f T] for fc >> (1/T)

Then mn = 0 for m n if f is an integer multiple of (1/2T),

and the FSK signals form an orthogonal signal set with

m(t) = (2/T)1/2cos(2 fct + 2 m f t), m = 0, 2, 3, .. , M-1,

@Y. C. Jenq 25

Optimal Receiver for M-ary Orthogonal Signals with Additive White Gaussian Noise

X

1(t)

X

2(t)

X

N(t)

T( )dt

T( )dt

T( )dt

Sampled at t = T

r1

r2

rN

ToDetector

ReceivedSignal

r(t)

@Y. C. Jenq 26


r(t) = sm(t) + n(t)

T r(t)k(t)(t)dt

= T [sm(t) + n(t)]k(t)(t)dt

=T sm(t)k(t)dt +

T n(t)k(t)dt

rk = smk + nk, k = 1, 2, 3, …,N

OR r = sm + n

Transmittedsignal sm(t) +

Receivbed signal r(t) = sm(t)+ n(t)

noise n(t)

@Y. C. Jenq 27

The Joint Conditional Probability Density Functions: f(rr|sm)

Given sm(t) is transmitted, we can express the received signal as

r(t) = k=1,N sm,k k(t) + k=1,N nk k(t) + n’(t)

= k=1,N rk k(t) + n’(t)

where n’(t) = n(t) - k=1,N nmk k(t)

It can be shown that the correlation E[n’(t) rk] = 0 for all k. Therefore n’(t) is irrelevant for the detection of sm(t)

It can also be shown that E[nk] = 0 for all k, and E[nk nm] = (No/2)km

Hence { nk } are zero-mean independent Gaussian random variables with a common variance = n

2 = No/2

@Y. C. Jenq 28

The Joint Conditional Probability Density Functions: f(rr|sm)

Since r = sm + n, i.e., rk = smk + nk k= 1, 2, …,N

Hence {rk, k= 1, 2, …,N} are independent Gaussian random

variables with E[rk] = smk, and r2

= n2 = No/2

Therefore

f(r r | sm) = 1/(No)N/2 EXP[- k=1,N (rk -sm,k)2/No ] m = 1, 2, …, M

Orf(r r | sm) = 1/(No)N/2 EXP[- (|r -sm,|)2/No ] m = 1, 2, …, M

@Y. C. Jenq 29

The Optimal (MAP) Detector

• MAP (Maximum A posterior Probability) Detector

The posterior probability that the signal sm was transmitted, given that the signal received is r:

P(sm |r) = Pr{ signal sm was transmitted | r }

= f(r r | sm)P(sm) / f(r), m= 1, 2, 3,…. , M

where P(sm) is the priori probability

and f(r) = m=1,M f(r r | sm)P(sm) (total probability theorem)

@Y. C. Jenq 30

The Optimal (ML) Detector

• ML (Maximum Likelihood) Detector

The conditional PDF, f(r r | sm), or any monotonic function of it is called the likelihood function (i.e., the likelihood that r is received if sm was transmitted).

If the priori probability is equally likely, i.e., P(sm) = 1/M , i.e., signals {sm, m = 1, 2, , …, M} are equi-probable, then maximizing the likelihood function is equivalent to maximizing the posterior probability. That is, the ML detector is the same as the MAP detector

@Y. C. Jenq 31

Optimal Detectors Minimize Probability of Error

Let Rm be the region in the N-dimensional space where we decide that signal, sm, was transmitted when the vectorr = (r1, r2, r3, …,rN) is received, then the probability of error

P(e) = m=1,M P(sm)P(e|sm)

= m=1,M P(sm)[1- Rm f(r|sm)dr ]

= 1 - m=1,M Rm f(r|sm)P(sm) dr

= 1 - m=1,M RmP(sm |r)f(r) dr (1)

For equi-probable transmitted signal set

P(e) = 1 – (1/M)m=1,M Rm f(r|sm)dr (2)

MAP detector minimizes (1) & ML detector minimizes (2)

@Y. C. Jenq 32

The Optimal (MD) Detector with Additive White Gaussian Noise

For the AWGN channel, let us use ln [f(r r | sm)]as the likelihood function, then

ln [f(r r | sm)] = -(N/2)ln(No) – (1/No)k=1,N (rk -sm,k)2

The maximum of ln [f(r r | sm)] over sm is equivalent to finding the signal sm that minimizes the Euclidean distance:

D(r, sm) = k=1,N (rk -sm,k)2

Therefore, for an AWGN channel and equi-probable transmitted signal set, {sm}, MAP detector = ML detector = MD (Minimum Distance) detector.

Example: Binary PAM System, s1=-s2= Eb, P(s1)=p.

@Y. C. Jenq 33


X

1(t)

X

2(t)

X

N(t)

T( )dt

T( )dt

T( )dt

Sampled at t = T

r1

r2

rN

ToDetector

ReceivedSignal

r(t)

@Y. C. Jenq 34

Probability of Error forM-ary Orthogonal Signals with AWGN

r = (Es+n1, n2 , n3 , n4 , ….. nM)

P(correct) = -

P(n2<r1, n3<r1 , n4<r1 , ….. n4<r1 |r1 )f(r1)dr1

= - [1-Q( 2r1

2 /No)](M-1)f(r1)dr1

ThereforePM = 1/(2) -

{1-[1-Q(x)](M-1)} e-(x-2Es/No)2 dx

@Y. C. Jenq 35

There are k bits in a symbol, and there are (2k-1) possible ways the symbol could be in error. If PM is the symbol error rate, then each possible way has a probability of PM/(2k-1) for occurring. Therefore, the average number of bits in error given that the symbol is in error is

n=1,k n(kn) PM/(2k-1) = k 2k-1 PM

/(2k-1)

Let Pb be the bit error rate

Then k Pb = k 2k-1 PM /(2k-1)

and Pb = 2k-1/(2k-1) PM PM/2 for large k

For Gray Coding Pb = PM/k

Symbol Error Rate &Bit Error Rate

@Y. C. Jenq 36

Let Pb be the bit error rate

Then Pb = P{e|symbol in error} PM

= 2k-1/(2k-1) PM

PM/2 for large k

For Gray Coding Pb = PM/k

Symbol Error Rate &Bit Error Rate

@Y. C. Jenq 37

Bi-orthogonal & Simplex Signals

Bi-orthogonal Signals

Simplex Signals – not Orthogonal but smaller energy

@Y. C. Jenq 38

Probability of Error for Binary PAM Signals with AWGNBinary PAM systems sm(t) = Am gT(t), m = 1, 2 0 t T

Let (t) = (1/Eg)gT(t) 0 t T

Then sm(t) = sm(t),

where sm = Am(Eg), A1 = 1, A2 = -1

Signal Energy: Em= sm2 = Eg

, m= 1, 2

Euclidean Distance: d12 =(|s1 – s2|2 ) = 2 Eg,

0- Eg Eg

d12 s1s2

@Y. C. Jenq 39

r = sm+ n = Eg + n, m = 1, 2

Where n is a zero mean Gaussain R.V. with variance No/2

Then f(r|s1) = 1/(No)1/2exp[-(r-Eg)2/No]

and f(r|s2) = 1/(No)1/2exp[-(r+Eg)2/No],

P(e|s1) = -0

f(r|s1)dr = (1/(2)(2Eg/No)e-r2/2 dr = Q[(2Eg/No)]

P(e|s2) = Q[(2Eg/No)]

and PB(e) = (1/2) P(e|s1)+ (1/2) P(e|s2) = Q[(2Eg/No)] f(r|s2) f(r|s1)

Eg- Eg

Probability of Error for Binary PAM Signals with AWGN

@Y. C. Jenq 40

Probability of Error for M-ary PAM Signals with AWGNM-ary PAM systems sm(t) = Am gT(t), m = 1, 2, 3, .. , M, 0 t T

Let (t) =(1/Eg) gT(t) 0 t T

Then sm(t) = sm(t),

where sm = Am(Eg), and Am = (2m-1-M), m= 1, 2, …, M

i.e., Am are: -(M-1), -(M-3), … -3, -1, 1, 3, …, (M-3), (M-1)

Signal energy: Em= sm2 = Eg Am

2 , m= 1, 2, …, M

Average signal energy: Eav= (1/M)Em = Eg(M2-1)/3

Average power: Pav = Eav /T

@Y. C. Jenq 41

Probability of Error for M-ary PAM Signals with AWGN

0d

… -5 -3 -1 1 3 5 (Eg1/2)…

r = sm+ n = AmEg + n, m = 1, 2, 3, … Mn is a zero mean Gaussian R.V. with variance No/2

PM(e) = (M-1)/M P{ |r- sm| > Eg}

= 2(M-1)/M Q[(2Eg/No)]

= 2(M-1)/M Q{(6PavT/[(M2-1)No)]}

= 2(M-1)/M Q{(6Eav/[(M2-1)No)]}

= 2(M-1)/M Q{(6[log2(M)Ebav]/[(M2-1)No)]}

@Y. C. Jenq 42

Probability of Error for M-ary PAM Signals with AWGN

-10 -5 0 5 10 15 20 2510-14

10-12

10-10

10-8

10-6

10-4

10-2

100

SNR/bit, dB

PM

(e),

Pro

ba

bil

ity

of

Sy

mb

ol

Err

or

M=2

M=4

M=8

M=16PAM Systems

@Y. C. Jenq 43

PSK (Phase-Shift Keying) signals:

um(t) =(2Es/T)cos(2 fct + 2 m/M), m = 1, 2, 3, .. , M,

0 t T

Let 1(t) = (2/T)cos(2 fct ), 0 t T

and2(t) = -(2/T)sin(2 fct ), 0 t T

Then

um(t) = Es cos(2 m/M)1(t) + Es sin(2 m/M)2(t)

Or um = [Es cos(2 m/M) Es sin(2 m/M)]

Probability of Error for Coherent PSK Signals with AWGN

@Y. C. Jenq 44

Assuming the noise process n(t) is AWGNn(t) = nc 1(t) + ns 2(t)

r = um + n = [r1,r2] = [Es cos(2 m/M) + nc Es sin(2 m/M) + ns]where nc and ns are independent zero mean Gaussian R.V.s with variances = No/2

Phase, r, of the received vector r is tan-1(r2/r1)


@Y. C. Jenq 45

Without loss of generality, let us assume that = 0 was transmitted, i.e., m = 0, or s0 = (Es,0) then

r1 = Es + nc & r2= ns

and fr(r1,r2) = 1/(No) exp{-[(r1-Es )2 +r2 2] /No }

Let V = (r1 2 +r2

2) and = tan-1(r2/r1)

Then fv,(v,) = v/(No) exp{-[v2+Es –2vEscos] /No }

and f() = 1/(2)e-sin20 -

ve-(v–(2cos )2/2 dv

where = Es/No


@Y. C. Jenq 46

And the probability of error is

PM = 1 - -/M

/M

f() d

For M=2 P2 = Q[(2Eb/No )]

For M=4 P4 = 1-(1- P2)2 = Q[(2Eb/No )]{2- Q[(2Eb/No )]}

For M>4 PM 2Q[(2Es/No ) sin(/M)]

=2Q[(2kEb/No ) sin(/M)], k=log2(M)


@Y. C. Jenq 47

-5 0 5 10 15 20 2510-14

10-12

10-10

10-8

10-6

10-4

10-2

100

SNR/bit, dB

PM

(e),

Pro

ba

bil

ity

of

Sy

mb

ol

Err

or

M=2

M=4

M=8

M=16

PSK Systems


@Y. C. Jenq 48

A coherent QAM system is simply a 2-channel PAM system,Hence

PQAM = 1 - (1-PM)2

where PM = 2(1 – 1/M) Q{[(6Eav/(M-1)No]}

PQAM 1 – (1 – 2 Q{[(6Eav/(M-1)No]})2

4 Q{[(3kEbav/(M-1)No]}

R = (PQAM /PPSK ) [3/(M-1)]/[2sin2(/M)]

R = 1 for M=4 & R > 1 for M >4, QAM is better?

Probability of Error for Coherent QAM Signals with AWGN

@Y. C. Jenq 49

-5 0 5 10 15 20 2510

-12

10-10

10-8

10-6

10-4

10-2

100

102

SNR/bit, dB

M=4

M=16

M=64

QAM Systems

PM

(e),

Pro

ba

bil

ity

of

Sy

mb

ol

Err

or

Probability of Error for Coherent QAM Signals with AWGN

@Y. C. Jenq 50

Probability of Error with AWGN QAM vs. PSK

R = (PQAM /PPSK ) [3/(M-1)]/[2sin2(/M)]

M 10log10R

8 1.65

16 4.20

32 7.02

64 9.95

R = 1 for M=4 & R > 1 for M >4,

QAM is better than PSK?

@Y. C. Jenq 51

For N-dimensional Orthogonal Signals

sm(t) = Es1/2

m(t), m = 1, 2, 3, .. , M, 0 t T

sm=( 0, 0, 0, … ,Es, …. ,0 )

Suppose that the signal s1 is transmitted, then the received signal vector is

r=(Es+n1, n2, n3, ……. , nM )

and

• PM = 1/(2){1-[1-Q(x)]M-1} e-[x- (2Es/No)]2/2dx

Probability of Error for M-ary Orthogonal Signals with AWGN

@Y. C. Jenq 52

The same as Orthogonal signals

• PM = 1/(2) {1-[1-Q(x)]M-1} e-[x- (2Es/No)]2/2dx

Except the saving of signal to noise ratio by

10 log10[M/(M-1)] dB

Probability of Error for M-ary Simplex Signals with AWGN

@Y. C. Jenq 53

For N-dimensional Orthogonal Signals

sm(t) = Es1/2

m(t), m = 1, 2, 3, .. , M, 0 t T

sm=( 0, 0, 0, … ,Es, …. ,0 )

Suppose that the signal s1 is transmitted, then the received signal vector is

r=(Es+n1, n2, n3, ……. , nM/2 )

and

• PM = 1/[2(2)] {1-[1-Q(x)]M/2-1} e-[x- (2Es/No)]2/2dx

Probability of Error for M-aryBi-orthogonal Signals with AWGN

@Y. C. Jenq 54

Union Bounds on The Probability of Error for

Orthogonal SignalsLet Em be the event that sm is received given that s1

is transmitted, thenPM = P(m=2,M Em) m=2,M P(Em)

(M-1)P2= (M-1)Q[(Es/No)] < M Q[(Es/No)]

Noting that Q[(Es/No)] < e-Es/2No

We have PM < M e-Es/2No = e-k(Eb/No-2ln2)/2, k=log2(M)As long As Eb/No > 2ln2 = 1.39 (1.42 dB), PM 0 for large M

Shannon Limit: Eb/No > ln2 = 0.693 (-1.6 dB)

@Y. C. Jenq 55

Union Bounds on The Probability of Error for Orthogonal Signals

0 2 4 6 8 10 1210

-14

10-12

10-10

10-8

10-6

10-4

10-2

100

SNR/bit, dB

M=2

M=4

M=8

M=16

PM

(e),

Pro

bab

ilit

y o

f S

ymb

ol

Err

or

Orthogonal Signals Union Bound

@Y. C. Jenq 56

X

1(t)

X

2(t)

X

N(t)

T( )dt

T( )dt

T( )dt

Sampled at t = T

r1

r2

rN

ToDetector

ReceivedSignal

r(t)

m(t) = (2/T)1/2cos(2 fct + 2 m f t),

Coherent Detection of FSK Signals with AWGN

@Y. C. Jenq 57

Non-coherent Detection of FSK Signals with AWGN

X

X

T( )dt

T( )dt

Sampled at t = T

r1c

r1s

ToDetector

ReceivedSignal

r(t)

(2/T)1/2cos(2 fct+)

(2/T)1/2sin(2 fct+ )

X

X

T( )dt

T( )dt

r2c

r2s

(2/T)1/2cos(2 ft+2ft+)

(2/T)1/2sin(2 fct+2ft+ )

@Y. C. Jenq 58

Non-coherent Detection of FSK Signals -The Envelope Detector

X

X

T( )dt

T( )dt

Sampled at t = T

ReceivedSignal

r(t)

(2/T)1/2cos(2 fct+)

(2/T)1/2sin(2 fct+ )

X

X

T( )dt

T( )dt

(2/T)1/2cos(2 ft+2ft+)

(2/T)1/2sin(2 fct+2ft+ )

+

(r1c)2

(r1s)2

+

(r2c)2

(r2s)2

@Y. C. Jenq 59

fr1(r1c ,r1s)= (1/22) e-(r1c2+r1s

2+Es)/22 I{[Es(r1c2+r1s

2)/2]}

frm(rmc ,rms) = (1/22) e-(rmc2+rms

2)/22 , m=2, 3,.., M

Let R = (r1c2+r1s

2)/ and = tan-1(rms /rmc), Then

fR1,1(R1 ,1)= (R1/2) e-(R1

2+2Es/No)/2 I{[R1(2Es/No)]}

fRm,m(Rm ,m)= (Rm/2) e-Rm

2/2 , m = 2, 3,…, M

PM = 1 - Pc = 1- P(R2<R1, R3<R1…, RM<R1|R1=x)fR1(x)dx

= 1- [P(R2<R1|R1=x)]M-1fR1(x)dx

PM = n=1,(M-1) (-1)n+1(n M-1)[1/(n+1)]e –nk(Eb/No)/(n+1) , k = log2(M)

Probability of Error for Non-coherent Detection of FSK Signals with AWGN

@Y. C. Jenq 60

Probability of Error for Non-coherent Detection of FSK Signals with AWGN

0 2 4 6 8 10 1210

-14

10-12

10-10

10-8

10-6

10-4

10-2

100

SNR/bit, dB

M=2

M=4

M=8

M=16

PM

(e),

Pro

bab

ilit

y o

f S

ymb

ol

Err

or

FSK Systems with Envelop Detector

@Y. C. Jenq 61

Comparison of Modulation Methods

PAM, QAM, PSK, and Orthogonal Signals

Symbol Duration: TBandwidth: WBit Rate: Rb

Normalized Data Rate: Rb/WSNR per Bit: Eb/No

Bandwidth Limited: Rb/W > 1Power Limited: Rb/W < 1

@Y. C. Jenq 62


PAM: Rb/W = 2log2(MPAM)

QAM: Rb/W = log2(MQAM)

PSK: Rb/W = log2(MPSK)

Orthogonal: Rb/W = 2log2(M)/M

@Y. C. Jenq 63


Digital Repeater: K repeater

Pb K Q(2Eb/No)

Analog Repeater

Pb Q(2Eb/KNo)

@Y. C. Jenq 64

Digital PAM Transmission through Band-limited Base-band Channels

TransmittingFilterGT(f)

ChannelC(f) +

ReceivingFilterGR(f)

InputData

Noise n(t)

r(t)

Symbol TimingEstimator

SamplerDetector

v(t) y(t)

y(mT)OutputData

H(f)= GT(f)C(f)

GR(f)=H*(f)e-j2fT

(S/N)o = 2Eh/No

@Y. C. Jenq 65


Example 8.1.1

gT(t) = (1/2)[1 + cos(2/T)(t-T/2)], 0<t<T

GT(f) = [sin(fT)/fT]/(1-f2T2) e-jfT

H(f) = C(f) GT(f) = GT(f), |f|<W= 0, otherwise

C(f)

-w wf

gT(t)

Tt

1

@Y. C. Jenq 66


v(t) = n an gT(t-nT), T is the symbol interval

r(t) = n an h(t-nT)+n(t), h=c*gT & n is AWGN

y(t) = n an x(t-nT)+u(t), x=gT*c*gR & u=n*gR

The receiver samples the received signal, y(t),Periodically, every T seconds, and the output isy(mT) = n an x(mT-nT) + u(mT), or in short,ym = n an xm-n + um

ym = xoam + nm an xm-n + um

@Y. C. Jenq 67



ChannelC(f) +


InputData

r(t)

Symbol TimingEstimator

SamplerDetector

v(t) y(t)

y(mT)OutputData


With a matched filterxo= h2(t)dt= |H(f)|2df= |C(f)|2|GT(f)|2df=Eh

n(t)

@Y. C. Jenq 68

PAM Systems with Inter-symbol Interference (ISI)


xo=Eh

The variance of um is m2= (No/2)Eh

nm an xm-n is the ISI

t

x(t)

@Y. C. Jenq 69

Inter-Symbol Interference

t

x(t)

t


@Y. C. Jenq 70

ISI and Eye Diagram

t

Noise margin

Timing error margin

Eye opening

Maximum ISI = nm |xm-n |

Optimal sampling time

@Y. C. Jenq 71

Probability of Error in PAM Systems with ISI and Additive Noise

Yih-Chyun Jenq, Bede Liu, & John B. Thomas

IEEE Transactions on Information TheoryVol. IT-23, No. 5, pp 575-582,

September 1977.

@Y. C. Jenq 72

Power Spectral Density of PAM Systems

v(t) = n=-

angT(t-nT)

E{V(t)} = n=-

E(an)gT (t-nT) = man=-

gT (t-nT)

RV(t+,t) = E{V*(t)V(t+)}

= m=-

Ra[m] n=-

gT (t-nT)gT(t+-nT-mT)

V(t) is cyclostationary with period T !

RV() = (1/T) -T/2

T/2

RV(t+,t)dt

SV(f) = (1/T) {m=-

Ra[m]e-jm2fT}|G(f)|2 = (1/T)Sa(f)|G(f)|2

@Y. C. Jenq 73


Example 1: {an} is an uncorrelated sequence

Ra[m] = a2+ma

2, m=0 & Ra[m] = ma2, m0

Sa(f) = a2+ma

2 m=-

e-jm2fT = a2+(ma

2)/T m=- (f-m/T)

SV(f) = (a2 )/T |GT(f)|2+(ma

2)/T m=-

|GT(m/T)|2 (f-m/T)

Example 1.1:

GT(f) = AT sin(fT)/(fT)e-jfT

|GT(f)|2 = (AT)2 sinc2(fT)

SV(f) = (a2)A2T sinc2(fT) + A2(ma

2)(f)

A

Tt

gT(t)

@Y. C. Jenq 74


Example 2: {an= bn+bn-1} and bn = ±1 are uncorrelated

Ra[m] = , m=0, Ra[m] = 1, m= ±1 & Ra[m] = 0, otherwisw

Sa(f) = 4cos2(fT) and SV(f) = (4/T) |GT(f)|2 cos2(fT)

-1/T -1/2T 1/2T 1/Tf

SV(f)

@Y. C. Jenq 75

Signal Design for Zero ISI

ym = x(0)am + nm an x(mT-nT) + um

We have zero ISI (with normalized x(0)=1)if and only if

x(0) = 1, and x(nT) = 0 for all n 0

and this is true if and only if

m X(f+m/T) = T

@Y. C. Jenq 76

Proof of Nyquist Theorem

x(nT) = -

X(f) ej2fnTdf

= -1/2T1/2T m X(f+m/T) ej2fnTdf

= -1/2T1/2T Z(f) ej2nTf df

Z(f) = m X(f+m/T) is a periodic function of f with period 1/T, and hence has a Fourier series expansion

Z(f) = n znej2nTf with zn=Tx(-nT)

For zero ISI, z0=T and zn=0 for all n≠0 Z(f) = T

@Y. C. Jenq 77

Examples of Zero-ISI Pulse Spectrum

-1/2T 1/2T-1/T 1/Tf

T

-1/2T 1/2T-1/T 1/Tf

T

-1/2T 1/2T-1/T 1/Tf

T

@Y. C. Jenq 78

Pulses with a Raised Cosine Spectrum

For 0 ≤ ≤ 1

Xrc(f) = T, 0 < |f| < (1-)/(2T)= (T/2)(1+cos{(T/)[|f|-(1-)/(2T)])= 0, |f| > (1+)/(2T)

x(t) = [sin(t/T)/(t/T)][cos(t/T)/(1-42t2/T2)]

-1/2T 1/2T-1/T 1/Tf

=0=1

TXrc(f)

@Y. C. Jenq 79

Raised Cosine Pulses

-4 -3 -2 -1 0 1 2 3 4-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

time, t

x(t)

, Rai

sed

Cos

ine

Pul

ses

= 0= 0.25= 0.5= 0.75= 1.00

@Y. C. Jenq 80

Controlled Inter-symbol Interference Partial Response Signals

Z(f) = n znej2nTf with zn=Tx(-nT)

A duobinary signal pulse:

Consider x(nT) = 1, n=0,1 and x(nT) = 0, otherwise

Then Z(f) = T+Te-j2Tf = 2Te-jTfcos(Tf)Choose X(f) = 2Te-jTfcos(Tf), |f| <1/2T

X(f) = 0, otherwiseand x(t) = sinc(t/T) + sinc[(t-T)/T]

@Y. C. Jenq 81

Controlled Inter-symbol Interference Partial Response Signals

Consider x(-T) = 1, x(T) = -1and x(nT) = 0, otherwise

Then Z(f) = T(ej2Tf - e-jTf)=2jTsin(Tf)

Choose X(f) = 2jTe-jTfsin(Tf), |f| <1/2T X(f) = 0, otherwise

and x(t) = sinc[(t+T)/T] - sinc[(t-T)/T]

Another possibility (no DC component)

@Y. C. Jenq 82

Symbol by Symbol Data Detection with Controlled ISI

Reveived duobinary signal (am = 1 or -1)ym = bm + vm = am + am-1 +vm

Decision feedback decoding Error Propagation

Pre-Coding: data sequence dm = 0 or 1

pm = dm – pm-1 (Mod 2) and am = 2 pm – 1bm = am + am-1 = 2(pm+pm-1 –1)

pm+pm-1 = bm/2 + 1

Decoding rule: dm= bm/2 – 1 (Mod 2)

@Y. C. Jenq 83

Pre-Coding for Duobinary Signals

dm 1 1 1 0 1 0 0 1 0 0 0 1 1 0 1

pm 0 1 0 1 1 0 0 0 1 1 1 1 0 1 1 0

am -1 1 -1 1 1 -1 -1 -1 1 1 1 1 -1 1 1 -1

bm 0 0 0 2 0 -2 -2 0 2 2 2 0 0 2 0

dm 1 1 1 0 1 0 0 1 0 0 0 1 1 0 1

@Y. C. Jenq 84

Probability of Error for Zero ISI

For ideal PAM cases, r = sm+ n = AmEg + n, m = 1, 2, … Mn is a zero mean Gaussian R.V. with variance No/2

PM(e) = 2(M-1)/M Q{(6[log2(M)Ebav]/[(M2-1)No)]}

For zero ISI cases, ym= x0 am+ vm,

where xo = -WW

|GT(f)|2df = Eg (|GR(f)|2= |GT(f)|2)and vm is is a zero mean Gaussian R.V. with the variance

v2 = Eg No/2

PM(e) = 2(M-1)/M Q{(6[log2(M)Ebav]/[(M2-1)No)]}

@Y. C. Jenq 85

Probability of Error for Partial Response Signals (PRS)

For partial response signals (consider binary cases), ym= am - am-1 + vm,

and vm is a zero mean Gaussian R.V. with the variance

v2 = No/2 -W

W |X(f)|df =2No/

where|X(f)| = |GR(f)|2 =|G*T(f)|2

Therefore the price for saving bandwidth is the decrease of the S/N by 10Log10(4/) = 2.1 dB!

@Y. C. Jenq 86

Digitally Modulated Signals with Memory

1 0 1 1 0 0 0 1 1 0 1

NRZ

NRZI

A

-A

@Y. C. Jenq 87

State Diagrams & Trellisof NRZI Signals

S1= 0 S2= 1

0/0 1/1

1/0

0/1

S1= 0 S2= 1

0/-A 1/A

1/-A

0/-A

@Y. C. Jenq 88

State Diagrams & Trellisof NRZI Signals

S1= 0

S2 =1

0/0 0/0 0/0

0/1 0/10/1

1/1

1/01/1 1/11/0 1/0

S1= 0

S2 =1

0/0 0/0 0/0

0/10/1

1/1 1/1 1/11/0 1/0

@Y. C. Jenq 89

Maximum Likelihood Sequence Detector – Viterbi Algorithm

Let r1, r2, r3, r4, …are the received signals, and sm1, sm2, sm3, sm4, …are the transmitted signals, and rk = smk + nk (for the m-th sequence)

For ML symbol by symbol detector Maximize f(rk|smk) for each individual k

For ML sequential detector Maximize f(r1, r2, r3, r4 …| sm1, sm2, sm3, sm4, …)

@Y. C. Jenq 90

Maximum Likelihood Sequence Detector – Viterbi Algorithm

S1= 0

S2 =1

0/0 0/0 0/0

0/10/1

1/1 1/1 1/11/0 1/0

t=T t=2T t=3T

Minimize the Euclidean Distance: k(rk-smk)2

@Y. C. Jenq 91

Probability of Error for PRS with ML Sequence Detector

t=T t=2T t=3T

S1= 1

S2 = -1

1/2 1/2 1/2

-1/-2

-1/0

1/0 1/0 1/0-1/0 -1/0

-1/-2 -1/-2

t=0

P(e) = Q{[(1.52/16)(2Eb/No)]1/2} 10log10(1.52/16) = -0.34 dB

@Y. C. Jenq 92

The Power Spectrum of Digital Signals with Memory

M symbols s1, s2, s3, ...., sM, M waveforms s1 (t), s2 (t), s3 (t), ...., sM (t)A Markov chian with M statesMxM state transition matrix P=[pij]Steady State Probabilities {p1, p2, s3, ...., pM }

v(t) = n=-

sIn(t-nT) (In = k with Probability pk )

RV(t+,t) = E{V*(t)V(t+)}

= m=-n=-

sIn

(t-nT)sIn+(t+-nT-mT)

V(t) is cyclostationary with period T !

@Y. C. Jenq 93

The Power Spectrum of Digital Signals with Memory

RV() = (1/T) -T/2

T/2

RV(t+,t)dt

= (1/T) m=-

i=1,Kj=1,K Rij(mT)pij[m]pi

where Rij(mT)= -

si(t)sj(t+)dt and pij[m]=Pm[i.j]

SV(f) = (1/T) i=1,Kj=1,K Sij(f) Pij(f)pi

where Sij(f)= -

Rij()e-j2fd

and Pij[f]= m=- pij[m] e-j2fmT

@Y. C. Jenq 94

System Design in the Presence of Channel Distortion


ChannelC(f) +


InputData

Noise n(t)

r(t)

v(t) y(t)

GT(f)C(f) GR(f)=Xrc(f)e-j2ft0

Output noise power spectral density Sv(f) = Sn(f)|GR(f)|2

For Zero ISI, ym= x0 am+ vm,

Assuming am= ± d and vm is zero mean Gaussian with variance v

2= Sn(f)|GR(f)|2df

@Y. C. Jenq 95

System Design in the Presence of Channel Distortion


ChannelC(f) +


InputData

Noise n(t)

r(t)

v(t) y(t)

Xrc(f) = GT(f) C(f) GR(f)

|GR(f) = K |Xrc(f)|1/2

|GT(f) C(f)| = (1/ K) |Xrc(f)|1/2

@Y. C. Jenq 96

Channel Equalizations


ChannelC(f) +


InputData

Noise n(t)

r(t)

v(t) y(t)

GT(f) C(f) GR(f)=Xrc(f)e-j2ft0

EqualizerGE(f)

GE(f)= 1/C(f)

v2

(f)= (No / 2) -W

W(|Xrc(f)|/ |C(f)|2)df

@Y. C. Jenq 97

Channel Equalizations

• Zero forcing Equalizer

• Mean Square Equalizer

• Minimum Probability Equalizer •(Jenq, Thomas & Liu: IEEE 1977)

• Decision Feedback Equalizer

• Automatic (Adaptive) Equalizer

@Y. C. Jenq 98

Linear Transversal Filter

X X XXX

Algorithm for Tap Gain Adjustment

c0 c1 c2c-1c-2

Input = y(t) = x(t)+n(t)

h(t)

@Y. C. Jenq 99

Linear Transversal Filter

z(t)= k= -

Bk h(t-k)

h(t)= j=-N

N cj y(t-j) = j=-N

N cj [x(t-j)+n(t-j)]

h(mT)= j=-N

N cj y(mT-j) or hm= j=-N

N cjym-j

ZFE – Zero Forcing Equalizer (hj= 0, -N j N, and h0= 1)

@Y. C. Jenq 100

Zero Forcing Equalizer

h(t)= j=-N

N cj x(t-j)

h(mT)= j=-N

N cj x(mT-j)

or hm= j=-N

N cjxm-j

hm = 1, m=00, m= -N, -(N-1), …-2, -1, 1, 2, …(N-1), N

@Y. C. Jenq 101

Zero Forcing Equalizer

Example: x(t)= [1+(2t/T)2]

@Y. C. Jenq 102

Mean Square Equalizer

h(t)= j=-N

N cj y(t-j) = j=-N

N cj [j=-

Bm x(t-j)+n(t-j)]

h(mT)= j=-N

N cj y(mT-j) or hm= j=-N

N cjym-j

E{hm –Bm}2 = j=-N

N k=-N

N cj ck RY(j-k)- 2 k=-N

N ck RBY(k) + E{Bm}2

where RY(j-k)=E{y(mT-j) y(mT-k) }and RBY(k)=E{y(mT-k) Bm}

j=-N

N cj RY(j-k) = RBY(k), -N k N

@Y. C. Jenq 103

Adaptive Equalizers

X X XXX

c0 c1 c2c-1c-2

input

MAC MAC MAC MACMAC

Detector++

_X

{ek}

yk

zk

ak

@Y. C. Jenq 104

Adaptive Equalizers

Error Function: e(cj, j=-N, …,-1,0,1,…,N)Gradient Vector: g = de/dc

Iterative Method:ck+1 = ck – gk =step size

MSE algorithm: Error function is Mean Square Error gk = -ek yk

ck+1 = ck + ek yk

@Y. C. Jenq 105

Symbol Synchronization

Receiver : To know When to Sample

1. Master clock distributed

2. Clock comes with data signals

3. Clock recovered from Data signal

1. Spectral line method2. Early-Late gate synchronizer

@Y. C. Jenq 106

Multi-Carrier Modulation & OFDM

xk(t) = Aksin (2fkt)k= 0,1,2,3, … ,(K-1)

If fk - fm = n/T, where n is an integer

Then sin (2fkt+k) sin (2fmt+m)dt = 0

Hence “orthogonal” among all K carriers

Multi-carrier signal

x(t) = k=0,(K-1) Aksin (2fkt)

@Y. C. Jenq 107

Multi-Carrier QAM Signals

xk(t) = Rkcos [k(2t] - Iksin [k(2t]

= Re{(Rk+jIk) exp[jk(2/T)t]}

= Re{Xkexp[jk(2/T)t]}

x(t) = k=0,(K-1) Re{Xkexp[jk(2/T)t]}

= Re{k=0,(K-1) Xkexp[jk(2/T)t] }

= (1/2){k=0,(K-1)Xkexp[jk(2/T)t] +

k=0,(K-1) Xk*exp[-jk(2/T)t] }

@Y. C. Jenq 108


2x(t) = k=0

(K-1)Xkexp[jk(2/T)t] +

k=0

(K-1) Xk

*exp[-jk(2/T)t]

For N=2K, and sampling 2x(t) at t = n(T/N), n = 0,1,2,.., (N-1)

We have

xn = k=0

(K-1)Xkexp[jkn(2/N)] +k=0

(K-1)Xk

*exp[-jkn(2/N)]

= k=0

(K-1)Xkexp[jkn(2/N)] +k=0

(K-1)Xk

*exp[j(N-k)n(2/N)]

@Y. C. Jenq 109


xn = k=0

(K-1)Xkexp[jkn(2/N)]

+k=0

(K-1)Xk

*exp[j(N-k)n(2/N)]

Now,

Let Yk = Xk & YN-k= Xk* for k = 1, 2,…, (K-1),

and let Y0 = 2Re(X0) & YN/2 = 0

We have xn = k=0

(N-1) Ykexp[jkn(2/N)]

That is, {xn}n=0, (N-1) is the Inverse DFT of {Yn}n=0, (N-1)

@Y. C. Jenq 110

Implementation of Multi-Carrier QAM Signals

• Given K QAM symbols X0, X1, X2, …, X(K-1)

• For N= 2K,

let Yk = Xk & YN-k= Xk* for k = 1, 2,…, (K-1),

and let Y0 = 2Re(X0) & YN/2 = 0

• Perform Inverse DFT on {Yn}n=0, (N-1) • to obtain {xn}n=0, (N-1)

• Feed {xn}n=0, (N-1) to a D/A converter

at the rate of (N/T) samples per second

@Y. C. Jenq 111

Another Formulation

Consider a complex multi-carrier OFDM signal

x(t) = k=0

(N-1)Xkexp[jk(2/T)t]

Taking samples at n(T/N) for n = 0, 1, 2, …, (N-1)We have

xn = k=0

(N-1)Xkexp[jkn(2/N)]

Necessary conditions for {xn}n=0, (N-1) to be real

are Xk = X*(N-k) for k = 1, 2,…, (K-1), and

X0 and XN/2 are real

@Y. C. Jenq 112

Another Implementation

• Given K QAM symbols X0, X1, X2, …, X(K-1)

• For N= 2K,

let Yk = Xk & YN-k= Xk* for k = 1, 2,…, (K-1), and

let Y0 = Re(X0) & YN/2 = Im(X0)

• Perform Inverse DFT on {Yn}n=0, (N-1) to obtain {xn}n=0, (N-1)

• Feed {xn}n=0, (N-1) to a D/A converter

at the rate of (N/T) samples per second

@Y. C. Jenq 113

Frequency Spectrum Interpretation

2/T

(N-1)2/T

Sampling FrequencyN2/T

Nyquist band

@Y. C. Jenq 114

TDMA, FDMA and CDMA

TDMA - Time Division Multiple AccessFDMA - Frequency Division Multiple AccessCDMA - Code Division Multiple Access

Multiple Access:Many users utilize a physical channelsimultaneously

@Y. C. Jenq 115

Time Division Multiple Access

1 2 3 4 N

Framing headerFrametime

Control/Access channel & Data channels

freq

uenc

y

1 2 3 4 N

@Y. C. Jenq 116

Frequency Division Multiple Access

1

time


freq

uenc

y

2

3

4

N

guardbands

@Y. C. Jenq 117

Code Division Multiple Access


1, 2, 3, 4, ….. N

time

freq

uenc

y

@Y. C. Jenq 118

DS-CDMA Systems

Direct Sequence, Code Division Multiple Access Systems

v(t) = n an gT(t-nTb)

c(t) = n cn p(t-nTc)

u(t) = Ac v(t) c(t) cos(2fct), c2(t) = 1

c(t) : pseudo-random noise (PN) sequence waveform

Tb : Bit interval Tc : Chip interval

@Y. C. Jenq 119

DS-CDMA Systems

t

t

t

v(t)

v(t)c(t)

c(t)

@Y. C. Jenq 120

Spectra of DS-CDMA Systems

f

f

f

V(f)

V(f)*C(f)

C(f)1/Tb

1/Tc

@Y. C. Jenq 121

Demodulation of DS-CDMA

X X

PN signalGenerator

r(t) = Ac v(t) c(t) cos(2fct)

gT(t)cos(2fct) c(t)

0

T()dt

@Y. C. Jenq 122

Narrowband Interference

X X

PN signalGenerator

r(t) = Ac v(t) c(t) cos(2fct) + AJ cos(2fjt)

gT(t)cos(2fct) c(t)

0

T()dt

Processing gain = Tb / Tc

@Y. C. Jenq 123

Wideband Interference

X X

PN signalGenerator

r(t) = Ac v(t) c(t) cos(2fct+) + nc(t)cos(2fct) - ns (t)sin(2fct)

gT(t)cos(2fct) c(t)

0

T()dt

Processing gain = Tb / Tc

@Y. C. Jenq 124

M-Sequence

1 2 3 4 . . . . . . m

+L=2m-1

R[m] = L, m=0= -1, otherwise

@Y. C. Jenq 125

M-Sequence

1 2 3 4

+

Gold Sequence & Kasami Sequence

@Y. C. Jenq 126

Walsh Coding

0 11 1

1 00 0

0 11 1

0 11 1

0 11 1

1 00 0

0 11 1

0 11 1

0 11 1

1 00 0

0 11 1

0 11 1

1 00 0

0 11 1

0 00 0

0 00 0

@Y. C. Jenq 127

Digital Cellular Communication Systems

The GSM System CDMA System – IS-95

Date post:	19-Dec-2015
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@Y. C. Jenq1 Digital Communication Theory and Systems Y. C. Jenq Department of Electrical & Computer...

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