Chapter 8course.sdu.edu.cn/Download2/20150625110127415.pdf · 2015. 6. 25. · 8.2 The Power Spectrum of Digitally Modulated Signals • Example 8.2.2 Consider a binary sequence {b

Copyrights Zhu, Weihong

School of Information Science and Engineering, Shandong University

Principles of the Communications

Chapter 8

Digital Transmission through Bandlimited AWGN Channels




8.1 Digital Transmission through Bandlimited Channel

2( ) ( ) j ftC f c t e dtπ∞ −

−∞= ∫

( ) 0, cC f f B= >

C(f) GR (f)GT (f)an h(t)

n(t)

y(t)

For baseband channel with bandlimited to Bc Hz,

So the channel bandwidth is W=Bc

Soppose that the input to the channel is a signal waveform gT (t). Then

( ) ( ) ( ) ( )( ) ( ) ( )

( ) T T

T

h t c g t d c t g t

or H f C f G f

τ τ τ∞

−∞= − = ∗

=∫





• Suppose the received filter is a matched filter. Then

( ) ( ) 02* j ftRG f H f e π−=

The signal component at the output of the matched filter at the sampling instant t=t0 is ( ) ( ) ( )

( ) ( )

( ) ( )

0 0

0

0

0 0

20

2 2

2 2

2

( )

( )

W j fts s t t R t tW

W j ft j ftt tW

W j ft j f t

W

W

hW

y t y t H f G f e df

H f H f e e df

H f H f e e df

H f df E

π

π π

π π

= =−

−∗=−

−∗

−

−

= =

=

=

= =

∫

∫∫∫

The noise component at the output of the matched filter has a zero mean and a power-spectral density

( ) 20 ( )2n

NS f h f=





Hence , the noise power at the output of the matched filter has a variance

( ) ( )2

2 0 0

2 2W W h

n nW W

N N ES f df H f dfσ− −

= = =∫ ∫

The SNR at the output of the matched filter at time instant t0 ( or the maximum SNR) is

2

0 0

2/ 2

h h

o h

E ESN N E N

⎛ ⎞ = =⎜ ⎟⎝ ⎠

The difference between this development with Chapter 7 is that the filter impulse response is matched to the received signal h(t) instead of the transmitted signal.





• Example 8.1.1The signal pulse gT (t) is depicted as the following equation. It is transmitted

through a baseband channel with frequency-response characteristic as shown in the figure. The channel output is corrupted by AWGN with power- spectral density N0 /2. Determine the matched filter to the received signal and the output SNR

( ) 1 21 cos2 2T

Tg t tTπ⎡ ⎤⎛ ⎞= + −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

0-W W

1C(f)

f

0 T

1 gT (t)

Solution ( ) 2 2

2 2

sin2 (1 )

sin2 (1 )

j ftT

j ft

T ftG f eft f T

T c ft ef T

π

π

ππ

π

−

−

=−

=−

( ) ( ) ( )( )

0

T

T

H f C f G f

G f f Wotherwise

=

⎧ <= ⎨⎩





( )2W

h TWE G f df

−= ∫

2 0

2h

nN Eσ =

Then, the signal compont at the output of the filter matched to H(f) is

The variance of the noise component is

The output SNR is0 0

2 hESN N

⎛ ⎞ =⎜ ⎟⎝ ⎠

- 4 - 3 - 2 - 1 0 1 2 3 40

0 . 1

0 . 2

0 . 3

0 . 4

0 . 5

0 . 6

0 . 7

0 . 8

0 . 9

1

( ) 2TG f





• 8.1.1 Digital PAM transmission through Bandlimited Baseband Channel

C(f) GR (f)GT (f)an h(t)

n(t)

y(t)sampler Detector

outputv(t) r(t) y(mT)

0,

m m n m n omn m n

y x a a x n∞

−≠ =−∞

= + +∑

Desired signal ISI Additive noise





• 8.1.1 Digital transmission through Bandlimited Bandpass ChannelFor PAM signal

( ) ( ) cos 2 cu t v t f tπ=

For QAM and PSK signal( ) ( ) cos 2 ( )sin 2

( ) ( )

( ) ( )

c c s c

c nc Tn

s ns Tn

u t v t f t v t f t

where v t a g t nT

v t a g t nT

π π∞

=−∞

∞

=−∞

= +

= −

= −

∑

∑Now we construct a complex signal to represent v(t) as

( ) ( ) ( ) ( ) ( ) ( )c s nc ns T n Tn n

v t v t jv t a ja g t nT a g t nT∞ ∞

=−∞ =−∞

= − = − − = −∑ ∑Then the corresponding bandpass QAM or PSK signal u(t) may be represented as

2( ) Re ( ) cj f tu t v t e π⎡ ⎤= ⎣ ⎦





• u(t) is called the equivalent lowpass signal which may have complex value.• When transmitted through the bandpass channel, the received bandpass

signal may be represented as2( ) Re ( ) cj f tw t r t e π⎡ ⎤= ⎣ ⎦

( ) ( ) ( )nn

r t a h t nT n t∞

=−∞

= − +∑

( ) ( ) ( )n on

y t a x t nT n t∞

=−∞

= − +∑




8.2 The Power Spectrum of Digitally Modulated Signals

• 8.2.1 The Power Spectrum of the Baseband Signal

( ) ( )n Tn

v t a g t nT∞

=−∞

= −∑Where v(t) can represented a digital PAM. PSK or QAM signal. {an } is the sequence of values corresponding to the information symbols from the source, and gT (t) is the impulse response of the transmitting filter.The power spectrum of V(t) can be determined as follows:1 the mean value of V(t) is

[ ] ( )( ) ( )

( )

n Tn

a Tn

E V t E a g t nT

m g t nT

∞

=−∞

∞

=−∞

= −

= −

∑

∑The mean value of V(t) is periodic with period T





• The autocorrelation function of V(t) is

( )( , ) ( ) ( ) ( ) ( )V n m T Tn m

R t t E V t V t E a a g t nT g t mTτ τ τ∞ ∞

∗ ∗

=−∞ =−∞

⎡ ⎤+ = + = − + −⎣ ⎦ ∑ ∑

If the information sequence {an} is wide-sense stationary with autocorrelation sequence

( ) ( )a n n mR n E a a∗ +=Then

( , ) ( ) ( ) ( )V a T Tn m

R t t R m n g t nT g t mTτ τ∞ ∞

=−∞ =−∞

+ = − − + −∑ ∑Let p=m-n

( )( , ) ( ) ( )V a T Tp n

R t t R p g t nT g t nT pTτ∞ ∞

=−∞ =−∞

+ = − − −∑ ∑

is a periodic function( , )VR t tτ+





• Thus the random process V(t) is cyclostationary• The average autocorrelation function of V(t) is

( )/ 2

/ 2

1 ( , )T

V VT

R R t t dtT

τ τ−

= +∫

( )/ 2

/ 2

1 ( ) ( )T

a T TTm n

R m g t nT g t nT mT dtT

τ∞ ∞

−=−∞ =−∞

= − + − −∑ ∑ ∫/ 2

/ 2

1( ) ( ) ( )nT T

a T TnT Tm n

R m g t g t mT dtT

τ∞ ∞ +

−=−∞ =−∞

= + −∑ ∑ ∫

( )1 ( ) ( )a T Tm

R m g t g t mT dtT

τ∞∞

=−∞ −∞

= + −∑ ∫

( )1 ( )a gm

R m R mTT

τ∞

=−∞

= −∑ where ( ) ( )( )g T TR g t g t mT dtτ τ∞

−∞= + −∫





• Thus the power spectrum of process V(t) is

( ) ( )

( ) ( )

( ) ( )

( ) ( )

2

2

2 2

2

1

1

1

j fV V

j fa g

m

j fmTT a

m

a T

S f R e d

R m R mT e dT

G f R m eT

S f G fT

π τ

π τ

π

τ τ

τ τ

∞ −

−∞

∞ ∞ −

−∞=−∞

∞−

=−∞

=

= −

=

=

∫

∑ ∫

∑

Where ( ) ( ) 2j fmT

a am

S f R m e π∞

−

=−∞

= ∑





If the information symbols in the sequence {an } are mutually uncorrelated. Then

2j fmT

m

e π∞

−

=−∞∑

( ) ( ) ( ) ( )( ) ( )

2 2

2

, 00,

n n n a a

a n m n

n m n a

E a a E a m mR m E a a

mE a E a m

σ∗

∗+ ∗

+

⎧ = = + =⎪= = ⎨ ≠=⎪⎩

( )2

2 2 2 2j fmT aa a a a

m m

m mS f m e fT T

πσ σ δ∞ ∞

−

=−∞ =−∞

⎛ ⎞= + = + −⎜ ⎟⎝ ⎠

∑ ∑

where is the Fourier series expression of function

1m

mfT T

δ∞

=−∞

⎛ ⎞−⎜ ⎟⎝ ⎠

∑





( ) ( )22 2

2

2a a

Vm

m m mS f G f G fT T T Tσ δ

∞

=−∞

⎛ ⎞ ⎛ ⎞= + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑Example 8.2.1Determine the power-spectral density when gT (t) is the rectangular pulse shown in following figureSolution

t

A

0 T

gT (t)

( ) sin j fTT

fTG f AT efT

πππ

−=

( ) ( )2

2 2 2 2sin( ) ( ) sinTfTG f AT AT c fT

fTπ

π⎛ ⎞

= =⎜ ⎟⎝ ⎠

( ) ( ) ( )2 2 2 2 2sinV a aS f A T c fT A m fσ δ= +





- 2 - 1 . 5 - 1 - 0 . 5 0 0 . 5 1 1 . 5 20

0 . 1

0 . 2

0 . 3

0 . 4

0 . 5

0 . 6

0 . 7

0 . 8

0 . 9

1

SV (f)





• Example 8.2.2Consider a binary sequence {bn }, from which we form the symbols

1n n na b b −= +The {bn } are assumed to be uncorrelated binary value (±1) random variables, each having a zero mean and a unit variance. Determine the power-spectral density of the transmitted signal.Solution The autocorrelation function of the sequence {an } is

[ ] ( )( )[ ]

1 1

1 1 1 1

( )

2, 01, 10,

a n n m n n n m n m

n n m n n m n n m n n m

R n E a a E b b b b

E b b b b b b b b

mm

otherwise

+ − + + −

+ + − − + − + −

⎡ ⎤= = + +⎣ ⎦= + + +

=⎧⎪= = ±⎨⎪⎩



Principles of the Communications- 2 - 1 . 5 - 1 - 0 . 5 0 0 . 5 1 1 . 5 2

0

0 . 5

1

1 . 5

2

2 . 5

3

3 . 5

4


Hence, the power-spectral density of the input sequence is

( ) ( ) 2

2 2 22 2(1 cos2 ) 4cos

j mTa a

m

j fT j fT

S f R m e

e e fT fT

π

π π π π

∞−

=−∞

−

=

= + + = + =

∑

The power spectrum of the signal is

( ) ( ) ( ) ( ) ( )2 2 21 4 cosV a T TS f S f G f G f fTT T

π= =

Sa (f)





• 8.2.2 The Power Spectrum of a Carrier-Modulated Signal A bandpass signal can be expressed as

( ) ( )cos2 cu t v t f tπ=

The autocorrelation function of u(t) is

[ ]( ) ( ) ( )

( )

[ ]

( , ) ( ) ( )

cos2 cos2

( , )cos2 cos21 ( , ) cos2 cos2 (2 )2

U

c c

V c c

V c c

R t t E U t U t

E V t V t f t f t

R t t f t f t

R t t f f t

τ τ

τ π π τ

τ π π τ

τ π τ π τ

+ = +

= + +⎡ ⎤⎣ ⎦= + +

= + + +

( )1( , ) cos22U V cR t t R fτ τ π τ+ =





• The power spectrum of the bandpass signal is

( ) ( ) ( )14U V c V cS f S f f S f f⎡ ⎤= − + +⎣ ⎦

This expression can be applied to PAM, PSK and QAM signals.Homework: problem 8.5




8.3 Signal Design For Bandlimited Channels

• Assumptions:02

0 1,( )

0,0

j ft f W f WC eC f

f W f W

π−⎧ ≤ ≤⎧= =⎨ ⎨> >⎩⎩

Thus 0m m n m n mn m

y x a a x n−≠

= + +∑

Experimental method to view ISI and noise: eye pattern (眼图)

-0 . 5 0 0 . 5-1 . 5

-1

-0 . 5

0

0 . 5

1

1 . 5

T im e

Am

plitu

de

E y e D ia g ra m

2PAM





-0 .5 0 0 .5-5

-4

-3

-2

-1

0

1

2

3

4

5

Tim e

Am

plitu

de

E y e D ia g ra m

-0.5 0 0.5-1.5

-1

-0.5

0

0.5

1

1.5

Time

Am

plitu

de

Eye Diagram for In-Phase Signal

-0.5 0 0.5-1.5

-1

-0.5

0

0.5

1

1.5

Time

Am

plitu

de

Eye Diagram for Quadrature Signal

4PAM

4PSKPlease see figure 8.7pp491





• 8.3.1 Design of Bandlimited Signals for Zero ISI – The Nyquist Criterion (奈奎斯特准则)

(0) ( ) ( )m m nn m

y x a a x mT nT n mT≠

= + − +∑For a zero ISI, we should let x(mT-nT)=0 for m≠n and x(0) ≠0. Assume x(0)=1, then we have

1, 0( )

0, 0n

x nTn=⎧

= ⎨ ≠⎩Theorem 8.3.1 [ Nyquist ]. A necessary and sufficient condition for x(t) to satisfy 1, 0

( )0, 0

nx nT

n=⎧

= ⎨ ≠⎩is that its Fourier transform X(f) satisfy

( )m

mX f TT

∞

=−∞

+ =∑





• Discussing: 1. T<1/2W, or 1/T>2W. There is no way that we can design a system

with no ISI

W-W 1/T1/T-W 1/T+W-1/T

2. T=1/2W. There is only one X(f) that results in Z(f)=T

,( ) ( ) sin

0,T f W tX f or x t c

Totherwise⎧ ≤ ⎛ ⎞= =⎨ ⎜ ⎟

⎝ ⎠⎩

1/2T-1/2T





This means that the smallest value of T for which transmission with zero ISI is possible is T=1/2W and for this value, x(t) has to be a sinc function.

- 2 - 1 . 5 - 1 - 0 . 5 0 0 . 5 1 1 . 5 2- 0 . 4

- 0 . 2

0

0 . 2

0 . 4

0 . 6

0 . 8

1

Since function is a noncausal function and the pulse shape’s rate of convergence to zero is 1/t, which is slow.





3. T>1/2W, Z(f) consists of overlapping replications of X(f) separated by 1/T. In this case, there exist numerous choices for X(f), such that Z(f)≡ T

W-1/T+W 1/T-W 1/T

A particular pulse spectrum, for the T>1/2W case, is the raise cosine spectrum. The raised cosine frequency characteristic is given as

( ), 0 1 / 21 1 1( ) 1 cos ( ,

2 2 210,2

T f TT TX f f f

T T T

fT

απ α α αα

α

⎧⎪ ≤ ≤ −⎪

− − +⎪ ⎡ ⎤= + − ≤ ≤⎨ ⎢ ⎥⎣ ⎦⎪⎪ +

>⎪⎩





2 2

cos( /( ) sin ( / )1 4 /

at Tx t c t Tt T

πα

=−

X(t)

0 2 0 4 0 6 0 8 0 1 0 0 1 2 0 1 4 0-0 . 4

-0 . 2

0

0 . 2

0 . 4

0 . 6

0 . 8

1a = 0a = 0 . 2 5a = 0 . 5a = 0 . 7 5





0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9-140

-120

-100

-80

-60

-40

-20

0

20

40

Normalized Frequency (×π rad/sample)

Mag

nitu

de (d

B)

Magnitude Response (dB)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9-160

-140

-120

-100

-80

-60

-40

-20

0

20

40


Mag

nitu

de (d

B)


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9-120

-100

-80

-60

-40

-20

0

20

40


Mag

nitu

de (d

B)


a=0.75

a=0.25

a=0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9-140

-120

-100

-80

-60

-40

-20

0

20

40


Mag

nitu

de (d

B)


a=1





• How to judge if a system could cause ISI or how to judge if a system could not cause ISI

Here, we should emphasis that the symbol interval is T, or the symbol rate is 1/T.Question: if the symbol interval is 2T, how to select the addition interval?Example: 下图所示系统可否以速率2/T 波特进行数据传输（无ISI）

1/2T-1/2T 3/2T-3/2T 5/2T-5/2T

3/2T-3/2T

f

f





• 8.3.2 Design of Bandlimited Signals with Controlled ISI-Partial Response Signals (部分响应信号）

In 1963, Adam Lender showed that it is possible to transmit 2W symbols/s with zero ISI, using the theoretical minimum bandwidth of WHz, without infinitely sharp filters. Lender used a technique called duobinary signaling (双二进制信号)， also referred to as correlative coding (相关编码)and partial response signaling. The basic idea behind the duobinary technique is to introduce some controlled amount of ISI into the data stream rather than trying to eliminate it completely. By introducing correlated interference betwween the pulses, and by changing the detection procedure, Lender, in effect, “cancelled out” the interference at the detector and thereby achieved the ideal symbol-rate packing of 2 symbols/s/Hz, an amount that had been considered unrealizable.





1. Duobinary SignalingTo understand how duobinary signaling introduces controlled ISI, let us

look at a model of the process. We can think of the duobinary coding operation as if it were implemented as shown in the following figure.

The symbol rate of Xk is 1/T. Therefore

1k k ky x x −= +

Delay T seconds

1/2T-1/2T

T

f

Ideal rectangular filter

Noise

Decoder {xk }

ˆ{ }kx

ykˆ ky





2. Duobinary decodingIf the binary digit xk is equal to ±1, then yk has one of three possible

values: +2,0 or -2. the duobinary code results in a three-level output: in general, for M-ary transmission, partial response signaling results in 2M-1 output levels. The decoding procedure involves the inverse of the coding procedure, namely, subtracting the xk-1 decision from the yk digit.

Example Duobinary Coding and Decoding sequence {xk }= 0010110. (the first bit of the sequence to be a startup digit)





Binary digit sequence {xk }: 0 0 1 0 1 1 0Bipolar amplitudes {xk }: -1 -1 +1 -1 +1 +1 -1Coding rule: yk= xk + xk-1 : -2 0 0 0 2 0Decoding decision rule: if =2, decide that = +1

if =-2, decide that = -1if =0, decide opposite of the previous decision

Decoded bipolar sequence { }: -1 +1 -1 +1 +1 -1Decoded binary sequence { }: 0 1 0 1 1 0One drawback of this detection technique is that once an error is made, it

tends to propagate, causing further errors, since present decisions depend on prior decisions.

Decoded bipolar sequence { }: -1 -1 +1 -1 +1 -1Decoded binary sequence { }: 0 0 1 0 1 0A means of avoiding error propagation is known as precoding

ˆkxˆkyˆkxˆky

ˆkyˆkxˆkx

ˆkxˆkx





3. PrecodingPrecoding is accomplished by first differentially encoding the {xk } binary

sequence into a new {wk } binary sequence by means of the equation:

wk = xk + wk-1 (mod 2)So: 0+0=0

0+1=11+0=11+1=0 (mod 2)

The {wk } binary sequence is then converted to a bipolar pulse sequence, and the coding operation proceeds in the same way as it did in the upper example. However, with precoding, the detection process is quite different from the detection of ordinary duobinary.





• Example: Duobinary PrecodingIllustrate the duobinary coding and decoding rules when using the

differential precoding. Binary digit sequence {xk }: 0 0 1 0 1 1 0Precoded sequence wk = xk + wk-1 : 0 0 1 1 0 1 1Bipolar sequence {wk } : -1 -1 +1 +1 -1 +1 +1Coding rule: yk= wk + wk-1 : -2 0 +2 0 0 +2Decoding decision rule: If =±2, decide that =0

if =0, decide that = 1Decoded binary sequence: 0 1 0 1 1 0

The differential precoding enables us to decode the{ } sequence by making a decision on each received sample singly, without resorting to prior decisions that could be in error. The major advantage is that in the event of a digit error due to noise, such an errors does not propagate to other digits. The startup bit is choosing arbitrary.

ˆkxˆky

ˆkxˆky

ˆky





4. Duobinary Equivalent Transfer Function

Precoded duobinary signaling

Delay T seconds

1/2T-1/2T

T

f

Ideal rectangular filter H2 (f)

Noise

Decoder {xk }

ˆ{ }kx

ykˆ ky

Modulo-2 adder

wk

Digital filter H1 (f)

21( ) 1 j fTH f e π−= +

2

1( ) 2

0T for f

H f Telsewhere

<⎧= ⎨⎩





• The overall equivalent transfer function of the digital filter cascaded with the ideal rectangular filter is then given by

1 2

2

1( ) ( ) ( )2

(1 )( )

12 cos( ) 2

0

e

j fT

j fT j fT j fT

e

H f H f H f for fT

e TT e e e

T fT for fH f T

elsewhere

π

π π π

π

−

− −

= <

= +

= +

⎧ <⎪= ⎨⎪⎩

( ) sin sinet t Th t c cT T

−⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠





-0 .2 5 -0 .2 -0 .1 5 -0 .1 -0 .0 5 0 0 .0 5 0 .1 0 . 1 5 0 .2 0 .2 50

0 .5

1

1 .5

2

2 .5

3

3 .5

4H (f)

-6 -4 -2 0 2 4 6 8-0 .2

0

0 .2

0 .4

0 .6

0 .8

1

1 .2

1 .4





-8 -6 -4 -2 0 2 4 6 8 10 12-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4t= 0t= Tt= 2Tt= -T





5. Comparison of Binary with Duobinary SignalingAlthough duobinary signaling accomplishes the zero ISI requirement with

minimum bandwidth, duobinary signaling also requires more power than binary signaling, for equivalent performance against noise. For a given probability of bit error, duobinary signaling requires approximately 2.5dB greater SNR than binary signaling, while using only 1/(1+r) the bandwidth that binary signaling requirs where r is the filter roll-off.

6. Another special case of partial response signals (Ⅳpartial response system)

Delay 2T seconds

1/2T-1/2T

T

f

Ideal rectangular filter H2 (f)

Noise

Decoder {xk }

ˆ{ }kx

ykˆ ky

Modulo-2 adder

wk

Digital filter H1 (f)

+

-





Binary digit sequence {xk }: 0 0 0 1 0 1 1 0Precoded sequence wk = xk + wk-2 : 0 0 0 1 0 0 1 0Bipolar sequence {wk } : -1 -1 -1 +1 -1 -1 +1 -1Coding rule: yk= wk - wk-2 : 0 +2 0 -2 +2 0Decoding decision rule: If =±2, decide that =1

if =0, decide that = 0Decoded binary sequence: 0 1 0 1 1 0•

ˆkxˆky

ˆkxˆky

Note: If the transmitting signal is not bipolar sequence, the decoding rule became easy. It is just modulo 2 if the input digit sequence is binary.If the input digit sequence is M-ary, the decoding rule is modulo M.





( ) sin ( ) / sin ( ) /( ) 2 sin 2

e

e

h t c t T T c t T TH f T fTπ

= + − −

=

-8 -6 -4 -2 0 2 4 6 8-1 . 5

-1

-0 . 5

0

0 . 5

1

1 . 5T = 2

-0 . 2 5 -0 . 2 -0 . 1 5 -0 . 1 -0 . 0 5 0 0 . 0 5 0 . 1 0 . 1 5 0 . 2 0 . 2 50

0 . 5

1

1 . 5

2

2 . 5

3

3 . 5

4

he (t)

H(f)





-8 -6 -4 -2 0 2 4 6 8 10 12-1.5

-1

-0.5

0

0.5

1

1.5




8.4 Probability of Error in Detection of Digital PAM

• 8.4.1 Probability of Error for Detection of Digital PAM with Zero ISIIf there is zero ISI, the probability of error for digital PAM, PSK or QAM

in a bandlimited additive white Gaussian noise channel is same to that of the ideal AWGN channel without band limited.




8.5 Equalization

• 8.5.1 Channel CharacterizationMany communication channels can be characterized as band-limited

linear filters with an impulse response hc (t) and frequency response( )( ) ( ) cj f

c cH f H f e θ=

In order to achieve ideal transmission characteristics over a channel, within a signal’s bandwidth W, │Hc (f)│must be constant and Θc (f) must be a linear function of frequency. If │Hc(f)│is not a constant within W, the effect is amplitude distortion. If Θc(f) ia not a linear function of frequency within W, the effect is phase distortion. They will cause ISI.In the broad sense, the name “equalization” refers to any signal processing or filtering technique that is designed to eliminate or reduce ISI.




8.5 Equalization

• Category of Equalization1. Maximum-likelihood sequence estimation ( MLSE, 最大似然序列

估计). This method entails making measurements of hc (t) and then providing a means for adjusting the receiver to the transmission environment. The goal of such adjustments is to enable the detector to make good estimates from the demodulated distorted pulse sequence. With an MLSE receiver, the distorted samples are not reshaped or directly compensated in any way; instead , the mitigating technique for the MLSE receiver is to adjust itself in such a way that it can better deal with the distorted samples.

2. Equalization with filters (均衡滤波器). This method uses filters to compensate the distorted pulse. In this category, the detector is presented with a sequence of demodulated samples that the equalizer has modified or “cleaned up” form the effects of ISI.




8.5 Equalization

Equalization with filters has a further partition:1. Linear equalizer or transversal equalizer (线性均衡器或横向均衡

器)2. Decision feedback equalizer(判决反馈均衡器)Or 1. Present (预制式)2. Adaptive(自适应)Or 1. Symbol spaced (码元间隔)2. Fractionally spaced （部分码元间隔）




8.5 Equalization

• If we add the equalizer at the receiver, the overall system transfer function can be written as

H(f)=Ht (f) Hc (f) Hr (f) He (f)In practical system, the channel’s frequency transfer function Hc (f) is not

known with sufficient precision to allow for a receiver design to yield zero ISI for all time. Usually, the transmit and receive filters are chosen to be matched so that

H(f)=Ht (f) Hr (f)Then, the equalizer transfer function needed to compensate for channel

distortion is simply the inverse of the channel transfer function:

( ) ( ) ( )( )1 1 j f

ec c

H f eH f H f

θ−= =

Sometimes s system frequency transfer function manifesting ISI at the sampling points is purposely chosen.




8.5 Equalization

• 8.5.2 Equalizer Filter Types• 8.5.2.1 Transversal Equalizer

T T T T

C-N C-N+1 CN-1CN

+

xk

Algorithm for coefficient adjustment

yk

Transversal filter




8.5 Equalization

• How the equalizer works?• The goal of the transversal filter

-6 -4 -2 0 2 4 6 8-0 .2

0

0 .2

0 .4

0 .6

0 .8

1

1 .2

1 .4

The equalizer filter could generate a set of canceling echoes.

In the transversal filter, N could be infinite and finite.




8.5 Equalization• In the transversal filter, the tap weights {cn } need to be chosen so as

to subtract the effects of interference from symbols adjacent in time to the desire symbol.consider that there are (2N+1) taps with weights c-N , c-N+1 ,…, cN . Output samples {z(k)} of the equalizer are then found by convolving the input samples {x(k)} and tap weights {cn } as follows:

( ) ( ) 2 ,..., 2 ,...N

nn N

z k x k n c k N N n N N=−

= − = − = −∑

In a matrix form, the upper equation can be written asz=xc

We can find c by solving the following equation if x is a square:c=x-1z

Notice that the size of the vector z and the number of rows in the matrix x may be chosen to be any value, because one might be interested in the ISI at sample points.




8.5 Equalization( 2 )

(0)

(2 )

z N

z

z N

−⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

z 0

N

N

c

c

c

−⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

c

( ) 0 0 0 0( 1) ( ) 0 0 0

( ) ( 1) ( 2) ( 1) ( )

0 0 0 ( ) ( 1)0 0 0 0 ( )

x Nx N x N

x N x N x N x N x N

x N x Nx N

−⎡ ⎤⎢ ⎥− + −⎢ ⎥⎢ ⎥⎢ ⎥= − − − + −⎢ ⎥⎢ ⎥⎢ ⎥

−⎢ ⎥⎢ ⎥⎣ ⎦

x




8.5 Equalization• The vectors z and c have dimensions 4N+1 and 2N+1.and the

matrix x is nonsquare with dimensions 4N=1 by 2N+1. ( This means there are more equations than unknowns). We can solve it in a deterministic way known as the zero-forcing solution or the minimum mean-square (MSE) error solution.

• Zero-Forcing solution. First, transform x into a dimension 2N+1 by 2N+1 matrixThen, transforming z into a vector of dimension 2N+1,and let z(0)=1,

and z(k)=0, k≠0.We can can get the coefficient c.Example a zero-forcing three-tap equalizer{x(k)}={0,0.2,0.9,-0.3,0.1}Find coefficients c




8.5 Equalization• Note : For a zero-forcing equalizer with finite length, the peak

distortion is guarantee to be minimized only if the eye pattern is initially open. Since the zero-forcing equalizer neglects the effect of noise , it is not always the best system solution.

• MMSE solution. A more robust equalizer is obtained if the {cn}tap weights are chosen to minimize the mean-square error of all the ISI terms plus the noise power at the output of the equalizer.

• MSE is defined as the expected value of the squared difference between the desired data symbol and the estimated data symbol.

• The MMSE solution of the coefficients is as follows-1xx xzc = R R

wherexz =

T TxxR x z R = x x




8.5 Equalization• 8.5.2.2 Decision Feedback Equalizer

Forward filter

Feedback filter

Detector +

-

Inputsignal

output

The basic limitation of a linear equalizer, such as transversal filter, is that it performs poorly on channels having spectral nulls. The DFE is nonlinear equalizer that uses previous detector decisions to eliminate the ISI on the pulses that are currently being demodulated.




8.5 Equalization• 8.5.2.3 preset and adaptive equalization• 8.5.2.4 filter Update Rate

Documents

Chapter 8course.sdu.edu.cn/Download2/20150625110127415.pdf · 2015. 6. 25. · 8.2 The Power Spectrum of Digitally Modulated Signals • Example 8.2.2 Consider a binary sequence {b