978-0072957167 Chapter 9 Part 2

21

The output of the matched filter is sampled at t=Tand the samples are passed to the detector.

The detector is a simple threshold device that decides if a binary 1 or 0 was transmitted depending

on the sign of the input samples. The following figure shows a block diagram of the optimum

bandpass coherent demodulator.

✂

.

❘❅

❅✲✲✲

device)

(Threshold

t=T

r(t)

gR(t)

matched filter

Problem 9.22

(a) The power spectral density of X(t) is given by (see (4-4-12))

Φx(f) = 1

TΦa(f)|GT(f)|2

The Fourier transform of g(t) is

GT(f) = F[g(t)] = AT sin πfT

πfT e−jπf T

(b) If g1(t) is used instead of g(t) and the symbol interval is T, then

(c) If we precode the input sequence as bn=an+αan−3, then





1 + α2m= 0

22

To obtain a null at f=1

3

(c) The answer to this question is no. This is because Φb(f) is an analytic function and unless it

Problem 9.23

The roll-off factor βis related to the bandwidth by the expression 1+β

T= 2W, or equivalently

R(1 + β) = 2W. The following table shows the symbol rate for the various values of the excess

Problem 9.24

The following table shows the precoded sequence, the transmitted amplitude levels, the received

signal levels and the decoded sequence, when the data sequence 10010110010 modulates a duobinary

transmitting filter.

Data seq. Dn: 1 0 0 1 0 1 1 0 0 1 0

Problem 9.25

23

The following table shows the precoded sequence, the transmitted amplitude levels, the received

signal levels and the decoded sequence, when the data sequence 10010110010 modulates a modified

duobinary transmitting filter.

Data seq. Dn: 1 0 0 1 0 1 1 0 0 1 0

Precoded seq. Pn: 0 0 1 0 1 1 1 0 0 0 0 1 0

Problem 9.26

Let X(z) denote the Z-transform of the sequence xn, that is

X(z) = X

n

xnz−n

Then the precoding operation can be described as

P(z) = D(z)

X(z)mod −M

Problem 9.27

24

(a) The frequency response of the RC filter is

C(f) =

1

j2πRCf

R+1

j2πRCf

=1

1 + j2πRCf

The amplitude and the phase spectrum of the filter are :

|C(f)|=1

1 + 4π2(RC)2f21

2

, θc(f) = arctan(−2πRCf)

A plot of τ(f) with RC = 10−6is shown in the next figure :

9.99

9.991

9.992

9.993

9.994

9.995

9.996

9.997

9.998

9.999

10x10-7

0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000

Frequency (f)

Tc(f)

(b) The following figure is a plot of the amplitude characteristics of the RC filter, |C(f)|. The

values of the vertical axis indicate that |C(f)|can be considered constant for frequencies up to 2000

Problem 9.28

Let GT(f) and GR(f) be the frequency response of the transmitting and receiving filter. Then, the

condition for zero ISI implies

GT(f)C(f)GR(f) = Xrc(f) = 









T, 0≤ |f| ≤ 1

4T

T

2[1 + cos(2πT (|f| − 1

T)],1

4T≤ |f| ≤ 3

4T

0,|f|>3

4T

Thus,





hT

1+0.3 cos 2πf T i1

2,0≤ |f| ≤ 1

4T

hT(1+cos(2πT (|f|− 1

1

Problem 9.29

A 4-PAM modulation can accomodate k= 2 bits per transmitted symbol. Thus, the symbol

interval duration is :

T=k

Then, the magnitude frequency response of the optimum transmitting and receiving filter is (see

(9-2-81))

4





1 + f

2400 21

4

,|f|<2400

Problem 9.30

We already know that

σ2

v=Z∞

−∞

Φnn(f)|GR(f)|2df

From these

σ2

v

d2=1

PavTZW

−W

Φnn(f)|GR(f)|2df ZW

−W

|Xrc(f)|2

|GR(f)|2|C(f)|2df (9.0.4)

The optimum |GR(f)|can be found by applying the Cauchy-Schwartz inequality

where |U1(f)|,|U2(f)|are defined as

|U1(f)|=|pΦnn(f)||GR(f)|

|U2(f)|=|Xrc(f)|

|GR(f)||C(f)|

Problem 9.31

In the case where the channel distortion is fully precompensated at the transmitter, the loss of

SNR is given by

10 log L1,with L1=ZW

−W

Xrc(f)

|C(f)|2

whereas in the case of the equally split filters, the loss of SNR is given by

10 log[L2]2,with L2=ZW

−W

Xrc(f)

|C(f)|

Assuming that 1/T =W, so that we have a raised cosine characteristic with β= 0, we have

Hence, the loss for the first type of filters is 10 log L1= 1.89 dB.

In a similar way,

1

[1+cos πf

W]

Problem 9.32

Suppose that am= +1 is the transmitted signal. Then the probability of error will be :

Pe|1=P(ym<0|am= +1)

=P(1 + nm+im<0)

Problem 9.33

(a) If the transmitted signal is :

r(t) = ∞

X

n=−∞

Inh(t−nT ) + n(t)

then the output of the receiving filter is :

X

n=−∞

where x(t) = h(t)⋆ h(t) and ν(t) = n(t)⋆ h(t). If the sampling time is off by 10%, then the samples

at the output of the correlator are taken at t= (m±1

10 )T. Assuming that t= (m−1

10 )Twithout

loss of generality, then the sampled sequence is :

X

and therefore, the SNR is :

2A2T

(b) Recall from part (a) that the sampled sequence is

1

29

The term Im−1A2T

10 expresses the ISI introduced to the system. If Im= 1 is transmitted, then the

probability of error is

P(e|Im= 1) = 1

2Q

N0

2Q

1022A2T

N0

Since the symbols of the binary PAM system are equiprobable the previous derived expression is

the probability of error when a symbol by symbol detector is employed. Comparing this with the

2Q

1022A2T

N0

2Q

N0

Problem 9.34

(a) Taking the inverse Fourier transform of H(f), we obtain :

h(t) = F−1[H(f)] = δ(t) + α

2δ(t−t0) + α

2δ(t+t0)

(b) If the signal s(t) is used to modulate the sequence {In}, then the transmitted signal is :

u(t) = ∞

X

n=−∞

Ins(t−nT )

w(t1) = ∞

X

n=−∞

InZ∞

−∞

s(τ−nT )s(τ−t1)dτ

+α

2

∞

X

InZ∞

−∞

s(τ−t0−nT )s(τ−t1)dτ

∞

X

2

n=−∞

2

n=−∞

(c) With t0=Tand k=nin the previous equation, we obtain :

wk=Ikx0+X

n6=k

Inxk−n

Problem 9.35

(a) Each segment of the wire-line can be considered as a bandpass filter with bandwidth W= 1200

Hz. Thus, the highest bit rate that can be transmitted without ISI by means of binary PAM is :

31

(b) The probability of error for binary PAM trasmission is :

P2=Q“r2Eb

N0#

N0

(c) The received power PRis related to the desired SNR per bit through the relation :

PR

N0

=1

TEb

N0

=REb

N0

Problem 9.36

xn=Z∞

−∞

h(t+nT )h∗(t)dt

vk=Z∞

z(t)h∗(t−kT )dt

Problem 9.37

In the case of duobinary signaling, the output of the matched filter is :

x(t) = sinc(2W t) + sinc(2W t −1)

n

Problem 9.38

(a) The output of the matched filter demodulator is :

y(t) = ∞

X

k=−∞

IkZ∞

−∞

gT(τ−kTb)gR(t−τ)dτ +ν(t)

=∞

X

k=−∞

Ikx(t−kTb) + ν(t)

Hence,

y(mTb) = ∞

X

k=−∞

Ikx(mTb−kTb) + v(mTb)

33

(b) In the next figure we show one trellis stage for the ML sequence detector. Since there is

postcursor ISI, we delay the received signal, used by the ML decoder to form the metrics, by one

sample. Thus, the states of the trellis correspond to the sequence (Im−1, Im), and the transition

✉

✏✏✏✏✏✏✏

◗

-1

1 1

Problem 9.39

(a) The output of the matched filter at the time instant mT is :

ym=X

k

Imxk−m+νm=Im+1

4Im−1+νm

The autocorrelation function of the noise samples νmis

If a symbol by symbol detector is employed and we assume that the symbols Im=Im−1=√Eb

have been transmitted, then the probability of error P(e|Im=Im−1=√Eb) is :

P(e|Im=Im−1=pEb) = P(ym<0|Im=Im−1=pEb)

4√Eb

m

N0dνm

If however Im−1=−√Eb, then :

P(e|Im=pEb, Im−1=−pEb) = P(3

4pEb+νm<0) = Q“3

4r2Eb

N0#

34

Since the two symbols √Eb,−√Ebare used with equal probability, we conclude that :

(b) In the next figure we plot the error probability obtained in part (a) (log10(P(e))) vs. the SNR

per bit and the error probability for the case of no ISI. As it observed from the figure, the relative

difference in SNR of the error probability of 10−6is 2 dB.

-7

-6.5

-6

-5.5

-5

-4.5

-4

-3.5

-3

-2.5

-2

log(P(e)

Problem 9.40

For the DFE we have that :

0

X

j=−K1

K2

X

j=1

We want to minimize J=EIk−ˆ

Ik

2.Taking the derivative of J, with respect to the real and

imaginary parts of cl=al+jbl,1≤l≤K2,we obtain :

k−lIk−ˆ

and similarly : ∂J

∂bl

= 0 ⇒EhIm nI∗

k−lIk−ˆ

Ikoi= 0

35

Since the information symbols are uncorrelated : E[IkI∗

l] = δkl.We also have :

E[Iku∗

l] = EhIkPL

m=0 f∗

mI∗

l−m+n∗

li

Hence, equation (1) gives :

EIkI∗

k−l=Ehˆ

IkI∗

k−li,1≤l≤K2⇒

which is the desired equation for the feedback taps.

Problem 9.41

(a) The equivalent discrete-time impulse response of the channel is :

h(t) =

1

X

n=−1

hnδ(t−nT ) = 0.3δ(t+T) + 0.9δ(t) + 0.3δ(t−T)

n=−1

which in matrix notation is written as :



0.9 0.3 0.



c−1



0



The coefficients of the zero-force equalizer can be found by solving the previous matrix equation.

Thus,



c−1



−0.4762



36

(b) The values of qmfor m=±2,±3 are given by

q2=

1

X

n=−1

cnh2−n=c1h1=−0.1429

1

X

n=−1

Problem 9.42

(a) The output of the zero-force equalizer is :

1

X

n=−1

With q0= 1 and qm= 0 for m6= 0, we obtain the system :



1.0 0.1−0.5



c−1



0



Solving the previous system in terms of the equalizer’s coefficients, we obtain :



c−1



0.000



37

(b) The output of the equalizer is :





0m≤ −4

c0x2+x1c1= 0.0098 m= 2

Hence, the residual ISI sequence is

and its span is 6 symbols.

Problem 9.43

(a) If {cn}denote the coefficients of the zero-force equalizer and {qm}is the sequence of the

equalizer’s output samples, then :

qm=

1

X

n=−1

cnxm−n

where {xk}is the noise free response of the matched filter demodulator sampled at t=kT . With

q−1= 0, q0=q1=Eb, we obtain the system :

0.1Eb0.9EbEb

c1

Eb

The solution to the system is :

(b) The set of noise variables {νk}at the output of the sampler is a gaussian distributed sequence

with zero-mean and autocorrelation function :



N0

2xk|k| ≤ 2