978-0072957167 Chapter 4 Part 2

21

At the sampling instant t=t0=T, the signal component at the output of the matched filter is

y(T) = Z∞

−∞

Y(f)ej2πfT df =Z∞

−∞

s(τ)g(T−τ)dτ

|S(f)|2

Problem 4.23

1. The number of bits per symbol is

k=4800

R=4800

2400 = 2

Thus, a 4-QAM constellation is used for transmission. The probability of error for an M-ary QAM

system with M= 2k, is

With PM= 10−5and k= 2 we obtain

N0#= 5 ×10−6=⇒Eb

N0

2. If the bit rate of transmission is 9600 bps, then

In this case a 16-QAM constellation is used and the probability of error is

3. If the bit rate of transmission is 19200 bps, then

k=19200

2400 = 8

In this case a 256-QAM constellation is used and the probability of error is

With PM= 10−5we obtain

N0

4. The following table gives the SNR per bit and the corresponding number of bits per symbol for

the constellations used in parts a)-c).

k2 4 8

Problem 4.24

1. Since m2(t) = −m3(t) the dimensionality of the signal space is two.

2. As a basis of the signal space we consider the functions:



1

√T0≤t≤T





1

√T0≤t≤T

2

T

The vector representation of the signals is:

m1= [√T , 0]

m2= [0,√T]

3. The signal constellation is depicted in the next figure :

23

✈

(0,√T)

4. The three possible outputs of the matched filters, corresponding to the three possible transmitted

signals are (r1, r2) = (√T+n1, n2), (n1,√T+n2) and (n1,−√T+n2), where n1,n2are zero-

r1> r2and r1>−r2

Similarly we find that R2is the set of points (r1, r2) that satisfy r2>0, r2> r1and R3is the

region such that r2<0 and r2<−r1. The regions R1,R2and R3are shown in the next figure.

❅❅❅❅❅❅

0

R3

R2

(e) If the signals are equiprobable then:

P(e|m1) = P(|r−m1|2>|r−m2|2m1) + P(|r−m1|2>|r−m3|2m1)

24

Since, n1,n2are zero-mean statistically independent Gaussian random variables, each with variance

N0

2, the random variables x=n1−n2and y=n1+n2are zero-mean Gaussian with variance N0.

Hence:

P(e|m1) = 1

√2πN0Z∞

√T

e−x2

2N0dx +1

√2πN0Z−√T

e−y2

2N0dy

When m2is transmitted then r= [n1, n2+√T] and therefore:

Similarly from the symmetry of the problem, we obtain:

P(e|m2) = P(e|m3) = Q“rT

N0#+Q“r2T

N0#

Since Q[·] is momononically decreasing, we obtain:

and therefore, the probability of error P(e|m1) is larger than P(e|m2) and P(e|m3). Hence, the

Problem 4.25

1. In QPSK, Pe≈2Q(p2Eb/N0). Here Eb=1

2E=1

4A2Tand Pe≈2Q(pA2T/2N0).

4. Here we have orthogonal signaling with Eb=B2T1/2. The rate is R=Rs= 1/T1. The rate

25

Problem 4.26

1. Signals have equal energy E=RT

0s2(t)dt =RT/2

04dt +RT

T/21dt = 5T/2. We also have

2. Binary equiprobable, hence Pe=Qqd2

2N0. We have

d2=ZT

0

(s1(t)−s2(t))2dt =ZT /2

0

(1)2dt +ZT

T/2

(−1)2dt =T

3. From part 1 and 2, we have Pe=Qq 1

2RN0=Qq1

5

Eb

N0=Qq1

10

2Eb

N0. Since for

4. We have Rnew =log24

T=2

T= 2R.

5. We need to find d2

min in the new constellation. Obviously, d2

34 =d2

12 =T. We also have

1. By inspection a two dimensional basis of the form

2. We have s1= (1,√2), s2= (−1,2√2), s3= (1,0), s4= (2,√2). The constellation is shown

below

26

−2 −1 0 1 2 3

−1

0

1

2

3

4

5

6

7

s2

s3

3. The perpendicular bisectors defining the decision regions are shown on the figure. s1has the

4. D1is given by

D1={(r1, r2) : r.s1−E1/2>r.s2−E2/2,r.s1−E1/2>r.s3−E3/2,r.s1−E1/2>r.s4−E4/2}

or

Problem 4.28

Using the Pythagorean theorem for the four-phase constellation, we find:

r2

1+r2

1=d2=⇒r1=d

√2

27

The radius of the 8-PSK constellation is found using the cosine rule. Thus:

Thus, the additional transmitted power needed by the 8-PSK signal is:

P= 10 log10

2d2

(2 −√2)d2= 5.3329 dB

Problem 4.29

For 4-phase PSK (M= 4) we have the following realtionship between the symbol rate 1/T , the

required bandwith Wand the bit rate R=k·1/T =log2M

T(see 5-2-84):

W=R

log2M→R=W log2M= 2W= 200 kbits/sec

Problem 4.30

1. The envelope of the signal is

|s(t)|=p|sc(t)|2+|ss(t)|2

2. The signal s(t) is equivalent to an MSK signal. A block diagram of the modulator for synthesizing

the signal is given in the next figure.

♥♥

✻

✲✲

×

a2n

3. A sketch of the demodulator is shown in the next figure.

♥

❅

✲

✻

❄

✲

t= 2Tb

×

0(·)dt

Threshold

Problem 4.31

29

A biorthogonal signal set with M= 8 signal points has vector space dimensionallity 4. Hence, the

P(E1)≤3 [P(|C2|>|C1|)] = 3 [2P(C2> C1)] = 6P(C2> C1) = 6Q rEs

N0!

where Ciis the correlation metric corresponding to the i-th vector space dimension; the probability

Problem 4.32

It is convenient to find first the probability of a correct decision. Since all signals are equiprobable,

P(C) =

M

X

1

MP(C|si)

30

Hence :

P(C|si) = Z∞

−d

f(n)dn!N

= 1−Z−d

2

−∞

f(n)dn!N

and therefore, the probability of error is given by :

P(e) = 1 −P(C) = 1 −

M

X

1

M1−Qd

√2N0N

i=1

the probability of error can be written as :

Problem 4.33

Consider first the signal :

y(t) =

n

X

k=1

ckδ(t−kTc)

The signal y(t) has duration T=nTcand its matched filter is :

g(t) = y(T−t) = y(nTc−t) =

n

X

ckδ(nTc−kTc−t)

n

X

n

X

31

the Fourier transform of the signal s(t) is :

n

X

k=1

and therefore, the Fourier transform of the signal matched to s(t) is :

H(f) = S∗(f)e−j2πf T =S∗(f)e−j2πf nTc

n

X

Thus, the matched filter H(f) can be considered as the cascade of a filter,with impulse response

p(−t), matched to the pulse p(t) and a filter, with impulse response g(t), matched to the signal

Problem 4.34

1. The inner product of si(t) and sj(t) is

Z∞

−∞

si(t)sj(t)dt =Z∞

−∞

n

X

k=1

cikp(t−kTc)

n

X

l=1

cjlp(t−lTc)dt

n

X

n

X

2. Using the results of Problem 5.35, we obtain that the filter matched to the waveform

si(t) =

n

X

k=1

cikp(t−kTc)

1. The optimal ML detector (see 5-1-41) selects the sequence Cithat minimizes the quantity:

D(r, Ci) =

n

X

k=1

(rk−pEbCik)2

The metrics of the two possible transmitted sequences are

w

X

n

X

k=1

k=w+1

Since the first term of the right side is common for the two equations, we conclude that the optimal

ML detector can base its decisions only on the last n−wreceived elements of r. That is

n

X

k=w+1

(rk−pEb)2−

n

X

k=w+1

(rk+pEb)2

C2

>

<

C1

0

k=w+1

<

C2

2. Since rk=√EbCik +nk, the probability of error P(e|C1) is

P(e|C1) = P pEb(n−w) +

n

X

k=w+1

nk<0!

X

Similarly we find that P(e|C2) = P(e|C1) and since the two sequences are equiprobable

3. The probability of error P(e) is minimized when Eb(n−w)

Problem 4.36

1. The noncoherent envelope detector for the on-off keying signal is depicted in the next figure.

34

♥

✻

✲

❅

rs

rc

(·)2

t=T

×

Rt

0(·)dτ

2. If s0(t) is sent, then the received signal is r(t) = n(t) and therefore the sampled outputs rc,

p(r|s0(t)) = r

σ2e−r2

N0

e−r2

If s1(t) is transmitted, then the received signal is :

r(t) = r2Eb

Tb

cos(2πfct+φ) + n(t)

Crosscorrelating r(t) by q2

Tcos(2πfct) and sampling the output at t=T, results in

where ncis zero-mean Gaussian random variable with variance N0

2. Similarly, for the quadrature

component we have :

rs=pEbsin(φ) + ns

3. For equiprobable signals the probability of error is given by:

P(error) = 1

2ZVT

−∞

p(r|s1(t))dr +1

2Z∞

VT

p(r|s0(t))dr

35

Since r > 0 the expression for the probability of error takes the form

2ZVT

0

σ2e−r2+Eb

σ2dr +1

2Z∞

VT

σ2e−r2

The optimum threshold level is the value of VTthat minimizes the probability of error. However,

when Eb

N0≫1 the optimum value is close to: √Eb

2and we will use this threshold to simplify the

analysis. The integral involving the Bessel function cannot be evaluated in closed form. Instead of

I0(x) we will use the approximation :

which is valid for large x, that is for high SNR. In this case :

2ZVT

0

σ2e−r2+Eb

σ2dr ≈1

2Z√Eb

0rr

2πσ2√Eb

This integral is further simplified if we observe that for high SNR, the integrand is dominant in the

vicinity of √Eband therefore, the lower limit can be substituted by −∞. Also

and therefore :

1

2

e−(r−√Eb)2/2σ2dr ≈1

2

≤1

2Q“rEb

2N0#+1

2e−Eb

Problem 4.37

1. In binary DPSK, the information bit 1 is transmitted by shifting the phase of the carrier by

πradians relative to the phase in the previous interval, while if the information bit is 0 then the

36

phase is not shifted. With this in mind :

Data : 1 1 0 1 0 0 0 1 0 1 1 0

2. We know that the power spectrum of the equivalent lowpass signal u(t) is :

T|G(f)|2Φii(f)

where G(f) = AT sin πfT

πfT ,is the spectrum of the rectangular pulse of amplitude Athat is used, and

Φii(f) is the power spectral density of the information sequence. It is straightforward to see that

the information sequence Inis simply the phase of the lowpass signal, i.e. it is ejπ or ej0depending

on the bit to be transmitted an(= 0,1).We have :

The statistics of Inare (remember that {an}are uncorrelated) :

E[In] = Eejπ Pkak=QkEejπak=Qk1

2ejπ −1

2ej0=Qk0 = 0

Eh|In|2i=Eejπ Pkake−jπ Pkak= 1

Hence, Inis an uncorrelated sequence with zero mean and unit variance, so Φii(f) = 1,and

Φuu(f) = 1

T|G(f)|2=A2Tsin πf T

πfT

A sketch of the signal power spectrum Φss(f) is given in the following figure :

−fc 0 fc

f

Power spectral density of s(t)

Problem 4.38

1. D=Re VmV∗

m−1where Vm=Xm+jYm.Then :

D=Re ((Xm+jYm)(Xm−1−jYm−1))

2. Vk=Xk+jYk= 2aE cos(θ−φ) + j2aE sin(θ−φ) + Nk,real +Nk,imag.Hence :

U1=Xm+Xm−1

The variance of U1is : E[U1−E(U1)]2=E1

the noise components are uncorrelated and have zero mean. Similarly for any i, j :cov(Ui, Uj) = 0

. The condition cov(Ui, Uj) = 0,implies that these random variables {Ui}are uncorrelated, and

since they are Gaussian, they are also statistically independent.

3. W1=U2

1+U2

2,with U1, U2being statistically independent Gaussian variables with means

2aE cos(θ−φ),2aE(sin θ−φ) and identical variances σ2= 2EN0.Then, W1follows a non-central

chi-square distribution with pdf given by (2-1-118):

p(w1) = 1

4EN0

e−(4a2E2+w1)/4EN0I0a

N0

√w1, w1≥0

Also, W2=U2

3+U2

4,with U3, U4being zero-mean Gaussian with the same variance. Hence, W1

follows a central chi-square distribution, with pfd given by (2-1-110) :

4.

Pb=P(D < 0) = P(W1−W2<0)

=R∞

0P(w2> w1|w1)p(w1)dw1

Problem 4.39

From 4.6-7, W=RN/2 log2M, therefor we have

1. BFSK, M= 2, N = 2, W =R.

Problem 4.40

39

1. These are given by Equations 4.2-32 and 4.2-33

rth =N0

2. When the receiver receives noise only with no trace of signal, it can make any decision and its

error probability will be 1/2 regardless of its decision. Therefore without any loss in optimality

Problem 4.41

1. Binary equiprobable, hence Pe=Qqd2

2N0, where

d2=Z(s1(t)−s2(t))2dt =Z2

1

42dt = 16

2. Note that the only relevant information is in the interval from 1 to 2. In the interval from 0

to 1 the two signals are equal and hence the received signals in that interval will be be equal

and the error probability (using symmetry) is

Pe=P(r1+r2>0|s2sent)

3. Here r1=Asm+n1and r2=sm+n2for m= 1,2, and Ais known to the detector. Again

here D1⇔p(r1, r2|s1)> p(r1, r2|s2), or

N0> Ke−(r1+2A)2+(r2+2)2

N0

This is the optimal detection scheme for the reciever who knows Aand hence it can compute

Ar1+r2. The error probability (for someone who does not know A) is

0

Therefore, we first find P(e|A) as

P(e|A) = P(r1A+r2>0|s2sent)

N0

1 + A2

Note that this relation for A= 0 and A= 1 reduces to the results of parts 1 and 2. Now we

0

N0

1 + A2

4. Again D1⇔p(r1, r2|s1)> p(r1, r2|s2) but since the detector does not know Awe are in a

situation like noncoherent detection, the difference being that here Ais random instead of φ.