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Signal & Linear System Analysis: Problem Solutions
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Chapter 2
Signal and Linear System Analysis
2.1
Problem Solutions
Problem 2.1
a. For the single-sided spectra, write the signal as
x1 (t) = 10 cos(4t + =8) + 6 sin(8t + 3=4)
= 10 cos(4t + =8) + 6 cos(8t + 3=4 =2)
= 10 cos(4t + =8) + 6 cos(8t + =4)
h
i
= Re 10ej(4t+=8) + 6ej(8t+=4)
For the double-sided spectra, write the signal in terms of complex exponentials using Eulerís
theorem:
x1 (t) = 5 exp[j(4t + =8)] + 5 exp[j(4t + =8)]
+3 exp[j(8t + 3=4)] + 3 exp[j(8t + 3=4)]
The spectra are plotted in Fig. 2.1.
b. Write the given signal as
h
i
x2 (t) = Re 8ej(2t+=3) + 4ej(6t+=4)
to plot the single-sided spectra. For the double-side spectra, write it as
x2 (t) = 4ej(2t+=3) + 4ej(2t+=3) + 2ej(6t+=4) + 2ej(6t+=4)
The spectra are plotted in Fig. 2.2.
1
2
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
c. Change the sines to cosines by subtracting =2 from their arguments to get
x3 (t) = 2 cos (4t + =8 =2) + 12 cos (10t =2)
= 2 cos (4t 3=8) + 12 cos (10t =2)
h
i
= Re 2ej(4t3=8) + 12ej(10t=2)
= ej(4t3=8) + ej(4t3=8) + 6ej(10t=2) + 6ej(10t=2)
Spectral plots are given in Fig. 2.3.
d. Use a trig identity to write
3 sin (18t + =2) = 3 cos (18t)
and get
x4 (t) = 2 cos (7t + =4) + 3 cos (18t)
h
i
= Re 2ej(7t+=4) + 3ej18t
= ej(7t+=4) + ej(7t+=4) + 1:5ej18t + 1:5ej18t
From this it is seen that the singe-sided amplitude spectrum consists of lines of amplitudes
2 and 3 at frequencies of 3.5 and 9 Hz, respectively, and the phase spectrum consists of
a line of height =4 at 3.5 Hz. The double-sided amplitude spectrum consists of lines of
amplitudes 1, 1, 1.5, and 1.5 at frequencies of 3.5, -3.5, 9, and -9 Hz, respectively. The
double-sided phase spectrum consists of lines of heights =4 and =4 at frequencies 3.5
Hz and 3:5 Hz, respectively.
e. Use sin (2t) = cos (2t =2) to write
x5 (t) = 5 cos (2t =2) + 4 cos (5t + =4)
h
i
= Re 5ej(2t=2) + 4ej(5t+=4)
= 2:5ej(2t=2) + 2:5ej(2t=2) + 2ej(5t+=4) + 2ej(5t+=4)
From this it is seen that the singe-sided amplitude spectrum consists of lines of amplitudes
5 and 4 at frequencies of 1 and 2.5 Hz, respectively, and the phase spectrum consists of
lines of heights =2 and =4 at 1 and 2.5 Hz, respectively. The double-sided amplitude
spectrum consists of lines of amplitudes 2.5, 2.5, 2, and 2 at frequencies of 1, -1, 2.5, and
-2.5 Hz, respectively. The double-sided phase spectrum consists of lines of heights =2,
=2, =4, and =4 at frequencies of 1, -1, 2.5, and -2.5 Hz, respectively.
2.1. PROBLEM SOLUTIONS
3
Single sided
Double sided
6
8
Amplitude
Amplitude
10
6
4
4
2
2
0
0
1
2
3
f, Hz
4
0
-5
5
0
f, Hz
5
0
f, Hz
5
0.5
0.6
Phase, rad
Phase, rad
0.8
0.4
0.2
0
-0.5
0
0
1
2
3
f, Hz
4
5
-5
f. Use sin (10t + =6) = cos (10t + =6 =2) = cos (10t =3) to write
x6 (t) = 3 cos (4t + =8) + 4 cos (10t =3)
h
i
= Re 3ej(4t+=8) + 4ej(10t=3)
= 1:5ej(4t+=8) + 1:5ej(4t+=8) + 2ej10t=3) + 2ej(10t=3)
From this it is seen that the singe-sided amplitude spectrum consists of lines of amplitudes
3 and 4 at frequencies of 2 and 5 Hz, respectively, and the phase spectrum consists of
lines of heights =8 and =3 at 2 and 5 Hz, respectively. The double-sided amplitude
spectrum consists of lines of amplitudes 1.5, 1.5, 2, and 2 at frequencies of 2, -2, 5, and -5
Hz, respectively. The double-sided phase spectrum consists of lines of heights =8, =8,
=3, and =3 at frequencies of 2, -2, 5, and -5 Hz, respectively.
4
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Single sided
Double sided
5
4
6
Amplitude
Amplitude
8
4
2
0
2
1
0
1
2
f, Hz
3
0
-4
4
-2
0
f, Hz
2
4
-2
0
f, Hz
2
4
1
Phase, rad
1
Phase, rad
3
0.5
0
0.5
0
-0.5
-1
0
1
2
f, Hz
3
4
-4
2.1. PROBLEM SOLUTIONS
5
Single sided
Double sided
6
Amplitude
Amplitude
10
5
0
0
2
4
4
2
0
6
-5
0
f, Hz
5
-5
0
f, Hz
5
f, Hz
0
Phase, rad
Phase, rad
1
-0.5
-1
0
-1
-1.5
0
2
4
f, Hz
6
6
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Problem 2.2
By noting the amplitudes and phases of the various frequency components from the plots,
the result is
x(t) = 4ej(8t+=2) + 4ej(8t+=2) + 2ej(4t=4) + 2ej(4t=4)
= 8 cos (8t + =2) + 4 cos (4t =4)
= 8 sin (8t) + 4 cos (4t =4)
Problem 2.3
a. Not periodic because f1 = 1= Hz and f2 = 3 Hz are not commensurable.
b. Periodic. To Önd the period, note that
6
30
= 3 = n1 f0 and
= 15 = n2 f0
2
2
Therefore
15
n2
=
3
n1
Hence, take n1 = 1, n2 = 5; and f0 = 3 Hz (we want the largest possible value for f0 with
n1 and n2 integer-valued).
c. Periodic. Using a similar procedure as used in (b), we Önd that n1 = 4, n2 = 21; and
f0 = 0:5 Hz.
d. Periodic. Using a similar procedure as used in (b), we Önd that n1 = 4, n2 = 7;
n3 = 11; and f0 = 0:5 Hz.
e. Periodic. We Önd that n1 = 17, n2 = 18; and f0 = 0:5 Hz.
f. Periodic. We Önd that n1 = 2, n2 = 3; and f0 = 0:5 Hz.
g. Periodic. We Önd that n1 = 7, n2 = 11; and f0 = 0:5 Hz.
h. Not periodic. The frequencies of the separate terms are incommensurable.
i. Periodic. We Önd that n1 = 19, n2 = 21; and f0 = 0:5 Hz.
j. Periodic. We Önd that n1 = 6, n2 = 7; and f0 = 0:5 Hz.
2.1. PROBLEM SOLUTIONS
7
Problem 2.4
a. The single-sided amplitude spectrum consists of a single line of amplitude 5 at 6 Hz
and the phase spectrum consists of a single line of height =6 rad at 6 Hz. The
double-sided amplitude spectrum consists of lines of amplitude 2.5 at frequencies 6
Hz. The double -sided phase spectrum consists of a line of height =6 at -6 Hz and
a line of height =6 at 6 Hz.
b. Write the signal as
x2 (t) = 3 cos(12t =2) + 4 cos(16t)
From this it is seen that the single-sided amplitude spectrum consists of lines of heights 3
and 4 at frequencies 6 and 8 Hz, respectively, and the single-sided phase spectrum consists
of a line of height =2 radians at frequency 6 Hz (the phase at 8 Hz is 0). The doublesided amplitude spectrum consists of lines of height 1.5 and 2 at frequencies of 6 and 8 Hz,
respectively, and lines of height 1.5 and 2 at frequencies 6 and 8 Hz, respectively. The
double-sided phase spectrum consists of a line of height =2 radians at frequency 6 Hz
and a line of height =2 radians at frequency 6 Hz.
c. Use the trig identity cos x cos y = 0:5 cos (x + y) + 0:5 cos (x y) to write
x3 (t) = 2 cos 20t + 2 cos 4t
From this we see that the single-sided amplitude spectrum consists of lines of height 2 at 2
and 10 Hz, and the single-sided phase spectrum is 0 at these frequencies. The double-sided
amplitude spectrum consists of lines of height 1 at frequencies of 10, 2, 2, and 10 Hz.
The double-sided phase spectrum is 0.
d. Use trig identies to get
x4 (t) = 4 sin (2t) [1 + cos (10t)]
= 4 sin (2t) 2 sin (8t + ) + 2 sin (12t)
= 4 cos (2t =2) + 2 cos (8t + =2) + 2 cos (12t =2)
h
i
= Re 4ej(2t=2) + 2ej(8t+=2) + 2ej(12t=2)
= 2ej(2t=2) + 2ej(2t=2) + ej(8t+=2) + ej(8t+=2) + ej(12t=2) + ej(12t=2)
From this we see that the single-sided amplitude spectrum consists of lines of heights 4,
2, and 2 at frequencies 1, 4, and 6 Hz, respectively and the single-sided phase spectrum is
=2 radians at 1 and 6 Hz and =2 radians at 4 Hz. The double-sided amplitude spectrum
8
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
consists of lines of height 2 at frequencies of 1 and 1 Hz and of height 1 at frequencies of
4, -4, 6, and -6 Hz. The double-sided phase spectrum is =2 radians at -1, 4, and -6 Hz and
=2 radians at 1, -4, and 6 Hz.
e. Clearly, from the form of the cosine sum, the single-sided amplitude spectrum has
lines of heights 1 and 7 at frequencies of 3 and 15 Hz, respectively. The single-sided
phase spectrum is zero. The double-sided amplitude spectrum has lines of heights 0.5,
0.5, 3.5, and 3.5 at frequencies of 3, -3, 15, and -15 Hz, respectfully. The double-sided
phase spectrum is zero.
f. The single-sided amplitude spectrum has lines of heights 1 and 9 at frequencies of
2 and 10.5 Hz, respectively. The single-sided phase spectrum is =2 radians at
10.5 Hz and 0 otherwise. The double-sided amplitude spectrum has lines of heights
0.5, 0.5, 4.5, and 4.5 at frequencies of 2, -2, 10.5, and -10.5 Hz, respectfully. The
double-sided phase spectrum is =2 radians at -10.5 Hz and =2 radians at 10.5 Hz
and 0 otherwise.
g. Convert the sine to a cosine by subtracting =2 from its argument. It then follows
that the single-sided amplitude spectrum is 2, 1, and 6 at frequencies of 2, 3, and
8.5 Hz and 0 otherwise. The single-sided phase spectrum is =2 radians at 8.5 Hz
and 0 otherwise. The double-sided amplitude spectrum is 1, 1, 0.5, 0.5, 3, and 3
at frequencies of 2, 2, 3, 3 8:5, and 8.5 Hz, respectively, and 0 otherwise. The
double-sided phase spectrum is =2 radians at a frequency of 8:5 Hz and =2
radians at a frequency of 8.5 Hz. It is 0 otherwise.
Problem 2.5
a. This function has area
Area =
=
Z1
1
1
Z1
1
sin(t=) 2
dt
(t=)
sin(u) 2
du = 1
(u)
where a tabulated integral has been used for sinc2 u. A sketch shows that no matter how
small is, the area is still 1. With ! 0; the central lobe of the function becomes narrower
and higher. Thus, in the limit, it approximates a delta function.
2.1. PROBLEM SOLUTIONS
9
b. The area for the function is
Area =
Z1
1
exp(t=)u (t) dt =
1
Z1
exp(u)du = 1
0
A sketch shows that no matter how small is, the area is still 1. With ! 0; the function
becomes narrower and higher. Thus, in the limit, it approximates a delta function.
R
R1
c. Area = 1 (1 jtj =) dt = 1 (t) dt = 1. As ! 0, the function becomes
narrower and higher, so it approximates a delta function in the limit.
Problem 2.6
1
a. Make use of the formula (at) = jaj
(t) to write (2t 5) = [2 (t 5=2)] =
1
5
2 t 2 and use the sifting property of the -function to get
1
Ia =
2
2
5
1
5
25 1
+ exp 2
=
+ exp [5] = 3:1284
2
2
2
8
2
b. Impulses at 10, 5, 0, 5, 10 are included in the integral. Use the sifting property
after writing the expression as the sum of Öve integrals to get
Ib = (10)2 + 1 + (5)2 + 1 + 02 + 1 + 52 + 1 + 102 + 1 = 255
c. Matching coe¢cients of like derivatives of -functions on either side of the equation
gives A = 5, B = 10, and C = 3.
1
d. Use (at) = jaj
(t) to write (4t + 3) = 14 t + 34 . The integral then becomes
1 3
I = 14 e4(3=4) + tan 10 34
= 4 e + tan (7:5) = 9:277 1013 .
e. Use property 5 of the unit impulse function to get
d2
d
Ie = (1)2 2 cos 5t + e3t t=2 =
5 sin 5t 3e3t t=2
dt
dt
h
i
2
3t
= (5) cos 5t + 9e
= (5)2 cos 10 + 9e6 = 246:73
t=2
10
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Problem 2.7
(a), (c), and (e) are periodic. Their periods are 2 s (fundamental frequency of 0.5 Hz),
2 s, and 3 s, respectively. The waveform of part (c) is a periodic train of triangles,
each 2 units wide, extending from -1 to 1 spaced by 2 s ((b) is similar except that it
is zero for t < 1 thus making it aperiodic). Waveform (d) is aperiodic because the
frequencies of its two components are incommensurable. The waveform of part (e) is a
doubly-inÖnite train of square pulses, each of which is one unit high and one unit wide,
centered at ; 6; 3; 0; 3; 6; . Waveform (f) is identical to (e) for t 1=2 but 0
for t < 1=2 thereby making it aperiodic.
Problem 2.8
a. The result is
h
i
j6t
j(10t=2)
j6t
j(10t=2)
x(t) = cos (6t)+2 cos (10t =2) = Re e
+Re 2e
= Re e
+ 2e
b. The result is
1
1
x(t) = ej(10t=2) + ej6t + ej6t + ej(10t=2)
2
2
c. The single-sided amplitude spectrum consists of lines of height 1 and 2 at frequencies
of 3 and 5 Hz, respectively. The single-sided phase spectrum consists of a line of height
=2 at frequency 5 Hz. The double-sided amplitude spectrum consists of lines of
height 1, 1/2, 1/2, and 1 at frequencies of 5; 3; 3; and 5 Hz, respectively. The
double-sided phase spectrum consists of lines of height =2 and =2 at frequencies
of 5 and 5 Hz, respectively.
Problem 2.9
a. Power. Since it is a periodic signal, we obtain
Z T0
Z T0
1
1
2
P1 =
4 cos (4t + 2=3) dt =
2 [1 + cos (8t + 4=3)] dt = 2 W
T0 0
T0 0
where T0 = 1=2 s is the period. The cosine in the above integral integrates to zero because
the interval of integratation is two periods.
b. Energy. The energy is
E2 =
Z 1
1
2t 2
e
u (t)dt =
Z 1
0
e2t dt =
1
J
2
2.1. PROBLEM SOLUTIONS
11
c. Energy. The energy is
E3 =
Z 1
2t 2
e
u (t)dt =
Z 0
e2t dt =
1
1
1
J
2
d. Energy. The energy is
E4 =
lim
T !1
Z T
dt
1
= lim 2
2 + t2 )
T
!1
(
T
Z T
T
dt
1 + (t=)2
T
1
1 1
1 t
= lim
tan
= lim
tan (T =) tan1 (T =)
T !1
T T !1
1 h i
=
= J
2
2
e. Energy. Since it is the sum of x2 (t) and x3 (t), its energy is the sum of the energies
of these two signals, or E5 = 1= J.
f. Energy. The energy is
E6 =
=
=
lim
Z T h
T !1 T
Z T h
lim
T !1 T
lim
Z T
i2
et u (t) e(t1) u (t 1) dt
i
e2t u2 (t) et e(t1) u (t) u (t 1) + e2(t1) u2 (t 1) dt
2t
0
dt e
Z T
2(t1)
1
e
dt +
Z T
2(t1)
e
dt
Z T
Z T 1
2t
2t0 0
2t0 0
= lim
e
dt e
e
dt +
e
dt
T !1
0
0
8 0
9
T 1
T 1
< e2t T
=
2t0
2t0
e
e
+ e
= lim
;
T !1 :
2 0
2
2
0
0
1
e
1
1
1
=
+
=
1 e J
2
2
2
2
T !1
e
1
Z T 1
12
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Problem 2.10
a. Power. Since the signal is periodic with period 2=!, we have
!
P =
2
Z 2=!
0
!
A jsin (!t + )j dt =
2
2
2
Z 2=!
0
A2
A2
f1 cos [2 (!t + )]g dt =
W
2
2
b. Neither. The energy calculation gives
Z T
Z T
(A )2 dt
(A )2 dt
p
p
p
E = lim
dt = lim
dt ! 1
T !1 T
T !1 T
+ jt jt
2 + t2
The power calculation gives
1
P = lim
T !1 2T
Z T
(A )2 dt
(A )2
p
dt = lim
ln
T !1 2T
2 + t2
T
c. Energy:
E=
Z 1
0
1
A t exp (2t= ) dt = A2
8
2 2
r
p
1 + T 2 = 2
p
1 + 1 + T 2 = 2
1+
!
=0W
W (use a table of integrals)
2
d. Energy: This is a "top hat" pulse which is height 2 for jtj =2, height 1 for
=2 < jtj , and 0 everywhere else. Making use of the even symmetry about t = 0,
the energy is
!
Z =2
Z
E=2
22 dt +
12 dt = 5 J
0
=2
e. Energy. The signal is a "house" two units wide and one unit up to the eves with a
equilateral triangle for a roof. Because of symmetry, the energy calculation need be
carried out for positive t and doubled. The calculation is
1
Z 1
2
2 28
14
2
3
E=2
(2 t) dt = (2 t) = +
=
J
3
3
3
3
0
0
f. Power. Since the two terms are harmonically related, we may add their respective
powers and get
A2 B 2
P =
+
W
2
2
2.1. PROBLEM SOLUTIONS
13
Problem 2.11
a. Using the fact that the power contained in a sinusoid is its amplitude squared divided
by 2, we get
22
P =
=2W
2
b. This is a periodic train of "box cars" 3 units high, 2 units wide, and occurring every
4 units (period of 4 seconds). The power calculation is
Z
1 1 2
32 2
P =
3 dt =
= 4:5 W
4 1
4
c. This is a train of triangles 1 unit high, 4 s wide, and occuring every 6 s. Using the
waveform period centered at 0, the power calculation is
2
Z
1 2
t 2
12
t 3
2
P =
1
dt =
1
= W
6 2
2
63
2
9
0
d. This is a train of "houses" each of which is 2 s wide, 1 unit high to the eves, with an
isoceles triangle on top for the roof. They are separated by 4 s (the period). Using
the even symmetry of each house, the power calculation is
1
Z
2 1
1 (2 t)3
1 1 23
7
2
Pd =
(2 t) dt =
= W
=
4 0
2
3
2 3
3
6
0
Problem 2.12
a. The energy is
Z 1
Z 1
(3+j4)t 2
E =
e(3+j4)t e(3j4)t dt
6e
dt = 36
0
0
Z 1
6t 1
e
=6J
= 36
e6t dt = 36
6 0
0
The power is 0 W.
b. This signal is a "top hat" pulse which is 2 for 2 t 4, 1 for 0 t < 2 and 4 < t 6,
and 0 everywhere else. It is clearly an energy signal with energy
E = 2 12 + 2 22 + 2 12 = 12 J
14
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Its power is 0 W.
c. This is a power signal with power
Z T
Z
1
49 T
49
j6t j6t
P = lim
49e
e
u (t) dt = lim
dt =
= 24:5 W
T !1 2T T
T !1 2T 0
2
Its energy is inÖnite.
2
d. This is a periodic signal with power P = 22 = 2 W. Its energy is inÖnite.
e. This is neither an energy nor a power signal. Its energy is inÖnite and its power is
T
Z T
1
1 t3
1 2T 3
2
P = lim
t dt = lim
=
lim
!1
T !1 2T T
T !1 2T 3 T
T !1 2T 3
f. This is neither an energy nor a power signal. Its energy is
Z 1
E=
t1 dt = ln (t)j1
1 !1
1
and its power is
1
P = lim
T !1 2T
Problem 2.13
Z T
1
ln (t)j1
1 =0
T !1 2T
t1 dt = lim
1
a. This is a cosine burst from t = 6 to t = 6 seconds. The energy is E1 =
R6
2 0 12 + 12 cos (12t) dt = 6 J
R6
6 cos
b. The energy is
E2 =
Z 1h
jtj=3
e
1
i2
dt = 2
1
= 2
=3J
2=3
e2t=3
Z 1
e2t=3 dt (by even symmetry)
0
0
Since the result is Önite, this is an energy signal.
c. The energy is
E3 =
Z 1
1
2
f2 [u (t) u (t 8)]g dt =
Z 8
0
4dt = 32 J
2 (6t) dt =
2.1. PROBLEM SOLUTIONS
15
Since the result is Önite, this is an energy signal.
d. Note that
r (t) ,
Z t
u () d =
1
0; t < 0
t; t 0
which is called the unit ramp. Thus the given signal is a triangle between 0 and 20. The
energy is
Z 1
Z 10
2 10 2000
2
E4 =
[r (t) 2r (t 10) + r (t 20)] dt = 2
t2 dt = t3 0 =
J
3
3
1
0
where the last integral follows because the integrand is a symmetrical triangle about t = 10.
Since the result is Önite, this is an energy signal.
Problem 2.14
a. This is a cosine burst nonzero between 0 and 2 seconds. Its power is 0. Its energy is
Z 2
Z
1 2
2
E1 =
cos (10t) dt =
[1 + cos (20t)] dt = 1 J
2 0
0
b. This is a periodic sequence of triangles of period 3 s. Its energy is inÖnite. Its power
is
Z
2 2
4
P2 =
(1 t=2)2 dt = J
3 0
9
c. This is an energy signal. Its power is 0. Using evenness of the integrand, its energy
is
Z 1
Z 1
Z 1
2t
2
2t
E3 = 2
e cos (2t) dt =
e dt +
e2t cos (4t) dt
0
=
0
0
1
2
+
J
2 4 + 16 2
.
d. This is an energy signal. Its energy is
1
14
2
3
E4 = 2
(2 t) dt = (2 t) =
J
3
3
0
0
Z 1
2
16
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Problem 2.15
a. Use the exponential representation of the sine to get the Fourier coe¢cients as (note
that the period = 12 f10 ).
j2f0 t
2
e
ej2f0 t
1
x1 (t) =
= ej4f0 t 2 + ej4f0 t
2j
4
from which we Önd that
1
1
X1 = X1 = ; X0 =
4
2
All other coe¢cients are zero.
b. Use the exponential representations of the sine and cosine to get
1
1
1
1
x2 (t) = ej2f0 t + ej2f0 t + ej4f0 t ej4f0 t
2
2
2j
2j
Therefore, the Fourier coe¢cients for this case are
X1 = X1 =
1
1
and X2 = X2
=
2
2j
All other coe¢cients are zero.
c. Use a trig identity to write this signal as
x3 (t) =
1
1
1
sin 8f0 t = ej8f0 t ej8f0 t
2
4j
4j
The fundamental frequency is 4f0 Hz. From this it follows that the Fourier coe¢cients are
X1 = X1
=
1
4j
All other coe¢cients are zero.
d. Use trig identities and the exponential forms of cosine to write this signal as
x4 (t) =
=
3
1
cos 2f0 t + cos 6f0 t
4
4
3 j2f0 t 3 j2f0 t 1 j6f0 t 1 j6f0 t
+ e
+ e
+ e
e
8
8
8
8
2.1. PROBLEM SOLUTIONS
17
The fundamental frequency is f0 Hz. It follows that the Fourier coe¢cients for this case
are
3
1
X1 = X1 = ; X3 = X3 =
8
8
All other Fourier coe¢cients are zero.
e. Use trig identies to write
x5 (t) =
=
1
1
1
sin (2f0 t) sin (6f0 t) + sin (10f0 t)
2
4
4
1 j2f0 t
1 j2f0 t
1 j6f0 t
1
1
1
e
e
e
+ ej6f0 t + ej10f0 t ej10f0 t
4j
4j
8j
8j
8j
8j
The fundamental frequency is f0 Hz. It follows that the Fourier coe¢cients for this case
are
j
j
j
X1
= X1 = ; X3
= X3 = ; X5
= X5 =
4
8
8
All other Fourier coe¢cients are zero.
f. Use trig identities to write
x6 (t) =
=
1
1
1
cos (6f0 t) cos (f0 t) cos (11f0 t)
2
4
4
1 j6f0 t 1 j6f0 t 1 jf0 t 1 jf0 t 1 j11f0 t 1 j11f0 t
e
+ e
e
e
e
e
4
4
8
8
8
8
The fundamental frequency is f0 =2 Hz. It follows that the Fourier coe¢cients for this case
are
1
1
1
X1
= X1 = ; X12
= X12 = ; X22
= X22 =
8
4
8
All other Fourier coe¢cients are zero.
Problem 2.16
The expansion interval is T0 = 4 so that f0 = 1=4 Hz. The Fourier coe¢cients are
Z
Z
1 2 2 jn(=2)t
2 2 2
Xn =
2t e
dt =
t (cos nt=2 j sin nt=2) dt =
4 2
4 2
Z
2 2 2
nt
=
2t cos
dt
4 0
2
which follows by the oddness of the second integrand and the eveness of the Örst integrand.
Let u = nt=2 to obtain the form
3 Z n
2
16
Xn =
u2 cos u du =
(1)n n 6= 0 (use a table of integrals)
2
n
(n)
0
18
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
If n = 0, the integral for the coe¢cients is
1
X0 =
4
Z 2
2t2 dt =
2
8
3
The Fourier series is therefore
8
x (t) = +
3
1
X
(1)n
n=1; n6=0
16 jn(=2)t
e
(n)2
Problem 2.17
Parts (a) through (c) were discussed in the text. For (d), break the integral for Xn up into
a part for t < 0 and a part for t > 0. Then use the odd half-wave symmetry condition.
The development follows:
Xn =
=
=
1
T0
1
T0
1
T0
1
=
T0
(
=
"Z
T0 =2
x (t) ej2nf0 t dt +
T0 =2
x (t) ej2nf0 t dt +
0
"Z
j2nf0 t
x (t) e
0
"Z
0
Z T0 =2
0
T0 =2
T0 =2
j2nf0 t
x (t) e
dt
#
x (t) ej2nf0 t dt
T0 =2
0
"Z
Z 0
Z T0 =2
0
n
dt (1)
0; n even
R T0 =2
2
x (t) ej2nf0 t dt; n odd
T0 0
#
0
j2nf0 (t0 +T0 =2)
x t + T0 =2 e
dt ; t0 = t + T0 =2
#
0 j2nf0 t0 jn
x t e
dt ; f0 = 1=T0
Z T0 =2
0
0
x t0 ej2nf0 t dt
#
2.1. PROBLEM SOLUTIONS
19
Problem 2.18
This is a matter of integration. Only the solution for part (b) will be given here.
integral for the Fourier coe¢cients is
Xn =
=
=
=
=
=
Z T0 =2
j!0 t
A
sin (! 0 t) e
dt =
e
ej!0 t ejn!0 t dt
2jT0 0
0
"Z
#
Z T0 =2
T0 =2
A
ej(1n)!0 t dt
ej(1+n)!0 t dt
2jT0 0
0
2
T0 =2
T0 =2 3
j(1n)!
t
j(1+n)!
t
0
0
A 4 e
e
5
2jT0 j (1 n) ! 0
j (1 + n) ! 0
0
0
"
#
A
ej(1n) 1 ej(1+n) 1
+
; n 6= 1 (! 0 T0 =2 = )
4
1n
1+n
A (1)n + 1 (1)n + 1
+
; n 6= 1
4
1n
1+n
(
0; n odd and n 6= 1
A
; n even
(1n2 )
A
T0
Z T0 =2
The
jn! 0 t
For n = 1, the integral is
X1 =
A
2jT0
=
A
2jT0
Z T0 =2
0
Z T0 =2
0
j!0 t
e
ej!0 t ej!0 t dt
jA
1 ej2!0 t dt =
= X1
4
This is the same result as given in Table 2.1.
Problem 2.19
a. Use Parsevalís theorem to get
Pjnf0 j 1= =
N
X
n=N
2
jXn j =
N
X
A 2
n=N
T0
sinc2 (nf0 )
where N is an appropriately chosen limit on the sum. We are given that only frequences for
which jnf0 j 1= are to be included. This is the same as requiring that jnj 1= ( f0 ) =
20
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
T0 = = 2: Also, for a pulse train, Ptotal = A2 =T0 and, in this case, Ptotal = A2 =2: Thus
2
Pjnf0 j 1=
2 X A 2
=
sinc2 (nf0 )
Ptotal
A2
2
n=2
2
=
1 X
sinc2 (nf0 )
2
n=2
=
=
1
1 + 2 sinc2 (1=2) + sinc2 (1)
2"
2 #
1
2
1+2
= 0:9053
2
b. In this case, jnj 5, Ptotal = A2 =5, and
Pjnf0 j 1=
Ptotal
5
1 X
=
sinc2 (n=5)
5
n=5
h
io
1n
=
1 + 2 (0:9355)2 + (0:7568)2 + (0:5046)2 + (0:2339)2
5
= 0:9029
c. In this case, jnj 10, Ptotal = A2 =10, and
Pjnf0 j 1=
Ptotal
10
1 X
sinc2 (n=10)
10
n=10
1
(0:9836)2 + (0:9355)2 + (0:8584)2 + (0:7568)2 + (0:6366)2
=
1+2
+ (0:5046)2 + (0:3679)2 + (0:2339)2 + (0:1093)2
10
= 0:9028
=
d. In this case, jnj 20, Ptotal = A2 =20, and
Pjnf0 j 1=
Ptotal
=
=
20
1 X
sinc2 (n=20)
20
n=20
(
)
20
X
1
1+2
sinc2 (n=20)
20
n=1
= 0:9028
2.1. PROBLEM SOLUTIONS
21
Problem 2.20
a. The integral for Yn is
Z
Z T0
1
1
jn! 0 t
Yn =
y (t) e
dt =
x (t t0 ) ejn!0 t dt; ! 0 = 2f0
T0 T0
T0 0
Let t0 = t t0 , which results in
Z T0 t0
0 jn!0 t0 0 jn!0 t0
1
Yn =
x t e
dt e
= Xn ej2nf0 t0
T0 t0
b. Note that
y (t) = A cos ! 0 t = A sin (! 0 t + =2) = A sin [! 0 (t + =2! 0 )]
Thus, t0 in the theorem proved in part (a) here is =2! 0 . By Eulerís theorem, a sine wave
can be expressed as
1
1
sin (! 0 t) = ej!0 t ej!0 t
2j
2j
1
1
Its Fourier coe¢cients are therefore X1 = 2j
and X1 = 2j
. According to the theorem
proved in part (a), we multiply these by the factor
ejn!0 t0 = ejn!0 (=2!0 ) = ejn=2
For n = 1, we obtain
Y1 =
For n = 1, we obtain
1 j=2 1
e
=
2j
2
1 j=2 1
e
=
2j
2
which gives the Fourier series representation of a cosine wave as
Y1 =
1
1
y (t) = ej!0 t + ej!0 t = cos ! 0 t
2
2
We could have written down this Fourier representation directly by using Eulerís theorem.
Problem 2.21
a. Use the Fourier series of a square wave (specialize the Fourier series of a pulse train)
with A = 1 and t = 0 to obtain the series
4
1 1 1
1=
1 + +
3 5 7
22
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Multiply both sides by 4 to get the series in the problem statement. Hence, the sum is 4 .
b. Use the Fourier series of a triangular wave as given in Table 2.1 with A = 1 and t = 0
to obtain the series
1=+
4
4
4
4
4
4
+
+
+
+
+
+
25 2 9 2 2 2 9 2 25 2
2
2
Multiply both sides by 8 to get the series in given in the problem. Hence, its sum is 8 .
Problem 2.22
a. In the expression for the Fourier series of a pulse train (Table 2.1), let t0 = T0 =8
and = T0 =4 to get
n
A
nf0
Xn = sinc
exp j
4
4
4
The spectra are shown in Fig. 2.4.
b. The amplitude spectrum is the same as for part (a) except that X0 = 3A
4 . Note that
this can be viewed as having a sinc-function envelope with zeros at multiples of 3T4 0 .
The phase spectrum can be obtained from that of part (a) by subtracting a phase shift
of for negative frequencies and adding for postitive frequencies (or vice versa).
The Fourier coe¢cients are given by
3A
3n
3nf0
Xn =
sinc
exp j
4
4
4
See Fig. 2.4 for amplitude and phase plots.
Problem 2.23
a. Use the rectangular pulse waveform of Table 2.1 specialized to
t T0 =4
xa (t) = 2A
A; jtj < T0 =2
T0 =2
and periodically extended. Hence, from Table 2.1, we have
2AT0 =2
nT0 =2
2nT0 =4
A
Xn =
sinc
exp j
T0
T0
T0
= Asinc (n=2) exp (jn=2) ; n 6= 0
sin (n=2)
= A
exp (jn=2) ; n 6= 0
n=2
2.1. PROBLEM SOLUTIONS
23
Part (a)
Part (b)
0.6
Amplitude
Amplitude
0.6
0.4
0.2
0
-5
0
f, Hz
0.2
0
5
-5
0
f, Hz
5
-5
0
f, Hz
5
2
Phase, rad
2
Phase, rad
0.4
0
-2
0
-2
-5
0
f, Hz
5
24
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
where the superscript A refers to xA (t). The dc component is 0 so X0A = 0. The Fourier
coe¢cients are therefore
X0A = 0
A
X1A = j2A=; X1
= j2A=
A
X2A = 0 = X2
A
X3A = j2A=3; X3
= j2A=3
b. Note that
xA (t) =
dxB (t)
dt
where A = T2B
= 4B
T0 obtained from matching the amplitude of xA (t) with the slope of
0 =2
xB (t) or B = T40 A. The relationship between spectral components is therefore
XnA = (jn! 0 ) XnB = j2nf0 XnB
or
XnB =
XnA
T0 XnA
=
j2nf0
j2n
where the superscript A refers to xA (t) and B refers to xB (t). For example,
X1B =
j2A=
2B
B
= 2 = X1
j2f0
Problem 2.24
a. This is a decaying exponential starting at t = 0 and its Fourier transform is
Z 1
Z 1
X1 (f ) = A
et= ej2f t dt = A
e(1= +j2f )t dt
0
0
1
(1=
+j2f
)t
Ae
A
=
=
1= + j2f
1= + j2f
0
=
A
1 + j2f
2.1. PROBLEM SOLUTIONS
25
b. Since x2 (t) = x1 (t) we have, by the time reversal theorem, that
X2 (f ) = X1 (f ) = X1 (f )
A
=
1 j2f
c. Since x3 (t) = x1 (t) x2 (t) we have, after some simpliÖcation, that
X3 (f ) = X1 (f ) X2 (f )
A
A
=
1 + j2f
1 j2f
j4Af
=
1 + (2f )2
d. Since x4 (t) = x1 (t) + x2 (t) we have, after some simpliÖcation, that
X4 (f ) = X1 (f ) + X2 (f )
A
A
=
+
1 + j2f
1 j2f
2A
=
1 + (2f )2
This is the expected result since x4 (t) is really a double-sided decaying exponential.
e. By part a and the delay theorem
X5 (f ) = X1 (f ) ej10f =
A ej10f
1 + j2f
f. By parts a and e and superposition
h
j10f
X6 (f ) = X1 (f ) 1 e
i
A 1 ej10f
=
1 + j2f
Problem 2.25
a. Using a table of Fourier transforms and the time reversal theorem, the Fourier transform of the given signal is
X (f ) =
1
1
+ j2f
j2f
26
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Note that x (t) ! sgn(t) in the limit as ! 0.
transform as ! 0, we deduce that
F [sgn (t)] =
Taking the limit of the above Fourier
1
1
1
=
j2f
j2f
jf
b. Using the given relationship between the unit step and the signum function and the
linearity property of the Fourier transform, we obtain
F [u (t)] =
=
1
1
F [sgn (t)] + F [1]
2
2
1
1
+ (f )
j2f
2
c. The same result as obtained in part (b) is obtained.
Problem 2.26
a. One di§erentiation gives
dxa (t)
= (t 0:5) (t 2:5)
dt
Two di§erentiations give
d2 xa (t)
= (t) (t 1) (t 2) + (t + 3)
dt2
Application of the di§erentiation theorem of Fourier transforms gives
(j2f )2 Xa (f ) = 1 1 ej2f 1 ej4f + 1 ej6f
=
ej3f ejf ejf + ej3f ej3f
= 2 (cos 3f cos f ) ej3f
where the time delay theorem and the Fourier transform of a unit impulse have been used.
Dividing both sides by (j2f )2 , we obtain
Xa (f ) =
cos f cos 3f j3f
2 (cos 3f cos f ) ej3f
=
e
2
2 2 f 2
(j2f )
2.1. PROBLEM SOLUTIONS
27
Use the trig identity sin2 x = 12 12 cos 2x or cos 2x = 1 2 sin2 x to rewrite this result as
Xa (f ) =
=
=
=
=
1 2 sin2 (0:5f ) 1 + 2 sin2 (1:5f ) j3f
e
2 2 f 2
sin2 (0:5f ) + sin2 (1:5f ) j3f
e
2f 2
sin2 (1:5f ) j3f sin2 (0:5f ) j3f
e
e
2f 2
2f 2
"
2
2 #
sin
1:5f
sin
0:5f
1:52
0:52
ej3f
1:5f
0:5f
2
1:5 sinc2 (1:5f ) 0:52 sinc2 (0:5f ) ej3f
This is the same result as would have been obtained by writing
t 1:5
t 1:5
xa (t) = 1:5
0:5
1:5
0:5
and using the Fourier transform of the triangular pulse along with the superposition and
time delay theorems.
b. Two di§erentiations give (sketch dxb (t) =dt to see this)
d2 xb (t)
= (t) 2 (t 1) + 2 (t 3) (t 4)
dt2
Application of the di§erentiation theorem gives
(j2f )2 Xb (f ) = 1 2ej2f + 2ej6f ej8f
Dividing both sides by (j2f )2 , we obtain
Xb (f ) =
1 2ej2f + 2ej6f ej8f
4 2 f 2
Further manipulation may be applied to this result to convert it to
h
i
Xb (f ) = sinc2 (f ) ej2f ej6f
h
i
= sinc2 (f ) ej2f ej2f ej4f
= 2j sin (2f ) sinc2 (f ) ej4f
which would have resulted from Fourier transforming the waveform written as
xb (t) = (t 1) (t 3)
28
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
c. Two di§erentiations give (sketch dxc (t) =dt to see this)
d2 xc (t)
= (t) 2 (t 1) + 2 (t 2) 2 (t 3) + (t 4)
dt2
Application of the di§erentiation theorem gives
(j2f )2 Xc (f ) = 1 2ej2f + 2ej4f 2ej6f + ej8f
Dividing both sides by (j2f )2 , we obtain
Xc (f ) =
1 2ej2f + 2ej4f 2ej6f + ej8f
(j2f )2
This result may be further arranged to give
2
h
j2f
Xc (f ) = sinc (f ) e
j6f
+e
i
= 2 cos (2f ) sinc2 (f ) ej4f
which would have resulted from Fourier transforming the waveform written as
xc (t) = (t 1) + (t 3)
d. Two di§erentiations give (sketch dxd (t) =dt to see this)
d2 xd (t)
= (t) 2 (t 1) + (t 1:5) + (t 2:5) 2 (t 3) + (t 4)
dt2
Application of the di§erentiation theorem gives
(j2f )2 Xd (f ) = 1 2ej2f + ej3f + ej5f 2ej6f + ej8f
Dividing both sides by (j2f )2 , we obtain
Xd (f ) =
1 2ej2f + ej3f + ej5f 2ej6f + ej8f
(j2f )2
This result may be further arranged to give
h
i
Xd (f ) = sinc2 (f ) ej2f + 0:5ej4f + ej6f
which would have resulted from Fourier transforming the waveform written as
xd (t) = (t 1) + 0:5 (t 2) + (t 3)
2.1. PROBLEM SOLUTIONS
29
Problem 2.27
See the solutions to Problem 2.26.
Problem 2.28
a. The steps in Önding the Fourier transform for (i) are as follows:
(t)
(t) exp [j4t]
(t 1) exp [j4 (t 1)]
! sinc (f )
! sinc (f 2)
! sinc (f 2) exp (j2f )
The steps in Önding the Fourier transform for (ii) are as follows:
(t)
(t) exp [j4t]
(t + 1) exp [j4 (t + 1)]
! sinc (f )
! sinc (f 2)
! sinc (f 2) exp (j2f )
b. The steps in Önding the Fourier transform for (i) are as follows:
(t)
! sinc (f )
(t 1)
! sinc (f ) exp (j2f )
(t 1) exp [j4 (t 1)]
=
(t 1) exp (j4t)
! sinc (f 2) exp [j2 (f 2)] = sinc (f 2) exp (j2f )
which follows because exp(jn2) = 1 where n is an integer. The steps in Önding the
Fourier transform for (ii) are as follows:
(t)
(t + 1)
(t + 1) exp [j4 (t + 1)]
! sinc (f )
! sinc (f ) exp (j2f )
=
(t + 1) exp (j4t)
! sinc (f 2) exp [j2 (f 2)] = sinc (f 2) exp (j2f )
30
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Problem 2.29
a. The time reversal theorem states that x (t)
so
xa (t)
!
=
! X (f ) 6= X (f ) if x (t) is complex,
1
1
1
1
X1 (f ) + X1 (f ) = sinc (f 2) exp (j2f ) + sinc (f 2) exp (j2f )
2
2
2
2
1
1
sinc (f 2) exp (j2f ) + sinc (f + 2) exp (j2f )
2
2
Note that
xa (t) =
=
1
1
(t 1) exp [j4 (t 1)] + (t 1) exp [j4 (t 1)]
2
2
1
1
(t 1) exp (j4t) + (t + 1) exp (j4t) (by the eveness of (u) )
2
2
b. Similarly to (a), we obtain
xb (t)
1
1
1
1
! X2 (f )+ X2 (f ) = sinc (f 2) exp (j2f )+ sinc (f + 2) exp (j2f )
2
2
2
2
2.1. PROBLEM SOLUTIONS
31
Problem 2.30
a. The result is
X1 (f ) = 2sinc (2f ) exp (j2f )
b. The result is
1
X2 (f ) = 2
2
f
exp (j2f )
2
c. The result is
X3 (f ) = 8sinc2 (8f ) exp (j4f )
d. The result is
X4 (f ) = 4 (4f ) exp (j6f )
e. The result is
1
f
1
f
X5 (f ) = 5
exp (j2f ) + 5
exp (j2f )
2
2
2
2
f
= 5
cos (2f )
2
f. The result is
X6 (f ) = 16sinc2 (8f ) exp (j4f ) + 16sinc2 (8f ) exp (j4f )
= 32sinc2 (8f ) cos (4f )
Problem 2.31
a. This is an odd signal, so its Fourier transform is odd and purely imaginary.
b. This is an even signal, so its Fourier transform is even and purely real.
c. This is an odd signal, so its Fourier transform is odd and purely imaginary.
d. This signal is neither even nor odd, so its Fourier transform is complex.
e. This is an even signal, so its Fourier transform is even and purely real.
f. This signal is even, so its Fourier transform is real and even.
32
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Problem 2.32
In the Poisson sum formula, we identify p (t) = (t=2) which has Fourier transform P (f ) =
2sinc(2f ). Thus, for this case, the Poisson sum formula becomes
1
X
m=1
p (t mTs ) =
or
1
X
m=1
1
X
m=1
t 4m
2
t 4m
2
1
1
X
1 X
2n j2(n=4)t
=
2sinc
e
= fs
P (nfs ) ej2nfs t
4 n=1
4
n=1
1
n
X
1
=
sinc
ej(n=2)t
2
2
n=1
The fundamental
are X0 = 1=2; X1 =
frequency is 0.25 Hz and the Fouriercoe¢cients
1
X1 = 12 sinc 12 = 1 , X2 = X2 = 0, X3 = X3 = 12 sinc 32 = 3
, etc.
Problem 2.33
a. The Fourier transform of this signal is
X1 (f ) =
10
2
=
5 + j2f
1 + j2f =5
Thus, the energy spectral density is
G1 (f ) =
4
1 + (2f =5)2
b. The Fourier transform of this signal is
X2 (f ) = 5
f
2
Thus, the energy spectral density is
f
f
X2 (f ) = 25
= 25
2
2
2
c. The Fourier transform of this signal is
3
X3 (f ) = sinc
2
so the energy spectral density is G3 (f ) = 94 sinc2
f
2
f
2
2.1. PROBLEM SOLUTIONS
33
d. The Fourier transform of this signal is
3
f 5
f +5
X4 (f ) =
sinc
+ sinc
4
2
2
so the energy spectral density is
9
f 5
f +5 2
G4 (f ) =
sinc
+ sinc
16
2
2
Problem 2.34
a. Use the transform pair
x1 (t) = et u (t)
!
1
+ j2f
Using Rayleighís energy theorem, we obtain the integral relationship
Z 1
Z 1
Z 1
Z 1
df
1
2
2
jX1 (f )j df =
jx1 (t)j dt =
e2t dt =
2 df =
2
2
1
1 + (2f )
1
0
b. Use the transform pair
1
x2 (t) =
Rayleighís energy theorem gives
Z 1
jX2 (f )j2 df
=
1
t
Z 1
1
=
Z 1
! sinc ( f ) = X2 (f )
2
sinc ( f ) df =
1 2
2
1
Z 1
1
jx2 (t)j2 dt
Z =2
t
dt
1
dt =
=
2
=2
c. Use the transform pair
ejtj
!
2
ejtj
or
2
2
2 + (2f )
!
1
2 + (2f )2
34
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
The desired integral, by Rayleighís energy theorem, is
I3 =
=
d. Use the transform pair
Z 1
2
Z 1 2jtj
1
e
df
=
dt
2
2
2
1 + (2f )
1 4
1
Z 1
2
1 e2t
1
2t
e
dt =
=
42 0
22 2 0
43
1
t
! sinc2 ( f )
The desired integral, by Rayleighís energy theorem, is
Z 1
Z 1
2
I4 =
jX4 (f )j df =
sinc4 ( f ) df
1
1
Z 1
Z
1
2
2
=
(t= ) dt = 2
[1 (t= )]2 dt
2 1
0
Z
1
2 1
2
1
2
2
2
=
(1 u) du =
1u
=
0
3
3
0
Problem 2.35
a. The convolution operation gives
8
t 1=2
< 0;
1
(t +1=2) ; 1=2 < t + 1=2
1
e
y1 (t) =
: 1 (t 1=2)
(t +1=2) ;
e
e
t > + 1=2
b. The convolution of these two signals gives
y2 (t) = (t) + trap (t)
where trap(t) is a trapezoidal function given by
8
0;
t < 3=2 or t > 3=2
>
>
<
1;
1=2 t 1=2
trap (t) =
3=2 + t; 3=2 t < 1=2
>
>
:
3=2 t;
1=2 t < 3=2
2.1. PROBLEM SOLUTIONS
35
c. The convolution results in
Z 1
Z t+1=2
jj
y3 (t) =
e
( t) d =
ejj d
t1=2
1
Sketches of the integrand for various values of t give the following cases:
8 R
t+1=2
>
e d;
t 1=2
>
< t1=2
R0
R t+1=2
y3 (t) =
e
d; 1=2 < t 1=2
t1=2 e d + 0
>
>
: R t+1=2 e d;
t > 1=2
t1=2
Integration of these three cases gives
8 1 (t+1=2)
e(t1=2) ;
t 1=2
< e
1
(t1=2)
(t+1=2)
y3 (t) =
e
e
; 1=2 < t 1=2
: 1 (t1=2)
(t+1=2) ;
e
e
t > 1=2
d. The convolution gives
y4 (t) =
Z t
x () d
1
Problem 2.36
a. The inverse FT of (f ) is sinc(t). By the time delay theorem, the inverse Fourier
transform of (f ) exp (j4f ) is sinc(t 2). The product of this and 2 cos (2f ) in
the frequency domain has an inverse Fourier transform which is the convolution of
their respective Fourier transforms. Thus
x1 (t) = sinc (t 2) [ (t 1) + (t + 1)]
= sinc (t 3) + sinc (t 1)
b. The inverse Fourier transform of (f =2) is 2sinc2 (2t). By the time delay theorem
x2 (t) = 2sinc2 [2 (t 2:5)]
c. The inverse Fourier transform of
By the modulation theorem,
(f =2)
is 2sinc(2t).
f +4
f 4
the inverse Fourier transform of 2 + 2
is sinc(2t) cos (8t). By the time
delay theorem
x3 (t) = sinc [2 (t 4)] cos [2 (t 4)]
36
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Problem 2.37
1
a. From before, the total energy is E1; total = 2
. The Fourier transform of the given
signal is
1
X1 (f ) =
+ j2f
so that the energy spectral density is
G1 (f ) = jX1 (f )j2 =
1
2 + (2f )2
By Rayleighís energy theorem, the normalized inband energy is
E1 (jf j W )
= 2
E1; total
Z W
df
2
= tan1
2
2
W + (2f )
2W
b. The total energy is E2; total = . The Fourier transform of the given signal and its
energy spectral density are, respectively,
X2 (f ) = sinc (f ) and G2 (f ) = jX2 (f )j2 = 2 sinc2 (f )
By Rayleighís energy theorem, the normalized inband energy is
E2 (jf j W )
1
=
E2; total
Z W
2 sinc2 (f ) df = 2
W
Z W
sinc2 (u) du
0
The integration must be carried out numerically.
c. The total energy is
Z 1h
Z 1h
i2
i2
E3; total =
et et dt =
e2t 2e(+)t + e2t dt
0
=
0
1
2
1
( + ) 4 + ( + )
( )2
+
=
=
2 + 2
2 ( + )
2 ( + )
The Fourier transform of the given signal and its energy spectral density are, respectively,
X3 (f ) =
1
1
+ j2f
+ j2f
2.1. PROBLEM SOLUTIONS
and
37
2
1
1
jX3 (f )j =
+ j2f
+ j2f
1
1
1
1
2 Re
+ 2
2
+ j2f j2f
2 + (2f )
+ (2f )2
1
j2f
+ j2f
1
2 2 Re
2 2
2 + 2
2
2
+ (2f )
+ (2f ) + (2f )
+ (2f )2
2
3
2
1
+ (2f ) 5
1
4 ( )j2f
+ 2
2 2 Re
2
2
2
2
+ (2f )
+ (2f )2
2 + (2f )
+ (2f )
2
G3 (f ) =
=
=
=
1
+ (2f )2
1
+ 2
2
2
2
2
2
2
+ (2f )
+ (2f )2
2 + (2f )
+ (2f )
=
The normalized inband energy is
E3 (jf j W )
1
=
E3; total
E3; total
Z W
G3 (f ) df
W
The Örst and third terms may be integrated easily as inverse tangents. The second term
may be integrated after partial fraction expansion:
where
2
+ (2f )2
A
B
=
2 + 2
2
2
2
2
+ (2f )
+ (2f )2
2 + (2f )
+ (2f )
A=2
Therefore
E3 (jf j W )
E3; total
=
=
=
=
=
1
A
B
1
+
df
E3; total W 2 + (2f )2 2 + (2f )2 2 + (2f )2 2 + (2f )2
" 1
#
1 2W A tan1 2W
1
tan
1 2W + 1 tan1 2W
E3; total B
tan
1
1
1
1 2W
1 2W
(1 A) tan
+ (1 B) tan
E3; total
1
1
2W
1
2W
tan1
tan1
E3; total +
2
2W
2W
tan1
tan1
( )
1
Z W
2
2
and
B
=
2
2 2
2 2
38
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Pulse
Fraction of total energy
1
E1/E1,tot
x 1(t)
1
0.5
0
-2
0
2
0.5
0
4
0
2
αt
E2/E2,tot
x 2(t)
0.5
0
2
0
4
0
2
6
4
6
τW
1
E3/E3,tot
1
x 3(t)
4
0.5
t/τ
β = 2α
0.5
0
-2
6
1
1
0
-2
4
W/ α
0
2
αt
4
0.5
0
0
2
W/ α
Plots of the signals and inband energy for all three cases are shown in Fig. 2.5
2.1. PROBLEM SOLUTIONS
39
Problem 2.38
a. By the modulation theorem
X (f ) =
=
AT0
T0
T0
sinc (f f0 )
+ sinc (f + f0 )
4
2
2
AT0
1 f
1 f
sinc
1 + sinc
+1
4
2 f0
2 f0
b. Use the superposition and modulation theorems to get
AT0
f
1
1 f
1 f
X (f ) =
sinc
+
sinc
2 + sinc
+2
4
2f0
2
2 f0
2 f0
c. In this case, p(t) = x(t) and P (f ) = X(f ) of part (a) and Ts = T0 : From part (a),
we have
AT0
n1
n+1
P (nf0 ) =
sinc
+ sinc
4
2
2
Using this in (2.149), we have the Fourier transform of the half-wave rectiÖed cosine waveform as
1
X
A
n1
n+1
X(f ) =
sinc
+ sinc
(f nf0 )
4
2
2
n=1
Note that sinc(x) = 0 for integer values of its argument and it is 1 for its argument 0. Also,
use sinc(1=2) = 2=; sinc(3=2) = 2=3, etc. to get
X (f ) =
A
A
A
(f ) + [ (f f0 ) + (f + f0 )] +
[ (f 2f0 ) + (f + 2f0 )]
4
3
A
[ (f 4f0 ) + (f + 4f0 )] +
15
Problem 2.39
Signals x1 (t), x2 (t), and x6 (t) are real and even. Therefore their Fourier transforms are
real and even. Signals x3 (t), x4 (t), and x5 (t) are real and odd. Therefore their Fourier
transforms are imaginary and odd. Using the Fourier transforms for a square pulse and a
triangle along with superposition, time delay, and scaling, the Fourier transforms of these
signals are the following.
a. X1 (f ) = 2sinc2 (2f ) + 2sinc(2f )
b. X2 (f ) = 2sinc(2f ) sinc2 (f )
40
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
c. X3 (f ) = sinc(f ) ejf sinc(f ) ejf = 2j sin (f )sinc(f )
d. X4 (f ) = sinc2 (f ) ej2f sinc2 (f ) ej2f = 2j sin (2f )sinc2 (f )
e. By duality sgn(t) ! j= (f ) so, by the convolution theorem of Fourier transforms,
X5 (f ) =sinc2 (f ) j= (f ). The convolution cannot be carried out in closed form,
but it is clear that the result is imaginary and odd.
f. By the modulation theorem X6 (f ) = 12 sinc2 (f 1) + 12 sinc2 (f + 1) which is real and
even.
Problem 2.40
a. Write
6 cos (20t) + 3 sin (20t) = R cos (20t )
= R cos cos (20t) + R sin sin (20t)
Thus
R cos = 6 cos (20t)
R sin = 3 sin (20t)
Square both equations, add, and take the square root to obtain
p
p
R = 62 + 32 = 45 = 6:7082
Divide the second equation by the Örst to obtain
tan = 0:5
= 0:4636 rad
So
x (t) = 3 +
p
45 cos (20t 0:4636)
Following Example 2.19, we obtain
Rx ( ) = 32 +
45
cos (20 )
2
b. Taking the Fourier transform of Rx ( ) we obtain
Sx (f ) = 9 (f ) +
45
[ (f 10) + (f + 10)]
4
2.1. PROBLEM SOLUTIONS
41
Problem 2.41
Use the facts that the power spectral density integrates to give total power, it must be even,
and contains no phase information.
a. The total power of this signal is 22 =2 = 2 watts which is distributed equally at the
frequencies 10 hertz. Therefore, by inspection we write
S1 (f ) = (f 10) + (f + 10) W/Hz
b. The total power of this signal is 32 =2 = 4:5 watts which is distributed equally at the
frequencies 15 hertz. Therefore, by inspection we write
S2 (f ) = 2:25 (f 15) + 2:25 (f + 15) W/Hz
c. The total power of this signal is 52 =2 = 12:5 watts which is distributed equally at the
frequencies 5 hertz. Therefore, by inspection we write
S3 (f ) = 6:25 (f 5) + 6:25 (f + 5) W/Hz
d. The power of the Örst component of this signal is 32 =2 = 4:5 watts which is distributed
equally at the frequencies 15 hertz. The power of the second component of this
signal is 52 =2 = 12:5 watts which is distributed equally at the frequencies 5 hertz.
Therefore, by inspection we write
S4 (f ) = 2:25 (f 15) + 2:25 (f + 15) + 6:25 (f 5) + 6:25 (f + 5) W/Hz
Problem 2.42
Since the autocorrelation function and power spectral density of a signal are Fourier transform pairs, we may write down the answers by inspection using the Fourier transform pair
A cos (2f0 t) ! A2 (f f0 ) + A2 (f + f0 ). The answers are the following.
a. R1 ( ) = 8 cos (30 ); Average power = R1 (0) = 8 W.
b. R2 ( ) = 18 cos (40 ); Average power = R2 (0) = 18 W.
c. R3 ( ) = 32 cos (10 ); Average power = R3 (0) = 32 W.
d. R4 ( ) = 18 cos (40 ) + 32 cos (10 ); Average power = R4 (0) = 50 W.
42
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Problem 2.43
The autocorrelation function must be (1) even, (2) have an absolute maximum at = 0,
and (3) have a Fourier transform that is real and nonnegative.
a. Acceptable - all properties satisÖed;
b. Acceptable - all properties satisÖed;
c. Not acceptable - none of the properties satisÖed;
d. Acceptable - all properties satisÖed;
e. Not acceptable - property (3) not satisÖed;
f. Not acceptable - none of the properties satisÖed.
Problem 2.44
2
Given that the autocorrelation function of x (t) = A cos (2f0 t + ) is Rx ( ) = A2 cos (2f0 )
(special case of Ex. 2.19), the results are as follows.
a. R1 ( ) = 2 cos (10 );
b. R2 ( ) = 2 cos (10 );
c. R3 ( ) = 5 cos (10 ) (write the signal as x3 (t) = Re 5 exp j tan1 (4=3) exp (j10t) );
2
2
d. R4 ( ) = 2 +2
cos (10 ) = 4 cos (10 ).
2
Problem 2.45
This is a matter of applying (2.151) by making the appropriate identiÖcations with the
parameters given in Example 2.20
Problem 2.46
Fourier transform both sides of the di§erential equation using the di§erentiation theorem
of Fourier transforms to get
[j2f + a] Y (f ) = [j2bf + c] X (f )
Therefore, the frequency response function is
H (f ) =
Y (f )
c + j2bf
=
X (f )
a + j2f
2.1. PROBLEM SOLUTIONS
43
a = 1; b = 2; c = 0
a = 1; b = 0; c = 3
2
3
mag(H)
mag(H)
1.5
1
0.5
0
1
0
-5
5
2
2
1
1
angle(H), rad
angle(H), rad
0
-5
2
0
-1
-2
-5
0
f, Hz
5
0
f, Hz
5
0
-1
-2
-5
5
0
The amplitude response function is
and the phase response is
q
c2 + (2bf )2
jH (f )j = q
a2 + (2f )2
1
arg [H (f )] = tan
2bf
c
1
tan
2f
a
Amplitude and phase responses for various values of the constants are plotted in Figure 2.6.
Problem 2.47
a. Use the transform pair
Aet u (t)
!
A
+ j2f
to Önd the unit impulse response as
h1 (t) = e7t u (t)
44
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
b. Long division gives
H2 (f ) = 1
Use the transform pair (t)
(a) to get
7
7 + j2f
! 1 along with superposition and the transform pair in part
h2 (t) = (t) 7e7t u (t)
c. Use the time delay theorem along with the result of part (a) to get
h3 (t) = e7(t3) u (t 3)
d. Use superposition and the results of parts (a) and (c) to get
h4 (t) = e7t u (t) e7(t3) u (t 3)
Problem 2.48
Use the transform pair for a sinc function to Önd that
f
f
Y (f ) =
2B
2W
a. If W < B, it follows that
b. If W > B, it follows that
f
Y (f ) =
2W
f
f
because 2B
= 1 throughout the region where 2W
is nonzero.
f
Y (f ) =
2B
f
f
because 2W
= 1 throughout the region where 2B
is nonzero.
c. Part (b) gives distortion because the output spectrum di§ers from the input spectrum.
In fact, the output is
y (t) = 2Bsinc (2Bt)
which clearly di§ers from the input for W > B.
2.1. PROBLEM SOLUTIONS
45
Problem 2.49
a. Replace the capacitors with 1=j!C which is their ac-equivalent impedance. Call
the junction of the input resistor, feedback resistor, and capacitors 1. Call the
junction at the positive input of the operational ampliÖer 2. Call the junction at the
negative input of the operational ampliÖer 3. Write down the KCL equations at these
three junctions. Use the constraint equation for the operational ampliÖer, which is
V2 = V3 , and the deÖnitions for ! 0 , Q, and K to get the given transfer function as
H (j!) = Vo (j!) =Vi (j!). The node equations are
V1 Vi
V1 V o
+ j!CV1 +
+ j!C(V1 V2 )
R
R
V2
j!C(V2 V1 ) +
R
V3
V3 Vo
+
Rb
Ra
(constraint on op amp input) V2
= 0
= 0
= 0
= V3
b. See plot given in Figure 2.7.
c. In terms of f , the transfer function magnitude is
K
(f =f0 )
jH(f )j = p s
2 2
2
2
1 ff0
+ Q12 ff0
It can be shown that
p the maximim of jH(f )j is at f = f0 . By substitution, this maximum
is jH(f0 )j = KQ= p2. To Önd the 3-dB bandwidth, we must Önd the frequencies for which
jH(f )j = jH(f0 )j= 2. This results in
Q
(f3 =f0 )
p = s
2 2
2
2
1 ff30
+ Q12 ff30
which reduces to the quadratic equation
f3
f0
4
1
2+ 2
Q
f3
f0
2
+1=0
46
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Using the quadratic formula, the solutions to this equation are
2
1
2
1
2+ 2
Q
s
1
=
2+ 2 4
Q
r
1
1
1
1
1
=
2+ 2
1+
1
2
Q
Q
4Q2
Q
r
1
f3
1
1
1
; Q >> 1
f0
Q
2Q
f3
f0
Therefore, the 3-dB bandwidth is, for large Q, approximately
1
1
f0
B = f3 f3 1 +
f0 1
f0 =
2Q
2Q
Q
d. Combinations of components giving
RC = 2:2508 104 seconds
and
Ra
= 2:5757
Rb
will work.
Problem 2.42
a. By voltage division, with the inductor replaced by j2f L, the frequency response
function is
R2 + j2f L
R2 =L + j2f
H1 (f ) =
=
R1 + R2 + j2f L
(R1 + R2 ) =L + j2f
By long division
R1 =L
H1 (f ) = 1 R1 +R2
+ j2f
L
Using the transforms of a delta function and a one-sided exponential, we obtain
R1
R1 + R2
h1 (t) = (t)
exp
t u (t)
L
L
2.1. PROBLEM SOLUTIONS
47
20
|H(f)|, dB
10
0
-10
f0 = 999.9959 Hz; B 3 dB = 300.0242 Hz
-20
2
10
3
10
4
10
angle(H(f)), radians
2
1
0
-1
-2
2
10
3
10
f, Hz
4
10
b. Substituting the ac-equivalent impedance for the inductor and using voltage division,
the frequency response function is
H2 (f ) =
=
R2 jj (j2f L)
j2f LR2
where R2 jj (j2f L) =
R1 + R2 jj (j2f L)
R2 + j2f L
R2
j2f L
R2
(R1 k R2 ) =L
=
1
R2
R1 + R2 RR1+R
R1 + R2
(R1 k R2 ) =L + j2f
+ j2f L
1
2
R2
where R1 k R2 = RR11+R
. Therefore, the impulse response is
2
R2
R 1 R2
R1 R2
h2 (t) =
(t)
exp
t u (t)
R1 + R2
(R1 + R2 ) L
(R1 + R2 ) L
2
Both have a high pass amplitude response, with the dc gain of the Örst circuit being R1R+R
2
and the second being 0; the high frequency gain of the Örst is 1 and that of the second is
R2
R1 +R2 .
Problem 2.51
Application of the Payley-Wiener criterion gives the integral
Z 1
f 2
I=
df
2
1 1 + f
48
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
which does not converge. Hence, the given function is not suitable as the frequency response
function of a causal LTI system.
Problem 2.52
a. The condition for stability is
Z 1
Z 1
jh1 (t)j dt =
jexp (jtj) cos (2f0 t)j dt
1
1
Z 1
Z 1
2
= 2
exp (t) jcos (2f0 t)j dt < 2
exp (t) dt = < 1
0
0
which follows because jcos (2f0 t)j 1. Hence this system is BIBO stable.
b. The condition for stability is
Z 1
Z 1
jh2 (t)j dt =
jcos (2f0 t) u (t)j dt
1
1
Z 1
=
jcos (2f0 t)j dt ! 1
0
which follows by integrating one period of jcos (2f0 t)j and noting that the total integral is
the limit of one period of area times N as N ! 1: This system is not BIBO stable.
c. The condition for stability is
Z 1
Z 1
1
jh3 (t)j dt =
u (t 1) dt
jtj
1
Z1
1
dt
=
= ln (t)j1
1 !1
t
1
This system is not BIBO stable.
d. The condition for stability is
Z 1
Z 1
jh4 (t)j dt =
jet u (t) e((t1)) u (t 1) jdt
1
Z1
1
et e(t1) dt = 1 + e < 1
0
This system is BIBO stable.
2.1. PROBLEM SOLUTIONS
49
e. The condition for stability is
Z 1
1
jh5 (t)j dt =
Z 1
1
t2 jdt = 1 < 1
This system is BIBO stable.
f. The condition for stability is
Z 1
1
jh6 (t)j dt =
Z 1
sinc (2t) dt =
1
1
<1
2
This system is BIBO stable.
Problem 2.53
The energy spectral density of the output is
Gy (f ) = jH (f )j2 jX (f )j2
where
jH (f )j2 =
X (f ) =
25
16 + (2f )2
1
1
; Gx (f ) = jX (f ) j2 =
3 + j2f
9 + (2f )2
Hence
Gy (f ) = h
25
ih
i
9 + (2f )2 16 + (2f )2
Plots of the input and output energy spectral densities are left to the student.
Problem 2.54
Using the Fourier coe¢cients of a half-rectiÖed sine wave from Table 2.1 and noting that
those of a half-rectiÖed cosine wave are related by
Xcn = Xsn ejn=2
50
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
The fundamental frequency is 10 Hz. The ideal rectangular Ölter passes all frequencies less
than 31 Hz and rejects all frequencies greater than 31 Hz. Therefore
A j j40t jA j=2 j20t A jA j=2 j20t
A j j40t
e e
+
e
e
+ +
e
e
e e
3
4
! 4
3
!
A A ej(20t=2) + ej(20t=2)
2A ej(40t) + ej(40t)
2
2j
3
2
y (t) =
=
=
=
A A
2A
sin (20t =2)
cos (40t )
2
3
A A
2A
+ cos (20t) +
cos (40t)
2
3
2.1. PROBLEM SOLUTIONS
51
Problem 2.55
a. The energy spectral density is given by
2
1
1
=
G1 (f ) = jX1 (f ) j =
2
+ j2f
+ (2f )2
2
1
Since Etotal = 2
, the 90% energy containment bandwidth is given by
Z B90
Z B90
0:9
1
2
1
2f
= 2
2 df; u =
2 df = 2
2
2
+ (2f )
0
0
1 + 2f
Z 2B90 =
1
du
1
=
=
tan1 (2B90 =)
2
0
1+u
or B90 =
tan (0:45) = 1:0049
2
b. For this case, using X2 (f ) = (f =2W ) and Etotal = 2W; we obtain
Z B90
0:9 (2W ) = 2
df = 2B90
0
or B90 = 0:9W
c. For this case, using X3 (f ) = sinc(f ) and E3 = , we obtain
Z B90
0:9 = 2
2 sinc2 (f ) df
0
Z B90
or 0:45 =
sinc2 (u) du
0
Numerical integration gives
B90 = 0:9=
R
d. For this case, using X4 (f ) = sinc2 (f ) and E4 = 2 0 (1 t= )2 dt = 2 =3, we
obtain
Z B90
1:8 =3 = 2
2 sinc4 (f ) df
0
Z B90
or 0:3 =
sinc4 (u) du
0
Numerical integration gives
B90 = 0:35=
52
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
R1
2
e. For this case, using X5 (f ) = 2 +(2f
and E5 = 2 0 exp (2t) dt = 1=, we obtain
)2
0:9= = 2
Z B90
0
or 0:225 =
2
2
df
2 + (2f )2
Z 2B90 =
0
du
(1 + u2 )2
; u = 2f =
Numerical integration gives 2B90 = = 1:18 or B90 = = 0:188 or B90 = 0:188:
Problem 2.56
The outputs are the inputs phase shifted by =2 radians for frequencies greater than 0
and =2 for frequencies less than 0 Hz.
a. y1 (t) = exp [j100t j=2] = j exp (j100t);
b. y2 (t) = 12 exp [j100t j=2] + 12 exp [j100t + j=2] = sin (100t);
1
1
c. y3 (t) = 2j
exp [j100t j=2] 2j
exp [j100t + j=2] = cos (100t);
d. The spectrum of the input is X4 (f ) = 2sinc(2f ). The spectrum of the output is
Y4 (f ) =
2sinc (2f ) exp (j=2) ; f > 0
2sinc (2f ) exp (j=2) ; f < 0
The time domain output signal is the inverse Fourier transform of this, which is
y4 (t) =
Z 1
0
= 2j
2sinc (2f ) exp (j=2) exp (j2f t) df +
Z 1
0
sinc (2f ) exp (j2f t) df + 2j
Z 0
Z 0
2sinc (2f ) exp (j=2) exp (j2f t) df
1
sinc (2u) exp (j2ut) (du) ; u = f
Z11
Z 1
= 2j
sinc (2f ) exp (j2f t) df + 2j
sinc (2u) exp (j2ut) du; sinc2u is even
0
0
Z 1
= 2
sinc (2f ) [j exp (j2f t) + j exp (j2f t)] df ; rewrite 2nd int in terms of f
Z0 1
= 4
sinc (2f ) sin (2f t) df
0
This requires numerical integration. A plot is given in Fig. 2.8.
2.1. PROBLEM SOLUTIONS
53
2
1.5
1
x 4(t), y 4(t)
0.5
0
-0.5
-1
-1.5
-2
-3
-2
-1
0
t, s
1
2
3
Problem 2.57
a. Amplitude distortion; no phase distortion.
The output for (a) is
ya (t) = 4 cos 48t 24
+ 10 cos 126t 63
150
150
1
1
= 4 cos 48 t
+ 10 cos 126 t
300
300
b. No amplitude distortion; phase distortion.
The output for (b) is
yb (t) = 2 cos 126t 63
+ cos (170t)
150
1
= 2 cos 126 t
+ cos (170t)
300
c. No amplitude distortion; no phase distortion.
54
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
The output for (c) is
1
1
yc (t) = 2 cos 126 t
+ 6 cos 144 t
300
300
d. No amplitude distortion; no phase distortion.
The output for (d) is
1
yd (t) = 4 cos 10 t
300
1
+ 16 cos 50 t
300
Problem 2.58
a. The frequency response function corresponding to this impulse response is
3
3
1 2f
H1 (f ) =
=q
exp j tan
5 + j2f
5
2
25 + (2f )
The group delay is
1 d
2f
tan1
2 df
5
1
1
2
2
2
5
1 + 2f
5
Tg1 (f ) =
=
=
5
25 + (2f )2
The phase delay is
Tp1 (f ) =
1 (f )
=
2f
tan1
2f
5
2f
b. The frequency response function corresponding to this impulse response is
5
2
H2 (f ) =
3 + j2f
5 + j2f
5 (5 + j2f ) 2 (3 + j2f )
=
(3 + j2f ) (5 + j2f )
19 + j6f
=
15 (2f )2 + j16f
q
h
i
361 + (6f )2 exp j tan1 6f
19
= rh
i2
h
i
16f
15 (2f )2 + (16f )2 exp j tan1 15(2f
)2
2.1. PROBLEM SOLUTIONS
55
Therefore
2 (f ) = tan1
6f
19
tan1
The group delay is
Tg2 (f ) =
=
=
=
16f
15 (2f )2
1 d
16f
1 6f
1
tan
tan
2 df
19
15 (2f )2
8
9
1
6
2
>
>
6f
19
=
1+( 19 )
1 <
2
16
15(2f
)
16f
[2(2f
)(2)]
(
)
1
>
2 >
2
: 1+ 16f 2
;
[15(2f )2 ]
15(2f )2
8
9
h
i
2
>
>
<
=
16 15 + (2f )
1
19 (6)
i2
2 h
>
2 >
: 361 + (6f )
15 (2f )2 + (16f )2 ;
h
i
2
8
15
+
(2f
)
57
+
361 + (6f )2 225 + 34 (2f )2 + (2f )4
The phase delay is
Tp2 (f ) =
2 (f )
=
2f
tan1
16f
15(2f )2
tan1
2f
6f
19
The group and phase delays for (a) and (b) are shown in Fig. 2.9.
c. The frequency response is
H (f ) =
f
2B
exp (j2t0 f )
The group delay is
1 d
[2t0 f ] ; B f B
2 df
= t0 ; B f B and 0 otherwise
Tg (f ) =
The phase delay is
Tp (f ) =
(2t0 f )
= t0 ; B f B and 0 otherwise
2f
56
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Tg1(f), Tp1(f), s
0.2
Group delay
Phase delay
0.15
0.1
0.05
0
-2
10
-1
0
10
10
f, Hz
1
10
2
10
Tg2(f), Tp2(f), s
0.5
0
-0.5
-1
-2
10
-1
0
10
10
f, Hz
1
10
d. The frequency response function is
H (f ) =
=
5
2
exp (j2t0 f )
3 + j2f
3 + j2f
2
[2:5 exp (j2t0 f )]
3 + j2f
The phase shift function is
(f ) =
The phase delay is
The group delay is
1
sin 2t0 f
1 2f
Tp (f ) =
tan
=
2f
3
2:5 cos 2t0 f
Tg (f ) =
=
Problem 2.59
sin 2t0 f
2f
tan1
2:5 cos 2t0 f
3
1 d
sin 2t0 f
1 2f
tan
=
2 df
3
2:5 cos 2t0 f
3
1 2:5 cos 2t0 f
2 +
9 + (2f )
(2:5 cos 2t0 f )2
2
10
2.1. PROBLEM SOLUTIONS
57
a. The amplitude response is
2jf j
q
64 + (2f )2 9 + (2f )2
jH (f ) j = r
b. The phase response is
(f ) = sgn (f ) tan1
2
f
4
1
tan
2f
3
c. The phase delay is
1
1 f
1 2f
Tp (f ) =
tan
+ tan
sgn (f )
2f
4
3
2
d. The group delay is
Tg (f ) =
1=8
1=3
1
2 +
2 4 (f )
1 + (f =4)
1 + (f =3)
Problem 2.60
In terms of the input spectrum, the output spectrum is
Y (f ) = X (f ) + 0:1X (f ) X (f )
f 10
f + 10
= 2
+
4
4
f 10
f + 10
f 10
f + 10
+0:4
+
+
4
4
4
4
f 10
f + 10
= 2
+
4
4
f 20
f
f + 20
+0:4 4
+ 8
+ 4
4
4
4
where (f ) is an isoceles triangle of unit height going from -1 to 1. The student should
sketch the output spectrum given the above analytical result.
Problem 2.61
a. The amplitude response is
jH(f )j = q
2jf j
(9 4 2 f 2 )2 + (0:3f )2
58
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
8
|H(f)|
6
4
2
0
-3
-2
-1
0
1
2
3
-2
-1
0
f, Hz
1
2
3
0:3f
9 4 2 f 2
θ(f), radians
2
1
0
-1
-2
-3
b. The phase response is
(f ) = sgn (f ) tan1
2
These are shown in Fig. 2.10.
c. The phase delay is
1
0:3f
1
Tp (f ) =
sgn (f ) tan
2 2
9 4 2 f 2
d. The group delay is
1 d
Tg (f ) =
2 df
2
=
=
sgn (f ) tan1
2
1 6
4 (f )
2
1+
1
0:3f
94 2 f 2
0:3f
9 4 2 f 2
3
2
2
2
0:3 9 4 f + (0:3f ) 8 f 7
5
2
(9 4 2 f 2 )2
1:35 + 0:6 2 f 2
1
(f )
2
2
4
4
81 71:91 f + 16 f
2
2.1. PROBLEM SOLUTIONS
59
Problem 2.62
Let u = 2t. We then have
y (t) = [cos (u) + cos (3u)]3
= cos3 (u) + 3 cos2 (u) cos (3u) + 3 cos (u) cos2 (3u) + cos3 (3u)
Use the trig identities
cos2 (z) =
cos3 (z) =
cos (w) cos (z) =
1
[1 + cos (2z)]
2
3
1
cos (z) + cos (3z)
4
4
1
1
cos (w z) + cos (w + z)
2
2
to get
3
1
3
cos (u) + cos (3u) + [1 + cos (2u)] cos (3u)
4
4
2
3
3
1
+ cos (u) [1 + cos (6u)] + cos (3u) + cos (9u)
2
4
4
5
3
3
1
= 3 cos (u) + cos (3u) + cos (5u) + cos (7u) + cos (9u)
2
2
4
4
5
3
3
1
= 3 cos (2t) + cos (6t) + cos (10t) + cos (14t) + cos (18t)
2
2
4
4
y (t) =
Problem 2.63
Write the transfer function as
j2f t0
H (f ) = H0 e
H0
f
2B
ej2f t0
Use the inverse Fourier transform of a constant, the delay theorem, and the inverse Fourier
transform of a rectangular pulse function to get
h (t) = H0 (t t0 ) 2BH0 sinc [2B (t t0 )]
Problem 2.64
a. The Fourier transform of this signal is
p
X (f ) = A 2b2 exp 2 2 2 f 2
60
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
By deÖnition, using a table of integrals,
Z 1
p
1
T =
jx (t)j dt = 2
x (0) 1
Similarly,
1
W =
2X (0)
Therefore,
Z 1
1
jX (f )j df = p
2 2
1
2 p
2W T = p
2 = 1
2 2
b. The Fourier transform of this signal is
X (f ) =
The pulse duration is
1
T =
x (0)
The bandwidth is
W =
Thus,
Z 1
1
2X (0)
2A=
1 + (2f =)2
1
Z 1
2W T = 2
jx (t)j dt =
1
2
jX (f )j df =
2
4
4
=1
Problem 2.65
a. The poles for a second order Butterworth Ölter are given by
!3
s1 = s2 = p (1 j)
2
where ! 3 is the 3-dB cuto§ frequency of the Butterworth Ölter.
function is
H (s) = h
Its s-domain transfer
!2
!2
i h3
i=
p 3
!3
!3
s2 + 2! 3 s + ! 23
s+ p
(1 j) s + p
(1 + j)
2
2
Letting ! 3 = 2f3 and s = j! = j2f , we obtain
H (j2f ) =
4 2 f 2 +
p
4 2 f32
f32
p
=
2 (2f3 ) (j2f ) + 4 2 f32
f 2 + j 2f3 f + f32
2.1. PROBLEM SOLUTIONS
61
b. If the phase response function of the Ölter is (f ), the group delay is
Tg (f ) =
1 d
[ (f )]
2 df
For the second-order Butterworth Ölter considered here,
!
p
2f
f
3
(f ) = tan1
f32 f 2
Therefore, the group delay is
Tg (f ) =
=
"
1 d
tan1
2 df
p
2f3 f
2
f3 f 2
!#
f f2 + f2
1 1 + (f =f3 )2
p 3 34
p
=
2 f3 + f 4
2f3 1 + (f =f3 )4
This is plotted in Fig. 2.11.
c. Use partial fraction expansion of H (s) =s and then inverse Laplace transform it to get
the given step response. The expansion is
!2
A
Bs + C
p 3
p
= +
2
2
2
s
s s + 2! 3 s + ! 3
s + 2! 3 s + ! 23
p
where A = 1; B = 1; C = 2! 3
H (s)
s
=
This allows H (s) =s to be written as
H (s)
s
=
=
=
=
=
p
1
s + 2! 3
p
s s2 + 2! 3 s + ! 23
p
1
s + 2! 3
p
s s2 + 2! 3 s + ! 23 =2 ! 23 =2 + ! 23
p
1
s + 2! 3
p 2
s
s + ! 3 = 2 + ! 23 =2
p
p
p
1 s + ! 3 = 2 ! 3 = 2 + 2! 3
p 2
s
s + ! 3 = 2 + ! 23 =2
p
p
1
s + !3= 2
!3= 2
p 2
p 2
s
s + ! 3 = 2 + ! 2 =2
s + ! 3 = 2 + ! 2 =2
3
3
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Group delay for a 2nd-order BW filter
0.35
Step response for a 2nd-order BW filter
1.4
0.3
1.2
0.25
1
0.2
0.8
y s (t)
f3Tg(f)
62
0.15
0.6
0.1
0.4
0.05
0.2
0
-1
10
0
10
f/f3
1
10
0
0
0.5
f3t
1
Using the s-shift theorem of Laplace transforms, this inverse transforms to the given expression for the step response (with ! 3 = 2f3 ). Plot it and estimate the 10% and 90%
times from the plot. From the MATLAB plot of Fig. 2.10, f3 t10% 0:08 and f3 t90% 0:42
so that the 10-90 % rise time is about 0:34=f3 seconds.
Problem 2.66
a. Slightly less than 0.5 seconds;
b & c. Use sketches to show.
Problem 2.67
a. The leading edges of the áat-top samples follow the waveform at the sampling instants.
b. The spectrum is
Y (f ) = X (f ) H (f )
where
X (f ) = fs
1
X
n=1
X (f nfs )
2.1. PROBLEM SOLUTIONS
63
and
H (f ) = sinc (f ) exp (jf )
The latter represents the frequency response of a Ölter whose impulse response is a square
pulse of width and implements áat top sampling. If W is the bandwidth of X (f ), very
little distortion will result if 1 >> W .
Problem 2.68
a. The sampling frequency should be large compared with the bandwidth of the signal.
b. The output spectrum of the zero-order hold circuit is
Y (f ) = sinc (Ts f )
1
X
n=1
X (f nfs ) exp (jf Ts )
where fs = Ts1 . For small distortion, we want Ts << W 1 .
Problem 2.69
Use trig identities to rewrite the signal as a sum of sinusoids:
x (t) = 10 cos2 (600t) cos (2400t)
= 5 [1 + cos (600t)] cos (2400t)
= 5 cos (2400t) + 2:5 cos (1800t) + 2:5 cos (3000t)
The lowpass recovery Ölter can cut o§ in the range 1.5+ kHz to 3 kHz where the superscript
+ means just above and the superscript means just below. The lower of these is the
highest frequency of x (t) and the larger is equal to the sampling frequency minus the highest
frequency of x (t). The minimum allowable sampling frequency is just above 3 kHz.
Problem 2.70
For bandpass sampling and recovery, all but (b) and (e) will work theoretically, although an
ideal Ölter with bandwidth exactly equal to the unsampled signal bandwidth is necessary.
For lowpass sampling and recovery, only (f) will work.
64
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Problem 2.71
The Fourier transform is
1
1
X (f f0 ) + X (f + f0 )
2
2
1
1
j=2
j=2
+ [jsgn (f ) X (f )] (f f0 ) e
+ (f + f0 ) e
2
2
1
1
=
X (f f0 ) [1 sgn (f f0 )] + X (f + f0 ) [1 + sgn (f + f0 )]
2
2
Y (f ) =
Noting that
and
1
[1 sgn (f f0 )] = u (f0 f )
2
1
[1 + sgn (f + f0 )] = u (f + f0 )
2
this may be rewritten as
Y (f ) = X (f f0 ) u (f0 f ) + X (f + f0 ) u (f + f0 )
Thus, if X (f ) = f2 (a unit-height rectangle 2 units wide centered at f = 0) and f0 = 10
Hz, Y (f ) would consist of unit-height rectangles going from 10 to 9 Hz and from 9 to
10 Hz.
Problem 2.72
a. x
ba (t) = cos (! 0 t =2) = sin (! 0 t), so
1
T !1 2T
lim
Z T
T
x (t) x
b (t) dt =
1
T !1 2T
lim
1
T !1 2T
Z T
sin (! 0 t) cos (! 0 t) dt
T
Z T
1
sin (2! 0 t) dt
T 2
1 cos (2! 0 t) T
= lim
=0
T !1 2T
4! 0
=
lim
T
b. Use trigonometric identities to express x (t) in terms of sines and cosines. Then
Önd the Hilbert transform of x (t) by phase shifting by =2. Multiply x (t) and
x
b (t) together term by term, use trigonometric identities for the product of sines and
cosines, then integrate. The integrand will be a sum of terms similar to that of part
(a). The limit as T ! 1 will be zero term-by-term.
2.1. PROBLEM SOLUTIONS
65
c. For this case x
b (t) = A exp (j! 0 t j=2) = jA exp (j! 0 t), take the product, integrate over time to get
Z T
Z T
1
1
lim
x (t) x
b (t) dt = lim
[A exp (j! 0 t)] [jA exp (j! 0 t)] dt
T !1 2T T
T !1 2T T
Z
jA2 T
= lim
exp (j2! 0 t) dt = 0
T !1 2T
T
by periodicity of the integrand
Problem 2.73
a. Note that F [jb
x(t)] = j [jsgn (f )] X (f ). Hence
2
1
2
1
x (t) + jb
x (t) ! X1 (f ) = X (f ) + j [jsgn (f )] X (f )
3
3
3
3
2 1
=
+ sgn (f ) X (f )
3 3
1
3 X (f ) ; f < 0
=
X (f ) ; f > 0
x1 (t) =
b. It follows that
3
3
x2 (t) =
x (t) + jb
x (t) exp (j2f0 t)
4
4
3
) X2 (f ) = [1 + sgn (f f0 )] X (f f0 )
4
0; f < f0
=
3
2 X (f f0 ) ; f > f0
c. This case has the same spectrum as part (a), except that it is shifted right by W Hz.
That is,
2
1
x3 (t) =
x (t) + jb
x (t) exp (j2W t)
3
3
2 1
! X3 (f ) =
+ sgn (f W ) X (f W )
3 3
d. For this signal
2
1
x4 (t) =
x (t) jb
x (t) exp (jW t)
3
3
2 1
! X4 (f ) =
sgn (f W=2) X (f W=2)
3 3
66
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Problem 2.74
The Hilbert transform of the given signal is
x
b(t) = 2 sin (52t)
The signal xp (t) is
xp (t) = x (t) + jb
x(t)
= 2 cos (52t) + j2 sin (52t)
= 2ej52t
a. We have, for f0 = 25 Hz,
x
~ (t) = xp (t) ej2f0 t = 2ej52t ej50t = 2ej2t = 2 cos (2t) + j2 sin (2t)
xR (t) = 2 cos (2t)
xI (t) = 2 sin (2t)
b. For f0 = 27 Hz,
x
~ (t) = 2ej52t ej54t = 2ej2t = 2 cos (2t) j2 sin (2t)
xR (t) = 2 cos (2t)
xI (t) = 2 sin (2t)
c. For f0 = 10 Hz,
x
~ (t) = 2ej52t ej20t = 2ej32t = 2 cos (32t) + j2 sin (32t)
xR (t) = 2 cos (32t)
xI (t) = 2 sin (32t)
d. For f0 = 15 Hz,
x
~ (t) = 2ej52t ej30t = 2ej22t = 2 cos (22t) + j2 sin (22t)
2.1. PROBLEM SOLUTIONS
67
xR (t) = 2 cos (22t)
xI (t) = 2 sin (22t)
e. For f0 = 30 Hz,
x
~ (t) = 2ej52t ej60t = 2ej8t = 2 cos (8t) j2 sin (8t)
xR (t) = 2 cos (8t)
xI (t) = 2 sin (8t)
f. For f0 = 20 Hz,
x
~ (t) = 2ej52t ej40t = 2ej12t = 2 cos (12t) + j2 sin (12t)
xR (t) = 2 cos (12t)
xI (t) = 2 sin (12t)
Problem 2.75
For t < =2, the output is zero. For jtj =2, the result is
y (t) =
q
=2
2 + (2f )2
n
o
cos [2 (f0 + f ) t ] e(t+ =2) cos [2 (f0 + f ) t + ]
For t > =2, the result is
y (t) =
q
(=2) et
2 + (2f )2
n
o
e =2 cos [2 (f0 + f ) t ] e =2 cos [2 (f0 + f ) t + ]
In the above equations, is given by
1
= tan
2f
68
2.2
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Computer Exercises
Computer Exercise 2.1
% ce2_1.m: Amplitude spectra and Fourier series synthesized
% for various periodic waveforms
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
clear all; clf
N = input(íNumber of harmonics in Fourier sum => í);
T = input(íPeriod of periodic waveform => í);
A = input(íAmplitude of waveform => í);
n = -N:1:N;
I_type = input(í1 = rect pulse train; 2 = HR sinewave; 3 = FR sinewave; 4 = triang
pulse train; 5 = triangle wave; 6 = sawtooth wave => í);
X = zeros(size(n));
if I_type == 1
tau = input(íWidth of rectangular pulse => í);
t0 = input(íDelay of rectangular pulse center => í);
d = tau/T;
X = A*d*sinc(n*d).*exp(-j*2*pi*n*t0/T);
elseif I_type == 2
I = Önd(n == 1 j n == -1);
II = Önd(rem(n, 2) == 0);
III = Önd(rem(abs(n), 2) == 1 & (n ~= 1 & n ~= -1));
X(I) = -0.25*j*n(I)*A;
X(II) = A./(pi*(1-n(II).*n(II)));
X(III) = 0;
elseif I_type == 3
X = 2*A./(pi*(1-4*n.*n));
elseif I_type == 4
tau = input(íHalf width of triangle pulse => í);
t0 = input(íDelay of triangle pulse => í);
d = tau/T;
X = A*d*sinc(n*d).*sinc(n*d).*exp(-j*2*pi*n*t0/T);
elseif I_type == 5
I = Önd(rem(abs(n), 2) == 1);
X(I) = 4*A./(pi^2*n(I).^2);
elseif I_type == 6
2.2. COMPUTER EXERCISES
69
I = Önd(n~=0);
X(I) = 2*A*(-exp(-j*2*pi*n(I))+(1 - exp(-j*2*pi*n(I)))./(j*2*pi*n(I)))./(j*2*pi*n(I));
end
subplot(2,1,1), stem(n, abs(X)),xlabel(íní), ylabel(íjX_njí), ...
if I_type == 1
title([íRectangular pulse train; period = í, num2str(T), í; delay = í, num2str(t0), í;
ampli = í, num2str(A)])
elseif I_type == 2
title([íHalf-rectiÖed sinewave; period = í, num2str(T),í ; ampli = í, num2str(A)])
elseif I_type == 3
title([íFull-rectiÖed sinewave; period = í, num2str(T),í ; ampli = í, num2str(A)])
elseif I_type == 4
title([íTriangle pulse train; í, num2str(2*N+1), í terms. A = í, num2str(A), í, ntau = í,
num2str(tau), í, T = í, num2str(T) í s; d = í, num2str(d), í; t_0 = í, num2str(t0), í sí])
elseif I_type == 5
title([íTriangle waveform; period = í, num2str(T),í; ampli = í, num2str(A)])
elseif I_type == 6
title([íSawtooth waveform; period = í, num2str(T),í ; ampli = í, num2str(A)])
end
fn = n./T;
t = -T:T/500:T;
x = real(X*exp(j*2*pi*fní*t));
subplot(2,1,2), plot(t, x), xlabel(ítí), ylabel(íx(t)í), ...
>> ce2_1
Number of harmonics in Fourier sum => 25
Period of periodic waveform => 2
Amplitude of waveform => 1
1 = rect pulse train; 2 = HR sinewave; 3 = FR sinewave; 4 = triang pulse train; 5 =
triangle wave; 6 = sawtooth wave => 6
>>
Computer Exercise 2.2
% ce2_2.m: Plot of line spectra for half-rectiÖed,
% full-rectiÖed sinewave, square wave, and triangle wave
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
clear all; clf
70
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Sawtooth waveform; period = 2 ; ampli = 1
0.4
|Xn|
0.3
0.2
0.1
0
-25
-20
-15
-10
-5
0
n
5
-0.5
0
t
0.5
10
15
20
25
1.5
2
2
x(t)
1
0
-1
-2
-2
-1.5
-1
1
waveform = input(íEnter type of waveform: 1 = HR sine; 2 = FR sine; 3 = square; 4
= triangle: í);
A = 1;
n_max = 13;
% maximum harmonic plotted; odd
n = -n_max:1:n_max;
if waveform == 1
X = A./(pi*(1+eps - n.^2)); % O§set 1 slightly to avoid divide by zero
for m = 1:2:2*n_max+1
X(m) = 0;
% Set odd harmonic lines to zero
X(n_max+2) = -j*A/4;
% Compute lines for n = 1 and n = -1
X(n_max) = j*A/4;
end
elseif waveform == 2
X = 2*A./(pi*(1+eps - 4*n.^2));
elseif waveform == 3
X = abs(4*A./(pi*n+eps));
for m = 2:2:2*n_max+1
X(m) = 0;
end
elseif waveform == 4
X = 4*A./(pi*n+eps).^2;
for m = 2:2:2*n_max+1
2.2. COMPUTER EXERCISES
71
X(m) = 0;
end
end
[arg_X, mag_X] = cart2pol(real(X),imag(X));
% Convert to magnitude and phase
if waveform == 1
for m = n_max+3:2:2*n_max+1
arg_X(m) = arg_X(m) - 2*pi;
% Force phase to be odd
end
elseif waveform == 2
m = Önd(n > 0);
arg_X(m) = arg_X(m) - 2*pi;
% Force phase to be odd
elseif waveform == 4
arg_X = mod(arg_X, 2*pi);
end
subplot(2,1,1),stem(n, mag_X),ylabel(íjX_njí)
if waveform == 1
title(íHalf-rectiÖed sine wave spectraí)
elseif waveform == 2
title(íFull-rectiÖed sine wave spectraí)
elseif waveform == 3
title(íSpectra for square wave with even symmetry í)
elseif waveform == 4
title(íSpectra for triangle wave with even symmetryí)
end
subplot(2,1,2),stem(n, arg_X),xlabel(ínf_0í),ylabel(íangle(X_n)í)
>> ce2_2
Enter type of waveform: 1 = HR sine; 2 = FR sine; 3 = square; 4 = triangle: 1
>>
Computer Exercise 2.3
% ce2_3.m: FFT plotting of line spectra for half-rectiÖed, full-rectiÖed sinewave, square
wave,
% and triangular waveforms
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
% User deÖned functions used: pls_fn( ); trgl_fn( )
%
clf
72
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Half-rectified sine wave spectra
0.4
|Xn|
0.3
0.2
0.1
0
-15
-10
-5
0
5
10
15
-10
-5
0
nf0
5
10
15
4
angle(Xn)
2
0
-2
-4
-15
I_wave = input(ëEnter type of waveform: 1 = positive squarewave; 2 = 0-dc level
triangular; 3 = half-rect. sine; 4 = full-wave sine: í);
T = 2;
del_t = 0.001;
t = 0:del_t:T;
L = length(t);
fs = (L-1)/T;
del_f = 1/T;
n = 0:9;
if I_wave == 1
x = pls_fn(2*(t-T/4)/T);
X_th = abs(0.5*sinc(n/2));
disp(ë í)
disp(ë 0 - 1 level squarewaveí)
elseif I_wave == 2
x = 2*trgl_fn(2*(t-T/2)/T)-1;
X_th = 4./(pi^2*n.^2);
X_th(1) = 0; % Set n = even coe¢cients to zero (odd indexed because of MATLAB)
X_th(3) = 0;
X_th(5) = 0;
X_th(7) = 0;
X_th(9) = 0;
disp(ë í)
2.2. COMPUTER EXERCISES
73
disp(ë 0-dc level triangular waveí)
elseif I_wave == 3
x = sin(2*pi*t/T).*pls_fn(2*(t-T/4)/T);
X_th = abs(1./(pi*(1-n.^2)));
X_th(2) = 0.25;
X_th(4) = 0;
% Set n = odd coe¢cients to zero (even indexed because of MATLAB)
X_th(6) = 0;
X_th(8) = 0;
X_th(10) = 0;
disp(ë í)
disp(ë Half-rectiÖed sinewaveí)
elseif I_wave == 4
x = abs(sin(pi*t/T));
% Period of full-rectiÖed sinewave is T/2
X_th = abs(2./(pi*(1-4*n.^2)));
disp(ë í)
disp(ë Full-rectiÖed sinewaveí)
end
X = 0.5*§t(x)*del_t; % Multiply by 0.5 because of 1/T_0 with T_0 = 2
f = 0:del_f:fs;
Y = abs(X(1:10));
Z = [ní Yí X_thí];
disp(ëMagnitude of the Fourier coe¢cientsí);
disp(ë í)
disp(ë n FFT Theoryí);
disp(ë __________________________í)
disp(ë í)
disp(Z);
subplot(2,1,1),plot(t, x), xlabel(ëtí), ylabel(ëx(t)í)
subplot(2,1,2),plot(f, abs(X),íoí),axis([0 10 0 1]),...
xlabel(ëní), ylabel(ëjX_njí)
% Unit-width pulse function
%
function y = pls_fn(t)
y = stp_fn(t+0.5)-stp_fn(t-0.5);
function y = stp_fn(t)
% Function for generating the unit step
%
y = zeros(size(t));
74
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
I = Önd(t >= 0);
y(I) = ones(size(I));
function y = trgl_fn(t)
%
This function generates a unit-high triangle centered
%
at zero and extending from -1 to 1
%
y = (1 - abs(t)).*pls_fn(t/2);
%
% End of script Öle
A typical run follows with a plot given in Fig. 2.13.
>> ce2_3
Enter type of waveform: 1 = positive squarewave; 2 = 0-dc level triangular; 3 = half-rect.
sine; 4 = full-wave sine: 3
Half-rectiÖed sinewave
Magnitude of the Fourier coe¢cients
n
FFT
Theory
__________________________
0
0.3183
0.3183
1.0000
0.2501
0.2500
2.0000
0.1062
0.1061
3.0000
0.0001
0
4.0000
0.0212
0.0212
5.0000
0.0001
0
6.0000
0.0091
0.0091
7.0000
0.0000
0
8.0000
0.0051
0.0051
9.0000
0.0000
0
Computer Exercise 2.4
Make the time window long compared with the pulse width.
Computer Exercise 2.5
% ce2_5.m: Finding the energy ratio in a preset bandwidth
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
% User deÖned functions used: pls_fn( ); trgl_fn( )
2.2. COMPUTER EXERCISES
75
x(t)
1
0.5
0
0
0.2
0.4
0.6
0.8
1
t
1.2
1.4
1.6
1.8
2
0
1
2
3
4
5
n
6
7
8
9
10
|Xn|
1
0.5
0
76
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
%
I_wave = input(íEnter type of waveform: 1 = rectangular; 2 = triangular; 3 = half-rect.
sine; 4 = raised cosine: í);
tau = input(íEnter pulse width: í);
per_cent = input(íEnter percent total desired energy: í);
clf
T = 20;
f = [];
G = [];
del_t = 0.001;
t = 0:del_t:T;
L = length(t);
fs = (L-1)/T;
del_f = 1/T;
% n = [0 1 2 3 4 5 6 7 8 9];
if I_wave == 1
x = pls_fn((t-tau/2)/tau);
disp(í í)
disp(í Rectangular pulseí)
elseif I_wave == 2
x = trgl_fn(2*(t-tau/2)/tau);
disp(í í)
disp(í Triangular pulseí)
elseif I_wave == 3
x = sin(pi*t/tau).*pls_fn((t-tau/2)/tau);
disp(í í)
disp(í Half sinewaveí)
elseif I_wave == 4
x = abs(sin(pi*t/tau).^2.*pls_fn((t-tau/2)/tau));
disp(í í)
disp(í Raised sinewaveí)
end
X = §t(x)*del_t;
f1 = 0:del_f*tau:fs*tau;
G1 = X.*conj(X);
NN = áoor(length(G1)/2);
G = G1(1:NN);
§ = f1(1:NN);
f = f1(1:NN+1);
E_tot = sum(G);
2.2. COMPUTER EXERCISES
77
E_f = cumsum(G);
E_W = [0 E_f]/E_tot;
test = E_W - per_cent/100;
L_test = length(test);
k = 1;
while test(k) <= 0
k = k+1;
end
B = k*del_f;
if I_wave == 2
tau1 = tau/2;
else
tau1 = tau;
end
subplot(3,1,1),plot(t/tau, x), xlabel(ít/ntauí), ylabel(íx(t)í), axis([0 2 0 1.2])
if I_wave == 1
title([íEnergy containment bandwidth for rectangular pulse of width í, num2str(tau),í
secondsí])
elseif I_wave == 2
title([íEnergy containment bandwidth for triangular pulse of width í, num2str(tau),í
secondsí])
elseif I_wave == 3
title([íEnergy containment bandwidth for half-sine pulse of width í, num2str(tau),í
secondsí])
elseif I_wave == 4
title([íEnergy containment bandwidth for raised cosine pulse of width í, num2str(tau),í
secondsí])
end
subplot(3,1,2),semilogy(§*tau1, abs(G./max(G))), xlabel(ífntauí), ylabel(íG(f)í), axis([0
10 1e-5 1])
subplot(3,1,3),plot(f*tau1, E_W,íñí), xlabel(ífntauí), ylabel(íE_Wí), axis([0 4 0 1.2])
legend([num2str(per_cent), í% bandwidth X (pulse width) = í, num2str(B*tau)],4)
%
This function generates a unit-high triangle centered
%
at zero and extending from -1 to 1
%
function y = trgl_fn(t)
y = (1 - abs(t)).*pls_fn(t/2);
% Unit-width pulse function
78
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Energy containment bandwidth for half-sine pulse of width 2 seconds
x(t)
1
0.5
0
0
0.2
0.4
0.6
0.8
1
t/τ
1.2
1.4
1.6
1.8
2
0
1
2
3
4
5
fτ
6
7
8
9
10
0
G(f)
10
EW
1
0.5
95% bandwidth X (pulse width) = 1.1
0
0
0.5
1
1.5
2
fτ
2.5
3
3.5
4
%
function y = pls_fn(t)
y = stp_fn(t+0.5)-stp_fn(t-0.5);
%
% End of script Öle
A typical run follows with a plot given in Fig. 2.14.
>> ce2_5
Enter type of waveform: 1 = positive squarewave; 2 = triangular; 3 = half-rect. sine; 4
= raised cosine: 3
Enter pulse width: 2
Enter percent total desired energy: 95
Computer Exercise 2.6
The program for this exercise is similar to that for Computer Exercise 2.5, except that the
waveform is used in the energy calculation.
Computer Exercise 2.7
Use Computer Example 2.2 as a pattern for the solution .
Chapter 3
Basic Modulation Techniques
3.1
Problems
Problem 3.1
The demodulated output, in general, is
() = Lp{ () 2 cos[ + ()]}
where Lp {•} denotes the lowpass portion of the argument. With
() = () cos [ + 0 ]
the demodulated output becomes
() = Lp {2 () cos [ + 0 ] cos [ + ()]}
。
⼀
Performing the indicated multiplication and taking the lowpass portion yields
() = () cos [ () − 0 ]
⼀
If () = 0 (a constant), the demodulated output becomes
() = () cos [0 − 0 ]
Letting = 1 gives the error
() = () [1 − cos (0 − 0 )]
⼀
( mct )
The mean-square error is
E
­ 2 ® D 2
() = () [1 − cos (0 − 0 )]2
⼀
1
-
yp (
t
) )
2
CHAPTER 3. BASIC MODULATION TECHNIQUES
where h·i denotes the time-average value. Since the term [1 − cos (0 − 0 )] is a constant,
we have
­ 2 ® ­ 2 ®
() = () [1 − cos (0 − 0 )]2
Note that for 0 = 0 , the demodulation carrier is phase coherent with the original modulation carrier, and the error is zero. For () = 0 we have the demodulated output
() = () cos ( 0 − 0 )
Letting = 1, for convenience, gives the error
() = () [1 − cos (0 − 0 )]
giving the mean-square error
E
­ 2 ® D 2
() = () [1 − cos ( 0 − 0 )]2
In many cases, the average of a product is the product of the averages. (We will say more
about this in Chapters 4 and 5). For this case
E
­ 2 ® ­ 2 ®D
() = () [1 − cos (0 − 0 )]2
Note that 1 − cos (0 − 0 ) is periodic. Taking the average over an integer number of
periods yields
D
E
­
®
[1 − cos ( 0 − 0 )]2
= 1 − 2 cos ( 0 − 0 ) + cos2 (0 − 0 )
= 1+
Thus
3
1
=
2
2
­ 2 ® 3­ 2 ®
() =
()
2
Problem 3.2
We are given that
() =
5
X
10
=1
sin (2 ) =
³
´
cos 2 −
2
5
X
10
=1
With a carrier of
() = 100 cos (200)
3.1. PROBLEMS
3
we have
() =
5
h
X
500 n h
i
io
cos (200 + 2 ) −
+ cos (200 − 2 ) +
2
2
=1
The complex exponential Fourier series is
() =
5
h
h
X
250 n
i
io
exp (200 + 2 ) −
+ exp (200 − 2 ) +
2
2
=1
5
h
h
X
250 n
i
io
exp (−200 − 2 ) +
+ exp (−200 + 2 ) −
+
2
2
=1
The top row represents the positive frequency terms and the second row represents the
negative frequency terms. The transmitted power is
¶
¶
µ
µ
5
X
250 2
1 1
1
1
=
2
= 125000 1 + + +
+
= 182960 kW
4 9 16 25
=1
Problem 3.3
A full-wave rectifier takes the form shown in Figure 3.1. The waveforms are shown in
Figure 3.2, with the half-wave rectifier on top and the full-wave rectifier on the bottom.
The message signal is the envelopes. Decreasing exponentials can be drawn from the peaks
of the waveform as depicted in Figure 3.3(b) in the text. It is clear that the full-wave
rectified () defines the message better than the half-wave rectified () since the carrier
frequency is effectively doubled.
Problem 3.4
Using
yields the table below.
Part
a
b
c
®
­
2 2 ()
=
1 + 2 h2 ()i
­ 2 ®
()
= 02
= 03
= 04
=7
=1
13
= 132% = 291% = 506% = 1404% = 25%
13
= 132% = 291% = 506% = 1404% = 25%
1
= 385% = 826% = 1379% = 3289% = 50%
4
CHAPTER 3. BASIC MODULATION TECHNIQUES
xC t
yD t
C
R
Figure 3.1: Full wave rectifier.
2
1.5
1
0.5
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
2
1.5
1
0.5
0
Figure 3.2: Output waveforms for a half-wave rectifier (top) and a full-wave rectifier (bottom).
3.1. PROBLEMS
5
mn t
1
t
0
T
-1
Figure 3.3: Normalized message signal for Problem 3.5.
Problem 3.5
By inspection, the normalized message signal is as shown in Figure 3.3.
Thus
2
() =
0≤≤
2
and
µ ¶ µ ¶
µ ¶
Z
­ 2 ®
2 2 2 2
2 2 21 3 1
=
=
() =
0
3 2
3
Also
[1 + ] = 40
[1 − ] = 10
This yields
or
40
1+
=
=4
1−
10
1 + = 4 − 4
5 = 3
Thus
= 06
Since the index is 06, we can write
[1 + 06] = 40
6
CHAPTER 3. BASIC MODULATION TECHNIQUES
This gives
=
40
= 25
16
This carrier power is
1
1
= 2 = (25)2 = 3125 Watts
2
2
The efficiency is
=
Thus
(06)2
¡1¢
036
3
2 ¡ 1 ¢ = 336 = 0107 = 107%
1 + (06) 3
= 0107
+
where represents the power in the sidebands and represents the power in the carrier.
The above expression can be written
= 0107 + 0107
This gives
=
0107
= 9748 Watts
10 − 0107
Problem 3.6
­
®
Using (3.14) with = 100, = 07 and 2 () = 1 gives
­ 2 ®
() = 05(100)2 + 05(100)2 (07)2
The power in the sidebands is
= 245 kW
The total power is 745 kW. Thus the efficiency is
=
245
= 03289 = 3289%
745
This may be checked by using
®
­
2 2 ()
(07)2
=
= 03289
=
1 + 2 h2 ()i
1 + (07)2
The modulation trapezoid is shown below.
3.1. PROBLEMS
7
70
30
‐30
‐70
Figure 3.4: Modulation trapezoid for Problem 3.6.
Problem 3.7
For the first signal
− + 5( − ) = 0
so that
=
Also
5
6
() = ()
∙
¸
­ 2®
30
1
2 5
2
(−1)
+ (5)
=
=5
=
6
6
6
For = 07 the efficiency is
®
­
2 2 ()
(07)2 (5)
=
= 07101 = 7101%
=
1 + 2 h2 ()i
1 + (07)2 (5)
and for = 1 the efficiency is
®
­
2 2 ()
(1)2 (5)
=
=
= 08333 = 8333%
1 + 2 h2 ()i
1 + (1)2 (5)
For the second signal
−5 + ( − ) = 0
8
CHAPTER 3. BASIC MODULATION TECHNIQUES
so that
=
Also
­ 2®
1
=
6
1
() = ()
5
µ ¶2 ∙
¸
1
5
2
2 5
+ (−5)
=
= 02
(1)
5
6
6
25
For = 07 the efficiency is
®
­
2 2 ()
(07)2 (02)
=
=
= 00893 = 893%
1 + 2 h2 ()i
1 + (07)2 (02)
and for = 1 the efficiency is
®
­
2 2 ()
(1)2 (02)
=
= 01667 = 1667%
=
1 + 2 h2 ()i
1 + (1)2 (02)
Problem 3.8
(a) With () = 9 cos(20) − 8 cos(60) we can use MATLAB and either plot the waveform or use a root finding algorithm to find the most negative value of (). Plotting the waveform shows that it varies between −129 and +129. Therefore () =
1
129 [9 cos(20) − 8 cos(60)]
i
­
® ¡ 1 ¢2 ¡ 1 ¢ h 2
2
= 04357
(b) 2 () = 129
+
(8)
(9)
2
£
¤
£
¤
(c) = (08)2 (04357) 1 + (08)2 (04357) = 02180 = 2180%
(d) The expression for () is
¸
∙
1
() = 110 1 +
(9 cos(20) − 8 cos(60)) cos 200
129
= −341085 cos(140) + 383721 cos(180)
+110 cos(200)
+383721 cos(220) − 341085 cos(260)
The spectrum is illustrated below.
The positive and negative terms dedined in the table below so that the magnitude spectrum
is even an the phase spectrum is odd.
3.1. PROBLEMS
9
f
0
AB
C DE
Figure 3.5: Spectrum of transmitted signal using AM.
Component Frequency (Hz) Magnitude Phase
A
70
341085
−
B
90
383721
0
C
100
110
0
D
110
383721
0
E
130
341085
-A
−70
341085
-B
−90
383721
0
-C
−100
110
0
-D
−110
383721
0
-E
−130
341085
−
Problem 3.9
(a) With () = 9 cos(20) + 8 cos(60) the most negative value of () = −17 and falls
1
[9 cos(20) − 8 cos(60)]
at = 005. Therefore () = 17
i
­ 2 ® ¡ 1 ¢2 ¡ 1 ¢ h 2
2
(b) () = 17
= 02509
+
(8)
(9)
2
£
¤
£
¤
(c) = (08)2 (02509) 1 + (08)2 (02509) = 01384 = 1384%
10
CHAPTER 3. BASIC MODULATION TECHNIQUES
(d) The expression for () is
¸
∙
1
() = 110 1 +
(9 cos(20) + 8 cos(60)) cos 200
17
= 258824 cos(140) + 291176 cos(180)
+110 cos(200)
+291176 cos(220) + 258824 cos(260)
The spectrum is as in the previous problem. Components are identified below. Note that
the change in sign makes phase spectrum everywhere zero.
Component Frequency (Hz) Magnitude Phase
A
70
258824
0
B
90
291176
0
C
100
110
0
D
110
291176
0
E
130
258824
0
-A
−70
258824
0
-B
−90
291176
0
-C
−100
110
0
-D
−110
291176
0
-E
−130
258824
0
Problem 3.10
The modulator output
() = 40 cos 2 (200) + 5 cos 2 (180) + 5 cos 2 (220)
can be written
() = [40 + 10 cos 2 (20) ] cos 2 (200)
or
¸
∙
10
cos 2 (20) cos 2 (200)
() = 40 1 +
40
By inspection, the modulation index is
=
10
= 025
40
Since the component at 200 Hertz represents the carrier, the carrier power is
=
1
(40)2 = 800 Watts
2
3.1. PROBLEMS
11
The components at 180 and 220 Hertz are sideband terms. Thus the sideband power is
1
1
(5)2 + (5)2 = 25 Watts
2
2
=
Thus, the efficiency is
=
25
=
= 03030 = 330%
+
800 + 25
Problem 3.11
The sideband power is
=
2 2
+
= 2
2
2
The efficiency is
=
2
2 2
=
=
+
(2 2) + 2
2 + 2 2
The carrier power is
=
Thus
=
2
= 200
2
√
400 = 20
This gives
=
2
=
= 03
+
200 + 2
Therefore
60 = 2 − 03 2 = 07 2
and
=
p
6007 = 92582
We determine the modulation index by writing
¸
∙
2
cos(2(20) cos(2(200))
() = 1 +
The modulation index is
=
Summarizing
2(92582)
2
=
= 09285
20
12
CHAPTER 3. BASIC MODULATION TECHNIQUES
= 20
= 92582
= 09258
Problem 3.12
The modulator output
() = 25 cos 2 (150) + 5 cos 2 (160) + 5 cos 2 (140)
is
¸
∙
10
cos 2 (10) cos 2 (150)
() = 25 1 +
25
Thus, the modulation index, , is
10
=
= 04
25
The carrier power is
1
= (25)2 = 3125 Watts
2
and the sideband power is
1
1
= (5)2 + (5)2 = 25 Watts
2
2
Thus, the efficiency is
25
=
= 00741
3125 + 25
Problem 3.13
(a) By plotting () or by using a root-finding algorithm we see that the minimum value
of () is = −3432. Thus
() = 05828 cos (2 ) + 02914 cos (4 ) + 05828 cos (10 )
The AM signal is
() = [1 + 08 ()] cos 2
= 02331 cos 2 ( − 5 )
+01166 cos 2 ( − 2 )
+02331 cos 2 ( − )
+ cos 2
+02331 cos 2 ( + )
+01166 cos 2 ( + 2 )
+02331 cos 2 ( + 5 )
3.1. PROBLEMS
13
The spectrum is drawn from the expression for (). It contains 14 discrete components
as shown.
Comp
1
2
3
4
5
6
7
(b) Since
the efficiency is
Freq
− − 5
− − 2
− −
−
− +
− + 2
− + 5
Amp
01166
00583
01166
05
01166
00583
01166
Comp
8
9
10
11
12
13
14
Freq
− 5
− 2
−
+
+ 2
+ 5
Amp
01166
00583
01166
05
01166
00583
01166
i
­ 2 ® 1h
(05838)2 + (02914)2 + (05838)2 = 03833
() =
2
­
®
2 2 ()
(08)2 (03833)
=
= 01970
=
1 + 2 h2 ()i
1 + (08)2 (03833)
Problem 3.14
(a) From Figure 3.35
() = () + cos
With the given relationship between () and () we can write
() = 4 { () + cos } + 2 { () + cos }2
which can be written
() = 4 () + 4 cos + 22 () + 4 () cos + 1 + 1 cos 2
The equation for () is more conveniently expressed
() = 1 + 4 () + 22 () + 4 [1 + ()] cos + cos 2
(b) The spectrum illustrating the terms involved in () is shown in Figure 3.6. The center
frequency of the filter is and the bandwidth must be greater than or equal to 2 . In
addition, − 2 or 3 , and + 2 . The last inequality states that
, which is redundant since we know that 3 .
14
CHAPTER 3. BASIC MODULATION TECHNIQUES
m t
0
W
Filter
Characteristic
m2 t
2W
fc W
fc
fc W
2 fc
f
Figure 3.6: Generation of an AM signal using a nonlinear operation.
(c) From the definition of () we have
() = ()
so that
() = 4 [1 + ()] cos
It follows that
= 01 =
Thus
= 01
(d) This method of forming a DSB signal avoids the need for an analog multiplier.
Problem 3.15
With
() = 4 cos (2 ) + cos (4 )
we have
Therefore
()
b
= 4 sin (2 ) + sin (4 )
() = 5 [4 cos (2 ) + cos (4 )] cos (2 )±5 [4 sin (2 ) + sin (4 )] sin (2 )
3.1. PROBLEMS
15
f
‐fc
0
f
Figure 3.7: Generation of upper-sideband SSB using sideband filtering.
This gives
() = 10 cos [2 ( + ) ] + 25 cos [2 ( + 2 ) ]
+10 cos [2 ( − ) ] + 25 cos [2 ( − 2 ) ]
∓10 cos [2 ( + ) ] + 25 cos [2 ( + 2 ) ]
±10 cos [2 ( − ) ] + 25 cos [2 ( − 2 ) ]
Note that by using the top, +, sign we have
() = 20 cos [2 ( − ) ] + 5 cos [2 ( − 2 ) ]
The upper-sideband (frquency components above ) cancel and we are left with only lowersideband terms (frquency components below ). Note that by using the bottom, +, sign
we have
() = 20 cos [2 ( + ) ] + 5 cos [2 ( + 2 ) ]
The lower-sideband (frquency components below ) cancel) and we are left with only uppersideband terms (frquency components above ). The spectrum in each case is obvious.
Problem 3.16
Consider the filter transfer function shown below
From the above figure, and comparing it with Figure 3.10 in the text, it is clear that
the uppersideband filter is given by
16
CHAPTER 3. BASIC MODULATION TECHNIQUES
1
1
[1 + sgn ( − )] + [1 − sgn ( + )]
2
2
Going through the same steps as in the text (pp 125-127) gives
( ) =
Problem 3.17
We have
1
1
() = () cos −
b () sin
2
2
2 () =
which is
∙
¸2
1
1
() cos ±
b () sin
2
2
1
1
1
b cos( ) sin( )+ 2
b 2 () [1 − cos(2 ]
2 () = 2 2 () [1 + cos(2 )]± 2 ()()
4
2
4
or
¤
1 £
b 2 () + 2 () cos(2 ) ± ()()
b sin( ) −
b 2 () cos(2 )
2 () = 2 2 () +
4
Therefore, yes, components are created at twice the carrier frequency.
Problem 3.18
We assume that the VSB waveform is given by
() =
1
cos ( − 1 )
2
1
+ (1 − ) cos ( + 1 )
2
1
+ cos ( + 2 )
2
We let () be () plus a carrier. Thus
() = () + cos
It can be shown that () can be written
() = 1 () cos () + 2 () sin
where
cos 1 + cos 2 +
2
¶ 2
µ
sin 1 − sin 2
2 () =
+
2
2
1 () =
3.1. PROBLEMS
17
f
A
‐A
f
‐W
0
W
Figure 3.8: Coherent demodulation of VSB.
Also
() = () cos ( + )
where () is the envelope and is therefore the output of an envelope detector. It follows
that
q
() = 12 () + 22 ()
For large, 1 () À 2 () for all so that 2 () can be neglected. Therefore, () = |1 ()|,
which is 12 () + , where is a dc bias. Thus if the detector is ac coupled, is removed
and the output () is () scaled by 12 .
Problem 3.19
Coherent demodulation translates the positive-frequency portion of the spectrum to = 0
and to = 2 . The negative-frequency portion of the spectrum is translated to = 0 and
to = −2 . The following figure illustrates the portion of the spectrum about = 0.
The portion of the spectrum is the portion between −A and A. Using the required symmetry properties for a VSB filter, the two portions of the translated spectra are conjugate
symmetric. Thus they add resulting in the spectrum shown in the bottom pane.
Problem 3.20
The result is shown in the following figure.
Problem 3.21
18
CHAPTER 3. BASIC MODULATION TECHNIQUES
Desired
Signal
1
Local
Oscillator
1 2
Signal at
Mixer
Output
2 IF
21 2
Image
Signal
21 2
Image Signal
at Mixer Output
3 2 21
2
Figure 3.9: Result for Problem 3.20.
3.1. PROBLEMS
19
Since high-side tuning is used, the local oscillator frequency is
= +
where , the carrier frequency of the input signal, varies between 10 and 30 MHz. The
ratio is
+ 30
=
+ 10
where is the frequency expressed in . We make the following table
MHz
04
06
08
10
50
100
29231
28868
28519
28182
23333
20000
A plot of as a function of is the required plot. However, we see that is nearly
constant and decreases as increases. Therefore, in this case, a large IF frequency should
be used consistant with hardware requirements and availability.
Problem 3.22
With = 455 kHz we have for high-side tuning
= + = 1100 + 455 = 1555 kHz
= + 2 = 1100 + 910 = 2010 kHz
and for low-side tuning we have
= − = 1100 − 455 = 665 kHz
= − 2 = 1100 − 910 = 210 kHz
With = 2500 we have for high-side tuning
= + = 1100 + 2500 = 3600 kHz
= + 2 = 1100 + 5000 = 6100 kHz
and for low-side tuning we have
= − = 1100 − 2500 = −1400 kHz
= 1400 kHz
= − 2 = 1100 − 5000 = −3900 kHz
= 3900 kHz
20
CHAPTER 3. BASIC MODULATION TECHNIQUES
In the preceding development for low-side tuning, recall that the spectra are symmetrical
about = 0.
Problem 3.23
We assume single-tone interference as in the body of the text. Therefore we assume
() = () cos (2 ) + cos (2 )
Therefore
2 () =
2 2
2
() [1 + cos (4 )] + 2 cos (2 ) cos (2 ) + [1 + cos (4 )]
2
2
Ignoring the dc terms, we have
2 () =
2 2
2
() cos (4 )+ [cos (2( + ) ]+ [cos (2( − ) ]+ cos (4 )
2
2
We therefore see that the squaring operation generates a component at twice the carrier
frequency and a component at twice the interference frequency. A pair of sidebands are also
generated about the carrier frequency, above and below the carrier frequency, and separated
from the carrier frequency by the interference frequency.
Problem 3.24
From (3.96) the transfer function of the holding operation is
( ) =
sin (2 )
exp (− )
2
Sampling reproduces this about = 0 and all harmonics of the sampling frequency. We
consider the terms about = 0 and = . Ignoring the phase, which is linear and therefore
only induces a delay, we have
1 ( ) =
sin (2 )
∗ [( ) + ( − )]
2
Finally we have
1 ( ) =
sin (2 )
sin (2( − ) )
+
2
2( − )
The equalizer will have the transfer function
( ) =
2
1
sin(2 )
3.1. PROBLEMS
21
1
where
In order for the frequency response to be bounded we must have 2 1 or 2
is the highest frequency in the message signal. We must also have 2 − 2 −1. To
plot this requires that be set as a parameter.
Problem 3.25
Let be the peak-to-peak value of the data signal. The peak error is 05% (±025%) and
the peak-to-peak error is 001 . The required number of quantizating levels is
= 100 ≤ 2 =
001
so we choose = 128 and = 7. The bandwidth is
= 2 log2 = 2 (7)
The value of is estimated by assuming that the speech is sampled at the Nyquist rate.
Then the sampling frequency is = 2 = 8 kHz. Each sample is encoded into = 7
pulses. Let each pulse be with corresponding bandwidth 1 . For our case
=
Thus the bandwidth is
1
1
=
2
1
= 2 = 2 log2
and so = 1. For = 1
= 2 (8 000) (7) = 112 kHz
Problem 3.26
The message signal is
() = 3 sin[2(10)] + 4 sin[2(20)]
The derivative of the message signal is
()
= 60 cos[2 (10) ] + 160 cos[2 (20)]
The maximum value of () is obviously 280 and the maximum occurs at = 0.
Thus
0
≥ 220
or
220
220
≥
=
= 4400
0
005
22
CHAPTER 3. BASIC MODULATION TECHNIQUES
x1 (BW W )
x2 (BW W )
x3 (BW 2W )
x4 (BW 4W )
x5 (BW 4W )
1
2
3
4
5
6
7
8
9
10
11
12
Output
Figure 3.10: Commutator configuration for Problem 3.27.
Thus, the minimum sampling frequency is 4400 Hz.
Problem 3.27
One possible commutator configuration is illustrated in Figure 3.10. The signal at the
point labeled “output” is the baseband signal. The minimum commutator speed is 2
revolutions per second. For simplicity the commutator is laid out in a straight line. Thus,
the illustration should be viewed as it would appear wrapped around a cylinder. After
taking the sample at point 12 the commutator moves to point 1. On each revolution, the
commutator collects 4 samples of 4 and 5 , 2 samples of 3 , and one sample of 1 and 2 .
The minimum transmission bandwidth is
X
= + + 2 + 4 + 4
=
= 12
Problem 3.28
If the speed of the comutator is doubled from 2W to 4W, the sampling frequency is obviously
doubled. The sampling frequencies are 4W, 4W, 8W, 16W, and 16W, respectively. The
required transmission freqency is doubled. However, the guard frequencies are doubled.
3.1. PROBLEMS
23
W
W
f
f1
f2
Figure 3.11: Single-sided spectrum for Problem 3.30.
Problem 3.29
The minimum baseband bandwidth is
= + + 2 + 5 + 7 = 16
A suitable scheme for this takes a bit of thought. However, the problem is not much different
than when we consider the requirement that a sum of sinusoids be periodic.To be periodic,
the frequencies must be commensurable (measurable by the same standard). In this case all
bandwidths must be factors of a common bandwidth, which is 70W. Thus,the communator
must have 70 distinct psitions.There are many possibilities one is shown in the Table below.
Single Bandwidth Communator Positions
7
10, 20, 30, 40, 50, 60, 70
5
1, 15, 29, 43, 57
2
2, 37
3
4
Note
Spacing = 70
7 = 10
70
Spacing = 5 = 14
Spacing = 70
2 = 35
Spacing = 70
1 = 70
70
Spacing = 1 = 70
The positions are arbitrary but the spacings are not. The communator rotates at a minimum rate of 2W. Note that there are many unused positions. These unused positions can be
used for other signals as long as the commensurable conditions hold.klmmmmmmmmmmmmmmmmmmmmmm
Problem 3.30
The single-sided spectrum for () is shown in Figure 3.11.
From the definition of () we have
() = 1 ( ) + 2 ( ) ∗ ( )
24
CHAPTER 3. BASIC MODULATION TECHNIQUES
Yf
a1
2a2W
a2W
f
f 2 f1
f1
f2
2 f2
f1 f 2
2 f2
Figure 3.12: Output spectrum for Problem 3.54.
The spectrum for ( ) is given in Figure 3.12. Demodulation can be a problem since it may
be difficult to filter the desired signals from the harmonic and intermodulation distortion
caused by the nonlinearity. As more signals are included in (), the problem becomes more
difficult. The difficulty with harmonically related carriers is that portions of the spectrum
of ( ) are sure to overlap. For example, assume that 2 = 21 . For this case, the harmonic
distortion arising from the spectrum centered about 1 falls exactly on top of the spectrum
centered about 2 .
3.2
Computer Exercises
Computer Exercise 3.1
The MATLAB program is
% File: ce3_1.m
t = 0:0.001:1;
fm = 1;
fc =10;
m = 4*cos(2*pi*fm*t-pi/9) + 2*sin(4*pi*fm*t);
[minmessage,index] = min(m);
mncoefs = [4 2]/abs(minmessage)
mn = m/abs(minmessage);
3.2. COMPUTER EXERCISES
25
plot(fm*t,mn,’k’),
grid, xlabel(’Normalized Time’), ylabel(’Amplitude’)
mintime = 0.001*(index-1);
dispa = [’The minimum of m(t) is ’, num2str(minmessage,’%15.5f’), ...
’and falls at ’,num2str(mintime,’%15.5f’), ’ s.’]; disp(dispa)
%
% Now we compute the efficiency.
%
mnsqave = mean(mn.*mn);
% mean-square value
a = 0.5;
% modulation index
asq = a^2;
efficiency = asq*mnsqave/(1+asq*mnsqave);
dispb = [’The efficiency is ’, num2str(efficency,’%15.5f’),’ .’]; disp(dispb)
%
% Now compute the carrier amplitude.
%
cpower = 50;
% carrier power
camplitude = sqrt(2*cpower); % carrier amplitude
%
% Compute the magnitude and phase spectra.
%
xct = camplitude*(1+a*mn).*cos(2*pi*fc*t);
Npts = 1000;
fftxct = fft(xct,Npts)/Npts;
s1 = [conj(fliplr(fftxct(2:25))) fftxct(1:25)];
mags1 = abs(s1);
angles1 = angle(s1);
for k=1:length(s1)
if mags1(k) = 0.01
angles1(k)=0;
end
end
%
% Plot the magnitude and phase spectra.
%
figure
subplot(2,1,1), stem(-24:24,mags1,’.’)
xlabel(’frequency’); ylabel(’magnitude’)
subplot(2,1,2), stem(-24:24,angles1,’.’)
xlabel(’frequency’); ylabel(’phase’)
26
CHAPTER 3. BASIC MODULATION TECHNIQUES
1.5
1
Amplitude
0.5
0
-0.5
-1
0
0.1
0.2
0.3
0.4
0.5
0.6
Normalized Time
0.7
0.8
0.9
Figure 3.13: Message signal.
% End of script file.
Executing the program gives:
» ce3_1
mncoefs =
0.9165 0.4583
The minimum of m(t) is -4.36424 and falls at 0.43500 s.
The efficiency is 0.11607 .
The program also generates Figures 3.13 and ??.
Computer Exercise 3.2
The MATLAB code written for Computer Exercise 3.2 follows.
% File: ce3_2.m
t = 0:0.001:1;
fm = 1;
1
3.2. COMPUTER EXERCISES
27
magnitude
6
4
2
0
-25
-20
-15
-10
-5
0
frequency
5
10
15
20
25
-20
-15
-10
-5
0
frequency
5
10
15
20
25
2
phase
1
0
-1
-2
-25
Figure 3.14: Spectra of AM signal.
28
CHAPTER 3. BASIC MODULATION TECHNIQUES
fc =10;
m = 2*cos(2*pi*fm*t) + cos(4*pi*fm*t);
mhil = 2*sin(2*pi*fm*t) + sin(4*pi*fm*t);
xctusb = 0.5*m.*cos(2*pi*fc*t)-0.5*mhil.*sin(2*pi*fc*t);
xctlsb = 0.5*m.*cos(2*pi*fc*t)+0.5*mhil.*sin(2*pi*fc*t);
%
% Plot time-domain signals.
%
subplot(2,1,1), plot(t,xctusb)
xlabel(’time’); ylabel(’USB signal’)
subplot(2,1,2), plot(t,xctlsb)
xlabel(’time’); ylabel(’LSB signal’)
%
% Determine and plot spectra.
%
Npts = 1000;
fftxctu = fft(xctusb,Npts)/Npts;
fftxctl = fft(xctlsb,Npts)/Npts;
sxctusb = [conj(fliplr(fftxctu(2:25))) fftxctu(1:25)];
magsusb = abs(sxctusb);
sxctlsb = [conj(fliplr(fftxctl(2:25))) fftxctl(1:25)];
magslsb = abs(sxctlsb);
figure
subplot(2,1,1), stem(-24:24,magsusb)
xlabel(’frequency’); ylabel(’USB magnitude’)
subplot(2,1,2), stem(-24:24,magslsb)
xlabel(’frequency’); ylabel(’LSB magnitude’)
% End of script file.
Executing the code gives Figures 3.15 and 3.16.
Computer Exercise 3.3
In this computer example we investigate the demodulation of SSB using carrier resertion
for several values of the constant .for an assumed message signal.
% File: ce3_3.m
t = 0:0.001:1;
fm = 1;
m = 2*cos(2*pi*fm*t) + cos(4*pi*fm*t);
mhil = 2*sin(2*pi*fm*t) + sin(4*pi*fm*t);
3.2. COMPUTER EXERCISES
29
USB signal
2
1
0
-1
-2
0
0.1
0.2
0.3
0.4
0.5
time
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
time
0.6
0.7
0.8
0.9
1
LSB signal
2
1
0
-1
-2
Figure 3.15: Time-domain waveforms.
30
CHAPTER 3. BASIC MODULATION TECHNIQUES
USB magnitude
0.8
0.6
0.4
0.2
0
-25
-20
-15
-10
-5
0
frequency
5
10
15
20
25
-20
-15
-10
-5
0
frequency
5
10
15
20
25
LSB magnitude
0.8
0.6
0.4
0.2
0
-25
Figure 3.16: Amplitude spectra.
3.2. COMPUTER EXERCISES
31
k = 1;
ydt = sqrt((k+m).*(k+m)+mhil.*mhil);
ydt1 = ydt-k;
k = 10;
ydt = sqrt((k+m).*(k+m)+mhil.*mhil);
ydt2 = ydt-k;
k = 100;
ydt = sqrt((k+m).*(k+m)+mhil.*mhil);
ydt3 = ydt-k;
subplot(3,1,1), plot(t,m,’k’,t,ydt1,’k--’)
ylabel(’k=1’)
subplot(3,1,2), plot(t,m,’k’,t,ydt2,’k--’)
ylabel(’k=10’)
subplot(3,1,3), plot(t,m,’k’,t,ydt3,’k--’)
ylabel(’k=100’)
% End of script file.
Executing the preceding MATLAB program gives the results illustrated in Figure 3.17. In
the three plots the message signal is the solid line and the dashed line represents the output
resulting from carrier reinsertion. It can be seen that the demodulated output improves for
increasing and that the distortion is negligible for = 100.
Computer Exercise 3.4
Two MATLAB programs are generated for this Computer Exercise, ce3_4a ancd ce3_4b.
The first of these two programs illustrates the message signal, the VSB signal and the VSB
signal after carrier reinsertion. The second program illustrates the spectrum of a VSB signal
The MATLAB code for the first program is
% File: ce3_4a.m
clear all
t = 0:0.001:1;
m = cos(2*pi*t)-cos(4*pi*t)+cos(6*pi*t);
e1 = 0.64; e2 = 0.78; e3 = 0.92;
fc = 25; A = 100;
ct = A*cos(2*pi*fc*t);
xct = e1*cos(2*pi*(fc+1)*t)+(1-e1)*cos(2*pi*(fc-1)*t)-...
e2*cos(2*pi*(fc+2)*t)-(1-e2)*cos(2*pi*(fc-2)*t)+...
e3*cos(2*pi*(fc+3)*t)+(1-e3)*cos(2*pi*(fc-3)*t)+...
cos(2*pi*(fc+4)*t)+cos(2*pi*(fc+5)*t)+cos(2*pi*(fc+6)*t);
32
CHAPTER 3. BASIC MODULATION TECHNIQUES
k=1
4
2
0
-2
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
k=10
4
2
0
-2
k=100
4
2
0
-2
Figure 3.17: Illustration of demodulation resertion for = 1, 10, and 100.
3.2. COMPUTER EXERCISES
33
m(t)
2
0
-2
-4
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
Time
0.6
0.7
0.8
0.9
1
xc(t)
5
0
xc(t)+c(t)
-5
102
100
98
Figure 3.18: Illustration of VSB (message signal, VSB signal, and VSB signal plus carrier
reinsertion).
xdem = abs(xct+ct);
subplot(3,1,1), plot(t,m), ylabel(’m(t)’)
subplot(3,1,2), plot(t,xct), ylabel(’xc(t)’)
subplot(3,1,3), plot(t,xdem), ylabel(’xc(t)+c(t)’)
xlabel(’Time’), axis([0 1 97 103])
% End of script file.
The result of demodulation using carrier reinsertion if shown in the bottom pane. The
message signal can be recovered using a dc-coupled envelope detection.
We now consider the spectrum of a VSB signal. We change the message signal to
include a carrier component and to include six components above the carrier. In addition,
all components in the message signal are set equal in order to illustrate the action of the
VSB filter. The MATLAB program is
% File: ce3_4b.m
clear all
t = 0:0.001:1;
e1 = 0.64; e2 = 0.78; e3 = 0.92;
fc = 25;
ct = 0.5*cos(2*pi*fc*t);
34
CHAPTER 3. BASIC MODULATION TECHNIQUES
1.4
VSB magnitude spectrum
1.2
1
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25
30
frequency
35
40
45
50
Figure 3.19: VSB spectrum.
xct = ct+e1*cos(2*pi*(fc+1)*t)+(1-e1)*cos(2*pi*(fc-1)*t)-...
e2*cos(2*pi*(fc+2)*t)-(1-e2)*cos(2*pi*(fc-2)*t)+...
e3*cos(2*pi*(fc+3)*t)+(1-e3)*cos(2*pi*(fc-3)*t)+...
cos(2*pi*(fc+4)*t)+cos(2*pi*(fc+5)*t)+cos(2*pi*(fc+6)*t);
Npts = 1000;
fftxctu = fft(xct,Npts)/Npts;
magvsb=2*abs(fftxctu(1:51));
freq=0:50;
stem(freq, magvsb)
xlabel(’frequency’); ylabel(’VSB magnitude spectrum’)
% End of script file.
The resulting spectrum follows. Note that there is a carrier component at = 25. There
are three attenuated components above the carrier and three non-attenuated components.
Below the carrier there are the complements of the three attenuated components above the
carrier.
Computer Exercise 3.5
The MATLAB program used appears below. Given the annotations, the code should be
understandable. (Thanks to Dr. John Tranter of the University of Minnesota for his help
on this problem.)
3.2. COMPUTER EXERCISES
35
% File CE3.5.m
clear all; close all;
f1 = 70; f2 = 300; % sine frequencies
fmin = min(f1,f2); % use lowest freq. for 1 cycle on time axis
A1 = 3; A2 = 1.5; % sine amplitudes
samp = [15 60]; % sample period (in MATLAB increments)
N = 1500; % discretize signal with N points
hl = 1; % staircase amplitude
t = 1/N/fmin:1/N/fmin:1/fmin; % x-axis in seconds
x = A1*sin(2*pi*f1*t)+A2*sin(2*pi*f2*t); % build our signal
for kk = 1:2
d(1:samp(kk))=0;
for ii=samp(kk)+1:samp(kk):N
dd(ii) = sign(x(ii) - d(ii-samp(kk))); % this is our delta
d(ii:ii+samp(kk)-1) = d(ii-samp(kk)) + hl*dd(ii);
end
d = d(1:N); % trim/pad extra samples (if sample
dd = [dd zeros(1,N-numel(dd))]; % period does not exactly divide N)
subplot(3,2,[kk kk+2]),plot(t,x,’k’),...
axis([t(1) t(N) min([x d]) max([x d])])
hold on,stairs(t,d,’r’),xlabel(’Time (s)’)
if (kk == 1)
str = ’no slope overload)’;
else
str = ’slope overload)’;
end
title(strcat(’Stairstep Approximation (’,str));
subplot(3,2,kk+4),stem(t(1:samp:end),dd(1:samp:end),’marker’,’none’)
axis([t(1) t(N) -1.2 1.2]),xlabel(’Time (s)’)
title(strcat(’Delta Modulation (’,str));
clear d; clear dd; % diff. sample rate = diff. dimensions...
end
% This is a neat trick that automatically maximizes your figure
screen_size = get(0, ’ScreenSize’);
set(figure(1), ’Position’, [0 0 screen_size(3) screen_size(4) ] );
% End of script file.
Executing the program yields the following output.
36
CHAPTER 3. BASIC MODULATION TECHNIQUES
Stairstep Approximation (no slope overload)
Stairstep Approximation (slope overload)
5
4
4
3
3
2
2
1
1
0
0
-1
-1
-2
-2
-3
-3
-4
2
4
6
8
10
12
Time (s)
Delta Modulation (no slope overload)
-4
14
2
4
-3
x 10
1
1
0.5
0.5
0
0
-0.5
-0.5
-1
6
8
10
Time (s)
Delta Modulation (slope overload)
12
14
-3
x 10
-1
2
4
6
8
Time (s)
10
12
14
-3
x 10
2
4
6
8
Time (s)
10
12
14
-3
x 10
Figure 3.20: Delta modulation without slope overload (left two panes) and with slope
overload (right two panes).
Computer Exercise 3.6
The MATLAB code follows.
% File ce3_6.m
clear all
t=0:0.001:(2-0.001);
m=zeros(1,2000);
x=zeros(1,2000);
m=cos(2*pi*t)-cos(4*pi*t)+cos(6*pi*t);
tau=5;
for k=1:2000
if rem(k,10)==0
x(k)=1;
end
end
xsamp=m.*x;
h=zeros(1,10);
for k=1:tau
h(k)=1;
end
xsh=conv(xsamp,h);
3.2. COMPUTER EXERCISES
37
Samp & Hold Output
2
0
-2
-4
0.5
0.55
0.6
0.65
0.7
0.75
Time
0.8
0.85
0.9
0.95
1
Approximation
1
0
-1
-2
0.4
Message
Reconstruction
0.6
0.8
1
Time
1.2
1.4
1.6
Figure 3.21: PAM and reconstruction. (Top pane - Sample and hold output. Bottom pane
- original message signal and reconstructured signal.)
[B,A]=butter(3,0.05);
y=filter(B,A,xsh);
subplot(2,1,1), plot(t(500:1000),xsh(500:1000))
axis([0.5 1 -4 2])
xlabel(’Time’); ylabel(’Samp & Hold Output’)
subplot(2,1,2), plot(t(500:1500),y(500:1500),’k’,...
t(500:1500),0.5*m(500:1500),’k--’)
xlabel(’Time’); ylabel(’Approximation’)
legend(’Message’,’Reconstruction’,’Location’,’South’)
% End of script file.
Note that for this example, the simulation sampling frequency is 1000 Hz. The message
signal is sampled every 10 simulation steps. Thus = 100 Hz. The width of the PAM
pulse is 5 simulation steps so = 0005 s. Thus = 05. The 3 dB frequency of the
Butterworth filter is 0.5(1000/2)=250 Hz. Other values for are easily tried.
Chapter 4
Angle Modulation and
Multiplexing
4.1
Problems
Problem 4.1
By definition
() = cos [ + ()] = cos [ + ( − 0 )]
The waveforms are shown below. The the top pane represents the unmodulated signal. The
middle pane is for is for = , and the bottom-right pane is for = −38. The plots
illustrate 2 s of data with a frequency 1 Hz. The parameter 0 = 0 for all plots.
Problem 4.2
Once again, by definition
() = cos [ + ()] = cos [ + ( − 0 )]
The waveforms are shown below. The the top pane represents the unmodulated signal. The
middle pane is for is for = −2, and the bottom-right pane is for = 38. The plots
illustrate 2 s of data with a frequency 1 Hz. The parameter 0 = 0 for all plots.
Problem 4.3
With
³
´
() = sin 2 +
6
1
2
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
Amplitude
1
0
-1
-1
-0.8
-0.6
-0.4
-0.2
0
Time
0.2
0.4
0.6
0.8
1
-0.8
-0.6
-0.4
-0.2
0
Time
0.2
0.4
0.6
0.8
1
-0.8
-0.6
-0.4
-0.2
0
Time
0.2
0.4
0.6
0.8
1
Amplitude
1
0
-1
-1
Amplitude
1
0
-1
-1
Figure 4.1: Time-domain plots for Problem 4.1.
Amplitude
1
0
-1
-1
-0.8
-0.6
-0.4
-0.2
0
Time
0.2
0.4
0.6
0.8
1
-0.8
-0.6
-0.4
-0.2
0
Time
0.2
0.4
0.6
0.8
1
-0.8
-0.6
-0.4
-0.2
0
Time
0.2
0.4
0.6
0.8
1
Amplitude
1
0
-1
-1
Amplitude
1
0
-1
-1
Figure 4.2: Time-domain plots for Problem 4.2.
4.1. PROBLEMS
3
we have for AM
h
³
´i
cos (2 )
() = cos 1 + sin 2 +
6
where is the modulation index. Thus () can be written
½µ
¾
¶
³
³
h
´
´i
1+
() = Re
exp 2 +
− exp −2 −
exp (2 )
2
6
6
or
() = Re
This gives
³
³
n³
o
h
´
´i´
1+
exp 2 + −
− exp −2 − +
exp (2 )
2
6
2
6
2
() = Re
³
³
n³
o
h
´
´i´
exp 2 −
− exp −2 +
exp (2 )
1+
2
3
3
The phasor diagram is the sama as in Figure 4.4 except that the vectors rpresenting the
modulation have inital angles ± 3
For FM we have
³
´
cos (2 )
() = cos (2 ) − sin 2 +
6
or
³
³
´
´
() = cos (2 ) + cos 2( − ) −
− cos 2( + ) +
2
6
2
6
The phasor diagram is obvious.
Problem 4.4
Let () = cos where = 2 . This gives
o
n
2 () = Re cos
Expanding a Fourier series gives
cos =
∞
X
=−∞
where
=
2
Z
−
cos −
4
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
With = , the Fourier coefficients become
Z
1
=
cos −
2 −
¡
¢
Since cos = sin + 2
Z
1
[ sin(+ 2 )−]
=
2 −
With = + 2 , the preceding becomes
1
2
Z 32
2
½
=
This gives
=
[ sin −+ 2 ]
−2
1
2
Z
[ sin −]
−
¾
where the limits have been adjusted by recognizing that the integrand is periodic with
period 2. Thus
= 2 ()
and
(
2 () = Re
Taking the real part yields
2 () =
∞
X
( +
2 )
()
=−∞
)
h
i
() cos ( + ) +
2
=−∞
∞
X
The amplitude spectrum is therefore the same as in the preceding problem. The phase
spectrum is the same as in the preceding problem except that
2 is added to each term.
Problem 4.5
Since sin() = cos( − 2 ) we can write
which is
Since
³
´
3 () = sin( + sin ) = cos − + sin
2
n
o
3 () = Re ( −2) sin
sin =
∞
X
=−∞
()
4.1. PROBLEMS
5
we have
(
( −2)
3 () = Re
Thus
∞
X
3 () =
3 () =
()
=−∞
=−∞
Thus
∞
X
)
o
n
() Re ( + −2)
h
i
() cos ( + ) −
2
=−∞
∞
X
Note that the amplitude spectrum of 3 () is identical to the amplitude spectrum for both
1 () and 2 (). The phase spectrum of 3 () is formed from the spectrum of 1 () by
adding −2 to each term.
For 4 () we write
³
´
4 () = sin( + cos ) = cos − + cos
2
Using the result of the preceding problem we write
(
)
∞
X
( +
( −2)
2 )
4 () = Re
()
=−∞
This gives
4 () =
∞
X
=−∞
Thus
4 () =
o
n
() Re ( + − 2 + 2 )
i
h
() cos ( + ) + ( − 1)
2
=−∞
∞
X
Compared to 1 (), 2 (), 3 () and 4 (), we see that the only difference is in the phase
spectrum.
Problem 4.6
From the problem statement
() = cos [2 (40) + 10 sin (2 (5) )]
Since = 40 and = 5 (it is important to note that is an integer multiple of )
there is considerable overlap of the portion of the spectrum centered about = −50 with
6
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
1
0.8
0.6
Magnitude
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
0
0.02
0.04
0.06
0.08
0.1
0.12
Frequency
0.14
0.16
0.18
0.2
Figure 4.3: Time-domain signal for Problem 4.6.
the portion of the spectrum centered about = 50 for large (such as 10). This can also
be seen in the time domain. We can see from the time domain plot (top pane) of Figure
??. Thus, this is not a true FM signal.
Problem 4.7
We are given 0 (5) = −0178 and 1 (5) = −0328. Using
+1 () =
with = 5 we have
2
() − −1 ()
2
+1 (5) = (5) + −1 (5)
5
With = 1,
2 (5) =
=
2
(1)1 (5) + 0 (5)
5
2
(−0328) + 0178 = 00468
5
With = 2,
3 (5) =
=
2
(2)2 (5) − 1 (5)
5
2
(2)00468 + 0328 = 03648
5
4.1. PROBLEMS
7
0.3
Magnitude
0.25
0.2
0.15
0.1
0.05
0
1000
1200
1400
1600
1800 2000 2200
Frequency
2400
2600
2800
3000
Figure 4.4:
Finally with = 3 we have
4 (5) =
=
2
(3)3 (5) − 2 (5)
5
2
(3) (03648) − 00468 = 0
5
Problem 4.8
The amplitude and phase spectra follow directly from the table of Fourier-Bessel coefficients,
or one can write a simple MATLAB program as was done here.. The single-sided magnitude
spectrum is shown in Figure ??. The magnitude spectrum is plotted assuming = 1.
The student should verify that this spectrum is correct and the student should plot the
phase spectrum using both a table of Bessel functions and the fft MATLAB function. The
table of Bessel functions approach is trivial. The other approach takes a bit of thought to
get as clean plot.
Problem 4.9
(a) Since the carrier frequency is 1000 Hertz, the general form of () is
£
¤
() = cos 2 (1000) + 40 sin(52 ) = cos [2 (1000) + ()]
8
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
The phase deviation, (), is therefore given by
() = 40 sin(52 )
rad
The frequency deviation is
= 400 cos(52 )
or
rad/sec
1
200
=
cos(52 )
2
Hz
(b) Since the carrier frequency is 1000 Hertz, the general form of () is
() = cos [2 (1000) − 2(400)] = cos [2 (1000) + ()]
The phase deviation is
() = −2 (400)
rad
(Note that we substracted the phase of the unmodulated carrier from the instantaneous
carrier.) The frequency deviation is
= −2 (400)
or
rad/sec
1
= −400
2
Hz
Problem 4.10
(a) Since the carrier frequency is 1000 Hertz, the general form of () is
£
¤
() = cos 2 (1000) − 2(1000) + 2(1200)2 = cos [2 (1000) + ()]
The phase deviation is
() = −2 (1000) + 2(1200)2
rad
and the frequency deviation is
= −2000 + 2(2400)
or
1
= −1000 + 2400
2
rad/sec
Hz
4.1. PROBLEMS
9
(b) Since the carrier frequency is 1000 Hertz, the general form of () is
h
√i
() = cos 2 (1000) − 2(100) + 10 = cos [2 (1000) + ()]
The phase deviation is
√
√
() = −2(100) + 10 − 2(1000) = −2(100) + 10
and the frequency deviation is
1
1
5
= −2(100) + (10) − 2 = −2(100) + √
2
or
1
5
= −100 + √
2
2
Hz
Problem 4.11
(a) The phase deviation is
() = 2(20)
Z
which is, for 0 ≤ ≤ 8,
() = 40
¸
1
4Π ( − 4)
8
Z
∙
4 = 160
0
Thus
() = 0,
0
() = 160
() = 160(8)
0≤≤8
8
(b) The frequency deviation is, by definition
∙
¸
1
1
= 80Π ( − 4) = 80
2
8
for 0 ≤ ≤ 8 and is zero for 0 and 8.
(c) The peak frequency deviation is 80 Hz.
(d) The peak phase deviation is 160(8) = 1280 radians.
(e) The transmitted power is
=
(100)2
= 5000
2
W
Hz
rad/sec
rad
10
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
Problem 4.12
(a) The phase deviation is
() = 0,
() =
=
() =
=
=
=
() =
3
µ
¶
Z
160 2 9
4
( − 3) =
− − 3 + 9
40
3
2
2
3 3
¡
¢
80 2
− 6 + 9
3≤≤6
3
¶
¶
Z µ
Z µ
4
4
4 − ( − 6) = 240 + 40
12 −
240 + 40
3
3
6
µ6
¶
41 2
240 + 40 12( − 6) −
( − 36)
32
µ
¶
2 2
240 + 40 − + 12 − 48
3
¢
80 ¡ 2
6≤≤9
240 +
− + 18 − 72
3
480
9
(b) The frequency deviation in Hz is
¸
1
() = 80Λ ( − 6)
3
∙
(c) The peak frequency deviation is 4 (20) = 80 Hz.
(d) The peak phase deviation is, since the area of Λ[(−6)3] = 3 is 2 (3) (4) = 24 (20) =
480
(e) The modulator output power is
1
1
= 2 = (100)2 = 5000 Watts
2
2
Problem 4.13
For all three message signals illustrated in Figure 4.37, the frequency deviation in Hz is
simply the given message signal with the ordinate values multiplied by 10.
For the first message signal (the top signal given in Figure 4.37) the phase deviation in
radians is given
Z
Z
() = 2
() = 20
()
4.1. PROBLEMS
11
For 0 ≤ ≤ 1, we have
() = 20
Z
2 = 202
0
For 1 ≤ ≤ 2
() = (1) + 20
Z
1
¡
¢
(5 − ) = 20 + 100 ( − 1) − 10 2 − 1
= 20 + 100 − 100 − 102 + 10 = −70 + 100 − 102
For 2 ≤ ≤ 3
() = (2) + 20
Z
3 = 90 + 60 ( − 2) = 60 − 30
Z
2 = 150 + 40( − 3) = 30 + 40
2
For 3 ≤ ≤ 4
() = (3) + 20
3
Finally, for 4 we recognize that () = (4) = 190. The required figure results by
plotting these curves.
The phase deviation for the second message signal (bottom right) is given again by
() = 2
Z
() = 20
For 0 ≤ ≤ 1, we have
() = 20
Z
Z
()
= 102
0
For 1 ≤ ≤ 2
Z
¡
¢
() = (1) + 20
( − 2) = 10 + 10 2 − 1 − 40( − 1)
1
¢
¡
= 10 2 − 4 + 4 = 10 ( − 2)2
For 2 ≤ ≤ 4
() = (2) + 20
Z
2
¡
¢
(6 − 2) = 0 + 20 (6) ( − 2) − 20 2 − 4
= −20(2 − 6 + 8)
12
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
Finally, for 4 we recognize that () = (4) = 0 (Note that the area under the curve
from 0 to 4 is obviously zero). The required figure follows by plotting these expressions.
Finally, the phase deviation in radians for the third message signal is again given by
Z
Z
() = 20
()
() = 2
For 0 ≤ ≤ 1, we have
() = 20
Z
0
For 1 ≤ ≤ 2
() = (1) + 20
Z
1
For 2 ≤ ≤ 25
() = (2) + 20
Z
2
(−2) = −202
2 = −20 + 40 ( − 1) = −60 + 40
¡
¢
(10 − 4) = 20 + 20 (10) ( − 2) − 20(2) 2 − 4
= 20(−22 + 10 + 7)
For 25 ≤ ≤ 3
() = (25) − 20
Z
25
2 = 15 − 40( − 25)
= (−250 + 500 + 140) − 40( + 25)
= 390 − 40 − 100 = 290 − 40
For 3 ≤ ≤ 4
() = (3) + 20
Z
3
(2 − 8) = 170 + 20(2 − 9) − 20(8)( − 3)
= 20(2 − 8 + 235)
Finally, for 4 we recognize that () = (4) = 20(16 − 32 + 235) = 150. The required
figure follows by plotting these expressions.
Problem 4.14
(a) The peak deviation is (20)(5) = 100 and = 10. Thus, the modulation index is
100
10 = 10.
(b) The magnitude spectrum, scaled to , is the Fourier-Bessel spectrum of Figure 4.5
with = 10
4.1. PROBLEMS
13
(c) Since is not ¿ 1, this is not narrowband FM. The bandwidth exceeds 2 .
(d) For phase modulation, (5) = 10 or = 2.
Problem 4.15
The results are given in the following table. The deviation ratio is found by definition and
the bandwidth follows from Carson’s rule.
Part
a
b
c
d
20
200
2000
20000
= 10 = 1500
0.0133
0.1333
1.3333
13.3333
Problem 4.16
From
() =
∞
X
= 2 ( + 1)
30.39 kHz
34 kHz
70 kHz
430 kHz
() cos[( + ) ]
=−∞
we obtain
∞
­ 2 ® 1 2 X
() =
2 ()
2 =−∞
Also
­ 2 ® ­ 2
®
() = cos2 [ + ()]
which, assuming that À 1 so that () has no dc component, is
­ 2 ® 1 2
() =
2
This gives
∞
1 2 1 2 X 2
=
()
2
2 =−∞
from which
∞
X
2 () = 1
=−∞
Problem 4.17
Since
1
() =
2
Z
−
−(− sin )
1
=
2
Z
−
( sin −)
14
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
we can write
1
() =
2
Z
1
cos ( sin − ) +
2
−
Z
−
sin ( sin − )
The imaginary part of () is zero, since the integrand is an odd function of and the
limits (− ) are even. Thus
Z
1
cos ( sin − )
() =
2 −
Since the integrand is even
1
() =
Z
0
cos ( sin − )
which is the first required result. With the change of variables = − , we have
Z
1
cos [ sin ( − ) − ( − )] (−1)
() =
0
Z
1
=
cos [ sin ( − ) − + ]
0
Since sin ( − ) = sin , we can write
Z
1
cos [ sin + − ]
() =
0
Using the identity
cos ( − ) = cos cos + sin sin
with
= sin +
and
=
yields
() =
1
+
Z
cos [ sin + ] cos ()
0
1
Z
sin [ sin + ] sin ()
0
Since sin () = 0 for all , the second integral in the preceding expression is zero. Also
cos () = (−1)
4.1. PROBLEMS
15
Thus
() = (−1)
However
1
− () =
1
Z
Z
cos [ sin + ]
0
cos [ sin + ]
0
Thus
() = (−1) − ()
or equivalently
− () = (−1) ()
Problem 4.18
(a) Peak frequency deviation = 10(8) = 80 Hz.
(b) Since
Z
() = 2(8)
10 cos(20) =
160
sin(20) = 8 sin(20)
20
the peak phase deviation is 8 rad.
(c) = 8
(d) The power at the filter input is
=
(10)2
2
=
= 50 W
2
2
The power at the filter output is determined by the number of spectral components passed
by the filter. The bandwith of the filter is 70 Hz, which is 35 Hz each sidfe of the carrier.
Since = 10 the filter passes 3 terms each side of the carrier. Thus the power at the filter
output is
#
"
3
3
X
2 X 2
2
2
2
=
(8) =
(8)
0 (8) + 2
2 =−3
2
=1
£
¤
= 50 (0172) + 2(0235)2 + 2(0113)2 + 2(0291)2
= 1674 W
(e) The input and output spectra are determined from the corresponding Fourier-Bessel
spectra.
Problem 4.19
16
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
We wish to find such that
= 02 (8) + 2
X
=1
02 (8) ≥ 080
This gives = 7, yielding a power ratio of = 08593. The bandwidth is therefore
= 2 = 2 (7) (250) = 3500 Hz
For ≥ 09, we have = 8 for a power ratio of 09592. This gives
= 2 = 2 (8) (250) = 4000 Hz
Problem 4.20
From the given data, we have
1 = 110
kHz
1 = 005
2 = (005) = 20
⼀
⼀
This gives
=
20
= 400
005
and
1 = (100)
kHz = 44 MHz
The two permissible local oscillator frequencies are
01 = 100 − 44 = 56
02 = 100 + 44 = 144
MHz
MHz
The center frequency of the bandpass filter must be = 100 MHz and the bandwidth is
¡ ¢
= 2 ( + 1) = 2 (20 + 1) (10) 103
⼀
or
= 420 kHz
Problem 4.21
For the circuit shown
( ) =
D =
( )
=
1
( )
+ 2 + 2
4.1. PROBLEMS
17
or
( ) =
where
1
³
´
1 + 2 − 21
10−3
=
= 10−6
3
10
¡ ¢¡
¢
= = 103 10−9 = 10−6
=
A plot of the amplitude response shows that the linear region extends from approximately
54 kHz to118 kHz. Thus an appropriate carrier frequency is
=
118 + 54
= 86
2
kHz
The slope of the operating characteristic at the operating point is measured from the amplitude response. The result is
¢
¡
∼
= 8 10−6
Problem 4.22
We can solve this problem by determining the peak of the amplitude response characteristic.
This peak falls at
1
= √
2
¡
¢
It is clear that 100 MHz. Let = 150 MHz and let = 0001 10−12 . This gives
=
1
(2)2 2
¡
¢
= 1126 10−3
We find the value of by trial and error using plots of the amplitude response. An
appropriate value for is found to be 1 Ω. With these values, the discriminator constant
is approximately
¢
¡
≈ 85 10−9
Problem 4.23
From (4.117)
¡
¢
0 2 + 2∆ + 2
= lim 3
→0 + 2 + +
18
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
1. For first-order PLL ( = 0, = 0) this is
¡
¢
0 2 + 2∆ + 2
0 2 + 2∆ + 2
= lim
=
lim
→0
→0
3 + 2
2 +
Thus
0 2 + 2∆ + 2
0 + (2∆ ) + (22 )
=
lim
→0
→0
2 +
1 + ( )
0
6= 0, ∆ = 0, and = 0:
= lim
=0
→0 1 + ( )
0 + (2∆ )
2∆
6
=
0, ∆ 6= 0, and = 0:
= lim
=
→0 1 + ( )
0 + (2∆ ) + (22 )
2
6
=
0, ∆ 6= 0, and 6= 0:
= lim
= lim
=∞
→0
→0
1 + ( )
= lim
For 0
For 0
For 0
2. For a second-order PLL ( 6= 0, = 0) this is
¡
¢
0 2 + 2∆ + 2
0 2 + 2∆ + 2
=
lim
= lim
→0
→0
3 + 2 +
2 + +
Thus
2
which is zero for = 0 and nonzero for 6= 0 for any finite values of 0 and ∆ .
3. For a third-order PLL ( 6= 0, 6= 0) this is
¡
¢
0 2 + 2∆ + 2
0
=
=0
= lim 3
2
→0 + + +
=
for any finite values 0 , ∆ , and .
Problem 4.24
The Costas PLL is shown in Figure 4.26. The output of the top multiplier is
() cos [2 cos ( + )] = () cos + () cos (2 + )
which, after lowpass filtering, is () cos . The quadrature multiplier output is
() cos [2 sin ( + )] = () sin + () sin (2 + )
which, after lowpass filtering, is () sin . The multiplication of the lowpass filter outputs
is
() cos () sin = 2 () sin 2
4.1. PROBLEMS
19
d / dt
A
Figure 4.5: Phase-plane plot for a Costas PLL.
as indicated. Note that with the assumed input () cos and VCO output 2 cos ( + ),
the phase error is . Thus the VCO is defined by
=
= − ()
This is shown in Figure 4.5.
Since the
intersection is on a portion of the curve with negative slope, the point A
at the origin is a stable operating point. Thus the loop locks with zero phase error and zero
frequency error.
Problem 4.25
¡ ¢
With () = cos(20 ), we desire 0 () = cos 2 73 0 . Assume that the VCO output
is a pulse train with frequency 13 . The pulse should be narrow so that the seventh harmonic
is relatively large. The spectrum of the VCO output consists of components separated by
1
3 0 with an envelope of sinc( ), where is the pulse width. The center frequency of the
bandpass filter is 73 0 and the bandwidth is on the order of 13 0 as shown in Figure.4.6.
Problem 4.26
The phase plane is defined by
= ∆ − sin ()
at = 0, = , the steady-state phase error. Thus
−1
= sin
µ
∆
¶
−1
= sin
µ
∆
2 (100)
¶
20
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
This component (the fundamental)
tracks the input signal
Bandpass filter
passband
0
1
f0
3
7
f0
3
f
Figure 4.6: Filtering scheme for Problem 3.44.
For ∆ = 2 (30)
30
100
¶
µ
50
100
¶
µ
80
100
¶
µ
¶
−1
= sin
µ
For ∆ = 2 (50)
−1
= sin
For ∆ = 2 (80)
−1
= sin
For ∆ = −2 (80)
−1
= sin
−80
100
= 1746 degrees
= 30 degrees
= 5313 defrees
= −5313 degrees
For ∆ = 2 (120), there is no stable operating point and the frequency error and the phase
error oscillate (PLL slips cycles continually).
Problem 4.27
From the definition
() = − ()
4.1. PROBLEMS
21
it is clear that in the limit as → ∞, (0) = ∞, and () = 0 for 6= 0. Also
Z ∞
Z ∞
¯∞
() =
− = −− −− ¯0 = 1
−∞
0
Therefore the given () satisfies all properties of an impulse function in the limit as →
∞.
Problem 4.28
From the definition of the transfer function
³
+
+
´
Θ ()
()
³
´
=
=
Φ ()
+ ()
+ +
which is
+
( + )
( + )
Θ ()
=
= 2
Φ ()
( + ) + ( + )
+ ( + ) +
Therefore
2 + 2 + 2 = 2 + ( + ) +
This gives
=
and
=
Problem 4.29
Since
p
+
√
2
Θ ()
( + )
=
Φ ()
( + ) + ( + )
we first determine
() =
so that
or
( + )
( + )
Ψ()
=1−
=
Φ ()
( + ) + ( + )
( + ) + ( + )
∙
0 2∆ 2
+ 2 + 3
Ψ() =
∙
¸∙
0 2 + 2∆ + 2
Ψ() =
2
( + )
( + ) + ( + )
¸∙
¸
( + )
( + ) + ( + )
¸
22
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
Multiplying by and taking the limit gives
¸∙
¸
∙
( + )
2
= lim 0 + 2∆ +
→0
( + ) + ( + )
or
¸∙ ¸
∙
2
= lim 2∆ +
→0
Thus
For 0 6= 0, ∆ = 0, and = 0:
= 0
For 0 6= 0, ∆ 6= 0, and = 0:
= lim
For 0 6= 0, ∆ 6= 0, and 6= 0:
2∆
2
= lim
=∞
→0
→0
Problem 4.30
Since the phase error is assumed small, so that cos [()] ≈ 1, the input from the top branch
(see Figure 4.26 in the text.) to the multiplier preceding the loop filter is simply the phase
error, () cos[()] ≈ (). The other input to the multiplier, for sufficiently small phase
error, is () sin[()] ≈ ()() Thus, the input to the loop filter preceding the VCO is
2 ()(). Since () is a step function
2 () = () = ( − 0 )
so that the input to the loop filter is ( − 0 )(), which is exactly the loop filter input to
a PLL. We therefore see that this problem is identical to Example 4.8 in the text and the
phase error as a function of time is that given by (4.140) with replaced by − 0 . The
parameter in the loop filter transfer function is = as defined by (4.136). This
problem illustrates the equivalence of a Costas PLL to a PLL in cases where the phase error
is small.
Problem 4.31
With the assumed VCO output represented by −() we have the output of the complex
phase detector defined as
()[()−()] = () cos[() − ()] + () sin[() − ()]
This is Figure 4.26 with the top branch representing the real component () cos[()]
and the bottom branch representing the quadrature, or imaginary, component () sin[()].
Costas PLLs are often simulated using a complex multiplication in this manner.
4.1. PROBLEMS
23
Problem 4.32
For = we can write, from (4.146),
() = [cos + cos ( + ) ]
which is
() = [(1 + cos ) cos − sin sin ]
This yields
() = () cos [ + ()]
where
¸
sin
() = tan
1 + cos
∙
¸
−1
= tan
tan
=
2
2
−1
This gives
1
() =
2
∙
µ
2
2
¶
1
=
2
For = − , we get
1
() = −
2
Finally, for À we see from the phasor diagram that
() ≈ () =
and
() =
(2 ) =
2
Problem 4.33
Let be the peak-to-peak value of the data signal. The peak error is 020% and the
peak-to-peak error is 0004 . The required number of quantizating levels is
= 250 ≤ 2 =
0004
so we choose = 256 and = 8. The bandwidth is
= 2 log2 = 2 (8)
24
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
W
W
f
f1
f2
Figure 4.7: Single-sided spectrum for Problem 3.54.
The value of is estimated by assuming that the speech is sampled at the Nyquist rate.
Then the sampling frequency is = 2 = 8 kHz. Each sample is encoded into = 8
pulses. Let each pulse be with corresponding bandwidth 1 . For our case
=
1
1
=
2
Thus the bandwidth is
1
= 2 = 2 log2 = 2 = 2
and so = 1. For = 1
= 2 (8 000) (8) = 112 kHz
Problem 4.34
The single-sided spectrum for () is shown in Figure 4.7.
From the definition of () we have
() = 1 ( ) + 2 ( ) ∗ ( )
The spectrum for ( ) is given in Figure 4.8. Demodulation can be a problem since it may
be difficult to filter the desired signals from the harmonic and intermodulation distortion
caused by the nonlinearity. As more signals are included in (), the problem becomes more
difficult. The difficulty with harmonically related carriers is that portions of the spectrum
of ( ) are sure to overlap. For example, assume that 2 = 21 . For this case, the harmonic
distortion arising from the spectrum centered about 1 falls exactly on top of the spectrum
centered about 2 .
4.2. COMPUTER EXERCISES
25
Yf
a1
2a2W
a2W
f
f 2 f1
f1
f2
2 f2
f1 f 2
2 f2
Figure 4.8: Output spectrum for Problem 3.54.
4.2
Computer Exercises
Computer Exercise 4.1
This Computer Exercise is a simple modification of Computer Exercise 4.1. Understanding
the relationship between Figure 4.7 and the result of this exercise is, however, important.
The MATLAB code follows.
% File: c4ce1.m
fs = 500;
delt = 1/fs;
t = 0:delt:1-delt;
npts = length(t);
fm = 20;
fd = [10 20 100];
for k=1:3
beta = fd(k)/fm;
cxce = exp(i*beta*sin(2*pi*fm*t));
as = (1/npts)*abs(fft(cxce));
evenf = [as(fs/2:fs) as(1:fs/2-1)];
fn = -fs/2:fs/2-1;
subplot(3,1,k); stem(fn,2*evenf,’.’)
ylabel(’Amplitude’)
end
26
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
Amplitude
2
1
0
-250
-200
-150
-100
-50
0
50
100
150
200
250
-200
-150
-100
-50
0
50
100
150
200
250
-200
-150
-100
-50
0
50
100
150
200
250
Amplitude
2
1
Amplitude
0
-250
1
0.5
0
-250
Figure 4.9: FM spectra with held constant and modulation indices of 0.5, 1 and 5.
% End of script file.
Executing the program gives the following spectral plots.
Computer Exercise 4.2
For a square-wave message signal, the phase deviation will be a triangular-wave. The
triangular-wave message signal is generated using a 6-term Fourier series. The MATLAB
code follows.
% File: c4ce2.m
fs = 1000;
% sampling frequency
delt = 1/fs;
% sampling increment
t = 0:delt:1-delt;
% time vector
npts = length(t);
% number of points
fn = (0:(npts/2))*(fs/npts); % frequency vector for plot
% Generate a triangular-wave signal as the phase deviation corresponding
% to a square-wave message signal using Fourier series and plot.
4.2. COMPUTER EXERCISES
27
fm = 5;
b = [1 -1/9 1/25 -1/49 1/81 -1/121];
c = 2*pi*fm*[1 3 5 7 9 11];
m = zeros(1,length(t));
for n=1:length(c)
m=m+b(n)*sin(c(n)*t);
end
m = m/max(m);
plot(t,m), xlabel(’Time’)
% Determine spectra.
fd1 = 0.10;
xct1 = sin(2*pi*250*t+fd1*m);
% modulated carrier
asxc1 = (2/npts)*abs(fft(xct1));
% amplitude spectrum
ampspec1 = asxc1(1:((npts/2)+1)); % positive frequency portion
fd2 = 1.0;
xct2 = sin(2*pi*250*t+fd2*m);
% modulated carrier
asxc2 = (2/npts)*abs(fft(xct2));
% amplitude spectrum
ampspec2 = asxc2(1:((npts/2)+1)); % positive frequency portion
fd3 = 10.0;
xct3 = sin(2*pi*250*t+fd3*m);
% modulated carrier
asxc3 = (2/npts)*abs(fft(xct3));
% amplitude spectrum
ampspec3 = asxc3(1:((npts/2)+1)); % positive frequency portion
% Plot spectra.
figure
% new figure
subplot(3,1,1)
stem(fn,ampspec1,’.k’);
ylabel(’Magnitude - fd=0.1’)
subplot(3,1,2)
stem(fn,ampspec2,’.k’);
ylabel(’Magnitude - fd=1’)
subplot(3,1,3)
stem(fn,ampspec3,’.k’);
xlabel(’Frequency’), ylabel(’Magnitude - fd=10’)
% End of script file.
Executing the program yields the results illustrated in Figures 4.10 and 4.11 , respectively.
Figure 4.10 shows the triangular-wave message signal. Figure 4.11 shows the spectrum. As
described in the text, all sum and difference frequencies are present. This non-linear effect
is most easily seen for = 10 since higher deviations spread the spectrum more than small
deviation constants.
28
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
0
0.1
0.2
0.3
0.4
0.5
Time
0.6
0.7
0.8
0.9
1
Figure 4.10: Triangular wave representing the phase deviation.
Computer Exercise 4.3
The MATLAB program for generating Figures 4.24 and 4.25 is given in the text in Computer
Example 4,4 (See page 192). The purpose of this Computer Exercise is to gain experience
in executing these simulations. Since the program is given, it will be executed for only one
case: = 10 Hz, ∆ = 40 Hz and = 0707. The results follow.
The other three cases are easily developed by simply entering the appropriate step sizes.
Computer Exercise 4.4
The labeling is straightforward. The appropriate model is illustrated in Figure 4.19. The
output of the phase detector is s1. The output of the sinusoidal nonlinearity is s2. The
next several lines of code implement the loop filter. The input to the VCO is vco_in and
the VCO output of the VCO output is vco_out.
The main sources of error are as follows: The output of the phase detector is not
the difference between the current input phase and current VCO output but the difference
between the current input phase and the previous VCO output. Thus, there is a one sample
delay induced in the simulation loop. This reduces the phase margin of the PLL. Another
error source is the approximation of integration by a discrete-time integrator. Sampling,as
we have seen, can induce errors.
Computer Exercise 4.5
Magnitude - fd=10
Magnitude - fd=1
Magnitude - fd=0.1
4.2. COMPUTER EXERCISES
29
1
0.5
0
0
50
100
150
200
250
300
350
400
450
500
0
50
100
150
200
250
300
350
400
450
500
0
50
100
150
200
250
300
Frequency
350
400
450
500
1
0.5
0
1
0.5
0
Figure 4.11: Spectra for FM signals with a square-wave message signal.
Input Frequency and VCO Freqeuncy
50
40
Frequency - Hertz
30
20
10
0
-10
0
0.1
0.2
0.3
0.4
0.5
0.6
Time - Seconds
0.7
0.8
0.9
1
Figure 4.12: Input frequency and VCO frequency.
30
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
Phase Plane
50
Frequency Error - Hz
40
30
20
10
0
-10
0
0.5
1
1.5
2
Phase Error / pi
2.5
3
3.5
Figure 4.13: Phase plane for Computer Exercise 4.3.
The MATLAB code for the PLL simulation with a line added for entering the sampling
frequency follows.
% File: c4ce5.m
% beginning of preprocessor
clear all
% be safe
fdel = input(’Enter frequency step size in Hz ’);
fn = input(’Enter the loop natural frequency in Hz ’);
zeta = input(’Enter zeta (loop damping factor) ’);
fs = 2000; % default sampling frequency
fs = input(’Enter the sampling frequency (default = 2000) ’);
npts = fs ; % default number of simulation point
T = 1/fs;
t = (0:(npts-1))/fs;
% time vector
nsettle = fix(npts/10);
% set nsettle time as 0.1*npts
Kt = 4*pi*zeta*fn;
% loop gain
a = pi*fn/zeta;
% loop filter parameter
filt_in_last = 0; filt_out_last=0;
vco_in_last = 0; vco_out = 0; vco_out_last=0;
% end of preprocesso - beginning of simulation loop
for i=1:npts
if insettle
4.2. COMPUTER EXERCISES
fin(i) = 0;
phin = 0;
else
fin(i) = fdel;
phin = 2*pi*fdel*T*(i-nsettle);
end
s1 = phin - vco_out;
s2 = sin(s1); % sinusoidal phase detector
s3 = Kt*s2;
filt_in = a*s3;
filt_out = filt_out_last + (T/2)*(filt_in + filt_in_last);
filt_in_last = filt_in;
filt_out_last = filt_out;
vco_in = s3 + filt_out;
vco_out = vco_out_last + (T/2)*(vco_in + vco_in_last);
vco_in_last = vco_in;
vco_out_last = vco_out;
phierror(i)=s1;
fvco(i)=vco_in/(2*pi);
freqerror(i) = fin(i)-fvco(i);
end
% end of simulation loop - beginning of postprocessor
kk = 0;
while kk == 0
k = menu(’Phase Lock Loop Postprocessor’,...
’Input Frequency and VCO Frequency’,...
’Phase Plane Plot’,...
’Exit Program’);
if k == 1
plot(t,fin,t,fvco), grid
title(’Input Frequency and VCO Freqeuncy’)
xlabel(’Time - Seconds’)
ylabel(’Frequency - Hertz’)
pause
elseif k == 2
plot(phierror/2/pi,freqerror), grid
title(’Phase Plane’)
xlabel(’Phase Error / 2*pi’)
ylabel(’Frequency Error - Hz’)
pause
31
32
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
Phase Plane
50
Frequency Error - Hz
40
30
20
10
0
-10
0
0.5
1
1.5
2
2.5
Phase Error / 2*pi
3
3.5
4
4.5
Figure 4.14: Phase-plane plot with = 400̇.
elseif k == 3
kk = 1;
end
end
% end of postprocessor
% End of script file.
The preceding program was executed twice; once for = 400 and once for = 1000. The
siumulation for = 400 is shown in Figure 4.14. We note that the curves, especially for
large frequency error, are not smooth, This is an indication that the samplingt frequency is
too low.
We therefore increase the sampling frequency from 400 to 1000. The results are illustrated in Figure 4.15.
We now see that the curve defining the phase plane is much smoother. In addition, for
= 400 we see that the PLL skips 4 cycles while acquiring lock. For = 1000 we see that
the PLL skips 3 cycles while acquiring lock. By executing the simulation a number of times
4.2. COMPUTER EXERCISES
33
Phase Plane
50
Frequency Error - Hz
40
30
20
10
0
-10
0
0.5
1
1.5
2
Phase Error / 2*pi
2.5
3
Figure 4.15: Phase-plane plot with = 1000
3.5
34
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
we see that additional increases in the sampling frequency have no effect on the results.
An effective way to determine an appropriate sampling frequency is to set the sampling
frequency at a high value and then reduce the sampling frequency until the results change
and the waveforms are distorted. At this point the sampling frequency is too low.
Computer Exercise 4.6
A rectangulator integrator is defined by
y[n] = y[n − 1] + Tx[n − 1]
so that, in the simulation program, the line
filt_out = filt_out_last + (T2) ∗ (filt_in + filt_in_last)
is replaced by
filt_out = filt_out_last + T ∗ filt_in_last
The VCO output is modified in exactly the same way.
The revised MATLAB code follows.
% File: c4ce6.m
% beginning of preprocessor
clear all % be safe
fdel = input(’Enter frequency step size in Hz ’);
fn = input(’Enter the loop natural frequency in Hz ’);
zeta = input(’Enter zeta (loop damping factor) ’);
npts = 2000; % default number of simulation points
fs = 2000; % default sampling frequency
T = 1/fs;
t = (0:(npts-1))/fs; % time vector
nsettle = fix(npts/10); % set nsettle time as 0.1*npts
Kt = 4*pi*zeta*fn; % loop gain
a = pi*fn/zeta; % loop filter parameter
filt_in_last = 0; filt_out_last=0;
vco_in_last = 0; vco_out = 0; vco_out_last=0;
% end of preprocesso - beginning of simulation loop
for i=1:npts
if insettle
fin(i) = 0;
phin = 0;
else
4.2. COMPUTER EXERCISES
fin(i) = fdel;
phin = 2*pi*fdel*T*(i-nsettle);
end
s1 = phin - vco_out;
s2 = sin(s1); % sinusoidal phase detector
s3 = Kt*s2;
filt_in = a*s3;
filt_out = filt_out_last + T*filt_in_last;
filt_in_last = filt_in;
filt_out_last = filt_out;
vco_in = s3 + filt_out;
vco_out = vco_out_last + T*vco_in_last;
vco_in_last = vco_in;
vco_out_last = vco_out;
phierror(i)=s1;
fvco(i)=vco_in/(2*pi);
freqerror(i) = fin(i)-fvco(i);
end
% end of simulation loop - beginning of postprocessor
kk = 0;
while kk == 0
k = menu(’Phase Lock Loop Postprocessor’,...
’Input Frequency and VCO Frequency’,...
’Phase Plane Plot’,...
’Exit Program’);
if k == 1
plot(t,fin,t,fvco), grid
title(’Input Frequency and VCO Freqeuncy’)
xlabel(’Time - Seconds’)
ylabel(’Frequency - Hertz’)
pause
elseif k == 2
plot(phierror/2/pi,freqerror), grid
title(’Phase Plane’)
xlabel(’Phase Error / pi’)
ylabel(’Frequency Error - Hz’)
pause
elseif k == 3
kk = 1;
end
35
36
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
Input Frequency and VCO Freqeuncy
50
40
Frequency - Hertz
30
20
10
0
-10
0
0.1
0.2
0.3
0.4
0.5
0.6
Time - Seconds
0.7
0.8
0.9
1
Figure 4.16: PLL behavior with a sampling frequency of 2000 Hz.
end
% end of postprocessor
% End of script file.
We compare the results of rectangular and trapezoidal integration are compared using
the result of Computer Exercise 4.3, which uses trapezoidal integration. The simulation
parameters used in Computer Exercise are: = 10 Hz, ∆ = 40 Hz and = 0707. The
result is as follows:
Reducing the sampling frequency to 500 Hz results in the following:
We see that with a reduced sampling frequency the PLL skips four cycles rather than
three and takes slightly longer to lock.. How would the PLL respond with a sampling frequency of 500 Hz and trapezoidal integration? This is left as an exercise for the student. In
general the integration algorithm should be as accurate as possible (trapezoidal integration
is better than rectanglar integration). Approximation accuracy becomes more important
as the sampling frequency is reduced.
Computer Exercise 4.7
The MATLAB program for the PLL with the limiting phase detector follows. We note
that the phase detector has been modified and a line has been added for entering the phase
detector constant .
4.2. COMPUTER EXERCISES
37
Input Frequency and VCO Freqeuncy
60
50
Frequency - Hertz
40
30
20
10
0
-10
0
0.1
0.2
0.3
0.4
0.5
0.6
Time - Seconds
0.7
0.8
0.9
1
Figure 4.17: PLL behavior with a sampling frequency of 500 Hz.
% File: c4ce7.m
% beginning of preprocessor
clear all % be safe
fdel = input(’Enter frequency step size in Hz ’);
fn = input(’Enter the loop natural frequency in Hz ’);
zeta = input(’Enter zeta (loop damping factor) ’);
AA = input(’Enter phase detector constant A ’);
npts = 2000; % default number of simulation points
fs = 2000; % default sampling frequency
T = 1/fs;
t = (0:(npts-1))/fs; % time vector
nsettle = fix(npts/10); % set nsettle time as 0.1*npts
Kt = 4*pi*zeta*fn; % loop gain
a = pi*fn/zeta; % loop filter parameter
filt_in_last = 0; filt_out_last=0;
vco_in_last = 0; vco_out = 0; vco_out_last=0;
% end of preprocessor - beginning of simulation loop
for i=1:npts
if insettle
fin(i) = 0;
phin = 0;
else
38
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
fin(i) = fdel;
phin = 2*pi*fdel*T*(i-nsettle);
end
s1 = phin - vco_out;
s2 = sin(s1); % sinusoidal phase detector
if s2-AA
s2=-AA;
elseif s2AA
s2=AA;
end
s3 = Kt*s2;
filt_in = a*s3;
filt_out = filt_out_last + (T/2)*(filt_in + filt_in_last);
filt_in_last = filt_in;
filt_out_last = filt_out;
vco_in = s3 + filt_out;
vco_out = vco_out_last + (T/2)*(vco_in + vco_in_last);
vco_in_last = vco_in;
vco_out_last = vco_out;
phierror(i)=s1;
fvco(i)=vco_in/(2*pi);
freqerror(i) = fin(i)-fvco(i);
end
% end of simulation loop - beginning of postprocessor
kk = 0;
while kk == 0
k = menu(’Phase Lock Loop Postprocessor’,...
’Input Frequency and VCO Frequency’,...
’Phase Plane Plot’,...
’Exit Program’);
if k == 1
plot(t,fin,t,fvco), grid
title(’Input Frequency and VCO Freqeuncy’)
xlabel(’Time - Seconds’)
ylabel(’Frequency - Hertz’)
pause
elseif k == 2
plot(phierror/2/pi,freqerror), grid
title(’Phase Plane’)
xlabel(’Phase Error / 2*pi’)
4.2. COMPUTER EXERCISES
39
Phase Plane
50
40
Frequency Error - Hz
30
20
10
0
-10
-20
0
2
4
6
Phase Error / 2*pi
8
10
12
Figure 4.18: PLL acquisition with = 05.
ylabel(’Frequency Error - Hz’)
pause
elseif k == 3
kk = 1;
end
end
% end of postprocessor
% End of script file.
We see that with = 05, the loop slips 5 cycles, whereas with ̇ 1 (limiter absent), the
PLL slips only three cycles with the parameters given. The acquisition time is therefore
increased.
Computer Exercise 4.8
The imperfect PLL is defined in Problem 4.28 as a PLL in which the loop filter is defined
as
+
() =
+
40
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
A non-zero value of moves the away from the origin so that the integration is no longer
"perfect".
In this example we use a more efficient and more general simulation model. The simulation model is developed in detail in Chapter 6 of the textbook: W. H. Tranter, K. S.
Shanmugan, T. S. Rappaport, and K. L. Kosbar, Principles of Communication Systems
Simulation with Wireless Applications, Printice Hall PTR, 2004. The MATALB code follows. (Note: The integration routine used here is the transposed direct form II realization
of a trapezoidal integrator.)
% File: c4ce8.m
clear all
% be safe
disp(’ ’) % insert blank line
fdel = input(’Enter the size of the frequency step in Hertz ’);
fn = input(’Enter the loop natural frequency in Hertz ’);
lambda = input(’Enter lambda, the relative pole offset ’);
disp(’ ’)
disp(’Accept default values:’)
disp(’ zeta = 1/sqrt(2) = 0.707,’)
disp(’ fs = 200*fn, and’)
disp(’ tstop = 1’)
dtype = input(’Enter y for yes or n for no ’,’s’);
if dtype == ’y’
zeta = 1/sqrt(2);
fs = 200*fn;
tstop = 1;
else
zeta = input(’Enter zeta, the loop damping factor ’);
fs = input(’Enter the sampling frequency in Hertz ’);
tstop = input(’Enter tstop, the simulation runtime ’);
end
npts = fs*tstop+1; % number of simulation points
t = (0:(npts-1))/fs; % default time vector
nsettle = fix(npts/10); % set nsettle time as 0.1*npts
tsettle = nsettle/fs; % set tsettle
%
% The next two lines establish the loop input frequency and phase
% deviations.
%
fin = [zeros(1,nsettle),fdel*ones(1,npts-nsettle)];
phin = [zeros(1,nsettle),2*pi*fdel*t(1:(npts-nsettle))];
4.2. COMPUTER EXERCISES
41
disp(’ ’) % insert blank line
%
w2b=0; w2c=0; s5=0; phivco=0;
%initialize
twopi = 2*pi;
% define 2*pi
twofs = 2*fs;
% define 2*fs
G = 2*pi*fn*(zeta+sqrt(zeta*zeta-lambda));
% set loop gain
a = 2*pi*fn/(zeta+sqrt(zeta*zeta-lambda));
% set filter parameter
a1 = a*(1-lambda); a2 = a*lambda;
% define constants
phierror = zeros(1,npts);
% initialize vector
fvco = zeros(1,npts);
% initialize vector
% beginning of simulation loop
for i = 1:npts
s1 = phin(i) - phivco;
% phase error
s2 = sin(s1);
% sinusoidal phase detector
s3 = G*s2;
s4 = a1*s3;
s4a = s4-a2*s5;
% loop filter integrator input
w1b = s4a+w2b;
% filter integrator (step 1)
w2b = s4a+w1b;
% filter integrator (step 2)
s5 = w1b/twofs;
% generate fiter output
s6 = s3+s5;
% VCO integrator input
w1c = s6+w2c;
% VCO integrator (step 1)
w2c = s6+w1c;
% VCO integrator (step 2)
phivco = w1c/twofs;
% generate VCO output
phierror(i) = s1;
% build phase error vector
fvco(i) = s6/twopi;
% build VCO input vector
end
% end of simulation loop
freqerror=fin-fvco;
% build frequency error vector
kk = 0;
while kk == 0
k = menu(’Phase Lock Loop Postprocessor’,...
’Input Frequency and VCO Frequency’,...
’Input Phase and VCO Phase’,...
’Frequency Error’,’Phase Error’,’Phase Plane Plot’,...
’Phase Plane and Time Domain Plots’,’Exit Program’);
if k == 1
plot(t,fin,’k’,t,fvco,’k’)
title(’Input Frequency and VCO Freqeuncy’)
xlabel(’Time - Seconds’);ylabel(’Frequency - Hertz’);
42
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
pause
elseif k ==2
pvco=phin-phierror;plot(t,phin,t,pvco)
title(’Input Phase and VCO Phase’)
xlabel(’Time - Seconds’);ylabel(’Phase - Radians’);
pause
elseif k == 3
plot(t,freqerror);title(’Frequency Error’)
xlabel(’Time - Seconds’);ylabel(’Frequency Error - Hertz’);
pause
elseif k == 4
plot(t,phierror);title(’Phase Error’)
xlabel(’Time - Seconds’);ylabel(’Phase Error - Radians’);
pause
elseif k == 5
ppplot
elseif k == 6
subplot(211);phierrn = phierror/pi;
plot(phierrn,freqerror,’k’);grid;
title(’Phase Plane Plot’);xlabel(’Phase Error /Pi’);
ylabel(’Frequency Error - Hertz’);subplot(212)
plot(t,fin,’k’,t,fvco,’k’);grid
title(’Input Frequency and VCO Freqeuncy’)
xlabel(’Time - Seconds’);ylabel(’Frequency - Hertz’);
subplot(111)
elseif k == 7
kk = 1;
end
end
% End of script file.
The routine is executed with the following dialog:
ce4_8
Enter the size of the frequency step in Hertz 40
Enter the loop natural frequency in Hertz 10
Enter lambda, the relative pole offset 0.1
Accept default values:
4.2. COMPUTER EXERCISES
43
Input Frequency and VCO Freqeuncy
50
40
Frequency - Hertz
30
20
10
0
-10
0
0.1
0.2
0.3
0.4
0.5
0.6
Time - Seconds
0.7
0.8
0.9
1
Figure 4.19: PLL acquisition with = 10 Hz, ∆ = 40 Hz, = 0707 and = 01
zeta = 1/sqrt(2) = 0.707,
fs = 200*fn, and
tstop = 1
Enter y for yes or n for no y
The result is shown in the following figure.
Comparing with the result of Computer Example 4.4 we see that the see that the number
of cycles slipped increases from three to five. The acquisition time is increased accordingly.
Computer Exercise 4.9
The loop filter for a third-order PLL is defined by (See Table 4.3)
() = (2 + + )2
This gives the transfer function
() =
2 + +
(2 + + )2
=
+ (2 + + )2
3 + 2 + +
The characteristic equation is
() = 3 + 2 + +
44
CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING
We see that for small loop gain the system exhibits a third-order pole at the origin.
As increases, a pair of these poles become conjugates and move into the right-hand
plane, which means that the system is unstable. As continues to increase the complexconjugate pair move back into the left-half plane and the system becomes stable. Therefore
a third-order PLL is unstable for small values of the loop gain but stable for larger values
of loop gain.
The MATLAB program follows.
% File c4ce9
clear all
clf
a = .70; b = .70; Kt0 = 0.01;
%D = s^3 +Kt*(s^2 + a*s^2 + b);
for k = 1:4000
Kt = k*Kt0;
dd = [1 Kt Kt*a Kt*b];
q = roots(dd);
if k == 1
dd
q
elseif k == 4000
dd
q
end
plot(q,’.’)
if k == 1
hold on
axis([-3 1 -2 2])
axis (’square’)
grid
xlabel(’real’), ylabel(’imaginary’)
end
end
The pole migration is illustrated in the following root-locus plot.
Computer Exercise 4.10
As mentioned in the text, there are many ways to work this problem. Rather than develop
the simulation we offer hints and point out pitfalls. Problem 4.25 illustrated one technique.
4.2. COMPUTER EXERCISES
45
2
1.5
1
imaginary
0.5
0
-0.5
-1
-1.5
-2
-3
-2.5
-2
-1.5
-1
real
-0.5
0
0.5
1
Figure 4.20: Root-locus plot for Computer Exercise 4.9.
¡ ¢
With () = cos(20 ), we desire 0 () = cos 2 75 0 . Assume that the VCO output
is a pulse train with frequency 15 . The pulse should be narrow so that the seventh harmonic
is relatively large. The spectrum of the VCO output consists of components separated by
1
5 0 with an envelope of sinc( ), where is the pulse width. There are two problems with
this technique. The first issue is that the output of the VCO, since it is a pulse, is rich in
harmonic content and getting the PLL to lock may be difficult. This issue can be solved by
having the VCO output not be a pulse but a sinusoid having a quiscent frequency of 15 0 .
We then have a second PLL output that is a pulse train having frequency 15 0 We desire
to track the 7th harmonic. Designing the filter can be difficult since adjacent harmonics
must be rejected. This problem can be addressed by incorporating a second PLL designed
to track the desired harmonic.
Chapter 5
Principles of Baseband Digital
Data Transmission
5.1
Problem Solutions
Problem 5.1
a. Split phase or bipolar RZ;
b. NRZ change or NRZ mark;
c. Split phase;
d. NRZ mark;
e. Polar RZ;
f. Unipolar RZ.
Problem 5.2
These are a matter of Ölling in the details given at the ends of Examples 5.1 through 5.5.
Problem 5.3
This is a matter of adapting Figures 5.2 top and bottom to the data sequence given in the
problem statement.
Problem 5.4
This is a matter of adapting Figure 5.2, second plot, to the data sequence given in the
problem statement.
1
2
CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION
Problem 5.5
This is a matter of adapting Figure 5.2, third - Öfth plots, to the data sequence given in
the problem statement.
Problem 5.6
a. 4 kbps;
b. 2 kbps;
c. 2 kbps;
d. 4 kbps for Örst null and 2 kbps for second null.
Problem 5.7
The step response is given in Problem 2.65c as
2f3
2f3
2f3
ys (t) = 1 exp p
cos p + sin p
u (t)
2
2
2
The speciÖed input is
x (t) = u (t) 2u (t T ) + u (t 2T )
Using the linear, time invariant properties of the system, we can immidiately write down
the output as
y (t) = ys (t) 2ys (t T ) + ys (t 2T )
Plots are given in Figure 5.1 for the two cases speciÖed in the problem statement.
:
Problem 5.8
The references should be to Equations (5:30) and (5:29), respectively. The result for
the response follows by noting that the response to Au (t) is A [1 exp (t=RC)] u (t) and
the responsse to Au (t T ) is A [1 exp ((t T )=RC)] u (t T ). When superposition is
applied, the result is that given by (5:30) :
5.1. PROBLEM SOLUTIONS
3
f3T = 2
2
y step(t)
1
0
-1
-2
0
0.5
1
1.5
t/T
f3T = 20
2
2.5
3
0
0.5
1
1.5
t/T
2
2.5
3
2
y step(t)
1
0
-1
-2
Problem 5.9
1
The reference should be to (5:37). Clearly, with = 0, the spectrum is T for jf j 1
2T = 2T
1
and 0 for jf j > 1+
2T = 2T :
4
CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION
Problem 5.10
Because PRC (f ) is even, it follows that
Z 1
1
pRC (t) = F [PRC (f )] = 2
PRC (f ) cos (2f t) df
0
Z (1)=2T
Z (1+)=2T
T
1
= 2T
cos (2f t) df + T
1 + cos
f
cos (2f t) df
2T
0
(1)=2T
sin 1
sin 1+
sin 1
T t
T t
T t
= 2T
+T
T
2t
2t
2t
Z (1+)=2T
T
1
+T
cos
f
cos (2f t) df
2T
(1)=2T
1+
sin 1
t
sin
t
T
T
= T
+T
2t
2t
Z
T (1+)=2T
T
1
T
1
+
cos
f
+ 2f t + cos
f
2f t df
2 (1)=2T
2T
2T
1+
sin 1
t
sin
t
T
T
= T
+T
2t
2t
h
i (1+)=2T
h
i (1+)=2T
1
1
T
T
sin
f
+
2f
t
sin
f
2f
t
2T
2T
T
T
+
+
T
T
2
2
+ 2t
2t
(1)=2T
(1)=2T
1
1+
sin T t
sin T t
= T
+T
2t
2t
h
i
h
i
1
1+
1
1
T 1+
T 1
sin
+
t
sin
+
t
2T
2T
T
2T
2T
T
T
T
+
2
T = + 2t
2
T = + 2t
h
i
h
i
1+
1
1+
1
1
1
T
T
sin
t
sin
t
2T
2T
T
2T
2T
T
T
T
+
2
T = 2t
2
T = 2t
sin (t=T ) cos (t=T )
1
1
1
t
t
=
+
cos
sin
t=T
2t=T T =2t 1 T =2t + 1
T
T
1
sin (t=T )
= 1
2 cos (t=T )
t=T
1 (T =2t)
"
#
(T =2t)2
sin (t=T )
cos (t=T )
=
=
sinc (t=T )
2 cos (t=T )
t=T
1 (T =2t)
1 (2t=T )2
5.1. PROBLEM SOLUTIONS
5
Problem 5.11
Nyquistís pulse shaping criterion is a su¢cient condition. Each of these spectra, being
built of square and triangular shapes, will have regularly spaced zero crossings in the time
domain.
a. This spectrum has the zero-ISI property. The corresponding pulse function is
p1 (t) = 2W sinc (2W t) + W sinc (W t)
k
The Örst sinc-function has zero crossings at tk = 2W
seconds (k an integer) and the second
k
one has zero crossings at tk = W seconds (k an integer). The appropriate sample interval
1
is ts = W
seconds for 0 ISI (both terms must contribute zero ISI).
b. This spectrum has the zero-ISI property. The corresponding pulse function is
p2 (t) = 2W sinc2 (2W t) + W sinc (W t)
k
The Örst sincñsquared function has zero crossings at tk = 2W
seconds (k an integer) and the
k
second one has zero crossings at tk = W seconds (k an integer). The appropriate sample
1
interval is ts = W
seconds for 0 ISI (both terms must contribute zero ISI).
c. This spectrum has the zero-ISI property. The corresponding pulse function is
p3 (t) = 4W sinc (4W t) W sinc2 (W t)
k
The Örst sincñsquared function has zero crossings at tk = 4W
seconds (k an integer) and the
k
second one has zero crossings at tk = W seconds (k an integer). The appropriate sample
1
interval is ts = W
seconds for 0 ISI (both terms must contribute zero ISI).
d. This spectrum has the zero-ISI property. By the modulation theorem, the corresponding pulse function is
p4 (t) = 2W sinc (W t) cos (2W t)
1
The zero crossings occur where the sinc-function is zero or at integer multiples of tk = W
seconds.
e. This spectrum has the zero-ISI property. The corresponding pulse function is
p5 (t) = 2W sinc2 (2W t) W sinc2 (W t)
6
CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION
k
The Örst sincñsquared function has zero crossings at tk = 2W
seconds (k an integer) and the
k
second one has zero crossings at tk = W seconds (k an integer). The appropriate sample
1
interval is ts = W
seconds
Problem 5.12
The MATLAB program is given below. The plot of the desired frequency response functions
follows the program.
% Problem 5_12
% Computation and plotting of transmitter and receiver frequency responses
% for raised cosine signaling through lowpass Butterworth Öltered channel
%
clear all
A = char(í-í,í:í,íñí,í-.í);
clf
f3dB = input(íEnter channel Ölter 3-dB frequency, Hz => í);
n_poles = input(íEnter number of poles for Butterworth Ölter repr channel => í);
Rb = input(íEnter data rate, bits/s => í);
beta0 = input(íEnter vector of betas (max no = 3) => í);
Lbeta = length(beta0);
for n = 1:Lbeta
beta = beta0(n);
f1 = 0:.1:(1-beta)*Rb/2;
HR1 = sqrt(1+(f1/f3dB).^(2*n_poles));
f2 = (1-beta)*Rb/2:.1:(1+beta)*Rb/2;
HR2 = sqrt(1+(f2/f3dB).^(2*n_poles)).*cos(pi*(f2-(1-beta)*Rb/2)./(2*beta*Rb));
f3 = (1+beta)*Rb/2:.1:Rb;
HR3 = zeros(size(f3));
f = [f1 f2 f3];
HR = [HR1 HR2 HR3];
plot(f,HR,A(n,:), íLineWidthí, 1.5),xlabel(í{nitf}, Hzí),...
ylabel(íj{nitH_R}({nitf})j or j{nitH_T}({nitf})jí)
if n == 1
hold on
end
end
title([íBit rate = í,num2str(Rb),í bps; channel Ölter 3-dB frequency = í,num2str(f3dB),
í Hz; no. of poles = í,num2str(n_poles)]),...
if Lbeta == 1
legend([ínbeta = í,num2str(beta0(1))],1)
elseif Lbeta == 2
5.1. PROBLEM SOLUTIONS
7
Bit rate = 5000 bps; channel filter 3-dB frequency = 5000 Hz; no. of poles = 1
1.4
β=1
β = 0.5
1.2
|HR(f )| or |HT(f )|
1
0.8
0.6
0.4
0.2
0
0
500
1000
1500
2000
2500
f, Hz
3000
3500
4000
4500
5000
legend([ínbeta = í,num2str(beta0(1))],[ínbeta = í,num2str(beta0(2))],1)
elseif Lbeta == 3
legend([ínbeta = í,num2str(beta0(1))],[ínbeta = í,num2str(beta0(2))],...
[ínbeta = í,num2str(beta0(3))],1)
end
8
CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION
Problem 5.13
Use (5:37) and set the maximum frequency equal to 7 kHz:
1+
= 7 kHz
2T
But T1 = 9 kbps, so
27
9
14
5
1 = = 0:556
9
9
1+ =
=
Problem 5.14
a. Consider P (f ) = 2 (f =2W ) (f =W ) which is a trapezoid 1 unit high going from
2W to 2W on its base and.going from W to W on its top. Clearly, superimposing
it with its translates by 1=T = 3W=2; 2=T = 3W; ::: produces a constant
height of 1 so Nyquistís pulse shaping criterion is satisÖed. The zero crossings of the
time domain pulse are spaced by 2=3W seconds.
b. Use the transform pair (t) $ sinc2 (f ) with duality to produce the transform pair
sinc2 (t) $ (f ). Then use scale change to get sinc2 (at) $ a1 (f =a). Apply this
to P (f ) to get
p (t) = 4W sinc2 (2W t) W sinc2 (W t)
Problem 5.15
a. The channel response matrix is
2
3
1:0
0:1 0:01
1:0
0:1 5
[Pc ] = 4 0:2
0:02 0:2
1:0
5.1. PROBLEM SOLUTIONS
Its inverse is
9
2
3
1:0217 0:0163 0:0209
[Pc ]1 = 4 0:0216 1:0423 0:0163 5
0:0626 0:0216 1:0217
The equalizer coe¢cients are the middle column:
[1 0 1 ] = [0:01631
1:0423
0:0216]
b. The output samples are given by
peq (mT ) =
1
X
n=1
n pc [(m n) T ]
[peq ] = [0:0121 0:0000 1:0000 0:0000 0:0214]
Problem 5.16
a. The sample values pc (4T ) = 0:001 and pc (4T ) = 0:003 should have been given.
The channel response matrix is
2
3
1:0
0:1
0:01 0:001 0:001
6 0:2
1:0
0:1
0:01 0:001 7
6
7
0:2
1:0
0:1
0:01 7
[Pc ] = 6
6 0:02
7
4 0:005 0:02
0:2
1:0
0:1 5
0:003 0:005 0:02
0:2
1:0
Its inverse is
2
3
1:0218 0:1067 0:0218 0:0046 0:0018
6 0:2111 1:0438 0:1111 0:0227 0:0046 7
6
7
7
[Pc ] = 6
6 0:0651 0:2179 1:0451 0:1111 0:0218 7
4 0:0234 0:0673 0:2179 1:0438 0:1067 5
0:0101 0:0234 0:0651 0:2111 1:0218
The equalizer coe¢cients are the middle column:
[2 1 0 1 2 ] = [0:0218 0:1111 1:0451 0:2179 0:0651]
10
CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION
b. The output samples are given by
2
X
peq (mT ) =
n=2
n pc [(m n) T ]
[peq ] = [0:0000 0:0000 1:0000 0:0000 0:0000]
Problem 5.17
a. The output in terms of the input is
y (t) = x (t) + x (t m )
Using the superposition and time delay theorems, its Fourier transform is
Y (f ) = X (f ) [1 + exp (j2 m f )]
The frequency response function is
Y (f )
= [1 + exp (j2 m f )]
X (f )
q
jH (f )j =
[1 + cos (2 m f )]2 + sin2 (2 m f )
q
=
1 + 2 cos (2 m f ) + 2
H (f ) =
b. From the block diagram
z (t) =
N
X
i=1
i y [t (i 1) ]
Using the superposition and time delay theorems, the Fourier transform of z (t) is
Z (f ) = Y (f )
N
X
i=1
i exp [j2 (i 1) ]
The frequency response function is
N
Z (f ) X
Heq (f ) =
=
i exp [j2 (i 1) ]
Y (f )
i=1
5.1. PROBLEM SOLUTIONS
11
c. From part a,
1
H (f )
1
1 + exp (j2 m f )
= 1 exp (j2 m f ) + 2 exp (j4 m f ) 3 exp (j6 m f ) + :::
=
Equate the above to Heq (f ) with m = : Clearly i = ()i1 ; i = 1; 2; :::; N:
Problem 5.18
a. The channel response matrix is
2
3
1:0 0:07 0:05
0:07 5
[Pc ] = 4 0:1 1:0
0:07 0:1
1:0
Its inverse is
2
3
0:9898 0:0639 0:0540
[Pc ]1 = 4 0:1031 0:9864 0:0539 5
0:0590 0:1031
0:9898
The equalizer coe¢cients are the middle column:
[1 0 1 ] = [0:0639
0:9864
0:1031]
b. The output samples are given by
peq (mT ) =
1
X
n=1
n pc [(m n) T ]
[peq ] = [0:0428 0:0000 1:0000 0:0000 0:0619]
Problem 5.19
a. The channel response matrix is
2
3
1:0
0:07 0:05 0:02 0:01
6 0:1
1:0
0:07 0:05 0:02 7
6
7
0:07
0:1
1:0
0:07 0:05 7
[Pc ] = 6
6
7
4 0:05 0:07
0:1
1:0
0:7 5
0:03 0:05 0:07
0:1
1:0
12
CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION
Its inverse is
2
3
0:9883 0:0617 0:0502 0:0248 0:0154
6 0:1054
0:9823 0:0569 0:0485 0:0248 7
6
7
1
6
0:9801 0:0569 0:0502 7
[Pc ] = 6 0::0621 0:1088
7
4 0:0307 0:0635 0:1088
0:9823 0:0617 5
0:0163 0:0370 0:0621 0:1054
0:9886
The equalizer coe¢cients are the middle column:
[2 1 0 1 2 ] = [0:0502 0:0569 0:9801 0:1088 0:0621]
b. The output samples are given by
peq (mT ) =
2
X
n=2
n pc [(m n) T ]
[peq ] = [0:0000 0:0000 1:0000 0:0000 0:0000]
Problem 5.20
a. For a timing jitter of zero
1
exp (z0 ) = 106
2
= ln 2 + 6 ln 10
PE0 =
or z0
= 13:12 = 11:18 dB
b. For a timing jitter of 5%
PE5% =
=
1
1
exp (z) + exp [z (1 2 0:05)]
4
4
1
1
exp (z) + exp (0:9z) = 106
4
4
Trial and error provides the value
z5% = 14:054 = 11:48 dB
Degradation = 11:48 11:18 = 0:3 dB
c.
1
exp (z0 ) = 104
2
= ln 2 + 4 ln 10
PE0 =
or z0
= 8:52 = 9:3 dB
5.1. PROBLEM SOLUTIONS
13
z5% = 9:07 = 9:6 dB
Degradation = 9:6 9:3 = 0:3 dB
Problem 5.21
a. The step response of a lowpass RC Ölter is
ys; RC (t) = [1 = exp (t=RC)] u(t)
Using superposition and time invariance properties of the Ölter, the output is
y (t) = ys; RC (t) 2ys; RC (t T ) + 2ys; RC (t 2T ) ys; RC (t 3T )
b. The output is the negative of the one found in part (a).
c. The output is the step response.
d. The output is the negatuve if the step response.
A plot is given in Fig. 5.3.
Problem 5.22
1+
1+
=
R = fmax
2T
2
so
1+
10 = 5
2
1+ = 1
= 0
Problem 5.23
We have, from the raised cosine spectrum expression, that
1+
1+
=
R = fmax
2T
2
or
=
2fmax
1
R
14
CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION
T/RC = 10
1
0.8
0.6
step response
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-1
We must have 0 so
-0.5
0
0.5
1
t, seconds
1.5
2
2.5
3
2fmax
10
R
or
R 2fmax
Note that requiring that 1 results in R fmax :
Problem 5.24
The modiÖed program follows with the spectrum shown in Fig. 5.2. The spectral line at
the bit rate does not appear to be much di§erent than that of Fig. 5.16 of the text.
% Problem 5.24
%
nsym = 1000; nsamp = 50; lambda = 0.7;
[b,a] = butter(3,2*lambda/nsamp);
l = nsym*nsamp;
% Total sequence length
y = zeros(1,l-nsamp+1);
% Initalize output vector
x =2*round(rand(1,nsym))-1; % Components of x = +1 or -1
for i = 1:nsym
% Loop to generate info symbols
k = (i-1)*nsamp+1;
y(k) = x(i);
Amplitude
5.1. PROBLEM SOLUTIONS
15
1
0
-1
0
100
200
300
400
500
600
0
100
200
300
400
500
600
Amplitude
1.5
1
0.5
0
Spectrum
1
0.5
0
0
200
400
600
800
1000 1200
FFT Bin
1400
1600
1800
2000
end
datavector1 = conv(y,ones(1,nsamp)); % Each symbol is nsamp long
subplot(3,1,1), plot(datavector1(1,200:799),íkí, íLineWidthí, 1.5)
axis([0 600 -1.4 1.4]), ylabel(íAmplitudeí)
Öltout = Ölter(b,a,datavector1);
% datavector2 = Öltout.*Öltout;
datavector2 = abs(Öltout);
subplot(3,1,2), plot(datavector2(1,200:799),íkí, íLineWidthí, 1.5), ylabel(íAmplitudeí)
y = §t(datavector2); yy = abs(y)/(nsym*nsamp);
subplot(3,1,3), stem(yy(1,1:2*nsym),ík.í)
xlabel(íFFT Biní), ylabel(íSpectrumí)
% End of script Öle.
They are about the same.
Problem 5.25
Given that NF F T = 5; 000. Since this corresponds to fs = 50 Hz, the spectral line at the
bit rate corresponds to 50 1000=5000 = 1 Hz:
Problem 5.26
16
CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION
Expand the signal as
h
i
xPSK (t) = Ac cos 2fc t + d (t)
h
i 2
h
i
= Ac cos d (t) cos (2fc t) sin d (t) sin (2fc t)
2
h 2 i
h
i
= Ac cos cos (2fc t) sin sin (2fc t)
h i2
h i 2
= Ac cos cos (2fc t) sin sin (2fc t)
2i
2h i
h
= Ac cos cos (2fc t) d (t) sin sin (2fc t)
2
2
The Örst term is clearly carrier and the second modulation. From the Örst line, the
1 2
total power
is PT = 2 Ac From the last line the power in the carrier component is PC =
1 2
2
2 Ac cos 2 . The fraction of total power in the carrier component is
h i
PC
= cos2
PT
2
We want
h i
h i p
p
2
cos2
= 0:1 or cos
= 0:1 or = cos1
0:1 = 0:7952
2
2
Problem 5.27
Equation (5:69) is
Z t
xFSK (t) = Ac cos 2fc t + kf
d () d
where d (t) = 1 in T = 1=1000 = 0:001 second intervals. The peak frequency deviation is
f
kf
5.2
1
kf dmax hertz = 10000 hertz
2
= 20000 radians/volt (or unit of d (t) )
=
Computer Exercises
Computer Exercise 5.1
5.2. COMPUTER EXERCISES
17
% ce5_1:
Simulation of a BB pulse transmission system Öltered
%
by Butterworth Ölter
%
N_order = input(íEnter order of Butterworth Ölter => í);
BWT_sym = input(íEnter bandwidth X sym dur product => í);
T_sym = 1;
BW = BWT_sym/T_sym;
sam_sym = 100;
[num den] = butter(N_order, 2*BW/sam_sym);
N_sym = 20;
ts = T_sym/sam_sym;
n_samp = N_sym*sam_sym;
t = 0:ts:(n_samp-1)*ts;
s = 3279;
rand(ístateí, s);
ss = sign(rand(1, N_sym) - 0.5);
% Bipolar
data = 0.5*(ss+1);
% Unipolar
x1 = zeros(size(t));
for n = 1:N_sym
x1 = x1 + ss(n)*unit_pulse((t-(n-1)*T_sym-T_sym/2)/T_sym);
%NRZ change
end
y1 = Ölter(num, den, x1);
x2 = zeros(size(t));
sgn = 1;
for n = 1:N_sym
if n == 1
x2 = x2 + ss(1)*unit_pulse((t-(n-1)*T_sym-T_sym/2)/T_sym); %NRZ mark
else
x2 = x2 + sgn*ss(1)*unit_pulse((t-(n-1)*T_sym-T_sym/2)/T_sym);
end
if n < N_sym
if data(n+1) == 1
sgn = -sgn;
elseif data(n+1) == 0
sgn = sgn;
end
end
end
y2 = Ölter(num, den, x2);
x3 = zeros(size(t));
18
CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION
for n = 1:N_sym
x3 = x3 + data(n)*unit_pulse(2*(t-(n-1)*T_sym-T_sym/4)/T_sym); %Unip RZ
end
y3 = Ölter(num, den, x3);
x4 = zeros(size(t));
for n = 1:N_sym
x4 = x4 + ss(n)*unit_pulse(2*(t-(n-1)*T_sym-T_sym/4)/T_sym); %Polar RZ
end
y4 = Ölter(num, den, x4);
x5 = zeros(size(t));
sgn = 1;
for n = 1:N_sym
x5 = x5 + sgn*data(n)*unit_pulse(2*(t-(n-1)*T_sym-T_sym/4)/T_sym); %Bip
RZ
if data(n) > 0
sgn = -sgn;
end
end
y5 = Ölter(num, den, x5);
x6 = zeros(size(t)); %Split phase
for n = 1:N_sym
x6 = x6 + ss(n)*(unit_pulse(2*(t-(n-1)*T_sym-T_sym/4)/T_sym) ...
-unit_pulse(2*(t-(n-1)*T_sym-3*T_sym/4)/T_sym));
end
y6 = Ölter(num, den, x6);
Ögure(1)
subplot(6,1,1), plot(t, x1, íLineWidthí, 1.5), axis([min(t) max(t) -1.2 1.2]), ...
ylabel(íNRZ changeí), title(íBaseband binary data formats; unÖlteredí)
subplot(6,1,2), plot(t, x2, íLineWidthí, 1.5), axis([min(t) max(t) -1.2 1.2]), ...
ylabel(íNRZ markí)
subplot(6,1,3), plot(t, x3, íLineWidthí, 1.5), axis([min(t) max(t) -1.2 1.2]), ...
ylabel(íUnipolar RZí)
subplot(6,1,4), plot(t, x4, íLineWidthí, 1.5), axis([min(t) max(t) -1.2 1.2]), ...
ylabel(íPolar RZí)
subplot(6,1,5), plot(t, x5, íLineWidthí, 1.5), axis([min(t) max(t) -1.2 1.2]), ...
ylabel(íBipolar RZí)
subplot(6,1,6), plot(t, x6, íLineWidthí, 1.5), axis([min(t) max(t) -1.2 1.2]), ...
ylabel(íSplit phaseí), xlabel(íTime, secondsí)
Ögure(2)
subplot(6,1,1), plot(t, y1, íLineWidthí, 1.5), axis([min(t) max(t) -1.2 1.2]), ...
5.2. COMPUTER EXERCISES
ylabel(íNRZ changeí), ...
title([íButterworth Ölter; order = í, num2str(N_order), í; ...
BW = í, num2str(BWT_sym), í/T_b_i_tí])
subplot(6,1,2), plot(t, y2, íLineWidthí, 1.5), axis([min(t) max(t) -1.2 1.2]), ...
ylabel(íNRZ markí)
subplot(6,1,3), plot(t, y3, íLineWidthí, 1.5), axis([min(t) max(t) -1.2 1.2]), ...
ylabel(íUnipolar RZí)
subplot(6,1,4), plot(t, y4, íLineWidthí, 1.5), axis([min(t) max(t) -1.2 1.2]), ...
ylabel(íPolar RZí)
subplot(6,1,5), plot(t, y5, íLineWidthí, 1.5), axis([min(t) max(t) -1.2 1.2]), ...
ylabel(íBipolar RZí)
subplot(6,1,6), plot(t, y6, íLineWidthí, 1.5), axis([min(t) max(t) -1.2 1.2]), ...
ylabel(íSplit phaseí), xlabel(íTime, secondsí)
% End of script Öle
function y = unit_step(t)
% Function for generating the unit step
%
y = zeros(size(t));
I = Önd(t >= 0);
y(I) = ones(size(I));
function y = unit_pulse(t)
% Unit rectangular pulse function
%
y = unit_step(t+0.5) - unit_step(t-0.5);
A typical run follows. Plots appesr in Figures 5.5 and 5.6.
>> ce5_1
Enter order of Butterworth Ölter => 2
Enter bandwidth X sym dur product => 1
19
CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION
Split phaseBipolar RZ Polar RZUnipolar RZNRZ markNRZ change
20
Baseband binary data formats; unfiltered
1
0
-1
0
2
4
6
8
10
12
14
16
18
0
2
4
6
8
10
12
14
16
18
0
2
4
6
8
10
12
14
16
18
0
2
4
6
8
10
12
14
16
18
0
2
4
6
8
10
12
14
16
18
0
2
4
6
8
10
12
Time, seconds
14
16
18
1
0
-1
1
0
-1
1
0
-1
1
0
-1
1
0
-1
Split phaseBipolar RZ Polar RZUnipolar RZNRZ markNRZ change
5.2. COMPUTER EXERCISES
21
Butterworth filter; order = 2; BW = 1/Tbit
1
0
-1
0
2
4
6
8
10
12
14
16
18
0
2
4
6
8
10
12
14
16
18
0
2
4
6
8
10
12
14
16
18
0
2
4
6
8
10
12
14
16
18
0
2
4
6
8
10
12
14
16
18
0
2
4
6
8
10
12
Time, seconds
14
16
18
1
0
-1
1
0
-1
1
0
-1
1
0
-1
1
0
-1
22
CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION
Computer Exercise 5.2
% ce5_2:
Computation and plotting of transmitter and receiver frequency responses
%
for raised cosine signaling through lowpass Butterworth Öltered channel
%
clf
clear all
A = char(í-í,í:í,íñí,í-.í,í-oí,í-xí);
clf
f3dB = input(íEnter channel Ölter 3-dB frequency, Hz => í);
n_poles = input(íEnter number of poles for BW Ölter representing channel => í);
Rb = input(íEnter data rate, bits/s => í);
beta0 = input(íEnter vector of betas => í);
delf = 100;
Lbeta = length(beta0);
for n = 1:Lbeta
beta = beta0(n);
f1 = 0:delf:(1-beta)*Rb/2;
HR1 = sqrt(1+(f1/f3dB).^(2*n_poles));
f2 = (1-beta)*Rb/2:delf:(1+beta)*Rb/2;
HR2 = sqrt(1+(f2/f3dB).^(2*n_poles)).*cos(pi*(f2-(1-beta)*Rb/2)./(2*beta*Rb));
f3 = (1+beta)*Rb/2:delf:Rb;
HR3 = zeros(size(f3));
f = [f1 f2 f3];
HR = [HR1 HR2 HR3];
plot(f,HR,A(n,:), íLineWidthí, 1.5)
if n == 1
hold on
xlabel(í{nitf}, Hzí),ylabel(íj{nitH_R}({nitf})j or j{nitH_T}({nitf})jí)
title([íBit rate = í,num2str(Rb),í bps; ch Ölter 3-dB freq = í,num2str(f3dB),í Hz;
no. of poles = í,...
num2str(n_poles) í; nbeta = í, num2str(beta0)])
end
end
A typical run follows with a plot given in Fig. 5.7.
>> ce5_2
Enter channel Ölter 3-dB frequency, Hz => 2000
Enter number of poles for Butterworth Ölter representing channel => 2
5.2. COMPUTER EXERCISES
23
Bit rate = 5000 bps; ch filter 3-dB freq = 2000 Hz; no. of poles = 2; β = 0
2
0.5
1.8
1.6
|HR(f )| or |HT(f )|
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
500
1000
1500
2000
2500
f, Hz
3000
3500
4000
4500
Enter data rate, bits/s => 5000
Enter vector of betas => [0 .5 1]
Computer Exercise 5.3
% ce5.3:
Design of an N-tap zero-forcing equalizer; input and
%
output plotted
%
pc = [-0.03 0.05 -0.1 0.02 -0.05 0.2 1 0.3 -0.07 0.03 -0.2 0.05 -0.1];
n_tap = input(íNumber of taps => í);
nn = áoor(n_tap/2)+1;
n_samp = length(pc);
n_center = áoor(n_samp/2)+1;
if n_tap > n_center
disp(íNot enough samples for speciÖed equalizer lengthí)
end
Pc = [];
for k = 1:n_tap
5000
1
24
CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION
Pc(:, k) = pc(n_center-k+1:n_center+n_tap-k)í;
end
Pcinv = inv(Pc);
A = Pcinv(:,nn);
disp(íThe equalizer tap vector is:í)
disp(A)
nT = [-(n_center-1):n_center-1];
peq = [];
n_eq = n_samp-n_tap+1;
nTc = [-áoor(n_eq/2):áoor(n_eq/2)];
for n = 1:n_eq
Nd = n-1;
pcd = pc(Nd+1:n_tap+Nd)í;
peq(n) = Aí*áipud(pcd);
end
subplot(2,1,1), stem(nT, pc), axis([min(nT) max(nT) -0.5 1.5]), ...
xlabel(í{nitn}í), ylabel(í{nitp_c}({nitn})í)
title([íInput sequence and zero-forced equalizer output for í, num2str(n_tap), í-tap
equalizerí])
subplot(2,1,2), stem(nTc, peq), axis([min(nT) max(nT) -0.5 1.5]), ...
xlabel(í{nitn}í), ylabel(í{nitp}_e_q({nitn})í)
%
End of script Öle
A typical run follows with a plot given in Fig. 5.8.
>> ce5_3
Number of taps => 5
The equalizer tap vector is:
0.1581
-0.3160
1.2068
-0.4405
0.2577
Computer Exercise 5.4: See Problem 5.24 solution
5.2. COMPUTER EXERCISES
25
Input sequence and zero-forced equalizer output for 5-tap equalizer
1.5
pc (n)
1
0.5
0
-0.5
-6
-4
-2
0
n
2
4
6
-4
-2
0
n
2
4
6
1.5
peq(n)
1
0.5
0
-0.5
-6
Chapter 6
Overview of Probability and
Random Variables
6.1
Problem Solutions
Problem 6.1
S = sample space is the collection of the 21 parts. Let Ai = event that the pointer stops
on the ith part, i = 1; 2; :::; 21. These events are exhaustive and mutually exclusive. Thus
P (Ai ) = 1=21
a. P (even number) = P (2 or 4 or or 20) = 10=21;
b. P (A21 ) = 1=21;
c. P (4 or 5 or 9) = 3=21 = 1=7;
d. P (number > 10) = P (11; 12; ; or 21) = 11=21:
Problem 6.2
a. Use a tree diagram similar to Fig. 6.2 to show that
P (3K; 2A) =
4
52
4
3 3 2
3
4 3 2
3 4 3 2
4
3 3 2
3
+
3
+
+
3
51 50 49 48
51 50 49 48
51 50 49 48
51 50 49 48
= 9:2345 106
1
2
CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES
b. Use the tree diagram of Fig. 6.2 except that we now look at outcomes that give 4 of a
kind. Since the tree diagram is for a particular denomination, we multiply by 13 and
get
4 3
2 1 48
48 3 2 1
48 4 3 2 1
P (4 of a kind) = 13
3
+
+
52 51
50 49 48
51 50 49 48
52 51 50 49 48
= 0:0002401
c. The Örst card can be anything. Given a particular suit on the Örst card, the probability
of suit match on the second card is 12=51; on the third card it is 11=50; on the fourth
card it is 10=49; on the Öfth and last card it is 9=48. Therefore,
P (all same suit) = 1
12 11 10 9
= 0:001981
51 50 49 48
d. For a royal áush
P (A; K; Q; J; 10 of same suit) =
4 1 1 1 1
= 1:283 108
52 51 50 49 48
e. The desired probability is
P (QjA; K; J; 10 not all of same suit) =
4
= 0:0833
48
Problem 6.3
P (A; B) = P (A) P (B)
P (A; C) = P (A) P (C)
P (B; C) = P (B) P (C)
P (A; B; C) = P (A) P (B) P (C)
Problem 6.4
a. According to (6.11) P (A \ B) = P (A) P (B) for statistical independence. For the
given probabilities, 0:2 0:5 = 0:1 6= 0:4 so they are not statistically independent.
6.1. PROBLEM SOLUTIONS
3
b. According to (6.6) P (A [ B) = P (A) + P (B) P (A \ B) = 0:2 + 0:5 0:4 = 0:3:
c. A and B being mutually exclusive implies that P (AjB) = P (BjA) = 0. A and B
being statistically independent implies that P (AjB) = P (A) and P (BjA) = P (B).
The only way that both of these conditions can be true is for P (A) = P (B) = 0.
Problem 6.5
a. The result is
2
P (AB) = 1 P (1 or more links broken) = 1 1 q 2 (1 q)
b. If link 4 is removed, the result is
P (ABj link 4 removed) = 1 1 q 2 (1 q)
c. If link 2 is removed, the result is
2
P (ABj link 2 removed) = 1 1 q 2
d. Removal of link 4 is more severe than removal of link 2.
Problem 6.6
Using Bayesí rule
P (AjB) =
P (BjA) P (A)
P (B)
where, by total probability
P (B) = P (BjA) P (A) + P BjA P A = P (BjA) P (A) + 1 P BjA P A
= (0:95) (0:45) + (0:35) (0:55) = 0:62
Therefore
P (AjB) =
Similarly,
with
(0:95) (0:45)
= 0:6895
0:62
P BjA P (A)
P AjB =
P B
P B = P BjA P (A) + P BjA P A = (0:05) (0:45) + (0:65) (0:55) = 0:38
4
CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES
(Note also that P B = 1 P (B).) Thus
(0:05) (0:45)
P AjB =
= 0:0592
0:38
Problem 6.7
a. P (A2 ) = 0:3; P (A2 ; B1 ) = 0:05; P (A1 ; B2 ) = 0:05; P (A3 ; B3 ) = 0:05; P (B1 ) =
0:15; P (B2 ) = 0:25; P (B3 ) = 0:6:
b. P(A3 jB3 ) = 0:083; P (B2 jA1 ) = 0:091; P (B3 jA2 ) = 0:333:
Problem 6.8
See the tables below for the results.
Outcomes
(1; 1)
(1; 2) ; (2; 1)
(1; 3) ; (3; 1) ; (2; 2)
(1; 4) ; (4; 1) ; (2; 3) ; (3; 2)
(1; 5) ; (5; 1) ; (2; 4) ; (4; 2) ; (3; 3)
(a)
(1; 6) ; (6; 1) ; (2; 5) ; (5; 2) ; (3; 4) ; (4; 3)
(2; 6) ; (6; 2) ; (3; 5) ; (5; 3) ; (4; 4)
(3; 6) ; (6; 3) ; (4; 5) ; (5; 4)
(4; 6) ; (6; 4) ; (5; 5)
(5; 6) ; (6; 5)
((6; 6))
(b)
Outcomes
(1; 1) ; (1; 3) ; (3; 1) ; (2; 2) ;
(1; 5) ; (5; 1) ; (2; 4) ; (4; 2) ; (3; 3) ;
(2; 6) ; (6; 2) ; (3; 5) ; (5; 3) ; (4; 4) ;
(4; 6) ; (6; 4) ; (5; 5) ; (6; 6)
(1; 2) ; (2; 1) ; (1; 4) ; (4; 1) ; (2; 3) ; (3; 2) ;
(1; 6) ; (6; 1) ; (2; 5) ; (5; 2) ; (3; 4) ; (4; 3) ;
(3; 6) ; (6; 3) ; (4; 5) ; (5; 4) ; (5; 6) ; (6; 5)
X1 =
2
3
4
5
6
7
8
9
10
11
12
P
spots up
X2
P (X2 = xi )
1
18=36 = 1=2
0
18=36 = 1=2
P (X1 = xi )
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
6.1. PROBLEM SOLUTIONS
5
Problem 6.9
First, make a table showing outcomes giving the two values of the random variable:
Outcomes (H = head, T = tail)
X P (X = xi )
TTH, THT, HTT, HHH (odd # of heads) 0
4=8 = 1=2
HHT, HTH, THH, TTT (even # of heads) 1
4=8 = 1=2
The cdf is zero for x < 0, jumps by 1/2 at x = 0, jumps another 1/2 at x = 1, and remains
at 1 out to 1. The pdf consists of an impulse of weight 1/2 at x = 0 and an impulse of
weight 1/2 at x = 1. It is zero elsewhere.
Problem 6.10
a. As x ! 1, the cdf approaches 1. Therefore B = 1. Assuming continuity of the cdf,
Fx (12) = 1 also, which says A 124 = 1 or A = 4:8225 105 .
b. The pdf is given by
dFX (x)
= 4 4:8225 105 x3 u (x) u (12 x)
dx
= 1:929 104 x3 u (x) u (12 x)
fX (x) =
The graph of this pdf is 0 for t < 0, the cubic 1:929 104 x3 for 0 x 12, and 0
for t > 12.
c. The desired probability is
P (X > 5) = 1 FX (5) = 1 4:8225 105 54 = 1 0:0303 = 0:9699
d. The desired probability is
P (4 X < 6) = FX (6) FX (4)
= 4:8225 105 64 4:8225 105 44
= 4:8225 105 64 44
= 0:0502
Problem 6.11
a. A = where the condition
b. B = where the condition
R1
0
R0
Aex dx = 1 must be satisÖed;
x
1 Be dx = 1 must be satisÖed;
6
CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES
c. C = e where the condition
d. D = 1= where the condition
R1
1
Cex dx = 1 must be satisÖed;
R
0 Ddx = 1 must be satisÖed.
Problem 6.12
a. Factor the joint pdf as
fXY (x; y) =
p
p
A exp [ jxj] A exp [ jyj]
With proper choice of A, the two separate factors are marginal pdfs. Thus X and Y
are statistically independent.
b. Note that the pdf is the volume between a plane which intersects the x and y coordinate axes one unit out and the fXY coordinate axis at C. First Önd C from the
integral
Z 1Z 1
Z 1 Z 1y
fXY (x; y) dxdy = 1 or
C (1 x y) dxdy = 1
1
1
0
0
This gives C = 6. Next Önd fX (x) and fY (y) as
(
Z 1x
3 (1 x)2 ; 0 x 1
fX (x) =
6 (1 x y) dy =
0; otherwise
0
and
fY (y) =
Z 1y
0
6 (1 x y) dx =
(
3 (1 y)2 ; 0 y 1
0; otherwise
Since the joint pdf is not equal to the product of the two marginal pdfs, X and Y are
not statistically independent.
Problem 6.13
RR
a.
fXY (x; y) dxdy = 1 gives C = 1=32;
b. fXY (1; 1:5) = 1+1:5
32 = 0:0781
c. fXY (x; 3) = 0 because y = 3 is in the domain where fXY is zero.
d. By integration, fY (y) = (1+2y)
; 0 y 2, so the desired conditional pdf is
8
fXjY (xjy) =
1
(1 + xy)
fXY (x; y)
(1 + xy)
= 32
=
; 0 x 4; 0 y 2
1
fY (y)
4 (1 + 2y)
8 (1 + 2y)
6.1. PROBLEM SOLUTIONS
7
Substitute y = 1 to get the asked for result:
fXjY (xj3) =
1+x
1+x
=
; 0x4
4 (1 + 2)
12
Problem 6.14
a. To Önd A, evaluate the double integral
Z 1Z 1
Z 1Z 1
fXY (x; y) dxdy = 1 =
Axye(x+y) dxdy
1 1
0
Z 10
Z 1
x
= A
xe dx
yey dy
0
0
= A
Thus A = 1.
b. Clearly, fX (x) = xex u (x) and fY (y) = yey u (y)
c. Yes, because the joint pdf factors into the product of the marginal pdfs.
Problem 6.15
a. Use normalization of the pdf to unity:
Z 1
Z 1
1
2
x u (x ) dx =
x2 dx = x1 = 1
1
Hence, this is a pdf for any value of .
b. The cdf is
FX (x) =
Z x
1
z
2
u (z ) dz =
0; x <
(1 =x) ; x >
c. The desired probability is
P (X 10) = 1 P (X < 10) = 1 FX (10) =
1; 10
=10; < 10
8
CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES
Problem 6.16
The result is
8
2
< expp(y=2 ) ; y 0
i 1
2 2 y
fY (y) = fX (x)jx=py + fX (x)jx=py p =
2 y :
0; y < 0
h
Problem 6.17
a. First note that
P (Y = 0) = P (X 0) = 1=2
For y > 0, transformation of variables gives
1
dg (y)
fY (y) = fX (x)
dy
x=g 1 (y)
Since y = g (x) = ax, we have g 1 (y) = y=a. Therefore
2
exp 2ay2 2
fY (y) = p
; y>0
2a2 2
For y = 0, we need to add 0:5 (y) to reáect the fact that Y takes on the value 0 with
probability 0.5. Hence, for all y, the result is
y2
exp
2
2
1
2a
fY (y) = (y) + p
u (y)
2
2a2 2
where u (y) is the unit step function.
b. Note that
2
y2
y2
exp 2
exp (y)
exp
2
2
2
2
2
p
fY (y) = p
+
=2 p
; y0
2
2
2
2
2 2
dy
and fY (y) = 0 otherwise, because Y is nonnegative. Also note that dx
= 1 for y 0
and y 0.
6.1. PROBLEM SOLUTIONS
9
Problem 6.18
a. The normalization of the pdf to unity provides the relationship
Z 1
fX (x) dx = A
1
Z 1
ebx dx = Ae2b =b = 1
2
Thus A = be2b .
b. The mean is
2b
Z 1
xebx dx
1 2b
2b 1
= be
2+
e
b
b
1
= 2+
b
E [X] = be
2
where integration by parts was used.
c. The second moment of X is
E X
2
2b
Z 1
x2 ebx dx
2
1
1
2b 2
2b
= be
2+
2+
e
b
b
b
1
1
= 2 2+
2+
b
b
= be
where integration by parts was used twice.
d. The variance of X is
2X
= E X 2 fE [X]g2
1
1
1 2
= 2 2+
2+
2+
b
b
b
1
= 2
b
10
CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES
Problem 6.19
h i2
h i2
R2
R 2 dx
1 x3
4
1 x2
4
2
a. E X 2 = 0 x2 dx
=
=
;
E
[X]
=
x
=
2
2 3 0
3
2
2 2 0 = 1; 3 > 1 so it is
0
true in this special case.
h i4
h i4
R4
R 4 dx
1 x3
16
1 x2
16
2
b. E X 2 = 0 x2 dx
=
=
;
E
[X]
=
x
=
4
4 3 0
3
4
4 2 0 = 2;
3 > 2 so it is
0
also true in this special case.
n
o
c. Use the fact that E [X E (X)]2 0 (0 only if X = 0 with probability one).
Expanding, we have E X 2 fE [X]g2 > 0 if X 6= 0.
Problem 6.20
Problem statement should refer to Table 6.4.
a. The mean and variance of the Rayleigh random variable can be obtained by using
tabulated deÖnite integrals. The mean is
r
Z 1 2
p
r
r2
1
1
2
2 =
E [R] =
exp
dr
=
2
2
2 2
2 22 (1=2 2 )
2
0
R1
by using a deÖnite integral for 0 x2n exp ax2 dx given in Appendix F. The mean
square is
Z 1 3
2
r
r2
r2
r
E R
=
exp
dr;
let
u
=
; du = 2 dr
2
2
2
2
2
0
Z 1
2
= 2
u exp (u) du
0
Z 1
1
2
= 2 u exp (u)j0 +
exp (u) du (integration by parts)
0
= 2 2
Thus, the variance is
var [R] = E R2 E 2 [R]
= 2 2 2
2
which is the same as given in Table 6.4.
6.1. PROBLEM SOLUTIONS
11
b. The mean and second moment of the single-sided exponential pdf are easily found in
terms of tabulated integrals or by integration by parts. The mean is
Z 1
E [X] =
x exp (x) dx
0
Z
1 1
=
u exp (u) du
0
1
=
The second moment is
E [X] =
=
=
=
and the variance is
Z 1
x2 exp (x) dx
0
Z 1
1
u2 exp (u) du
2 0
Z 1
1
1
2
u exp (u) 0 + 2
u exp (u) du
2
0
2
2
1
var [X] = E X 2 E 2 [X] = 2
c. The mean of the hyperbolic pdf is zero by virtue of the evenness of the pdf, . For the
variance of the hyperbolic pdf, consider
Z 1 2
Z 1 2
x (m 1) hm1 dx
x (m 1) hm1 dx
E X2 =
dx
=
dx
m
2 (jxj + h)
(x + h)m
1
0
where the second integral follows by evenness of the integrand in the Örst integral,
and the absolute value on x is unnecessary because the integration is for positive x.
Integrate by parts twice to obtain
("
#1 Z
)
1
2
x2 (h + h)m1
2x (x + h)m+1
m1
E X = (m 1) h
+
dx
m + 1
m1
0
0
The Örst term is zero if m 3. Therefore
Z 1
E X 2 = 2hm1
x (x + h)m+1 dx =
0
The variance is the same since its mean is zero.
2h2
; m4
(m 2) (m 3)
12
CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES
d. The mean of a Poisson random variable is
1
X
k
E [K] =
k exp ()
k!
k=0
1
X
k1
= exp ()
; let n = k 1
(k 1)!
k=1
= exp ()
1
X
n
n=0
=
= exp () exp ()
n!
Its mean-square value is
1
X
k
E K2 =
k 2 exp ()
k!
k=0
= exp ()
1
X
k1
(k 1)!
k
k=1
= exp ()
= exp ()
1
X
(n + 1)
n=0
"1
X
"
n
n!
1
n X n
n
+
n!
n!
n=1
1
X
= exp ()
n=1
n=0
#
1
X n
+
(n 1)!
n!
n1
n=0
#
= exp () [ exp () + exp ()] = 2 +
Hence, var[K] = E K 2 E 2 [K] = :
e. For the geometric distribution, the mean is
E [K] =
1
X
kpq k1
k=1
Di§erentiate the sum
P1
k=0 q
k =
1
1q ; jqj < 1 to get the sum
1
X
k=1
kq k1 =
1
(1 q)2
6.1. PROBLEM SOLUTIONS
13
Apply this to E [K] to get
E [K] =
p
1
2 = p
(1 q)
A similar procedure results in
1
X
E K2 =
k 2 pq k1 =
k=1
p
1
q
1
2 + p 2 = p2 + p2
(1 q)
Problem 6.21
1
a. A = b 1 ebB
;
b. The cdf is 0 for x < 0 and 1 for x > B. For 0 x B, the result is (A=b) 1 ebx ;
c. The mean is
1
ebB
E [X] =
1 bB
b
1 ebB
d. The mean-square is
2 2A 1
ebB
AB 2 bB
E X =
(1
+
bB)
e
b b2
b2
b
where A must be substituted from part (a.).
e. For the variance, subtract the result of part (c) squared from the result of part (d).
Problem 6.22
a. The integral evaluates as follows:
Z 1 p
Z
2
2n ey2
ex2=2
2n+1 2n 1 2n y
p dy = p
x2n p
dx = 2
2 2y
y e dy
2 2
1
0
0
where y = x= 21=2 . Using a table of deÖnite integrals, the given result is obtained.
E X 2n =
Z 1
b. The integral is zero by virtue of the oddness of the integrand.
14
CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES
Problem 6.23
E [X] =
=
E X2 =
=
1
1
x
(x 5) + [u (x 4) u (x 8)] dx
2
8
1
Z 1
Z
1
1 8
1
1 82 42
11
x (x 5) dx +
xdx = (5) +
=
2 1
8 4
2
8
2
2
Z 1
1
1
x
(x 5) + [u (x 4) u (x 8)] dx
2
8
1
Z 1
Z
1
1 8 2
1
1 83 43
187
2
x (x 5) dx +
x dx = (25) +
=
2 1
8 4
2
8
3
6
Z 1
2
187 121
11
2X = E X 2 E 2 [X] =
=
6
4
12
Problem 6.24
Regardless of the correlation coe¢cient, the mean of Z is
E fZg = E f3X 4Y g = 3E fXg 4E fY g = 3 (1) 4 (3) = 9
The variance of Z is
n
o
var (Z) = E [Z E (Z)]2
n
o
= E [3X 4Y 3E(X) + 4E(Y )]2
n
2 o
= E 3 X X 4(Y Y )
= E 9 X X 24 X X (Y Y ) + 16(Y Y )2
= 9 2X + 16 2Y 24 X Y XY
Putting in numbers, the results for the variance of Z are: (a) 148; (b) 122.6; (c) 59.1; (d)
21.0.
6.1. PROBLEM SOLUTIONS
15
Problem 6.25
The conditional pdf fXjY (xjy) is given by
fXjY (xjy) =
=
=
=
=
=
=
fXY (x; y)
2 2
=
fY (y)
p
1
p
1
p
h 2
i
x 2xy+y 2
exp
2
2
2 (1 )
2
1
exp(y 2 =(2 2 ))
p
2 2
2
x 2xy + y 2
exp
y2
+ 2
2
2 2 (1 2 )
1
x2 2xy
y2
y2
p
p
exp 2
+
2 (1 2 ) 2 2 2 2 (1 2 )
2 2 1 2
1
x2 2xy
y2
1
p
p
exp 2
+
1
2 (1 2 ) 2 2
1 2
2 2 1 2
1
x2 2xy
y2
2
p
p
exp 2
+
2 (1 2 ) 2 2
1 2
2 2 1 2
2
1
x 2xy + 2 y 2
p
p
exp
2 2 (1 2 )
2 2 1 2
"
#
1
(x y)2
p
p
exp 2
2 (1 2 )
2 2 1 2
2 2
1 2
Note that the conditional pdf of X given
with conditional mean E [XjY ] = Y
Y is Gaussian
and conditional variance var(XjY ) = 1 2 2 .
Problem 6.26
Divide the joint Gaussian pdf of two random variables by the Gaussian pdf of Y , collect all
exponential terms in a common exponent, complete the square of this exponent which will
be quadratic in x and y, and the desired result is a Gaussian pdf with
E fXjY g = mX +
X
(Y mY ) and var (XjY ) = 2X 1 2
Y
The derivation follows closely that of Problem 6.25 except for the presence of nonzero means
mX and mY :
Problem 6.27
Convolve the two component pdfs. A sketch of the two component pdfs making up the
integrand shows that there is no overlap until z > 1, and the overlap ends when z > 7. The
16
CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES
result, either obtained graphically or analytically, is given by (a trapezoid)
8
0; z < 1
>
>
>
>
< (z 1) =8; 1 z < 3
1=4; 3 z < 5
fZ (z) =
>
>
(z 7) =8; 5 z 7
>
>
:
0; z > 7
Problem 6.28
a. From Table 6.4,
E [X] = 0; E X 2 = 1=32 = var [X]
b. By transformation of random variables, the pdf of Y is
dx
fY (y) = fX (x)jx=(y4)=5
dy
4 8jy4j=5
=
e
5
c. Using the transformation from X to Y , it follows that
E [Y ] = E [4 + 5X] = 4 + 5E [X] = 4;
h
i
E Y 2 = E (4 + 5X)2
= E 16 + 40X + 25X 2 = 16 + 40E [X] + 25E X 2
816
51
= 16 + 25=32 =
=
= 25:5;
32
2
2 = E Y 2 E 2 [Y ] = 25:5 42 = 9:5
Problem 6.29
a. The characteristic function is
MX (jv) =
a
a jv
6.1. PROBLEM SOLUTIONS
17
b. The mean and mean-square values are
@MX (jv)
E [X] = j
@v
and
= 1=a
v=0
2
2
2
2 @ MX (jv)
E X = (j)
= 2
@v 2
a
v=0
respectively.
d. var[X] = E X 2 E 2 [X] = 1=a2 .
Problem 6.30
The required distributions are
n k
P (k) =
p (1 p)nk (Binomial)
k
2
P (k) =
P (k) =
e(knp) =[2np(1p)]
p
(Laplace)
2np (1 p)
(np)k np
e
(Poisson)
k!
Comparison tables are given below:
n; p
3,1/5
k
0
1
2
3
Binomial
0.512
0.384
0.096
0.008
Laplace
0.396
0.487
0.075
0.0014
Poisson
0.549
0.329
0.099
0.020
n; p
3,1/10
k
0
1
2
3
Binomial
0.729
0.243
0.027
0.001
Laplace
0.650
0.310
0.004
1106
Poisson
0.741
0.222
0.033
0.003
a.
b.
18
CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES
n; p
10,1/5
k
0
1
2
3
4
5
6
7
8
9
Binomial
0.107
0.268
0.302
0.201
0.088
0.026
0.005
7.9104
7.4105
4.1106
Laplace
0.090
0.231
0.315
0.231
0.090
0.019
0.002
1.3104
4.1106
7.1108
n; p
10,1/10
k
0
1
2
3
4
5
6
7
8
9
Binomial
0.348
0.387
0.194
0.057
0.011
1.5103
1.4104
3.6107
9.0109
11010
Laplace
0.241
0.420
0.241
0.046
0.003
6105
3.9107
6.31013
1.51016
1.21020
c.
d.
Poisson
0.135
0.271
0.271
0.180
0.090
0.036
0.012
3.4103
8.6104
1.9104
Poisson
0.368
0.368
0.183
0.061
0.015
3.1103
5.1104
3.4103
1106
1107
Problem 6.31
a. P (5 or 6 heads in 10 trials) =
P
k=5;6 [10!=k! (10 k)!] 2
10 = 10! 1 = 0:2461
5!5! 1024
b. P (Örst head at trial 15) = (1=2)(1=2)4 = 1=32
P
100!
100 = 0:1193; P (Örst head at
c. P (50 to 60 heads in 100 trials) = 60
k=50 k!(100k)! 2
trial 50) = (1=2)(1=2)49 = 8:89 1016 .
Problem 6.32
a. Np = 26 (25) (24) (23) = 358; 800;
b. Np = (26)4 = 456; 976;
6.1. PROBLEM SOLUTIONS
19
c. The answers are the reciprocals of the numbers found in (a) and (b), or 2:79 106
and 2:19 106 ; respectively.
Problem 6.33
a. The desired probability is
P (fewer than 3 heads in 20 coins tossed)
=
20
2
X
20
1
k=0
k
2
= (1 + 20 + 190) 220 = 2:0123 104
b. var(N ) = npq = 20 (1=2) (1=2) = 5; E [N ] = np = 10. The Laplace-approximated
probability is
P (< 3 heads in 20 coins tossed)
2
2
X
e(k10) =10
p
=
10
k=0
1
= p
e10 + e81=10 + e64=10
10
= 3:587 104
c. The percent error is
% error =
3:587 104 2:0123 104
100 = 78:25%
2:0123 104
Problem 6.34
a. The probability of exactly one error in 105 digits, by the binomial distribution, is
5
99;999 5
10
5
P 1 error in 10
=
1 105
10
1
= 105 0:3679 105 = 0:3679
By the Poisson approximation, it is
P 1 error in 105 = exp (1) = 0:3679
20
CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES
b. The Poisson approximation gives
105 105
P 2 errors in 105 =
exp 105 105 = 0:18395
2!
c. Use the Poisson approximation:
P (more than 5 errors in 105 ) = 1 P (0; 1; 2; 3; 4; or 5 errors)
5
X
(np)k
= 1
exp (np)
k!
k=0
1 1
1
1
= 1 exp (1) 1 + 1 + + +
+
2 6 24 120
= 1 0:9994 = 0:0006
Problem 6.35
a. The marginal pdf for X is
"
(x 1)2
fX (x) = p
exp
24
2 (4)
1
#
"
#
1
(x 1)2
= p exp
8
8
The marginal pdf for Y is of the same form except for Y and y in place of X and x,
respectively.
b. The joint pdf is
fXY (x; y) =
=
=
8
9
x1 2
x1 y1 y1 2 >
>
<
=
2
(0:5)
+
2
2
2
2
1
p
exp
>
>
2 (1 0:52 )
2 (2) (2) 1 0:52
:
;
(
)
1
(x 1)2 (x 1) (y 1) + (y 1)2
p
exp
4 2 (1 0:52 )
8 0:75
2
1
x x xy + y 2 y + 1
p
exp
6
8 0:75
6.1. PROBLEM SOLUTIONS
21
The conditional pdf of X given Y = y is
fXjY (xj y) =
=
=
=
=
fXY (x; y)
fY (y)
p1
exp
8 0:75
n
o
2
2
(x1) (x1)(y1)+(y1)
6
h
i
(y1)2
1
p
exp
8
8
(
)
1
(x 1)2 (x 1) (y 1) + (y 1)2 (y 1)2
p exp
+
6
8
6
(
)
1
(x 1)2 (x 1) (y 1) (y 1)2
p exp
6
24
6
(
)
(
)
1
[x 1 0:5 (y 1)]2
1
[x 0:5y 0:5]2
p exp
= p exp
6
6
6
6
c. See Problem 6.26 to deduce that
E [Xj Y ] = 1 + 0:5 (Y 1)
and var [Xj Y ] = 4 1 0:52 = 3
Problem 6.36
a. K = 1=;
b. E [X] is not deÖned, but one could argue that it is zero from the oddness of the
integrand for this expectation. If one writes down the integral for the second moment,
it clearly does not converge (the integrand approaches a constant as jxj ! 1).
c. The answer is given;
d. Compare the form of the characteristic function with the answer given in (c) and do
a suitable redeÖnition of variables.
22
CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES
Problem 6.37
a. The characteristic function is found from
n Pn
o
2
MY (jv) = E ejvY = E ejv i=1 Xi = E
( n
Y
i=1
jvXi2
e
)
But
Z 1
n
o
x2 =2 2
2
2e
E ejvXi
=
ejvx p
dx
2 2
1
Z 1 (1=22 jv)x2
e
p
=
dx
2 2
1
1=2
= 1 j2v 2
which follows by redeÖning the variance of the integrand to get a Gaussian function.
Thus
N=2
MY (jv) = 1 j2v 2
b. The given pdf follows easily by letting = 2 2 in the given Fourier transform pair.
c. The mean of Y is
E [Y ] = N 2
by taking the sum of the expectations of the separate terms of the sum deÖning Y .
The mean-square value of Xi2 is
h
i
2
2 1=2
h i
d
1
j2v
2
E Xi2
= (j)2
= 3 4
2
dv
v=0
Thus,
var
h
Xi2
2 i
= 3 4 4 = 2 4
Since the terms in the sum deÖning Y are independent, var[Y ] = 2N 4 . The mean
and variance of Y can be put into the deÖnition of a Gaussian pdf to get the desired
approximation.
d. The cases N = 2; 4; 8; and N = 16 are shown in Figure 6.1. The approximations
appear to get better as N increases.
Note the extreme di§erence between exact and approximation due to the central limit
theorem not being valid for N = 2; 4.
6.1. PROBLEM SOLUTIONS
0.5
23
0.2
chi-sq; σ 2 = 1; N = 2
Gauss approx
0.4
chi-sq; σ 2 = 1; N = 4
Gauss approx
0.15
fY (y)
fY (y)
0.3
0.2
0.05
0.1
0
0.1
0
10
20
0
30
0
10
y
chi-sq; σ 2 = 1; N = 8
Gauss approx
0.12
20
30
y
chi-sq; σ 2 = 1; N = 16
Gauss approx
0.08
0.1
0.06
fY (y)
fY (y)
0.08
0.06
0.04
0.04
0.02
0.02
0
0
10
20
30
0
0
y
10
20
30
y
e. Evident by direct substitution.
Problem 6.38
R 1 exp(t2 =2)
exp(x2 =2)
p
p
This is a matter of plotting Q (x) = x
dx
and
Q
(x)
=
on the same
a
2
2x
set of axes.
Problem 6.39
By deÖnition, the cdf of a Gaussian random variable is
h
i
h
i
2
2
Z x exp (um)
Z 1 exp (um)
2
2
2
2
p
p
FX (x) =
du = 1
du
2
2
2 2
1
x
Change variables to
v=
um
with the result that
2
exp u2
xm
p
FX (x) = 1
du = 1 Q
2
(xm)=
Z 1
24
CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES
A plot may easily be obtained byR a MATLAB
2 program. It is suggested that you use the
1
2
p
MATLAB function erfc(x) = x exp t dt.
Problem 6.40
Consider the product of integrals
Z 1
Z 1
exp u2 =2
exp v 2 =2
p
p
I (x) =
du
dv
2
2
0
x
Z
1 1 exp v 2 =2
1
p
=
dv = Q (x)
2 x
2
2
where the 1/2 comes from the fact that the Örst integral is the integral of the right half of
a Gaussian pdf. Now write the product of the two integrals as an iterated integral to get
2
Z 1Z 1
1
u + v2
I (x) =
exp
dudv
2 0
2
x
Transform the integral to polar coordinates via u = r cos ; v = r sin ; and dudv = rdrd
to get
Z =2 Z 1
1
1
I (x) =
Q (x) =
exp r2 =2 rdrd
2
2 0
x= sin
Z =2
1
x2
=
exp
d
2 0
2 sin2
Multiplying through by 2 gives the answer given in the problem statement.
Problem 6.41
The following relationships involving the Q-function will prove useful:
Z b
exp u2 =2
p
du = Q (a) Q (b) ; b > a;
2
a
Q (x) = 1 Q (jxj) ; x < 0; Q (0) = 0:5
a. The probability is
Z 15
2
e(x10) =50
p
dx
50
15
Z 1
exp u2 =2
x 10
p
=
du; u = p
2
25
5
= Q (5) Q (1) = 1 Q (5) Q (1)
P (jXj 15) =
= 0:8413
6.1. PROBLEM SOLUTIONS
25
b. This probability is
Z 20
2
e(x10) =50
p
dx
P (10 < X 20) =
50
10
Z 2
exp u2 =2
x 10
p
=
du; u = p
2
25
0
= Q (0) Q (2) = 0:5 Q (2)
= 0:4772
c. The desired probability is
Z 25
2
e(x10) =50
p
dx
50
5
Z 3
exp u2 =2
x 10
p
=
du; u = p
2
25
1
= Q (1) Q (3) = 1 Q (1) Q (3)
P (5 < X 25) =
= 0:8400
d. The result is
Z 30
2
e(x10) =50
p
P (20 < X 30) =
dx
50
20
Z 4
exp u2 =2
x 10
p
=
du; u = p
2
25
2
= Q (2) Q (4) = 0:0227
Problem 6.42
a. Observe that
2
=
Z 1
Z1
(x m)2 fX (x) dx
jxmj>k
Z
jxmj>k
2 2
(x m)2 fX (x) dx
k 2 2 fX (x) dx
= k P fjX mj > kg = k 2 2 f1 P [jX mj > k]g
26
CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES
or
P [jX mj < k] 1
1
k2
b. Note that for this random variable mX = 0 and 2X = 22 =12 = 1=3: The actual
probability is
h
p i
P [jX mX j < k X ] = P jXj < k= 3
( R p
p
p
k= 3 dx
p
=
k=
3;
k
<
3
k= 3p2
=
1; k 3
A comparison is given in the table below:
k Actual prob.
Ch. bound
0 0 p
1 1= 3 = 0:577 0
2 1
0.75
3 1
0.889
Note that in all cases the bound exceeds the actual probability.
Problem 6.43
The desired probability is
P [jXj > k] = P [X < k] + P [X > k] ; k = 1; 2; 3
Z k
Z 1
exp x2 =2 2
exp x2 =2 2
p
p
=
dx +
dx
2 2
2 2
1
k
= 2Q (k)
a. P [jXj > ] = 0:3173;
b. P [jXj > 2] = 0:0455;
c. P [jXj > 3] = 0:0027:
Problem 6.44
6.1. PROBLEM SOLUTIONS
27
a. The mean is 0 by even symmetry of the pdf. Thus, the variance is
2 = var [X] = E X 2 (because the mean is 0)
Z 1
a
=
x2
exp (ajxj) dx
2
1
a Z 1
= 2
x2 exp (ax) dx
2 0
Z 1
y 2
dy
= 2a
exp (y)
a
a
0
1
1
= 2a
= 1=a2
a3
2
Thus, in terms of the variance, the pdf is
fX (x) =
1
exp ( jxj =)
2
b. The desired probability is
P [jXj > k] = P [X < k] + P [X > k] ; k = 1; 2; 3
Z k
Z 1
1
1
=
exp (x=) dx +
exp (x=) dx
2
k 2
Z1
1
=
exp (u) du (by eveness of the integrand and u = x=)
k
= exp (k)
Thus, the required numerical values are
P [jXj > ] = 0:3679;
P [jXj > 2] = 0:1353;
P [jXj > 3] = 0:0498
Problem 6.45
Since both X and Y are Gaussian and the transformation is linear, Z is Gaussian. Thus,
all we need are the mean and variance of Z. Since the means of X and Y are 0, so is the
28
CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES
mean of Z. Therefore, its variance is
h
i
2Z = E Z 2 = E (X + 2Y )2
= E X 2 + 4XY + 4Y 2
= E X 2 + 4E [XY ] + 4E Y 2
= 2X + 4 X Y XY + 4 2Y (because the means are 0)
p p
= 3 + (4)
3
4 (0:4) + (4) (4)
p
= 19 3:2 3 = 13:457
The pdf of Z is
exp z 2 =26:915
p
fZ (z) =
26:915
Problem 6.46
Since both X and Y are Gaussian and the transformation is linear, Z is Gaussian. Thus,
all we need are the mean and variance of Z. The mean of Z is
mZ
= E [3X + Y ] = 3mX + mY
= (3) (1) + 2 = 5
Its variance is
2Z
h
i
h
i
= E (Z mZ )2 = E (3X + Y 3mX mY )2
h
i
= E (3X 3mX + Y mY )2
h
i
h
i
= 9E (X mX )2 + 6E [(X mX ) (Y mY )] + E (Y mY )2
= 9 2X + 6 X Y XY + 2Y
p p
= 9 (3) + (6)
3
2 (0:2) + 2
p
= 29 + 1:2 6 = 31:939
The pdf of Z is
Problem 6.47
h
i
exp (z 5)2 =63:879
p
fZ (z) =
63:879
6.1. PROBLEM SOLUTIONS
a. fX (x) =
exp[(x5)2 =2]
p
;
2
b. fXY (x; y) =
29
fY (y) =
exp[(y3)2 =4]
p
4
exp[(x5)2 =2] exp[(y3)2 =4]
p
p
2
4
c. E (Z1 ) = mX + mY = 5 + 3 = 8; E (Z2 ) = mX mY = 5 3 = 2
d. The variances are
h
i
h
i
= E (Zi mZi )2 = E (X Y mX mY )2
h
i
= E ((X mX ) (Y mY ))2
h
i
h
i
= E (X mX )2 + 2E (X mX ) E (Y mY ) + E (Y mY )2
2Zi
= 2X + 0 + 2Y
= 1+2=3
e. The desired pdf is
h
i
exp (z1 8)2 =6
p
fZ1 (z1 ) =
6
f. The desired pdf is
h
i
exp (z2 2)2 =6
p
fZ2 (z2 ) =
6
Problem 6.48
a. fX (x) =
exp[(x4)2 =6]
p
;
6
b. fXY (x; y) =
fY (y) =
exp[(y2)2 =10]
p
10
exp[(x4)2 =6] exp[(y2)2 =10]
p
p
6
10
c. E (Z1 ) = 3mX + mY = (3) (4) + 2 = 14; E (Z2 ) = 3mX mY = (3) (4) 2 = 10
d. The variances are
h
i
h
i
2Zi = E (Zi mZi )2 = E (3X Y mX mY )2
h
i
= E (3 (X mX ) (Y mY ))2
h
i
h
i
= 9E (X mX )2 + 6E (X mX ) E (Y mY ) + E (Y mY )2
= 9 2X + 0 + 2Y
= (9) (3) + 5 = 32
30
CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES
e. The pdf of Z1 is
h
i
exp (z1 14)2 =64
p
fZ1 (z1 ) =
64
f. The pdf of Z2 is
h
i
exp (z2 10)2 =64
p
fZ2 (z2 ) =
64
Problem 6.49
a. The probability of mean exceedance for a uniformly distributed random variable is
Z b
1
dx
P X > (a + b) =
2
b
(a+b)=2 a
b
1
1
=
x
=
b a (a+b)=2 2
b. The probability of mean exceedance for a Rayleigh distributed random variable is
r
Z 1
r
r2
r2
P X>
= p
exp
dr;
v
=
2
2
2 2
2 2
=2
Z 1
=
ev dv = e=4 = 0:4559
=4
c. The probability of mean exceedance for a one-sided exponential distributed random
variable is
Z 1
1
P X>
=
exp (x) dx
1=
= e1 = 0:3679
6.2
Computer Exercises
Computer Exercise 6.1
% ce6_1.m: Generate an exponentially distributed random variable
%
with parameter a from a uniform random variable in [0, 1)
%
clf
6.2. COMPUTER EXERCISES
31
a = input(íEnter the parameter of the exponential pdf: f(x) = a*exp(-a*x)*u(x) => í);
N = input(íEnter number of exponential random variables to be generated => í);
U = rand(1,N);
V = -(1/a)*log(U);
[M, X] = hist(V);
disp(í í)
disp(íNo. of random variables generatedí)
disp(N)
disp(í í)
disp(íBin centersí)
disp(X)
disp(í í)
disp(íNo. of random variable counts in each biní)
disp(M)
disp(í í)
norm_hist = M/(N*(X(2)-X(1)));
plot(X, norm_hist, íoí)
hold
plot(X, a*exp(-a*X), íñí), xlabel(íxí), ylabel(íf_X(x)í),...
title([íTheoretical pdf and histogram for í,num2str(N),...
í computer generated exponential random variablesí])
legend([íHistogram pointsí],[íExponential pdf; a = í, num2str(a)])
A typical run follows with a plot given in Fig. 6.2.
>> ce6_1
Enter the parameter of the exponential pdf: f(x) = a*exp(-a*x)*u(x) => 2
Enter number of exponential random variables to be generated => 5000
No. of random variables generated
5000
Bin centers
Columns 1 through 6
0.2094
0.6280
1.0467
1.4653
1.8839
2.3026
Columns 7 through 10
2.7212
3.1398
3.5585
3.9771
No. of random variable counts in each bin
Columns 1 through 5
2880
1200
543
215
90
Columns 6 through 10
38
16
9
5
4
32
CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES
Theoretical pdf and histogram for 5000 computer generated exponential random variables
1.4
Histogram points
Exponential pdf; a = 2
1.2
1
fX(x)
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
x
2.5
3
3.5
4
Current plot held
Computer Exercise 6.2
% ce6_2.m: Generate pairs of Gaussian random variables from Rayleigh and uniform
RVs
%
clf
m = input(íEnter the mean of the Gaussian random variables => í);
sigma = input(íEnter the standard deviation of the Gaussian random variables => í);
N = input(íEnter number of Gaussian random variable pairs to be generated => í);
U = rand(1,N);
V = rand(1,N);
R = sqrt(-2*log(V));
X = sigma*R.*cos(2*pi*U)+m;
Y = sigma*R.*sin(2*pi*U)+m;
disp(í í)
disp(íCovarance matrix of X and Y vectors:í)
6.2. COMPUTER EXERCISES
33
disp(cov(X,Y))
disp(í í)
[MX, X_bin] = hist(X, 20);
norm_MX = MX/(N*(X_bin(2)-X_bin(1)));
[MY, Y_bin] = hist(Y, 20);
norm_MY = MY/(N*(Y_bin(2)-Y_bin(1)));
gauss_pdf_X = exp(-(X_bin - m).^2/(2*sigma^2))/sqrt(2*pi*sigma^2);
gauss_pdf_Y = exp(-(Y_bin - m).^2/(2*sigma^2))/sqrt(2*pi*sigma^2);
subplot(2,1,1), plot(X_bin, norm_MX, íoí)
hold
subplot(2,1,1), plot(X_bin, gauss_pdf_X, íñí), xlabel(íxí), ylabel(íf_X(x)í),...
title([íTheoretical pdfs and histograms for í,num2str(N),í ...
computer generated independent Gaussian RV pairsí])
legend([íHistogram pointsí],[íGauss pdf; nsigma, nmu = í, num2str(sigma),í, í, num2str(m)])
subplot(2,1,2), plot(Y_bin, norm_MY, íoí)
hold
subplot(2,1,2), plot(Y_bin, gauss_pdf_Y, íñí), xlabel(íyí), ylabel(íf_Y(y)í)
A typical run follows with a plot given in Fig. 6.3.
>> ce6_2
Enter the mean of the Gaussian random variables => 2
Enter the standard deviation of the Gaussian random variables => 3
Enter number of Gaussian random variable pairs to be generated => 5000
Covarance matrix of X and Y vectors:
9.0928
0.2543
0.2543
9.0777
Current plot held
CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES
fX(x)
Theoretical pdfs and histograms for 5000 computer generated independent Gaussian RV pairs
0.2
Histogram points
0.15
Gauss pdf; σ , µ = 3, 2
0.1
0.05
0
-10
-5
0
5
10
15
5
10
15
x
0.2
0.15
fY (y)
34
0.1
0.05
0
-10
-5
0
y
6.2. COMPUTER EXERCISES
35
Computer Exercise 6.3
% ce6_3.m: Generate a sequence of Gaussian random variables with speciÖed correlation
%
between adjacent RVs
%
clf
m = input(íEnter the mean of the Gaussian random variables => í);
sigma = input(íEnter the standard deviation of the Gaussian random variables => í);
rho = input(íEnter correlation coe¢cient between adjacent Gaussian random variables
=> í);
N = input(íEnter number of Gaussian random variables to be generated => í);
var12 = sigma^2*(1 - rho^2);
X = [];
X(1) = sigma*randn(1)+m;
for k = 2:N
m12 = m + rho*(X(k-1) - m);
X(k) = sqrt(var12)*randn(1) + m12;
end
[MX, X_bin] = hist(X, 20);
norm_MX = MX/(N*(X_bin(2)-X_bin(1)));
gauss_pdf_X = exp(-(X_bin - m).^2/(2*sigma^2))/sqrt(2*pi*sigma^2);
subplot(2,1,1), plot(X_bin, norm_MX, íoí)
hold
subplot(2,1,1), plot(X_bin, gauss_pdf_X, íñí), xlabel(íxí), ylabel(íf_X(x)í),...
title([íTheoretical pdf and histogram for í,num2str(N),í computer generated Gaussian
RVsí])
legend([íHistogram pointsí],[íGauss pdf; nsigma, nmu = í, num2str(sigma),í, í, num2str(m)],2)
Z = X(1:50);
subplot(2,1,2), plot(Z, íxí), xlabel(íkí), ylabel(íX(k)í)
title([íGaussian sample values with correlation í, num2str(rho), í between samplesí])
A typical run follows with a plot given in Fig. 6.4.
>> ce6_3
Enter the mean of the Gaussian random variables => 0
Enter the standard deviation of the Gaussian random variables => 2
Enter correlation coe¢cient between adjacent Gaussian random variables => 0.8
Enter number of Gaussian random variables to be generated => 250
Current plot held
CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES
Theoretical pdf and histogram for 250 computer generated Gaussian RVs
0.4
Histogram points
fX(x)
0.3
Gauss pdf; σ , µ = 2, 0
0.2
0.1
0
-5
-4
-3
-2
-1
0
1
2
3
4
x
Gaussian sample values with correlation 0.8 between samples
5
5
10
15
20
50
5
X(k)
36
0
-5
0
25
k
30
35
40
45
6.2. COMPUTER EXERCISES
37
Computer Exercise 6.4
% ce6_4.m: Testing the validity of the Central Limit Theorem with sums of uniform
% random numbers
%
clf
N = input(íEnter number of Gaussian random variables to be generated => í);
M = input(íEnter number of uniform random variables to be summed to generate one
GRV => í);
% Mean of uniform = 0; variance = 1/12
Y = rand(M,N)-0.5;
X_gauss = sum(Y)/(sqrt(M/12));
disp(í í)
disp(íEstimated variance of the Gaussian RVs:í)
disp(cov(X_gauss))
disp(í í)
[MX, X_bin] = hist(X_gauss, 20);
norm_MX = MX/(N*(X_bin(2)-X_bin(1)));
gauss_pdf_X = exp(-X_bin.^2/2)/sqrt(2*pi);
plot(X_bin, norm_MX, íoí)
hold
plot(X_bin, gauss_pdf_X, íñí), xlabel(íxí), ylabel(íf_X(x)í),...
title([íTheoretical pdf & histogram for í,num2str(N),í ...
Gaussian RVs each generated as the sum of í, num2str(M), í uniform RVsí])
legend([íHistogram pointsí],[íGauss pdf; nsigma = 1, nmu = 0í],2)
A typical run follows with a plot given in Fig. 6.5.
>> ce6_4
Enter number of Gaussian random variables to be generated => 1000
Enter number of uniform random variables to be summed to generate one GRV => 10
Estimated variance of the Gaussian RVs:
0.9091
Current plot held
CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES
Theoretical pdf & histogram for 1000 Gaussian RVs each generated as the sum of 10 uniform RVs
0.45
Histogram points
0.4
Gauss pdf; σ = 1, µ = 0
0.35
0.3
0.25
fX(x)
38
0.2
0.15
0.1
0.05
0
-3
-2
-1
0
1
x
2
3
4
Chapter 7
Random Signals and Noise
7.1
Problem Solutions
Problem 7.1
The various sample functions are as follows. Sample functions for case (a) are horizontal
lines at levels 2A; 0; 2A, each case of which occurs equally often (with probability 1/3).
Sample functions for case (b) are horizontal lines at levels 3A; 2A; A; A; 2A; 3A,
each case of which occurs equally often (with probability 1/6). Sample functions for case
(c) are horizontal lines at levels 4A; 2A; 2A; 4A; or oblique straight lines of slope A or
A, each case of which occurs equally often (with probability 1/6).
Problem 7.2
a. For case (a) of problem 7.1, since the sample functions are constant with time and
are less than or equal 2A, the probability is one. For case (b) of problem 7.1, since
the sample functions are constant with time and 5 out of 6 are less than or equal to
2A, the probability is 5/6. For case (c) of problem 7.1, the probability is 5/6 because
5 out of 6 of the sample functions will be less than 2A at t = 2.
b. The probabilities are now 2/3, 1/2, and 1/2, respectively.
c. The probabilities are 1, 5/6, and 5/6, respectively.
Problem 7.3
a. The sketches would consist of squarewaves of random delay with respect to t = 0.
1
2
CHAPTER 7. RANDOM SIGNALS AND NOISE
b. X (t) takes on only two values A and A, and these are equally likely. Thus
1
1
fX (x) = (x A) + (x + A)
2
2
Problem 7.4
a. The sample functions at the integrator output are of the form
Z t
A
(i)
Y (t) =
A cos (! 0 ) d =
sin ! 0 t
!0
where A is Gaussian.
b. Y (i) (t0 ) is Gaussian with mean zero and standard deviation
Y = a
jsin (! 0 t0 )j
!0
c. Not stationary (the second moment, for example, depends on time) and not ergodic.
Problem 7.5
a. By inspection of the sample functions, the mean is zero. Considering the time average
correlation function, deÖned as
1
R () = lim
T !1 2T
Z T
1
x (t) x (t + ) dt =
T0
T
Z
x (t) x (t + ) dt
T0
where the last expression follows because of the periodicity of x (t), it follows from a
sketch of the integrand that
A2 (1 4=T0 ) , 0 T0 =2
R ( ) =
A2 (1 + 4=T0 ) , T0 =2 0
Since R ( ) is periodic, this deÖnes the autocorrelation function for all .
b. Yes, it is wide sense stationary. The random delay, , being over a full period, means
that no untypical sample functions occur.
7.1. PROBLEM SOLUTIONS
3
Problem 7.6
a. The time average mean is zero. The statistical average mean is
E [X (t)] =
d
2A
A cos (2f0 t + )
=
sin (2f0 t + )
=2
=2
=2
Z
2A
[sin (2f0 t + ) sin (2f0 t + =2)]
2A
=
[ sin (2f0 t) cos (2f0 t)]
p
2 2A
=
[sin (2f0 t + =4)]
=
The time average variance is A2 =2. The statistical average second moment is
E X 2 (t) =
=
=
=
Z
d
A2 cos2 (2f0 t + )
=2
=2
"Z
#
"
#
Z
A2
A2
1
d +
cos (4f0 t + 2) d =
+ sin (4f0 t + 2)
2
2
=2
=2
=2
A2 A2
+
[sin (4f0 t + 2) sin (4f0 t + )]
2
2
A2 A2
A2 A2
+
[sin (4f0 t) + sin (4f0 t)] =
+
sin (4f0 t)
2
2
2
The statistical average variance is this expression minus E 2 [X (t)].
b. The time average autocorrelation function is
hx (t) x (t + )i =
A2
cos (2f0 )
2
4
CHAPTER 7. RANDOM SIGNALS AND NOISE
The statistical average autocorrelation function is
Z
d
R ( ) =
A2 cos (2f0 t + ) cos (2f0 (t + ) + )
=2
=2
Z
A2
=
[cos (2f0 ) + cos (2f0 (2t + ) + 2)] d
=2
A2
A2
=
cos (2f0 ) +
sin (2f0 (2t + ) + 2)
2
2
=2
=
=
=
A2
A2
cos (2f0 ) +
[sin (2f0 (2t + ) + 2) sin (2f0 (2t + ) + )]
2
2
A2
A2
cos (2f0 ) +
[sin 2f0 (2t + ) + sin 2f0 (2t + )]
2
A2
2A2
cos (2f0 ) +
sin 2f0 (2t + )
2
c. No. The random phase, not being uniform over (0; 2), means that for a given time,
t, certain phases will be favored over others.
Problem 7.7
a. The time-average mean is zero. The time-average autocorrelation function is
Z 2
1
A
hY (t) Y (t + )i =
sin (! 0 t) sin [! 0 (t + )] dt
T0 T0 ! 0
2 Z
1
A
=
fcos (! 0 ) cos [! 0 (2t + )]g dt
2T0 ! 0
T0
1 A 2
=
cos (! 0 )
2 !0
b. Again, the mean is zero because the mean of A is zero. The ensemble-average autocorrelation function is
R ( ) = E [Y (t; A) Y (t + ; A)]
Z 1 2
exp a2 =2 2a
a
p
=
sin (! 0 t) sin [! 0 (t + )]
da
2 2a
1 ! 0
=
=
2a
sin (! 0 t) sin [! 0 (t + )]
! 20
2a
fcos (! 0 ) cos [! 0 (2t + )]g
2! 20
7.1. PROBLEM SOLUTIONS
5
c. No, it is not wide sense stationary because the ensemble-average autocorrelation function is time dependent (not a function only of time di§erence).
Problem 7.8
From the measurements given, assuming ergodicity, we can infer that the mean is 6 V and
the second moment is 62 + 49 = 85 V2 (we assume that the true-rms meter is AC coupled).
Thus the variance is 49 V2 . Hence, the pdf of the noise voltage is
2
2
e(x6) =249
e(x6) =98
p
fX (x) = p
=
2 49
98
Problem 7.9
a. Suitable;
b. Suitable;
c. Not suitable because the Fourier transform is not everywhere nonnegative;
d. Not suitable because autocorrelation functions must be even;
e. Suitable;
f. Suitable (note the A missing on the RHS in the Örst printing).
Problem 7.10
Use the Fourier transform pair
N0 W sinc (2W ) $
N0
(f =2W )
2
with W = 103 Hz and N20 = 2 105 V2 /Hz to get the autocorrelation function as
RX ( ) = N0 W sinc (2W ) = 4 105 103 sinc 2 103
= 0:04 sinc 2 103
6
CHAPTER 7. RANDOM SIGNALS AND NOISE
Problem 7.11
a. The r ( ) function, (7.55), is
Z
1 1
r ( ) =
p (t) p (t + ) dt
T 1
Z
1 T =2
(t + )
=
cos (t=T ) cos
dt, 0 T =2
T T =2
T
1
=
(1 =T ) cos
, 0 T =2
2
T
Now r ( ) = r ( ) and r ( ) = 0 for j j > T =2. Therefore it can be written as
1
r ( ) = ( =T ) cos
2
T
Hence, by (7.54)
A2
Ra ( ) =
( =T ) cos
2
T
The power spectral density is
A2 T
1
1
2
2
Sa (f ) =
sinc T f
+ sinc T f +
4
2T
2T
b. The autocorrelation function now becomes
Rb ( ) = A2 [2r ( ) + r ( + T ) + r ( T )]
The corresponding power spectrum is
Sb (f ) = A2 T sinc2 (T f 0:5) + sin c2 (T f + 0:5) cos2 (f T )
c. The plots are left for the student.
Problem 7.12
a. The autocorrelation functions are
RX ( ) = E [X (t) X (t + )] = Rn ( ) + E A2 cos (! 0 t + ) cos [! 0 (t + ) + ]
= Rn ( ) +
A2
A2
cos (! 0 ) = B ( = 0 ) +
cos (! 0 )
2
2
7.1. PROBLEM SOLUTIONS
7
and
RY ( ) = E [Y (t) Y (t + )] = Rn ( ) + E A2 sin (! 0 t + ) sin [! 0 (t + ) + ]
= Rn ( ) +
A2
A2
cos (! 0 ) = B ( = 0 ) +
cos (! 0 )
2
2
b. The cross-correlation function is
RXY ( ) = E [X (t) Y (t + )] = Rn ( ) + E A2 cos (! 0 t + ) sin [! 0 (t + ) + ]
= B ( = 0 ) +
A2
sin (! 0 )
2
Problem 7.13
a. By deÖnition
RZ ( ) = E [Z (t) Z (t + )]
= E [X (t) X (t + ) Y (t) Y (t + )]
= E [X (t) X (t + )]E[Y (t) Y (t + )]
= RX ( ) RY ( )
b. Since a product in the time domain is convolution in the frequency domain, it follows
that
SZ (f ) = SX (f ) SY (f )
c. Use the transform pairs
2W sinc (2W )
and
cos (2f0 )
! (f =2W )
1
1
! (f f0 ) + (f + f0 )
2
2
Also,
RY ( ) = E f4 cos (50t + ) cos [50 (t + ) + ]g = 2 cos (50 )
Using the Örst transform pair, we have
RY ( ) = 500 sinc (100 )
8
CHAPTER 7. RANDOM SIGNALS AND NOISE
This gives
RZ ( ) = [500 sinc (100 )] [2 cos (50 )]
= 1000 sinc (100 ) cos (50 )
and
SZ (f ) = 5 10
3
Y
f 25
200
+
Y f + 25
200
Problem 7.14
a. E X 2 (t) = R (0) = 12 W; E 2 [X (t)] = lim !1 R ( ) = 9 W;
2X = E X 2 (t) E 2 [X (t)] = 12 9 = 3 W.
b. DC power = E 2 [X (t)] = 9 W.
c. Total power = E X 2 (t) = 12 W.
d. S (f ) = 9 (f ) + 15 sinc2 (5f ).
Problem 7.15
a. The autocorrelation function is
RX ( ) = E [Y (t) Y (t + )]
= E f[X (t) + X (t T )][X (t + ) + X (t + T )]g
= E [X (t) X (t + )] + E[X (t) X (t + + )]
+E [X (t T ) X (t + )] + E[X (t T ) X (t + T )]
= 2RX ( ) + RX ( T ) + RX ( + T )
b. Application of the time delay theorem of Fourier transforms gives
SY (f ) = 2SX (f ) + SX (f ) [exp (j2f T ) + exp (j2f T )]
= 2SX (f ) + 2SX (f ) cos (2f T )
= 4SX (f ) cos2 (f T )
7.1. PROBLEM SOLUTIONS
9
c. Use the transform pair
RX ( ) = 6 ( )
! SX (f ) = 6 sinc2 (f )
and the result of (b) to get
SY (f ) = 24 sinc2 (f ) cos2 (f =4)
Problem 7.16
a. The student should carry out the sketch.
R 0+
b. DC power = 0 S (f ) df = 10 W.
R1
c. Total power = 1 S (f ) df = 10 + 25=5 + 5 + 5 = 25 W
d. Power for jf j 0:2 Hz = 10 + 0:9 (25=5) = 14:5 W. Fraction of total power =
14:5=25 = 0:58 = 58%.
Problem 7.17
a. This is left for the student.
b. Using the Fourier transform of a two-sided decaying exponential and the modulation
theorem, the Fourier transform of the Örst is
SX1 (f ) =
4
4
+
2 + [2 (f 1)]2 2 + [2 (f + 1)]2
The Fourier transform of the second is
SX2 (f ) =
4
+ 2 (f b) + 2 (f + b)
2 + (2f )2
Using the Fourier transform of the Gaussian pulse, the Fourier transform of the third is
2
5p
SX3 (f ) =
exp 2
2
4
All Fourier transforms are non-negative functions of frequency.
10
CHAPTER 7. RANDOM SIGNALS AND NOISE
c. None of the Fourier transforms above has a unit impulse at f = 0 so no DC power.
d. Evaluate the correlation function at = 0 to get total power. The results are 4, 6,
and 5 W, respectively, for (a)-(c).
e. Only the second one has a periodic component at b Hz.
Problem 7.18
a. The output power spectral density is
Sout (f ) = 106 f =5 105
Note that the rectangular pulse function does not need to be squared because its
amplitude squared is unity.
b. Use the Fourier transform pair 2W sinc(2W ) $ (f =2W ). The autocorrelation
function of the output is
Rout ( ) = 5 105 106 sinc 5 105 = 0:5 sinc 5 105
c. The output power is 0:5 W, which can be found from Rout (0) or the integral of
Sout (f ).
Problem 7.19
a. By assuming a unit impulse at the input, the impulse response (i.e., the response to
a unit impulse) is
1
h (t) = [u (t) u (t T )]
T
b. Use the time delay theorem of Fourier transforms and the Fourier transform of a
rectangular pulse to get
H (f ) = sinc (f T ) ejf T
c. The output power spectral density is
S0 (f ) =
N0
sinc2 (f T )
2
d. Use F [ ( = 0 )] = 0 sinc2 ( 0 f ) to get
R0 ( ) =
N0
( =T )
2T
7.1. PROBLEM SOLUTIONS
11
e. By deÖnition of the equivalent noise bandwidth
2
E Y 2 = Hmax
BN N0
The output power is also R0 (0) = N0 =2T . Equating the two results and noting that
Hmax = 1, gives BN = 1= (2T ) Hz.
f. Evaluate the integral of the power spectral density over all frequency:
Z 1
Z
2
N0
N0 1
N0
2
E Y =
sinc (f T ) df =
sinc2 (u) du =
= R0 (0)
2T 1
2T
1 2
Problem 7.20
a. The output power spectral density is
S0 (f ) =
N0 =2
1 + (f =f3 )4
b. The autocorrelation function of the output, by inverse Fourier transforming the output
power spectral density, is (a table of deÖnite integrals is necessary)
Z 1
N0 =2
j2f
R0 ( ) = F 1 [S0 (f )] =
df
4e
1
+
(f
=f
)
1
3
Z 1
Z 1
cos (2f )
cos (2f3 x)
= N0
dx
4 df = N0 f3
1 + x4
1 + (f =f3 )
0
0
p
f3 N0 p2f3
=
e
cos
2f3 =4
2
c. Yes. R0 (0) = f32N0 = N0 BN ; BN = f3 =2
Problem 7.21
We have
jH (f )j2 =
(2f )2
(2f )4 + 5; 000
Thus
jH (f )j = q
j2f j
(2f )4 + 5; 000
This could be realized with a second-order Butterworth Ölter in cascade with a di§erentiator.
12
CHAPTER 7. RANDOM SIGNALS AND NOISE
Problem 7.22
a. The power spectral density is
N0
(f =2B)
2
SY (f ) =
The autocorrelation function is
RY ( ) = N0 B sinc (2B )
b. The transfer function is
H (f ) =
A
+ j2f
The power spectral density and autocorrelation function are, respectively,
A2
B
2
2
+ (2f ) 1 + (2f )2
A2 =2
B
A2 B=2
1
2 2
=
1 2 2 1 + (2f =)2 1 + (2f )2
1 + (2f =)2 1 + (2f )2
SY (f ) = jH (f )j2 SX (f ) =
=
Use the transform pair exp (j j= 0 )
autocorrelation function is
RY (t) =
h
A2 B=
2 1 ()2
2 0
! 1+(2f
to Önd that the corresponding
)2
0
h
i
i ej j ej j= , 6= 1=
Problem 7.23
a. E [Y (t)] = 0 because the mean of the input is zero.
b. The frequency response function of the Ölter is
H (f ) =
1
10 + j2f
The output power spectrum is
SY (f ) = Sx (f ) jH (f )j2
1
= 1
100 + (2f )2
0:01
2=10
=
2 = 0:05
1 + (2f =10)
1 + (2f =10)2
which is obtained applying (7.89).
7.1. PROBLEM SOLUTIONS
13
c. Use the transform pair exp (j j= 0 )
2 0
! 1+(2f
to Önd the power spectrum as
)2
0
RY ( ) = 0:05e10j j
d. Since the input is Gaussian, so is the output. Also, E [Y ] = 0 and var[Y ] = RY (0) =
0:05, so
exp y 2 =2 2Y
q
fY (y) =
2 2Y
where 2Y = 0:05:
e. Use (7.189) with
x = y1 = Y (t1 ) , y = y2 = Y (t2 ) , mx = my = 0, and 2x = 2y = 0:05
Also
( ) =
RY ( )
= e10j j
RY (0)
Set = 0:03 to get (0:03) = 0:741. Put these values into (7.189).
Problem 7.24
Use the integrals
I1 =
and
Z 1
b0 ds
jb0
=
a0 a1
1 (a0 s + a1 ) (a0 s + a1 )
b0 s2 + b1 ds
b0 + a0 b1 =a2
I2 =
= j
2
2
a0 a1
1 (a0 s + a1 s + a2 ) (a0 s a1 s + a2 )
Z 1
to get the following results for BN .
Filter Type
First Order
Chebyshev
fc =2
Butterworth
2 fc
Second Order
p
fc =2
qp
1+1=2
1+1=2 1
p f
2 2 c
Problem 7.25
The transfer function is
H (s) =
=
!2
p 3
p
p
s + ! 3 = 2 + j! 3 = 2 s + ! 3 = 2 j! 3 = 2
A
A
p
p +
p
p
s + ! 3 = 2 + j! 3 = 2
s + ! 3 = 2 j! 3 = 2
p
14
CHAPTER 7. RANDOM SIGNALS AND NOISE
p
where A = j! 3 = 2 with ! 3 the 3-dB frequency in rad/s. This is inverse Fourier transformed,
with s = j!, to yield
p
p
!3t
h (t) = 2! 3 exp ! 3 t= 2 sin p
u (t)
2
for the impulse response. Now
Z
1 1
f3
!3
jh (t)j2 dt = p = p
2 1
4 2
2 2
R1
after some e§ort. Also note that 1 h (t) dt = H (0) = 1: Therefore, the noise equivalent
bandwidth, from (7.108), is
f3
250
BN = p Hz = p Hz
2 2
2
Problem 7.26
a. Note that Ha; max = 2, Ha (f ) = 2 for 1 f 1; Ha (f ) = 1 for 2 f 1 and
1 f 2 and is 0 otherwise. Thus
Z 1
Z 1
Z 2
1
1
2
2
2
BN =
jHa (f )j df =
2 df +
1 df
2
Ha;
4
0
1
max 0
1
=
(4 + 1) = 1:25 Hz
4
b. For this case Hb; max = 2 and the frequency response function is a 2-unit high isoceles
triangle centered at 0 and 100 Hz wide so that
Z 1
Z
1
1 50
2
BN =
jH
(f
)j
df
=
[2 (1 f =50)]2 df ; v = f =50
b
4 0
Hb;2 max 0
1
Z 1
50
2
3
= 50
(1 v) dv = (1 v) = 16:67 Hz
3
0
0
c. For this case Hc; max = 1 and for the given frequency response function
Z 1
Z
1
1 1
100
2
BN =
df
jH
(f
)j
df
=
c
Hc;2 max 0
1 0 100 + (2f )2
Z 1
1
=
df ; v = 2f =10
1 + (2f =10)2
0
1
Z
10 1 1
5
1
=
dv = tan v
2
2 0 1 + v
0
5
=
= 2:5 Hz
2
7.1. PROBLEM SOLUTIONS
15
d. This frequency response function can be written in terms of piecewise functions as
8
0; f < 5
>
>
<
2 + f =5; 5 f < 0
Hd (f ) =
2
f =5; 0 f 5
>
>
:
0; f > 5
That is, it is a unit-high isoceles triangle of total width 10 centered at 0 on top of a unit-high
rectangle centered at 0 of total width 10. Thus Hd; max = 2 and for the given frequency
response function
Z 1
Z
1
1 5
2
BN =
jH
(f
)j
df
=
(2 f =5)2 df; v = f =5
d
2
2
2
Hd;
0
max 0
1
Z 1
5
51
5 3
2
3
=
(2 v) dv =
(2 v) =
1 23
4 0
43
12
0
= 35=12 = 2:92 Hz
Problem 7.27
By deÖnition
BN
Z
1 1
f 500 2
=
jH (f )j df =
2
df
2
Hmax
4 0
100
0
Z 500
Z 600
f 500 2
f 500 2
f 500
=
1+
df +
1
df ; u =
100
100
100
400
500
Z 0
Z 1
= 100
(1 + u)2 du + 100
(1 u)2 du
1
Z 1
1
2
0
0
1
100
100
100 100
3
3
=
(1 + u)
(1 u) =
+
3
3
3
3
1
0
= 66:67 Hz
Problem 7.28
a. Use the impulse response method. First, use partial fraction expansion to get the
impulse response:
10
1
1
10 2t
Ha (f ) =
, ha (t) =
e
e25t u (t)
23 j2f + 2 j2f + 25
23
16
CHAPTER 7. RANDOM SIGNALS AND NOISE
Thus
BN
=
=
=
=
=
R1
10 2 R 1 2t
2
e
e25t dt
23
0
hR
i2 = 10 R 1
2
1
2 23 0 (e2t e25t ) dt
2 1 h (t) dt
R1
1 0 e4t 2e27t + e50t dt
1
1 25t 1 2
2
2 e2t + 25
e
0
1 4t
2 27t
1 50t 1
+ 27 e
50 e
1 4e
0
2
(1=2 1=25)2
1 50 2 1
2
1
1 50 2
529
+
=
2 23
4 27 50
2 23
27 100
0:463 Hz
2
1 jh (t)j dt
b. Again use the impulse response method. Use the transform pair t exp (at) u (t) $
1
to get
(a+j2f )2
hb (t) = 100t exp (10t) u (t)
Thus, the equivalent noise bandwidth is
R1
R1
2
2
1 jh (t)j dt
0 [100t exp (10t)] dt
BN =
=
hR
i2
hR
i2
1
1
2 1 h (t) dt
2 1 100t exp (10t) dt
R1 2
R
3 1 u2 exp (u) du
t
exp
(20t)
dt
20
0
0
=
hR
i2 = h
i2 ; u = 20t; v = 10t
R1
1
2
2 1 t exp (10t) dt
2 10
v
exp
(v)
dv
1
R1 2
Z 1
4
10 0 u exp (u) du
=
xn ex dx = n! for n integer
hR
i2 ; use the integral
1
0
2 8 103 1 v exp (v) dv
=
10 2!
20
=
= 1:25 Hz
2
16 (1!)
16
Problem 7.29
a. Choosing f0 = f1 moves f1 right to f = 0 and f1 left to f = 0. The lowpass
Öltered version of the superposition of these two spectra is the power spectrum of
nc (t)h or ns (t). It
iis a v-shaped spectrum centered at f = 0 given by SLP (f ) =
f
1
2 N0 1 f2 f1
7.1. PROBLEM SOLUTIONS
17
b. Choosing f0 = f2 moves f2 right to f = 0 and
f2 left
to f = 0. Thus, the baseband
f
1
spectrum can be written as SLP (f ) = 2 N0 f2 f1 :
c. For this case both triangles (left
and
right) are centered around the origin and they
f
1
add to give SLP (f ) = 2 N0 f2 f1 .
d. They are not uncorrelated for any case for an arbitrary delay. However, all cases give
quadrature components that are uncorrelated at the same instant.
Problem 7.30
a. By inverse Fourier transformation, the autocorrelation function is
Rn ( ) =
j j
e
2
where K = =2.
b. Use the result from part (a) and the modulation theorem to get
Rnp ( ) =
j j
e
cos (2f0 )
2
c. The result is
Snc (f ) = Sns (f ) =
2
2 + (2f )2
and
Snc ns (f ) = 0
Problem 7.31
f
a. The equivalent lowpass power spectral density is given by Snc (f ) = Sns (f ) = N0 f2 f
:
1
The cross-spectral density is Snc ns (f ) = 0:
b. For this case Snc (f ) = Sns (f ) = N20 2(f2ff1 ) and the cross-spectral density is
Snc ns (f ) =
N20 ; (f2 f1 ) f 0
N0
2 ; 0 f (f2 f1 )
18
CHAPTER 7. RANDOM SIGNALS AND NOISE
c. For this case, the lowpass equivalent power spectral densities are the same as for part
(b) and the cross-spectral density is the negative of that found in (b).
d. For part (a)
Rnc ns ( ) = 0
For part (b)
Rnc ns ( ) =
=
=
=
=
"Z
#
Z f1 f2
N0 j2f
N0 j2f
j
e
df +
e
df
2
2
(f2 f1 )
0
"
0
(f f ) #
N0 ej2f
ej2f 2 1
j
+ j2
j2
2
(f2 f1 )
0
N0
1 + ej2(f2 f1 ) + e j2(f2 f1 ) 1
4
N0
f2 2 cos [2 (f2 f1 ) ]g
4
N0
sin2 [ (f2 f1 ) ]
h
i
0
= N0 (f2 f1 )2 sinc2 [(f2 f1 ) ]
For part (c), the cross-correlation function is the negative of the above.
Problem 7.32
The result is
1
1
Sn2 (f ) = Sn (f f0 ) + Sn (f + f0 )
2
2
so take the given spectrum, translate it to the right by f0 and to the left by f0 , add the two
translated spectra, and divide the result by 2.
Problem 7.33
DeÖne N1 = A + Nc . Then
R=
Note that the joint pdf of N1 and Ns is
q
N12 + Ns2
i
1
1 h
2
2
fN1 Ns (n1 , ns ) =
exp 2 (n1 A) + ns
2 2
2
Thus, the cdf of R is
FR (r) = Pr (R r) =
Z Z
p 2
N1 +Ns2 r
fN1 Ns (n1 , ns ) dn1 dns
7.1. PROBLEM SOLUTIONS
19
Change variables in the integrand from rectangular to polar with
n1 = cos and n2 = sin
Then
i
1 h
2
2
FR (r) =
exp 2 ( cos A) + ( sin )
dd
2 2
2
0
0
Z r
1 2
A
2
=
exp 2 + A
I0
d
2
2
2
0
Z r Z 2
Di§erentiating with respect to r, we obtain the pdf of R:
r
1 2
rA
2
fR (r) = 2 exp 2 r + A
I0
, r0
2
2
Problem 7.34
The suggested approach is to apply
Sn (f ) = lim
T !1
n
o
E j= [x2T (t)]j2
2T
where x2T (t) is a truncated version of x (t). Thus, let
x2T (t) =
N
X
k=N
nk (t kTs )
The Fourier transform of this truncated waveform is
= [x2T (t)] =
N
X
k=N
nk ej2kTs
20
CHAPTER 7. RANDOM SIGNALS AND NOISE
Thus,
8
2 9
N
< X
=
n
o
E j= [x2T (t)]j2
= E
nk ej2kTs
:
;
k=N
( N
)
N
X
X
j2kTs
j2lTs
= E
nk e
nl e
k=N
=
N
X
N
X
l=N
E [nk nl ] ej2(kl)Ts
k=N l=N
=
N
N
X
X
Rn (0) kl ej2(kl)Ts
k=N l=N
=
N
X
Rn (0) = (2N + 1) Rn (0)
k=N
But 2T = 2N Ts so that
Sn (f ) =
=
lim
T !1
n
o
E j= [x2T (t)]j2
2T
Rn (0)
Ts
= lim
N !1
(2N + 1) Rn (0)
2N Ts
Problem 7.35
a. The output is
1
y (t) =
T
Z t+T
t
x () x ( ) d
Therefore
Z
1 t+T
E [y (t)] = E
x () x ( ) d
T t
Z
1 t+T
=
E [x () x ( )] d
T t
Z
1 t+T
=
Rx ( ) d = Rx ( )
T t
7.1. PROBLEM SOLUTIONS
21
b. Obtain the result by writing y 2 (t) as an iterated integral:
Z t+T
Z t+T
1
2
y (t) =
x () x ( ) d
x () x ( ) d
T2 t
t
Z t+T Z t+T
1
=
x () x () x ( ) x ( ) dd
T2 t
t
Take the expectation:
Z t+T Z t+T
2
1
E y (t) =
E
x () x () x ( ) x ( ) dd
T2
t
t
Z t+T Z t+T
1
=
E [x () x () x ( ) x ( )] dd; identify x1 = x () etc.
T2 t
t
Z t+T Z t+T fE [x () x ()] E [x ( ) x ( )]
1
+E [x () x ( )] E [x () x ( )] dd
=
T2 t
t
+E [x () x ( )] E [x () x ( )]g
Z t+T Z t+T
2
1
2
=
R
(
)
+
R
(
)
+
R
(
)
R
(
)
dd
x
x
x
x
T2 t
t
Problem 7.36
a. In the expectation for the cross correlation function, write the derivative as a limit:
dy (t + )
Ryy_dotp ( ) = E y (t)
dt
y (t + + ) y (t + )
= E y (t) lim
!0
1
= lim fE [y (t) y (t + + )] E [y (t) y (t + )]g
!0
Ry ( + ) Ry ( )
= lim
!0
dRy ( )
=
d
b. The variance of Y = y (t) is 2Y = N0 B. The variance for Z = dy
dt is found from its
power spectral density. Using the fact that the transfer function of a di§erentiator is
j2f we get
N0
SZ (f ) = (2f )2
(f =2B)
2
22
CHAPTER 7. RANDOM SIGNALS AND NOISE
so that
2Z
2
= 2 N0
=
B
f 3
f df = 2 N0
3 B
B
Z B
2
2
4 2
N0 B 3
3
dR ( )
y
The two processes are uncorrelated because Ryy_dotp (0) = d
= 0 because the
derivative of the autocorrelation function of y (t) exists at = 0 and therefore must be 0.
Thus, the random process and its derivative are independent. Hence, their joint pdf is
exp 2 =2N0 B exp 2 =2:67 2 N0 B 3
p
p
fY Z (; ) =
2N0 B
2:67 3 N0 B 3
c. No, because the derivative of the autocorrelation function at = 0 of such noise does
not exist (it is a double-sided exponential).
7.2. COMPUTER EXERCISES
7.2
23
Computer Exercises
Computer Exercise 7.1
% ce7_1.m:
Computes approximations to mean and mean-square ensemble and time
%
averages of cosine waves with phase uniformly distributed between 0 and 2*pi
%
clf
f0 = 1;
A = 1;
theta = 2*pi*rand(1,100);
t = 0:.1:1;
X = [];
for n = 1:100
X = [X; A*cos(2*pi*f0*t+theta(n))];
end
EX = mean(X);
AVEX = mean(Xí);
EX2 = mean(X.*X);
AVEX2 = mean((X.*X)í);
disp(ë í)
disp(ëSample means (across 100 sample functions)í)
disp(EX)
disp(ë í)
disp(ëTypical time-average means (sample functions 1, 20, 40, 60, 80, & 100)í)
disp([AVEX(1) AVEX(20) AVEX(40) AVEX(60) AVEX(80) AVEX(100)])
disp(ë í)
disp(ëSample mean squaresí)
disp(EX2)
disp(ë í)
disp(ëTypical time-average mean squaresí)
disp([AVEX2(1) AVEX2(20) AVEX2(40) AVEX2(60) AVEX2(80) AVEX2(100)])
disp(ë í)
for n = 1:5
Y = X(n,:);
subplot(5,1,n),plot(t,Y),ylabel(ëX(t,ntheta)í)
if n == 1
title(ëPlot of Örst Öve sample functionsí)
end
if n == 5
24
CHAPTER 7. RANDOM SIGNALS AND NOISE
xlabel(ëtí)
end
end
A typical run follows with a plot given in Fig. 7.1.
>> ce7_1
Sample means (across 100 sample functions)
Columns 1 through 6
-0.0121
0.0226
0.0486
0.0560
0.0421
0.0121
Columns 7 through 11
-0.0226
-0.0486
-0.0560
-0.0421
-0.0121
Typical time-average means (sample functions 1, 20, 40, 60, 80, & 100)
-0.0232
-0.0654
0.0255
-0.0344
0.0055
0.0245
Sample mean squares
Columns 1 through 6
0.4613
0.4538
0.5101
0.5525
0.5223
0.4613
Columns 7 through 11
0.4538
0.5101
0.5525
0.5223
0.4613
Typical time-average mean squares
0.4604
0.5015
0.4617
0.4676
0.4549
0.4611
Computer Exercise 7.2
This is a matter of changing the statement
theta = 2*pi*rand(1,100);
in the program of Computer Exercise 7.1 to
theta = (pi/2)*(rand(1,100) - 0.5);
The time-average means will be the same as in Computer Exercise 7.1, but the ensembleaverage means will vary depending on the time at which they are computed.
Computer Exercise 7.3
This was done in Computer exercise 6.2 by printing the covariance matrix. The diagonal terms give the variances of the X and Y vectors and the o§-diagonal terms give the
correlation coe¢cients. Ideally, they should be zero.
% ce7_3:
%
computes covariance matrix of two vectors of
Gaussian random variables that are independent
7.2. COMPUTER EXERCISES
25
Plot of first five sample functions
X(t,θ)
1
0
-1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
t
0.6
0.7
0.8
0.9
1
X(t,θ)
1
0
-1
X(t,θ)
1
0
-1
X(t,θ)
1
0
-1
X(t,θ)
1
0
-1
26
CHAPTER 7. RANDOM SIGNALS AND NOISE
4
3
2
1
Y
0
-1
-2
-3
-4
-4
-3
-2
-1
0
X
1
%
clf
X = randn(1,1000);
Y = randn(1,1000);
disp(íCovariance matrix:í)
disp(cov(X, Y))
plot(X, Y, í+í), axis square, xlabel(íXí), ylabel(íYí)
A typical run follows with a plot given in Fig. 7.2.
>> ce7_3
Covariance matrix:
1.0553 0.0235
0.0235 1.0341
2
3
4
7.2. COMPUTER EXERCISES
27
Computer Exercise 7.4
% ce7_4.m: plot of Ricean pdf for several values of K
%
clf
A = char(ë-í,ëñí,ë-.í,ë:í,ëñ.í,ë-..í);
sigma = input(ëEnter desired value of sigma => í);
r = 0:.1:15*sigma;
n = 1;
KdB = [];
for KK = -10:5:15;
KdB(n) = KK;
K = 10.^(KK/10);
fR = (r/sigma^2).*exp(-(r.^2/(2*sigma^2)+K)).*besseli(0, sqrt(2*K)*(r/sigma));
plot(r, fR, A(n,:))
if n == 1
hold on
xlabel(írí), ylabel(íf_R(r)í)
grid on
end
n = n+1;
end
legend([ëK = í, num2str(KdB(1)),ë dBí],[ëK = í, num2str(KdB(2)),ë dBí],
[ëK = í, num2str(KdB(3)),ë dBí],
[ëK = í, num2str(KdB(4)),ë dBí],[ëK = í, num2str(KdB(5)),ë dBí],
[ëK = í, num2str(KdB(6)),ë dBí])
title([ëRicean pdf for nsigma = í, num2str(sigma)])
A typical run follows with a plot given in Fig. 7.3.
>> ce7_4
Enter desired value of sigma => 2
28
CHAPTER 7. RANDOM SIGNALS AND NOISE
Ricean pdf for σ = 2
0.35
K = -10 dB
K = -5 dB
K = 0 dB
K = 5 dB
K = 10 dB
K = 15 dB
0.3
0.25
fR(r)
0.2
0.15
0.1
0.05
0
0
5
10
15
r
20
25
30
Chapter 8
Noise in Modulation Systems
8.1
Problems
Problem 8.1
Degrees kelvin (degrees absolute) are converted to to the Celsius scale by substracting 273
(actually 273.15). Degrees Celsius are converted to degrees Fahrenheit by the familiar
expression
9
degrees F = degrees C × + 32
5
Thus
9
degrees F = (kelvin − 273) + 32
5
Thus, 3 degrees on the kelvin scale is
(5 − 273)
9
+ 32 = −4504 degrees F
5
and 290 degrees on the kelvin scale is
(290 − 273)
9
+ 32 = 626 degrees F
5
Problem 8.2
The signal power at the output of the lowpasss filter is . The noise power is 0 ,
where is the noise-equivalent bandwidth of the filter. From (6.116), we know that the
noise-equivalent bandwidth of an order Butterworth filter with 3 dB bandwidth is
() =
2
sin (2)
1
2
CHAPTER 8. NOISE IN MODULATION SYSTEMS
Thus, the signal-to-noise ratio at the filter output is
=
sin (2)
= ()
=
0
2 0
0
so that () is given by
() =
sin (2)
2
Thus we have the values given in the following table.
1
3
5
10
()
06366
09549
09836
09959
SNR
06366 0
09549 0
09836 0
09959 0
Note that for large , sin(2)(2) ≈ 1, so that the SNR→ 0 as → ∞. As
the order of the Butterworth filter increases, the amplitude response approaches that of an
ideal filter. Thus the output SNR approaches that of the input SNR as increases.
Problem 8.3
The signal power is
=
The noise power is
=
(5)2
= 125
2
Z 8
1
0 = 80
−8 2
The SNR is
=
125
80
or
125 −3010
125
(10
= 00015625 W/Hz
)=
8
8000
Any larger value of 0 will result in an SNR30db.
0 =
Problem 8.4
We express () as
¸
¸
∙
∙
1
1
() = () cos 2 ± (2 ) + + () sin 2 ± (2 ) +
2
2
8.1. PROBLEMS
3
Snc f , Sns f
N0
1
W
2
1
W
2
f
Figure 8.1: Plots of ( ) and ( ) for Problem 8.4.
where we use the plus sign for the and the minus sign for the . The received
signal is
b () sin (2 + )]
() = [ () cos (2 + ) ∓
Multiplying () + () by 2 cos (2 + ) and lowpass filtering yields
() = () + () cos ( ) ± () sin ( )
From this expression, we can show that the postdetection signal and noise powers are given
by
= 2 2
= 0
= 2
= 0
This gives a detection gain of one. The power spectral densities of () and () are
illustrated in Figure 8.1.
Problem 8.5
Expanding the noise about ± 2 gives
() = () cos [2 ± + ] + () sin [2 ± + ]
where we use the plus sign for the and the minus sign for the . The received
signal is
b () sin (2 + )]
() = [ () cos (2 + ) ∓
Multiplying () + () by 2 cos (2 + ) and lowpass filtering yields
() = () + () cos ( ) ± () sin ( )
4
CHAPTER 8. NOISE IN MODULATION SYSTEMS
We see that the postdetection signal and noise powers are, as given in the text,
= 2 2
= 0
2
= 0
=
This gives a detection gain of one. The power spectral densities of () and () are
easily plotted.
Problem 8.6
The message signal is
() = 10 cos (8)
Thus,
() = cos (8)
so that
2 =
1
2
The efficiency is therefore given by
=
(05) (05)2
= 01111 = 1111%
1 + (05) (05)2
From (8.36), the detection gain is
(SNR)
= 2 = 02222 = −65256 dB
(SNR)
Relative to baseband, the output SNR is
(SNR)
= −65256 dB
0
If the modulation index in increased to 08, the efficiency becomes
=
(05) (08)2
= 02424 = 2424%
1 + (05) (08)2
This gives a detection gain of
(SNR)
= 2 = 04848 = −31443 dB
(SNR)
which, relative to baseband, is
(SNR)
= −3882 dB
0
8.1. PROBLEMS
5
This represents an improvement of 276 dB.
Problem 8.7
The first step is to determine the value of . Since
{− − } + { } = 001
First we compute
Z ∞
{ } =
µ ¶
1
− 2 22
√
=
= 0005
2
This gives
= 2576
Thus,
= 2576
and
() =
()
2576
Thus, since 2 = 2
2 =
2
2
=
= 01507
68358 2
(2576)2
Since = 08, we have
=
01507 (08)2
= 0088 = 802%
1 + 01507 (08)2
and the detection gain is
(SNR)
= 2 = 01759
(SNR)
Problem 8.8
The output of the predetection filter is
() = [1 + ()] cos [ + ] + () cos [ + + ()]
The noise function () has a Rayleigh pdf. Thus
( ) =
−2 2
6
CHAPTER 8. NOISE IN MODULATION SYSTEMS
where is the predetection noise power. This gives
1
1
= 2 + 2 = 0
2
2
From the definition of threshold
099 =
Z
0
−2 2
which gives
2
099 = 1 − − 2
Thus,
−
2
= ln (001)
2
which gives
2 = 921
The predetection signal power is
i 1
1 h
= 1 + 2 2 ≈ 2 [1 + 1] = 2
2
2
which gives
2
= = 921 ≈ 964 dB
Problem 8.9
Since () is a sinusoid, 2 = 12 , and the efficiency is
=
1 2
2
2
1 2 = 2 + 2
1 + 2
and the output SNR is
(SNR) = =
In dB we have
(SNR) = 10 log10
µ
2
2
2 + 0
2
2 + 2
¶
+ 10 log10
For = 03,
(SNR) = 10 log10
− 13659
0
0
8.1. PROBLEMS
7
For = 05,
(SNR) = 10 log10
− 9542
0
(SNR) = 10 log10
− 8166
0
(SNR) = 10 log10
− 6154
0
For = 06,
For = 08,
The plot of (SNR) as a function of 0 in dB is linear having a slope of one. The
modulation index only biases (SNR) down by an amount given by the last term in the
preceding four expressions.
Problem 8.10
Let the predetection filter bandwidth be and let the postdetection filter bandwidth be
. The received signal (with noise) at the predetection filter output is represented
() = [1 + ()] cos + () cos − () sin
which can be represented
() = { [1 + ()] + ()} cos − () sin
The output of the square-law device is
() = { [1 + ()] + ()}2 cos2
− { [1 + ()] + ()} () cos ()
+2 () sin2 cos ()
Assuming that the postdetection filter removes the double frequency terms, () can be
written
1
1
() = { [1 + ()] + ()}2 + 2 ()
2
2
Except for a factor of 2, this is (8.53). We now express the preceding expression as
1
1
1
() = 2 [1 + 2 () + 2 2 ()] + [1 + ()] () + 2 () + 2 ()
2
2
2
Let () = cos( ) + cos(2 ), which clearly has a maximum value of 2. Thus
() =
1
1
cos( ) + cos(2 )
2
2
8
CHAPTER 8. NOISE IN MODULATION SYSTEMS
and
2 () =
1 1
1 1
1
+ cos(2 ) + + cos(4 ) + cos( ) cos(2 )
8 8
8 8
2
or
1 1
1
1
1
+ cos( ) + cos(2 ) + cos(3 ) + cos(4 )
4 4
8
4
8
Putting this together gives
¶
µ
1 2
2
1+
+
() =
2
4
¶
¶
∙µ
µ
¸
1
1
2
+
cos( ) +
+
cos(2 ) +
2 8
2 16
¸
∙
1
1
2 2
cos(3 ) +
cos(4 ) +
8
16
() +
∙
¸
1
1
cos( ) + cos(2 ) () +
2
2
1
1 2
() + 2 ()
2
2
2 () =
The PSD therefore consists of 6 basic terms as follows:
Term 1: A dc term ( = 0) which is an impulse.
Term 2: Impulse functions at frequencies = ± and = ±2 . The weights are given
on line 2 of the preceding equation when squared and divided by two. Note that the terms
do not have the same weight due to intermodulation distortion. Thus the demodulated
output is distorted.
Term 3: Impulse functions at frequencies = ±3 and = ±4 resulting from
harmonic distortion.
Term 4: The PSD of () (flat) about = 0 and weighted by .
Term 5: The PSD of () (flat) translated to frequencies = ± and = ±2 and
weighted by 2.
Term 6: The PSD of 2 () (triangular) about = 0 The PSD of 2 () is identical to
the PSD of 2 ()
Problem 8.11
Equation (8.59) is, suppressing ,
22 +2 = {[2 + 2 ]2 } − 2 {2 + 2 }
The first term is
{[2 + 2 ]2 } = {4 } + {4 } + 2{2 }{2 }
8.1. PROBLEMS
9
where the last term follows from the independence of and . Since (See Problem 6.22)
{4 } = {4 } = 3 4
and
{2 } = {2 } = 2
we have
{[2 + 2 ]2 } = 3 4 + 3 4 + 2 2 2 = 8 4
Also
Thus
£
¤2 £
¤2 £
¤2
2 {2 + 2 } = {2 + 2 } = {2 + 2 } = 2 2 = 4 4
£
¤2
22 +2 = 2 {2 + 2 } − {2 + 2 } = 8 4 − 4 4 = 4 4
Problem 8.12
By definition
() =
and
2 =
so that
()
2
2
1
=
= 2
2
2
2
( )
¡ ¢2
2
(SNR) = 2
=
¡ ¢2
0
0
1+
or
(SNR) =
2
()2 + 1 0
The plot is easily generated. We conclude that for À
(SNR) ≈ 2
and for ¿
Problem 8.13
³ ´2
0
(SNR) ≈ 2
0
10
CHAPTER 8. NOISE IN MODULATION SYSTEMS
Assume sinusoidal modulation since sinusoidal modulation was assumed in the development
of the square-law detector. For = 1 and () = cos (2 ) so that 2 = 05, we have,
for linear envelope detection (since 0 À 1, we use the coherent result),
(SNR) =
1
2 2
1
=
= 2 1
=
0
3 0
1 + 2 0
1 + 2 2 0
For square-law detection we have, from (6.61) with = 1
(SNR) = 2
µ
2 + 2
¶2
2
=
0
9 0
Taking the ratio
(SNR)
(SNR)
2
= 91 0 =
3 0
2
= −176 dB
3
This is approximately −18 dB.
Problem 8.14
The transfer function of the RC highpass filter is
( ) =
2
=
1
1 + 2
+ 2
so that
|( )|2 =
(2 )2
1 + (2 )2
Thus, The PSD at the output of the ideal lowpass filter is
(
2
0 (2 )
| |
2,
2
1+(2
)
( ) =
| |
0,
The noise power at the output of the ideal lowpass filter is
=
Z
( ) = 0
−
Z
(2 )2
1 + (2 )2
0
with = 2 , the preceding expression becomes
0
=
2
Z 2
0
2
1 + 2
8.1. PROBLEMS
11
Since
2
1
=1−
1 + 2
1 + 2
we can write
0
=
2
or
=
which is
½Z 2
0
−
Z 2
0
1 + 2
¾
¢
0 ¡
2 − tan−1 (2 )
2
¸
∙
0 tan−1 (2 )
2 − tan−1 (2 )
= 0
= 0 −
2
2
The output signal power is
1
2 (2 )2
= 2 | ( )|2 =
2
2 1 + (2 )2
Thus, the signal-to-noise ratio is
(2 )2
2
2
=
20 1 + (2 )2 2 − tan−1 (2 )
→ 0.
Note that as → ∞
Problem 8.15
We can write
¶
µ
¶
µ
¶
∙
µ
¶¸
µ
7
5
5
7
= cos 20 +
= cos 20 +
sin 20 + = cos 20 − +
4
2
4
4
80
so that the receiver input is
∙
µ
¶¸
5
() = 5 cos 20 +
+ 02 cos (60) + ()
80
The signal power is
52
= 125
2
and the power in the components orthogonal to the signal (noise) is
=
=
022
+ 012 = 002 + 001 = 005
2
12
CHAPTER 8. NOISE IN MODULATION SYSTEMS
The SNR is
125
= 250 = 2398 dB
005
The delay is negative as shown. The delay cannot be negative and so we add two periods,
which is 220 s. Therefore the dlay is
SNR =
Delay =
5
3
8
−
=
s
80 80
80
Problem 8.16
From (8.101) and (8.107), we have
2
SSB
2
3 ¡ 2 ¢2 2
+ 2 DSB
4
006 = 2 +
006 =
Thus we have the following signal-to-noise ratios
2
2
=
2
2
=
1
006 − 2
1
006 − 34
SSB
³ ´2
2
DSB
The SSB characteristic has an asymptote defined by
2 = 006
and the DSB result has an asymptote defined by
or
3 ¡ 2 ¢2
= 006
4
2 = 02828
The curves are shown below. The appropriate operating regions are above and to the left
of the curves. It is clear that DSB has the larger operating region.
Problem 8.17
This problem is identical to the previus problem except the upper bound on the meansquare-error is 0.1. Therefore
8.1. PROBLEMS
13
1200
USB
SSB
Signal-to-Noise Ratio
1000
800
600
400
200
0
0
0.05
0.1
0.15
0.2
phase error variance
0.25
0.3
0.35
Figure 8.2: Operating regions for Problem 8.16.
2
2
=
2
2
=
The result is shown below.
1
01 − 2
1
SSB
³ ´2
01 − 34 2
DSB
Problem 8.18
The phasor diagram for SNR À 1 is illustrated in Figure 8.4. The phasor diagram for
SNR¿1 is shown in Figure 8.5. We see that the angle (), which determines the demodulated output, is approximately (), whereas for SNR À 1 the angle () is approximately
(). For () ≈ () the error in the demodulated output is small.
Problem 8.19
The single-sided spectrum of a stereophonic FM signal and the noise spectrum is shown in
Figure 8.6. The two-sided noise spectrum is given by
( ) =
2
0 ,
2
−∞ ∞
14
CHAPTER 8. NOISE IN MODULATION SYSTEMS
25
USB
SSB
Signal-to-Noise Ratio
20
15
10
0
0.05
0.1
0.15
0.2
phase error variance
0.25
0.3
0.35
Figure 8.3: Operating regions for Problem 8.17.
R(t )e j (t )
rn (t )e jn ( t )
rn (t )sin[ (t ) (t )] rn (t )sin[e (t )]
rn (t ) cos[ (t ) (t )] rn (t ) cos[e (t )]
Ac e j ( t )
Figure 8.4: Phasor diagram for SNR À 1.
8.1. PROBLEMS
15
rn (t )sin[ (t ) (t )] rn (t )sin[e (t )]
rn (t ) cos[ (t ) (t )] rn (t ) cos[e (t )]
rn (t )e jn ( t )
R(t )e j (t )
Ac e j (t )
Figure 8.5: Phasor diagram for SNR ¿ 1
The predetection noise powers are easily computed. For the + channel
Z 15000 2
2
¡ 12 ¢
2
+ = 2
=
225
10
0
0
2
2
0
For the − channel
− = 2
Z 53000
23000
2
2
¡
¢
0 2 = 9114 1012
0
2
2
Thus, the noise power on the − channel is over 40 times the noise power in the +
channel. After demodulation, the difference will be a factor of 20 because of 3 dB detection
gain of coherent demodulation. Thus, the main source of noise in a stereophonic system
is the − channel. Therefore, in high noise environments, monophonic broadcasting is
preferred over stereophonic broadcasting.
Problem 8.20
The received FDM spectrum is shown in Figure 8.7. The k th channel signal is given by
() = () cos 21
Since the guardbands have spectral width 3 , 11 = + 3 + = 5 , 21 = 1 + +
3 + = 10 and so on. Clearly 1 = 5 and the k th channel occupies the frequency
band
(5 − 1) ≤ ≤ (5 + 1)
16
CHAPTER 8. NOISE IN MODULATION SYSTEMS
Pilot
L f R f
0
15
Noise Spectrum
SnF f
L f R f
23
53
f kHz
Figure 8.6: Sterophonic FM baseband signal and noise.
Since the noise spectrum is given by
=
2
0 2 ,
2
| |
The noise power in the k th channel is
Z (5+1)
i
3 h
(5 + 1)3 − (5 − 1)3
=
2 =
3
(5−1)
where is a constant. This gives
=
¢
3 ¡
150 2 + 2 ∼
= 50 3 2
3
The signal power is proportional to 2 . Thus, the signal-to-noise ratio can be expressed as
µ ¶2
2
(SNR) = 2 =
where is a constant of proportionality and is a function of , , ,0 . If all channels
are to have equal (SNR) , must be a constant, which means that is a linear
function of . Thus, if 1 is known, = 1 = 1 2 7. 1 is fixed by setting the
SNR of Channel 1 equal to the SNR of the unmodulated channel. The noise power of the
unmodulated channel is
Z
2 = 3
0 =
3
0
8.1. PROBLEMS
17
Noise
PSD
f
0
f1
f2
f3
f4
f5
f6
f7
Figure 8.7: FDM signal and noise for Problem 7.23.
yielding the signal-to-noise ratio
0
(SNR) = 3
3
where 0 is the signal power.
For 3 = 21 and = 15, the value of is
¶
µ ¶2 µ
21
15
−1 15
=3
− tan
= 0047
15
21
21
Expressed in dB this is
= 10 log10 (0047) = −133 dB
The improvement resulting from the use of preemphasis and deemphasis is therefore 133 dB.
Neglecting the tan−1 (3 ) gives an improvement of 21 − 875 = 1225 dB. The difference
is approximately 1 dB.
Problem 8.21
With de-emphasis we have
2
=
∙
µ ¶¸
0 2
3 1 −
tan−1
3
3
Without de-emphasis we have
1 2 2 0
=
3
18
CHAPTER 8. NOISE IN MODULATION SYSTEMS
Since
2 2 2
=
For we have
=3
µ
3
¶2 ∙
µ ¶¸
3
−1
tan
1−
3
With 3 = 21 kHz and = 15 kHz we have
=3
or
µ
21
15
¶2 ∙
µ ¶¸
21
15
−1
tan
1−
15
21
£
¤
= 00588 1 − 0140 tan−1 (71429)
This gives
= 00446
The result with pre-emphasis from Example 8.3 is
(SNR) =
or
(SNR) = 12755
µ
3
¶2
2
1
1 − 3 tan−1
³
3
´
1
= 12755 ∗ 1251
1 − 0140 tan−1 (71429)
The first term in the above epression, 127.55, is the result from Example 8.3. The second
term represents the result of not making the approximation that
µ ¶
3
−1
tan
¿1
3
Not making the approximation increases (SNR) by approximately 25%. The plot of
as a function of 3 follows.
Problem 8.22
From the plot of the signal spectrum it is clear that
=
Thus the signal power is
=2
Z
0
2
2
2
=
2
3
8.1. PROBLEMS
19
1
0.9
0.8
0.7
R
0.6
0.5
0.4
0.3
0.2
0.1
0
0
2
4
6
8
10
W/f3
12
14
16
18
20
Figure 8.8: Ratio as a function of 3 .
The noise power is 0 . This yields the signal-to-noise ratio
(SNR)1 =
2
3 0
If is reduced to , the SNR becomes
(SNR)2 =
2
3 0
This increases the signal-to-noise ratio by a factor of .
Problem 8.23
Since the signal is integrated twice and then differentiated twice, the output signal is equal
to the input signal. The output signal is
() = cos 2
and the output signal power is
1
= 2
2
| |
For | | , the output signal power and the SNR are zero. The noise power spectral
density at () is determined as in the previous problem. Therefore
( ) =
0
(2 )4
2
20
CHAPTER 8. NOISE IN MODULATION SYSTEMS
so that the noise power is
0
(2)4
=
2
Z
0
=
(2)4 (2)
2
−
4
Z
4 =
0
16
0 4 5
5
Thus, the signal-to-noise ratio at () is
(SNR) =
2
5
5
=
4
5
4
4
2 160
16 0
| |
Problem 8.24
The signal power is
= 2
Z
0
= 23
1 + ( 3 )2
Z 3
0
= 23 tan−1
1 + 2
µ
3
¶
Since
= 0
the signal-to-noise ratio at () is
(SNR) =
2
0
µ
3
¶
tan−1
3
The normalized (SNR) , defined by (SNR) (20 ) is illustrated in Figure 8.9.
Problem 8.25
The instantaneous frequency deviation in Hz is
= () = ()
where () is a zero-mean Gaussian process with variance 2 = 2 2 . Thus,
Z ∞
||
2
2
√
| | = || =
− 2
2
−∞
Recognizing that the integrand is even gives
r
r
Z ∞
2 1
2
−2 22
| | =
=
0
Therefore,
8.1. PROBLEMS
21
1
0.9
0.8
Norm alized S NR
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
1
2
3
4
5
W/f3
6
7
8
9
10
Figure 8.9: (SNR)D for Problem 8.24.
| | =
r
2
³
´2
This gives
(SNR) =
√
1 + 2 3
2 0
¸
q
h
i
2
−2
2
+
6
exp
0
20
3
∙q
The preceding expression can be placed in terms of the deviation ratio, , by letting
=
and
= 2 ( + 1)
Problem 8.26
The input to the PLL is
() = cos(2 + ) + () cos(2 + ) − () sin((2 + )
22
CHAPTER 8. NOISE IN MODULATION SYSTEMS
which can be written
() cos[2 + + ()] = { + ()} cos(2 + ) − () sin(2 + )
The phase deviation is
−1
() = tan
∙
()
+ ()
¸
Since the signal-to-noise ratio is sufficiency large to justify the linear model, the phase
deviation is written
()
() =
We know that the PSD of the quadrature noise is 0 for | | 2, where is the
predetection bandwidth. Thus, the PSD of the input signal to the baseband (lowpass)
model of the PLL is
0
| | 2
( ) = 2
Using the PLL transfer function (The linear model is assumed valid since the input SNR is
assumed large, which yields small phase errors.) gives
( ) = ( ) |( )|2 =
0
|( )|2
2
where ( ) is the transfer function relating the phase error Ψ( ) to the input phase Φ( )
(See (3.221)). The variance of the phase error is
Z
Z
0 ∞
20 ∞
|( )|2 = 2
|( )|2
2 = 2
−∞
0
The integral represents the equivalent noise bandwidth of the PLL based on the closedloop transfer function ( ). We have obviously assumed ¿ , which is typically the
case. Thus
20
0
=
2 =
2
where = 2 2 is the input signal power. The input noise is assumed Gaussian. The
phase error is therefore Gaussian since it is a linear transformation, through the transfer
function ( ), of the input noise. The preceding expression shows that the phase error
variance is the reciprocal of the signal-to-noise ratio, where the noise power is measured in
the bandwidth of the transfer function ( ).
Problem 8.27
8.1. PROBLEMS
23
PPM was defined in Chapter 3. Assume that the pulse position is represented by
() = + 1 ()
where is the nominal pulse position for () = 0 and 1 establishes the peak offset from
the nominal pulse position and is adjusted so that the pulse does not deviate outside of its
assigned slot. We assume a zero-mean signal so that
() = 0
for which = . The pulse is detected when the received pulse crosses a threshold. When
noise is added to the pulse we can approximate the error in detecting the threshold crossing
as
(1 )
(1 )
=
where is the pulse risetime, 1 is the point at which the pulse would cross the threshold
in the absence of noise, and represents the pulse amplitude. Thus, in general
µ ¶2
µ ¶2
2
2
=
=
As we saw in Chapter 2, the relationship between risetime and bandwidth gives = 12 ,
where is the transmission bandwidth. We let
¶
µ
1
1
=
0 = 2 = 2
2
This gives
2 =
µ
¶2
=
=
2
1
2
42
The average transmitted signal power is
where is the sampling frequency or 12 assuming Nyquist rate sampling. Using this
in the preceding equation by substituting for 2 gives
2 =
4
Since 1 () represents the signal, the received signal power is
= 21 2
24
CHAPTER 8. NOISE IN MODULATION SYSTEMS
we have
(SNR)D =
21 2
2
= 21 2
or
(SNR)D = 21 2
µ
µ
4
¶
= 21 2
¶µ
¶
= 2 21 2 2
2
µ
1
2
¶µ
µ
4
¶
¶
where we have used = 1 . The peak-to-peak pulse displacement is 2 1 = = 12
so that 1 = 14 . Substituting this into the preceding equation yields
1
(SNR)D = 2
8
µ
¶2 µ
¶
Therefore with all the assumptions made = 18.
Note: We have made many assumptions in this problem including Nyquist rate sampling.
It is useful, however, for illustrating the thought process for applying the concepts of this
chapter to modulation processes other than those considered in this chapter.
Problem 8.28
The peak value of the signal is 75 (15 peak-to-peak) so that the signal power is
=
1
(75)2 = 5625
2
If the A/D converter has a wordlength , there are 2 quantizing levels. Since these span
the peak-to-peak signal range, the width of each quantization level is
=
This gives
2 =
¡
¢
25
= 25 2−
2
¡
¢
1 2
1
=
(25)2 2−2 = (5208)2−2
12
12
This results in the signal-to-noise ratio
=
2
=
¡ ¢
5625 ¡ 2 ¢
= 108 22
2
5208
8.2. COMPUTER EXERCISES
8.2
25
Computer Exercises
Computer Exercise 8.1
From the expression for the normalized error
2 = 2 +
2
2
the following MATLAB program results:
% File: ce8_1a.m
snrdb = [0 1 2];
snr = 10.^(snrdb./10);
varphas = 0:0.1:1;
hold on
for k=1:3
nerrqpsk = varphas + (1/snr(k));
plot(varphas,nerrqpsk’)
end
hold off
grid, xlabel(’Phase Error Variance’), ylabel(’Normalized MSE (QDSB)’)
% End of script file.
Executing the program gives the results illustrated in Figure 8.10. We see from the defining
equation that the mse is linear in the phase error variance for a given SNR. The SNR simply
provides an offset (SNR = 1 (0 dB) curve is on top).
For the DSB system we have the MATLAB program
% File: ce8_1b.m
snrdb = [0 1 2];
snr = 10.^(snrdb./10);
varphas = 0:0.1:1;
hold on
for k=1:3
nerrqpsk = 0.75*varphas.*varphas + (1/snr(k));
plot(varphas,nerrqpsk’)
end
hold off
26
CHAPTER 8. NOISE IN MODULATION SYSTEMS
2
Normalized MSE (QDSB)
1.8
1.6
1.4
1.2
1
0.8
0.6
0
0.1
0.2
0.3
0.4
0.5
0.6
Phase Error Variance
0.7
0.8
0.9
1
Figure 8.10: Normalized mse for QDSB and SSB with the SNR as a parameter.
8.2. COMPUTER EXERCISES
27
2
1.8
Normalized MSE (DSB)
1.6
1.4
1.2
1
0.8
0.6
0
0.1
0.2
0.3
0.4
0.5
0.6
Phase Error Variance
0.7
0.8
0.9
1
Figure 8.11: Normalized mse for DSB with the SNR as a parameter.
grid, xlabel(’Phase Error Variance’), ylabel(’Normalized MSE (DSB)’)
% End of script file.
This yields the result shown in Figure 8.11. The mse is now quadradic in the phase
error variance for a given SNR. The SNR again simply provides an offset (SNR = 1 (0 dB)
curve is on top).
Computer Exercise 8.2
The MATLAB program for solving this computer exercise follows.
% File: ce8_2.m
zdB = 0:50;
z = 10.^(zdB/10);
beta = [1 5 10 20];
hold on
for j=1:length(beta)
bta = beta(j);
% predetection SNR in dB
% predetection SNR
% modulation index vector
% hold for plots
% current index
28
CHAPTER 8. NOISE IN MODULATION SYSTEMS
80
70
20
Postdetection SNR in dB
60
50
5
40
1
30
20
10
0
-10
0
5
10
15
20
25
30
35
Predetection SNR in dB
40
45
50
Figure 8.12: Plot for Computer Exercise 8.2.
a1 = exp(-(0.5/(bta+1)*z)); % temporary constant
a2 = q(sqrt(z/(bta+1)));
% temporary constant
num = (1.5*bta*bta)*z;
den = 1+(4*sqrt(3)*(bta+1))*(z.*a2)+(12/pi)*bta*(z.*a1);
result = num./den;
resultdB = 10*log10(result);
plot(zdB,resultdB,’k’)
end
hold off
xlabel(’Predetection SNR in dB’)
ylabel(’Postdetection SNR in dB’)
% End of script file.
Executing the program gives the results shown in Figure 8.12. We see that the threshold
effect is less pronounced. A smaller value of yields a smaller bandwidth and decreased
detection gain.
Computer Exercise 8.3
8.2. COMPUTER EXERCISES
29
The MATLAB program which solves this computer exercise is a modification of the program
used in Computer Example 8.5 and follows:
% File: ce8_3.m
zdB = 0:0.1:40;
% predetection SNR in dB
z = 10.^(zdB/10);
% predetection SNR
beta = [2 4 6 8 10 15 20]; % modulation index vector
hold on
% hold for plots
for j=1:length(beta)
a2 = exp(-(0.5/(beta(j)+1)*z));
a1 = q(sqrt((1/(beta(j)+1))*z));
r1 = 1+(4*sqrt(3)*(beta(j)+1)*a2);
r2 = r1+((12/pi)*beta(j)*a2);
semilogy(zdB,r2,’k’)
axis([10 30 0 4])
grid
end
hold off
% release
xlabel(’Predetection SNR in dB’)
ylabel(’1+D_2+D_3’)
% End of script file.
This program gives the results shown in Figure 8.13 for = 2, 4, 6, 8 10, 15, and 20. We
see that denominator has a value of two for = 2 at approximately 13 dB. This increases
to approximately 23.5 dB at an index of 20. These values can be taken off Figure 8.13
and plotted if desired. It is clear that the predetection SNR for which the denominator of
(7.148) is equal to 2 increases as the modulation index increases.
Computer Exercise 8.4
The MATLAB program for solving this computer exercise follows. The strategy used is
to plot two curves for each modulation index; one curve with the modulation not included
and one curve with the modulation included. In order to keep the results from being too
cluttered only two modulation indexes are used; = 5 and = 20.
% File: ce8_4.m
zdB = 0:50;
z = 10.^(zdB/10);
beta = [5 20];
hold on
% predetection SNR in dB
% predetection SNR
% modulation index vector
% hold for plots
30
CHAPTER 8. NOISE IN MODULATION SYSTEMS
4
3.5
3
1+D2+D3
2.5
4
6
2
8
2
10
15
1.5
20
1
0.5
0
10
12
14
16
18
20
22
24
Predetection SNR in dB
26
28
30
Figure 8.13: Values of 0 for which 1 + 2 + 3 = 2.
8.2. COMPUTER EXERCISES
31
for j=1:length(beta)
bta = beta(j);
% current index
a1 = exp(-(0.5/(bta+1)*z));
% temporary constant
a2 = q(sqrt(z/(bta+1)));
% temporary constant
num = (1.5*bta*bta)*z;
den1 = 1+(4*sqrt(3)*(bta+1))*(z.*a2);
den2 = 1+(4*sqrt(3)*(bta+1))*(z.*a2)+(12/pi)*bta*(z.*a1);
result1dB = 10*log10(num./den1); % w/o modulation
result2dB = 10*log10(num./den2); % with modulation
plot(zdB,result1dB,’k’,zdB,result2dB,’k-.’)
end
hold off
legend(’w/o modulation’,’with modulation’,4)
xlabel(’Predetection SNR in dB’)
ylabel(’Postdetection SNR in dB’)
% End of script file.
Executing the program gives the results shown in Figure 8.14. We see, as expected, that
above threshold the results are identical. This results since the above-threshold performance
is only a function of the predetection SNR and the modulation index. Below threshold,
a region in which we typically have little interest, the inclusion of modulation reduces the
postdetection SNR, since the denominator of the postdetection SNR expression is increased.
Computer Exercise 8.5
(Note: Since Computer Exercise 8.4 was changed for the new edition, the first sentence
for Computer Exercise 8.5 should be deleted.) The input to the PLL is assumed to be a
sinusoid plus bandlimited white noise. Thus the PLL input is modeled as
() = cos(2 + ) + () cos(2 + ) − () sin(2 + )
or
() = [ + ()] cos(2 + ) − () sin(2 + )
This can be represented as
() = () cos[2 + + ()]
where
−1
() = tan
∙
¸
∙
¸
()
()
−1 ()
≈ tan
=
+ ()
32
CHAPTER 8. NOISE IN MODULATION SYSTEMS
80
70
Postdetection SNR in dB
60
50
40
30
20
10
0
w/o modulation
with modulation
0
5
10
15
20
25
30
35
Predetection SNR in dB
40
45
50
Figure 8.14: FM performance results with and without modulation.
The approximation results since the input SNR is assumed large. We therefore use () as
the input to the baseband PLL model. The SNR of the input signal is
=
2
2 2
=
2
2 2
where 2 = 2 . The variance of () is
2 =
2
1
=
2
2
In the following program we set the phase error variance. Note that this does not determine
the SNR. One can set 2 at some value (such as 1) and then adjust to give the required
SNR.
% File: ce8_5.m
% beginning of preprocessor
clear all
% be safe
fn = 10; zeta = 0.707;
varn = input(’Enter noise variance ’);
npts = 50000;
% number of simulation points
fs = 2000;
% sampling frequency
T = 1/fs;
8.2. COMPUTER EXERCISES
33
t = (0:(npts-1))/fs;
% time vector
Kt = 4*pi*zeta*fn;
% loop gain
a = pi*fn/zeta;
% loop filter parameter
filt_in_last = 0; filt_out_last=0;
vco_in_last = 0; vco_out = 0; vco_out_last=0;
% end of preprocessor - beginning of simulation loop
for i=1:npts
phin = sqrt(varn)*randn(1);
s1 = phin - vco_out;
aa(1,i) = s1;
s2 = sin(s1);
% sinusoidal phase detector
s3 = Kt*s2;
filt_in = a*s3;
filt_out = filt_out_last + (T/2)*(filt_in + filt_in_last);
filt_in_last = filt_in;
filt_out_last = filt_out;
vco_in = s3 + filt_out;
vco_out = vco_out_last + (T/2)*(vco_in + vco_in_last);
vco_in_last = vco_in;
vco_out_last = vco_out;
phierror(i)=s1;
end
% end of simulation loop - beginning of postprocessor
[N_samp,x] = hist(aa,40);
% get histogram parameters
bar(x,N_samp,1)
xlabel(’Histogram Bin’), ylabel(’Number of Samples’)
% end of postprocessor
% End of script file.
Executing the program gives the histogram illustrated in Figure 8.15. The histogram suggests that the pdf of the phase error is closely gaussian.
Computer Exercise 8.6
The MATLAB program for this computer exercise is a simple modification of the program
used for Computer Example 8.6. The only change is for the bit error probabilities for FSK
(frequency-shift keying) and BPSK (binary phase-shift keying). In order to keep the results
from being cluttered results are computed for = 8. The MATLAB program follows.
% File:
ce8_6.m
34
CHAPTER 8. NOISE IN MODULATION SYSTEMS
4500
4000
Number of Samples
3500
3000
2500
2000
1500
1000
500
0
-2
-1.5
-1
-0.5
0
0.5
Histogram Bin
1
1.5
2
Figure 8.15: Histogram of phase error.
n = 8;
% wordlength
snrtdB = 0:0.1:30;
% predetection snr in dB
snrt = 10.^(snrtdB/10);
% predetection snr
Pb1 = q(sqrt(snrt));
% bit error probability for FSK
Pb2 = q(sqrt(2*snrt));
% bit error for BPSK
Pw1 = 1-(1-Pb1).^n;
% Pw for FSK
Pw2 = 1-(1-Pb2).^n;
% Pw for BPSK
a = 2^(-2*n);
% temporary constant
snrd1 = 1./(a+Pw1*(1-a));
snrddB1 = 10*log10(snrd1); % postdetection snr (FSK)
snrd2 = 1./(a+Pw2*(1-a));
snrddB2 = 10*log10(snrd2); % postdetection snr (BPSK)
plot(snrtdB,snrddB1,’k’,snrtdB,snrddB2,’k-.’)
legend(’FSK’,’BPSK’)
xlabel(’Predetection SNR in dB’)
ylabel(’Postdetection SNR in dB’)
% End of script file.
The results are shown in Figure 8.16. We see that above threshold, the postdetection SNR is
independent of the modulation scheme. This, of course, was expected since above threshold
8.2. COMPUTER EXERCISES
35
50
FSK
BPSK
45
Postdetection SNR in dB
40
35
30
25
20
15
10
5
0
0
5
10
15
20
Predetection SNR in dB
25
30
Figure 8.16: PCM performance for FSK and BPSK with = 8
the postdetection SNR is a function only of wordlength. We also note that the use of
BPSK extends the threshold over that of FSK due to the 3-dB improvement in bit error
probability.
Computer Exercise 8.7
A little thought shows that since the SNR estimate is very sensitive to the peak of the
crosscorrelation function, if the peak is not sampled, a large error can occur. As a simple
test let
() = 5 cos(2)
and let the reference signal be
h
i
() = 4 cos 2 −
2
Consider the following MATLAB program:
% File: ce8_7a
fs=1/32;
t=1:6400;
x=5*cos(2*pi*fs*t);
y=4*cos(2*pi*fs*t-(pi/2));
36
CHAPTER 8. NOISE IN MODULATION SYSTEMS
[gain,delay,px,py,rxy,rho,snrdb] = snrmse(x,y);
a=[’The gain estimate is ’,num2str(gain),’.’]; disp(a)
b=[’The delay estimate is ’,num2str(delay),’.’]; disp(b)
c=[’The correlation coefficient estimate is ’,num2str(rho),’.’]; disp(c)
d=[’The signal-to-noise ratio estimate is ’,num2str(snrdb),’ dB.’]; disp(d)
% End of script file.
Executing the program gives:
The gain estimate is 0.8.
The delay estimate is 8.
The correlation coefficient estimate is 1.
The signal-to-noise ratio estimate is 150.0228+13.64376i dB.
These values are easily understood. From the definition of the signals () and () the
gain is 45 = 08. Note that the period is 1. The quantity is one when () = 32 since
the sampling frequency is 1/32. The period of each waveform is 32 samples. The delay is
2, which is 1/4 of a period or 8 sample periods. Therefore the cross-correlation function
is sampled precisely at it’s peak. The correlation coefficient is
=
( )
This gives, for (8.78)
SNR =
1−
Therefore = 1 gives an SNR of ∞, which is correct since () is simply an amplitude
scaled and delayed version of (). A complex number for the signal-to-noise ratio in dB
appears strange. However, this is easily explained. If is displayed to a sufficient number of
decimal places, it will be found that , due to roundoff and quantization of the calculations,
is very slightly negative. When the logarithm of a negative number is taken to express SNR
in dB, the result is a complex number results. Thus, all computed values are correct.
In order to illustrate the result when the cross-correlation function is not sampled at
the peak, the delay of () is set so that that the peak value of the crosscorrelation falls
equidistant between samples. consider the following MATLAB program.
% File: ce8_7b.m
fs=1/32;
t=1:6400;
x=5*cos(2*pi*fs*t);
y=4*cos(2*pi*fs*t-(pi/2)-(pi/32));
8.2. COMPUTER EXERCISES
37
[gain,delay,px,py,rxy,rho,snrdb] = snrmse(x,y);
a=[’The gain estimate is ’,num2str(gain),’.’]; disp(a)
b=[’The delay estimate is ’,num2str(delay),’.’]; disp(b)
c=[’The correlation coefficient estimate is ’,num2str(rho),’.’]; disp(c)
d=[’The signal-to-noise ratio estimate is ’,num2str(snrdb),’ dB.’]; disp(d)
% End of script file.
Executing this program gives:
The gain estimate is 0.79615.
The delay estimate is 9.
The correlation coefficient estimate is 0.99518.
The signal-to-noise ratio estimate is 20.132 dB.
Note the very large drop in the SNR estimate. This issue is mitigated by using a much higher
sampling frequency. This significantly increases executation time, especially for complicated
systems. However in problems of this type (simulation) there is typically a trade-off between
accuracy and execution time.
Computer Exercise 8.8
Since the signal is a sinusoid, we can use an inverse sin function as a compressor characteristic. Note that this actually expands the signal but we will use the term ‘compressor’
since this is the first operation performed. The compressor output is a triangular wave so
that the compressor output has a uniform pdf ranging from −2 to 2. Thus, with 3 bit
words we have 8 quantization levels given by
1 = 0 ≤ ≤ 8
2 = 8 ≤ ≤ 4
3 = 4 ≤ ≤ 38
4 = 38 ≤ ≤ 2
5 = −8 ≤ ≤ 0
6 = −4 ≤ ≤ −8
7 = −38 ≤ ≤ −4
8 = −2 ≤ ≤ −38
The expander characteristic has a sinusoidal characteristic to restore the original sinusiodal
signal. The compressor and expander characteristics are given in Figure 8.17 and the original
signal, the compressed signal, and the expanded signal are shown in Figure 8.18.
38
CHAPTER 8. NOISE IN MODULATION SYSTEMS
2
1
0.8
1.5
0.6
Expander Characteristic
Compressor Characteristic
1
0.5
0
-0.5
0.4
0.2
0
-0.2
-0.4
-1
-0.6
-1.5
-0.8
-2
-1
-0.5
0
Time
0.5
-1
-2
1
-1
0
Time
1
2
Figure 8.17: Compressor and expander characteristics.
Orig. sig.
1
0
Compressed sig.
-1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
Time
0.6
0.7
0.8
0.9
1
2
0
-2
Expanded sig.
1
0
-1
Figure 8.18: Original signal, compressed signal, and expanded signal.
8.2. COMPUTER EXERCISES
39
Computer Exercise 8.9
The computer exercise, as stated in the text, does not make sense since the compressor
characteristic, as given, has no frequeny dependence. Please substitute the following.
A compressor characteristic is often modeled as
out () = tanh[in ()]
The purpose of this Computer Exercise is to show that this simple expression can be used
to model a vast number of compressor characteristics. Consider three basic models.
(a. Model 1) A compressor characteristic is often modeled as
out () = (1) tanh[in ()]
Using three separate values of , namely = 03, = 06, = 10, and = 15, explore
the characteristics of this compressor by plotting out as a function of in over a sufficiently
wide range to fully define the compressor characteristic.
(b. Model 2) A compressor characteristic is often modeled as
out () = (1) tanh[(1) in ()]
Once again, using three separate values of , namely = 03, = 06, = 10, and = 15,
explore the characteristics of this compressor by plotting out as a function of in over a
sufficiently wide range to fully define the compressor characteristic.
(c. Model 3) A compressor characteristic is often modeled as
out () = tanh[in ()]
This model is a little different. We wish the asympotic values of out to be ±1. Select the
value of so that this condition is satisfied. Then using the same three separate values
of , namely = 03, = 06, = 10, and = 15, explore the characteristics of this
compressor by plotting out as a function of in over a sufficiently wide range to fully define
the compressor characteristic.
Solution
The characteristic for (a. Model 1) is developed using the following MATLAB code:
%File ce8_9a.m
a=[0.3 0.6 1.0 1.5];
40
CHAPTER 8. NOISE IN MODULATION SYSTEMS
4
y1
y2
y3
y4
3
Output signal level
2
1
0
-1
-2
-3
-4
-10
-8
-6
-4
-2
0
2
Input signal level
4
6
8
10
Figure 8.19: Compressor characteristic for Model 1.
x=-10:0.001:10;
for j=1:4
for i=1:length(x)
y1(i)=(1/a(1))*tanh(a(1)*x(i));
y2(i)=(1/a(2))*tanh(a(2)*x(i));
y3(i)=(1/a(3))*tanh(a(3)*x(i));
y4(i)=(1/a(4))*tanh(a(4)*x(i));
end
end
plot(x,y1,’k-’,x,y2,’k—’,x,y3,’k-.’,x,y4,’k:’)
grid
xlabel(’Input signal level’)
ylabel(’Output signal level’)
legend(’y1’,’y2’,’y3’,’y4’)
% End of script file.
The result is shown below.
The characteristic for (b. Model 2) is developed using the following MATLAB code:
%File ce8_9b.m
a=[0.3 0.6 1.0 1.5];
x=-10:0.001:10;
8.2. COMPUTER EXERCISES
41
4
y1
y2
y3
y4
3
Output signal level
2
1
0
-1
-2
-3
-4
-10
-8
-6
-4
-2
0
2
Input signal level
4
6
8
10
Figure 8.20: Compressor characteristic for Model 2.
for j=1:6
for i=1:length(x)
y1(i)=(1/a(1))*tanh((1/a(1))*x(i));
y2(i)=(1/a(2))*tanh((1/a(2))*x(i));
y3(i)=(1/a(3))*tanh((1/a(3))*x(i));
y4(i)=(1/a(4))*tanh((1/a(4))*x(i));
end
end
plot(x,y1,’k-’,x,y2,’k—’,x,y3,’k-.’,x,y4,’k:’)
grid
xlabel(’Input signal level’)
ylabel(’Output signal level’)
legend(’y1’,’y2’,’y3’,’y4’)
% End of script file.
The result for Model 2 is shown below.
The characteristic for (c. Model 3) is developed using the following MATLAB code:
% File ce8_9c.m
a=[0.3 0.6 1.0 1.5];
x=-10:0.001:10;
42
CHAPTER 8. NOISE IN MODULATION SYSTEMS
1
y1
y2
y3
y4
0.8
0.6
Output signal level
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-10
-8
-6
-4
-2
0
2
Input signal level
4
6
8
10
Figure 8.21: Compressor characteristic for Model 3.
for j=1:6
for i=1:length(x)
y1(i)=tanh(a(1)*x(i));
y2(i)=tanh(a(2)*x(i));
y3(i)=tanh(a(3)*x(i));
y4(i)=tanh(a(4)*x(i));
end
end
plot(x,y1,’k-’,x,y2,’k—’,x,y3,’k-.’,x,y4,’k:’)
grid
xlabel(’Input signal level’)
ylabel(’Output signal level’)
legend(’y1’,’y2’,’y3’,’y4’)
% End of script file.
The result is shown below.
Chapter 9
Principles of Digital Data
Transmission in Noise
9.1
Problem Solutions
Problem 9.1
The signal-to-noise ratio is
z=
A2 T
A2
=
N0
N0 R
p
Trial and error using the asymptotic expression Q (x) exp x2 =2 = 2x (or use of a
Q-function program in MATLAB) shows that
PE = Q
p
2z = 106 for z = 11:297 = 10:53 dB
Thus
A2
= 11:297
N0 R
or
A =
=
p
11:297N0 R
p
11:297 106 20000
= 0:4753 V
A rough estimate of the required bandwidth is BW = R = 20000 Hz.
1
2
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
Problem 9.2
The bandwidth should be equal to the data rate to pass the main lobe of the signal spectrum.
By trial and error, solving the following equation for z,
q
PE jdesired = Q
2 zjrequired ; z = Eb =N0
we Önd that zjrequired = 6:78 dB (4.76 ratio); 8:39 dB (6.90 ratio), 9:58 dB (9.08 ratio),
10:53 dB (11.3 ratio) for PE jdesired = 103 , 104 , 105 , 106 , respectively. The required
signal power is
Ps, required = Eb, required =Tsymbol = Eb, required R = zjrequired N0 R
For example, for PE = 103 ; N0 = 103 ; and R = 1000 bits/second, we get
Ps, required = zjrequired N0 R = 4:76 103 1000 = 4:76 W
Similar calculations allow the rest of the table to be Ölled in, giving the following results.
R, bps
1; 000
10; 000
100; 000
PE = 103 ; A2 and BW
4:76 W; 1 kHz
47:6 W; 10 kHz
476:0 W; 100 kHz
PE = 104
6:9 W
69 W
690:0 W
PE = 105
9:08 W
90:8 W
908:0 W
PE = 106
11:3 W
113:0 W
1130:0 W
Problem 9.3
a. For PE jdesired = 104 we have zjrequired = 8:39 dB (6.91 ratio). For R = B = 5; 000
bps, we have
signal power = A2 = zjrequired N0 R
= 6:91 106 5; 000
= 6:91 5 103 = 0:0346 W
b. For R = B = 10; 000 bps, the result is
signal power = A2 = zjrequired N0 R
= 6:91 106 10; 000 = 0:0691 W
c. For R = B = 100; 000 bps, we get
signal power = A2 = 6:91 106 100; 000 = 0:69 W
9.1. PROBLEM SOLUTIONS
3
d. For R = B = 1; 000; 000 bps, the result is
signal power = A2 = 6:91 106 1; 000; 000 = 6:91 W
Problem 9.4
The decision criterion is
V
> ", choose + A
V
< ", choose A
where
V = AT + N
N is a Gaussian random variable of mean zero and variance 2 = N0 T =2. Because
P (A sent) = P (A sent) = 1=2, it follows that
1
1
1
PE = P (AT + N > " j A sent) + P (AT + N < " j A sent) = (P1 + P2 )
2
2
2
where
2s
3
s
Z 1
2 =N T
2
2
0
e
2A T
2" 5
p
P1 =
d = Q 4
+
N
N
N0 T
0
0T
AT +"
= Q
Similarly
P2
Use the approximation
p
2z +
p
A2 T
2= ; = N0 T; z =
N0
Z AT +"
Z 1
2
2
e =N0 T
e =N0 T
p
p
=
d =
d
N0 T
N0 T
1
AT "
2s
3
s
p
2
2
p
2A T
2" 5
= Q4
=Q
2z 2=
N0
N0 T
2
Q (u) =
to evaluate
eu =2
p
u 2
1
(P1 + P2 ) = 106
2
by solving iteratively for z for various values of "=. The results are given in the table
below.
"= zdB for PE = 106 "= zdB for PE = 106
0
10.53
0.6
11.34
0.2
10.69
0.8
11.67
0.4
11.01
1.0
11.98
4
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
Problem 9.5
Use the zjrequired values found in the solution of Problem 9.2 along with zjrequired = A2 =N0 R
h
i
to get R = A2 = zjrequired N0 : Substitute N0 = 105 V2 /Hz, A = 40 mV to get R =
16 104 105 = zjrequired = 160= zjrequired .
below.
PE
zjrequired , dB (ratio) R, bps
104
105
106
8.39 (6.9)
9.58 (9.08)
10.53 (11.3)
This results in the values given in the table
23.19
17.62
14.16
Problem 9.6
a. Since the amplitudes A1 and A2 each occur 1/2 the time on average, the average
probability of error is
0s
1
PE = Q @
2
1
2A21 T A
N0
0s
1
+ Q@
2
1
2A22 T A
N0
But the average signal energy is
1 + 2 2
A21 T + A22 T
1 + A22 =A21 2
E=
=
A1 T =
A1 T
2
2
2
Thus
A21 T =
2
E
1 + 2
and
A22 T = 2 A21 T =
22
E
1 + 2
b. Substitution of these into the average probability of error expression gives the stated
result. A magniÖed plot of the probability of error is shown below. The curves
starting at the lefthand curve are for 2 = 1; 0:8; 0:6; 0:4. The respective values
of E=N0 for the crossing of the PE = 106 level are approximately 10.5, 10.8, 11.5,
and 12.7 dB giving degradations of 0, 0.3, 1, and 2.2 dB for 2 = 1; 0:8; 0:6; 0:4,
respectively..
9.1. PROBLEM SOLUTIONS
5
-3
10
-4
PE
10
-5
10
-6
10
10
10.5
11
11.5
E/N0, dB
12
12.5
13
6
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
0
10
∆ T/T = 0
∆ T/T = 0.1
∆ T/T = 0.2
∆ T/T = 0.3
-1
10
-2
PE
10
-3
10
-4
10
-5
10
-6
10
0
2
4
6
8
10
z, dB
12
14
16
18
20
Problem 9.7
a. For the pulse sequences (A; A) and (A; A), which occur half the time,
no degrap
dation results from timing error, so the error probability is Q
2Eb =N0 . The error
probability for the sequences (A; A) and
hp (A; A), which occur ithe other half the
time, result in the error probability Q
2Eb =N0 (1 2 jT j =T ) (sketches of the
pulse sequences with the integration interval o§set from the transition between pulses
is helpful here). Thus, the average error probability is the given result in the problem
statement. Plots are shown in Figure 9.2 for the giving timing errors.
b. Blowing up the plots of Figure 9.2 show that the intercepts of PE = 104 are about
8.4, 9.9, 12.4, and 15.9 dB, respectively, for timing errors of 0, 0.1, 0.2, and 0.3. The
corresponding degradations are 1.5, 4, and 7.5 dB for timing errors of 0.1, 0.2, and
0.3, respectively
9.1. PROBLEM SOLUTIONS
7
Problem 9.8
As in the derivation in the text for antipodal signaling,
1
var [N ] = N0 T
2
The output of the integrator at the sampling time is
AT + N; A sent
V =
N; 0 sent
The probabilities of error, given these two signaling events, are
1
P (error j A sent) = P AT + N < AT
2
1
= P N < AT
2
Z AT =2 2 =N0 T
e
p
=
d
N0 T
1
2s
3
2T
A
5
= Q4
2N0
Similarly,
P (error j 0 sent) = P
1
N > AT
2
Z 1
2
e =N0 T
p
=
d
N0 T
AT =2
2s
3
2
A T5
= Q4
2N0
Thus
1
1
P (error j A sent) + P (error j 0 sent)
22
2
3
s
A2 T 5
= Q4
2N0
PE =
Therefore, the average probability of error is
"r
PE = Q
Eave
N0
#
8
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
Problem 9.9
a. The impulse response of the Ölter is
h (t) = (2f3 ) e2f3 t u (t)
and the step response is the integral of the impulse response, or
a (t) = 1 e2f3 t u (t)
Thus, for a step of amplitude A, the signal out at t = T is
s0 (T ) = A 1 e2f3 T
The power spectral density of the noise at the Ölter output is
Sn (f ) =
N0 =2
1 + (f =f3 )2
so that the variance of the noise component at the Ölter output is
Z 1
2
N0 f3
E N =
Sn (f ) df =
= N0 BN
2
1
where BN is the noise equivalent bandwidth of the Ölter. Thus
2
2A2 1 e2f3 T
s20 (T )
SNR =
=
E [N 2 ]
N0 f3
b. To maximize the SNR, di§erentiate with respect to f3 and set the result equal to 0.
Solve the equation for f3 . This results in the equation
4f3 T e2f3 T 1 + e2f3 T = 0
Let = 2f3 T to get the equation
(2 + 1) e = 1
A numerical solution results in 1:25: Thus
f3, opt =
1:25
= 0:199=T = 0:199R
2T
9.1. PROBLEM SOLUTIONS
9
Problem 9.10
Substitute (9.27) and (9.28) into
PE = pP (E j s1 (t)) + (1 p) P (E j s2 (t))
h
i
h
i
Z 1 exp (v s01 )2 =2 2
Z k exp (v s02 )2 =2 2
0
0
p
p
= p
dv + (1 p)
dv
2 20
2 20
k
1
Use Leibnitzís rule to di§erentiate with respect to the threshold, k, set the result of the
di§erentiation equal to 0, and solve for k = kopt : The di§erentiation gives
or
or
h
i
h
i
exp (k s01 )2 =2 20
exp (k s02 )2 =2 20
p
p
p
+ (1 p)
2 20
2 20
=0
k=ko p t
h
i
h
i
exp (kopt s01 )2 =2 20
exp (kopt s02 )2 =2 20
p
p
p
= (1 p)
2 20
2 20
2
2
(kopt s01 ) + (kopt s02 )
(s01 s02 ) kopt +
s202 s201
2
kopt
1p
=
p
1p
2
= 0 ln
p
2
0
1p
s02 + s01
=
ln
+
s01 s02
p
2
2 20 ln
Problem 9.11
a. Denote the output due to the signal as s0 (t) and the output due to noise as n0 (t).
We wish to maximize
R
2
1
j2f t0 df
S
(f
)
H
(f
)
e
2
m
s (t)
2 1
R1
0 2 =
N0
E n0 (t)
1 Hm (f ) Hm (f ) df
10
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
By Schwartzís inequality, we have
R1
R1
Z 1
s20 (t)
2 1 S (f ) S (f ) df 1 Hm (f ) Hm (f ) df
2
R1
=
jS (f )j2 df
(f ) ej2f t0 df
N0
N
H
(f
)
H
E n20 (t)
0
m
1
m
1
where the maximum value on the right-hand side is attained if
Hm (f ) = S (f ) ej2f t0
b. The matched Ölter impulse response is hm (t) = s (t0 t) by taking the inverse Fourier
transform of Hm (f ) employing the time reversal and time delay theorems.
c. The realizable matched Ölter has zero impulse response for t < 0.
d. By sketching the components of the integrand of the convolution integral, we Önd the
following results:
For t0 = 0; s0 (t0 ) = 0;
For t0 = T =2; s0 (t0 ) = A2 T =2;
For t0 = T; s0 (t0 ) = A2 T ;
For t0 = 2T; s0 (t0 ) = A2 T:
Problem 9.12
a. These would be x (t) and y (t) reversed and then shifted to the right so that they are
nonzero for t > 0.
p
b. A = 7B:
c. The outputs for the two cases are
So1 (t) = A2 [(t t0 ) = ] = 7B 2 [(t t0 ) = ]
So2 (t) = 7B 2 [(t t0 ) =7 ]
d. The shorter pulse gives the more accurate time delay measurement.
e. Peak power is lower for y (t).
Problem 9.13
a. The matched Ölter impulse response is given by
hm (t) = s2 (T t) s1 (T t)
9.1. PROBLEM SOLUTIONS
11
b. Using (9.55) and a sketch of s2 (t) s1 (t) for an arbitrary t0 , we obtain
2
Z 1
2
[s2 (t) s1 (t)]2 dt
N0 1
2
2
2
2 T
A t0 + 4A t0 + A
t0
N0
2
2 A2 T
T
+ 4A2 t0 ; 0 t0
N0
2
2
=
=
=
This result increases linearly with t0 and has its maximum value for for t0 = T =2, which is
5A2 T =N0 :
c. The probability of error is
PE = Q
p
2 2
To minimize it, use the maximum value of , which is given in (b).
d. The optimum receiver would have two correlators in parallel, one for s1 (t) and one
for s2 (t). The outputs for the correlators at t = T are di§erenced and compared
with the threshold k calculated from (9.31), assuming the signals are equally likely.
Problem 9.14
Use 2max = 2Es =N0 for a matched Ölter, where Es is the signal energy. The required values
for signal energy are: (a) A2 T ; (b) 3A2 T =8; (c) A2 T =2; (d) A2 T =3
Problem 9.15
The required formulas are
1
(E1 + E2 )
2
Z
1 T
=
s1 (t) s2 (t) dt
E 0
1
=
(E2 E1 )
22
3
s
(1 R12 ) E 5
= Q4
N0
E =
R12
kopt
PE
12
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
The energies for the three given signals are
Z T
EA =
A2 dt = A2 T
0
Z T
Z T
2
2 t
2
2 t
EB =
B cos
dt =
B sin
dt
T
2
T
0
0
Z
B2 T
2t
B2T
=
1 cos
dt =
2 0
T
2
2
Z T 2
C
2t
EC =
1 + cos
dt
4
T
0
Z T 2
C
2t 2
=
1 cos
dt
4
T
0
Z
C2 T
2t
2 2t
=
1 2 cos
+ cos
dt
4 0
T
T
Z
C2 T 3
2t
1
4t
3C 2 T
=
2 cos
+ cos
dt =
4 0 2
T
2
T
8
a. For s1 = sA and s2 = sB , the average signal energy is
1
A2 T
B2
1+
E = (EA + EB ) =
2
2
2A2
The correlation coe¢cient is
R12 =
=
=
(t T =2)
AB cos
dt
T
0
Z
1 T
t
2ABT
AB sin
dt =
E 0
T
E
4AB
(A2 + B 2 =2)
1
E
Z T
The optimum threshold is
1
1
kopt = (EB EA ) =
2
2
B2
2
A T
2
The probability of error is
2s
3
"s
2
#
2
(1
R
)
E
1
4AB
A
T
B
12
5=Q
PE = Q 4
1
1+
N0
N0
(A2 + B 2 =2)
2
2A2
"s
"s
#
#
2
T
4AB
B
T
B 2 4AB
2
2
= Q
1
A +
=Q
A +
2N0
(A2 + B 2 =2)
2
2N0
2
9.1. PROBLEM SOLUTIONS
13
b. s1 = sA and s2 = sC :
1
1 2
(EA + EC ) =
A + 3C 2 =8 T
2
2
AC
R12 =
A2 + 3C 2 =8
1
1 3C 2
2
kopt = (EC EA ) =
A T
2
2
8
"s
#
2
T
3C
PE = Q
A2 +
AC
2N0
8
E =
c. s1 = sB and s2 = sC :
1
1
3C 2
2
E =
(EB + EC ) =
B +
T
2
4
4
16BC
R12 =
3 (B 2 + 3C 2 =4)
1
1 3C 2
2
kopt = (EC EB ) =
B T
2
4
4
"s
#
2
T
3C
16BC
PE = Q
B2 +
4N0
4
d. s1 = sB and s2 = sB :
E = EB = B 2 T =2
R12 = 1
e. s1 = sC and s2 = sC :
kopt = 0
2s
3
2T
B
5
PE = Q 4
N0
E = EC = 3C 2 T =8
R12 = 1
kopt = 0
14
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
2s
PE = Q 4
3
3C 2 T 5
8N0
Problem 9.16
a. The energies are
EA =
Z T
A2 dt = A2 T
0
EB =
Z T =2
2
A dt +
EC
=
0
(A)2 dt = A2 T
T =2
0
Z T =4
Z T
2
A dt +
Z T =2
2
(A) dt +
T =4
Z 3T =4
2
A dt +
T =2
Z T
(A)2 dt = A2 T
3T =4
b. Sketch the combinations (A; B) ; (B; C) ; and (A; C) to show that their respective
products have as much area above the t-axis as below it. Therefore, RAB , RBC , and
RAC are all zero because they are directly proportional to the integral of the products
of the respective signals. The optimum threshold for each case is zero because it is
proportional to the di§erence of the respective signal energies.
c. The probability of error for each case is
0s
PE = Q @
A2 T
N0
1
A
Problem 9.17
Candidate modulation schemes are noncoherent FSK and DPSK. The bandwidth e¢ciency
for NFSK is 0.4 and for DPSK is 0.5, so either one will provide the desired data rate for
the given bandwidth. For NFSK, the average probability of error is
1
1
Eb
1
A2 T
PE; NFSK = exp (z=2) = exp
= exp
2
2
2N0
2
4N0
2
where Pave = A2 is the average signal power. We want PE 106 , so solve
1
A2 T
exp
= 106
2
4N0
9.1. PROBLEM SOLUTIONS
or
15
A2 T
= 2 ln 2 106
2N0
or
A2
2
= 2N0 R ln 2 106 ; R = 1=T = data rate
= 2 107 4 103 ln 2 106
= 0:0105 watts (normalized)
= 10:5 mW
For DPSK, the average probability of error is
1
1
Eb
1
A2 T
PE; DPSK = exp (z) = exp
= exp
2
2
N0
2
2N0
2
where Pave = A2 is the average signal power. We want PE 106 , so solve
1
A2 T
exp
= 106
2
2N0
or
A2 T
= ln 2 106
2N0
or
A2
= N0 R ln 2 106 ; R = 1=T = data rate
2
= 107 4 103 ln 2 106
= 5:25 mW
From the standpoint of simplicity, the NFSK receiver is simpler than the optimum DPSK
receiver. If a suboptimum delay-and-multiply DPSK receiver is used then DPSK wins in
terms of simplicity with some loss in performance. In this case, we must solve
0s
1
2T
A
A = 106
Q@
2N0
s
A2 T
= 4:754
2N0
A2
2
= N0 R (4:754)2
= 107 4 103 (4:754)2
= 9 mW
16
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
which is still less than for NFSK.
Problem 9.18
Numerical integration of (9.83) with (9.81) and (9.82) substituted shows that the degradation for 2 = 0:01 ( = 0:1 rad) is about 0.045 dB; for 2 = 0:05 ( = 0:224 rad) it is
about 0.6 dB; for 2 = 0:1 ( = 0:316 rad) it is about 9.5 dB. For the constant phase
error model, the degradation is
Dconst = 20 log10 [cos (const )]
As suggested in the problem statement, we set the constant phase error equal to the standard
deviation of the Gaussian phase error for comparison purposes; thus, we consider constant
phase errors of const = 0:1; 0:224; and 0.316 radians. The degradations corresponding to
these phase errors are 0:044, 0:219, and 0:441 dB, respectively. The constant phase error
degradations are much less serious than the Gaussian phase error degradations.
Problem 9.19
p
a. For ASK, PE = Q [ z] = 105 gives z = 4:2642 = 18:182 or z = 12:596 dB.
b. For BPSK, PE = Q
p
2z = 105 gives z = 4:2642 =2 = 9:091 or z = 9:586 dB.
c. Binary FSK bit error probability is the same as ASK.
d. The degradation of BPSK with a phase error of 5 degrees is Dconst = 20 log10 [cos (5o )] =
0:033 dB, so the required SNR to give a bit error probability of 105 is 9:586+0:033 =
9:619 dB.
hp
i
p
p
e. For PSK with m = 1= 2, PE = Q
2 (1 1=2) z = Q [ z] = 105 gives z = 12:596
dB.
f. Assuming that the separate e§ects are additive, we add the degradation of part (d)
to the SNR found in part (e) to get 12:596 + 0:033 = 12:629 dB.
Problem 9.20
a. Without frequency uncertainty, the bandwidth of the input Ölters is BT ' 2=T Hz.
With frequency uncertainty it must be BT + jf j Hz. Therefore, the noise power
9.1. PROBLEM SOLUTIONS
17
-1
10
-2
10
-3
PE
10
-4
10
-5
10
-6
10
10
11
12
13
E/N0, dB
14
15
16
out of the Ölters is (BT + jf j) N0 watts instead of BT N0 watts. The probability of
error is therefore
0 1
1
A2
PE =
exp z =2 = exp
2
2
4 (BT + jf j) N0
2
1
A
=
exp
2
4 (1 + jf j =BT ) N0 BT
1
1
A2
=
exp
2
2 (1 + jf j =BT ) 2N0 BT
1
z
1
=
exp
2
2 1 + jf j =BT
b. The probability of error is plotted in Figure 9.3 with the leftmost curve being for no
frequency error.The required value of z to provide a desired value of error probability,
PE; 0 , is given by
1
z0
1
exp
= PE; 0
2
2 1 + jf j =BT
18
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
or
z0 = 2 (1 + jf j =BT ) ln (2PE; 0 )
In dB, this is
z0; dB = 10 log10 [2 (1 + jf j =BT ) ln (2PE; 0 )]
c. The degradation is the di§erence in dB between z0; dB with frequency uncertainty and
the value without it:
Degradation = D = 10 log10 [2 (1 + jf j =BT ) ln (2PE; 0 )] 10 log10 [2 ln (2PE; 0 )]
= 10 log10 [(1 + jf j =BT )] dB
Thus, the degradation is seen to be independent of error probability. A table of degradation
values is given below.
jf j =BT D, dB
0
0
0.1
0.414
0.2
0.792
0.3
1.139
0.4
1.461
Problem 9.21
We have that
1 m2 =
Use this in
PE = Q
% modulation
100
"r
Eb
2 (1 m2 )
N0
#
This results in the plot below:
From this plot, we estimate the following values for the percent mudulation to produce a
probabillity of error of 104 for each value of z = Eb =N0 speciÖed.
Eb =N0 , dB % modulation to give PE = 104
8:5
62
9:0
70
9:5
78
10:0
87
10:5
98
9.1. PROBLEM SOLUTIONS
19
-2
10
-3
PE
10
-4
10
Eb/N0 = 8.5
9
9.5
10
10.5dB
-5
10
-6
10
50
55
60
65
70
75
80
% modulation
85
90
95
100
Problem 9.22
a. For BPSK, an SNR of 10.53 dB is required to give an error probability of 106 (see
the solution to Problem 9.2). For ASK and FSK it is 3.01 dB more, or 13.54 dB. For
BPSK and ASK, the required bandwidth is 2R Hz where R is the data rate in bps.
For FSK with minimum tone spacing of 0:5R Hz, the required bandwidth is 2:5R Hz.
Hence, for BPSK and ASK, the required bandwidth is 100 kHz. For FSK, it is 125
kHz.
b. For BPSK, the required SNR is 9.588 dB.for an error probability of 105 : For ASK
and FSK it is 3.01 dB more or 12.598 dB. The required bandwidths are now 1 MHz
for BPSK and ASK; it is 1.25 MHz for FSK.
Problem 9.23
20
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
The correlation coe¢cient is
R12
Z
1 T
=
s1 (t) s2 (t) dt
E 0
Z
1 T 2
=
A cos (! c t) cos [(! c + !) t] dt
E 0
= sinc (2f T )
where f is the frequency separation between the FSK signals in Hz. Using a calculator,
we Önd that the sinc-function takes on its Örst minimum at an argument value of about
1.4, or f = 0:7=T . The minimum value at this argument value is about 0:216. The
improvement in SNR is 10 log10 (1 R12 ) = 10 log10 (1 + 0:216) = 0:85 dB.
Problem 9.24
The encoded bit streams, assuming a reference 1 to start with, are: (a) 1 111 110 100 010;
(b) 1 100 111 100 111; (c) 1 111 111 111 111; (d)1 010 101 010 101; (e) 1 111 111 010 101;
(f) 1 110 000 011 011; (g) 1 100 101 010 110; (h) 1 100 001 000 010.
9.1. PROBLEM SOLUTIONS
21
Problem 9.25
The di§erentially encoded bit stream is 1 000 011 001 111 (1 assumed for the starting
reference bit). The phase of the transmitted carrier is 000 00 0 . Integrate the
product of the received signal and a 1-bit delayed signal over a (0; T ) interval. Each time
the integrated productRis greater than 0, decide 1; each time it is less than 0, decide 0. The
T
Örst integration gives 0 A2 cos (! 0 t + ) cos (! 0 t) dt = A2 T =2. The resulting decision is
RT
that a 0 was sent. The next integration gives 0 A2 cos (! 0 t) cos (! 0 t) dt = A2 T =2. The
resulting decision is that a 1 was sent. Continuing in this manner, using the previous signal
for a reference in the integration for the current signal and using the stated decision criterion,
we obtain the original bit sequence. If the signal is inverted, the same demodulated signal
is detected as if it werenít inverted.
Problem 9.26
Note that
E [n1 ] = E
Z 0
n (t) cos (! c t) dt =
T
Z 0
E [n (t)] cos (! c t) dt = 0
T
which follows because E [n (t)] = 0: Similarly for n2 , n3 , and n4 . Consider the variance of
each of these random variables:
var [n1 ] = E n21 = E
=
=
Z 0 Z 0
T
n (t) n () cos (! c t) cos (! c ) dtd
T
N0
2
E [n (t) n ()] cos (! c t) cos (! c ) dtd
T T
Z 0 Z 0
T
=
Z 0 Z 0
N0
(t ) cos (! c t) cos (! c ) dtd
T 2
Z 0
T
cos2 (! c t) dt =
N0 T
4
where the fact that E [n (t) n ()] = N20 (t ) has been used along with the sifting property
of the -function to reduce the double integral to a single integral. Similarly for n2 , n3 ,
and n4 . Therefore, w1 has zero mean and variance
1
1
N0 T
var [w1 ] = var [n1 ] + var [n2 ] =
4
4
8
Similarly for w2 , w3 , and w4 .
22
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
Problem 9.27
a. The noise power out of the input Ölter with frequency uncertainty is (2R + jf j) N0
watts instead of 2RN0 watts. Thus, the probability of error is
PE
p
z0 = Q
= Q
s
A2
2 (2R + jf j) N0
s
!
A2
= Q
2 (1 + jf j =2R) (2RN0 )
r
z
= Q
1 + jf j =2R
!
b. The required value of z to provide a desired value of error probability, PE; 0 , is given
by
r
z0
Q
= PE; 0
1 + jf j =2R
or
In dB, this is
2
z0 = (1 + jf j =2R) Q1 (2PE; 0 )
n
2 o
z0; dB = 10 log10 (1 + jf j =2R) Q1 (2PE; 0 )
c. The degradation is the di§erence in dB between z0; dB with frequency uncertainty and
the value with out it:
n
n
2 o
2 o
Degradation = D = 10 log10 (1 + jf j =2R) Q1 (2PE; 0 )
10 log10 Q1 (2PE; 0 )
= 10 log10 [(1 + jf j =2R)] dB
Thus, the degradation is seen to be independent of error probability. A table of degradation
values is given below.
jf j =R D, dB
0
0
0.2
0.414
0.4
0.792
0.6
1.139
0.8
1.461
9.1. PROBLEM SOLUTIONS
23
.
Problem 9.28
a. By plotting sequences of two successive bit waveforms, it is seen that no energy cancellation for the cases 11 or 00 (i.e., a cosinusoid followed by a cosinusoid of the same
phases, or the negation of this situation). The probability of error for these cases is
PE; 1 = Q
r
Eb
N0
!
However, for the cases 10 or 01, there is energy loss in the amount of 1 jT j =R (i.e., a
cosinusoid followed by its negation or vice versa). The probability of error for these cases
is
s
!
Eb
jT j
PE; 1 = Q
1
N0
T
For equally likely 1s and 0s in the data stream, each case occurs half the time, giving
PE; jT j =
1
Q
2
=
1
Q
2
r
r
Eb
N0
Eb
N0
!
!
1
+ Q
2
1
+ Q
2
s
r
Eb
N0
jT j
1
T
!
!
Eb
(1 jT j R)
N0
b. A plot of magniÖed error probability is given below for jT j R = 0; 0:1; 0:2; 0:3; 0:4
starting with the leftmost curve. The values of Eb =N0 giving PE; jT j = 106 can be
estimated as 13.5, 13.8, 14.2, 14.8, and 15.5 dB for these values of jT j R, respectively.
The value for 13.5 dB is for no delay error. The degradation values are 13:8 13:5 =
0:3 dB, 14:2 13:5 = 0:7 dB, 14:8 13:5 = 1:3 dB, and 15:5 13:5 = 2 dB for
jT j =R = 0:1; 0:2; 0:3; 0:4, respectively.
24
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
-2
10
-3
10
-4
10
-5
10
-6
10
11
11.5
12
12.5
13
13.5
14
14.5
15
15.5
16
9.1. PROBLEM SOLUTIONS
25
Problem 9.29
Bandwidths are given by
BBPSK = BDPSK = 2R; RBPSK = RDPSK = BBPSK, DPSK =2 = 50 kbps
BCFSK = 2:5R; RCFSK = BCFSK =2:5 = 40 kbps
BNCFSK = 4R; RNCFSK = BNCFSK =4 = 25 kbps
Problem 9.30
Consider the bandpass Ölter, just wide enough to pass the ASK signal, followed by an
envelope detector with a sampler and threshold comparator at its output. The output of
the bandpass Ölter is
u (t) = Ak cos (! c t + ) + n (t)
= Ak cos (! c t + ) + nc (t) cos (! c t + ) ns (t) sin (! c t + )
= x (t) cos (! c t + ) ns (t) sin (! c t + )
where Ak = A if signal plus noise is present at the receiver input, and A = 0 if noise alone
is present at the input. In the last equation
x (t) = Ak + nc (t)
Now u (t) can be written in envelope-phase form as
u (t) = r (t) cos [! c t + + (t)]
where
r (t) =
p
x2 (t) + n2s (t) and (t) = tan1
ns (t)
x (t)
The output of the envelope detector is r (t), which is sampled each T seconds and compared
with the threshold. Thus, to compute the error probability, we need the pdf of the envelope
for noise alone at the input and then with signal plus noise at the input. For noise alone,
it was shown in Chapter 4 that the pdf of r (t) is Rayleigh, which is
fR (r) =
r r2 =2N
e
; r 0; noise only
R
where N is the variance (mean-square value) of the noise at the Ölter output. For signal
plus noise at the receiver input, the pdf of the envelope is Ricean. Rather than use this
complex expression, however, we observe that for large SNR at the receiver input
r (t) A + nc (t) ; large SNR, signal present
26
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
which follows by expanding the argument of the square root, dropping the squared terms,
and approximating the square root as
p
1
1 + 1 + ; jj << 1
2
Thus, for large SNR conditions, the envelope detector output is approximately Gaussian
for signal plus noise present. It has mean A and variance N (recall that the inphase and
quadrature lowpass noise components have variance the same as the bandpass process).
The pdf of the envelope detector output with signal plus noise present is approximately
2
e(rA) =2N
p
fR (r) =
; large SNR
2N
For large SNR, the threshold is approximately A=2. Thus, for noise alone present, the
probability of error is exactly
Z 1
r r2 =2N
2
P (E j 0) =
e
dr = eA =8N
R
A=2
For signal plus noise present, the probability of error is approximately
Z A=2
P (E j S + N present) =
p
e(rA) =2N
p
dr = Q
A2 =2N
2N
2
1
2
eA =4N
Now the average signal-to-noise ratio is
z=
p
=N A
1 A2 =2
A2
=
2 N0 BT
4N
which follows because the signal is present only half the time.
probability of error is
Therefore, the average
1
1
ez
1
PE = P (E j S + N ) + P (E j 0) p
+ ez=2
2
2
4z 2
Problem 9.31
The integral to be integrated is
z
PE = e
Z 1
0
r r2 =N
e
I0
N
Ar
N
dr
9.1. PROBLEM SOLUTIONS
The Ricean pdf, from (7.150), is
(
fR (r) =
r
exp
2
0; r < 0
27
n h 2
io p
r
r
2
+
K
I
2K
0
2
; r 0
Since it is a pdf, its integral over -1 < r < 1 is 1. The trick is to get the integrand of the
expression
for PE to look like a Ricean pdf. Obviously, a start is to identify N = 2 2 and
p
A
2K
A2
A2
which can be solved to give K = 4N
= z2 because z = 2N
. The the integral for
N =
PE becomes
Z
ez 1 r r2 =22 p
r
PE =
e
I
2K
dr
0
2 0 2
We multiply by 1 = eK eK to get
z Z 1
Ke
r (r2 =22 +K ) p
r
e
I
2K
dr
0
2 0 2
ez
1
1
= eK
= ez=2 ez = ez=2
2
2
2
PE = e
where the integral is 1 because the integrand is a Ricean pdf.
Problem 9.32
The binary and Gray encoded numbers are given in the table below:
Decimal Binary Gray Decimal Binary Gray
0
00000
00000 16
10000
11000
1
00001
00001 17
10001
11001
2
00010
00011 18
10010
11011
3
00011
00010 19
10011
11110
4
00100
00110 20
10100
11110
5
00101
00111 21
10101
11111
6
00110
00101 22
10110
11101
7
00111
00100 23
10111
11100
8
01000
01100 24
11000
10100
9
01001
01101 25
11001
10101
10
01010
01111 26
11010
10111
11
01011
01110 27
11011
10110
12
01100
01010 28
11100
10010
13
01101
01011 29
11101
10011
14
01110
01001 30
11110
10001
15
01111
01000 31
11111
10000
28
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
Problem 9.33 Problem statement should be "Show that (9.138) is the average
energy in terms of for M -ary antipodal PAM."
Evaluation is facilitated by means of the following sums:
m
X
1 = m;
k=1
m
X
m
k=
k=1
m (m + 1) X 2 m (m + 1) (2m + 1)
;
k =
2
6
k=1
The expression to be evaluated is
Eave =
=
M
M
1 X
1 X
M 1 2
Ej =
(j 1)
M
M
2
j=1
j=1
"
#
M
M
2 X
M 1 2 2 X
(M 1)2
2
(j 1)
=
(j 1) (M 1) (j 1) +
M
2
M
4
j=1
=
=
=
=
=
j=1
M
M
M
2 X
2 X
(M 1)2 2 X
(j 1)2 (M 1)
(j 1) +
1
M
M
4
M
j=1
j=1
j=1
2
3
M
1
M
1
M
2 X
2
X
X
4
(M 1)
k 2 (M 1)
k+
15
M
4
j=1
k=1
k=1
"
#
2
2
(M
1)
(2M
1)
(M
1)
(M
1)
2
+
6
2
4
"
#
(M 1) (2M 1) (M 1)2
2
2
=
(M 1) [2 (2M 1) 3 (M 1)]
6
4
12
2
M 1 2
2
(M 1) (M + 1) =
12
12
Problem 9.34
a. Appy (9.142) since baseband is speciÖed:
BP AM =
R
20
or 5 =
) M = 16
log2 M
log2 M
b. Apply (9.141):
2 (M 1)
Pb; antip. PAM =
Q
M log2 M
r
6 log2 M Eb
M 2 1 N0
!
9.1. PROBLEM SOLUTIONS
29
For M = 16 and a BEP of 106 this becomes
2 (16 1)
Q
16 log2 16
Q
r
r
6 log2 16 Eb
162 1 N0
6 log2 16 Eb
162 1 N0
!
!
= 106
=
32
106
15
Trial and error with a Q-function routine in MATLAB yields
r
6 log2 16 Eb
Eb
255
' 4:598 )
=
4:5982 = 23:515 dB
162 1 N0
N0
24
For M = 16 and a BEP of 105 we have
Q
r
6 log2 4 Eb
42 1 N 0
Problem 9.35
a. BPSK or DPSK, 10.53 dB;
b. 4-PAM (RF), 14.4 dB;
c. 8-PAM (RF), 18.8 dB;
d. 16-PAM (RF), 23.52 dB;
e. 32-PAM (RF), 28.51 dB.
!
32
105
15
r
6 log2 16 Eb
)
' 4:0925
162 1 N0
Eb
255
)
=
4:09252 = 22:503 dB
N0
24
=
30
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
First 3 answers from Table 9.5. Solve equation (9.141) for 16-PAM and 32-PAM:
!
!
r
r
2 (16 1)
6 log2 16 Eb
30
24
E
b
Q
= 106 =) Q
= 106
16 log2 16
162 1 N0
64
255 N0
!
r
Eb
=) Q
0:0941
= 2:13 106
N0
r
Eb
0:0941
= 4:598
=)
N0
Eb
=) 0:0941
= 21:14
N0
Eb
=)
= 224:67 = 23:52 dB
N0
2 (32 1)
Q
32 log2 32
r
6 log2 32 Eb
322 1 N0
!
=
=)
=)
=)
=)
!
r
62
30
E
b
106 =)
Q
= 106
160
1023 N0
!
r
Eb
Q
0:0293
= 2:58 106
N0
r
Eb
0:0293
= 4:56
N0
Eb
0:0293
= 20:79
N0
Eb
= 709:586 = 28:51 dB
N0
Problem 9.36
For BPSK, the BEP is found from
p
p
Pb; BPSK = Q
2Eb =N0 = 104 ) 2Eb =N0 = 3:719; Eb =N0 = 8:4 dB
The data rate in terms of bandwidth for BPSK is
RBPSK =
BRF
100 kHz
=
= 50 kbps
2
2
For DPSK
1
exp (Eb =N0 ) = 104 ) Eb =N0 = 10 log10 [ ln 2 104 ] = 9:3 dB
2
9.1. PROBLEM SOLUTIONS
31
The data rate is the same as for BPSK.
For coherent FSK, the required Eb =N0 is 3.01 dB more than for BPSK, or Eb =N0 =
8:4 + 3:01 = 11:41 dB. The data rate is
BRF
100 kHz
=
= 40 kbps
2:5
2:5
RCFSK =
For noncoherent FSK
1
exp (Eb =2N0 ) = 104 ) Eb =N0 = 10 log10 [2 ln 2 104 ] = 12:31 dB
2
The data rate is
RNCFSK =
BRF
100 kHz
=
= 25 kbps
4
4
For antipodal 4-ary PAM,
2 (4 1)
Q
4 log2 4
r
6 log2 4 Eb
42 1 N 0
!
= 104 )
Eb
= 11:3 dB
N0
The data rate is:
1
1
R = BRF , PAM log2 M = 100 log2 4 = 100 kbps
2
2
For antipodal 8-ary PAM,
2 (8 1)
Q
8 log2 8
r
6 log2 8 Eb
82 1 N 0
!
= 105 )
Eb
= 16:5dB
N0
The data rate in terms of bandwidth can be obtained from (9.141):
1
1
R = BRF , PAM log2 M = (100) log2 8 = 150 kbps
2
2
Problem 9.37
For = 0:25,
P (f ) =
8
< T;
:
0;
T
2
T
0:75
1 + cos 0:25 jf j 2T
0 jf j 0:75
2T
1:25
; 0:75
jf
j
Hz
2T
2T
jf j > 0:75
2T
The optimum transmitter and receiver transfer functions are
1=4
jHT (f )jopt =
A jP (f )j1=2 Gn (f )
jHC (f )j1=2
32
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
and
jHR (f )jopt =
K jP (f )j1=2
1=4
Gn (f ) jHC (f )j1=2
Setting all scale factors equal to 1, these become
q
1 + (f =fC )2
1=2 1=4
jHT (f )jopt =
jP (f )j
q
1=2 1=4
jHR (f )jopt = jP (f )j
q
1=4
1 + (f =f3 )2
q
2 1=4
1 + (f =f3 )
1 + (f =fC )2
A MATL:AB program for computing the optimum transfer functions is given below. Plots
for the various cases are then given in Figures 8.3 - 8.5.
% Problem 9.37: Optimum transmit and receive Ölters for raised-cosine pulse shaping,
% Örst-order Butterworth channel Öltering, and Örst-order noise spectrum
%
A = char(í-í,í:í,íñí,í-.í);
clf
beta00 = input(íEnter starting value for beta: >=0 and <= 1 í);
del_beta = input(íEnter step in beta: starting value + 4*step <= 1 í);
R = 1;
f_3 = R/2;
f_C = R/2;
delf = 0.01;
HT_opt = [];
HR_opt = [];
for n = 1:4
beta = (n-1)*del_beta+beta00;
beta0(n) = beta;
f1 = 0:delf:(1-beta)*R/2;
f2 = (1-beta)*R/2:delf:(1+beta)*R/2;
f3 = (1+beta)*R/2:delf:R;
f = [f1 f2 f3];
P1 = ones(size(f1));
P2 = 0.5*(1+cos(pi/(R*beta)*(f2-0.5*(1-beta)*R)));
P3 = zeros(size(f3));
Gn = 1./(1+(f/f_3).^2);
HC = 1./sqrt(1+(f/f_C).^2);
P = [P1 P2 P3];
HT_opt = sqrt(P).*(Gn.^.25)./sqrt(HC);
9.1. PROBLEM SOLUTIONS
33
|HR,opt(f)||
R = 1 bps; channel filter 3-dB frequency = 0.5 Hz; noise 3-dB frequency = 0.5 Hz
1.5
1
0.5
0
0
0.1
0.2
0.3
0.4
0.5
f, Hz
0.6
0.7
0.8
0.9
1
|HT,opt(f)|
1.5
β = 0.2
β = 0.3
1
β = 0.4
β = 0.5
0.5
0
0
0.1
0.2
0.3
0.4
0.5
f, Hz
0.6
0.7
0.8
0.9
1
HR_opt = sqrt(P)./((Gn.^.25).*sqrt(HC));
subplot(2,1,1),plot(f,HR_opt,A(n,:)),xlabel(íf, Hzí),ylabel(íjH_R_,_o_p_t(f)jjí)
title([íR = í,num2str(R),í bps; channel Ölter 3-dB frequency = í,num2str(f_C),í
Hz; noise 3-dB frequency = í,num2str(f_3),í Hzí])
if n == 1
hold on
end
subplot(2,1,2),plot(f,HT_opt,A(n,:)),xlabel(íf, Hzí),ylabel(íjH_T_,_o_p_t(f)jí)
if n == 1
hold on
end
end
legend([ínbeta = í,num2str(beta0(1))],[ínbeta = í,num2str(beta0(2))],
[ínbeta = í,num2str(beta0(3))],[ínbeta = í,num2str(beta0(4))],1)
a. f3 = fC = 1=2T
b. fC = 2f3 = 1=T
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
|HR,opt(f)||
R = 1 bps; channel filter 3-dB frequency = 1 Hz; noise 3-dB frequency = 0.5 Hz
1.5
1
0.5
0
0
0.1
0.2
0.3
0.4
0.5
f, Hz
0.6
0.7
0.8
0.9
1
|HT,opt(f)|
1
β = 0.2
β = 0.3
β = 0.4
β = 0.5
0.5
0
0
0.1
0.2
0.3
0.4
0.5
f, Hz
0.6
0.7
0.8
0.9
1
|HR,opt(f)||
R = 1 bps; channel filter 3-dB frequency = 0.5 Hz; noise 3-dB frequency = 1 Hz
1.5
1
0.5
0
0
0.1
0.2
0.3
0.4
0.5
f, Hz
0.6
0.7
0.8
0.9
1
1.5
|HT,opt(f)|
34
β = 0.2
β = 0.3
1
β = 0.4
β = 0.5
0.5
0
0
0.1
0.2
0.3
0.4
0.5
f, Hz
0.6
0.7
0.8
0.9
1
9.1. PROBLEM SOLUTIONS
35
c. f3 = 2fC = 1=T
Problem 9.38
b
a
This a simple matter of sketching P (f ) = ba
(f =b) ba
(f =a) and P (f B), where
1
B = 2 (a + b), side by side and showing that they add to a constant in the overlap region.
Problem 9.39
a. The optimum transmitter and receiver transfer functions are
1=4
jHT (f )jopt =
A jP (f )j1=2 Gn (f )
jHC (f )j1=2
and
jHR (f )jopt =
K jP (f )j1=2
1=4
Gn (f ) jHC (f )j1=2
where P (f ) is the signal spectrum and A and K are arbitrary constants. From the problem
1=4
1=4
statement, Gn (f ) = 1011
= 1011=4 and
1
jHC (f )j = q
1 + (f =4800)2
The signal spectrum for = 1 and T1 = 9600 bps is, from (5.137),
T
T
1
1
1+
P (f ) =
1 + cos
f
;
= 0 jf j
= 9600 Hz
2
2T
2T
2T
1
jf j
=
1 + cos
; 0 jf j 9600 Hz
2 9600
9600
1
jf j
2
=
cos
; 0 jf j 9600 Hz
9600
19200
Thus,
s
i1=4
1
jf j h
cos2
1 + (f =4800)2
; 0 jf j 9600 Hz
9600
19200
i1=4
1011=4 A
jf j h
2
=
cos
1
+
(f
=4800)
; 0 jf j 9600 Hz
97:98
19200
s
i1=4
1
jf j h
11=4
2
= 10
K
cos
1 + (f =4800)2
; 0 jf j 9600 Hz
9600
19200
i1=4
1011=4 K
jf j h
2
=
cos
1
+
(f
=4800)
; 0 jf j 9600 Hz
97:98
19200
jHT (f )jopt = 1011=4 A
jHR (f )jopt
36
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
11=4
11=4
where A and K may be chosen so that 1097:98 A = 1097:98K = 1.
b. From (9.149) Q (A=) = 104 results in A= ' 3:72.
c. Matching arguments of (9.151) and (9.155) we have
A p
3:72 '
= ET
"Z
1
1=2
Gn (f ) P (f )
df
jHC (f )j
1
#1
Therefore, we need to evaluate the integral
Z 1
Z 1 1=2
1=2
Gn (f ) P (f )
Gn (f ) P (f )
I =
df = 2
df
jHC (f )j
jHC (f )j
1
0
q
Z
2 1011=4 9600
f
2
=
cos
1 + (f =4800)2 df; v = f =4800
9600
19200
0
Z 2
v p
= 1011=4
cos2
1 + v 2 dv ' 1:213 1011=4
4
0
Thus
3:72 =
p
ET
) ET = 6:44 105 J
11=4
1:213 10
Problem 9.40
The MATLAB program is given below and Figure 9.28 shows the probability of error curves.
% Plots for Prob. 9.40
%
clf
A = char(í-í,íñí,í-.í,í:í,íñ.í,í-..í);
delta = input(í Enter value for delta: í)
taum_T0 = input(í Enter vector of values for tau_m/T: í)
L_taum = length(taum_T0);
z0_dB = 0:.1:15;
z0 = 10.^(z0_dB/10);
for ll = 1:L_taum
ll
taum_T = taum_T0(ll)
P_E = 0.5*qfn(sqrt(2*z0)*(1+delta))+0.5*qfn(sqrt(2*z0)*((1+delta)-2*delta*taum_T));
semilogy(z0_dB, P_E,A(ll,:))
if ll == 1
9.1. PROBLEM SOLUTIONS
37
PE versus z 0 in dB for δ = 0.5
0
10
-1
10
-2
PE
10
-3
10
-4
10
τm/T = 0.2
-5
10
τm/T = 0.6
τm/T = 1
-6
10
0
5
10
15
z 0, dB
hold on; grid on; axis([0, inf, 10^(-6), 1]); xlabel(íz_0, dBí); ylabel(íP_Eí)
end
end
title([íP_E versus z_0 in dB for ndelta = í, num2str(delta)])
if L_taum == 1
legend([íntau_m/T = í,num2str(taum_T0)],3)
elseif L_taum == 2
legend([íntau_m/T = í,num2str(taum_T0(1))],[íntau_m/T = í,num2str(taum_T0(2))],3)
elseif L_taum == 3
legend([íntau_m/T = í,num2str(taum_T0(1))],[íntau_m/T = í,num2str(taum_T0(2))],
[íntau_m/T = í,num2str(taum_T0(3))],3)
elseif L_taum == 4
legend([íntau_m/T = í,num2str(taum_T0(1))],[íntau_m/T = í,num2str(taum_T0(2))],
[íntau_m/T = í,num2str(taum_T0(3))],[íntau_m/T = í,num2str(taum_T0(4))],3)
end
Problem 9.41
The program given in Problem 9.40 can be adapted for making the asked for plot.
38
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
Problem 9.42
The appropriate equations are:
BPSK:
(9.74) with m = 0 for nonfading; (9.205) for fading;
DPSK:
(9.112) for nonfading; (9.207) for fading;
CFSK:
(9.88) for nonfading; (9.206) for fading;
NFSK:
(9.122) for nonfading; (9.208) for fading.
Degradations for PE = 104 are:
Modulation type
BPSK
DPSK
Coh. FSK
Noncoh. FSK
Fading SNR, dB
33.99
37.00
37.00
40.01
Nonfaded SNR, dB
8.41
9.32
11.42
12.33
Fading Margin, dB
25.58
27.68
25.58
27.68
Problem 9.43
Comparing the exponents in (9.203) and (9.204) it is apparent that
1
1
1
Z
=1+
or 2 2w =
=
1
2
2 w
1+
Z
1+Z
Z
so that (9.203) becomes
PE =
=
=
p
Z
2 2w 1 exp w2 =2 2w
p
p
dw
2 2w
0
p
p
2 2 1
1
1
p w =
1 2 2w
2
2
2
0
1
s
1@
Z A
1
2
1+Z
1
2
Problem 9.44
The expressions with Q-functions can be integrated by parts as shown in the text (BPSK
and coherent FSK). The other two cases involve exponential integrals which are easily
integrated (DPSK and noncoherent FSK).
9.1. PROBLEM SOLUTIONS
Problem 9.45
The computer program given below is used for the solution.
clf
pc = input(íEnter sample values for channel pulse response (odd #): í);
L_pc = length(pc);
disp(íMaximum number of zeros each side of decision sample: í);
disp((L_pc-1)/4)
N = input(íEnter number of zeros each side of decision sample desired: í);
mid_samp = Öx(L_pc/2)+1
Pc = [];
for m = 1:2*N+1
row = [];
for n = 2-m:2*N+2-m
row = [row pc(mid_samp-1+n)];
end
Pc(:,m) = rowí;
end
disp(í í)
disp(íPc:í)
disp(í í)
disp(Pc)
Pc_inv = inv(Pc);
disp(í í)
disp(íInverse of Pc:í)
disp(í í)
disp(Pc_inv)
coef = Pc_inv(:,N+1);
disp(í í)
disp(íEqualizer coe¢cients:í)
disp(í í)
disp(coef)
disp(í í)
steps = (L_pc-length(coef))+1;
p_eq = [];
for m = mid_samp-áoor(steps/2):mid_samp+áoor(steps/2)
p_eq(m) = sum(coefí.*áiplr(pc(m-N:m+N)));
end
p_eq_p = p_eq(N+1:mid_samp+áoor(steps/2));
disp(í í)
disp(íEqualized pulse trainí)
39
40
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
disp(í í)
disp(p_eq_p)
disp(í í)
disp(íNoise enhancement factor, dBí)
N_enh = coefí*coef;
N_enh_dB = 10*log10(N_enh);
disp(N_enh_dB)
t=-5:.01:5;
y=zeros(size(t));
subplot(2,1,1),stem(pc),ylabel(íCh. output pulseí),...
axis([1 L_pc -1 1.5]),...
title([íCh. pulse samp. = [í,num2str(pc),í]í])
subplot(2,1,2),stem(p_eq_p),xlabel(íní),ylabel(íEqual. output pulseí),...
axis([1 L_pc -1 1.5]),...
title([íOutput equalized with í,num2str(N),í zero samples either sideí])
pause
% Equalizer freq response
%
clf
alpha = coefí;
T = 1;
L = length(alpha);
f = -0.5/T:.001:0.5/T;
%Efreq = [exp(j*2*pi*2*f); exp(j*2*pi*f); exp(j*2*pi*0*f); exp(-j*2*pi*f); exp(-j*2*pi*2*f)];
nn = -(L-1)/2:1:(L-1)/2;
Efreq = [];
for n = 1:L
Efreq = [Efreq; exp(j*2*pi*nn(n)*f)];
end
freq_resp = alpha*Efreq;
plot(f, abs(freq_resp)), xlabel(í{nitfT}í), ylabel(í{nitH}_e_q({nitf})í)
title([ínalpha = [í, num2str(alpha), í]í])
a. The given sample values give a singular sample matrix for a three-tap equalizer.
b. Equalizer coe¢cients:
1:7217
0:1565
2:0696
9.1. PROBLEM SOLUTIONS
Ch. output pulse
Ch. pulse samp. = [0
1.5
41
-0.11111
0.5
-1
0.5
-1
0.5
-0.11111
3
4
5
6
7
8
9
8
9
1
0.5
0
-0.5
-1
1
2
Output equalized with 2 zero samples either side
Equal. output pulse
1.5
1
0.5
0
-0.5
-1
1
2
3
4
5
n
6
7
0:1565
1:7217
Plots of the unequalized and equalized pulses are shown below:
The frequency response function and noise enhancement are given below:
Problem 9.46
a. The equalizer coe¢cients are:
1:3333
0:6667
0
b. The equalizer coe¢cients are:
Problem 9.47
3:7877
1:5498
3:4334
0:0064
3:7906
0]
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
Noise enhancement factor = 10.1118 dB
2
1
Eq. freq. response
0
-1
-2
-3
-4
-5
-6
-0.5
-0.4
Ch. output pulse
Ch. pulse samp. = [0
1.5
-0.3
-0.2
-0.1
0
fT
0.1
0.2
0.3
0.4
0.5
-0.11111
0.1
-1
0.5
-1
0.5
-0.11111
3
4
5
6
7
8
9
8
9
1
0.5
0
-0.5
-1
1
2
Output equalized with 1 zero samples either side
1.5
Equal. output pulse
42
1
0.5
0
-0.5
-1
1
2
3
4
5
n
6
7
0]
9.1. PROBLEM SOLUTIONS
43
Noise enhancement factor = 3.4679 dB
1
Eq. freq. response
0.5
0
-0.5
-1
-1.5
-2
-0.5
-0.4
Ch. output pulse
Ch. pulse samp. = [0
1.5
-0.3
-0.2
-0.1
0
fT
0.1
0.2
0.3
0.4
0.5
-0.11111
0.1
-1
0.5
-1
0.5
-0.11111
3
4
5
6
7
8
9
8
9
1
0.5
0
-0.5
-1
1
2
Output equalized with 2 zero samples either side
Equal. output pulse
1.5
1
0.5
0
-0.5
-1
1
2
3
4
5
n
6
7
0]
44
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
Noise enhancement factor = 16.3252 dB
12
10
Eq. freq. response
8
6
4
2
0
-2
-4
-6
-0.5
-0.4
-0.3
-0.2
-0.1
0
fT
0.1
0.2
0.3
0.4
0.5
a. The solution is similar to Example 9.12 except that
Snn (f ) =
N0 =2
1 + (f =f3 )2
so that
Rnn ( ) =
f3 N0
exp (2f3 j j)
2
The desired normalized autocorrelation matrix, with = Tm = T and f3 = 1=T , is
2
1 + b2 z + 2
N0 4
bz + 2 e2
[Ryy ] =
T
4
2e
b. The following matrix is needed:
2
2
bz + 22e
1 + b z + 2
bz + 2 e2
4
2e
2
bz
+
2e
2
1 + b z + 2
3 2
3
Ryd (T )
bA
[Ryd ] = 4 Ryd (0) 5 = 4 A 5
Ryd (T )
0
3
5
9.1. PROBLEM SOLUTIONS
45
Multiply the matrix given in (a) by the column matrix of unknown weights to get the
equations
2
32 0 3 2
3
4
1 + b2 z + 2 bz + 2e2
e
a1
bA
2
N0 4
bz + 2 e2
1 + b2 z + 2 bz + 2e2 5 4 a00 5 = 4 A 5
T
4
bz + 2 e2
1 + b2 z + 2
a01
0
2e
or
2
1 + b2 z + 2
4 bz + e2
2
4
2e
2
bz + 22e
1 + b z + 2
bz + 2 e2
4
2e
2
bz
+
2e
2
1 + b z + 2
32
3 2
3 2
3
a1
bA2 T =N0
bz
5 4 a0 5 = 4 A2 T =N0 5 = 4 z 5
a1
0
0
where the factor N0 =T has been normalized out so that the right hand side is in terms of
the signal-to-noise ratio, z (the extra factor of A needed mulitplies both sides of the matrix
equation and on the left hand side is lumped into the new coe¢cients a1 , a0 , and a1 .
Invert the matrix equation above to Önd the weights. A MATLAB program for doing so is
given below. For z = 10 dB and b = 0:5, the weights are a1 = 0:0773; a0 = 0:7821; and
a1 =.0:2781
c. According to (9.238) the minimum MSE is
Normalized minimum MSE
[Ryd ]T [A]opt
= E d2 (t)
(z is for normalization)
z2
3
a1
1
bz z 0 4 a0 5
= 1
z
a1
2
3
0:0783
1
0:5 10 10 0 4 0:7821 5
= 1
10
0:2781
= 0:1793
% Solution for Problem 9.47
%
z_dB = input(íEnter the signal-to-noise ratio in dB í);
z = 10^(z_dB/10);
b = input(íEnter multipath gain í);
R_yy(1,1)=(1+b^2)*z+pi/2;
46
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
R_yy(2,2)=R_yy(1,1);
R_yy(3,3)=R_yy(1,1);
R_yy(1,2)=b*z+(pi/2)*exp(-2*pi);
R_yy(1,3)=(pi/2)*exp(-4*pi);
R_yy(3,1)=R_yy(1,3);
R_yy(2,1)=R_yy(1,2);
R_yy(2,3)=R_yy(1,2);
R_yy(3,2)=R_yy(1,2);
disp(íR_yyí)
disp(R_yy)
B = inv(R_yy);
disp(íR_yy^-1í)
disp(B)
R_yd = [b*z z 0]í
A = B*R_yd
MSE = 1 - (1/z)*R_ydí*C
9.1. PROBLEM SOLUTIONS
47
Problem 9.48
In normalized form, from (9:245):
2
3
1 + 2 (2Eb =N0 ) + 1
2E
0
b =N0
5
[Ryy ] = 4
2Eb =N0
1 + 2 (2Eb =N0 ) + 1
2E
b =N0
2
0
2Eb =N0
1 + (2Eb =N0 ) + 1
2
3
2Eb =N0
4
2Eb =N0 5
[Ryd ] =
0
The steepest descent tap weight adjustment algorithm is
h
i
[A](k+1) = [A](k) + [Ryd ] [Ryy ] [A](k)
48
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
or
2
3
2 k 3
ak+1
a1
1
4 ak+1 5 = 4 ak0 5
0
ak1
ak+1
1
2
3
8
9
2Eb =N0
>
>
>
>
>
>
4 2Eb =N0 5
>
>
>
>
>
>
>
>
>
>
0
>
>
>
>
2
3
>
>
2
>
>
1
+
(2E
=N
)
+
1
2E
=N
0
<
=
0
0
b
b
2
4
5
+
2Eb =N0
1 + (2Eb =N0 ) + 1
2E
=N
0
b
>
>
>
>
0
2E
1 + 2 (2Eb =N0 ) + 1
>
>
0
b =N3
>
>
2
>
>
k
>
>
>
>
a
>
>
1
>
>
>
>
k
4
5
>
>
a0
>
>
:
;
k
a1
2 k 3
a1
= 4 ak0 5
ak1
2
3
8
9
2Eb =N0
>
>
>
>
>
>
4
5
>
>
2E
=N
>
>
0
b
>
>
<
=
0
2
3
+
k
1 + 2
(2Eb =N0) + 1 ak1 + (2E
>
>
b =N0 ) a0
>
>
>
2
k
k
> 4 (2Eb =N0 ) a + 1 + (2Eb =N0 ) + 1 a + (2Eb =N0 ) ak 5 >
>
>
>
1
0
1
>
>
k
:
;
2
k
(2Eb =N0 ) a0 + 1 + (2Eb =N0 ) + 1 a1
2
3
Eb
2 N
6 Eb0 7
= 4 2N
5
0
0
2 k 3
8
9
a1
>
>
>
>
>
>
4 ak0 5
>
>
>
>
>
>
<
=
k
a
1
2
3
+
k
1 + 2
(2Eb =N0) + 1 ak1 + (2E
>
>
>
>
kb =N0 ) a0
>
>
2
k
k 5 >
>
4
>
>
(2E
=N
)
a
+
1
+
(2E
=N
)
+
1
a
+
(2E
=N
)
a
0
0
b
b 0
b
1
0
1
>
>
:
;
2
k
k
(2Eb =N0 ) a0 + 1 + (2Eb =N0 ) + 1 a1
9.1. PROBLEM SOLUTIONS
49
The separate update equations are
ak+1
1
ak+1
0
ak+1
1
Eb
k
2 2Eb
= 2
1 a0 + 1 1 +
+ 1 ak1
N0
N0
2Eb
Eb
Eb k
= 2
+ 1 1 + 2
+ 1 ak0 2
a1 + ak1
N0
N0
N0
2Eb
Eb
=
1 1 + 2
+ 1 ak1 2
ak0
N0
N0
Problem 9.49
From Example 9.12, the autocorrelation matrix equation with numerical values substituted, from (9.246), is
2
3
26 10 0
Ryy = 4 10 26 10 5
0 10 26
Using MATLAB its eigenvalues are
1 = 11:86
2 = 26:00
3 = 40:14
An acceptable range for is therefore
0<<
2
= 0:0498
40:14
a. Equation (9.246) scaled by 10 is
2
32
3 2 3
2:6 1
0
c1
1
4 1 2:6 1 5 4 c0 5 = 4 2 5
0
1 2:6
c1
0
The inverse of the new Ryy is
2
3
0:4654 0:2101 0:0808
1
Ryy
= 4 0:2101 0:5462 0:2101 5
0:0808 0:2101 0:4654
50
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
The recomputed weights are
2
3 2
32 3 2
3
c1
0:4654 0:2101 0:0808
1
0:0452
4 c0 5 = 4 0:2101 0:5462 0:2101 5 4 2 5 = 4 0:8824 5
c1
0:0808 0:2101 0:4654
0
0:3394
which are the same as obtained in Example 9.12.
b. The eigenvalues of the scaled Ryy are 1.1858, 2.6000, and 4.0142 which are smaller by
a factor of 10 than those found ifor the unscaled case.
Problem 9.50
The MATLAB program below computes the coe¢cients among other quantities.
output for an SNR of 20 dB and = 0:1 is also given.
% Solution for Problem 9.50
%
z_dB = input(íEnter the signal-to-noise ratio in dB í);
z = 10^(z_dB/10);
beta = input(íEnter multipath gain í);
R_yy(1,1)=(1+beta^2)*2*z+1;
R_yy(2,2)=R_yy(1,1);
R_yy(3,3)=R_yy(1,1);
R_yy(1,2)=2*beta*z;
R_yy(1,3)=0;
R_yy(3,1)=R_yy(1,3);
R_yy(2,1)=R_yy(1,2);
R_yy(2,3)=R_yy(1,2);
R_yy(3,2)=R_yy(1,2);
disp(íR_yyí)
disp(R_yy)
B = inv(R_yy);
disp(íR_yy^-1í)
disp(B)
R_yd = [b*z z 0]í
C = B*R_yd
Eigen = eig(R_yy)
mu_max = 2/max(Eigen)
>> pr9_50
Enter SNR in dB => 13.01
Its
9.2. COMPUTER EXERCISES
Enter multipath gain, beta => 0.1
R_yy
41.3972
3.9997
0
3.9997
41.3972
3.9997
0
3.9997
41.3972
R_yy^-1
0.0244
-0.0024
0.0002
-0.0024
0.0246
-0.0024
0.0002
-0.0024
0.0244
R_yd =
1.9999
19.9986
0
C=
0.0012
0.4875
-0.0471
Eigen =
35.7407
41.3972
47.0537
mu_max =
0.0425
9.2
Computer Exercises
Computer Exercise 9.1
% ce9_1.m: Simulation of BEP for antipodal signaling
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
clear all; clf
n_sim = input(íEnter number of bits for simulation => í);
zdB = input(íEnter z = A^2*T/N_0 vector in dB => í);
z = 10.^(zdB/10);
Lz = length(z);
A = 1;
T = 1;
51
52
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
N0 = (A^2*T)./z;
S = A*T*sign(rand(1,n_sim)-.5);
PE = [];
ET = [];
for k = 1:Lz
N = sqrt(0.5*N0(k)*T)*randn(1,n_sim);
Y = S+N;
Z = sign(Y);
V = 0.5*(1+Z);
W = 0.5*(1+sign(S));
E = xor(V,W);
ET(k) = sum(E);
PE(k) = ET(k)/length(Y);
end
PEth = 0.5*erfc(sqrt(z));
disp(í í)
disp(í SNR, dB Errors P_E est. P_E theoryí)
disp(í ________________________________________í)
disp(í í)
disp([zdBí ETí PEí PEthí])
disp(í í)
semilogy(zdB, PE, íoí), xlabel(íz in dBí), ylabel(íP_Eí), grid
hold on
semilogy(zdB, PEth)
title([íBER simulated with í, num2str(n_sim), í bits per SNR value í])
legend([íSimulatedí], [íTheoreticalí])
%End of script Öle
A typical run is given below:
>> ce9_1
Enter number of bits for simulation => 20000
Enter z = A^2*T/N_0 vector in dB => [2 4 6 8]
SNR, dB
Errors
P_E est.
P_E theory
______________________________________
2.0000
704.0000
0.0352
0.0375
4.0000
238.0000
0.0119
0.0125
6.0000
44.0000
0.0022
0.0024
8.0000
4.0000
0.0002
0.0002
9.2. COMPUTER EXERCISES
53
A plot also appears showing the estimated and theoretical probabilities of error plotted
versus signal-to-noise ratio.
54
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
Computer Exercise 9.2
% ce9_2: Compute degradation in SNR due to bit timing error
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
clear all; clf
A = char(í-í,í-.í,í:í,íñí);
for m = 1:4
PE0 = 10^(-m);
delT = [];
deg = [];
Eb1 = [];
k = 0;
for Y = 0:.025:.45;
k = k+1;
delT(k) = Y;
Eb1(k) = fzero(@PE1, 2, [], delT(k), PE0);
end
Eb = fzero(@PE, 2, [], PE0);
deg = Eb1 - Eb;
plot(delT, deg, A(m,:)), ...
if m == 1
hold on
axis([0 .45 0 20])
xlabel(íFractional timing error, nDeltaT/Tí), ylabel(íDegradation in dBí),...
title([íDegradation in SNR for bit synchronization timing errorí]), grid
end
end
legend([íP_E = 10^-^1í], [í = 10^-^2í], [í = 10^-^3í], [í = 10^-^4í], 2),
%End of script Öle
The output plot is shown in Fig. 9.7.
9.2. COMPUTER EXERCISES
55
Degradation in SNR for bit synchronization timing error
20
PE = 10-1
18
Degradation in dB
= 10-2
16
= 10-3
14
= 10-4
12
10
8
6
4
2
0
0
0.05
0.1
0.15
0.2
0.25
0.3
Fractional timing error, ∆ T/T
0.35
0.4
0.45
56
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
Computer Exercise 9.3
%
ce9_3.m; Evaluates e§ect of Gaussian
% phase error in BPSK; Öxed variance.
%
%
Uses subprogram pb_phase_pdf
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
clear all; clf
sigma_psi2 = input(íEnter 3-vector of variances of phase error in radians^2 => í);
L_sig = length(sigma_psi2);
Eb_N0_dB_min = input(íEnter minimum Eb/N0 in dB desired => í);
Eb_N0_dB_max = input(íEnter maximum Eb/N0 in dB desired => í);
A = char(í-.í,í:í,íñí,í-..í);
a_mod = 0;
for m = 1:L_sig
Eb_N0_dB = [];
Eb_N0 = [];
PE = [];
k = 1;
for snr = Eb_N0_dB_min:.5:Eb_N0_dB_max
Eb_N0_dB(k) = snr;
Eb_N0(k) = 10.^(Eb_N0_dB(k)/10);
sigma_psi = sqrt(sigma_psi2(m));
PE(k) = quadl(ípb_phase_pdfí,-pi,pi,[],[],Eb_N0(k),sigma_psi,a_mod);
k = k+1;
end
semilogy(Eb_N0_dB, PE,A(m,:)),xlabel(íE_b/N_0, dBí),ylabel(íP_Eí),...
axis([Eb_N0_dB_min Eb_N0_dB_max 10^(-7) 1]),...
if m == 1
hold on
grid on
end
end
P_ideal = .5*erfc(sqrt(Eb_N0));
semilogy(Eb_N0_dB, P_ideal)
legend([ínsigma_ntheta_e^2 = í,num2str(sigma_psi2(1))],[ínsigma_ntheta_e^2 = í,num2str(sigma_psi2(2
= í,num2str(sigma_psi2(3))],[ínsigma_ntheta_e^2 = 0í],3)
title(íE§ect of Gaussian phase error on detection of BPSKí)
9.2. COMPUTER EXERCISES
57
Effect of Gaussian phase error on detection of BPSK
0
10
-1
10
-2
10
-3
PE
10
-4
10
2
σ θe = 0.05
-5
10
2
σ θe = 0.1
2
-6
σ θe = 0.5
10
2
σ θe = 0
-7
10
2
3
4
5
6
Eb/N0, dB
7
8
9
10
function XX = pb_phase_pdf(psi,Eb_N0,sigma_psi,a)
arg = Eb_N0*(1-a^2)*(cos(psi)-a/sqrt(1-a^2)*sin(psi)).^2;
T1 = .5*erfc(sqrt(arg));
T2 = exp(-psi.^2/(2*sigma_psi^2))/sqrt(2*pi*sigma_psi^2);
XX = T1.*T2;
%End of script Öle
A plot from the program is given in Fig. 9.8.
Computer Exercise 9.4
% ce9_4: For a given data rate and error probability, Önd the required
% bandwidth and Eb/N0 in dB; baseband, BPSK/DPSK, coherent FSK
% and noncoherent FSK modulation may be speciÖed.
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
clear all; clf
I_mod = input(íEnter desired modulation: 0 = baseband; 1 = BPSK; 2 = DPSK; 3 =
CFSK; 4 = NCFSK: í);
58
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
I_BW_R = input(íEnter 1 if bandwidth speciÖed; 2 if data rate speciÖed: í);
if I_BW_R == 1
BkHz = input(íEnter vector of desired bandwidths in kHz => í);
elseif I_BW_R == 2
Rkbps = input(íEnter vector of desired data rates in kbps => í);
end
PE = input(íEnter desired BEP => í);
N0 = 1;
if I_BW_R == 1
if I_mod == 0
Rkpbs = BkHz; % Rate for baseband
elseif I_mod == 1 j I_mod == 2
Rkbps = BkHz/2; % Rate in kbps for BPSK or DPSK
elseif I_mod == 3
Rkbps = BkHz/3; % Rate in kbps for CFSK
elseif I_mod == 4
Rkbps = BkHz/4; % Rate in kbps for NFSK
end
elseif I_BW_R == 2
if I_mod == 0
BkHz = Rkbps; % Transmission bandwidth for baseband in kHz
elseif I_mod == 1 j I_mod == 2
BkHz = 2*Rkbps; % Transmission bandwidth for BPSK or DPSK in kHz
elseif I_mod == 3
BkHz = 3*Rkbps; % Transmission bandwidth for CFSK in kHz
elseif I_mod == 4
BkHz = 4*Rkbps; % Transmission bandwidth for NFSK in kHz
end
end
R = Rkbps*1000;
B = BkHz*1000;
% PE = 0.5*erfc(sqrt(z))
if I_mod == 0 j I_mod == 1
sqrt_z = erÖnv(1-2*PE);
z = sqrt_z^2;
elseif I_mod == 2
z = -log(2*PE);
elseif I_mod == 3
sqrt_z_over_2 = erÖnv(1-2*PE);
z = 2*sqrt_z_over_2^2;
9.2. COMPUTER EXERCISES
59
elseif I_mod == 4
z = -2*log(2*PE);
end
Eb_N0_dB = 10*log10(z); % This is the required Eb/N0 in dB
if I_mod == 0
A = sqrt(R*N0*z); % z = A^2/(R*N0) for baseband
elseif I_mod == 1 j I_mod == 2 j I_mod == 3 j I_mod == 4
A = sqrt(2*R*N0*z); % z = A^2/2*R*N0 for BPSK, DPSK, CFSK, NFSK; A = modulated carrier amplitude
end
disp(í í)
disp(íDesired P_E and required E_b/N_0 in dB: í)
format long
disp([PE Eb_N0_dB])
format short
disp(í í)
disp(í R, kbps BW, Hz A, voltsí)
disp(í ___________________________í)
disp(í í)
disp([Rkbpsí BkHzí Aí])
disp(í í)
%End of script Öle
60
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
A typical run is given below:
>> ce9_4
Enter desired modulation: 0 = baseband; 1 = BPSK; 2 = DPSK;
3 = CFSK; 4 = NCFSK: 2
Enter 1 if bandwidth speciÖed; 2 if data rate speciÖed: 2
Enter vector of desired data rates in kbps => [5 10 50 100]
Enter desired BEP => 1e-6
Desired P_E and required E_b/N_0 in dB:
0.000001000000000 11.180120598361940
R, kbps
BW, Hz
A, volts
___________________________
1.0e+003 *
0.0050
0.0100
0.3622
0.0100
0.0200
0.5123
0.0500
0.1000
1.1455
0.1000
0.2000
1.6200
Computer Exercise 9.5
% Öle: ce9_5.m
% Simulation of suboptimum bandpass Ölter/delay-and-multiply demodulator
% with integrate-and-dump detection for DPSK; input bandpass Ölter is
% Butterworth with selectable bandwidth-bit period product
%
% User deÖned subprograms di§_enc( ) and int_and_dump( ) called
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
clear all; clf
Eb_N0_dB_max = input(íEnter maximum Eb/N0 in dB í);
Eb_N0_dB_min = input(íEnter minimum Eb/N0 in dB í);
samp_bit = input(íEnter number of samples per bit used in simulation í);
n_order = input(íEnter order of Butterworth detection Ölter í);
BWT_bit = input(íEnter Ölter bandwidth normalized by bit rate í);
N_bits = input(íEnter total number of bits in simulation í)
clf
ss = sign(rand(1,N_bits)-.5);
% Generate sequence of random +-1s
data = 0.5*(ss+1);
% Logical data is sequence of 1s and
0s
data_di§_enc = di§_enc(data);
% Di§erentially encode data for DPSK
9.2. COMPUTER EXERCISES
61
s = 2*data_di§_enc-1;
% Generate bipolar data for modulation
T_bit = 1;
% Arbitrarily take bit time as 1
second
BW = BWT_bit/T_bit;
% Compute Ölter bandwidth
from BW*T_bit
[num,den] = butter(n_order,2*BW/samp_bit);
% Obtain Ölter num/den coe¢cients
Eb_N0_dB = [];
% Ensure that plotting arrays are
empty
Perror = [];
k = 0;
for kk = Eb_N0_dB_min:Eb_N0_dB_max
% Loop simulates detector for
each desired Eb/N0
k = k+1;
Eb_N0_dB(k) = kk;
Eb_N0 = 10^(Eb_N0_dB(k)/10);
% Convert desired Eb/N0 from dB to
ratio
Eb = T_bit;
% Bit energy is T_bit, assuming
ampl = 1
N0 = Eb/Eb_N0;
% Compute noise spectral density from Eb/N0
del_t = T_bit/samp_bit;
% Compute sampling interval
sigma_n = sqrt(N0/(2*del_t));
% Compute standard deviations of
noise samples
t=0:del_t:N_bits*T_bit-del_t;
% Set up time axis
sig = [];
sig = s(ones(samp_bit,1),:);
% Build array whose columns are samp_bit
long
sig = sig(:);
% Convert matrix where bit samples
occupy columns to vector
zi = [];
% Make sure various arrays are empty
bits_out = [];
y = [];
y_det = [];
y_ref = [];
noise = sigma_n*randn(size(sig));
% Form sequence of Gaussian noise samples
[y,zf] = Ölter(num,den,sig+noise,zi);
% Filter signal plus noise with chosen Ölter
zi = zf;
% Save Önal values for future initial
conditions in a block simulation
n_tot = length(y);
62
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
y_ref = [y(n_tot-samp_bit+1:n_tot,1); y(1:n_tot-samp_bit,1)]; % Reference signal for multiply is 1-bit delayed signal + noise
y_mult = y.*y_ref;
% Multiply received signal plus noise
with 1 bit delayed version
bits_out = int_and_dump(y_mult,samp_bit,N_bits+1); % di§_enc appends a
1 bit to start of bit stream
bits_out = bits_out(2:N_bits+1);
error_array = abs(bits_out-data);
% Compare detected bits with actual input
data stream; errors = 1s
error_array(1:5) = 0;
% Donít include Örst 5 bits due to transients
ss = sum(error_array);
% Sum to get total number of errors
Perror(k) = ss/(N_bits-5);
% Subtract 5 from denominator to account
for initial errors set = 0
end
% End of Eb/N0 loop
disp(íE_b/N_0, dB; P_Eí)
% Display computed probability of error values versus Eb/N0
disp([Eb_N0_dBí Perrorí])
% Plot simulated bit error probabilities versus Eb/N0
semilogy(Eb_N0_dB, Perror,íñí), grid, xlabel(íE_b/N_0; dBí), ylabel(íP_Eí), hold
title(íSimulation of BEP for delay-and-multiply detector with Butterworth preÖlter for
DPSKí)
% Plot theoretical bit error probability for optimum DPSK detector
semilogy(Eb_N0_dB, 0.5*exp(-10.^(Eb_N0_dB/10)))
% Plot approximate theoretical result for suboptimum detector
semilogy(Eb_N0_dB, qfn(sqrt(10.^(Eb_N0_dB/10))),í-.í)
legend([íSimulation; BT = í, num2str(BWT_bit),í; í, num2str(N_bits),í bitsí],íTheory;
optimum di§erential detectorí,íTheory; delay/multiply detectorí, 3)
% Functions
output = di§_enc(input)
L_in = length(input)
output = [];
for k==1:L_in
if k==1
output(k)=not(bitxor(input(k),1));
else
output(k)=not(bitxor(input(k),output(k-1)));
end
end
function y_out=int_and_dump(input, samp_bit, N_bits)
9.2. COMPUTER EXERCISES
63
samp_array=reshape(input, samp_bit, N_bits) % Reshape input array so that bits
occupy columns
integrate=sum(samp_array);
bits_out=(sign(integrate)+1/2)/2;
% End of script Öle
Computer Exercise 9.6
% ce9_6: program to plot Figure 9-28
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
clf
z0dB = 0:1:30;
z0 = 10.^(z0dB/10);
for delta = -0.9:0.3:0.6;
if delta > -eps & delta < eps
delta = 0;
end
for taum_over_T = 0:1:1
T1 = 0.5*qfn(sqrt(2*z0)*(1+delta));
T2 = 0.5*qfn(sqrt(2*z0)*(1+delta)-2*delta*taum_over_T);
PE = T1 + T2;
if taum_over_T == 0
semilogy(z0dB, PE, í-í)
elseif taum_over_T == 1
semilogy(z0dB, PE, íñí)
end
if delta <= -.3
text(z0dB(10), PE(10), [índelta = í, num2str(delta)])
elseif delta > -0.3 & delta <= 0.3
text(z0dB(8), PE(8), [índelta = í, num2str(delta)])
elseif delta > 0.3
text(z0dB(5), PE(5), [índelta = í, num2str(delta)])
end
if delta == -0.9
hold on
ylabel(íP_Eí), xlabel(íz_0 in dBí)
axis([0 30 1E-5 1])
grid
64
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
Bit error probability for two-ray multipath channel
0
10
τm/T = 0
δ = -0.9
δ = -0.9
-1
τm/T = 1
10
δ = -0.6
δ = -0.6
-2
10
PE
δ = 0.6
-3
10
δ = -0.3
δ = -0.3
δ=0
δ = 0.6
-4
10
δ = 0.3
δ = 0.3
0
5
10
15
z 0 in dB
end
end
end
legend([íntau_m/T = 0í],[íntau_m/T = 1í])
The output plot is shown in Fig. 9.9.
20
25
30
9.2. COMPUTER EXERCISES
65
Computer Exercise 9.7
Use Equations (9.74) (with m = 0), (9.88), (9.112), (9.122), and (9.205) - (9.208). Solve
them for the SNR in terms of the error probability. Express the SNR in dB and subtract
the nonfading result from the fading result. The MATLAB program below does this for
the modulation cases of BPSK, coherent FSK (CFSK), DPSK, and noncoherent (NCFSK).
% ce9_7.m: Program to evaluate degradation due to áat
% Rayleigh fading for various modulation schemes
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
mod_type = input(íEnter type of modulation: 1 = BPSK; 2 = CFSK; 3 = DPSK; 4 =
NCFSK: í);
PE = input(íEnter vector of desired probabilities of error: í);
disp(í í)
if mod_type == 1
disp(í BPSKí)
Z_bar = (0.25*(1 - 2*PE).^2)./(PE - PE.^2);
z = (erÖnv(1 - 2*PE)).^2; % MATLAB has an inv. error function
elseif mod_type == 2
disp(í CFSKí)
Z_bar = (0.5*(1 - 2*PE).^2)./(PE - PE.^2);
z = 2*(erÖnv(1 - 2*PE)).^2;
elseif mod_type == 3
disp(í DPSKí)
Z_bar = 1./(2*PE) -1;
z = -log(2*PE);
elseif mod_type == 4
disp(í NCFSKí)
Z_bar = 1./PE -2;
z = -2*log(2*PE);
end
Z_bar_dB = 10*log10(Z_bar);
z_dB = 10*log10(z);
deg_dB = Z_bar_dB - z_dB;
disp(í í)
disp(í P_E Degradation, dBí)
disp(í_________________________í)
disp(í í)
disp([PEí deg_dBí])
66
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
disp(í í)
%End of script Öle
A typical run is given below:
>> ce9_7
Enter type of modulation: 1 = BPSK; 2 = CFSK; 3 = DPSK; 4 = NCFSK: 3
Enter vector of desired probabilities of error: [1e-2 1e-3 1e-4]
DPSK
P_E Degradation, dB
_________________________
0.0100
10.9779
0.0010
19.0469
0.0001
27.6859
Computer Exercise 9.8
The solution to this exercise consists of two MATLAB programs, one for the zero-forcing
case and one for the MMSE case.
% ce9_8a.m: Plots unequalized and equalized sample values for ZF equalizer. The
MATLAB program for the MMSE case is ce9_9b.
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%:
%
clf
pc = input(íEnter sample values for channel pulse response (odd #): í);
L_pc = length(pc);
disp(íMaximum number of zeros each side of decision sample: í);
disp((L_pc-1)/4)
N = input(íEnter number of zeros each side of decision sample desired: í);
mid_samp = Öx(L_pc/2)+1;
Pc = [];
for m = 1:2*N+1
row = [];
for n = 2-m:2*N+2-m
row = [row pc(mid_samp-1+n)];
end
Pc(:,m) = rowí;
end
disp(í í)
9.2. COMPUTER EXERCISES
disp(íPc:í)
disp(í í)
disp(Pc)
Pc_inv = inv(Pc);
disp(í í)
disp(íInverse of Pc:í)
disp(í í)
disp(Pc_inv)
coef = Pc_inv(:,N+1);
disp(í í)
disp(íEqualizer coe¢cients:í)
disp(í í)
disp(coef)
disp(í í)
steps = (L_pc-length(coef))+1;
p_eq = [];
for m = mid_samp-áoor(steps/2):mid_samp+áoor(steps/2)
p_eq(m) = sum(coefí.*áiplr(pc(m-N:m+N)));
end
p_eq_p = p_eq(N+1:mid_samp+áoor(steps/2));
disp(í í)
disp(íEqualized pulse trainí)
disp(í í)
disp(p_eq_p)
disp(í í)
t=-5:.01:5;
y=zeros(size(t));
subplot(2,1,1),stem(pc),ylabel(íCh. output pulseí),...
axis([1 L_pc -.5 1.5]),...
title([íCh. pulse samp. = [í,num2str(pc),í]í])
subplot(2,1,2),stem(p_eq_p),xlabel(íní),ylabel(íEqual. output pulseí),...
axis([1 L_pc -.5 1.5]),...
title([íOutput equalized with í,num2str(N),í zero samples either sideí])
pause
clf
n = [-N:1:N]í;
fT = -0.5:0.01:0.5;
Heq = coefí*exp(-j*2*pi*n*fT);
sigmaN2 = coefí*coef
plot(fT, Heq), xlabel(ífTí), ylabel(íEq. freq. responseí), ...
67
68
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
title([íNoise enhancement factor = í, num2str(10*log10(sigmaN2)), í dBí])
%End of script Öle
A typical run is given below:
>> ce9_8a
Enter sample values for channel pulse response (odd #): [.01 -.05 .2 -.3 1 -.2 .15 -.25
.07]
Maximum number of zeros each side of decision sample:
2
Enter number of zeros each side of decision sample desired: 2
mid_samp =
5
Pc:
1.0000
-0.3000
0.2000
-0.0500
0.0100
-0.2000
1.0000
-0.3000
0.2000
-0.0500
0.1500
-0.2000
1.0000
-0.3000
0.2000
-0.2500
0.1500
-0.2000
1.0000
-0.3000
0.0700
-0.2500
0.1500
-0.2000
1.0000
Inverse of Pc:
1.0688
0.3050
-0.1338
-0.0441
0.0181
0.1482
1.1107
0.2824
-0.1388
-0.0441
-0.0631
0.1309
1.1285
0.2824
-0.1338
0.2381
0.0073
0.1309
1.1107
0.3050
0.0193
0.2381
-0.0631
0.1482
1.0688
Equalizer coe¢cients:
-0.1338
0.2824
1.1285
0.1309
-0.0631
Equalized pulse train
-0.0000
-0.0000
1.0000
0.0000
0.0000
%The MATLAB program for the MMSE case is given below:
% ce9.8b
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
z_dB = input(íEnter the signal-to-noise ratio in dB: í);
z = 10^(z_dB/10);
9.2. COMPUTER EXERCISES
b = input(íEnter multipath gain: í);
R_yy(1,1)=(1+b^2)*z+pi/2;
R_yy(2,2)=R_yy(1,1);
R_yy(3,3)=R_yy(1,1);
R_yy(1,2)=b*z+(pi/2)*exp(-2*pi);
R_yy(1,3)=(pi/2)*exp(-4*pi);
R_yy(3,1)=R_yy(1,3);
R_yy(2,1)=R_yy(1,2);
R_yy(2,3)=R_yy(1,2);
R_yy(3,2)=R_yy(1,2);
disp(í í)
disp(íR_yy:í)
disp(í í)
disp(R_yy)
B = inv(R_yy);
disp(í í)
disp(íR_yy^-1:í)
disp(í í)
disp(B)
R_yd = [0 z b*z]í;
disp(í í)
disp(íR_yd:í)
disp(í í)
disp(R_yd)
A = B*R_yd;
disp(í í)
disp(íOptimum coe¢cients:í)
disp(A)
disp(í í)
%End of script Öle
A typical run is given below:
>> ce9_8b
Enter the signal-to-noise ratio in dB: 7
Enter multipath gain: 0.6
R_yy:
8.3869
3.0101
0.0000
3.0101
8.3869
3.0101
0.0000
3.0101
8.3869
R_yy^-1:
69
70
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
0.1399
-0.0576
-0.0576
0.1606
0.0207
-0.0576
R_yd:
0
5.0119
3.0071
Optimum coe¢cients:
-0.2267
0.6316
0.1319
0.0207
-0.0576
0.1399
Computer Exercise 9.9
% ce9_9.m: Simulation of multipath channel and adaptive MMSE equalizer
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
clear all; clf
Eb_N0_dB0 = input(íEnter starting Eb/N0 in dB => í);
%beta = input(íEnter multipath constant, beta => í);
N_bits = input(íEnter number of bits in simulation => í);
N_bits_train = input(íEnter number of training bits => í);
mu = input(íEnter feedback constant, mu => í);
samp_bit = 10; T = 1;
A = 1; beta = 0.5; delta = T;
delt = T/samp_bit;
Eb = A^2*T;
kk = 0;
for Eb_N0_dB = Eb_N0_dB0:.5:Eb_N0_dB0+12
kk = kk + 1;
Eb_N0plt(kk) = Eb_N0_dB;
Eb_N0 = 10.^(Eb_N0_dB/10);
N0 = Eb/Eb_N0;
sigma_n = sqrt(N0/(2*delt));
data = zeros(size(N_bits));
data_10 = zeros(size(N_bits));
d = zeros(size(samp_bit, N_bits));
y = zeros(1, N_bits*samp_bit);
data = sign(rand(1, N_bits)-0.5);
data_10 = 0.5*(data +1);
9.2. COMPUTER EXERCISES
d = data(ones(samp_bit, 1), :);
d = d(:)í;
t = 0:delt:N_bits*T-delt;
tau = length(0:delt:delta);
y = A*d + beta*delay1(d, tau);
noise = sigma_n*randn(size(y));
z = y + noise;
bits_out = int_and_dump(z, samp_bit, N_bits);
bits_outs = 2*bits_out - 1;
bits = bits_outs(ones(samp_bit, 1), :);
bits = bits(:)í;
error_array = abs(bits_out - data_10);
error_array(1:10) = 0;
errors = sum(error_array);
PE_ue(kk) = errors/(N_bits - 10);
% Equalizer
z_eq = []; zd = []; zdd = []; a1 = []; a2 = []; a3 = [];
epsilon = [];
a1(1) = 1; a2(1) = 1; a3(1) = 1;
zd = delay1(z, tau);
zdd = delay1(z, 2*tau);
z_eq(1) = a1(1)*z(1) + a2(1)*zd(1) + a3(1)*zdd(1);
epsilon(1) = z_eq(1) - d(1);
for k = 1:samp_bit*N_bits-1
a1(k+1) = a1(k) - mu*z(k)*epsilon(k);
a2(k+1) = a2(k) - mu*zd(k)*epsilon(k);
a3(k+1) = a3(k) - mu*zdd(k)*epsilon(k);
z_eq(k+1) = a1(k+1)*z(k+1) + a2(k+1)*zd(k+1) + a3(k+1)*zdd(k+1);
if k <= samp_bit*N_bits_train-1
epsilon(k+1) = z_eq(k+1) - d(k+1);
elseif k > samp_bit*N_bits_train-1
epsilon(k+1) = z_eq(k+1) - bits(k+1);
end
end
bits_out_eq = int_and_dump(z_eq, samp_bit, N_bits);
error_array_eq = abs(bits_out_eq - data_10);
error_array_eq(1:10) = 0;
errors_eq = sum(error_array_eq);
PE_e(kk) = errors_eq/(N_bits - 10);
PE_gauss(kk) = qfn(sqrt(2*Eb_N0));
71
72
CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE
aa(:, kk) = [a1(k) a2(k) a3(k)]
end
semilogy(Eb_N0plt, PE_ue, íñí, Eb_N0plt, PE_e, í-.í, Eb_N0plt, PE_gauss, í-í), grid,
...
xlabel(í{nitE_b/N}_0, dBí), ylabel(í{nitP_b}í), axis([min(Eb_N0plt) max(Eb_N0plt)
1e-6 1]), ...
legend([íUnequalized; í, num2str(N_bits), í bitsí], [íEqualized, nmu = í, num2str(mu)],
íGauss noise, theoryí, 1)
pause %Enter to view tap weights
%subplot(6,1,1), plot(d), axis([0 500 -1.1 1.1]), ylabel(íin dataí), ...
% legend([íT_b_i_t = í, num2str(T),í; No. train bits = í, num2str(N_bits_train)])
%subplot(6,1,2), plot(bits_out_eq), axis([0 500 -1.1 1.1]), ylabel(íeq_dataí)
subplot(4,1,1), plot(t, epsilon), axis([0 200 -2.1 2.1]), ylabel(íerrorí)
subplot(4,1,2), plot(t, a1), axis([0 200 -1.1 1.1]), ylabel(ía_1í)
subplot(4,1,3), plot(t, a2), axis([0 200 -1.1 1.1]), ylabel(ía_2í)
subplot(4,1,4), plot(t, a3), axis([0 200 -1.1 1.1]), ylabel(ía_3í), xlabel(ítí)
>> ce9_9
Enter starting Eb/N0 in dB => 4
Enter number of bits in simulation => 10000
Enter number of training bits => 500
Enter feedback constant, mu => .01
9.2. COMPUTER EXERCISES
73
0
10
Unequalized; 10000 bits
Equalized, µ = 0.01
Gauss noise, theory
-1
10
-2
Pb
10
-3
10
-4
10
-5
10
-6
10
4
6
8
10
Eb /N0, dB
12
14
16
error
2
0
-2
0
20
40
60
80
100
120
140
160
180
200
0
20
40
60
80
100
120
140
160
180
200
0
20
40
60
80
100
120
140
160
180
200
0
20
40
60
80
100
t
120
140
160
180
200
a1
1
0
-1
a2
1
0
-1
a3
1
0
-1
Chapter 10
Advanced Data Communications
Topics
10.1
Problem Solutions
Problem 10.1
Use the relationship
Rb = (log2 M ) Rs bps
where Rb is the data rate, Rs is the symbol rate, and M is the number of possible signals
per signaling interval. In this case, Rs = 2; 000 symbols per second. For M = 4, Rb =
2 2; 000 = 4; 000 bps, for M = 8, Rb = 6; 000 bps, for M = 16, Rb = 8; 000 bps, for
M = 32, Rb = 10; 000 bps, and for M = 64, Rb = 12; 000 bps. The bit rare is linear as a
function of log2 M:
Problem 10.2
a. 5,000 symbols per second.
b. Recall that
y (t) = A [d1 (t) cos (! c t) d2 (t) sin (! c t)] =
p
1
2A cos [! c t + i (t)] ; i (t) = tan
d2 (t)
d1 (t)
Alternating every other bit of the given data sequence, 110110 010111 011011, between
d1 (t) and d2 (t) gives d1 (t) = 1; 1; 1; 1; 1; 1; 1; 1; 1 and d2 (t) = 1; 1; 1; 1; 1; 1; 1; 1; 1
which results in the table below for the phases:
d2 (t) 1
1
1 1
1
1 1
1 1
d1 (t)
1 1 1
1 1 1 1 1
1
3
7
3
3
3
7
i (t)
4
4
4
4
4
4
4
4
4
1
2
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
For QPSK the symbol switching points occur each Ts = 2Tb seconds.
c. Now the switching instants are each Tb seconds. Start with d1 (t) = 1. Then d2 (t)
is staggered, or o§set, by Tb seconds. So the Örst phase shift is for d1 (t) = 1 and
d2 (t) = 1 or 1 (t) = =4: After Ts = 2Tb seconds d1 (t) takes a 1-value, but d2 (t)
is still 1, so 2 (t) = 3=4: At 3Tb seconds, d2 (t) takes on a second 1-value while
d1 (t) = 1, so 3 (t) = 3=4: At 4Tb seconds, d1 t) = 1, so 4 (t) = =4, etc. The
phase sequence is shown in the table below:
d2 (t)
d1 (t)
i (t)
1
1
1
4
3
4
1
1
1
4
3
4
7
4
1
5
4
1
Problem 10.3
For QPSK,
PE; symbol = 2Q
3
4
1
3
4
1
1
1
1
3
4
4
4
3
4
1
1
1
3
4
4
7
4
1
1
7
4
p
Es =N0 = 104
Trial and error using the asymptotic approximation for the Q-function gives Q (x) = 5
105 for x = 3:89 or [Es =N0 ]reqíd 3:892 = 15:14 = 11:8 dB. If the quadrature-carrier
amplitudes are A, then the amplitude of the envelope-phase-modulated
q form of the carrier
p
p 2
2
is 2A, and Es =N0 =
2A T = (2N0 ) = A =N0 R: Hence, Areqíd = N0 R [Es =N0 ]reqíd =
p
p
(1011 ) (15:14) R = 1:23 105 R: The answers are as follows: (a) 8:7 104 V; (b)
0.0012 V; (c) 0.0028 V; (d) 0.0039 V; (e) 0.0087 V; (f) 0.012 V.
Problem 10.4
Take the expectation of the product after noting that the average values are 0 because
E[n (t)] = 0:
Z Ts
Z Ts
E [N1 N2 ] = E
n (t) cos (! c t) dt
n (t) sin (! c t) dt
0
=
Z Ts Z Ts
0
=
=
=
E [n (t) n ()] cos (! c t) sin (! c ) dtd
0
Z Ts Z Ts
0
0
0
Z Ts
N0
(t ) cos (! c t) sin (! c ) dtd
2
N0
cos (! c t) sin (! c t) dt
2 0
Z
N0 Ts
sin (2! c t) dt = 0
4 0
4
10.1. PROBLEM SOLUTIONS
3
Problem 10.5
a. The integrator outputs, without noise, are (the signs are determined by the data
polarities)
V10 =
=
=
=
V20 =
=
=
=
Z Ts
xc (t) cos (2fc t) dt
Z Ts
A
cos 2fc t +
sin 2fc t
cos (2fc t) dt
2
2
0
Z
A Ts
cos
sin
+ dbl freq terms dt
2 0
2
2
ATs
cos
sin
2
2
2
Z Ts
xc (t) sin (2fc t) dt
0
Z Ts
A
cos 2fc t +
sin 2fc t
sin (2fc t) dt
2
2
0
Z
A Ts
sin
cos
+ dbl freq terms dt
2 0
2
2
ATs
sin
cos
2
2
2
0
b. This follows from the outline given in the problem statement.
c. Note that Es = 2Eb and that the expression given in the problem statement reduces
to (10.12) for = 0: A plot is given in Figure 10.1. The degradation at an error
probability of 106 for 10 degrees phase imbalance is about 1.1 dB.
Problem 10.6
a. Both are equivalent in terms of symbol error probabilities.
b. QPSK is 3 dB worse in terms of symbol error probability for equal transmission
bandwidths, but it handles twice as many bits per second.
c. Choose QPSK over BPSK in terms of performance; however, other factors might favor
BPSK, such as simpler implementation.
4
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
-3
10
β = 0 deg
β = 2.5 deg
β = 5 deg
β = 7.5 deg
β = 10 deg
-4
PE, qaud ch
10
-5
10
-6
10
8
8.5
9
9.5
10
Eb/N0, dB
10.5
11
11.5
12
10.1. PROBLEM SOLUTIONS
5
Problem 10.7
For the data stream 101011 010010 100110 110011, the quadrature data streams are
d1 (t) = 1, 1, 1, -1, -1, 1, 1, -1, 1, 1, -1, 1
d2 (t) = -1, -1, 1, 1, -1, -1, -1, 1, -1, 1, -1, 1
Each 1 above means a positive rectangular pulse Ts seconds in duration and each -1 above
means a negative rectangular pulse Ts seconds in duration. For QPSK, these pulse sequences are aligned and for OQPSK, the one corresponding to d2 (t) is delayed by Ts =2 = Tb
seconds with respect to d1 (t). Type I MSK corresponds to the OQPSK waveforms d1 (t) and
d2 (t) multiplied, respectively, by cosine and sine waveforms with periods of 2Ts and type II
MSK corresponds to d1 (t) and d2 (t) multiplied, respectively, by absolute-value cosine and
sine waveforms with periods of Ts (one half cosine or sine per Ts pulse).
Problem 10.8
For QPSK, the excess phase corresponds to a stepwise function that may change values
each Ts seconds. The phase deviation is given by
i (t) = tan1
d2 (t)
d1 (t)
so the sequence of phases for QPSK computed from d1 (t) and d2 (t) given in Problem 10.7
is 7=4, 7=4, =4, 3=4, 5=4, 7=4, 7=4, 3=4, 7=4, =4, 5=4, =4 radians. For
OQPSK, the phase can change each Ts =2 = Tb seconds because d2 (t) is delayed by Ts =2
seconds with respect to d1 (t). The maximum phase change is =2 radians. For MSK,
the excess phase trajectories are straight lines of slopes =2Tb radians/second.
Problem 10.9
Write the sinc-functions as sin(x) =x functions. Use appropriate trigonometric identities to
reduce the product of the sinc-functions to the given form. For positive frequencies
SM SK (f ) = jG (f )j2 SBP SK (f )
A2 T b
=
sinc2 [(f fc ) Tb 0:25] sinc2 [(f fc ) Tb + 0:25]
2
A2 Tb sin2 [(f fc ) Tb 0:25] sin2 [(f fc ) Tb + 0:25]
=
2 f [(f fc ) Tb 0:25]g2 f [(f fc ) Tb + 0:25]g2
A2 Tb sin2 [(f fc ) Tb 0:25] sin2 [(f fc ) Tb + 0:25]
=
h
i2
2 4
(f fc )2 Tb2 1=16
6
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
Use sin [(f fc ) Tb 0:25] = p12 sin (f fc ) Tb p12 cos (f fc ) Tb to get
SM SK (f ) =
=
=
=
A2 Tb [sin (f fc ) Tb cos (f fc ) Tb ]2 [sin (f fc ) Tb + cos (f fc ) Tb ]2
h
i2
8 4
2 2
(f fc ) Tb 1=16
2
2
A2 Tb sin (f fc ) Tb cos2 (f fc ) Tb
h
i2
8 4
(f fc )2 Tb2 1=16
A2 Tb cos2 2 (f fc ) Tb
h
i2
8 4
(f fc )2 Tb2 1=16
32A2 Tb cos2 2 (f fc ) Tb
h
i2
4
1 16 (f fc )2 Tb2
Problem 10.10
If d (t) is a sequence of 1s or 0s, the instantaneous frequency is 1=4Tb Hz above the carrier
(with the deÖnition of i (t) given by (10.20) and (10.21)). If it is a sequence of alternating
1s and 0s, the instantaneous frequency is 1=4Tb Hz below the carrier. (a) The instantaneous
frequency is 10; 000; 000 50; 000=4 = 9; 987; 500 Hz. (b) The instantaneous frequency is
10; 000; 000 + 50; 000=4 = 10; 012; 500 Hz.
10.1. PROBLEM SOLUTIONS
7
Problem 10.11
The Fourier transform of h (t) is
Z 1 r
2
2 2 B 2 2
H (f ) =
B
exp
t exp (j2f t) dt
ln 2
ln 2
1
r
Z 1
2
2 2 B 2 2 jf ln 2
= B
exp
t
t dt; complete square in exponent
ln 2 1
ln 2
B 2
"
r
!#
Z 1
2
2 2 B 2 2 jf ln 2
jf ln 2 2
jf ln 2 2
= B
exp
t
t
+
dt
ln 2 1
ln 2
B 2
2B 2
2B 2
"
r
Z 1
#
2
ln 2 2
2 2 B 2
jf ln 2 2
jf ln 2
= B
exp 2 f
exp
t
dt; u = t
2
ln 2
2B
ln 2
2B
2B 2
1
"
#
r
Z 1
2
ln 2 f 2
2 2 B 2 2
1
2 2 B 2
= B
exp
exp
u du;
=
ln 2
2
B
ln 2
2 2
ln 2
1
"
#
r
2 p
Z 1
2
ln 2 f
1
u2
p
= B
exp
2 2
exp 2 du; integral is a Gauss pdf
ln 2
2
B
2
2 2
1
"
r
2 # r
2
ln 2 f
ln 2
= B
exp
2 2 ; substitute for 2 2 and simplify
ln 2
2
B
2 B
"
#
ln 2 f 2
= exp
2
B
Problem 10.12
p
a. The signal points lie on a circle of radius Es centered at the origin equally spaced
at angular intervals of 22.5 degrees (360=16). The decision regions are pie wedges
centered over each signal point.
b. For M = 16, the upper bound (10.50) well approximates the symbol error probability.
Thus
"r
#
2Es
PE, 16-PSK ' 2Q
sin
N0
16
2s
3
2
log
(16)
E
b
2
5
= 2Q 4
sin
N0
16
8
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
0
10
Symbol error probability
Bit error probability
-1
10
-2
PE, Pb; 16-PSK
10
-3
10
-4
10
-5
10
-6
10
0
2
4
6
8
10
12
Eb/N0, dB
14
16
18
20
c. The bit error probability, for Gray bit encoding, is well approximated by
Pb; 16-PSK =
=
PE, 16-PSK
log2 16
"r
#
1
8Eb
Q
sin
2
N0
16
A plot comparing the two error probabilities is given in Figure 10.2.
Problem 10.13
a. The bounds on symbol error probability are given by
P1 PE; symbol 2P1
where
P1 = Q
"r
#
2Es
sin (=M )
N0
10.1. PROBLEM SOLUTIONS
9
For moderate signal-to-noise ratios and M 8, the actual error probability is very close to
the upper bound. Assuming Gray encoding, the bit error probability is given in terms of
symbol error probability as
PE; symbol
PE; bit '
log2 (M )
and the energy per bit-to-noise spectral density ratio is
Eb
1
Es
=
N0
log2 (M ) N0
Thus, we solve
2
Q
log2 (M )
"r
#
Eb
2 log2 (M )
sin (=M ) = 104
N0
or
Q
"r
8
1:5 104 , M = 8
>
>
<
Eb
105
2 104 , M = 16
2 log2 (M )
sin (=M ) =
log2 (M ) =
2:5 104 , M = 32
>
N0
2
>
:
3 104 , M = 64
#
The arguments of the Q-function to give these various probabilities are found, approximately, by trial and error (either using MATLAB and a program for the Q-function or a
calculator and the asymptotic expansion of the Q-function):
8
>
> 3:62; M = 8
<
3:54; M = 16
Q-function argument =
3:48; M = 32
>
>
:
3:43; M = 64
Thus, for M = 8, we have
r
Eb
2 log2 (8)
sin (=8) = 3:62
N0
r
Eb
6
0:3827 = 4:17
N0
Eb
N0
"
#
1
3:62 2
= 10 log10
= 11:74 dB
6 0:3827
10
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
For M = 16; we have
r
Eb
2 log2 (16)
sin (=16) = 3:54
N0
r
Eb
8
0:1951 = 3:54
N0
Eb
N0
"
#
1
3:54 2
= 10 log10
= 16:14 dB
8 0:1951
For M = 32; we have
r
Eb
2 log2 (32)
sin (=32) = 3:48
N
r 0
Eb
10
0:098 = 3:48rQAM
N0
"
#
Eb
1
3:48 2
= 10 log10
= 21:00 dB
N0
10 0:098
For M = 64; we have
r
Eb
2 log2 (64)
sin (=64) = 3:43
N
r 0
Eb
12
0:049 = 3:43
N0
Eb
N0
= 10 log10
"
1
12
3:43
0:049
2 #
= 26:11 dB
b. For QAM the values can be estimated from Figure 10.14(c): for M = 16 we get
Eb =N0 12:4 dB to give Pb = 104 :
Problem 10.14
Let the coordinates of the received data vector, given signal labeled 1111 was sent, be
(X; Y ) = (a + N1 ; a + N2 )
where N1 and N2 are zero-mean, uncorrelated Gaussian random variables with variances
N0 =2. Since they are uncorrelated, they are also independent. Thus,
P (C j I) = P (0 X 2a) P (0 Y 2a)
Z 2a (xa)2 =N0 Z 2a (ya)2 =N0
e
e
p
p
=
dx
dy
N0
N0
0
0
10.1. PROBLEM SOLUTIONS
11
Let
u=
p
2 (x a)
and v =
N0
p
2 (y a)
N0
This results in
Z p2a2 =N0
Z p2a2 =N0 v2 =2
2
eu =2
e
p
p
P (C j I) =
du p
dv
p 2
2
2
2a =N0
2a2 =N0
Z p2a2 =N0 u2 =2 Z p2a2 =N0 v2 =2
e
e
p
p
= 4
du
dv (by symmetry of integrands)
2
2
0
0
2
0s
132
2
2a A5
= 41 2Q @
N0
Similar derivations can be carried out for the type II and type III regions.
12
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
Problem 10.15
The symbol error probability expression is
1
1
1
Ps = 1 P (C j I) + P (C j II) + P (C j III)
4
2
4
where
2
P (C j I) = [1 2Q (A)] ; A =
s
2a2
N0
P (C j II) = [1 2Q (A)] [1 Q (A)]
P (C j III) = [1 Q (A)]2
Thus
PE
1
= 1
M
=
=
=
( p
M 2
2
[1 2Q (A)]2 + 4
p
)
M 2 [1 2Q (A)] [1 Q (A)]
+4 [1 Q (A)]2
( p
)
2
p
1
M 2 [1 4Q (A) + ] + 4
M 2 [1 3Q (A) + ]
1
M
+4 [1 2Q (A) + ]
p
(
)
p
1
M 4 M + 4 [1 4Q (A)] + 4
M 2 [1 3Q (A)]
1
; neglect Q2 (A)
M
+4 [1 2Q (A)]
n
o
p
1
1
1
M 4 M M Q (A) = 1 1 4 1 p
Q (A)
M
M
0s
1
1
2a2 A
4 1 p
Q@
N0
M
Problem 10.16
The expression for a follows by computing the average symbol energy in terms of a. To
accomplish this, the sums
m
X
i=1
m
i=
X
m (m + 1)
m (m + 1) (2m + 1)
and
i2 =
2
6
i=1
are convenient. The average symbol energy is (the overbar indicates an average over the
signal constellation)
!
Z Ts
Bi2
2 A2i
2
Es =
si (t) dt =
+
Ts = A2i + Bi2
T
2
2
s
0
10.1. PROBLEM SOLUTIONS
13
p
where Ai ; Bi = a; 3a; ;
M 1 a with M = 22n , n integer.
value of Ai ; Bi equally likely, these averages become
A2i
=
Assuming each
p
2
Bi2 = p
M
M =2
X
i=1
(2i 1)2 a2 (the factor of 2 accounts for negative amplitudes)
2 p
3
p
p
M =2
M =2
M =2
2
X
X
X
2a2 X
2a
= p
(2i 1)2 = p 44
i2 4
i+
15
M i=1
M
i=1
i=1
i=1
p
p
p
p
2 p
3
M =2
M =2 + 1
M +1
M =2
M =2 + 1
p
2a2 4
= p
4
4
+ M =25
6
2
M
p
p
3
2 p
M =2 + 1
M +1
M =2 + 1
= a2 44
4
+ 15
6
2
p
M =2
p
p
i
a2 hp
M +2
M +1 3
M +2 +3
3
a2
p
p
a2
=
M +3 M +23 M 6+3 =
(M 1)
3
3
Therefore
2a2
Es = A2i + Bi2 =
(M 1)
3
and the given relationship follows.
=
Problem 10.17
Use
M
Es
Eb
PE; symbol and
= log2 (M )
2 (M 1)
N0
N0
A tight bound for the symbol error probability for coherent M -ary FSK is
r !
Es
PE; symbol (M 1) Q
N0
PE; bit =
so
M
PE; bit '
Q
2
r
Eb
log2 (M )
N0
For M = 2 the exact expression is
PE; binary = Q
r
Eb
N0
!
!
14
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
a. Coherent FSK. Use the asymptotic expression for the Q-function or iteration on a
calculator to get the following results:
p
For M = 2, PE; bit = Q
Eb =N0 = 103 for Eb =N0 = 9:49 = 9:77 dB;
p
2Eb =N0 = 103 for Eb =N0 = 5:42 = 7:34 dB;
For M = 4, PE; bit = 2Q
p
For M = 8, PE; bit = 4Q
3Eb =N0 = 103 for Eb =N0 = 4:10 = 6:13 dB;
p
For M = 16, PE; bit = 8Q
4Eb =N0 = 103 for Eb =N0 = 3:35 = 5:25 dB;
p
For M = 32, PE; bit = 16Q
5Eb =N0 = 103 for Eb =N0 = 2:94 = 4:69 dB;
p
For M = 64, PE; bit = 32Q
6Eb =N0 = 103 for Eb =N0 = 2:67 = 4:27 dB.
b. Noncoherent FSK. Values can be estimated from Figure 10.15(b).
For M = 2, PE; bit = 103 for Eb =N0 10:8 dB;
For M = 4, PE; bit = 103 for Eb =N0 8:4 dB; etc.
Problem 10.18
For M -ary noncoherent FSK, the symbol error probability was given in (10.68) as
M
1
k+1
X
k Es
M 1 (1)
PE; symbol =
exp
k
k+1
k + 1 N0
k=1
The derived result, given by (10.89), is
M 1
k
Es X
Es
M 1 (1)
PE; symbol = 1 exp
exp
k
N0
k+1
(k + 1) N0
Es
Take the exp N
0
PE; symbol =
=
=
=
k=0
inside the sum and break out the Örst term of the sum:
M
1
X
(1)k
Es
Es
11
exp
exp
k+1
(k + 1) N0
N0
k=1
M
1
k
X
Es
Es
M 1 (1)
exp
k
k+1
(k + 1) N0 N0
k=1
M
1
k
X
Es
1
M 1 (1)
exp
1
k
k+1
N0 k + 1
k=1
M
1
k+1
X
k Es
M 1 (1)
exp
k
k+1
k + 1 N0
k=1
M 1
k
10.1. PROBLEM SOLUTIONS
15
which shows they are equivalent.
Problem 10.19
a. The bandwidth e¢ciencies are given by
8
> 0:5 log2 (M ) ; M -PSK and M -QAM
R < 2 log2 (M )
=
M +3 ; coherent M -FSK
B >
: log2 (M ) ; noncoherent M -FSK
2M
For a data rate of 4 Mbps and a channel bandwidth of 1 MHz, the only candidate schemes
are PSK, QAM or DPSK for which
R
= 4 = 0:5log2 (M ) or log2 (M ) = 8 or M = 28 = 256
B
Eb
24 dB to give a bit error probability of 106 (Figure 10.14(c)).
b. 256-QAM requires N
0
256-PSK and 256-DPSK require considerably more and are of the same order in
complexity as 256-QAM.
c. Based purely of simplicity of implementation, the choice would probably be DPSK.
However, QAM gives quite an advantage in terms of required power. It does not have
constant envelope which might be another consideration.
Problem 10.20
The bandwidth e¢ciencies of coherent and noncoherent FSK areve
(
2 log2 (M )
R
M +3 ; coherent M -FSK
=
log
(M )
2
B
2M ; noncoherent M -FSK
R
2 (M )
a. The required bandwidth e¢ciency is B
= 0:25 = 2 log
M +3 . For M = [2 4 8 16 32 64]
R
we get B
= [0:4000 0:5714 0:5455 0:4211 0:2857 0:1791] bits/s/Hz. Thus, M = 2 32
Eb
will work. From Fig. 10.15(a) we get Pb = 104 for N
= [11:5 8:6 7:2 6:4 5:7] dB
0
b. Figure 10.15(a) gives Eb =N0 = 5:7 dB for M = 32.
R
2 (M )
c. Solve B
= 0:25 log2M
which gives M = [2 4].
d Solve 0:5 exp (Eb =2N0 ) = 104 or Eb =N0 = 2ln 2 104 = 17:04 = 12:3 dB for
M = 2: For M = 4 Fig. 10.15(b) gives Eb =N0
= 9:5 dB to give Pb = 104 .
16
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
e. In this case the cost of a noncoherent demodulator is 9:5 5:7 = 3:8 dB in power
(over a factor of 2.4).
Problem 10.21
Use Figure 10.16 noting that the abscissa is normalized baseband bandwidth. RF bandwidth is twice as large. The 90% power containment bandwidth is deÖned by the ordinate
= -10 dB. (a) B90; BPSK = 1:6=Tb = 1:6Rb ; (b) B90; QPSK, OQPSK = 0:8=Tb = 0:8Rb ; (c)
B90; MSK = 0:8=Tb = 0:8Rb ; (d) The di§erence between (10.114) for BPSK and (10.118)
for QAM as far as bandwidth is concerned is the factor of log2 M for the latter in the
argument of the sinc-function. For 16-QAM, this is a factor of 4 so 16-QAM is a factor
of 4 better than BPSK as far as bandwidth e¢ciency is concerned. For a bit rate of 50
kbps, the resulting bandwidths are B90; BPSK = 80 kHz, B90; QPSK, OQPSK, MSK = 40 kHz,
B90; 16-QAM = 20 kHz.
Problem 10.22
The baseband power spectrum is
G (f ) = 2A2 Tb [log2 (M )] sinc2 [log2 (M ) Tb f ]
where
A2 = Es cos2 i = Es sin2 i
M
M
1 X
1 X 2 2 (i 1)
2 2 (i 1)
=
cos
=
sin
M
M
M
M
i=1
i=1
The 10% power containment bandwidth is deÖned by (see Problem 10.21)
log2 (M ) B90; MPSK = 0:8=Tb = 0:8Rb
B90; MPSK
8
< 0:267Rb Hz; M = 8
0:25Rb Hz; M = 16
= 0:8Rb = log2 (M ) =
:
0:16Rb Hz; M = 32
10.1. PROBLEM SOLUTIONS
17
Problem 10.23
a. From the given expression, B10% QAM, 4 = Rb = 20 kbps.
b. B10% QAM, 16 = Rb =2 = 20 kHz; Rb = 40 kbps.
c. B10% QAM, 64 = Rb =3 = 20 kHz; Rb = 60 kbps.
d. B10% QAM, 256 = Rb =4 = 20 kHz; Rb = 80 kbps.
Problem 10.24
a. Find the autocorrelation function and Fourier transform it to obtain the power spectral
density. By deÖnition
RASK ( ) = E [sASK (t) sASK (t + )]
1 2
=
A E f[1 + d (t)] [1 + d (t + )] cos (! c t + ) cos [! c (t + ) + ]g
4
1 2
=
A E f[1 + d (t)] [1 + d (t + )]g E fcos (! c t + ) cos [! c (t + ) + ]g
4
1 2
=
A E f[1 + d (t)] [1 + d (t + )]g cos (! c )
8
1 2
=
A E f1 + d (t) + d (t + ) + d (t) d (t + )g cos (! c )
8
1 2
=
A f1 + E [d (t) d (t + )]g cos (! c )
8
1 2
=
A f1 + Rd ( )g cos (! c )
8
The Fourier transform is
SASK (f ) =
1 2
1
A [ (f fc ) + (f + fc )] + A2 [Sd (f fc ) + Sd (f + fc )]
16
16
where
Sd (f ) = F [Rd ( )] = T sinc2 (T f )
b. Write this as
sP SK (t) = Am sin (! c t + ) + Ad (t)
p
1 m2 cos (! c t + )
18
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
Following a procedure similar to that used in (a), we have
RASK ( ) = E [sP SK (t) sP SK (t + )]
8
9
h
i
p
2 cos (! t + )
<
=
m
sin
(!
t
+
)
+
d
(t)
1
m
c
c
h
i
= A2 E
p
: m sin (! c (t + ) + ) + d (t + ) 1 m2 cos (! c (t + ) + ) ;
2
m sin (! c t + )sin (! c (t + ) + ) + cross terms in d (t)
2
= A E
+d (t) d (t + ) 1 m2 cos (! c t + ) cos (! c (t + ) + )
(cross term average are 0)
(10.1)
1 2 2
1
=
m A cos (! c ) +
1 m2 A2 E [d (t) d (t + )] cos (! c )
2
2
The power spectral density is
1
1
SP SK (f ) = m2 A2 [ (f fc ) + (f + fc )] +
1 m2 A2 [Sd (f fc ) + Sd (f + fc )]
4
4
where Sd (f ) = T sinc2 (T f ) as in (a).
Problem 10.25
To show:
cos
t
2Tb
t
2Tb
!
4Tb cos (2Tb f )
1 (4Tb f )2
Use the modulation theorem together with the Fourier transform of a rectangular pulse to
get
t
t
cos
! Tb sinc [2Tb (f 1=4Tb )] + Tb sinc [2Tb (f + 1=4Tb )]
2Tb
2Tb
Rewrite the RHS as
RHS = Tb
sin (2Tb f =2) sin (2Tb f + =2)
+
2Tb f =2
2Tb f + =2
Note that
sin (2Tb f =2) = cos (2Tb f )
Then
RHS =
=
2Tb
1
1
+
cos (2Tb f )
1 4Tb f
1 + 4Tb f
4Tb cos (2Tb f )
1 (4Tb f )2
10.1. PROBLEM SOLUTIONS
19
L = Rs /BL = 5
L = Rs /BL = 10
3
1.5
PLL
Costas
Data est.
σ
2
1
σ
2
2
PLL
Costas
Data est.
1
0
0.5
0
5
10
15
z, dB
L = Rs /BL = 100
20
0
0
5
10
z, dB
15
20
0.2
PLL
Costas
Data est.
σ
2
0.15
0.1
0.05
0
0
5
10
z, dB
15
20
Problem 10.26
This would involve an 8th power device which would produce a frequency component of
810 = 80 MHz at its output. This would be tracked by a PLL and then frequency divided
to 10 MHz and used as the reference in a coherent demodulator for the 8-PSK signal.
Problem 10.27
A plot is given in Figure 10.3.
Problem 10.28
The dB di§erence is
2
58
2
10 log10 ; SL = ; AV = 10 log10
= 0:9691 dB
32
20
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
Problem 10.29
See the table below. As many shifts as will Öt on the page are given. Shift 5 is within
Hamming distance 1
HD
Data
Shft0
Shft1
Shft2
Shft3
Shft4
Shft5
Shft6
Shft7
Shft8
Shft9
Shft10
Shft11
Shft12
Shft13
Shft14
1
1
0
0
1
1
1
0
1
1
1
1
0
1
0
1
1
1
0
1
1
0
1
1
1
0
1
0
0
0
1
1
1
0
1
1
0
0
0
1
1
1
0
1
1
0
0
0
1
1
1
0
1
0
0
0
0
1
1
1
0
1
0
0
0
0
1
1
1
0
1
0
0
0
0
1
1
1
0
1
0
0
0
0
1
1
1
0
1
1
0
0
0
1
1
1
0
1
1
0
0
0
1
1
1
0
1
1
0
0
0
1
1
1
0
0
0
0
0
1
1
1
1
0
0
0
1
1
0
0
0
0
1
1
0
0
0
0
0
0
Problem 10.30
See Example 2.20 for the special case of N = 7 for more details.
Problem 10.31
a. N = 2no. of stages 1 = 26 1 = 63.
b. Each clock pulse corresponds to one chip, so Tone seq period = 63 104 = 6:3 milliseconds.
c. The autocorrelation function consists of a periodic train of equilateral triangles with
bases of 2 milliseconds between 1=64 and 1 unit high spaced by 6.3 milliseconds.
d. The spectral lines are spaced by f = 1/period = 1=6:3 103 = 158:73 Hz.
e. The weight of the impulse at f = 0 in the power spectral density is (1=63)2 = 2:52
104 V2 . The DC level of the m-sequence is 1=63 V.
f. The Örst null in the envelope of the power spectrum is at 63f = 10 kHz.
3
5
5
2
3
1
5
6
6
5
4
2
5
3
4
10.1. PROBLEM SOLUTIONS
21
Problem 10.32
a. The two boxes with arrows below are modulo-2 summed and the sum fed back to box
1.
x1
x2
x3
x4
#
x5
x6
x7
#
b. Assuming an all-1s initial load the states are as follows. The output is the last column.
1
0
0
0
0
1
1
1
0
1
1
1
1
0
0
1
0
1
1
0
0
1
0
0
1
0
0
0
0
0
1
1
0
0
0
0
1
1
1
0
1
1
1
1
0
0
1
0
1
1
0
0
1
0
0
1
0
0
0
0
1
1
1
0
0
0
0
1
1
1
0
1
1
1
1
0
0
1
0
1
1
0
0
1
0
0
1
0
0
0
1
1
1
1
0
0
0
0
1
1
1
0
1
1
1
1
0
0
1
0
1
1
0
0
1
0
0
1
0
0
1
1
1
1
1
0
0
0
0
1
1
1
0
1
1
1
1
0
0
1
0
1
1
0
0
1
0
0
1
0
1
1
1
1
1
1
0
0
0
0
1
1
1
0
1
1
1
1
0
0
1
0
1
1
0
0
1
0
0
1
1
1
1
1
1
1
1
0
0
0
0
1
1
1
0
1
1
1
1
0
0
1
0
1
1
0
0
1
0
0
0
1
0
0
0
1
0
0
1
1
0
0
0
1
0
1
1
1
0
1
0
1
1
0
1
1
0
0
0
0
0
0
1
0
0
0
1
0
0
1
1
0
0
0
1
0
1
1
1
0
1
0
1
1
0
1
1
0
0
0
0
0
0
1
0
0
0
1
0
0
1
1
0
0
0
1
0
1
1
1
0
1
0
1
1
0
1
1
0
0
0
0
0
0
1
0
0
0
1
0
0
1
1
0
0
0
1
0
1
1
1
0
1
0
1
1
0
1
1
0
0
0
0
0
0
1
0
0
0
1
0
0
1
1
0
0
0
1
0
1
1
1
0
1
0
1
1
0
1
1
0
0
0
0
0
0
1
0
0
0
1
0
0
1
1
0
0
0
1
0
1
1
1
0
1
0
1
1
0
1
1
0
0
0
0
0
0
1
0
0
0
1
0
0
1
1
0
0
0
1
0
1
1
1
0
1
0
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
0
1
1
1
0
0
1
1
1
1
0
1
1
0
1
0
0
0
0
0
1
1
0
0
1
1
0
1
0
1
0
0
1
1
1
0
0
1
1
1
1
0
1
1
0
1
0
0
0
0
0
1
1
0
0
1
1
0
1
0
1
0
0
1
1
1
0
0
1
1
1
1
0
1
1
0
1
0
0
0
0
0
1
1
0
0
1
1
0
1
0
1
0
0
1
1
1
0
0
1
1
1
1
0
1
1
0
1
0
0
0
0
0
1
1
0
0
1
1
0
1
0
1
0
0
1
1
1
0
0
1
1
1
1
0
1
1
0
1
0
0
0
0
0
1
1
0
0
1
1
0
1
0
1
0
0
1
1
1
0
0
1
1
1
1
0
1
1
1
1
0
0
0
0
0
1
1
0
0
1
1
0
1
0
1
0
0
1
1
1
0
0
1
1
1
1
0
1
0
1
0
1
0
1
0
1
1
1
1
1
0
1
0
0
1
0
1
0
0
0
1
1
0
1
1
1
0
0
0
0
1
0
1
0
1
0
1
1
1
1
1
0
1
0
0
1
0
1
0
0
0
1
1
0
1
1
1
0
0
0
0
1
0
1
0
1
0
1
1
1
1
1
0
1
0
0
1
0
1
0
0
0
1
1
0
1
1
1
0
0
0
0
1
0
1
0
1
0
1
1
1
1
1
0
1
0
0
1
0
1
0
0
0
1
1
0
1
1
1
0
0
0
0
1
0
1
0
1
0
1
1
1
1
1
0
1
0
0
1
0
1
0
0
0
1
1
0
1
0
1
0
0
0
0
1
0
1
0
1
0
1
1
1
1
1
0
1
0
0
1
0
1
0
0
0
1
1
0
1
0
1
0
0
0
0
1
0
1
0
1
0
1
1
1
1
1
0
1
0
0
1
0
1
0
0
0
1
1
22
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
0
1
1
1
1
1
1
1
0
0
1
1
1
1
1
1
0
0
0
1
1
1
1
1
1
0
0
0
1
1
1
1
1
1
0
0
0
1
1
1
1
1
1
0
0
0
1
1
0
1
1
1
0
0
0
1
c. The two boxes with arrows below are modulo-2 summed and the sum fed back to box
1.
x1
x2
x3
x4
x5
x6
#
x7
#
d. Assuming an all-1s initial load the states are as follows. The output is the last column.
10.1. PROBLEM SOLUTIONS
1
0
0
0
0
0
0
1
0
0
0
0
0
1
1
0
0
0
0
1
0
1
0
0
0
1
1
1
1
0
1
1
0
0
0
0
0
0
1
0
0
0
0
0
1
1
0
0
0
0
1
0
1
0
0
0
1
1
1
1
1
1
1
0
0
0
0
0
0
1
0
0
0
0
0
1
1
0
0
0
0
1
0
1
0
0
0
1
1
1
1
1
1
1
0
0
0
0
0
0
1
0
0
0
0
0
1
1
0
0
0
0
1
0
1
0
0
0
1
1
1
1
1
1
1
0
0
0
0
0
0
1
0
0
0
0
0
1
1
0
0
0
0
1
0
1
0
0
0
1
1
1
1
1
1
1
0
0
0
0
0
0
1
0
0
0
0
0
1
1
0
0
0
0
1
0
1
0
0
0
1
1
1
1
1
1
1
0
0
0
0
0
0
1
0
0
0
0
0
1
1
0
0
0
0
1
0
1
0
0
0
1
1
1
1
1
1
1
1
0
1
1
1
1
1
1
0
1
0
1
1
1
1
1
1
0
1
0
1
1
1
1
0
1
0
1
0
1
1
1
1
0
1
0
1
0
1
1
0
1
0
1
0
1
0
1
0
1
0
0
0
1
0
1
1
0
0
1
1
1
0
1
0
1
0
0
1
1
1
1
1
0
1
0
0
0
23
0
0
1
0
0
0
1
0
1
1
0
0
1
1
1
0
1
0
1
0
0
1
1
1
1
1
0
1
0
0
1
0
0
1
0
0
0
1
0
1
1
0
0
1
1
1
0
1
0
1
0
0
1
1
1
1
1
0
1
0
1
1
0
0
1
0
0
0
1
0
1
1
0
0
1
1
1
0
1
0
1
0
0
1
1
1
1
1
0
1
1
1
1
0
0
1
0
0
0
1
0
1
1
0
0
1
1
1
0
1
0
1
0
0
1
1
1
1
1
0
1
1
1
1
0
0
1
0
0
0
1
0
1
1
0
0
1
1
1
0
1
0
1
0
0
1
1
1
1
1
0
1
1
1
1
0
0
1
0
0
0
1
0
1
1
0
0
1
1
1
0
1
0
1
0
0
1
1
1
1
0
1
1
1
0
0
0
1
0
0
1
0
0
1
1
0
1
1
0
1
0
1
1
0
1
1
1
1
0
1
0
0
1
1
1
0
0
0
1
0
0
1
0
0
1
1
0
1
1
0
1
0
1
1
0
1
1
1
1
0
0
0
0
1
1
1
0
0
0
1
0
0
1
0
0
1
1
0
1
1
0
1
0
1
1
0
1
1
1
1
0
0
0
0
1
1
1
0
0
0
1
0
0
1
0
0
1
1
0
1
1
0
1
0
1
1
0
1
1
1
1
0
0
0
0
1
1
1
0
0
0
1
0
0
1
0
0
1
1
0
1
1
0
1
0
1
1
0
1
1
0
1
0
0
0
0
1
1
1
0
0
0
1
0
0
1
0
0
1
1
0
1
1
0
1
0
1
1
0
1
1
0
1
0
0
0
0
1
1
1
0
0
0
1
0
0
1
0
0
1
1
0
1
1
0
1
0
1
1
0
1
0
0
0
1
1
0
1
0
0
1
0
1
1
1
0
1
1
1
0
0
1
1
0
0
1
0
1
0
1
1
1
0
0
0
1
1
0
1
0
0
1
0
1
1
1
0
1
1
1
0
0
1
1
0
0
1
0
1
0
0
1
1
0
0
0
1
1
0
1
0
0
1
0
1
1
1
0
1
1
1
0
0
1
1
0
0
1
0
1
1
0
1
1
0
0
0
1
1
0
1
0
0
1
0
1
1
1
0
1
1
1
0
0
1
1
0
0
1
0
1
1
0
1
1
0
0
0
1
1
0
1
0
0
1
0
1
1
1
0
1
1
1
0
0
1
1
0
0
1
1
1
1
0
1
1
0
0
0
1
1
0
1
0
0
1
0
1
1
1
0
1
1
1
0
0
1
1
0
0
1
1
1
1
0
1
1
0
0
0
1
1
0
1
0
0
1
0
1
1
1
0
1
1
1
0
0
1
1
0
24
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
Problem 10.33
a. Lengths 11 and 13 are not included (reference should be to Table 10.12).
N =2
Barker code
Delay = 0
Delay = 1
N =4
Barker code
Delay = 0
Delay = 1
Delay = 2
Delay = 3
N =5
Barker code
Delay = 0
Delay = 1
Delay = 2
Delay = 3
Delay = 5
N =7
Barker code
Delay = 0
Delay = 1
Delay = 2
Delay = 3
Delay = 5
Delay = 6
Delay = 7
1
1
1
1
1
1
1
1
0
0
1
1
1
1
1
1
1
1
1
1
NA NU
NA NU
N
2
-1
1
-1/2
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
0
1
1
0
0
1
1
1
0
0
1
1
1
1
0
1
1
1
0
1
1
1
0
0
0
1
1
1
1
0
1
0
1
1
1
1
0
0
1
1
1
NA NU
NA NU
N
4
1
0
1
1
0:25
0
0:25
1
1
0
1
0
0
1
0
0
1
1
1
1
0
0
1
0
0
1
1
NA NU
NA NU
N
5
0
1
0
1
1
0
0:2
0
0:2
1
0
1
0
0
1
0
1
0
0
0
1
0
0
1
0
NA NU
NA NU
N
7
0
1
0
1
0
1
1
0
0:143
0
0:143
0
0:143
b. The m-sequence is 111100010011010. Autocorrelation results for delays up to 9 are
given.
10.1. PROBLEM SOLUTIONS
25
NA NU
N
m-seq
D=0
D=1
D=2
D=3
D=5
D=6
D=7
D=8
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
1
1
1
1
0
0
0
1
1
1
1
0
0
0
0
1
1
1
1
1
1
0
0
0
1
1
1
1
0
0
1
0
0
0
1
1
1
0
0
0
1
0
0
0
1
1
1
1
0
0
1
0
0
0
1
1
1
1
0
0
1
0
0
0
0
0
1
1
0
0
1
0
0
1
1
0
1
1
0
0
1
0
0
0
1
0
1
1
0
0
1
0
1
0
1
1
0
0
0
1
0
1
1
0
0
1
0
1
1
0
1
0
1
0
1
0
0
1
0
Problem 10.34
Note that the expectation of Ng is zero because n (t) has zero mean. Write the expectation
of the square of Ng as an iterated integral. The expectation can be taken inside.
Z
Z
2
E Ng
= E 4
n (t) c (t) cos (! c t + ) dt
n () c () cos (! c + ) d
Tb
Tb
Z Z
= E 4
n (t) n () c (t) c () cos (! c t + ) cos (! c + ) dtd
Tb Tb
Z Z
= 4
E [n (t) n ()] E [c (t) c ()] E [cos (! c t + ) cos (! c + )] dtd
ZTb ZTb
N0
= 4
(t ) E [c (t) c ()] E [cos (! c t + ) cos (! c + )] dtd
Tb Tb 2
Z
= 2N0
E c2 (t) E cos2 (! c t + ) dt = N0 Tb
Tb | {z }
1
(10.2)
Use has been made of the fact that
E [n (t) n ()] =
N0
(t )
2
to reduce the double integral to a single integral. The integral of the resulting cosine
squared is Tb =2. The result for the variance (same as the mean square) is then found to be
N0 Tb .
Problem 10.35
The random variable to be considered is
Z Tb
NI =
AI c (t) cos (!t + ) dt
0
1
0
0:08
0
0:09
0:4
0:11
0:25
26
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
where is assumed uniform in[0; 2). Clearly, the expectation of NI is zero due to
this random phase. The variance is the same as the mean square, which is given by
Z Tb Z Tb
var (NI ) = E
A2I c (t) c () cos (!t + ) cos (! + ) dtd
0
Z Tb Z Tb
0
A2I E [c (t) c ()] cos (!t + ) cos (! + ) dtd
0
0
Z Tb Z Tb
1
1
= A2I Tc
[(t ) =Tc ] cos [! (t )] dtd
Tc
2
0
0
Z Tb
2
A Tc Tb
1
= A2I Tc
dt = I
2
2
0
=
Problem 10.36
a. The processing gain is
Gp =
Tb
Rc
=
or Rc = Gp Rb
Tc
Rb
For Gp = 1000 and a data rate of 10 kbps, Rc = 103 104 = 10 megachips per second;
b. BRF = 2Rc = 20 MHz;
c. Use (10.155) and (10.158) to get the following:
SNR
For a JSR of 5 dB = 3:162 and SNR = 10 dB = 10, then 1+(SNR)(JSR)=G
=
p
p
10
4
9:694 = Q (3:113) = 9:25 10 :
1+103:162=1000 = 9:694 and PE = Q
SNR
For a JSR of 10 dB = 10 and SNR = 10 dB = 10, then 1+(SNR)(JSR)=G
=
p
p
10
3
9:091 = Q (3:015) = 1:3 10 :
1+1010=1000 = 9:091 and PE = Q
SNR
For a JSR of 15 dB = 31:62 and SNR = 10 dB = 10, then 1+(SNR)(JSR)=G
=
p
p
10
3
7:598 = Q (2:756) = 2:9 10 :
1+1031:62=1000 = 7:598 and PE = Q
SNR
For a JSR of 30 dB = 1000 and SNR = 10 dB = 10, then 1+(SNR)(JSR)=G
=
p
p
10
0:909 = Q (0:954) = 0:17:
1+101000=1000 = 0:909 and PE = Q
Problem 10.37
The limit of the arguement of the Q-function in (10.155) is
s
r
Gp
SNR
lim
=
SNR!1
1 + (SNR) (JSR) =Gp
JSR
To get PE = Q (A) = 105 trial and error in MATLAB requires A = 4:264.
10.1. PROBLEM SOLUTIONS
a. For JSR = 30 dB = 1000;
1:82 104 = 42:6 dB;
27
p
Gp =JSR =
p
Gp =1000 = 4:264 or Gp = 1000 4:2642 =
p
p
b. For JSR = 25 dB = 316:23;
Gp =JSR = Gp =316:23 = 4:264 or Gp = 316:23
4:2642 = 5:75 103 = 37:6 dB;
p
p
c. For JSR = 20 dB = 100;
Gp =JSR = Gp =100 = 4:264 or Gp = 100 4:2642 =
1:82 103 = 32:6 dB
Problem 10.38
p
In the limit at SNR Eb =N0 ! 1, the argument of the Q-function for PE = Q
SNR
becomes
r
p
3N
SNR =
3:08 to give PE = 103
K 1
Thus,
3 255
+1
(3:08)2
= 81:64
K =
Round this down to K = 81 users since the number of users must be an integer.
Problem 10.39
With a propagation delay uncertainty of 1:5 ms the chip uncertainty becomes
C = 2 2 1:5 103 s 3 106 chips/s
= 18; 000 half chips
Equation (10.158) becomes
2 Pd
1
Tacq = 17; 999 (1 + 1000Pfa )
+
Ti
2Pd
Pd
a. Tacq = 27:197 s;
b. Tacq = 18:114 s;
c. Tacq = 14:112 s.
28
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
Problem 10.40
In this case T = 10 5 s= 5 105 s, Tb = 1=Rb = 1=500 103 = 2 106 s, and M = 4
so (10.161) becomes
5 105 = log2 (4) N 2 106
5 105
or N =
= 12:5 subcarriers
4 106
We round this up to the next power of 2, or N = 16.
Problem 10.41
The equation for the e§ective rate is
Re§ =
Ndaata SC log2 M
(Code rate)
TFFT + Tguard
For example, the Örst set of values gives
Re§ =
48 log2 2 1
= 6 Mbps
4 s
2
Taking all possible combinations of parameter values, we get the following results:
10.1. PROBLEM SOLUTIONS
Mod. type
BPSK
BPSK
BPSK
BPSK
BPSK
BPSK
QPSK
QPSK
QPSK
QPSK
QPSK
QPSK
16-QAM
16-QAM
16-QAM
16-QAM
16-QAM
16-QAM
64-QAM
64-QAM
64-QAM
64-QAM
64-QAM
64-QAM
Ndaata SC
48
48
48
48
48
48
48
48
48
48
48
48
48
48
48
48
48
48
48
48
48
48
48
48
Code rate
1/2
1/2
1/2
3/4
3/4
3/4
1/2
1/2
1/2
3/4
3/4
3/4
1/2
1/2
1/2
3/4
3/4
3/4
1/2
1/2
1/2
3/4
3/4
3/4
29
TFFT + Tguard , s
4
8
16
4
8
16
4
8
16
4
8
16
4
8
16
4
8
16
4
8
16
4
8
16
Re§ , Mbps
6
3
1.5
9
4.5
2.25
12
6
3
18
9
4.5
24
12
6
36
18
9
36
18
9
54
27
13.5
30
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
Problem 10.42
a. The number of users per reuse pattern remain the same at 120. For 20 dB minimum
signal-to-interference ratio and a propagation power law of = 4, we have
Dco
20 = 10(4) log10
1 7:7815
dA
or
log10
Dco
20 + 7:7815
1
=
= 0:6945
dA
40
Dco
= 100:6945 + 1
dA
p
= 5:9492 = 3N
or
N = d11:798e = 12 (i = j = 2)
The e¢ciency is
v =
120
12
6 MHz
10
5
=
= voice circuits per base station per MHz
6
3
b. For 14 dB minimum signal-to-interference ratio and a propagation power law of = 4,
we have
Dco
14 = 10(4) log10
1 7:7815
dA
Dco
14 + 7:7815
log10
1
=
= 0:5445
dA
40
Dco
= 100:5445 + 1
dA
p
= 4:5038 = 3N
or
N = d6:7614e = 7 (i = 1; j = 2)
The e¢ciency is
v =
=
120
7
6 MHz
17
= 2:83 voice circuits per base station per MHz
6
10.1. PROBLEM SOLUTIONS
31
Problem 10.43
The number of users per reuse pattern remain the same at 120. For 10 dB minimum
signal-to-interference ratio and a propagation power law of = 3:5, we have
10 = 10(3:5) log10
Dco
1 7:7815
dA
or
log10
Dco
10 + 7:7815
1
=
= 0:508
dA
35
Dco
= 100:508 + 1
dA
p
= 4:2214 = 3N
or
N = d5:94e = 7 (i = 1; j = 2)
The e¢ciency is
v =
=
120
7
6 MHz
17
= 2:83 voice circuits per base station per MHz
6
Problem 10.44
a. Consider a hexigon made up of six equalateral triangles having sides d.. Pick one of
these triangles; it consists of two right triangles each having hypoyenuse d and sides
d=2 and 1=2. Thus
2 2
1
d
+
= d2
2
2
or
p
d = 1= 3
b. This is obvious from (10:166).
32
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
10.2
Computer Exercises
Computer Exercise 10.1
% ce10_1.m: Bit error probabilities for coh. and noncoh. M-ary FSK
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
clf
z_dB = 3:.1:15;
z = 10.^(z_dB/10);
for j = 1:5
M=2^j;
A = M/(2*(M-1));
if j == 1
Pbc = qfn(sqrt(z));
elseif j >= 2
Psc = (M-1)*qfn(sqrt(log2(M)*z));
Pbc = A*Psc;
end
Psnc = zeros(size(z));
for k = 1:M-1
B = (prod(1:(M-1))/(prod(1:k)*prod(1:(M-1-k))))*(-1)^(k+1)/(k+1);
alpha=k*log2(M)/(k+1);
Psnc = Psnc + B*exp(-alpha*z);
end
Pbnc = A*Psnc;
subplot(1,2,1), semilogy(z_dB, Pbc, íLineWidthí, 1.5), ...
axis([min(z_dB) max(z_dB) 1e-6 1]), ylabel(í{nitP_b}í), xlabel(í{nitE_b/N}_0í), ...
if j == 1
hold on; grid
text(z_dB(45)+.2, Pbc(45), [í{nitM} = í, num2str(M)]),...
title(í{nitM}-ary CFSKí)
else
text(z_dB(45)+.2, Pbc(45), [í= í, num2str(M)])
end
subplot(1,2,2), semilogy(z_dB, Pbnc, íLineWidthí, 1.5), ...
axis([min(z_dB) max(z_dB) 1e-6 1]), ...
ylabel(í{nitP_b}í), xlabel(í{nitE_b/N}_0í), ...
if j == 1
10.2. COMPUTER EXERCISES
33
hold on; grid
text(z_dB(45)+.2, Pbnc(45), [í{nitM} = í, num2str(M)]),...
title(í{nitM}-ary NCFSKí)
else
text(z_dB(45)+.2, Pbnc(45), [í= í, num2str(M)])
end
end
% End of script Ölw
The curves obtained are identical to Fig. 10.15
Computer Exercise 10.2
% ce10_2.m: Out-of-band power for M-ary PSK, QPSK (OQPSK), and MSK
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
clf
A = char(í-í,íñí,í-.í,í:í,íñ.í,í-..í);
Tbf = 0:.025:7;
LTbf = length(Tbf);
A_BPSK = [];
A_QPSK = [];
A_MSK = [];
for k = 1:LTbf
A_BPSK(k) = quadl(íG_BPSKí, Tbf(k), 14);
A_QPSK(k) = quadl(íG_QPSKí, Tbf(k), 7);
A_MSK(k) = quadl(íG_MSKí, Tbf(k), 7);
end
OBP_BPSK = 10*log10(A_BPSK/A_BPSK(1));
OBP_QPSK = 10*log10(A_QPSK/A_QPSK(1));
OBP_MSK = 10*log10(A_MSK/A_MSK(1));
plot(Tbf, OBP_BPSK, A(1,:)), ylabel(íRelative out-of-band power, dBí), xlabel(íT_bfí),...
axis([0 5 -40 0]),grid,...
hold on
plot(Tbf, OBP_QPSK, A(2,:))
plot(Tbf, OBP_MSK, A(3,:))
legend(íBPSKí, íQPSK/OQPSKí, íMSKí)
%
%
Function to compute area under BPSK spectrum
34
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
0
BPSK
QPSK/OQPSK
MSK
Relative out-of-band power, dB
-5
-10
-15
-20
-25
-30
-35
-40
0
0.5
1
1.5
2
2.5
Tbf
3
3.5
4
function y = G_BPSK(Tbf)
y = 2*(sinc(Tbf)).^2;
%
Function to compute area under MSK spectrum
%
function y = G_MSK(Tbf)
y = (32/pi^2)*((cos(2*pi*Tbf)).^2)./(1-(4*Tbf+eps).^2).^2;
%
Function to compute area under QPSK spectrum
%
function y = G_QPSK(Tbf)
y = 4*(sinc(2*Tbf)).^2;
% End of script Öle
A typical plot generated by the program is shown in Fig. 10.4.
4.5
5
10.2. COMPUTER EXERCISES
Computer Exercise 10.3
% ce10_3.m: Computes bit error probability curves for jamming in DSSS.
% For a desired P_E and given JSR and Gp, computes the required Eb/N0
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
clf
A = char(í-í,íñí,í-.í,í:í,íñ.í,í-.xí,í-.oí);
Gp_dB = input(íEnter desired processing gain in dB í);
Gp = 10^(Gp_dB/10);
z_dB = 0:.5:30;
z = 10.^(z_dB/10);
k = 1;
JSR0 = [];
for JSR_dB = 5:5:25;
JSR0(k) = JSR_dB;
JSR = 10^(JSR_dB/10);
arg = z./(1+z*JSR/Gp);
PE = 0.5*erfc(sqrt(arg));
semilogy(z_dB, PE, A(k+1,:))
if k == 1
hold on
axis([0 30 1E-15 1])
grid on
xlabel(íE_b/N_0, dBí), ylabel(íP_Eí)
title([íBEP for DS BPSK in jamming for proc. gain = í,
num2str(Gp_dB),í dBí])
end
k = k+1;
end
PEG = 0.5*qfn(sqrt(2*z));
semilogy(z_dB, PEG), text(z_dB(30)-5, PEG(30), íNo jammingí)
legend([íJSR = í, num2str(JSR0(1))], [íJSR = í, num2str(JSR0(2))],
[íJSR = í, num2str(JSR0(3))], [íJSR = í, num2str(JSR0(4))],
[íJSR = í, num2str(JSR0(5))], 3)
disp(íStrike ENTER to continueí);
pause
PE0 = input(íEnter desired value of P_E í);
JSR_dB_0 = input(íEnter given value of JSR in dB í);
35
36
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
Gp_dB_0 = input(íEnter given value of processing gain in dB í);
JSR0 = 10^(JSR_dB_0/10);
Gp0 = 10^(Gp_dB_0/10);
PELIM = 0.5*erfc(sqrt(Gp0/JSR0));
disp(í í)
disp(íInÖnite Eb/N0 BEP limití)
disp(PELIM)
if PELIM >= PE0
disp(íFor given choices of G_p, JSR, & desired P_E,
a solution is not possible í);
Gp0_over_JSR0 = (erÖnv(1-2*PE0))^2;
Gp0_over_JSR0_dB = 10*log10(Gp0_over_JSR0);
disp(íThe minimum required value of G_p over JSR in dB is:í)
disp(Gp0_over_JSR0_dB)
else
arg0 = (erÖnv(1-2*PE0))^2;
SNR0 = 1/(1/arg0 - JSR0/Gp0);
SNR0_dB = 10*log10(SNR0);
end
disp(í í)
disp(íTo give BEP of:í)
disp(PE0)
disp(íRequires GP, JSR, and Eb/N0 in dB of:í)
disp([Gp_dB JSR_dB SNR0_dB])n
% End of script Öle
A typical run follows, with both a plot given and a speciÖc output generated:
>> ce10_3
Enter desired processing gain in dB: 30
Strike ENTER to continue
Enter desired value of P_E: 1e-3
Enter given value of JSR in dB: 20
Enter given value of processing gain in dB: 30
InÖnite Eb/N0 BEP limit
3.8721e-006
To give BEP of:
1.0000e-003
Requires GP, JSR, and Eb/N0 in dB of:
30.0000
25.0000
9.6085
10.2. COMPUTER EXERCISES
37
BEP for DS BPSK in jamming for processing gain = 30 dB
0
10
-5
PE
10
-10
10
JSR = 5
JSR = 10
JSR = 15
JSR = 20
JSR = 25
-15
10
0
5
No jamming
10
15
Eb/N0, dB
20
25
30
38
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
Computer Exercise 10.4
% ce10_4.m: plots waveforms or spectra for gmsk and msk
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
A = char(í-í,í-.í,íñí,í:.í);
samp_bit = input(íEnter number of samples per bit used in simulation: í);
N_bits = input(íEnter total number of bits: í);
GMSK = input(íEnter 1 for GMSK; 0 for normal MSK: í);
if GMSK == 1
BT_bit = input(íEnter vector of bandwidth X bit periods: í);
N_samp = 50;
end
I_plot = input(íEnter 1 to plot waveforms; 2 to plot spectra and out-of-band power í);
if I_plot == 1
f0 = 1;
elseif I_plot == 2
f0 = 0;
end
clf
kp = pi/2;
T_bit = 1;
LB = length(BT_bit);
for k = 1:LB
B = BT_bit(k)/T_bit;
alpha = 1.1774/B; % Note that B is two-sided bandwidth
del_t = T_bit/samp_bit;
fs = 1/del_t;
data = 0.5*(sign(rand(1,N_bits)-.5)+1);
s = 2*data - 1;
L = length(s);
t=0:del_t:L*T_bit-del_t;
s_t = s(ones(samp_bit,1),:); % Build array whose columns are samp_bit long
s_t = s_t(:)í; % Convert matrix where bit samples occupy columns to vector
L_s_t = length(s_t);
tp=0:del_t:N_samp*T_bit-del_t;
L_t = length(t);
L_tp=length(tp);
t_max = max(tp);
10.2. COMPUTER EXERCISES
if GMSK == 1
hG = (sqrt(pi)/alpha)*exp(-pi^2*(tp-t_max/2).^2/alpha^2);
freq1 = kp*del_t*conv(hG,s_t);
L_hG = length(hG);
freq = freq1(L_hG/2:length(freq1)-L_hG/2);
elseif GMSK == 0
freq = kp*s_t;
end
phase = del_t*cumsum(freq);
if GMSK == 0
y_mod = exp(j*2*pi*f0*t+j*phase);
elseif GMSK == 1
y_mod = exp(j*2*pi*f0*t+j*phase);
end
if I_plot == 1
subplot(3,1,1),plot(t,s_t),axis([min(t), max(t), -1.2, 1.2]), ylabel(íDataí)
if GMSK == 0
title(íMSK waveformsí)
elseif GMSK == 1
title([íGMSK for BT_b = í, num2str(BT_bit(k))])
end
subplot(3,1,2),plot(t,phase), ylabel(íExcess phase, rad.í)
for nn = 1:11
if nn == 1
hold on
end
subplot(3,1,2),plot(t,-nn*(pi/2)*ones(size(t)),ímí)
subplot(3,1,2),plot(t,nn*(pi/2)*ones(size(t)),ímí)
end
subplot(3,1,3),plot(t,real(y_mod)), xlabel(ítí), ylabel(íModulated signalí)
elseif I_plot == 2
Z=PSD(real(y_mod),2048,fs);
% MATLAB supplied function
ZN = Z/max(Z);
ET = sum(ZN);
OBP = 1-cumsum(ZN)/ET;
LZ = length(Z);
del_FR = fs/2/LZ;
FR = [0:del_FR:fs/2-del_FR];
subplot(1,2,1),semilogy(FR,ZN,A(k,:)), axis([0 3 10^(-8) 1])
if k == 1
39
40
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
hold on
grid
ylabel(íPSD, J/Hzí), xlabel(ífT_b_i_tí)
title(íPower spectra and out-of-band power for GMSKí)
end
subplot(1,2,2),semilogy(FR,OBP,A(k,:)), axis([0 3 10^(-8) 1])
if k == 1
hold on
grid
ylabel(íFraction of out-of-band powerí), xlabel(ífT_b_i_tí)
end
if LB == 1
legend([íBT_b_i_t = í,num2str(BT_bit(1))],1)
elseif LB == 2
legend([íBT_b_i_t = í,num2str(BT_bit(1))],
[íBT_b_i_t = í,num2str(BT_bit(2))],1)
elseif LB == 3
legend([íBT_b_i_t = í,num2str(BT_bit(1))],
[íBT_b_i_t = í,num2str(BT_bit(2))],
[íBT_b_i_t = í,num2str(BT_bit(3))],1)
elseif LB == 4
legend([íBT_b_i_t = í,num2str(BT_bit(1))],
[íBT_b_i_t = í,num2str(BT_bit(2))],
[íBT_b_i_t = í,num2str(BT_bit(3))],
[íBT_b_i_t = í,num2str(BT_bit(4))],1)
end
end
end
% End of script Öle
Two typical runs follow - one for plotting waveforms and one for plotting spectra.
>> ce10_4
Enter number of samples per bit used in simulation: 20
Enter total number of bits: 50
Enter 1 for GMSK; 0 for normal MSK: 1
Enter vector of bandwidth X bit periods: .5
Enter 1 to plot waveforms; 2 to plot spectra and out-of-band power: 1
>> ce10_4
Enter number of samples per bit used in simulation: 10
10.2. COMPUTER EXERCISES
41
GMSK for BTb = 0.5
Data
1
0
Modulated signal
Excess phase, rad.
-1
0
5
10
15
20
25
30
35
40
45
0
5
10
15
20
25
30
35
40
45
50
0
5
10
15
20
25
t
30
35
40
45
50
20
0
-20
1
0
-1
42
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
Power0 spectra and out-of-band power for GMSK 0
10
10
-1
10
10
-2
Fraction of out-of-band power
-3
10
PSD, J/Hz
BTbit = 1
BTbit = 1.5
-2
10
-4
10
-5
10
-6
10
-7
10
-3
10
-4
10
-5
10
-6
10
-7
10
10
-8
10
BTbit = 0.5
-1
-8
0
1
2
fTbit
3
10
0
1
2
fTbit
Enter total number of bits: 2000
Enter 1 for GMSK; 0 for normal MSK: 1
Enter vector of bandwidth X bit periods: [.5 1 1.5]
Enter 1 to plot waveforms; 2 to plot spectra and out-of-band power: 2
Computer Exercise 10.5
% ce10_5.m: Mimo Alamouti system with receiver diversity
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
clf; clear all
sigma_n = 1;
N_sim = input(íNo. of bits in simulation => í);
SNRdB = 0:5:25;
SNR = 10.^(SNRdB/10);
k_max = length(SNRdB);
for k = 1:k_max
3
10.2. COMPUTER EXERCISES
sigma_n = 1/sqrt(SNR(k));
h0 = randn(1, N_sim) + j*randn(1, N_sim);
h1 = randn(1, N_sim) + j*randn(1, N_sim);
h2 = randn(1, N_sim) + j*randn(1, N_sim);
h3 = randn(1, N_sim) + j*randn(1, N_sim);
s0 = sign(rand(1, N_sim) - 0.5);
data0 = (s0 + 1)/2;
s1 = sign(rand(1, N_sim) - 0.5);
data1 = (s1 + 1)/2;
n0 = sigma_n*(randn(1, N_sim)+j*randn(1, N_sim));
n1 = sigma_n*(randn(1, N_sim)+j*randn(1, N_sim));
n2 = sigma_n*(randn(1, N_sim)+j*randn(1, N_sim));
n3 = sigma_n*(randn(1, N_sim)+j*randn(1, N_sim));
r0 = h0.*s0 +h1.*s1 + n0;
r1 = -h0.*conj(s1) + h1.*conj(s0) + n1;
r2 = h2.*s0 +h3.*s1 + n2;
r3 = -h2.*conj(s1) + h3.*conj(s0) + n3;
y0 = conj(h0).*r0 +h1.*conj(r1);
y1 = conj(h1).*r0 - h0.*conj(r1);
y2 = conj(h2).*r2 +h3.*conj(r3);
y3 = conj(h3).*r2 - h2.*conj(r3);
s0_hat2 = real(y0+y2);
s1_hat2 = real(y1+y3);
s0_hat1 = real(y0);
s1_hat1 = real(y1);
s0_out1 = (sign(s0_hat1) + 1)/2;
s1_out1 = (sign(s1_hat1) + 1)/2;
error01 = abs(s0_out1 - data0);
error11 = abs(s1_out1 - data1);
error_tot1 = sum(error01 + error11);
Perror1(k) = error_tot1/(2*N_sim);
s0_out2 = (sign(s0_hat2) + 1)/2;
s1_out2 = (sign(s1_hat2) + 1)/2;
error02 = abs(s0_out2 - data0);
error12 = abs(s1_out2 - data1);
error_tot2 = sum(error02 + error12);
Perror2(k) = error_tot2/(2*N_sim);
end
SNRdB
Perror1
43
44
CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS
Perror2
Perror_1rec_tr = 0.5*(1 - sqrt(SNR./(1 + SNR)));
% semilogy(SNRdB, Perror1, íLineWidthí, 1.5, SNRdB, Perror2, í-.í, SNRdB, Perror_1rec_tr, íñí); xlabel(íSNR; dBí),...
semilogy(SNRdB, Perror1, íLineWidthí, 1.5), xlabel(íSNR; dBí),...
ylabel(íBit error probabilityí), grid, hold on, ...
title([íAlamouti-type diversity; í, num2str(N_sim), í bits simulatedí])
semilogy(SNRdB, Perror2, í-.í, íLineWidthí, 1.5)
semilogy(SNRdB, Perror_1rec_tr, íñí, íLineWidthí, 1.5)
legend(í2 trans, 1 recí, í2 trans, 2 recí, í1 trans, 1 recí)
A run produces a plot like Figure 10.31.
Chapter 11
Optimum Receivers and Signal
Space Concepts
Problem 11.1
a. Given H1 , Z = N , so
fZ (zjH1 ) = fN (n) jz=n = 10e10z u (z)
Given H2 , Z = S + N , where S and N are independent. Thus, the resulting pdf under
H2 is the convolution of the separate pdfs of S and N :
Z 1
fZ (zjH2 ) =
fS (z ) fN () d
1
Z 1
=
2e2(z) u (z ) 10e10 u () d
1
Z z
2z
= 20e
e8 d, z 0
0
z
1
= 20e2z e8 , z 0
8
0
2z
8z
= 2:5e
1e
u (z)
2z
10z
= 2:5 e
e
u (z)
which is the result given in the problem statement.
b. The likelihood ratio is
2:5 e2Z e10Z
fZ (ZjH2 )
(Z) =
=
; Z0
fZ (ZjH1 )
10e10Z
= 0:25 e8Z 1 , Z 0
1
2
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
Note that an uppercase Z is used because we substitute data to carry out the test.
c. The threshold is
=
Pr (H1 true) (c21 c11 )
(1=3) (7 0)
1
=
=
Pr (H2 true) (c12 c22 )
2
(2=3) (7 0)
d. The likelihood ratio test becomes
H2
8Z
> 1
0:25 e 1
< 2
H1
This can be reduced to
H2
> ln (4=2 + 1)
Z
= 0:1373
<
8
H1
e. The probability of detection is
Z 1
PD = 1 PM =
2:5 e2z e10z dz
0:1373
1 2z
1 10z 1
= 2:5 e
+ e
2
10
0:1373
1 0:2746
1 1:373
= 2:5 e
e
= 0:6991
2
10
The probability of false alarm is
Z 1
1
PF =
10e10z dz = e10z 0:1373 = e1:373 = 0:2533
0:1373
Therefore, the risk is
Risk
= Pr (H1 ) c21 + Pr (H2 ) c22 + Pr (H2 ) (c12 c22 ) PM
Pr (H1 ) (c21 c11 ) (1 PF )
1
2
2
1
=
7 + 0 + (7 0) (1 0:6991) (7 0) (1 0:2533)
3
3
3
3
7 14
7
=
+
0:3009 0:7467 = 1:9952
3
3
3
3
f. Consider the threshold set at . Then
PF = e10 103 ) 0:3 ln (10) = 0:6908
to give PF 103 . Also, from part (e),
Z 1
2z
PD = 2:5
e
e10z dz
= 1:25e2 0:25e10 ; 0
For values of 0, PD is a monotonically decreasing function with maximum value
of 1 at = 0. For = 0:6908, PD = 0:3137.
g. From part (f),
PF = e10 and PD = 1:25e2 0:25e10
A plot of PD versus PF for various values of constitutes the operating characteristic
of the test.
Problem 11.2
a. The likelihood ratio is
ejZj =2
p =
(Z) =
eZ 2 =2 = 2
r
Z 2 =2jZj
e
2
b. The decision regions are determined by the set of inequalities
r
ejZj =2
Z 2 =2jZj H2
p =
(Z) =
e
?
2
2
eZ =2 = 2
H1
or
H2
Z 2 =2 jZj ? ln
H1
1
ln =
2 2
Thus, solve the equation
Z 2 =2 jZj =
for Z > 0 and Z < 0 for the decision boundary points:
Z > 0 : Z 2 =2 Z = 0 ) Z1;2 = 1
p
1 + 2
p
Z < 0 : Z =2 + Z = 0 ) Z3;4 = 1 1 + 2
2
4
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
For example, for = 0:1
p
Z1;2 = 1 1 + 0:2 = 1 1:0954 = 2.0954 ; 0:0954; Z > 0
p
Z3;4 = 1 1 + 0:2 = 1 1:0954 = 0:0954; -2.0954 ; Z < 0
So the decision regions in this particular case are
R1 :
R2 :
1 < Z < 2:0954; 2:0954 < Z < 1
2:0954 Z 2:0954
Note that the equality sign can be associated with either region.
Problem 11.3
Under both hypotheses, the statistic Z is zero mean and Gaussian. Under hypothesis
H1 (noise alone) Z is zero-mean Gaussian with variance 2n and under hypothesis
H2 (signal plus noise) it is zero-mean Gaussian with variance 2s + 2n : Thus, the
likelihood ratio test is
s
2n exp Z 2 =2 2s + 2n H2
Pr (H1 ) (c21 c11 )
(Z) =
?=
2s + 2n
exp [Z 2 =2 2n ]
Pr (H2 ) (c12 c22 )
H1
or
H2
exp Z 2 =2 2s + 2n + Z 2 =2 2n ?
H1
or
Z2
or
1
1
2n 2s + 2n
= Z2
2s
2
n ( 2s + 2n )
s
2s + 2n
2n
H2
? 2 ln () + ln
H1
2
H2 2 2 + 2
s + 2n
n
s
n
2
Z ?
2 ln () + ln
2s
2n
H1
2s + 2n
2n
For the rest of the problem, we assume that 2s = 3; 2n = 1 so the LRT becomes
H2 4
Z2 ?
H1 3
[2 ln () + ln (4)]
q
The boundaries of the decision regions are at Z1;2 = 43 [2 ln () + ln (4)] which
results in the decision regions
r
r
4
4
R1 :
[2 ln () + ln (4)] Z
[2 ln () + ln (4)]
3
3
5
and
r
R2 : 1 < Z <
4
[2 ln () + ln (4)];
3
r
4
[2 ln () + ln (4)] < Z < 1
3
a. For c11 =qc22 = 0, c21 = c12 , and Pr (H1 ) = Pr (H2 ) = 1=2 we have = 1 so that
Z1;2 =
4
3 ln (4) = 1:3596.
b. For c11 = c22 = 0, c21 = c12 , p0 = Pr (H1 )q
= 1=4, and q0 = 1 p0 = Pr (H2 ) = 3=4
1 4
1
we have = 4 3 = 3 so that Z1;2 = 43 [2 ln (1=3) + ln (4)] which is imaginary
(i.e., Z 2 is being compared with a negative number in the LRT) which means that
the decision is in favor of H2 (signal plus noise).
c. For c11 = c22 = 0, c21 = c12 =2, p0 = Pr (H1 ) = Pr (H2 ) = 1=2 we have = cc21
= 12 so
12
q
that Z1;2 = 43 [2 ln (1=2) + ln (4)] = 0 so the decision is in favor of H2 .
d. For c11 = c22 = 0, c21 = 2c12 , p0 = Pr (H1 ) = Pr (H2 ) = 1=2 we have = cc21
= 2 so
12
q
that Z1;2 = 43 [2 ln (2) + ln (4)] = 1:9227.
Problem 11.4
For 2n = 9; 2s = 16 the LRT is
H2 9 (25)
25
Z2 ?
2 ln () + ln
= 28:125 ln () + 14:367
16
9
H1
p
The boundaries of the decision regions are at Z1;2 = 28:125 ln () + 14:367 which
results in the decision regions
p
p
R1 : 28:125 ln () + 14:367 Z 28:125 ln () + 14:367
and
p
p
R2 : 1 < Z < 28:125 ln () + 14:367; 28:125 ln () + 14:367 < Z < 1
a. For c11 =pc22 = 0, c21 = c12 , and Pr (H1 ) = Pr (H2 ) = 1=2 we have = 1 so that
Z1;2 = 14:367 = 3:7904 giving decision regions of
R1 :
R2 :
3:7904 Z 3:7904
1 < Z < 3:7904; 3:7904 < Z < 1:
6
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
The probability of false alarm is the probability that a decision is made in favor of
signal plus noise when noise alone is present, and is given in this case by
Z 3:7904
Z 3:7904
exp z 2 =2 9
exp z 2 =2 9
z
p
p
PF =
dz = 2
dz; u =
3
2
9
2
9
3:7904
"0 Z
#
2
Z 3:7904=3
1
exp u =2
exp u2 =2
1
p
p
= 2
du = 2
du
2
2
2
0
3:7904=3
= 1 2Q (1:2635) = 0:7936
The probability of detection is the probability that a decision is made in favor of signal
plus noise when signal plus noise present, and is given in this case by
Z 3:7904
Z 1
exp z 2 =2 25
exp z 2 =2 25
p
p
PD =
dz +
dz
2 25
2 25
1
3:7904
Z 1
exp z 2 =2 25
z
p
= 2
dz; u =
5
2 25
3:7904
2
Z 1
exp u =2
p
= 2
du
2
3:7904=5
= 2Q (0:7581) = 0:4484
The risk is
Risk
= Pr (H1 ) c21 + Pr (H2 ) c22 + Pr (H2 ) (c12 c22 ) PM
Pr (H1 ) (c21 c11 ) (1 PF )
1
1
1
1
=
1 + 0 + 1 (1 0:4484) 1 (1 0:7936)
2
2
2
2
= 0:6726
b. For c11 = c22 =
p0, c21 = c12 = 1, Pr (H1 ) = 1=4; and Pr (H2 ) = 3=4 we have = 1=3 so
that Z1;2 = 28:125 ln (1=3) + 14:367 = imaginary which says Z 2 is being compared
with a negative number in the LRT which means that the decision is in favor of H2
(signal plus noise). The probability of false alarm is the probability that a decision is
made in favor of signal plus noise when noise alone is present, and is 0 in this case.
Since the decision is always in favor of H2 the probability of detection is 1. The risk
is
Risk
= Pr (H1 ) c21 + Pr (H2 ) c22 + Pr (H2 ) (c12 c22 ) PM
Pr (H1 ) (c21 c11 ) (1 PF )
1
3
3
1
=
1 + 0 + 1 (1 1) 1 (1 0)
4
4
4
4
= 0
7
c. For c11 = c22 = p
0, c21 = c12 =2 = 1=2, and Pr (H1 ) = Pr (H2 ) = 1=2 we have = 1=2
so that Z1;2 = 28:125 ln (1=2) + 14:367 =imaginary giving which gives PF = 0 and
PD = 1. The risk is
Risk
= Pr (H1 ) c21 + Pr (H2 ) c22 + Pr (H2 ) (c12 c22 ) PM
Pr (H1 ) (c21 c11 ) (1 PF )
1 1 1
1
1 1
=
+ 0 + 1 (1 1) (1 0)
2 2 2
2
2 2
= 0
d. For c11 =pc22 = 0, c21 = 2c12 = 2, and Pr (H1 ) = Pr (H2 ) = 1=2 we have = 2 so that
Z1;2 = 28:125 ln (2) + 14:367 = 5:8191 giving decision regions of
R1 :
R2 :
5:8191 Z 5:8191
1 < Z < 5:8191; 5:8191 < Z < 1:
The probability of false alarm is the probability that a decision is made in favor of
signal plus noise when noise alone is present, and is given in this case by
Z 5:8191
Z 5:8191
exp z 2 =2 9
exp z 2 =2 9
z
p
p
PF =
dz = 2
dz; u =
3
2 9
2 9
5:8191
"0 Z
#
2
Z 5:8191=3
2
1
exp u =2
exp u =2
1
p
p
= 2
du = 2
du
2
2
2
0
5:8191=3
= 1 2Q (1:9397) = 0:9476
The probability of detection is the probability that a decision is made in favor of signal
plus noise when signal plus noise present, and is given in this case by
Z 5:8191
Z 1
exp z 2 =2 25
exp z 2 =2 25
p
p
PD =
dz +
dz
2 25
2 25
1
5:8191
Z 1
exp z 2 =2 25
z
p
= 2
dz; u =
5
2 25
5:8191
2
Z 1
exp u =2
p
= 2
du
2
5:8191=5
= 2Q (1:1638) = 0:2445
The risk is
Risk
= Pr (H1 ) c21 + Pr (H2 ) c22 + Pr (H2 ) (c12 c22 ) PM Pr (H1 ) (c21 c11 ) (1 PF )
1
1
1
1
=
2 + 0 + 1 (1 0:2445) 2 (1 0:9476)
2
2
2
2
= 1:3254
8
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
Problem 11.5
DeÖne A = (a1 ; a2; a3 ) and B = (b1 ; b2 ; b3 ). DeÖne scalar multiplication by (a1 ; a2; a3 ) =
(a1 ; a2; a3 ) and vector addition by (a1 ; a2; a3 ) +(b1 ; b2 ; b3 ) = (a1 + b1 ; a2 + b2 ; a3 + b3 ).
The properties listed under îStructure of Signal Spaceî in the text then become:
1. 1 A + 2 B = (1 a1 + 2 b1 ; 1 a2 + 2 b2 ; 1 a3 + 2 b3 ) R3 ;
2. (A + B) = (a1 + b1; a2 + b2 ; a3 + b3 )
= (a1 + b1; a2 + b2 ; a3 + b3 ) = (a1 ; a2; a3 ) + (b1; b2 ; b3 )
= A + B R3 ;
3. 1 (2 A) = (1 2 A) R3 (follows by writing out in component form);
4. 1 A = A (follows by writing out in component form);
5. The unique element 0 is (0; 0; 0) so that A + 0 = A;
6. A = (a1 ; a2; a3 ) so that A + (A) = 0:
Problem 11.6
p
p
p
p
p
p
a. jAj = A A = 12 + 32 + 22 = 14; jBj = B B = 52 + 12 + 32 = 35;
AB
p p
cos = jAjjBj
= 15+31+23
= 0:6325.
14 35
p
p
p
p
p
p
b. jAj = A A = 62 + 22 + 42 = 56; jBj = B B = 22 + 22 + 22 = 12;
AB
p p
cos = jAjjBj
= 62+22+42
= 0:9258.
56 12
p
p
p
p
p
p
c. jAj = A A = 42 + 32 + 12 = 26; jBj = B B = 32 + 42 + 52 = 50;
AB
p p
cos = jAjjBj
= 43+34+15
= 0:8043.
26 50
p
p
32 + 32 + 22 =
p
p
d. jAj = A A =
22; jBj = B B =
p
3(1)+3(2)+23
AB
p p
14; cos = jAjjBj =
= 0:1709.
22 14
q
(1)2 + (2)2 + 32 =
9
Problem 11.7
Consider
(x; y) = lim
Z T
x (t) y (t) dt
Z T
T !1 T
Then
(x; y) = lim
Z T
y (t) x (t) dt =
(x; y) = lim
Z T
x (t) y (t) dt = lim
T !1 T
Also
T !1 T
lim
T !1 T
x (t) y (t) dt
Z T
T !1 T
= (x; y)
x (t) y (t) dt = (x; y)
and
(x + y; z) =
lim
Z T
T !1 T
Z T
=
lim
T !1 T
[x (t) + y (t)] z (t) dt
x (t) z (t) dt + lim
T !1 T
= (x; z) + (y; z)
Finally
(x; x) = lim
Z T
T !1 T
Z T
x (t) x (t) dt = lim
Z T
T !1 T
y (t) z (t) dt
jx (t)j2 dt 0
The scalar product for power signals is considered in the same manner.
Problem 11.8
a. This scalar product is
(x1 ; x2 ) = lim
Z T
T !1 0
1
2
2e4t dt =
b. Use the energy signal scalar product deÖnition as in (a):
(x1 ; x2 ) = lim
Z T
T !1 0
e(7j2)t dt =
1
7 2j
c. Use the power signal product deÖnition:
1
(x1 ; x2 ) = lim
T !1 2T
Z T
T
cos (2t) cos (4t) dt = 0
10
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
d. Use the power signal scalar product deÖnition:
1
T !1 2T
(x1 ; x2 ) = lim
Z T
5 cos (2t) dt = 0
0
Problem 11.9
Since the signals are assumed real,
kx1 + x2 k
2
=
=
=
lim
Z T
T !1 T
Z T
lim
[x1 (t) + x2 (t)]2 dt
x21 (t) dt + 2
lim
Z T
T !1 T
T !1 T
2
kx1 k + 2 (x1 ; x2 ) + kx2 k2
x1 (t) x2 (t) dt + lim
Z T
T !1 T
x22 dt
From this it is seen that
kx1 + x2 k2 = kx1 k2 + kx2 k2
if and only if (x1 ; x2 ) = 0.
Problem 11.10
By straight forward integration, it follows that
kx2 k2
kx3 k2
Z 1
p
kx1 k = 2
1
r
Z 1
2
2
2
=
t dt = or kx2 k =
3
3
1
r
Z 1
8
8
2
=
(1 + t) dt = or kx3 k =
3
3
1
kx1 k2 =
12 dt = 2 or
Also
(x1 ; x2 ) = 0 and (x3 ; x1 ) = 2
Since (x1 ; x2 ) = 0, they are orthogonal. Choose them as basis functions (not normalized). Clearly, x3 is their vector sum.
11
Problem 11.11
It follows that
kx1 k2 =
=
=
and similarly kx2 k2 =
X
X
an n ;
n
XX
bm m
m
!
an bm (n ; m )
n m
N
X
n=1
N
X
n=1
jan j2 because (n ; m ) = mn
jbn j2
Also, in a similar fashion,
(x1 ; x2 ) =
N
X
an bn
n=1
Thus we must show that
N
2
N
N
X
X
X
2
an bn
jan j
jbn j2
n=1
n=1
n=1
Writing the both sides out we get
XX
n
m
an am bn bm
XX
n
an an bm bm
m
This is equivalent to
or
XX
n
m
XX
n
m
(2an an bm bm 2an am bn bm ) 0
(an an bm bm 2an am bn bm + am am bn bn )
0
The inequality is therefore equivalent to
XX
m
n
jan bm am bn j2 0
which is seen to be true because each term in the sum is nonnegative.
12
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
Problem 11.12
p
p
p
p
p
p
2 + 12 + 32 =
a. jAj = A A = 12 + 32 + 22 = 14; jBj = B B =
5
35;
p p
A B = 1 5 + 3 1 + 2 3 = 14; jA Bj = 14 jAj jBj = 14 35 = 22:1359 which
is true.
p
p
p
p
p
p
2 + 22 + 22 =
b. jAj = A A = 62 + 22 + 42 = 56; jBj = B B =p 2p
12;
A B = 6 2 + 2 2 + 4 2 = 24; jA Bj = 24 jAj jBj = 56 12 = 25:923 which
is true.
p
p
p
p
p
p
2 + 42 + 52 =
c. jAj = A A = 42 + 32 + 12 = 26; jBj = B B =
3
50;
p p
A B = 4 3 + 3 4 + 1 5 = 29; jA Bj = 29 jAj jBj = 26 50 = 36:0555 which
is true.
q
p
p
p
p
d. jAj = A A = 32 + 32 + 22 = 22; jBj = B B = (1)2 + (2)2 + 32 =
p
p p
14; A B = 3 (1) + 3 (2) + 2 3 = 3; jA Bj = 3 jAj jBj = 22 14 =
17:5499 which is true.
Problem 11.13
a. A set of normalized basis functions is
1
p s1 (t)
2
r
2
1
2 (t) =
s2 (t) s1 (t)
3
2
p
2
2
3 (t) =
3 s3 (t) s1 (t) s2 (t)
3
3
1 (t) =
b. Evaluate (s1 ; 1 ), (s2 ; 1 ), and (s3 ; 1 ) to get
s1 =
p
1
21 ; s2 = p 1 +
2
r
p
3
2 ; s3 = 21 +
2
r
2
( + 3 )
3 2
Note that a basis function set could have been obtained by inspection. For example,
three unit-height, unit-width, nonoverlapping pulses spanning the interval could have
been used.
13
Problem 11.14
Normalize the Örst vector to form the Örst orthonormal vector:
x1
3bi + 2b
jb
k
=q
jx1 j
32 + 22 + (1)2
1 b
= p
3i + 2b
jb
k
14
= 0:8018bi + 0:5345b
j 0:2673b
k
1 =
Take the second vector and subtract its component along 1 :
1
1 b
3i + 2b
jb
k 2bi + 5b
j+b
k p
3bi + 2b
jb
k
2bi + 5b
j+b
k p
14
14
= 2:6429bi + 4:5714b
j + 1:2143b
k
v2 =
Normalize v2 to form the second orthonormal vector:
2 =
q
2:6429bi + 4:5714b
j + 1:2143b
k
(2:6429)2 + (4:5714)2 + (1:2143)2
= 0:4878bi + 0:8437b
j + 0:2241b
k
Take the third vector and subtract its components along 1 and 2 :
v3 =
6bi 2b
j + 7b
k
0:8018bi + 0:5345b
j 0:2673b
k 6bi 2b
j + 7b
k 0:8018bi + 0:5345b
j 0:2673b
k
0:4878bi + 0:8437b
j + 0:2241b
k 6bi 2b
j + 7b
k 0:4878bi + 0:8437b
j + 0:2241b
k
= 3:0146bi 0:4307b
j + 8:1825b
k
Normalize:
3:0146bi 0:4307b
j + 8:1825b
k
p
3:01462 + 0:43072 + 8:18252
= 0:3453bi 0:0493b
j + 0:9372b
k
3 =
Finally, take the fourth vector and subtract components of the other three along it:
14
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
v4 =
3bi + 8b
j 3b
k
0:8018bi + 0:5345b
j 0:2673b
k 3bi + 8b
j 3b
k 0:8018bi + 0:5345b
j 0:2673b
k
0:4878bi + 0:8437b
j + 0:2241b
k 3bi + 8b
j 3b
k 0:4878bi + 0:8437b
j + 0:2241b
k
0:3453bi 0:0493b
j + 0:9372b
k 3bi + 8b
j 3b
k 0:3453bi 0:0493b
j + 0:9372b
k
= 0bi + 0b
j + 0b
k (actually of the order of 1014 )
Therefore there are only three orthornormal vectors.
Problem 11.15
a. A basis set is
r
2
cos (2fc t) and
T
r
2
2 (t) =
sin (2fc t) for 0 t T
T
1 (t) =
b. The coordinates of the signal vectors are
xi =
Z T
0
and
yi =
Z T
0
si (t) 1 (t) dt =
p
E cos
p
si (t) 2 (t) dt = E sin
i
4
i
4
where E = A2 T . Thus
p
p
i
i
si (t) = E cos
1 (t) E sin
2 (t) ; i = 0; 1; 2; 3; 4; 5; 6; 7
4
4
p
The signal point for i = 0 is at
E; 0 with the rest at multiples of =4 radians on
p
a circle of radius E centered at the origen.
15
Problem 11.16
a. Normalize x1 (t) to produce 1 (t):
2
jjx1 jj
=
Z 1
0
t 2
e
dt =
Z 1
0
1 1
1
e2t dt = e2t 0 =
2
2
x1 (t) p
) 1 (t) = p = 2et u (t)
1= 2
Take the second signal and subtract its component along 1 :
Z 1p
p
2et e2t dt 2et u (t)
"0 p
#1
p t
2
= e2t u (t)
e3t
2e u (t)
3
2t
v2 (t) = e
u (t)
0
2
= e2t u (t) et u (t)
3
Normalize v2 (t) to produce 2 (t):
2
jjv2 jj
Z 1
Z 1
4 3t 4 2t
2 t 2
4t
=
e
e
dt =
e
e
+ e
dt
3
3
9
0
0
1 4t 4 3t 2 2t 1
1
=
e
+ e
e
=
4
9
9
36
0
v2 (t) 2t
) 2 (t) =
= 6e
4et u (t)
1=6
2t
Take the third signal and subtract its components along 1 and 2 :
3t
v3 (t) = e
u (t)
Z 1
Z 1p
p
2et e3t dt 2et u (t)
0
6e2t 4et e3t dt 6e2t u (t) 4et u (t)
0
6 2t
3 t
3t
=
e
e
+ e
u (t)
5
10
16
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
Normalize v3 (t) to produce 3 (t):
2
jjv3 jj
Z 1
6 2t
3 t 2
=
e
e
+ e
dt
5
10
0
Z 1
12 5t 51 4t 18 3t
9 2t
6t
=
e
e
+ e
e
+
e
dt
5
25
25
100
0
1
=
600
p
v3 (t)
p = 6 10e3t 12e2t + 3et u (t)
) 3 (t) =
1=10 6
3t
b. There does not appear to be a general pattern developing.
Problem 11.17
a. Normalize x1 (t) to produce 1 (t):
2
jjx1 jj
Z 1
1 3 1
2
t 1 =
3
3
1
r
x1 (t)
3
) 1 (t) = p
=
t; 1 t 1
2
2=3
=
t2 dt =
Take the second signal and subtract its component along 1 :
r
Z 1r
3 2
3
2
v2 (t) = t
tt dt
t
2
2
1
1
3 t4
2
= t
t
2 4 1
= t2 ; 1 t 1
Normalize v2 (t) to produce 2 (t):
2
jjv2 jj
=
=
Z 1
2 2
t dt =
1
5 1
t
5
1
=
Z 1
t4 dt
1
2
5
x2 (t)
) 2 (t) = p
=
2=5
r
5 2
t ; 1 t 1
2
17
Take the third signal and subtract its components along 1 and 2 :
r
r
Z 1r
Z 1r
3 3
3
5 2 3
5 2
3
v3 (t) = t
tt dt
t
t t dt
t
2
2
2
2
1
1
1
1
3 t5
5 t6
3
= t
t
t2
2 5 1
2 6 1
3
= t3 t; 1 t 1
5
Normalize v3 (t) to produce 3 (t):
jjv3 jj2
=
Z 1
1
3
t3 t
5
2
dt =
Z 1
1
6
9
t6 t4 + t2 dt
5
25
7
1
t
6 t5
9 t3
2 62
9 2
8
=
+
=
+
=
7
55
25 3 1 7 5 5 25 3
175
r
v3 (t)
7 5 3 3
) 3 (t) = p
=
t t ; 1 t 1
2 2
2
8=175
Take the fourth signal and subtract its components along 1 , 2 , and 3 :
r
r
Z 1r
Z 1r
3 4
3
5 2 4
5 2
4
v4 (t) = t
tt dt
t
t t dt
t
2
2
2
2
1
1
r
r
Z 1
7 5 3 3
7 5 3 3
4
t t t dt
t t
2 2
2
2 2
2
1
6 1
7 1
8
1
3 t
5 t
7 5t
3 t6
5 3 3
4
2
= t
t
t
t t
2 6 1
2 7 1
2 28
2 6 1 2
2
5
= t4 t2 ; 1 t 1
7
Normalize v4 (t) to produce 4 (t):
2
jjv4 jj
=
Z 1
1
9
t
5
t t2
7
4
10 t7
2
dt =
1
25 t5
Z 1
1
10 6 25 4
t t + t dt
7
49
8
2 10 2 25 2
8
+
=
9
7 7
49 5 1 9
7 7 49 5
9 49
xv4 (t)
27
5
) 4 (t) = p
= p
t4 t2 ; 1 t 1
7
2 2
8= (9 49)
=
+
=
18
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
b. The pattern is not apparent.
Problem 11.18
Normalize s1 (t) to produce 1 (t):
Z 2
2
jjs1 jj =
12 dt = 2
0
s1 (t)
1
) 1 (t) = p = p ; 0 t 2
2
2
Take the second signal and subtract its component along 1 :
Z 2
1
1
p cos (t) dt p
v2 (t) = cos (t)
2
2
0
2
1 1
= cos (t)
sin (t)
2
0
= cos (t) ; 0 t 2
Normalize it:
2
jjv2 jj
1 1
=
cos (t) dt =
+ cos (2t) dt = 1
2 2
0
0
) 2 (t) = s2 (t) = cos (t) ; 0 t 2
Z 2
2
Z 2
A similar operation with the third signal shows that
3 (t) = s3 (t) = sin (t) ; 0 t 2
Take the fourth signal and subtract its components along 1 and 3 :
Z 2
Z 2
1
1
2
2
p
p
v4 (t) = sin (t)
sin (t) dt
cos (t) sin2 (t) dt cos (t)
2
2
0
0
Z 2
sin (t) sin2 (t) dt sin (t)
0
2
= sin (t)
1
1 1
1
= cos (2t)
2
2 2
2
1
= cos (2t) ; 0 t 2
2
Normalize it:
Z 2
Z
1
1 2 1 1
1
2
2
jjv4 jj =
cos (2t) dt =
+ cos (4t) dt =
4
4
2
2
4
0
0
v4 (t)
) 4 (t) =
= cos (2t) ; 0 t 2
1=2
19
From these results we Önd that
p
s1 (t) =
21 (t)
s2 (t) = 2 (t)
s3 (t) = 3 (t)
1 1
1
1
s4 (t) =
cos (2t) = p 1 (t) + 4 (t)
2 2
2
2
Problem 11.19
By the time delay and modulation theorems,
1
1
(f ) =
sinc f
+ sinc f +
ej2k f
2
2
2
The total energy in the kth signal is
E=
Z =2
2
cos (t= ) dt =
=2
Z =2
=2
1 1
+ cos (2t= ) dt =
2 2
2
The energy for jf j W is
EW
2
1
1 2
=
sinc f
+ sinc f +
df
2
2
W 4
Z W
1
1 2
=
sinc v
+ sinc v +
dv
2
2
W 4
Z W
1
1 2
= 2
sinc v
+ sinc v +
dv (even integrand)
4
2
2
0
Z W
Thus
EW
=
E
Z W
1
1 2
sinc v
+ sinc v +
dv
2
2
0
The MATLAB program below computes EW =E:
% pr11_19
%
disp(í W_tau E_W/Eí)
for tau_W = 0.7:.1:1.3
v = 0:0.01:tau_W;
y = (sinc(v-0.5)+sinc(v+0.5)).^2;
20
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
EW_E = trapz(v, y);
disp([tau_W, EW_E])
end
The results are given below:
>> pr11_19
W_tau E_W/E
0.7000 0.8586
0.8000 0.9106
0.9000 0.9467
1.0000 0.9701
1.1000 0.9838
1.2000 0.9909
1.3000 0.9939
The product 0:8 W 0:9 for EW =E 11
12 = 0:9167.
Problem 11.20
a. By inspection of the given waveform set it follows that a satisfactory set of orthonormal
functions is
1 (t) =
2 (t) =
3 (t) =
4 (t) =
p
p
p
p
2 cos (2t) ; 0 t 1
2 sin (2t) ; 0 t 1
2 cos (4t) ; 0 t 1
2 sin (4t) ; 0 t 1
For i = 0; 1; 2; 3
si (t) = A cos (i=2) cos (2t) A sin (i=2) sin (2t)
p
=
E [cos (i=2) 1 (t) sin (i=2) 2 (t)]
For i = 4; 5; 6; 7
si (t) = A cos ((i 4) =2) cos (4t) A sin ((i 4) =2) sin (4t)
p
=
E [cos ((i 4) =2) 3 (t) sin ((i 4) =2) 4 (t)]
b. The receiver looks like Figure 11.5 with four multipliers and integrators, one for each
21
orthonormal function. The signal coordinates are
A = [Aij ; i = 0; 1; 2; 3; 4; 5; 6; 7; j = 1; 2; 3; 4]
8 hp
i
>
E; 0; 0; 0; 0; 0; 0; 0 ; i = 0
>
>
>
h
i
>
p
>
>
0;
E;
0;
0;
0;
0;
0;
0
; i=1
>
>
>
h
i
>
p
>
>
0; 0; E; 0; 0; 0; 0; 0 ; i = 2
>
>
>
h
i
>
p
>
< 0; 0; 0; E; 0; 0; 0; 0 ; i = 3
h
i
=
p
>
0;
0;
0;
0;
E;
0;
0;
0
; i=4
>
>
>
h
i
>
p
>
>
0; 0; 0; 0; 0; E; 0; 0 ; i = 5
>
>
>
h
i
>
p
>
>
0;
0;
0;
0;
0;
0;
E;
0
; i=6
>
>
>
h
>
p i
>
: 0; 0; 0; 0; 0; 0; 0; E ; i = 7
From (11:96) the decision rule is
Choose H` so that d2 =
4
X
j=1
(Zj A`j )2 = minumum
where
Zj = Aij + Nj =
Z T =1
0
y (t) j (t) dt =
Z 1
0
[si (t) + n (t)] j (t) dt; j = 1; 2; 3; 4
This can be done suboptimally in two stages: (1) Correlate with two tones, one at
frequency 1 Hz and the other at frequency 2 Hz and choose the output having the
largest envelope (or squared envelope) (note that this will involve a correlation with
both a sine and a cosine at each frequency); (2) Taking the pair of outputs from the
correlators (one sine and one cosine) having the largest output envelope, determine
in which =2 radian range the output pair lies (i.e., 0 to =2, =2 to , to 3=2, or
3=2 to 2 radians).
c. As before E [Ni Nj ] = N20 ij . According to the suboptimum strategy given above, we
make the comparison
(y; 1 )2 + (y; 2 )2
>
(y; 3 )2 + (y; 4 )2
<
Suppose s0 (t) was really transmitted. Then the probability of a correct decision on
22
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
frequency is
Pcf
= Pr
= Pr
p
p
E + N1
E + N1
2
2
+ N22 > N32 + N42
> N32 + N42 N22
The probability of correct decision on signal phase, from the analysis of quadriphase
in Chapter 10 resulting in (10:15), is
Pcp
= 1 2Q
r
E
N0
!
By symmetry, these expressions hold for either frequency or any phase. The overall
probability of correct reception is Pc = Pcf Pcp for an overall probability of error of
PE = 1 Pcf Pcp
Problem 11.21
Write
Z 1
2
ey
p
Pc =
1
"Z
Pc =
1
p
E=N0
1
and let z = x y. This gives
Z pE=N0
y+
2
ez =2
p
"Z
#
2
ex
p dx dy
#
2
e2(y+z=2)
p
dy dz
1
1
Complete the square in the exponent
p of the inside integral and use a table of deÖnite
integrals to show that is evaluates to 1= 2. The result then reduces to
Pc = 1 Q
which gives PE = Q
hp
p
E=N0
i
E=N0 , the desired result.
23
Problem 11.22
a. The space is three-dimensional with signal points at the eight points
p
p
p
E=3 ; E=3 ; E=3
The optimum partitions are planes determined by the coordinate axes taken two at
a time (three planes). Thus the optimum decision regions are the eight quadrants of
the signal space.
b. Consider S1 . Given the partitioning discussed in part (a), we make a correct decision
only if
kZ S1 k2 < kZ S2 k2
and
kZ S1 k2 < kZ S4 k2
and
kZ S1 k2 < kZ S8 k2
where S2 , S4 , and S8 are the nearest-neighbor signal points topS1 . These are the
most
Z = (Z1 ; Z2 ; Z3p
) and S1 = E=3 (1; 1; 1) ; S2 =
p probable errors. Substituting
p
E=3 (1; 1; 1; ) ; S4 = E=3 (1; 1; 1), and S8 = E=3 (1; 1; 1), the above conditions become
2
2
2
2
p
p
p
p
Z1 + E=3
<
Z1 E=3 ; Z2 + E=3 < Z2 E=3 ;
2
2
p
p
and Z3 + E=3
<
Z3 E=3
by canceling like terms on each side. For S1 , these reduce to
Z1 > 0; Z2 > 0; and Z3 > 0
to deÖne the
decision region for S1 . Therefore, the probability of correct decision is
P [correct dec.js1 (t)] = Pr (Z1 > 0; Z2 > 0; Z3 > 0) = [Pr (Zi > 0)]3
because the noises along each coordinate axis are independent. Note that E [Zi ] =
(Es =3)1=2 , all i. The noise variances, and therefore the variances of the Zi s are all
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
0
10
-1
10
-2
10
Ps
24
-3
10
n=4
n=3
n=2
n=1
-4
10
-5
10
-6
10
0
5
10
15
Eb/N0
N0 =2. Thus
Z 1
2 3
p
1
1
exp
y Es =3
dy
N0
N0 0
"
r
2 #
Z 1
p
1
u
2
p
=
exp
du ; u =
y Es =3
p
2
N0
2 2Es =3N0
h
p
i3
= 1Q
2Es =3N0
P [correct dec.js1 (t)] =
p
Since this is independent of the signal chosen, this is the average probability of correct
detection. The symbol error probability is 1 Pc . Generalizing to n dimensions, we
have
h
p
in
h
p
in
PE = 1 1 Q
2Es =nN0
=1 1Q
2Eb =N0
where n = log2 M . Note that Eb = Es =n since there are n bits per dimension and
M = 2n .
c. The symbol error probability is plotted in Fig. 11.1.
25
Problem 11.23
a. Integrate the product of the i 1 and i signals over [0; Ts ]:
Z Ts
=
=
=
'
A2 cos f2 [fc + (i 1) f ] tg cos [2 (fc + if ) t] dt
Z
A2 Ts
fcos [2 (2fc + (2i 1) f ) t]g dt +
fcos [2 (f ) t]g dt
2 0
2 0
A2 sin [2 (2fc + (2i 1) f ) t] sin [2 (f ) t] Ts
+
2
2 (2fc + (2i 1) f )
2 (f )
0
2
A
sin [2 (2fc + (2i 1) f ) Ts ] sin [2 (f ) Ts ]
+
2
2 (2fc + (2i 1) f )
2 (f )
2
2
A Ts sin [2f Ts ]
A Ts
=
sinc (2f Ts )
2
2f Ts
2
0
Z
A2 Ts
where the Örst term has been neglected because it is small for typical values of fc .
The remaining term is 0 for its argument equal to a nonzero integer. The smallest
such integer is 1, which gives
2f Ts = 1 or f =
1
2Ts
b. We need 1=2Ts Hz of bandwidth per signal, so
M=
W
M
= 2W Ts or W =
1=2Ts
2Ts
c. For vertices-of-a-hypercube signaling M = 2n where n is the dimensionality (number
of orthogonal functions) which is 2W Ts , so
log2 M = 2W Ts or W =
log2 M
2Ts
Problem 11.24
In the equation
P (EjH1 ) =
Z 1 Z 1
0
r1
fR2 (r2 jH1 ) dr2 fR1 (r1 jH1 ) dr1
26
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
substitute
fR1 (r1 jH1 ) =
2
2
2r1
er1 =(2E +N0 ) ; r1 0
2
2E + N0
and
fR2 (r2 jH1 ) =
The inside integral becomes
2r2 r22 =N0
e
; r2 > 0
N0
2
I = er1 =N0
When substituted in the Örst equation, it can be reduced to
Z 1
Z 1
2r1
2r1
2
r12 =(2E 2 +N0 ) r12 =N0
P (EjH1 ) =
e
e
dr1 =
er1 =K dr1
2+N
2+N
2E
2E
0
0
0
0
where
N0 2E 2 + N0
K=
2 (E 2 + N0 )
The remaining integral is then easily integrated since its integrand is a perfect di§erential
if written as
Z 1
K
2r1 r12 =K
P (EjH1 ) =
e
dr1
2
2E + N0 0 K
In the integral, let v = r12 =K to get
Z 1
Z 1
2r1 r12 =K
e
dr1 =
ev dv = 1
K
0
0
so that Önally
P (EjH1 ) =
=
N0 2E 2 + N0
K
1
=
2E 2 + N0
2E 2 + N0 2 (E 2 + N0 )
N0
1
=
2
2 (E 2 + N0 )
2 1 + 1 2E
2 N0
which is the same as (10.142).
Problem 11.25
a. By deÖnition, the new signal set is
s0i (t) = si (t)
M 1
1 X
sj (t)
M
j=0
27
The signal energy for the ith signal in the new set is
2
32
Z Ts
M
1
X
4si (t) 1
Es0 =
sj (t)5 dt
M
0
j=0
2
3
Z Ts
M
1
M
1 M
1
X
X
X
2
1
4s2i (t)
=
si (t)
sj (t) + 2
sj (t) sk (t)5 dt
M
M
0
j=0
=
Z Ts
0
s2i (t) dt
= Es
M 1
2
M
j=0 k=0
M
1 Z Ts
X
j=0
0
M 1 M 1 Z Ts
1 X X
si (t) sj (t) dt + 2
M
j=0 k=0
sj (t) sk (t) dt
0
M 1 M 1
2 X
1 X X
Es ij + 2
Es jk ; jk = 0; j 6= k and 1; j = k
M
M
j=0
j=0 k=0
M
1
X
2
1
2
1
Es + 2
Es = Es
Es +
Es
M
M
M
M
j=0
1
M 1
= Es 1
= Es
M
M
= Es
b. The correlation coe¢cient between two di§erent signals is
2
3"
#
Z Ts
M
1
M
1
X
X
1
1
1
4sm (t)
mn =
sj (t)5 sn (t)
sk (t) dt; m 6= n
Es0 0
M
M
j=0
k=0
#
P
P 1
Z Ts "
M
1
1
1
sm (t) sn (t) M
sn (t) j=0 sj (t) M
sm (t) M
sk (t)
1
k=0
P 1 PM 1
=
dt
Es0 0
+ M12 M
j=0
k=0 sj (t) sk (t)
2 RT
3
R
s
1 PM 1 Ts
s
(t)
s
(t)
dt
s
(t)
s
(t)
dt
m
n
n
j
j=0
0
RM
1 6 0
7
1 PM 1 Ts
=
s
(t)
s
4
5
m
k (t) dt
k=0
M
0 R
Es0
P
P
1
M 1 Ts
+ M12 M
s
(t)
s
(t)
dt
j
k
j=0
k=0 0
2
3
M 1
M 1
M 1 M 1
1 4
1 X
1 X
1 X X
=
Es mn
Es nj
Es nk + 2
Es jk 5
Es0
M
M
M
j=0
j=0
k=0
k=0
1
Es Es Es
M
Es
=
0
+
=
Es0
M
M
M
Es (M 1)
M
1
=
; m 6= n
M 1
c. Simply solve for Es in the above expression and substitute for Es in the orthogonal
28
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
signaling result. This gives
s
!#)M 1 v2 =2
Z 1( "
2Es0
M
e
p
Pc; simplex = 1
Q v+
dv
N
M
1
2
0
1
d. The union bound approximation for the symbol error probability for coherent orthogonal signaling is given by (10.67) as
r !
Es
Ps; orthog ' (M 1) Q
N0
Using the substitution used in part c, we get
Ps; simplex ' (M 1) Q
s
Es0
N0
M
M 1
!
The bit error probability can be approximated as
s
!
M
log2 (M ) Eb
M
Pb; simplex '
Q
2
N0
M 1
Plots are given in Fig. 11.2.
Problem 11.26
a. Use the basis set
r
2
cos ! i t; 0 t T
T
r
2
si (t) =
sin ! i t; 0 t T; i = 1; 2; ; M
T
ci (t) =
Base the decision on Z = (Zc1 ; Zs1 ; :::; ZcM ; ZsM ) where Zci = (y; ci ) and Zsi =
(y; si ). Suppose that the ith hypothesis is true. Then the components of Z are
p
Zci =
Ej Gcj + Ncj ; i = j
Zsi
= Ncj ; i 6= j
p
=
Ej Gsj + Nsj ; i = j
= Nsj ; i 6= j
29
M-ary CFSK
0
M-ary Simplex
0
10
10
-1
-1
10
10
-2
10
-3
10
=4
10
-3
M= 2
Pb
Pb
10
-2
M= 2
-4
10
-4
=8
10
= 16
10
=4
=8
-5
10
-5
= 16
-6
-6
= 32
10
10
-7
10
= 32
-7
5
10
Eb /N0
15
10
5
10
Eb /N0
15
30
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
where Gcj = Gj cos and Gsj = Gj sin are independent Gaussian random variables
with variances 2 and Ncj ; Nsj are independent Gaussian random variables with
variances N0 =2. The pdf of Z given hypothesis Hi is true is
fZjHi (zjHi ) = exp
z2 + z2
ci 2 si
Ei + N 0
0
exp @
1
N0
M
X
j=1; j6=i
1
2
2 A
zci + zsi
= M Ei 2 + N0 N0M
Decide hypothesis i is true if
fZjHi (zjHi ) fZjHj ; all j
This constitutes the maximum likelihood decision rule since the hypotheses are equally
probable. Alternatively, we can take the natural log of both sides and use that as a
test. This reduces to computing
2
Ei 2
2
zci + zsi
2
Ei + N 0
and choosing the signal corresponding to the largest. If the signals have equal energy,
then the optimum receiver is seen to be an M -signal replica of the one shown in Fig.
10.8a.
q
2 + Z 2 is Rayleigh distributed
b. Assume that the Ei s are equal. Note that Zi = Zci
si
given Hi . Without loss of genarality assume that H1 is given. Then
fZ1 jH1 (z1 j H1 ) =
fZi jH1 (zi j H1 ) =
exp z12 =2 E 2 + N0 ; z1 0
0
2
zi
exp zi =2N0 ; zi 0; i 6= 1
N0
z1
2
E + N
31
The probability of correct decision is
Pc = E fPr [Zi < Z1 ; all i = 2; 3; :::; M ]g
Z z1
M 1
Z 1
2 2
2
z1
zi
=
exp z1 =2 E + N0
exp zi =2N0 dzi
dz1
E 2 + N0
N0
0
0
Z 1
M 1
z1 exp z12 =2 E 2 + N0
=
1 exp z12 =2N0
dz1 , use binomial theorem
2
E + N0
0
Z 1
z1 exp z12 =2 E 2 + N0
=
E 2 + N0
0
"M 1
#M 1
X
M 1
exp (M 1 i) z12 =2N0
dz1
i
i=0
Z 1
M
1
X
1
M 1
2
=
z
exp
K
z
dz1
1
i
1
i
E 2 + N0
0
i=0
M
1
X
N0 E 2 + N0
1
M 1
=
i
E 2 + N0
N0 + (E 2 + N0 ) (M 1 i)
i=0
M
1
X
M 1
N0
=
2
i
N0 + (E + N0 ) (M 1 i)
i=0
M
1
X
M 1
1
=
= 1 PE
2
i
1 + (E =N0 + 1) (M 1 i)
i=0
where
Ki ,
1
M 1i
+
2 (E 2 + N0 )
2N0
Problem 11.27
a. It can be shown that under hypothesis H1 ; Y1 is the sum of squares of 2N independent
Gaussian random variables each having mean zero and variance
211 = 2
E 2 N0
+
N
2
and Y2 is the sum of squares of 2N independent Gaussian random variables each
having zero mean and variance
N0
221 =
2
32
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
Therefore, from Problem 5.37, the pdfs under hypothesis H1 are
2
y N 1 ey1 =211
fY1 (y1 jH1 ) = 1 N N
; y1 0
2 11 (N )
and
fY2 (y2 jH1 ) =
y2N 1 ey2 =N0
2N (N0 =2)N (N )
; y2 0
b. The probability of error is
PE = Pr (EjH1 ) = Pr (EjH2 ) = Pr (Y2 > Y1 jH1 )
Z 1 Z 1
=
fY2 (y2 jH1 ) dy2 fY1 (y1 jH1 ) dy1
y1
0
Using integration by parts, it can be shown that
I (x; a) =
Z 1
n az
z e
ax
dz = e
x
n
X
i=0
n!
i!an+1i
xi
This can be used to get the formula given in the problem statement.
c. Plots are given for N = 1 13 in steps of 2 in Fig. 11.3.
Problem 11.28
a. The characteristic function is found and used to obtain the moments. By deÖnition,
Z 1
j!
m m1
(j!) = E e
=
ej!
e
d
(m)
0
To integrate, note that
Z 1
0
ej! e d =
1
j!
Di§erentiate both sides m 1 times with respect to . This gives
Z 1
(1)m1 (m 1)!
ej! (1)m1 m1 e d =
( j!)m
0
Multiply both sides by m and cancel like terms to get
Z 1
m m1
m
ej!
e
d =
(m)
( j!)m
0
33
PE for FSK signaling with diversity in Rayleigh fading: N = 1 - 13 paths in steps of 2
0
10
-1
10
-2
10
-3
PE
10
-4
10
-5
10
-6
10
-7
10
0
2
4
6
8
10
12
SNR, dB
14
16
18
20
34
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
Thus
(j!) =
m
( j!)m
Di§erentiating this with respect to !, we get
E [] = j0 (0)
The variance is
m
m (m + 1)
and E 2 = j 2 00 (0) =
m
Var () = E 2 E 2 () = 2
b. Use Bayesí rule. But Örst we need fZ (z), which is obtained as follows:
Z 1
fZ (z) =
fZj (zj) f () d
Z0 1
m m1
=
ez
e
d
(m)
0
Z 1
m+1
m m
(+z) ( + z)
=
e
m d
(m + 1)
( + z)m+1 0
m m
=
( + z)m+1
where the last integral is evaluated by noting that the integrand is a pdf and therefore
integrates to unity. Using Bayesí rule, we Önd that
fjZ (jz) =
m ( + z)m+1 e(+z)
(m + 1)
Note that this is the same pdf as in part (a) except that has been replaced by + z
and m has been replaced by m + 1. Therefore, we can infer the moments from part
(a) as well. They are
E [jZ] =
m+1
m+1
and Var [jZ] =
+Z
( + Z)2
c. Assume the observations are independent. Then
fz1 z2 j (z1; z2 j) = 2 e(z1 +z2 ) ; z1 ; z2 0
The integration to Önd the joint pdf of Z1 and Z2 is similar to the procedure used to
Önd the pdf of Z above with the result that
fz1 z2 (z1; z2 ) =
m (m + 1) m
( + z1 + z2 )m+2
35
Again using Bayesí rule, we Önd that
fjz1 z2 (jz1 ; z2 ) =
m+1 e(+z1 +z2 ) ( + z1 + z2 )m+2
(m + 2)
Since this is of the same form as in parts (a) and (b), we can extend the results there
for the moments to
E [jZ1 ; Z2 ] =
m+2
m+2
and V ar [jZ1 ; Z2 ] =
+ Z1 + Z2
( + Z1 + Z2 )2
d. By examining the pattern developed in parts. (a), (b), and (c), we conclude that
m+1+K e(+z1 +z2 +:::+zK ) ( + z1 + z2 + : : : + zK )m+K
fjz1 z2 :::zK (jz1 ; z2; : : : zK ) =
(m + 2)
and
E [jZ1 ; Z2 ; : : : ; ZK ] =
var [jZ1 ; Z2 ; : : : ; ZK ] =
m+2
+ Z1 + Z2 + : : : + ZK
m+K
( + Z1 + Z2 + : : : + ZK )2
Problem 11.29
The conditional mean and Bayesí estimates are the same if:
1. C (x) is symmetric;
2. C (x) is convex upward;
3. fAjZ is symmetric about the mean.
Armed with this, we Önd the following to be true:
a. Same - all conditions are satisÖed.
b. Same - all conditions are satisÖed;
c. Condition 3 is not satisÖed;
d. Condition 2 is not satisÖed.
36
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
Problem 11.30
The mean of the individual samples is A as is the mean of a
^ML (Z). Therefore
3
2
!
2
K
X
1
var [^
aML (Z)] = E 4
(Zk A) 5
K
k=1
2
3
K X
K
X
1 4
=
E
(Zk A) (Zj A)5
K2
k=1 j=1
K
=
K
1 XX
E [(Zk A) (Zj A)]
K2
k=1 j=1
But, by the independence of samples and the fact that Zk A = Nk
0; k 6= j
E [(Zk A) (Zj A)] =
2n ; k = j
Thus
K
K
1 XX 2
n kj
K2
var [^
aML (Z)] =
k=1 j=1
K
1 X 2
n
K2
=
k=1
2n
K
=
Problem 11.31
a. The conditional pdf of the noise samples is
2
2 K=2
fZ z1 ; z2 ; : : : ; zK j n = 2 n
"
exp
K
X
i=1
zi2 =2 2n
#
The maximum likelihood estimate for 2n maximizes this expression. It can be found
by di§erentiating the pdf with respect to 2n and setting the result equal to zero.
Solving for 2n , we Önd the maximum likelihood estimate to be
K
b2n =
1 X 2
Zi
K
i=1
37
b. The variance of the estimate is
c. Yes.
PK
2
d.
i=1 Zi is a su¢cient statistic.
2 2 4n
var
bn =
K
Problem 11.32
We base our estimate on the observation Z0 , where Z0 = M0 + N where N is a
Gaussian random variable of mean zero and variance N0 =2. The sample value M0 is
Gaussian with mean m0 and variance 2m . The Ögure of merit is mean-squared error.
Thus, we use a MAP estimate. It can be shown that the estimate for m0 is the mean
value of M0 conditioned on Z0 because the random variables involved are Gaussian.
That is,
m
b 0 = E (M0 jZ0 = z0 ) = z0
where
= p
E (M0 Z0 )
m
p
=q
var (M0 ) var (Z0 )
2 + N0
m
2
It is of interest to compare this with the ML estimate, which is c
m
b 0;ML = Z0
which results if we have no a priori knowledge of the parameter. Note that the MAP
estimate reduces to the ML estimate when 2m ! 1. Note also that for N0 ! 1
(i.e., small signal-to-noise ratio) we donít use the observed value of Z0 , but instead
estimate m0 as zero (its mean).
Problem 11.33
a. Use the basis function set
1 (t) =
r
2
cos ! c t and 2 (t) =
T
r
2
sin ! c t
T
and trigonometric expansion of the received data as
H1 :
H2 :
y(t) = A cos cos ! c t A sin sin ! c t + n (t)
y(t) = A cos cos ! c t + A sin sin ! c t + n (t)
38
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
to write the data in vector form as
H1 :
Z = (Z1 ; Z2 ) =
H2 :
Z = (Z1 ; Z2 ) =
where
N1 =
Z T
0
!
r
T
T
A cos + N1 ;
A sin + N2
2
2
!
r
r
T
T
A cos + N1 ;
A sin + N2
2
2
r
n (t) 1 (t) dt and N2 =
Z T
0
n (t) 2 (t) dt
Note that E [N1 ] = [N2 ] = 0 and var[N1 ] = var[N2 ] = N0 =2. Thus
r
T
A cos , A1
2
r
T
E [Z2 j H1 ] =
A sin , A2
2
r
T
E [Z1 j H2 ] =
A cos , A1
2
r
T
E [Z1 j H2 ] =
A sin , A2
2
E [Z1 j H1 ] =
Hence the conditional pdfs of the data, given and the particular hypothesis, are
fZj;H1 (z1 ; z2 j ; H1 ) =
fZj;H2 (z1 ; z2 j ; H2 ) =
i
1
1 h
exp
(z1 A1 )2 + (z2 + A2 )2
N0
N0
i
1
1 h
2
2
exp
(z1 + A1 ) + (z2 A2 )
N0
N0
b. Assume equally likely hypotheses so that
fZj (z1 ; z2 j ) =
=
1
1
fZj;H1 (z1 ; z2 j ; H1 ) + fZj;H2 (z1 ; z2 j ; H2 )
2
2
2
n
h
io 3
2
2
1
1 A1 ) + (z2 + A2 )
1 4 exp n N0 (z
h
io 5
2N0 + exp 1 (z1 + A1 )2 + (z2 A2 )2
N0
39
Expanding the exponents and taking out common terms results in
1
1
T 2
2
2
fZj (z1 ; z2 j ) =
exp
z1 + z2 + A
2N0
N0
2
2
2
exp
(A1 z1 A2 z2 ) + exp
(A1 z1 A2 z2 )
N0
N0
1
1
T 2
2
2
2
=
exp
z1 + z2 + A
cosh
(A1 z1 A2 z2 )
N0
N0
2
N0
The maximum likelihood estimator for satisÖes (the data variables are capitalized
to signify actual data values)
@
ln fZj (Z1 ; Z2 j )= =0
ML
@
From above
1
T 2
2
2
ln fZj (Z1 ; Z2 j ) = ln (N0 )
Z1 + Z2 + A
N0
2
2
+ ln cosh
(A1 Z1 A2 Z2 )
N0
Taking the derivative with respect to and setting the result equal to 0 results in the
condition
"r
#r
T 2A
T 2A
tanh
(Z1 cos Z2 sin )
(Z1 sin Z2 cos ) = 0
2 N0
2 N0
Using deÖnitions for Z1 and Z2 , which are
r
r
Z T
Z T
2
2
Z1 =
y (t)
cos ! c t dt and Z2 =
y (t)
sin ! c t dt
T
T
0
0
this can be put into the form
4z
tanh
AT
Z T
0
4z
y (t) cos (! c t + ) dt
AT
Z T
0
y (t) sin (! c t + ) dt = 0
where z = A2 T =2N0 . This implies a Costas loop type structure shown in the problem
statement with integrators in place of the low pass Ölters and a tanh function in the
leg with the cosine multiplication. Note that for z small tanh (x) ' x and the phase
estimator becomes a Costas loop if the integrators are viewed as low pass Ölters.
40
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
c. The variance of the phase estimate is lower bounded by applying the Cramer-Rao
bound which is
"
#1
@ 2 fZj
var (ML ) E
@2
The Örst derivative, from above, is
Z T
Z
@fZj
2A
2A T
=
tanh
y (t) cos (! c t + ) dt
y (t) sin (! c t + ) dt
@
N0
N0 0
0
Assume low SNR so that tanh (x) ' x. Under this assumption the second derivative,
after some work, can be shown to be
2 Z T Z T
@ 2 fZj
2A
y (t) y t0 cos ! c t t0 dtdt0
2 =
N0
@
0
0
After some development, it can be shown that
E y (t) y t0
= Pr [H1 ] E y (t) y t0 j H1 + Pr [H2 ] E y (t) y t0 j H2
1
E [A cos (! c t + ) + n (t)] A cos ! c t0 + + n t0
=
2
1
+ E [A cos (! c t + ) + n (t)] A cos ! c t0 + + n t0
2
2
N0
A
=
cos ! c t t0 +
t t0
2
2
so that
"
#
2 Z T Z T
@ 2 fZj
2A
E
=
E y (t) y t0 cos ! c t t0 dtdt0
2
N0
@
0
0
2 Z T Z T 2
N0
2A
A
0
0
=
cos ! c t t +
t t cos ! c t t0 dtdt0
N0
2
2
0
0
2 Z T Z T 2
2A
A
N0 T
=
cos2 ! c t t0 dtdt0 +
N0
2
2
0
0
2 2 2
2A
A T
N0 T
=
+
N0
4
2
2
T
Since the SNR is z = A
2N0 this result becomes
"
#
@ 2 fZj
E
= 4z (z + 1)
@2
41
Thus, the variance of the phase estimate, for low SNR, is bounded by
var (M L )
1
4z (z + 1)
Reference should
to Table 10.7 where the tracking loop error variance is
have been 2made
2
1
given as = L
1=z + 0:5=z = L1 z+0:5
. the best that can be said of the comparison
z2
is that both are proportional to 1=z 2 in the limit of small SNRs.
Problem 11.34
1=2
a. Note that cos cos1 m = m and sin cos1 m = 1 m2
to write
s (t) =
p
2P m sin (! c t + )
Use the basis functions
1 (t) =
r
p
2P
p
1 m2 cos (! c t + ) ; sign due to data
2
cos ! c t and 2 (t) =
T
r
2
sin ! c t
T
to write
p
p
p
p
s (t) = P T m sin 1 m2 cos 1 (t)+ P T m cos 1 m2 sin 2 (t)
b. Use the above bases functions and base a decision on the vector
2 p
3
p
P T m sin 1 m2 cos + N1 ;
5
Z = [Z1 ; Z2 ] = 4 p
p
2
P T m cos 1 m sin + N2
, [S1 + N1 ; S2 + N2 ]
where N1 and N2 are independent Gaussian random variables with zero means and
variances N0 =2 The pdf of Z conditioned on the signal sign and is
i
1
1 h
2
2
fZj; (Z1 ; Z2 j ; ) =
exp
(Z1 S1 ) + (Z2 S2 )
N0
N0
To Önd the pdf conditioned only on use the fact the Pr(+ bit) = Pr ( bit) = 1=2
to get
1
1
fZj (Z1 ; Z2 j ) = fZj; + (Z1 ; Z2 j ; +) + fZj; (Z1 ; Z2 j ; )
2
2
42
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
After considerable simpliÖcation we obtain
" p
#
2m P T
fZj (Z1 ; Z2 j ) = C exp
(Z1 sin + Z2 cos )
N0
" p
#
p
2 1 m2 P T
cosh
(Z1 cos Z2 sin )
N0
where C includes all terms independent of . The log-likelihood function is
p
2m P T
L () =
(Z1 sin + Z2 cos )
N0
" p
#
p
2 1 m2 P T
+ ln cosh
(Z1 cos Z2 sin )
N0
Note that
Z1 sin + Z2 cos =
Z T
y (t)
0
Z1 cos Z2 sin =
Z T
0
y (t)
r
r
2
sin (! c t + ) dt
T
2
cos (! c t + ) dt
T
so that
p
Z
2m 2P T
L () =
y (t) sin (! c t + ) dt
N0
0
" p
#
p
Z
2 1 m2 2P T
+ ln cosh
y (t) cos (! c t + ) dt
N0
0
The maximum likelihood estimate satisÖes @L()
= 0.
@
espression for L () gives
@L ()
@
Di§erentiating the above
p
Z
2m 2P T
= 0=
y (t) cos (! c t + ML ) dt
N0
0
" p
#
p
Z
2 1 m2 2P T
tanh
y (t) cos (! c t + ML ) dt
N0
0
" p
#
p
Z
2 1 m2 2P T
y (t) sin (! c t + ML ) dt
N0
0
11.1. COMPUTER EXERCISES
43
c. The block diagram follows the development of the one for Problem 11.33. The block
diagram of the phase estimator has another arm that adds into the feedback to the
VCO that acts as a phase-lock loop for the carrier component.
Problem 11.35
Write the impulse response as
1
h (t) =
T
t T =2
T
and use the delay theorem to get
H (f ) =
1
T sinc (T f ) ejf T = sinc (T f ) ejf T
T
Thus the equivalent noise bandwidth is
BN
11.1
1
Z 1
jH (f )j2 df
2
Hmax
Z 1 0
=
sinc2 (T f ) df
0
Z
1 1
=
sinc2 (u) du
T 0
1
=
2T
=
Computer Exercises
Computer Exercise 11.1
% Öle: ce11_1.m ROC; Pd near 1, PFA small
%
% PF = Q(A) < 1e-3 => A > 3.08; B = ln(eta)/d + d/2
% PD = Q(B) > 0.95 => B < -1.65; A = ln(eta)/d -d/2
% d = A - B > 4.73;
% ln(eta) = (A^2 - B^2)/2 > 3.38 => eta > 29.37
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
44
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
%
clf
clear all; clf
d = [4.73 5 5.5 6];
log_eta = 3.38:0.01:9;
lend = length(d);
hold on
for j = 1:lend
dj = d(j);
af = log_eta/dj + dj/2;
ad = log_eta/dj - dj/2;
pf = qfn(af);
pd = qfn(ad);
semilogy(pf, pd)
text(pf(20), pd(20)-.01, [íd = í, num2str(dj)])
end
grid on
hold o§
xlabel(íProbability of False Alarmí)
ylabel(íProbability of Detectioní)
title(íReceiver operating characteristicí)
%
This function computes the Gaussian Q-function
%
function Q=qfn(x)
Q = 0.5*erfc(x/sqrt(2));
%
% End of script Öle
11.1. COMPUTER EXERCISES
45
Receiver operating characteristic
1
d=6
d = 5.5
d=5
0.95
Probability of Detection
d = 4.73
0.9
0.85
0.8
0.75
0.7
0.65
0
0.2
0.4
0.6
0.8
Probability of False Alarm
1
1.2
-3
x 10
46
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
Computer Exercise 11.2
% Öle: ce11_2.m
%
% From (11.171), sigmap^2/sigman^2 = (K + sigman^2/sigmaA^2)^-1
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
clf
clear all;
sigmaA2_sigman2 = 0.1:0.5:2.1;
Ls = length(sigmaA2_sigman2);
K = 1:25;
hold on
for l=1:Ls
sigma_ratio = sigmaA2_sigman2(l);
sigmap2_sigman2 = 1./(K + 1./sigma_ratio);
plot(K, sigmap2_sigman2)
text(K(1), sigmap2_sigman2(1),[ínsigma_A^2/nsigma_n^2 = í, num2str(sigma_ratio)])
end
hold o§
xlabel(íKí),ylabel(ínsigma_p^2/nsigma_n^2í)
%
% End of script Öle
Computer Exercise 11.3
% Öle: ce11_3.m
% Simulation of a digital PLL
%
% R. Ziemer & W. Tranter, Principles of Communications, 7th edition
%
clf
clear all;
AA = char(í-í,í:í,íñí,í-.í);
K = 1000;
epsilon = input(íEnter feedback constant epsilon => í);
theta0 = 0.5*pi;
T = 1;
A = 1;
11.1. COMPUTER EXERCISES
0.7
2
2
2
2
2
2
2
2
2
σ A /σ n = 2.1
0.6
σ A /σ n = 1.6
0.5
σ A /σ n = 1.1
0.4
2
σ A /σ n = 0.6
2
σ p/σ n
2
47
0.3
0.2
0.1
0
σ A /σ n = 0.1
0
5
10
15
K
20
25
48
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
kk = 1:K;
sigma_n = [0.1 0.5 1 2];
Ln = length(sigma_n);
hold on
for n = 1:Ln
sigma = sigma_n(n);
Z1(1) = sqrt(T/2)*A*cos(theta0) + sigma*randn(1);
Z2(1) = -sqrt(T/2)*A*sin(theta0) + sigma*randn(1);
theta(1) = theta0;
for k = 1:K-1
theta(k+1) = theta(k) + epsilon*atan2(Z2(k), Z1(k));
Z1(k+1) = sqrt(T/2)*A*cos(theta(k+1)) + sigma*randn(1);
Z2(k+1) = -sqrt(T/2)*A*sin(theta(k+1)) + sigma*randn(1);
end
plot(kk, theta, AA(n,:))
end
hold o§
xlabel(íkí),ylabel(íntheta(k), radí)
legend([ínsigma_n = í, num2str(sigma_n(1))], [ínsigma_n = í, num2str(sigma_n(2))],
[ínsigma_n = í, num2str(sigma_n(3))], [ínsigma_n = í, num2str(sigma_n(4))])
title([íFirst-order digital phase tracking device performance for nepsilon = í, num2str(epsilon)])
%
% End of script Öle
11.1. COMPUTER EXERCISES
49
First-order digital phase tracking device performance for ε = 0.01
1.6
σ n = 0.1
1.4
σ n = 0.5
1.2
σn = 1
1
θ(k), rad
0.8
0.6
0.4
0.2
0
-0.2
-0.4
0
100
200
300
400
500
k
600
700
800
900
1000
CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS
2
First-order digital phase tracking device performance for ε = 0.1
σ n = 0.1
σ n = 0.5
1.5
σn = 1
1
θ(k), rad
50
0.5
0
-0.5
-1
-1.5
0
100
200
300
400
500
k
600
700
800
900
1000
Chapter 12
Information Theory and Coding
12.1
Problem Solutions
Problem 12.1
The transition probabilities for the 3-hop system are (The first matrix represents the 2-hop
system as defined by equations (12.19)-(12.22))
( |) =
This gives
∙
=
∙
11 12
21 22
∙
1 1 + 2 3 1 2 + 2 4
3 1 + 4 3 3 2 + 4 4
¸
¸∙
1 2
3 4
¸
1 1 1 + 2 3 1 + 1 2 3 + 2 4 3 1 1 2 + 2 3 2 + 1 2 2 + 2 4 2
3 1 3 + 4 3 1 + 3 2 3 + 4 4 3 3 1 2 + 4 3 2 + 3 2 4 + 4 4 4
¸
Tracing all paths from the two inputs to the two outputs is consistent with the above. Note
that there are four paths from a given input to a given output. Colored pencils help since
there are a tptal of 16 paths.
Problem 12.2
(a) () = − log2 (52) = 57004 bits
(b) () = − log2 [(52)(52)] = 114009 bits
(c) () = − log2 [(52)(51)] = 113729 bits
Problem 12.3
1
2
CHAPTER 12. INFORMATION THEORY AND CODING
The entropy is
() = −030 log2 (030) − 025 log2 (025)
−020 log2 (020) − 015 log2 (015)
−010 log2 (010) = 22282 bits/symbol
The maximum entropy is
() = log2 (5) = 23219 bits/symbol
Problem 12.4
The entropy is
() = −030 log2 (030) − 025 log2 (025) − 020 log2 (020)
−010 log2 (010) − 010 log2 (010) − 005 log2 (005)
= 23660 bits/symbol
Problem 12.5
(a) The channel diagram is shown in Figure 12.1.
(b) The output probabilities are
(1 ) =
(2 ) =
(3 ) =
1
1
(06 + 02 + 02) =
3
3
1
1
(03 + 05 + 02) =
3
3
1
1
(01 + 03 + 06) =
3
3
(c) Since all columns sum to one, equal input probabilities yield equal output probabilities.
(d) The joint probability matrix is
⎤⎡ 6
⎤
⎡ 1
3
1
0
0
3
10
10
10
5
3 ⎦
2
[ (; )] = ⎣ 0 13 0 ⎦ ⎣ 10
10
10
2
6
2
1
0 0 3
10
10
10
which gives
⎡
[ (; )] = ⎣
6
30
2
30
2
30
3
30
5
30
2
30
1
30
3
30
6
30
⎤
⎦
12.1. PROBLEM SOLUTIONS
3
0.6
0.1
0.2
0.3
0.2
0.3
0.5
0.2
0.6
Figure 12.1: Channel diagram for Problem 12.5
Note that the column sum gives the output probabilities [ ( )], and the row sum gives
the input probabilities [ ()]
Problem 12.6
This problem may be solved by raising the channel matrix
∙
¸
0993 0007
=
0007 0993
which corresponds to an error probability of 0007, to increasing powers and seeing where
the error probability reaches the critical value of 002. Consider the MATLAB program
a = [0.993 0.007; 0.007 0.993];
n = 1;
a1 = a;
while a1(1,2)0.02
n=n+1;
a1=a^n;
end
n-1
% channel matrix
% initial value
% save a
% display result
Executing the program yields − 1 = 22. Thus we compute (MATLAB code is given)
À a^2
4
CHAPTER 12. INFORMATION THEORY AND CODING
ans =
0.9861 0.0139
0.0139 0.9861
This gives an error probability of 0.0139, which is less than 0.02. Next we compute
a^3
ans =
0.9793 0.0207
0.0207 0.9793
Thus 2 cascaded channels meets the specification for 002. However cascading 3
channels yields 002 and the specification is not satisfied.
Problem 12.7
For the erasure channel we determine
(; ) = () − (| )
where
(| ) = −
2 X
3
X
=1 =1
( ) log2 ( | ) = −
2 X
3
X
=1 =1
( | )( ) log2
∙
( | )( )
( )
¸
Let (1 ) = and (2 ) = 1 − . This gives
(1 ) = (1 − )
(2 ) = + (1 − ) =
(3 ) = (1 − )(1 − )
Thus
¸
∙ ¸
(1 − )
− log2
(| ) = −(1 − ) log2
(1 − )
¸
¸
∙
∙
(1 − )
(1 − )(1 − )
−(1 − ) log2
− (1 − )(1 − ) log2
(1 − )(1 − )
∙
This is
(| ) = − log2 () − (1 − ) log2 (1 − ) = − [ log2 () + (1 − ) log2 (1 − )]
Since
() = − log2 () − (1 − ) log2 (1 − )
12.1. PROBLEM SOLUTIONS
5
we have
(; ) = () − () = ()(1 − )
Therefore
=1−
bits/symbol
and capacity is achieved for equally likely inputs.
Problem 12.8
The cascade of the two channels has the channel matrix
¸∙
¸
∙
1 − 2 2
1
0
1 − 1
( |) =
1
1 − 1
0
2 1 − 2
This gives
( |) =
∙
(1 − 1 ) (1 − 2 ) (1 − 1 ) 2 + 1 2
1 (1 − 2 )
1 2 + (1 − 1 ) 2 (1 − 1 ) (1 − 2 )
1 (1 − 2 )
which can be written
( |) =
∙
(1 − 1 ) (1 − 2 ) 2
1 (1 − 2 )
2 (1 − 1 ) (1 − 2 )
1 (1 − 2 )
¸
¸
The output probability for 2 is
(2 ) = 2
Thus, the probability of 2 is independent of the input so that the cascade is a more general
erasure channel in that (1 |2 ) and (3 |1 ) are not zero. We will encounter a channel like
this when feedback channels are encountered.
Problem 12.9
The capacity of the channel described by the transition probability matrix
⎡
⎤
0 0
⎢ 0 0 ⎥
⎢
⎥
⎣ 0 0 ⎦
0 0
is easily computed by maximizing
(; ) = ( ) − ( |)
The conditional entropy ( |) can be written
X
XX
( ) ( | ) log2 ( | ) =
( ) ( | )
( |) = −
6
CHAPTER 12. INFORMATION THEORY AND CODING
where
( | ) = −
X
( | ) log2 ( | )
For the given channel
( | ) = − log2 − log2 =
so that ( | ) is a constant independent of . This results since each row of the matrix
contains the same set of probabilities although the terms are not in the same order. For
such a channel
X
( ) =
( |) =
and
(; ) = ( ) −
We see that capacity is achieved when each channel output occurs with equal probability.
We now show that equally likely outputs result if the inputs are equally likely. Assume
that ( ) = 14 for all . Then
X
( ) =
( ) ( | )
For each
( ) =
1
1
1X
1
( | ) = ( + ) = [ + (1 − )] =
4
4
4
4
so that
1
( ) =
4
all
Since ( ) = 14 for all , ( ) = 2, and the channel capacity is
= 2 + log2 + (1 − ) log2 (1 − )
or
= 2 − ()
This is shown in Figure 12.2.
This channel is modeled as a pair of binary symmetric channels as shown. If = 1 or
= 0, each input uniquely determines the output and the channel reduces to a noiseless
channel with four inputs and four outputs. This gives a capacity of log2 4 = 2 bits/symbol.
If = = 12 , then each of the two subchannels (Channel 1 and Channel 2) have zero
capacity. However, 1 and 2 can be viewed as a single input and 3 and 4 can be viewed
as a single input as shown in Figure 12.3. The result is equivalent to a noiseless channel
with two inputs and two outputs. This gives a capacity of log2 2 = 1 bit/symbol.
12.1. PROBLEM SOLUTIONS
7
C
2
1
0
1/2
p
1
Figure 12.2: Channel capacity for Problem 12.9.
p
x1
q
Channel 1
q
x2
p
x3
p
q
Channel 2
x4
q
p
y1
y2
y3
y4
Figure 12.3: Additional illustration for Problem 12.9.
8
CHAPTER 12. INFORMATION THEORY AND CODING
Note that this channel is an example of a general symmetric channel. The capacity of
such channels are easily computed. See Gallager (Information Theory and Reliable Communications, Wiley, 1968, pages 91-94) for a discussion of these channels.
Problem 12.10
To show (12.30), write ( ) in (12.29) as ( | ) ( ) and recall that
log2 ( ) = log2 ( | ) + log2 ( )
Evaluating the resulting sum yields (12.30). The same method is used to show (12.31)
except that ( ) is written ( | ) ( )
Problem 12.11
The entropy is maximized when each quantizing region occurs with probability 025. Thus
Z 1
1
− = 1 − −1 =
4
0
which gives
−1 =
or
1
1 = ln
In a similar manner
Z 2
1
µ ¶
4
= 02877
3
− = −1 − −2 =
which gives
−2 =
or
2 =
Also
3
4
1
2
1
ln (2) = 06931
Z 3
− =
2
gives
3 = 13863
Problem 12.12
3
1
− −2 =
4
4
1
4
12.1. PROBLEM SOLUTIONS
9
The thresholds are now parameterized in terms of . We have
Z 1
−2 22
1
2
2
= 1 − −1 =
2
4
0
which gives
µ 2¶
∙ ¸
1
3
= ln(43)
= − ln
2
4
Thus
1 =
For 2 we have
Z 2
1
p
log(43) = 05364
−2 22
3
1
2
2
2
2
2
2
= −1 − −2 = − −2 =
2
4
2
so that
2 =
p
ln(2) = 06931
Finally, for 3 we have
Z 3
1
−2 22
3 1
2
2
2
2
2
2
= −2 − −3 = −2 = − =
2
4
4
2
2
so that
3 =
As a check we compute
Z ∞
3
p
ln(2) = 11774
−2 22
1
2
2
2
= −3 − 0 = −(11774) =
2
4
which is the correct result.
Problem 12.13
Five thresholds are needed to define the six quantization regions. The five thresholds are
−2 , −1 , 0, 1 , and 2 . The thresholds are selected so that the six quantization regions
are equally likely. For 1 we have
µ ¶
Z 1
Z ∞
1
1
1
1
1
1
− 2 2 2
−2 22
=√
= − √
= −
6
2
2
2 0
2 1
so that
µ
1
¶
=
1 1
1
− =
2 6
3
10
CHAPTER 12. INFORMATION THEORY AND CODING
inding the inverse Q-function yields
−1
For 2 we have
1
1
=√
6
2
µ ¶
1
1
= 04307 =
3
Z ∞
− 2 2 2
=
2
µ
2
¶
Taking the inverse function
−1
µ ¶
1
2
= 09674 =
6
and
1 = 04307
and
2 = 09674
This gives the quantizing rule
Quantizer Input
−∞ −09674
−09674 −04307
−04307 0
0 04307
04307 09674
09674 ∞
Quantizer Output
0
1
2
3
4
5
Problem 12.14
The transition probabilities for the cascade of Channel 1 and Channel 2 are
∙
¸ ∙
¸∙
¸ ∙
¸
11 12
09 01
075 025
07 03
=
=
21 22
01 09
025 075
03 07
Therefore, the capacity of the overall channel is
= 1 + 07 log2 07 + 03 log2 03 = 01187 bits/symbol
The capacity of Channel 1 is
= 1 + 09 log2 09 + 01 log2 01 = 05310 bits/symbol
and the capacity of Channel 2 is
= 1 + 075 log2 075 + 025 log2 025 = 01887 bits/symbol
12.1. PROBLEM SOLUTIONS
11
It can be seen that the capacity of the cascade channel is less than the capacity of either
channel taken alone, which is to be expected since each channel causes errors and the error
probability on the cascade channel is greater than the error probability of either of the
individual channels.
Problem 12.15
The source entropy is
5 3
3
5
() = − log2 − log2 = 00944
8
8 8
8
and the entropy of the fourth-order extension is
¡ ¢
4 = 4 () = 4(09544) = 38177
The second way to determine the entropy of the fourth-order extension, is to determine the
probability distribution of the extended source. We have
µ ¶4
5
625
=
1 symbol with probability
8
4096
µ ¶3
3 5
375
4 symbols with probability
=
8 8
4096
µ ¶2 µ ¶2
3
5
225
6 symbols with probability
=
8
8
4096
µ ¶3
135
5
3
=
4 symbols with probability
8
8
4096
µ ¶4
81
3
=
1 symbol with probability
8
4096
Thus
¡
4
¢
Problem 12.16
µ
¶
µ
¶
375
225
625
375
225
625
log2
−4
log2
−6
log2
= −
4096
4096
4096
4096
4096
4096
µ
¶
135
81
135
81
−4
log2
−
log2
4096
4096
4096
4096
= 38177 bits/symbol
12
CHAPTER 12. INFORMATION THEORY AND CODING
For the fourth-order extension we have
Source Symbol
AAAA
BAAA
ABAA
AABA
AAAB
AABB
ABAB
BAAB
ABBA
BABA
BBAA
ABBB
BABB
BBAB
BBBA
BBBB
P (·)
0.6561
0.0729
0.0729
0.0729
0.0729
0.0081
0.0081
0.0081
0.0081
0.0081
0.0081
0.0009
0.0009
0.0009
0.0009
0.0001
Codeword
0
100
101
110
1110
111100
1111010
1111011
1111100
1111101
1111110
111111100
111111101
111111110
1111111110
1111111111
The average wordlength: = 19702 which is 4 = 197024 = 04926. The efficiencies
are given in the following table:
1
2
3
4
1
1.29
1.598
1.9702
( )
0.4690
0.9380
1.4070
1.8760
Efficiency
46.90%
72.71%
88.05%
95.22%
Problem 12.17
For the Shannon-Fano code we have
Source Symbol P( · ) Codeword
1/7
00
1
1/7
010
2
1/7
011
3
4
1/7
100
1/7
101
5
1/7
110
6
1/7
111
7
12.1. PROBLEM SOLUTIONS
13
11
000
0
001
1
010
0
011
100
1
0
101
1
2/7
0
2/7
2/7
1
0
1/7
1
0
1
Figure 12.4: Huffman coding for Problem 12.17.
The entropy is
() = −7
µ
1
1
log2
7
7
¶
= log2 (7) = 28074
The average wordlength is
µ ¶
20
1
(2 + 3 + 3 + 3 + 3 + 3 + 3) =
= 28571
=
7
7
The efficiency is therefore given by
()
28074
= 09826 = 9826%
=
28571
(12.1)
The diagram for determining the Huffman code is illustrated in Figure ??. This diagram
yields the codewords shown. Note that we have six codewords of length three and one
codeword of length two. This average wordlength is the same as for the Shannon-Fano code.
Thus, the Huffman code and the Shannon-Fano code have the same average wordlength and
the same efficiency.
Problem 12.18
The codewords for the Shannon-Fano and Huffman codes are summarized in the following
14
CHAPTER 12. INFORMATION THEORY AND CODING
table:
Source Symbol
1
2
3
4
5
Codewords
Shannon-Fano Huffman
00
1
01
000
10
001
110
010
111
011
P (·)
0.40
0.20
0.17
0.13
0.10
The average wordlengths are:
= 225
(Shannon-Fano code)
= 222
(Huffman code)
Note that the Huffman code gives the shorter average wordlength and therefore the higher
efficiency.
Problem 12.19
The entropy is
() = −085 log2 085 − 015 log2 015
= 060984 bits/symbol
We know that
lim
→∞
≤ 350
and that
= ()
→∞
lim
Therefore, for large
() ≤ 350
or
≤
350
350
=
()
060984
which gives
≤ 573921 symbols/s
Since we cannot transmit partial symbols, the maximum symbol rate is 573 symbols/s.
Problem 12.20
The Shannon-Fano and Huffman codes are as follows. (Note that the codes are not unique.)
12.1. PROBLEM SOLUTIONS
15
Source Symbol
1
2
3
4
5
6
7
8
9
10
11
Shannon-Fano
Codeword
001
0000
0001
011
0100
0101
1000
1001
110
1110
1111
Huffman
Codeword
010
0110
0111
0000
0001
0010
0011
100
101
110
111
The source entropy is
() = log2 11 = 34594
For the Shannon-Fano code the average wordlength
=
1
[3(3) + 8(4)] = 37273
11
and the efficiency is
=
()
34594
= 9281%
=
37273
For the Huffman code the average wordlength
=
1
[5(3) + 6(4)] = 35455
11
and the efficiency is
=
()
34594
= 9757%
=
35455
We see that the Huffman code is better in this case.
Problem 12.21
The probability of message is
( ) =
Z 01
01(−1)
2 = 001 (2 − 1)
16
CHAPTER 12. INFORMATION THEORY AND CODING
A Huffman code yields the codewords in the following table:
Source Symbol
10
9
8
7
6
5
4
3
2
1
P (·)
0.19
0.17
0.15
0.13
0.11
0.09
0.07
0.05
0.03
0.01
Codeword
11
000
010
011
100
101
0011
00100
001010
001011
The source entropy is given by
() = 30488 bits/source symbol
and the average wordlength is
= 31000 binary symbols/source symbol
A source symbol rate of 250 symbols (samples) per second yields a binary symbol rate of
= 250(31000) = 775 binary symbols/second
and a source information rate of
= () = 250(30488) = 752 bits/second
Problem 12.22
The source entropy is
() = −035 log2 035 − 025 log2 025 − 02 log2 02 − 015 log2 015 − 005 log2 005
= 21211 bits/source symbol
and the entropy of the second-order extension of the source is
¡ ¢
2 = 2 () = 42422
The second-order extension source symbols are shown in the following table along with the
Shannon-Fano code.
12.1. PROBLEM SOLUTIONS
17
Source Symbol Probability
1 1
1 2
2 1
1 3
3 1
2 2
1 4
4 1
3 2
2 3
3 3
4 2
2 4
4 3
3 4
4 4
5 1
1 5
5 2
2 5
5 3
3 5
4 5
5 4
5 5
01225
00875
00875
00700
00700
00625
00525
00525
00500
00500
00400
00375
00375
00300
00300
00225
00175
00175
00125
00125
00100
00100
00075
00075
00025
Shannon-Fano
Codeword
000
001
0100
0101
0110
0111
1000
1001
1010
10110
10110
11000
11001
11010
11011
111000
111001
111010
111011
111100
111101
1111100
1111101
1111110
1111111
The preceding table defnes the 25 codewords of the second-order extension of the source.
Also given in the table are the probabilities of the 25 symbols of the second-order extension.
This allows the average wordlength and efficiency to be easily computed. For the Huffman
code, note that Computer Exercise 12.2 calls for the development of a MATLAB program to
generate the Huffman code. After Computer Exercise 12.2 is worked, this problem offers an
excellent set of source symbol probabilities for testing the problem generated in Computer
Exercise 12.2. If this is done it will be found that the Huffman code offers a slight increase
in efficency.
Problem 12.23
18
CHAPTER 12. INFORMATION THEORY AND CODING
6
9
x 10
8
7
Capacity
6
5
4
3
2
1
0
2
10
4
6
10
10
Bandwidth
8
10
10
10
Figure 12.5: Capacity for the AWGN channel given in Problem 12.24.
The set of wordlengths from Table 12.3 is {1 3 3 3 5 5 5 5}. The Kraft inequality gives
8
X
=1
¢
¡
¢ 1 3
¡
4
2 = 2−1 + 3 2−3 + 4 2−5 = + +
=1
2 8 32
Thus the Kraft inequality is satisfied.
Problem 12.24
The Shannon-Hartley law with = 60 and = 10−5 gives
Ã
¡ ¢!
¶
µ
60 105
= log2 1 +
= log2 1 +
This is illustrated in the following figure. It can be seen that as → ∞ the capacity
approaches a finite constant. For sufficiently small , ln(1 + ) ≈ . Therefore
lim =
→∞
60(105 )
= 86562 × 106
ln(2)
as we see from the illustration that follows.
Problem 12.25
12.1. PROBLEM SOLUTIONS
19
4
10
x 10
9
8
Capacity - bits/s
7
6
5
4
3
2
1
0
0
10
20
30
40
50
60
Signal Power - Watts
70
80
90
100
Figure 12.6: Capacity as a function of bandwidth for Problem 12.25.
We now determine
Ã
¡ ¢!
¡ 4¢
¡ ¢
105
= 10 log2 1 +
= 104 log2 (1 + 10 )
4
10
The plot is given in Figure 12.6. For sufficiently large ,
¡ ¢
≈ 105 log2 ( )
and, therefore, → ∞ as → ∞.
Problem 12.26
Consider the ratio of the error probability of the coded system to the error probability of
the uncoded system for high SNR. Since we are considering À 1 we use the asympotic
20
CHAPTER 12. INFORMATION THEORY AND CODING
approximation for the Gaussian function. Therefore, for the coded system
Ãr !
r
2
1 −(√2)2
1 −2
1
√
√
=
≈p
=
2
2
2 2
For the uncoded system we simply let = 1. Thus
s
¶¸
∙ µ
2 exp (−2) √
1
=
= exp 2 1 −
2 exp (−2)
√
Note that for large the ratio becomes , where is a constant defined by exp(2).
For fixed , the loss in symbol error probability is therefore exponential with respect to the
SNR , which is a greater loss than can be compensated by the error-correcting capability
√
of the repetition code. Note that for fixed , the corresponding losss is proportional to .
Problem 12.27
The parity-check matrix will have 15 − 11 = 4 rows and 15 columns. Since rows may be
permuted, the parity-check matrix is not unique. One possible selection is
⎡
⎤
0 0 0 0 1 1 1 1 1 1 1 1 0 0 0
⎢ 0 1 1 1 0 0 0 1 1 1 1 0 1 0 0 ⎥
⎥
[] = ⎢
⎣ 1 0 1 1 0 1 1 0 0 1 1 0 0 1 0 ⎦
1 1 0 1 1 0 1 0 1 0 1 0 0 0 1
The corresponding generator matrix is
⎡
1 0 0 0 0
⎢ 0 1 0 0 0
⎢
⎢ 0 0 1 0 0
⎢
⎢ 0 0 0 1 0
⎢
⎢ 0 0 0 0 1
⎢
⎢ 0 0 0 0 0
⎢
⎢
⎢ 0 0 0 0 0
⎢
[] = ⎢ 0 0 0 0 0
⎢
⎢ 0 0 0 0 0
⎢
⎢ 0 0 0 0 0
⎢
⎢ 0 0 0 0 0
⎢
⎢ 0 0 0 0 1
⎢
⎢ 0 1 1 1 0
⎢
⎣ 1 0 1 1 0
1 1 0 1 1
0
0
0
0
0
1
0
0
0
0
0
1
0
1
0
0
0
0
0
0
0
1
0
0
0
0
1
0
1
1
0
0
0
0
0
0
0
1
0
0
0
1
1
0
0
0
0
0
0
0
0
0
0
1
0
0
1
1
0
1
0
0
0
0
0
0
0
0
0
1
0
1
1
1
0
⎤
0
0 ⎥
⎥
0 ⎥
⎥
0 ⎥
⎥
0 ⎥
⎥
0 ⎥
⎥
⎥
¸
0 ⎥ ∙
11
⎥
0 ⎥=
11
⎥
0 ⎥
⎥
0 ⎥
⎥
1 ⎥
⎥
1 ⎥
⎥
1 ⎥
⎥
1 ⎦
1
12.1. PROBLEM SOLUTIONS
21
where 11 is the 11×11 identity matrix and 11 represents the first 11 columns of the paritycheck matrix. From the structure of the parity-check matrix, we see that each parity symbol
is the sum of 7 information symbols. Since 7 is odd, the all-ones information sequence yields
the parity sequence 1111. This gives the codeword
[ ] = [111111111111111]
A single error in the third position yields a syndrome which is the third column of the
parity-check matrix. Thus, for the assumed parity-check matrix, the syndrome is
⎡ ⎤
0
⎢ 1 ⎥
⎥
[] = ⎢
⎣ 1 ⎦
0
Problem 12.28
The generator matrix corresponding to the given parity-check matrix is
⎤
⎡
1 0 0 0
⎢ 0 1 0 0 ⎥
⎥
⎢
⎢ 0 0 1 0 ⎥
⎥
⎢
⎥
[] = ⎢
⎢ 0 0 0 1 ⎥
⎢ 0 1 1 1 ⎥
⎥
⎢
⎣ 1 0 1 1 ⎦
1 1 0 1
The code words are of the form
£
1 2 3 4 1 2 3
⎡
⎤
1 = 2 ⊕ 3 ⊕ 4
⎣ 2 = 1 ⊕ 3 ⊕ 4 ⎦
3 = 1 ⊕ 2 ⊕ 4
¤
where
Thus, the 24 = 16 code words are (read veritcally)
a1
a2
a3
a4
0
0
0
0
0
0
0
1
0
0
1
0
0
0
1
1
0
1
0
0
0
1
0
1
0
1
1
0
0
1
1
1
1
0
0
0
1
0
0
1
1
0
1
0
1
0
1
1
1
1
0
0
1
1
0
1
1
1
1
0
1
1
1
1
c1
c2
c3
0
0
0
1
1
1
1
1
1
0
0
1
1
0
1
0
1
0
0
0
1
1
0
0
0
1
1
1
0
0
1
0
1
0
0
0
1
0
0
0
0
1
0
0
0
1
1
1
22
CHAPTER 12. INFORMATION THEORY AND CODING
Problem 12.29
We first determine [ ] = [] [ ] for = 1 2. The generator matrix for the previous problem
is used. For [1 ] we have
⎡ ⎤
0
⎡ ⎤ ⎢ 1 ⎥
⎢ ⎥
0
⎢ ⎥
⎢ 1 ⎥ ⎢ 1 ⎥
⎥
⎢
⎥
[] [1 ] = [] ⎣ ⎦ = ⎢
⎢ 1 ⎥ = [1 ]
1
⎢ 1 ⎥
⎢ ⎥
1
⎣ 0 ⎦
0
and for [2 ] we have
⎤
1
⎡ ⎤ ⎢ 0 ⎥
⎢ ⎥
1
⎢ ⎥
⎢ 0 ⎥ ⎢ 1 ⎥
⎥
⎢
⎥
[] [2 ] = [] ⎣ ⎦ = ⎢
⎢ 0 ⎥ = [2 ]
1
⎢ 1 ⎥
⎢ ⎥
0
⎣ 0 ⎦
⎡
1
We note that [1 ] ⊕ [2 ]
Problem 12.30
For a Hamming code with = − parity check symbols, the wordlength is
= 2 − 1
and
= 2 − − 1
The rate is
2 − − 1
=
2 − 1
If is large, is large. As → ∞ both and approach 2 . Thus
=
lim = lim =
→∞
Problem 12.31
→∞
2
=1
2
12.1. PROBLEM SOLUTIONS
23
The codeword is of the form (1 2 3 4 1 2 3 ) where
1 = 1 ⊕ 2 ⊕ 3
2 = 2 ⊕ 3 ⊕ 4
3 = 1 ⊕ 2 ⊕ 4
This gives the generator matrix
⎡
1
⎢ 0
⎢
⎢ 0
⎢
[] = ⎢
⎢ 0
⎢ 1
⎢
⎣ 0
1
0
1
0
0
1
1
1
0
0
1
0
1
1
0
⎤
0
0 ⎥
⎥
0 ⎥
⎥
1 ⎥
⎥
0 ⎥
⎥
1 ⎦
1
Using the generator matrix, we can determine the codewords. The codewords are (read
vertically)
a1
a2
a3
a4
0
0
0
0
1
0
0
0
1
1
0
0
1
1
1
0
1
1
1
1
0
0
0
1
0
0
1
1
0
1
1
1
1
0
1
1
1
1
0
1
1
0
0
1
0
1
1
0
0
1
0
0
0
0
1
0
0
1
0
1
1
0
1
0
c1
c2
c3
0
0
0
1
0
1
0
1
0
1
0
0
1
1
1
0
1
1
1
0
1
0
1
0
0
0
0
0
0
1
1
1
0
0
0
1
1
1
1
1
1
0
1
0
0
0
1
1
Let denote the -th code word and let → indicate that a cyclic shift of yields
. The cyclic shifts are illustrated in the following figure.
Problem 12.32
The parity-check matrix for the coder in the previous problem is
⎡
⎤
1 1 1 0 1 0 0
[] = ⎣ 0 1 1 1 0 1 0 ⎦
1 1 0 1 0 0 1
For the received sequence
[] = [1101001]
24
CHAPTER 12. INFORMATION THEORY AND CODING
1
5
2
3
12
9
15
14
6
4
8
7
11
13
16
10
Figure 12.7: Illustration of cyclic property.
the syndrome is
⎡
⎤
0
[] = [] [] = ⎣ 0 ⎦
0
This indicates no errors, which is in agreement with Figure 12.15. For the received sequence
[] = [1101011]
the syndrome is
⎡
⎤
0
[] = [] [] = ⎣ 1 ⎦
0
This syndrome indicates an error in the 6th position. Thus the assumed transmitted codeword is
[ ] = [1101001]
which is also in agreement with Figure 12.15.
Problem 12.33
The probability of two errors in a 7 symbol codeword is
µ ¶
7
{2 errors} =
(1 − )5 2 = 212 (1 − )5
2
and the probability of three errors in a 7 symbol codeword is
µ ¶
7
{3 errors} =
(1 − )4 3 = 353 (1 − )4
3
12.1. PROBLEM SOLUTIONS
25
The ratio of {3 errors} to {2 errors} is
=
353 (1 − )4
5
Pr {3 errors}
=
=
5
2
Pr {2 errors}
3 1 −
21 (1 − )
We now determine the coded error probability . Since BPSK modulation is used
Ãr !
2
=
7
where is 0 . This gives us the following table
8 dB
9 dB
10 dB
11 dB
12 dB
13 dB
14 dB
00897
00660
00455
00290
00167
00085
00037
errors}
= Pr{3
Pr{2 errors}
01642
01177
00794
00497
00278
00141
00062
It is clear that as the SNR, , increases, the value of becomes negligible.
Problem 12.34
For the convolutional coder shown in Figure 12.24 we have the three outputs
1 = 1 ⊕ 2 ⊕ 3
2 = 1
2 = 1 ⊕ 2
Shifting in a 0 gives, in terms of the initial shift register contents
1 = 0 ⊕ 1 ⊕ 2 = 0 ⊕
2 = 0
3 = 0 ⊕ 1
and shifting in a binary 1 gives
1 = 1 ⊕ 1 ⊕ 2 = 1 ⊕
2 = 1
3 = 1 ⊕ 1
26
CHAPTER 12. INFORMATION THEORY AND CODING
where = 1 ⊕ 2 . Clearly 1 = 0 ⊕ or 1 = 1 ⊕ are complements for = 0 or 1. The
output 2 for a 0 or 1 input are obviously complements. Also 1 = 0 ⊕ 1 or 1 = 1 ⊕ 1 are
complements for 1 = 0 or 1
Problem 12.35
For the convolutional coder shown in Figure 12.24 we have the three outputs
1 = 1 ⊕ 2 ⊕ 3 ⊕ 4
2 = 1 ⊕ 2 ⊕ 4
Shifting in a 0 gives, in terms of the initial shift register contents
1 = 0 ⊕ 1 ⊕ 2 ⊕ 3 = 0 ⊕
2 = 0 ⊕ 1 ⊕ 3 = 0 ⊕
and shifting in a binary 1 gives
1 = 1 ⊕ 1 ⊕ 2 ⊕ 3 = 1 ⊕
2 = 1 ⊕ 1 ⊕ 3 = 1 ⊕
where = 1 ⊕ 2 ⊕ 3 and = 1 ⊕ 3 . Clearly 1 = 0 ⊕ or 1 = 1 ⊕ are complements
for = 0 or 1. In the same manner 2 = 0 ⊕ or 2 = 1 ⊕ are complements for = 0 or 1.
Problem 12.36
For the encoder shown, the constraint span is = 4. We first need to compute the states
for the encoder. The output is 1 2 where 1 = 1 ⊕ 2 ⊕ 3 ⊕ 4 and 2 = 1 ⊕ 2 ⊕ 4 .
We now compute the state transitions and the output using the state diagram shown in
12.1. PROBLEM SOLUTIONS
27
Figure 12.8. This gives the following table.
State
A
Previous
1 2 2
000
B
001
C
010
D
011
E
100
F
101
G
110
H
111
Input
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
1 2 3 4
0000
1000
0001
1001
0010
1010
0011
1011
0100
1100
0101
1101
0110
1110
0111
1111
Current
State
A
E
A
E
B
F
B
F
C
G
C
G
D
H
D
H
Output
00
11
11
00
10
01
01
10
11
00
00
11
01
10
10
01
Problem 12.37
For the encoder shown, we have 1 = 1 ⊕ 2 ⊕ 3 and 2 = 1 ⊕ 3 . The trellis diagram
is illustrated in Figure 12.9. The state transitions, and the output corresponding to each
transition appears in the following table.
State
A
1 2
00
B
01
C
10
D
11
Input
0
1
0
1
0
1
0
1
1 2 3
000
100
001
101
010
110
011
111
State
A
C
A
C
B
D
B
D
Output
00
11
11
00
10
01
01
10
Problem 12.38
The source rate is = 5000 symbols/second. The channel symbol rate is
³´
= = 5000
28
CHAPTER 12. INFORMATION THEORY AND CODING
(0 0)
A
(1 1)
(1 1)
(1 0)
C
B
(0 1)
(1 1)
(0 0)
(0 1)
E
(0 0)
D
(0 0)
(0 1)
(1 0)
(1 1)
G
F
(1 0)
(1 0)
H
(0 1)
Figure 12.8: State diagram for Problem 12.36.
If the burst duration is , then
³´
= 5000
channel symbols can be affected. Assuming that = 02 s, the burst length, expressed in
symbols, is
³´
³´
= = (5000) (02)
= 1000
symbols affected by the burst. The interleaving array contains rows with each row
representing a ( ) codeword. We now must now determine and .
Since , it is clear that the burst affects more than 1000 code symbols. We also
know that the rate of a Hamming code approaches 1 as gets large, so that for large .
/
=
12.1. PROBLEM SOLUTIONS
(00)
(11)
29
(00)
(00)
(11)
(11)
(10)
(00)
(10)
(01)
(01)
(00)
(00)
(11)
(11)
(11)
(10)
(01)
(01)
(10)
Termination
Steady-State
Transitions
Figure 12.9: Trellis diagram for Problem 12.37.
will exceed 1000 by only a small amount. We therefore try a (1023,1013) Hamming code
having − = 10 parity symbols. The number of symbols affected by the burst is
= 1000
µ
1023
1013
¶
= 10099
Since must be an integer lower bounded by 1009.9, we let = 1010. The full interleaving
array contains symbols, since may be received in error (worst case) ( − 1) must
be received correctly. Since the symbol rate on the channel is 5000 this gives a required
error-free span of
1010 1022
( − 1)
=
= 20443 s
5000
5000 10231013
This is probably not very practical but it illustrates the process and the problem of using
long codes with low error-correcting capabilities for interleaving.
Problem 12.39
30
CHAPTER 12. INFORMATION THEORY AND CODING
The source rate is = 5000 symbols/second. The channel symbol rate is
µ ¶
23
= = 5000
12
since we have a (23,12) code. If the burst duration is , then
µ ¶
23
= = 5000
12
channel symbols can be affected by the burst. The burst can affect 3 columns, containing
3 symbols, of the interleaving table. Assuming that = 02 s, the burst length, expressed
in symbols, is
µ ¶
23
= 19167 symbols
3 = = (5000) (02)
12
Thus
19167
= 63889
3
We therefore set = 639, which gives an interleaving array with 639 rows and 23 columns.
Since 3 columns can contain errors we must have 20 error-free columns representing 20(639)
symbols. This represents an error-free span of
=
20
20(639)
=
= 13336
5000(2312)
s
We see that the more powerful code results in a much shorter required error-free span and
a smaller interleaving array.
Problem 12.40
The probability of transmissions is the probability of − 1 retransmissions. This is 2−1 .
The expected value of is therefore
(∞
)
X
−1
{ } =
2
=1
From
∞
X
=
=0
1
1−
|| 1
we can obtain, by differentiating both sides of the above wrt ,
∞
X
=1
−1 =
1
(1 − )2
12.1. PROBLEM SOLUTIONS
31
Figure 12.10: Detection gain characteristice for Problem 12.41.
There the expected value of is
{ } =
1
(1 − 2 )2
Problem 12.41
The required plot is shown in Figure 12.10.
Problem 12.42
The expected number of transmissions, , is
= 1 Pr( = 1) + 2 Pr( = 2) + 3 Pr( = 3) · · · Pr( = ) · · ·
32
CHAPTER 12. INFORMATION THEORY AND CODING
or
=
∞
X
2−1
=1
where is the probability of a decision on the transmission, which is 1 ( − 1 erasures
plus a decision). Thus, with = − 1 we have
=
∞
X
2−1 =
=1
12.2
1
(1 − 2 )2
Computer Exercises
Computer Exercise 12.1
For two messages the problem is easy since there is only a single independent probability.
The MATLAB program follows
% File: ce12_1a.m
a = 0.001:0.001:0.999;
b =1-a;
lna = length(a);
H = zeros(1,lna);
H=-(a.*log(a)+b.*log(b))/log(2);
plot(a,H);
xlabel(’a’)
ylabel(’Entropy’)
The plot is illustrated in Figure 12.11.
For three source messages we have two independent probabilities. Two of them will be
specified and the remaining probability can then be determined. The resulting MATLAB
code follows:
% File: ce12_1b.m
n = 50;
nn = n+1;
H = zeros(nn,nn);
z = (0:n)/n;
z(1) = eps;
% number of points in vector
% increment for zero
% initialize entropy array
% probability vector
% prevent problems with log(0)
12.2. COMPUTER EXERCISES
33
1
0.9
0.8
0.7
E ntropy
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
a
0.6
0.7
0.8
0.9
1
Figure 12.11: Entropy for binary source.
for i=1:nn
for j=1:nn
c = 1-z(i)-z(j);
% calculate third probability
if c0
xx = [z(i) z(j) c];
H(i,j)= -xx*log(xx)’/log(2); % compute entropy
end
end
end
colormap([0 0 0])
% b&w colormap for mesh
mesh(z,z,H);
% plot entropy
view(-45,30);
% set view angle
xlabel(’Probability of message a’)
ylabel(’Probability of message b’)
34
CHAPTER 12. INFORMATION THEORY AND CODING
Figure 12.12: Entropy for three-symbol source.
zlabel(’Entropy’)
figure
contour(z,z,H,10)
xlabel(’Probability of message a’)
ylabel(’Probability of message b’)
% new figure
% plot contour map
The resulting plot is illustrated in Figure 12.12. One should experiment with various viewing
locations.
Next we draw a countour map with 10 countour lines. The result is illustrated in Figure
10.17. Note the peak at 0.33, 0.33, indicating that the maximum entropy occurs when all
three source messages occur with equal probability
Computer Exercise 12.2
The solution of this computer example is, in large part, taken from the book by Proakis,
Salehi, and Bauch (Comtemporary Communication Systems using MATLAB, Thomson-
12.2. COMPUTER EXERCISES
35
Brooks/Cole, 2004, ISBN 0-534-40617-3).1 Modifications made to the ortiginally published
program include changing the program from a function to a script, allowing the user to
determine probability distributions and source codes for extended sources, allowing the user
to enter the source distribution and desired extension order interactively and, in addition,
the entropy of the source, the entropy of the extended source, the average word length, and
the efficiency are displayed as the program is executed. This was done since a typical use of
source coding is to continue incrementing the source extension until the efficiency reaches
a point where transmission can be accomplished. The resulting MATLAB code follows.
% File: ce12_2.m
clear all
msgprobs = input(’Enter message probabilities as a vector, ...
e.g., [0.5 0.25 0.25] ’);
ext = input(’Enter desired extension (1 if no extension is desired) ’);
entp = -sum(msgprobs.*log(msgprobs))/log(2); % original source entropy
dispa = [’The source entropy is ’, ...
num2str(entp,’%15.5f’),’ bits/symbol.’]; disp(dispa)
%
% Determine probabilities and entropy of extended source.
%
if ext1
b = msgprobs;
% working array
for k=2:ext
b = kron(b,msgprobs);
% Kronecker tensor product
end
msgprobs = b;
% probabilities for extended source
entp = -sum(msgprobs.*log(msgprobs)) ...
/log(2);
% entropy of extended source
dispaa = [’The extended source entropy is ’, ...
um2str(entp,’%15.5f’),’ bits/symbol.’]; disp(dispaa)
end
%
% Continue
1
This book is highly recommended as a supplemental text in basic communications courses. We have
found it useful as a resource for in-class demonstrations and for homework assignments extending material
covered in class.
This particular computer example, as originally developed by Proakis, Salehi, and Bauch, provides an
excellent example of several MATLAB programming techniques. It has been said that a key to understanding
efficient MATLAB programming techniques is to understand the use of the colon and the find command.
This program makes efficient use of both.
36
CHAPTER 12. INFORMATION THEORY AND CODING
%
n = length(msgprobs);
% length of source (or extended source)
q = msgprobs;
% copy of msgprobs
m = zeros(n-1,n);
% initialize
for i=1:n-1
[q,l] = sort(q);
% order message probabilities
m(i,:) = [l(1:n-i+1),zeros(1,i-1)];
q = [q(1)+q(2),q(3:n),1]; % combine messages in column vector
end
%
% Code words are estabilished in the c array.
%
for i = 1:n-1
c(i,:) = blanks(n*n);
% initialize character array
end
c(n-1,n) = ’0’; c(n-1,2*n) = ’1’;
for i=2:n-1
c(n-i,1:n-1) = c(n-i+1,n*(find(m(n-i+1,:)==1)) ...
(n-2):n*(find(m(n-i+1,:)==1)));
c(n-i,n) = ’0’;
c(n-i,n+1:2*n-1) = c(n-i,1:n-1);
c(n-i,2*n) = ’1’;
for j=1:i-1
c(n-i,(j+1)*n+1:(j+2)*n) = ...
c(n-i+1,n*(find(m(n-i+1,:)==j+1)-1)+1:n*find(m(n-i+1,:)==j+1));
end
end
%
% Establish code words, average word length, and entropy.
%
for i=1:n
% determine codewords and wordlengths
codewords(i,1:n) = c(1,n*(find(m(1,:)==i)-1)+1:find(m(1,:)==i)*n);
wrdlengths(i) = length(find(abs(codewords(i,:))~=32));
end
avewl = sum(msgprobs.*wrdlengths);
% average word length
codewords
% display code words
disp3 = [’The average wordlength is ’, ...
num2str(avewl,’%15.5f’),’ symbols.’]; disp(disp3)
disp4 = [’The efficiency is ’, ...
num2str(entp*100/avewl,’%15.5f’),’ %.’]; disp(disp4)
12.2. COMPUTER EXERCISES
37
% End of function file.
We will use the preceding MATLAB program to work Problem 12.28. The MATLAB
dialog follows.
Computer Exercise 12.3
The MATLAB code for generating the performance curves illustrated in Figure 11.28 in the
text follows.
% File: ce12_3.m
zdB = 0:30;
z = 10.^(zdB/10);
qa = 0.5*exp(-0.5*z);
qa3 = 0.5*exp(-0.5*z/3);
qa7 = 0.5*exp(-0.5*z/7);
pa7 = 35*((1-qa7).^3).*(qa7.^4)+21*((1-qa7).^2).*(qa7.^5)...
+7*(1-qa7).*(qa7.^6)+(qa7.^7);
pa3 = 3*(1-qa3).*(qa3.^2)+(qa3.^3);
qr = 0.5./(1+z/2);
qr3 = 0.5./(1+z/(2*3));
qr7 = 0.5./(1+z/(2*7));
pr7 = 35*((1-qr7).^3).*(qr7.^4)+21*((1-qr7).^2).*(qr7.^5)...
+7*(1-qr7).*(qr7.^6)+(qr7.^7);
pr3 = 3*(1-qr3).*(qr3.^2)+(qr3.^3);
semilogy(zdB,qr,zdB,pr3,zdB,pr7,zdB,qa,zdB,pa3,zdB,pa7)
axis([0 30 0.0001 1])
xlabel(’Signal-to-Noise Ratio, z - dB’)
ylabel(’Probability ’)
Executing the code yields the results illustrated in Figure 12.13 in the text. The curves can
be identified from Figure 10.18 in the text.
Computer Exercise 12.4
For the rate 0.5 BCH codes, we use the following MATLAB program.
% File: ce12_4a.m
zdB = 0:0.5:10;
% set Eb/No in dB
z = 10.^(zdB/10);
% convert to linear scale
ber1 = q(sqrt(4*2*z/7));
% SER for (7,4) BCH code
ber2 = q(sqrt(7*2*z/15)); % SER for (15,7) BCH code
ber3 = q(sqrt(16*2*z/31)); % SER for (31,16) BCH code
38
CHAPTER 12. INFORMATION THEORY AND CODING
0
10
-1
P robability
10
-2
10
-3
10
-4
10
0
5
10
15
20
Signal-to-Noise Ratio, z - dB
25
30
Figure 12.13: Results for Computer Exercise 12.3.
ber4 = q(sqrt(30*2*z/63)); % SER for (63,30) BCH code
berbch1 = ser2ber(2,7,3,1,ber1);
% BER for (7,4) BCH code
berbch2 = ser2ber(2,15,5,2,ber2); % BER for (15,7) BCH code
berbch3 = ser2ber(2,31,7,3,ber3); % BER for (31,16) BCH code
berbch4 = ser2ber(2,64,13,6,ber4); % BER for (63,30) BCH code
semilogy(zdB,berbch1,’k’,zdB,berbch2,’k’,zdB,berbch3,’k’,zdB,berbch4,’k’)
xlabel(’E_b/N_o in dB’) % label x axis
ylabel(’Bit Error Probability’) % label y axis
% End of script file.
Executing the code for the rate 0.5 code yields the results shown in Figure 12.14. For SNRs
exceeding 5 dB, the order of the plots is in order of word length , with = being the top
curve (least error correcting capability) and = 63 being the bottom curve (greatest error
correcting capability).
12.2. COMPUTER EXERCISES
39
0
10
-2
10
-4
Bit Error Probability
10
-6
10
-8
10
-10
10
-12
10
-14
10
0
1
2
3
4
5
6
Eb/No in dB
7
8
9
10
Figure 12.14: Rate 1/2 BCH code results for Computer Exercise 12.4.
40
CHAPTER 12. INFORMATION THEORY AND CODING
For the rate 0.75 BCH codes, we use the following MATLAB program.
% File: ce12_4b.m
zdB = 0:0.5:10;
% set Eb/No in dB
z = 10.^(zdB/10);
% convert to linear scale
ber1 = q(sqrt(11*2*z/15));
% SER for (15,11) BCH code
ber2 = q(sqrt(21*2*z/31));
% SER for (31,21) BCH code
ber3 = q(sqrt(45*2*z/63));
% SER for (63,45) BCH code
ber4 = q(sqrt(99*2*z/127)); % SER for (127,99) BCH code
berbch1 = ser2ber(2,15,3,1,ber1); % BER for (15,11) BCH code
berbch2 = ser2ber(2,31,5,2,ber2); % BER for (31,21) BCH code
berbch3 = ser2ber(2,63,7,3,ber3); % BER for (63,45) BCH code
berbch4 = ser2ber(2,127,9,4,ber4); % BER for (127,99) BCH code
semilogy(zdB,berbch1,zdB,berbch2,zdB,berbch3,zdB,berbch4)
xlabel(’E_b/N_o in dB’) % label x axis
ylabel(’Bit Error Probability’) % label y axis
% End of script file.
Executing the code for the rate 0.75 code yields the results shown in Figure 12.15. For SNRs
exceeding 5 dB, the order of the plots is in order of word length , with = 15 being the
top curve (least error correcting capability) and = 127 being the bottom curve (greatest
error correcting capability).
Computer Exercise 12.5
In order to illustrate the problems associated with using nchoosek with large arguments
we compute
» nchoosek(1000,500)
Warning: Result may not be exact.
In C:\MATLABR11\toolbox\matlab\specfun\nchoosek.m at line 50
ans =
2.7029e+299
The warning indicates a possible with numerical percision. A possible way to mitigate this
is to perform the calculation using logarithms as illustrated in the function below.
%File: nkchoose.m
Function out=nkchoose(n,k)
% Computes n!/k!/(n-k)!
a = sum(log(1:n));
% ln of n!
12.2. COMPUTER EXERCISES
41
0
10
Bit Error Probability
-5
10
-10
10
-15
10
0
1
2
3
4
5
6
Eb/No in dB
7
8
9
10
Figure 12.15: Rate 3/4 BCH code results for Computer Exercise 12.4.
42
CHAPTER 12. INFORMATION THEORY AND CODING
b = sum(log(1:k));
c = sum(log(1:(n-k)));
out = round(exp(a-b-c));
% End of function file.
% ln of k!
% ln of (n-k)!
% result
The main program for evaluating the performance of the (511,385) BCH code is shown
below.
% File: ce12_5.m
zdB = 0:0.5:10;
% set Eb/No in dB
z = 10.^(zdB/10);
% convert to linear scale
ber1 = q(sqrt(z));
% FSK result
ber2 = q(sqrt(385*z/511));
% SER for (511,385) BCH code
ber3 = q(sqrt(768*z/1023));
% SER for (1023,768) BCH code
berbch1 = ser2ber1(2,511,29,14,ber2);
% BER for (511,385) BCH code
berbch2 = ser2ber1(2,1023,53,26,ber3); % BER for (1023,768) BCH code
semilogy(zdB,ber1,’k-’,zdB,berbch1,’k--’,zdB,berbch2,’k-.’)
xlabel(’E_b/N_o in dB’) % label x axis
ylabel(’Bit Error Probability’) % label y axis
legend(’Uncoded’,’(511,385) BCH code’,’(1023,768) BCH code’,3)
% End of script file.
Note that this program calls ser2ber1 rather than ser2ber. In order to evaluate the
performance of the (1023,768) BCH code, it is necessay to modify the routine for mapping
the code symbol error rate so that the calculation is performed using logarithms as the
following MATLAB code indicates.
% File: ser2ber1.m
function [ber] = ser2ber(q,n,d,t,ps)
lnps = length(ps);
% length of error vector
ber = zeros(1,lnps);
% initialize output vector
for k=1:lnps
% iterate error vector
ser = ps(k);
% channel symbol error rate
sum1 = 0; sum2 = 0;
% initialize sums
for i=(t+1):d
lnterm = log(nkchoose(n,i))+(i*log(ser))+(n-i)*log((1-ser));
term = exp(lnterm);
sum1 = sum1+term;
end
for i=(d+1):n
12.2. COMPUTER EXERCISES
43
0
10
-5
Bit Error Probability
10
-10
10
-15
10
Uncoded
(511,385) BCH code
(1023,768) BCH code
-20
10
0
1
2
3
4
5
6
Eb/No in dB
7
8
9
10
Figure 12.16: Performance results for Computer Exercise 12.5.
lnterm = log(i)+log(nkchoose(n,i))+(i*log(ser))+(n-i)*log((1-ser));
term = exp(lnterm);
sum2 = sum2+term;
end
ber(k) = (q/(2*(q-1)))*((d/n)*sum1+(1/n)*sum2);
end
% End of function file.
Executing the program yields the results shown in Figure 12.16.
Computer Exercise 12.6
In order to work this computer exercise we write a MATLAB program to determine the
outputs for each of the 16 tree branches illustrated in Figure 12.25. The inputs for each
44
CHAPTER 12. INFORMATION THEORY AND CODING
tree branch are illustrated in the right-hand column of Figure 12.25. The outputs for the
portion of the tree are generated by inputting the appropriate binary vector corresponding
the portion of the tree of interest. The MATLAB code follows.
% File: ce12_6.m
clear all
invector = input(’Enter input as vector (in brackets with spaces) ’);
d = [’The input sequence is ’,num2str(invector),’.’];
disp(d)
sr = [0 0 0];
for k=1:4
sr = [invector(1,k) sr(1,1) sr(1,2)];
v1 = dec2bin(rem(sr(1,1)+sr(1,2)+sr(1,3),2));
v2 = dec2bin(sr(1,1));
v3 = dec2bin(rem(sr(1,1)+sr(1,2),2));
c = [v1 v2 v3];
out(1,k) = bin2dec(c);
end
out = dec2bin(out)
% End of script file.
The program is illustrated by generating the outputs for the path labeled ’A’ in Figure
12.25., for which the input bit stream is 1010. The MATLAB dialog is as follows:
» ce12_6
Enter input sequence as a vector (in brackets with spaces) [1 0 1 0]
The input sequence is 1 0 1 0.
out =
111
101
011
101
The four outputs are represented as the rows following ‘out’ in the preceding dialog. This
corresponds to the sequence 111 101 011 101, which is Path A as shown in Figure 12.25 in
the text. (Note: If all outputs are 0, the sequence 000 is represented by a single 0.)
Computer Exercise 12.7
This Computer Exercise is worked by modifying the MATLAB code in the text (c12ce6.m)
to allow the calculation of results for = 01 and = 03 and plotting the results together
so that they are easily compared. The revised MATLAB code follows.
12.2. COMPUTER EXERCISES
45
% File: ce12_7.m
g = [0.1 0.3];
% gamma
zdB = 0:0.1:10;
% z in dB
z = 10.^(zdB/10);
% vector of z values
qt = Q(sqrt(2*z));
% gamma=0 case
for k=1:2
q1 = Q((1-g(k))*sqrt(2*z));
q2 = Q((1+g(k))*sqrt(2*z));
p2 = q1-q2;
% P2
p1 = q2;
% P1
pe(k,:) = p1./(1-p2); % probability of error
N(k,:) = 1./((1-p2).*(1-p2));
end
subplot(2,2,1)
semilogy(zdB,pe(1,:),zdB,qt)
title(’gamma = 0.1’)
xlabel(’z - dB’); ylabel(’Error Probability’); grid
subplot(2,2,2)
semilogy(zdB,pe(2,:),zdB,qt)
title(’gamma = 0.3’)
xlabel(’z - dB’); ylabel(’Error Probability’); grid
subplot(2,2,3)
plot(zdB,N(1,:))
xlabel(’z - dB’); ylabel(’Ave. Transmissions/Symbol’);
axis([0 10 1 1.4]); grid
subplot(2,2,4)
plot(zdB,N(2,:))
xlabel(’z - dB’); ylabel(’Ave. Transmissions/Symbol’);
axis([0 10 1 1.4]); grid
% End of script file.
Executing the program gives the results shown in the illustration below. As expected, as
is increased, the error probability deceases at the cost of an increase in the average number
of transmissions. The key, however, is that increased performance is achieved with only a
moderate increase in the rate of retransmissions. This is especially true at higher SNRs.
Computer Exercise 12.8
We start with Computer Example 12.7 and add the required code to give the numbers of
matching elments. The MATLAB code follows.
46
CHAPTER 12. INFORMATION THEORY AND CODING
gamma = 0.1
0
Error Probability
-5
-10
10
-10
0
5
z - dB
10
1.4
1.3
1.2
1.1
1
-5
10
0
5
z - dB
10
10
Ave. Transmissions/Symbol
Error Probability
10
10
Ave. Transmissions/Symbol
gamma = 0.3
0
10
0
5
z - dB
10
0
5
z - dB
10
1.4
1.3
1.2
1.1
1
Figure 12.17: Performance results for feedback channel.
12.2. COMPUTER EXERCISES
47
% File: ce12_8.m
elements = 40;
clear all;
for k=1:elements
z(k)=rand(1);
if z(k)0.3
z(k)=’B’;
else
z(k)=’W’;
end
end
mask = [’W’ ’B’];
t = [z(2:end) z(end)];
diff = z-t;
% subtract string with its offset
trans = [0 find(diff ~= 0)]; % vector with transition indexes
if (trans(end) ~= numel(z))
trans = [trans numel(z)] % nasty case (no transition at end)
end
r = trans(2:end) - trans(1:end-1); % lengths of consecutive B’s and W’s
count = [sum(r(1:2:end)) sum(r(2:2:end))];
% number of B’s and W’s
if(z(1)==’W’) % if sequence starts with ’W’
count = count([2 1]);
% swap count and mask
mask = mask([2 1]);
end
disp(sprintf(’Sequence: %s’,z));
disp(sprintf(’B count: %i W count: %i’,count));
% Display results
BWstr = [];
for k=1:numel(r)
BWstr = strcat(BWstr,num2str(r(k)),mask(mod(k,2)+1));
end
disp(sprintf(’BW string: %s’,BWstr));
% End of script file.
Executing the program gives (Keep in mind that separate executions will give separate
results.):
CE12_8
Sequence: WWWWBWWWWWBWWBBWWWWBWBWWWWWBBBWBWBWWBBWW
B count: 13 W count: 27
48
CHAPTER 12. INFORMATION THEORY AND CODING
BW string:
4W1B5W1B2W2B4W1B1W1B5W3B1W1B1W1B2W2B2W
Acknowledgement: Thanks to Dr. John Tranter at the University of Minnesota for his help
on this Computer Exercise.
0
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