rice m digital communications a discretetime approach soluti
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2.1
(a)
E 2T
P0
(b)
E 10T1 2T2
P0
(c)
E
1
P
2
(d)
E
P
A2
2
(e)
E
3
P
8
(f)
E T
P0
(g)
1
4a
P0
E
(h)
A2T 3
E
4
P0
(i)
E
1
P lim
T T
3 T
t
3
33 T 3 3 3
1/3
2/3t
dt lim
T
T
3 T
3 lim
1/3
T
T
0
2.2
(a)
8
(b)
0
(c)
4
2
(d)
1
2.3
(a)
X ( s)
5
s 1
(b)
x(t ) 5t eln(5
1
X ( s)
s ln(5)
(c)
5
s 1
(d)
5
s 1
6
)
e t ln(5)
2.4
(a)
2
X (s)
(b)
2 4 4 2 4s 4s
s3 s 2 s
s3
2
x(t ) t 2 e 3t u (t ) t 2 4t 4 e 3t u (t )
2
X ( s)
(c)
x(t ) t 2 u (t ) t 2 4t 4 u (t )
2
s 3
3
4
s 3
2
4
26 20 s 4 s 2
3
s3
s 3
x(t ) cos 0t u (t )
4
cos cos 0t sin sin 0t u (t )
4
4
1
1
cos 0t
sin 0t u (t )
2
2
0
s
1
1
1 s 0
X (s)
2
2
2
2
2
2
2 s 0
2 s 0
2 s 0
(d)
x(t ) e 3t cos 0t u (t )
3
e 3t cos cos 0t sin sin 0t u (t )
3
3
1
3
sin 0t u (t )
e 3t cos 0t
2
2
X ( s)
0
1
3
1 s 3 30
s3
2
2
2
2
2 s 3 0
2 s 3 0 2 s 32 02
2.5
(a)
1
1
j
j
X (s) 2
2
s2 j s2 j
1
1
x(t ) j e (2 j ) t j e (2 j ) t u (t )
2
2
1
1
e 2t e jt je jt e jt je jt u (t )
2
2
e 2t cos(t ) 2sin(t ) u (t )
5e 2t cos t 63.4 u (t )
(b)
X ( s)
4
4
4
s 2 s 1 s 12
x(t ) 4e 2t 4e t 4te t u (t ) 4 e2t t 1 e t u (t )
(c)
X ( s)
4
4
4
2
s 2 s 2
s 1
x(t ) 4e2t 4te2t 4e t u (t ) 4 e t t 1 e2t u (t )
2.6
(a)
(b)
3
2
1
s 3 s 2 s 1
x(t ) 3e 3t 2e 2t e t u (t )
X ( s)
X (s)
229
54
216
108
135
13
2
3
2
s 3 s 3 s 3 s 2 s 2
s 1
x(t ) 229 135t 26t 2 e 3t 216 108t e 2t 13e t u (t )
(c)
X (s)
32
31
30
21
20
1
2
3
2
s 3 s 3 s 3 s 2 s 2
s 1
x(t ) 32 31t 30t 2 e 3t 21 20t e 2t e t u (t )
2.7
(a)
y(t ) a1 y(t ) a2 y (t ) b1 x(t ) b0 x(t ) initial conditions
(b)
The poles are at s a1 a12 4a0
(i)
For real and distinct poles, we require a12 4a0 . In this case, the poles
are s a1 a12 4a0 .
(ii)
For real and repeated poles, we require a12 4a0 . In this case, the poles
are s a1 , s a1 .
(iii)
For complex conjugate poles, we require a12 4a0 . In this case, the poles are
s a1 j 4a0 a12
(c)
(i)
n a0
a1
2 a0
In general, the poles are at s n n 2 1
(ii)
For real and distinct poles, we require 1 . In this case, the poles
are s n n 2 1
(iii)
For real and repeated poles, we require 1 . In this case, the poles
are s n , s n .
(iv)
For complex conjugate poles, we require 0 1 . In this case, the poles
are s n jn 1 2
(v)
This form is preferred because the nature of the poles is determined by a single
parameter.
(d)
(i)
b0
H (s)
h(t )
(ii)
H ( s)
b0
s 2 2n s n2
2n 1
n
e
2 1
n
e
2 1
b0 ent
2n
b0
n 1
2
s n 2 1
b0
2n
b0
2
2 1 t
2 1 t
e
e
2n 2 1
s n 2 1
u (t )
n 2 1 t
u (t )
n 2 1 t
e n t sinh n 2 1 t u (t )
b0
s n
2
h(t ) b0te nt u (t )
(iii)
b0
H (s)
h(t )
(e)
b0
j 2n 1
2
s n j 1 2
b0
j 2n
e n t e jn
2
1
b0 e n t
n 1
2
j 2n 1 2
s n j 1 2
1 2 t
e n t e jn
1 2 t
u (t )
sin n 1 2 t u (t )
An overdamped system does not display any oscillations in its impulse response, whereas
the underdamped system does display oscillations. Hence, the term damped refers to
oscillations: if oscillations are present, the oscillations have not been damped; if there are
no oscillations, the oscillations have been damped.
2.8
(a)
6
6 5 5
5
X ( s)
s6
s6 s s s6
6 t
y (t ) 5 1 e u (t )
Y (s)
(b)
x(t)
y(t)
6
4
2
0
(c)
(d)
0
0.1
0.2
0.3
0.4
0.5
t
0.6
0.7
0.8
0.9
1
6s
6s 5
30
X (s)
s6
s6 s s6
6 t
v(t ) 30e u (t )
V ( s)
30
v(t)
20
10
0
0
0.1
0.2
0.3
0.4
0.5
t
0.6
0.7
0.8
0.9
1
2.9
(a)
6
6 5
5 5/6 5/6
X (s)
2
2
s6
s6 s
s
s
s6
5
y (t ) 5t 1 e6t u (t )
6
Y (s)
(b)
x(t)
y(t)
6
4
2
0
(c)
V ( s)
0
0.1
0.2
0.3
0.4
0.5
t
0.6
0.7
0.8
0.9
1
6s
6s 5
30
5
5
X (s)
2
s6
s6 s
s ( s 6) s s 6
v(t ) 5 1 e 6t u (t )
(d)
5
v(t)
4
3
2
1
0
0
0.1
0.2
0.3
0.4
0.5
t
0.6
0.7
0.8
0.9
1
2.10
(a)
6
6
5 5 15
10
X ( s) 2
s 5s 6
s 5s 6 s s s 2 s 3
2 t
3t
y (t ) 5 1 3e 2e u (t )
Y (s)
2
(b)
6
5
4
3
2
x(t)
y(t)
1
0
(c)
(d)
0
0.5
1
1.5
2
2.5
t
3
3.5
4
4.5
5
6s
6s
5
30
30
X ( s) 2
s 5s 6
s 5s 6 s s 2 s 3
2 t
3t
v(t ) 30 e e u (t )
V ( s)
2
5
v(t)
4
3
2
1
0
0
0.5
1
1.5
2
2.5
t
3
3.5
4
4.5
5
2.11
(a)
25 15
10
6
6
5
5
Y (s) 2
X (s) 2
2 6 2 3
2
s 5s 6
s 5s 6 s
s
s s2 s3
2
5 3
y (t ) 5 t e 2t e 3t u (t )
3
6 2
(b)
25
20
15
10
5
0
(c)
(d)
x(t)
y(t)
0
0.5
1
1.5
2
2.5
t
3
3.5
4
4.5
5
6s
6s
5 5 15
10
X ( s) 2
2
s 5s 6
s 5s 6 s
s s2 s3
v(t ) 5 1 3e 2t 2e 3t u (t )
V ( s)
2
5
v(t)
4
3
2
1
0
0
0.5
1
1.5
2
2.5
t
3
3.5
4
4.5
5
2.12
(a)
10 j 5
10 j 5
125
125
5 5
4
4
Y (s) 2
X (s) 2
s 10 s 125
s 10 s 125 s s s 5 j10 s 5 j10
10 j 5 j10t 10 j 5 j10t
y (t ) 5 e 5t
e
e
u (t )
4
4
5 5t
1
5 1 e 5t 2 cos 10t sin 10t u (t ) 5 1
e cos 10t 26.6 u (t )
2
2
(b)
6
5
4
3
2
x(t)
y(t)
1
0
(c)
(d)
0
0.2
0.4
0.6
0.8
1
t
1.2
1.4
1.6
1.8
2
125
125
125s
125s
5
j4
j4
V ( s) 2
X ( s) 2
s 10s 125
s 10s 125 s s 5 j10 s 5 j10
125 5t j10t j10t
125 5t
u (t )
v(t )
e e e
e sin 10t u (t )
2
j4
40
v(t)
30
20
10
0
-10
0
0.2
0.4
0.6
0.8
1
t
1.2
1.4
1.6
1.8
2
2.13
(a)
2
4 j3
4 j3
125
5
5
125
20
20
X (s) 2
5 2
Y (s) 2
s 10 s 125
s 10 s 125 s 2
s
s
s 5 j10 s 5 j10
2
4 j 3 j10t 4 j 3 j10t
y (t ) 5t e 5t
e
e
u (t )
5
20
20
2
3
2 1
4
5t e 5t cos 10t sin 10t u (t ) 5t e 5t cos 10t 36.9 u (t )
5
10
5 2
10
(b)
10
8
6
4
2
0
(c)
x(t)
y(t)
0
0.2
0.4
0.6
0.8
1
t
1.2
1.4
1.6
1.8
2
10 j 5
10 j 5
125s
125s
5 5
4
4
V (s) 2
X ( s) 2
2
s 10s 125
s 10s 125 s
s s 5 j10 s 5 j10
10 j 5 j10t 10 j 5 j10t
v(t ) 5 e 5t
e
e
u (t )
4
4
5 5t
1
5 1 e 5t 2 cos 10t sin 10t u (t ) 5 1
e cos 10t 26.6 u (t )
2
2
(d)
8
v(t)
6
4
2
0
0
0.2
0.4
0.6
0.8
1
t
1.2
1.4
1.6
1.8
2
2.14
Y ( s)
k
. This system has one pole at s k . The system is
X ( s) s k
stable if the pole is in the left-half plane. The pole is in the left-half plane when k 0 .
The transfer function is H ( s )
2.15
The system transfer function is H ( s )
Y (s)
1
a a 2 4
2
. The poles are s
.
X ( s ) s as 1
2
(a)
The system is stable when the poles are in the left-half plane. The poles are in the left-half
plane for a 0 .
(b)
The system impulse response exhibits oscillations when the poles have a non-zero
imaginary part. The poles have a non-zero imaginary part when a 2 4 0 , which implies
2 a 2 . (Note that the system response is oscillatory and stable only for 0 a 2 .)
These results are summarized by the plot below. This plot, known as a “root-locus” plot, plots
the location of the poles as a function of a.
p1
1
p2
1
a=0
0
0.5
a -> infinity
a -> -infinity
-0.5
Imag(s)
Imag(s)
0.5
a -> -infinity
0
a -> infinity
-0.5
a=0
-1
-10
-5
0
Real(s)
5
10
-1
-10
-5
0
Real(s)
5
10
2.16
The system transfer function is H ( s )
a a a 4
Y ( s)
a
2
. The poles are s
.
2
X ( s ) s as a
(a)
The system is stable when the poles are in the left-half plane. The poles are in the left-half
plane for a 0 .
(b)
The system impulse response exhibits oscillations when the poles have a non-zero
imaginary part. The poles have a non-zero imaginary part when a 4 0 and a 0 ,
which implies 0 a 4 . (Note that the system response is oscillatory and stable for these
values of a.)
These results are summarized by the plot below. This plot, known as a “root-locus” plot, plots
the location of the poles as a function of a.
p2
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
a -> infinity
a=0
a -> -infinity
-0.2
Imag(s)
Imag(s)
p1
1
a -> -infinity
0
-0.4
-0.6
-0.6
-0.8
-0.8
-5
0
Real(s)
5
10
a -> infinity
-0.2
-0.4
-1
-10
a=0
-1
-10
-5
0
Real(s)
5
10
2.17
x t A cos 0t
(a)
X j
(b)
Xf
A j 0t A j 0t A j j0t A j j0t
e
e e e e
e
2
2
2
2
A j
A
e 2 0 e j 2 0
2
2
j
A e 0 A e j 0
A j
A
e f f 0 e j f f 0
2
2
2.18
Xf
0
e
at j 2 ft
e
dt e at e j 2 ft dt
0
1
j 2 f a
2a
2
a 4 2 f 2
X
0
e
at jt
e
1
j 2 f a
dt e at e jt dt
0
1
j a
2a
2
a 2
1
j a
2.19
The easiest way to compute the Fourier transform is to recognize that x t , plotted below
T
T
can be produced by convolving the function plotted below with itself.
1
T
T
2
T
2
Fourier transform
T
sin fT
fT
Thus, the Fourier transform may be computed as follows:
Xf T
sin fT
fT
T
sin fT
fT
T
sin 2 fT
2 f 2T 2
Similarly,
T
T
sin
sin
2 T
2 T
X j T
T
T
2
2
T
sin 2
2
2
T
2
2.20
The easiest way to compute the Fourier transform is to recognize that x t is produced by the
convolution of the two functions shown below.
A
T2 T1
T2 T1
2
T2 T1
2
T2 T1
2
Fourier transform
T2 T1
T2 T1
2
Fourier transform
sin f T2 T1
T2 T1
f T2 T1
sin f T2 T1
f T2 T1
Multiplying the two Fourier transforms produces
X f A T2 T1
sin f T2 T1 sin f T2 T1
f T2 T1
f T2 T1
Similarly
T T
T T
sin 2 1 sin 2 1
2
2
X j A T2 T1
T T
T T
2 1
2 1
2
2
2.21
(a)
X f 1 e j 2 f
(b)
Let e j . Then X f 1 e j 2 f
X f X f X f 1 e j 2 f e j 2 f
2
2
1 2 cos 2 f
2
(c)
Xf
2
1
1
2
1
2
f
0
2.22
j
1 j
(a)
Y j 3e j 5
(b)
Y j e j 3 5
(c)
(d)
j 5
1 j 5
The symmetry properties of X j show that x t is real. Hence,
Y j
j
1 j
Y j
2
1 j
2.23
(a)
j
x t j 5e j 2 1000t 3e 6 e j 2 60t 16 3e
j
6
e j 2 60t j 5e j 2 1000t
j 2 60t 6
j 2 60 t
6
16 3 e
e
j 5 e j 2 1000t e j 2 1000t
16 6 cos 2 60t 10sin j 2 1000t
6
(b)
1 1
1
Yes, x t is periodic. The period is LCM ,
.
60 1000 10
2.24
Using the identity
1
e 2t u (t ) and the property e j 2 ft0 X f x t t0 for the last
2 j 2 f
term, we have
1 j 2 2 t 1 j 2 2 t
e
e
e 2( t 2)u t 2
2
2
cos 2 2t e 2( t 2)u t 2
x t
2.25
(a)
Xf
B3
(b)
Xf
B3
(c)
1
a j 2 f
2
a 2 2 f
2
a
2
2a
a 2 2 f
a
2
X f e f
B3
1
Xf
2
2a
a 2 2 B3
2
1 2
2 a
2 1
2
ln(2)
2
X f e 2 f
2
2
2
e 2 B3
1
2
1
a 2 2 B3
2
1
2a 2
2.26
Using the transform pair
2T
set T
sin 2 fT
2 fT
1 T t T
,
0 otherwise
1
and apply Parseval’s theorem:
2
sin 2 f
f
2
df
1
2
1
1
2
2
dt 1
2.27
(a)
E
Xf
x 2 (t )dt e 2 at dt
0
1
a j 2 f
2
0
0
2 at
2 at
e dt e dt
x 2 (t )dt
Xf
2a
a 2 2 f
B90
2
E
x 2 (t )dt
Xf e
Xf
e
t 2
2
df
Now, using e
0.45
u2
2
e
B90
2 f 2
2 2
2 B90
2 tan 1
a
a
2
du
df
1
4
1
df
4
e
u2
2
e 2 f df e 2 f df
2
du
e
u2
2
4 B90
1
4
1
du
2
2
0
2
1
and Q z
2
2
0
2
B90
0
0
2 f
a 2 2 B90
2
2
0
2
dt 1
0
2 f 2
a
4 B90
X f e 2 f
B90
e
4a 2
f 2
0.9 2 e 2 f df
1
a
2
2
0.9
4a
2
2
a
0
a 2 2 f
(c)
2
a
0.9
tan
2
a
E
a 2 2 f
2 B90
tan 1
a
a
0.9
df
2 2
2
2a
0 a 2 f
(b)
1
Xf
B90
B90
1
2a
e
u2
2
e
2 f 2
df
B90
du the two intergrals are
z
2
1
2
2 2
1
2
4 B90
e
u2
2
du
1
Q
2
4 B90
Putting this together produces, we have that B90 is determined from
0.45 2 0.5 Q
4 B90
2.28
(a)
A2
A3
cos 2 0t
cos3 0t
2
3
2
3
2
A
A
A
A3
A
cos
t
cos
2
t
cos 30t
0
0
4
4
4
12
y t A cos 0t
(b)
Yf
6
A
576
−3000
(c)
A3
4 f 1000 f 1000
2
A
2
A
f
4
A2
A3
f 2000 f 2000
f 3000 f 3000
8
24
A4
64
−2000
A3
A
4
4
2
A3
A
4
4
4
A
16
−1000
0
1000
2
A4
64
2000
A6
576
3000
A6
A6 A4 A4
1
9 A 4 A6
576
576
64
64
THD
100
100
2
2
9 16 A2 8 A4 A6
A3
A3
A
A
4
4
4
4
A plot of the THD is shown below. Note that the THD increases as the amplitude of the
sinusoid increases.
2
10
0
THD (%)
10
-2
10
-4
10
0
1
2
A
3
4
2.29
(a)
A
cos 0t A cos 1t
4
17 A2 A2
A2
A2
A2
x2 t
cos 20t
cos 21t
cos 1 0 t
cos 1 0 t
32
32
32
4
4
A3
A3
99 A3
27 A3
x3 t
cos 0t
cos 1t
cos 30t
cos 31t
256
32
256
4
3 A3
3 A3
cos 1 20 t
cos 1 20 t
64
64
3 A3
3 A3
cos 21 0 t
cos 21 0 t
16
16
17 A2 A 99 A3
3 A3
A2
A2
cos
t
A
cos
t
cos
2
t
cos 21t
y t
0
1
0
64 4 768
32
64
4
x t
A3
A3
A2
A2
cos 30t
cos 31t
cos 1 0 t
cos 1 0 t
768
12
8
8
A3
A3
cos 1 20 t
cos 1 20 t
64
64
A3
A3
cos 21 0 t
cos 21 0 t
16
16
(b)
Yf
A 99 A3
17 A2
f
f 60 f 60
64
8 1536
A 3 A3
f 7000 f 7000
2 64
A2
A2
f 120 f 120
f 14000 f 14000
128
8
A3
A3
f 180 f 180
f 21000 f 21000
1538
24
A2
A2
f 6740 f 6740
f 7060 f 7060
16
16
A3
A3
f 6880 f 6880
f 7120 f 7120
128
128
A3
A3
f 13940 f 13940
f 14060 f 14060
32
32
2
A3
24
2
21000
0
60 60
14000
14060 13940
2
7000
7060 6880
120
6740
7120
180
2
2
120
180
2
7000
6880
7060
6740
7120
2
2
13940
2
2
2
2
2
2
A2 A3
A3
A2
A3 A2
2
2
2
128 1538
128
16
32 8
2
2
A 99 A2 A 3 A2
8 1536 2 64
2
A plot of the ID is shown below. Note that the ID is a function of A.
0
10
-2
10
-4
10
-6
10
0
1
2
3
A
4
5
2
21000
14060
A2
A3
A3
A2
A2
A3
A3
A2
A3
2
2
2
2
2
2
2 2
2
128
1538
128
16
16
128
32
8
32
ID
2 2
2 2
A 99 A
A 3A
2
2
8 1536
2 64
2
A3
24
14000
ID
(c)
2
A 99 A3 A 99 A3
2 2
8 1536 8 1536 A 3 A
2 2
17 A
2 64
2
2
2
A3 A2 2 64 A2 2 A3 2
A2
A2
2
2
2
2
2
A2 A3 2 8 A3 2
A3 8 A3 A 128 128
128 128
2
2
16
3
3
A
A
16 32
32
32
32
2
3 2
A3
1538
1538
A
128
128
2
f
2.30
x(t )
ce
k
ck
1
T0
T0
2
0
k
j
2 k
t
T0
A
k
sin
2 k
2
k
j
t
A
2 j 2
T0
0
Ae
dt
e
k
2
A
k
2
j
k
k
Xf
k
k
k
ck f
T0
5
T0
Xf
A
A
A
3
A
5
A
7
7
T0
k , odd
k
0
A
2
k , even
c f T
Xf
k 0
3
T0
A
3
1
T0
0
1
T0
3
T0
A
5
5
T0
A
7
7
T0
f
2.31
x(t )
ce
k
1
ck
T0
Xf
k
T0
2
Ae
2 k
t
T0
j
2 k
t
T0
0
T0
1
dt
T0
T0
2
k
0
sin
2 k
k
t
j
j
k
2
T0
2
0
Ae
dt A
1 e
k
2A
k
2
j
k
k 0
k , even
k , odd
k
0
c f T
k
Xf
j
k
k
k
ck f
T0
Xf
7
T0
5
T0
2A
2A
3
2A
5
2A
7
2A
3
T0
2A
3
1
T0
0
1
T0
3
T0
2A
5
5
T0
2A
7
7
T0
f
2.32
x(t )
ce
k
ck
j
k
T1
1
T0
Ae
2 k
t
T0
2 k
t
j
T0
0
Xf
j k T1
e T0
k
0
c f T
k
Xf
k T1
sin
T
2 T0
dt A 1
k T1
T0
2 T0
k
k
k
ck f
T0
A
T1
T0
Xf
T
A 1
T0
fT1
sin
2
fT1
2
8
T0
7
T0
6
T0
5
T0
4
T0
3
T0
2
T0
1
T0
0
1
T0
2
T0
3
T0
4
T0
5
T0
6
T0
7
T0
8
T0
f
2.33
x(t )
ce
k
ck
1
T0
Xf
T0
2
0
k
2 k
t
T0
A
2
2 k
2 k
T
2 A j T0 t
1 0
2 A j T0 t
A
k
dt 2 2 1 1 0
te
dt 2 A
t e
T0
T0 T0
T0
k
2A
2
2 2
k
k
k
2A
49 2
7
T0
k , even
k , odd
k
0
k
ck f
T0
A
2
k 0
c f T
k
Xf
j
2A
25 2
5
T0
Xf
2A
2A
2
2
2A
9 2
3
T0
2A
9 2
1
T0
0
1
T0
3
T0
2A
25 2
5
T0
2A
49 2
7
T0
f
2.34
x(t )
ce
k
k
1
ck
T0
T0
4
T0
4
j
2 k
t
T0
4A
te
T0
j
2 k
t
T0
1
dt
T0
3T0
4
T0
4
4A j
2
A
t e
T0
2 k
t
T0
dt j
4A
k
sin
2 2
k
2
0
k 0
0
k , even
k
4A
j 2 2 sin
k , odd
k
2
Xf
k
0
c f T
k
Xf
k
k
k
ck f
T0
Xf
4A
49 2
7
T0
4A
25 2
5
T0
4A
4A
2
2
4A
9 2
3
T0
4A
9 2
1
T0
0
1
T0
3
T0
4A
25 2
5
T0
4A
49 2
7
T0
f
2.35
x(t )
ce
k
k
1
ck
T0
Xf
2 k
t
T0
T0
A
0 T0 te
j
2 k
t
T0
A
2
dt
j A
2 k
k 0
k 0
k
0
c f T
k
Xf
j
k
k
k
ck f
T0
A
2
A
2
A
14
7
T0
A
12
6
T0
5
T0
4
T0
3
T0
A
2
A
4
A
6
A
8
A
10
Xf
2
T0
A
4
1
T0
0
1
T0
2
T0
A
6
3
T0
A
8
4
T0
A
10
5
T0
A
12
6
T0
A
14
7
T0
f
2.36
x(t )
ce
k
k
ck
1
T0
j
2 k
t
T0
T0
2
t e
0
A sin T
0
A
j A
4
0
2A
k2 1
Xf
k 1
k , odd, k 1
k , even
k
0
c f T
k
Xf
A
k
k 2 1 1 1 k 1
dt
A
j
k 1
4
k 0
2 k
j
t
T0
k
k
k
ck f
T0
2A
A
4
Xf
A
4
2A
3
2A
35
6
T0
2A
3
2A
15
4
T0
2A
15
2
T0
1
T0
0
1
T0
2
T0
4
T0
2A
35
6
T0
f
2.37
x(t )
ce
k
k
2
ck
T0
Xf
4 k
t
T0
T0
2
0 A sin T0
j
t e
4 k
t
T0
dt
2A
4k 2 1
2k
0
cf T
k
Xf
j
k
k
2k
ck f
T0
2A
Xf
2A
3
2A
35
6
T0
2A
3
2A
15
4
T0
2A
15
2
T0
0
2
T0
4
T0
2A
35
6
T0
f
2.38
p t
t nT is periodic with period T. Hence it can be represented by a Fourier series
n
p t
ce
k
1
ck
T
T
2
k
j
2 k
t
T
where the Fourier series coefficients are
t nT e
j
T n
2
1 j
Thus, p t e
T k
2 k
t
T
1
dt
T
2
T
k
t e
T
2
j
2 k
t
T
dt
1
T
2 k
t
T
Now using the transform pair e
P j
T
2
2 k
.
T
j
2 k
t
T
2 k
2
we have
T
2.39
The key to solving this problem is to understand what H f does to cos 2 f1t . The Fourier
1
1
f f1 f f1 . When this is the input, the Fourier
2
2
j
j
transform of the output of H f is f f1 f f1 whose inverse Fourier
2
2
j
j
transform is e j 2 f1t e j 2 f1t sin 2 f1t . Armed with this relationship, we may now
2
2
compute y t :
transform of cos 2 f1t is
y t cos 2 f 0t cos 2 f1t cos 2 f 2t sin 2 f 0t sin 2 f1t sin 2 f 2t
1
1
1
1
cos 2 f 0 f1 t cos 2 f 0 f1 t cos 2 f 0 f 2 t cos 2 f 0 f 2 t
2
2
2
2
1
1
1
1
cos 2 f 0 f1 t cos 2 f 0 f1 t cos 2 f 0 f 2 t cos 2 f 0 f 2 t
2
2
2
2
cos 2 f 0 f1 t cos 2 f 0 f 2 t
Yf
1
2
1
2
1
2
1
2
1162
1161
−1161
−1162
f (kHz)
2.40
The Fourier transform of the upper input to the adder is
1165
1160
1155
−1155
−1160
−1165
f (kHz)
The Fourier transform of the output of H f is
j
0
−5
5
f (kHz)
j
Now, if z t Z f are a Fourier transform pair, then
1
1
Z f f0
Z f f0
j2
j2
Applying this relationship to the output of H f produces the Fourier transform of the lower
z t sin 2 f 0t
input to the adder:
1165
1160
1155
−1155
−1160
−1165
f (kHz)
The result is
Yf
1165
1160
−1160
−1165
f (kHz)
2.41
The key to solving this problem is to understand what H f does to cos 2 f1t . The Fourier
1
1
f f1 f f1 . When this is the input, the Fourier
2
2
j
j
transform of the output of H f is f f1 f f1 whose inverse Fourier
2
2
j
j
transform is e j 2 f1t e j 2 f1t sin 2 f1t . Armed with this relationship, we may now
2
2
compute y t :
transform of cos 2 f1t is
y t cos 2 f 0 t cos 2 f1t cos 2 f 2 t sin 2 f 0 t sin 2 f1t sin 2 f 2 t
1
1
1
1
cos 2 f 0 f1 t cos 2 f 0 f1 t cos 2 f 0 f 2 t cos 2 f 0 f 2 t
2
2
2
2
1
1
1
1
cos 2 f 0 f1 t cos 2 f 0 f1 t cos 2 f 0 f 2 t cos 2 f 0 f 2 t
2
2
2
2
cos 2 f 0 f1 t cos 2 f 0 f 2 t
Yf
1
2
1
2
1
2
1
2
1159
1158
−1158
−1159
f (kHz)
2.42
The Fourier transform of the upper input to the adder is
1165
1160
1155
−1155
−1160
−1165
f (kHz)
The Fourier transform of the output of H f is
j
0
−5
5
f (kHz)
j
Now, if z t Z f are a Fourier transform pair, then
1
1
Z f f0
Z f f0
j2
j2
Applying this relationship to the output of H f produces the Fourier transform of the lower
z t sin 2 f 0t
input to the adder:
1165
1160
1155
−1155
−1160
−1165
f (kHz)
The result is
Yf
1160
1155
−1155
−1160
f (kHz)
2.43
Xf
(a)
3
3
3
2
3
2
−1000
1000
2000
3000
1000
2000
3000
f
−3000
−2000
Hf
f
−3000
−2000
−1000
Yf
1
1
1
1
−2000
−1000
1000
2000
f
−3000
3000
Y f 2 cos 2 1000t 2 cos 2 2000t
Xf
(b)
2
f
−3000
−2000
−1000
1000
2000
3000
1000
2000
3000
Hf
f
−3000
−2000
−1000
Yf
2
2
3
2
3
f
−3000
−2000
−1000
1000
2000
3000
2.44
Xf
(a)
1
1
1
2
1
2
−440
440
f
−3000
−2000
2000
3000
Hf
2
1
f
−3000
−2000
−1000
1000
2000
3000
Yf
1
1
1
1
−2000
−440
440
2000
f
−3000
3000
y t 2 cos 2 440t 2 cos 2 2000t
Xf
(b)
1
4
1
2
1
2
1
4
f
−3000
−880 −440
440 880
3000
Hf
2
1
f
−3000
−2000
−1000
1000
2000
3000
Yf
1
2
1
1
1
2
f
−3000
−880 −440
440
880
y t 2 cos 2 440t cos 2 880t
3000
Xf
(c)
1
1
1
2
1
2
−600
600
f
−2600
2600
Hf
2
1
f
−3000
−2000
−1000
1000
2000
3000
Yf
1
1
1
1
−2600
−600
600
2600
f
y t 2 cos 2 600t 2 cos 2 2600t
(d)
Xf
A
2
A
8
A
8
−60
60
A
2
f
−7000
7000
Hf
2
1
f
−3000
−2000
−1000
1000
2000
3000
Yf
A
4
A
4
−60
60
f
−3000
y t
A
cos 2 60t
2
3000
Xf
(e)
3
4
3
4
3
4
3
4
−3600
−1800
1800
3600
f
Hf
2
1
f
−3000
−2000
−1000
1000
2000
3000
Yf
3
4
3
4
−1800
1800
f
y t
3
cos 2 1800t
2
2.45
H f
xt
y t
cos2 70t
cos2 70t
1
4
1
4
1
2
f
−141 −140 −139
−1
0
1
139 140 141
Hf
4
f
−1
0
1
2
f
−1
0
1
Yf
1
1
f
−71 −70 −69
69
70
71
2.46
H f
xt
G f
y t
cos2 3930t
cos2 5930t
1
2
1
2
1
2
1
2
f
−11930
−120
−70
−20
20
70
120
11930
Hf
2
2
f
−100 −70 −40
40
70 100
1
1
f
−100 −70 −40
1
2
1
2
40
70 100
1
2
1
2
f
−4030 −4000 −3970
−3860
2
3860
G f
3970 4000 4030
2
f
−4050 −4000 −3950
1
3950
Yf
4000
4050
1
f
−4030 −4000 −3970
3970 4000 4030
2.47
H f
xt
G f
y t
cos2 3930t
cos2 5930t
1
2
1
2
1
2
1
2
f
−11930
−120
−70
−20
20
70
120
11930
Hf
2
2
f
−120
−70
−20
20
70
1
120
1
f
−120
1
2
1
2
−70
−20
20
70
120
1
2
1
2
f
−4050 −4000 −3950
2
−3860
3860
G f
3950
4000 4050
2
f
−4050 −4000 −3950
1
3950
Yf
4000
4050
1
f
−4050 −4000 −3950
3950
4000 4050
2.48
Yf
(a)
5
4
71
70
69
69
0
70
71
f (MHz)
71
f (MHz)
Yf
(b)
5
4
71
(c)
70
69
69
0
70
r t cos 2 f 0 t I t cos 2 f1t Q t sin 2 f1t cos 2 f 0 t
I t
2
y t
cos 2 f1 f 0 t
Q t
I t
2
I t
2
sin 2 f1 f 0 t
cos 2 f1 f 0 t
2
In part (a), f 0 930 so that
y t
In part (b), f 0 1070 so that
y t
I t
2
I t
cos 2 f1 f 0 t
Q t
Q t
2
cos 2 70t
cos 2 70t
2
sin 2 f1 f 0 t
sin 2 f1 f 0 t
Q t
2
Q t
sin 2 70t
sin 2 70t
2
2
The difference is the sign of the “quadrature term” (the term involving sin 2 ft ).
2.49
(a)
Xf
9
0.1
(b)
0.1
f
Yf
1
0.1
0.1
f
2.50
(a)
Fourier transform of input to H(f)
H(f)
f
Fourier transform of input to G(f)
G(f)
f
sum of “house” and “half-ellipse”
Hf
f2
455
f1
G f
f1
455
f2
0 f1 450
460 f 2 1850
f
5
5
f
(b)
Fourier transform of input to H(f)
f
Note the overlapping spectra here.
This is bad because there is not
way to use a filter to isolate the
desired spectrum from the
combination.
It is not possible to design an H(f) and G(f) to produce the desired output. The signal centered at 250
Hz must be removed before the mixer preceding the filter H(f). The typical solution is to add an
additional filter as follows:
This filter needs to remove the spectral content that interferes with the
desired signal. In tunable systems (i.e., systems that need to isolate
different channels), this BPF may need to be tunable.
r t
image
rejection
BPF
H f
G f
cos2f 0t
cos2f1t
f 0 705 kHz
f1 455 kHz
xt
2.51
(a)
Fourier transform of input to H(f)
H(f)
f
Fourier transform of input to G(f)
G(f)
f
sum of “house” and “half-ellipse”
Hf
f2
455
f1
G f
f1
455
f2
0 f1 450
460 f 2 1850
f
5
5
f
(b)
Fourier transform of input to H(f)
f
Note the overlapping spectra here.
This is bad because there is not
way to use a filter to isolate the
desired spectrum from the
combination.
It is not possible to design an H(f) and G(f) to produce the desired output. The signal centered at 250
Hz must be removed before the mixer preceding the filter H(f). The typical solution is to add an
additional filter as follows:
This filter needs to remove the spectral content that interferes with the
desired signal. In tunable systems (i.e., systems that need to isolate
different channels), this BPF may need to be tunable.
r t
image
rejection
BPF
Hf
G f
cos2f 0t
cos2f1t
f 0 705 kHz
f1 455 kHz
xt
2.52
(a)
Xf
0
15 19 23
38
(b)
l t r t
LPF 1
x t
BPF 1
2r t
l t r t
cos 4 f 0t
×2
LPF 1
LPF 2
1
−15
2l t
LPF 2
BPF 2
f
53
15
2
f
−15
f
15
BPF 1
1
f
−53
−38
23
−23
38
53
BPF 2
1
f
−21
−17
17
(c) l(t) + r(t) is available at the output of LPF 1.
21
2.53
(a)
Using x t e j 2 f0t X f f 0 , we have
1
1
x t cos 2 f 0 t x t e j 2 f0t e j 2 f0t
2
2
1
1
x t e j 2 f 0 t x t e j 2 f 0 t
2
2
1
1
X f f0 X f f0
2
2
(b)
Ac
2
Ac
2
f
fc B
(c)
(d)
fc fc B
fc B
fc
fc B
fc B
The minimum theoretical approach would be to space the carriers 2B Hz apart. But, this
would require ideal low-pass filters. Real filters require a transition band thus
necessitating a larger spacing. In commercial broadcast AM, for example, the channel
assignments are every 2B = 10 kHz, but broadcast stations are not assigned to adjacent
channels in any given geographical area. Instead, the closest spacing is every other
channel slot.
2.54
Mf
(a)
Am
2
Am
2
fm
(b)
f
fm
s t Am cos 2 f m t Ac cos 2 f c t
Am Ac
A A
cos 2 f c f m t m c cos 2 f c f m t
2
2
Am Ac
f f c f m f f c f m f f c f m f f c f m
S f
4
S f
(c)
Ac Am
4
Ac Am
4
Ac Am
4
Ac Am
4
f
fc f m fc fc fm
fc f m
fc
fc fm
2.55
(a)
i.
R f
Ac
2
Ac
2
f
fc B
fc fc B
fc B
fc
fc B
Xf
Ac2
2
Ac2
4
2 f c B
ii.
2 f c
2 f c B
B
Ac2
4
B
2 fc B
2 fc
f
2 fc B
The filter must satisfy the conditions illustrated below.
Hf
2
Ac2
B
(b)
i.
ii.
B
A2
A2
x t
m t
cos 4 f c t
2
2
The filter removes the double frequency term and scales the baseband term by
Thus the filter output is
y t m t
(c)
i.
f
Ac2
Ac2
x t
m t cos
m t cos 4 f c t
2
2
y t m t cos
2
.
A2
ii.
The phase offset scales the output by the constant cos . This does not cause
any distortion, especially when is small. As
, the result is disastrous.
2
There are only two solutions: either must be known (or estimated from the
received signal) or the modulation format should be altered to allow non-coherent
detection.
2.56
(a)
Ac m t cos 2 f c t Ac Am cos 2 f m t cos 2 f c t
envelope Ac Am cos 2 f m t
(b)
envelope output
t
(c)
The envelope detector does not preserve sign information. The solution is to reformat the
signal so that the envelope is never negative.
2.57
(a)
Ac
2
AAc
2
S f
AAc
2
Ac
2
f
fc B
(b)
fc fc B
fc B
fc
fc B
Write the modulated signal as
s t A m t Ac cos 2 f c t AAc cos 2 f c t Ac m t cos 2 f c t
The first term is a replica of the unmodulated carrier. Because this replica is part of the
transmitted signal, the term “transmitted carrier” is used.
(c)
The envelope detector output is Ac A m t . Assuming A m t 0 for all t, the output
may be written as Ac A Ac m t . The term Ac A needs to be subtracted from the envelope
detector output to produce an appropriately scaled version of the desired signal. (In other
words, the envelope detector needs to be AC-coupled.)
The condition that guarantees correct output is A min m t 0 .
2.58
(a)
A Am
(b)
A f Ac A f
(c)
S f
Ac Am
AA
f fm c m f fm
2
2
1
1
A f fc A f fc
2
2
AA
AA
AA
c f fc c m f fc f m c m f fc f m
2
2
2
Ac A
AA
AA
f fc c m f fc f m c m f fc f m
2
2
2
(d)
S f
Ac A
2
Ac Am
4
Ac Am
4
Ac A
2
Ac Am
4
Ac Am
4
f
fc f m fc fc fm
fc f m
fc
fc fm
2.59
s t Ac Am cos 2 f m t cos 2 f c t
(a)
Ac Am
AA
cos 2 f c f m t c m cos 2 f c f m t
2
2
2 2
2 2
A A
A A
s 2 t c m c m cos 4 f c f m t
8
8
2 2
A A
A 2 A2
c m cos 4 f c t c m cos 4 f m t
4
4
2 2
2 2
A A
A A
c m c m cos 4 f c f m t
8
8
1
Pm
Tm
Tm
Ac2 Am2
0 s t dt 4
2
Ptot Pm
power ratio
(b)
Pm
1
Ptot
s t Ac A cos 2 f c t Ac Am cos 2 f m t cos 2 f c t
Ac A cos 2 f c t
s2 t
Ac Am
AA
cos 2 f c f m t c m cos 2 f c f m t
2
2
Ac2 A2 Ac2 A2
cos 4 f c t
2
2
A 2 A2 A 2 A2
A 2 A2 A 2 A2
c m c m cos 4 f c f m t c m c m cos 4 f c f m t
8
8
8
8
2
2
A AA
A AA
c m cos 2 2 f c f m t c m cos 4 f m t
2
2
2
2
A AA
A AA
c m cos 2 2 f c f m t c m cos 4 f m t
2
2
2 2
2 2
A A
A A
c m cos 4 f c t c m cos 4 f m t
4
4
Pm
Ac Am
4
1
Ptot
Tm
Tm
(from part (a))
2
s t dt
0
Ac2 A2 Ac2 Am2
2
4
Ac2 Am2
P
1
power ratio m 2 2 4 2 2
2
Ptot
Ac A
A A
c m 2 A 1
4
2
Am
A Am power ratio
1
3
2.60
(a)
1
s(t)
0.5
0
-0.5
-1
0
0.5
1
1.5
2
t
2.5
3
3.5
4
0
0.5
1
1.5
2
t
2.5
3
3.5
4
(b)
1
s(t)
0.5
0
-0.5
-1
2.61
(a)
t
(t ) 2f Am cos 2 f m d
0
Am f
, then (t ) sin 2 f m t and s (t ) Ac cos 2 f c t sin 2 f m t .
fm
Let
(b)
Am f
sin 2 f m t
fm
Using the identity e jX cos( X ) j sin( X ) , we see that cos( X ) Re e jX . Thus,
Ac cos 2 f c t (t ) Ac Re e
j 2 f c t ( t )
A Ree
c
j ( t )
e j 2 fct . Substitute
(t ) sin 2 f m t to obtain the desired result.
(c)
1
ck
T0
T0
s(t )e
j 2 kf m t
0
1
dt
T0
T0
e
j sin 2 f m t j 2 kf m t
e
dt
0
Using the substitution x 2 f m t , the integral becomes
ck
1
2
2
e
j sin x kx
dx J k
0
Thus s (t ) Ac
J cos 2 f
k
k
c
kf m t
(d)
Ac
J k f fc kf m f fc kf m
2 k
The bandwidth is infinite.
(e)
The power at the carrier frequency is determined by the k 0 in the answer to part (d).
The k 0 term is
Ac
J 0 f f c f f c
2
The condition of zero power at the carrier frequency occurs when J 0 0 . Thus the
S f
curious and interesting situation occurs at the values of corresponding to the zeros of
J 0 · . The first five zeros are 2.4048, 5.5201, 8.6537,11.7915,14.9309 .
2.62
(a)
S f
S f
2
Ac
2
Ac2
4
P
J f f
k
k
J f f
k
2
k
S f df
2
2
c
A
4
Ac2
4
c
kf m f f c kf m
kf m f f c kf m
c
k
J k2 f f c kf m f f c kf m df
2
c
A
J 2 2 J
k
2
k
2
k
k
(b) The table of the Bessel functions is
| k=0
k=1
k=2
k=3
k=4
k=5
k=6
k=7
k=8
------+-------------------------------------------------------------------------------B = 1 | 0.58553 0.19364 0.01320 0.00038 0.00001 0.00000 0.00000 0.00000 0.00000
------+-------------------------------------------------------------------------------B = 2 | 0.05013 0.33261 0.12449 0.01663 0.00116 0.00005 0.00000 0.00000 0.00000
------+-------------------------------------------------------------------------------B = 5 | 0.03154 0.10731 0.00217 0.13310 0.15306 0.06819 0.01717 0.00285 0.00034
------+--------------------------------------------------------------------------------
The table of the power ratio R for different values of K is
K
R J 02 2 J k2
k 1
| K=0
K=1
K=2
K=3
K=4
K=5
K=6
K=7
K=8
------+-------------------------------------------------------------------------------B = 1 | 0.58553 0.97282 0.99922 0.99999 1.00000 1.00000 1.00000 1.00000 1.00000
------+-------------------------------------------------------------------------------B = 2 | 0.05013 0.71535 0.96433 0.99759 0.99990 1.00000 1.00000 1.00000 1.00000
------+-------------------------------------------------------------------------------B = 5 | 0.03154 0.24616 0.25049 0.51670 0.82282 0.95921 0.99356 0.99926 0.99993
------+--------------------------------------------------------------------------------
The bold numbers indicate the smallest value of K which captures 98% of the power. The values
are
K 98 2 for 1
K 98 3
for
K 98 6
for
2
5
from which we deduce that K 98 1 . Thus we have
BW 2 K 98 f m 2 1 f m
2.63
(a)
d
Ac cos 2 f c t (t ) Ac 2 f c (t ) sin 2 f c t (t )
dt
Ac 2 f c 2fm(t ) sin 2 f c t (t )
(b)
The envelope detector output is Ac 2 f c 2fm(t ) . To produce the desired output, we
need f c fm(t ) for all t.
(c)
Ac 2 f c fAm
envelope
detector
output
Ac 2 f c
Ac 2 f c fAm
t
Ac 2fAm
y t
t
Ac 2fAm
2.64
(a)
(b)
(c)
H FM ( s)
sk p F ( s)
s k0 k p F ( s )
V ( s) H FM ( s)( s) s( s ) v(t )
sk p F ( s)
s F ( s)
s k0 k p F ( s )
The filter is a differentiator.
k0 1 should be avoided.
d
(t ) which is the desired result.
dt
s
k p 1 k0
2.65
(a)
1 1
1
h( n) u ( n)
2 2
2
(b)
1 1
1
h( n) u ( n)
3 2
2
(c)
1
h( n)
2
(d)
h(n) (n 1) 2 (n 2) 3 (n 3) 4 (n 4)
n
n
n 1
1
u (n 1) (n) 1
2
n
n 1
u (n 1)
n2
u (n 2)
n 1
u (n 1)
2.66
(a)
(b)
1
h( n) 2
2
H z
n 1
n
1
u (n 1) u (n 1)
2
3 j 4 z 1
1
1
1 j z 1
4
3
1
1
h( n) 3 j 4 j
4
3
1 5 j
1
j e
4 12
3
3 j 4 z 1
1
1
1 j z 1
4
3
n 1
1
1
u (n 1) 3 j 4 j
4
3
3 j 4 5e
1 5 j
1
j e
4 12
3
j
2
3 j 4 5e
j
2
n 1
u (n 1)
j 5e j
j 5e j
3
tan 1
4
5
h(n) 10
12
(c)
n 1
sin n u (n 1)
3 1
3 1
1 j z
1 j z
4
4
H z
1
1
1
1
1 j z 1 1 j z 1
3
3
4
4
3 1
1
h( n) 1 j j
4 4
3
n 1
3 1
1
u (n 1) 1 j j
4 4
3
1 5 j
1
j e
3 12
4
3 5 j
5
1 j e 2 j e j
4 4
4
1 5 j
1
j e
3 12
4
3 5 j
5
1 j e 2 j e j
4 4
4
4
tan 1
3
5 5
h( n)
2 12
n 1
sin n u (n 1)
n 1
u (n 1)
(d)
H z
2 z 2
2
2 z 1
1
1 z 1
4
1 1
1 z
2
n 1
n 1
1
1
h n 4 n 1 2 u n 1
2
4
2.67
(a)
Y ( z ) z 1Y ( z ) z 2Y ( z ) z 1
(b)
z 1
Y ( z)
1 z 1 z 2
(c)
Y ( z)
z
z
z z 1
1 5
1 5
z
z
2
2
2
5 5
5 5
10
10
1 5
1 5
z
z
2
2
5 5 1
5 5 1
z
z
10
10
1 5 1
1 5 1
z
z
1
1
2
2
5 5 1 5 n 1 5 5 1 5 n 1
y ( n)
u (n 1)
10 2
10 2
n
n
1 1 5 1 5
u (n 1)
5 2 2
2.68
(a)
1
−3 −2 −1
(b)
5
X z z
0
n
n 0
(c)
1
2
3
4
5
6
7
3
4
5
6
7
1 z 6
1 z 1
1
−3 −2 −1
0
1
2
−1
(d)
G z 1 z 6
(e)
G z X z z 1 X z 1 z 1 X z 1 z 1
This answer is the same as that from part (d).
(f)
1
1 z 6
G
z
1 z 1
1 z 1
This answer is the same as that from part (b).
X z
1 z 6
1 z 6
1
1 z
2.69
(a)
H z
1
1
1 z 1
3
(b)
Imaginary Part
1
0.5
0
-0.5
-1
-1
(c)
-0.5
0
0.5
Real Part
1
n
1
h n u n
3
y n h n x n
h n n n 1 n 2
h n h n 1 h n 2
0
1
4
3
13 1 n 2
9 3
n0
n0
n 1
n2
2.70
(a)
H z
1
1
1 z 1
4
(b)
Imaginary Part
1
0.5
0
-0.5
-1
-1
(c)
-0.5
0
0.5
Real Part
1
n
1
h n u n
4
y n h n x n
h n n n 1 n 2
h n h n 1 h n 2
0
1
3
4
13 1 n 2
16 4
n0
n0
n 1
n2
2.71
(a)
IIR
(b)
FIR
(c)
H z H1 z H 2 z
(e)
(f)
1
1 1 1
z z 2
6
7 3
z
3
1
z 3
6
IIR
y ( n)
11
1
7
y (n 1) y (n 2) y (n 3) 6 x(n 1) 8 x(n 2) x(n 3)
6
6
3
2
2 1 2
2 1
z 1 1
z
1
z
3 6 3 6
H z
1 1
1 z 1 1 12 z 1
1 z
3
1
0.5
Imaginary Part
(d)
6 z 1 8 z 2
0
-0.5
-1
-1
-0.5
0
Real Part
0.5
1
(g)
H z
(h)
y ( n) h( n) x ( n)
z 1
1 z 1
2 z 1
3 z 1
1
1
1 z 1 1 z 1
2
3
n 1
n 1
1
1
h n 1 2 3 u (n 1)
2
3
h(n) (n) (n 1) (n 2)
h(n) h(n 1) h(n 2)
0
6
9
n4
n4
3 7 1 13 1
4 2
9 3
n0
n 1
n2
n3
12
10
y(n)
8
6
4
2
0
0
5
10
n
15
20
2.72
(a)
H z 1 z 8
1
H z has 8 zeros equally spaced on a circle of radius 8 and 8 poles at the origin as
shown below.
Imaginary Part
1
0.5
8
0
-0.5
-1
-1
-0.5
0
0.5
Real Part
1
The ROC is the entire z-plane except for z = 0.
(b)
G z
1
H z
1
1 z 8
1
8
G z has 8 poles equally spaced on a circle of radius and 8 zeros at the origin as
1
8
shown below. The ROC for a causal stable system is z .
Imaginary Part
1
0.5
8
0
-0.5
-1
-1
(c)
n
g n
0
-0.5
0
0.5
Real Part
n a multiple of 8
otherwise
1
2.73
(a)
(b)
y n a1 y n 1 a2 y n 2 b0 x n b1 x n 1 b2 x n 2
The poles are z
i.
a2
a1
1 1 4 2
2
a1
a22
a
a
. The poles are z 1 1 1 4 22
4
2
a1
ab a
a1b0
a
a
2 2 b0 1 0 2 b0 1 4 22
a1
a1
2
a1 1
2
z
a2
1 4 2
a1
H z b0
a
a
1 1 1 1 4 22 z 1
2
a1
For real, distinct poles we require a2
h n b0 n
ii.
.
ab a
a1b0
a
a
2 2 b0 1 0 2 b0 1 4 22
a1
a1
2
a1 1
2
z
a2
1 4 2
a1
a
a
1 1 1 1 4 22 z 1
2
a1
a1b0
ab a
a
a
2 2 b0 1 0 2 b0 1 4 22
a1
a1
2
a1
2
a
1 4 22
a1
a1 1 1 4 a2
2
a12
n 1
a1b0
ab a
a
a
2 2 b0 1 0 2 b0 1 4 22
2
a1
a1
a1
2
a
1 4 22
a1
a1 1 1 4 a2
2
a12
n 1
For real, repeated poles, we require a12 4a2 . The poles are z
a1 a1
,
2 2
u n 1
u n 1
1
b0 a1 z 1 b0 a12 z 2
4
b0 2b0
H z b0
2
a1 1
1 2 z
n
a1 1
a1 1
z
z
a1 1
2
2
b0 z
2
2
2
a1 1
a1 1
1
1
z
z
2
2
a
a
a
h n b0 n 2b0 n 1 u n b0 1 n 1 1
2
2
2
iii.
n 1
u n 1
For complex conjugate poles, we require a12 4a2 0 . Let 2 a12 4a2 , then the
a
poles are z 1 j .
2
2
a1b0
a12 b0 a2 b0 1 a1b0
a12 b0 a2 b0 1
j
j
z
z
2
2
2
2
H z b0
1
a1
a
1 j z
1 1 j z 1
2
2
2
2
n 1
a1b0
a12 b0 a2 b0 a1
j
j u n 1
h n b0 n
2
2
2
2
n 1
a1b0
a12 b0 a2 b0 a1
j
j u n 1
2
2
2
2
(c)
i.
The poles are z
a1
1
j
a12 4a2 re j from which we obtain
2
2
r a2
tan 1 1 4
ii.
a2
a12
B z 1 2r cos z 1 r 2 z 2
r
iii.
r cos
1
times the coefficient of z 1 .
2
The distance from the origin is the square root of the coefficient of z 2 .
The location of the pole intersects the real axis at
iv.
H z b0 b0
jre j 2 1
z
2sin
jre j 2 1
z
2sin
b0
1 re j z 1
1 re j z 1
rn
h n b0 n b0
sin n 1 u n 1
sin
(d)
The case of real, repeated poles corresponds to 0 . In this case, r a2
Also note that lim
sin n 1
0
h n b0 n b0
sin
n 1 . Now, the answer to part (c)-iv becomes
rn
sin n 1 u n 1
sin
b0 n b0 n 1 r n u n 1
b0 n b0 n 1 2 r n u n 1
b0 n 2b0 r n u n b0 r n 1 r n 1u n 1
Using r
a1
gives the answer from part (c)-iv.
2
a12 a1
.
4
2
2.74
(a) F ( z ) 0.5
0.5 z 1
0.5 z 1
1
z 1
z 1
Y ( z)
X ( z)
1 0.5 z 1
1 0.5 z 1 1 z 1 1 z 1 1 0.5 z 1
y (n) 1 (0.5) n 1 u (n 1)
2
u(n)
y(n)
1
0
-1
0
2
4
6
8
10
n
(b) F ( z ) 1
Y ( z ) z 1 X ( z )
z 1
1 z 1
y (n) u (n 1)
2
u(n)
y(n)
1
0
-1
0
2
4
6
n
8
10
(c) F ( z ) 1.5
1.5 z 1
1.5 z 1
1
z 1
0.5 z 1
Y ( z)
X ( z)
1 0.5 z 1
1 0.5 z 1 1 z 1 1 z 1 1 0.5 z 1
y (n) 1 0.5(0.5) n 1 u (n 1)
2
u(n)
y(n)
1
0
-1
0
2
4
6
8
10
n
(d) Unlike the case with continuous-time systems, a first-order system can show an oscillatory
transient reponse.
2.75
(a)
1
1
1 z 1
1
2
V z 2
1
1
1 z
1
1 z 1
1 1 z 1
2
2
n
11
v n u n
22
2
u(n)
y(n)
1.5
1
0.5
0
-0.5
-1
0
2
4
6
8
10
n
(b)
V z
1 1 z 1
1 1 1 z
1
1
1
1 z 1
v n n
2
u(n)
y(n)
1.5
1
0.5
0
-0.5
-1
0
2
4
6
n
8
10
(c)
3
3
1 z 1
1
2
V z 2
1
1
1 z
3
1 z 1
1 1 z 1
2
2
v n
n
3 1
u n
2 2
2
u(n)
y(n)
1.5
1
0.5
0
-0.5
-1
0
2
4
6
8
n
(d)
Even though this is a first-order system, it can still exhibit oscillations.
10
2.76
(a)
1 1
z
4
H z
1
1
1 z 1 z 2
4
8
1
(b)
1
Imaginary Part
0.5
0
-0.5
-1
-1
(c)
-0.5
0
Real Part
0.5
1
7
9 7 1 7
9 7 1
6 1
j
j
z
z
z
308
308
308
308
11
Y z 1
1 z 1
1
1
7 1
7 1
1 j
1 j
z
z
8
8
8
8
n 1
7
9 7 1
7
j
j
308
8
8
308
n 1
7
1
7
9 7
j
y n (n)
j
u n 1
308
8
308
8
6
11
2.77
The closed loop transfer function is
1 0.25 z 1
Y ( z)
X ( z ) 1 0.75 K z 1 0.25 Kz 2
z z 0.25
z 0.75 K z 0.125K
2
This system has two poles at
p1 , p2
0.75 K
0.75 K
2
K
2
The goal is to find the values of K for which both poles are inside the unit circle. That is, find the
values of K for which | p1 | 1 and | p2 | 1 . The following Matlab script plots the poles for
0 K 3:
%% ex2_77
K = 0:0.001:3;
p1 = 0.5*(K-0.75 + sqrt((0.75-K).^2-K));
p2 = 0.5*(K-0.75 - sqrt((0.75-K).^2-K));
plot(real(p1),imag(p1),'b-',real(p2),imag(p2),'r.-',...
real(p1(1)),imag(p1(1)),'bo',...
real(p1(end)),imag(p1(end)),'bs',...
real(p2(1)),imag(p2(1)),'ro',...
real(p2(end)),imag(p2(end)),'rs','LineWidth',2);
legend('p_1','p_2');
grid on;
%% plot unit circle for reference
hold on; plot(exp(j*2*pi*[0:0.01:1]),'k:'); hold off;
This script produces the following plot
1
p1
p2
0.5
0
-0.5
-1
-1
-0.5
0
0.5
1
1.5
2
From this plot, we see that p2 is always inside the unit circle and that p1 is outside the unit circle
when K is too large. Using the data vector p1 produced by the script, we see that p1 is outside the
unit circle for K 2.334 .
Repeating the same script for K 0 produces the following plot
1
p1
0.5
p2
0
-0.5
-1
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
Form this plot, we see that p1 is always inside the unit circle and that p2 is outside the unit
circle when K gets too negative. Using the data vector p2 produced by the script we see that p2
is outside the unit circle for K 0.2
Putting this all together, the system is stable for 0.2 K 2.334
2.78
(a)
H z
(b)
1 1
z
2
1
1 1 1 2
z z
2
4
There is one zero at z 0 and two poles at z
1
3
j
4
4
Imaginary Part
1
0.5
0
-0.5
-1
-1
(c)
-0.5
0
0.5
Real Part
1
2 1
1 j 3 1
1 j 3 1
z
z
z
3
12
12
Y z
1
1
1
3 1
3 1 1 z
1 j
1 j
z
z
4
4
4
4
n 1
n 1
1 j 3 1
1 j 3 1
3
3
2
y n
j
j
u n 1
4
4
3
12
4
12
4
n 1
1 1
2
cos n 1 u n 1
3
3 2
3
tan 3
tan 1
1
2.79
H z
Y z
X z
Kz 1
1 K 1 z 1
The poles are at z
1 K
K 2
z
2
1 K
2
2K
2
.
The position of the poles in the z-plane (as a function of K) is illustrated below.
pole 1
1
imag. part
0.5
K
0
K -
-0.5
-1
-1.5
-1
-0.5
0
0.5
real part
1
1.5
2
pole 2
1
imag. part
0.5
0
K
K -
-0.5
-1
-2
-1.5
-1
-0.5
real part
0
0.5
1
Several observations are in order
The poles are real and distinct for K 2 3 and K 2 3
The poles are real and repeated for K 2 3 and K 2 3
The poles are complex conjugates for 2 3 K 2 3 .
The pole at z
The pole at z
1 K
1 K
2
2K
2
1 K
1 K
2
The system is stable for 0 K 2
2
2K
is inside the unit circle for 0 K 2 and K 4 .
is inside the unit circle for K 2 .
2.80
(a)
2
z 1
H z 3
2
1 z 1
3
(b)
2
Imaginary Part
1
0
-1
-2
-2
-1
0
Real Part
1
2
(c)
2
e j
H e j 3
2
1 e j
3
(d)
2
2
4 2 j 2 j
e j e j
e e 1
j
3
3
9
3
3
H e
1
2
2
2
2
4
1 e j 1 e j
1 e j e j
3
3
3
3
9
The system is an all pass filter!
2
2.81
(a)
(b)
(c)
X e j e j
X e
j
e j e j 4
1 e j
1 1 j n j
x n e 8 e 4
2 2
n
e
j
1
j
e
2
e
4
e j
j
4
2
8
j 8
4
j n j
e
e
e 8 e 4
2 2
2
X e j
1
e
j
2
8
e j
j
1
2 j
cos cos
e
4 2
8
1
1 cos e j e j 2
4
8
(d)
9
sin
2
X e j
sin
2
(e)
X e j 1 2 l 1 2 l
(f)
X e j
n
j j
4
e e 8
2 2
l
5
5
3 2 l 3 2 l
7
7
l
2 l j
2 l
j
3
3
n
2.82
(a)
3 n
sin
3
8
x n
8 3 n
8
(b)
1
sin n
2
x n
1
n
2
(c)
x n 10 n n 1 2 n 2 3 n 3
2.83
The DTFT of the sequence x n is
X e j
j 2
0.5
Y e j
j
1
2
1
2
j
1
2
Y e j
Y e j
1
2
0.5
1
2
0.5
j2
j2
1
0.5
Y e j
j
j
1
2
(d)
0.5
j
1
2
1
0.5
j
j2
1
0.5
j2
(c)
j
1
2
0.5
(b)
0.5
(a)
j
j 2
1
0.5
(e)
j
1
2
j
Y e j
1
2
j2
1
1
2
0.5
Y e j
j2
j
1
2
0.5
(f)
j
1
0.5
0.5
2.84
(a)
Y e j
1
2
0.5
0.5
0.5
(b)
Y e j
1
0.5
(c)
Y e j
1
2
1
2
0.5
0.5
Y e j
(d)
1
0.5
0.5
Y e j
(e)
1
2
0.5
0.5
Y e j
(f)
1
0.5
0.5
2.85
From the input/output relation we have
12Y e j 7e jY e j e j 2 Y e j 12 X e j 5e j X e j
from which we obtain
12 5e
X e 12 7e e
Y e j
j
j
j
j 2
From the block diagram we have
H e H e
X e
Y e j
j
and we are given H1 e j
j
1
j
2
1
3
.
1 j 3 e j
1 e
3
Putting this all together gives,
3
12 5e j
H1 e j H 2 e j
H 2 e j
j
j 2
j
3e
12 7e e
j
12 5e
3
8
H 2 e j
j
j 2
j
12 7e e
3e
4 e j
n
1
h2 n 2 u n
4
2
1
1 e j
4
2.86
1
1
1 z 1
3
1
1 z 1
4
X z
1
1 z 1
2
Y z
2
3 1
1 1
z 1
z
z
2
H z
1 3
4
1
1
X z 1 1 1 1
1 z 1 1 z 1
1 z 1 z
3
4
3 4
1
Y z
21
h n n
33
n 1
31
u n 1
44
n 1
u n 1
The frequency response is
H e j
H e j
2
1
1 e j
2
1 j 1 j
1 e 1 e
3
4
5
cos
4
97 91
1
cos cos 2
72 72
6
2.87
(a)
1
a
a
1
H e j
2
a re j
1 j 1 j
1
e e 2
a
a
a
1 a e j ae j a
2
1 1 j
e
a r
1 2
cos
2
r
r
1
2
1 r 2r cos
1
1 1
z
a
H z
1 az 1
1
(b)
a
1
a
H e
j
2
a re j
1 1 j
e
a r
1 r 2 2r cos
1 a e j ae j a
1
1 j 1 j
1
1 2
1 e e 2 1 2 cos
a
r
a
r
a
H z
2
1 a z 1
1
1 z 1
a
(c)
1
a
a
1
H e
j
2
| a |
2
a re j
1 j 1 j
1
e e 2
a
a
a
1 a e j ae j a
2
1 1 j
e
a r
1 r 2 2r cos
1 r 2 2r cos
1
1 1
z
a
H z a
1 az 1
1
(d)
1
a
a
H e
j
2
H z
a re j
| a |2 a e j ae j 1
1 a e j ae j a
2
1 1 j
e
a r
1 r 2 2r cos
1 r 2 2r cos
1
a z 1
1 az 1
Both the systems of parts (c) and (d) are all-pass systems. However, only the system in part (d)
has a z-transform in the standard form.
2.88
(a)
8
which means the frequency response
9
has a large amplitude in the vicinity of . Hence this is a high-pass filter as shown.
The pole-zero plot (below) shows a pole at z
100
2
80
|H(ej |2
Imaginary Part
1
0
60
40
-1
20
-2
-2
0
Real Part
1
2
0
-0.5
0
0.5
normalized frequency (cycles/sample)
8
8
and a zero at z which means
9
9
the frequency response has a large amplitude in the vicinity of 0 and a small
amplitude at . Hence this is a low-pass filter as shown.
The pole-zero plot (below) shows two poles at z
4
2.5
x 10
2
2
|H(ej |2
1
Imaginary Part
(b)
-1
2
0
1.5
1
-1
0.5
-2
-2
-1
0
Real Part
1
2
0
-0.5
0
0.5
normalized frequency (cycles/sample)
8
The pole-zero plot (below) shows complex-conjugate poles at z j and a zero at the
9
origin, which means the frequency response has a large amplitude in the vicinity of
2
. Hence this is a band-pass filter as shown.
25
2
20
1
|H(ej |2
Imaginary Part
(c)
2
0
15
10
-1
5
-2
-2
-1
0
Real Part
1
2
0
-0.5
0
0.5
normalized frequency (cycles/sample)
2.89
(a)
2
x(n)
1.5
1
0.5
0
-4
(b)
-2
0
2
n
4
6
8
X e 6 4 cos 2 cos 2 4 cos 3
X e j 2 e j e j 3
j
2
20
|X(ej |2
15
10
5
0
-0.5
(c)
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
frequency (cycles/sample)
2
x(n)
1.5
1
0.5
0
-4
(d)
-2
0
2
n
X [0] 2 1 1 4
X [1] 2 e j /2 e j 3 /2 2
X [2] 2 e j 3 j 3 0
X [3] 2 e j 3 /2 e j 9 /2 2
4
6
8
0.3
0.4
0.5
(e)
X e 2 e
X e 2 e
X e
2e
X e0 2 1 1 4
j /2
j
j 3 /2
j /2
j
e j 3 /2 2
3 j 3 0
j 3 / 2
e j 9 / 2 2
These numbers are the same as those obtained in part (d).
(f)
2
x(n)
1.5
1
0.5
0
-10
-5
0
5
n
(g)
X [0] 2 1 1 4
X [1] 2 e j /4 e j 3 /4 2 j 2
X [2] 2 e j /2 e j 3 /2 2
X [3] 2 e j 3 /4 e j 9 /4 2 j 2
X [4] 2 e j 3 j 3 0
X [5] 2 e j 5 /4 e j15 /4 2 j 2
X [6] 2 e j 3 /2 e j 9 /2 2
X [7] 2 e j 7 /4 e j 21 /4 2 j 2
10
15
(h)
X e 2 e
X e
2e
X e
2e
X e
2e
X e
2e
X e
2e
X e
2e
X e j0 2 1 1 4
j /4
j /4
e j 3 /4 2 j 2
j 2 /4
j /2
e j 3 /2 2
j 3 /4
j 3 /4
j 4 /4
j
j 5 /4
j 5 /4
e j15 /4 2 j 2
j 6 /4
j 3 /2
e j 9 /2 2
j 7 /4
j 7 /4
e j 21 /4 2 j 2
e j 9 /4 2 j 2
3 j 3 0
These numbers are the same as those obtained in part (g).
(i)
The values obtained in part (d) are a subset of those obtained in part (g). This is because
the DFT length of part (g) is divisible by the length of the DFT length of part (d).
2.90
Start with the sequence x1 n
x1 n
2
1
1
0 1 2 3
4 5
6
7 8
9
n
2
Draw the length-4 periodic extension of x1 n – call it x1,4 n – and delay it by 3. This
produces the sequence
x1,4 n 3
2
1
1
0 1 2 3 4 5
6 7 8
9
n
2
Note that the samples at n 0,1, 2,3 are the same as those of x2 n .
Now the DTFT of x1,4 n 3 is e
X 1,4 [k ] e
3
j k
2
j3
2
4
X 1 [k ]
and
e
j
3
k
2
X 1 [k ] X 2 [k ]
Thus, X 2 [k ] e
j
3
k
2
X 1[k ] X 1 [k ]
times the DTFT of x1 n . Thus we have
2.91
(a)
decimal
numbers
3-bit
binary
equivalent
3-bit binary
equivalent
index of input
0
000
000
1
001
100
2
010
010
3
011
110
4
100
001
5
101
101
6
110
011
7
111
111
(b) The 3-binary equivalent of the indexes of the inputs are reversed (bit reversed) versions of the
indexes associated with the natural (temporal) order.
2.92
(a)
X d e j
100
(b)
9 8
10 10
8
10
9
10
The negative frequency component of X c f shows up as the positive frequency
component of X d e j , and the positive frequency component of X c f shows up as the
negative frequency component of X d e j .
2.93
(a)
X d e j
102
(b)
51
51
51
51
42
51
42
51
The negative and positive frequency components of X c f are preserved on the negative
and positive frequency components of X d e j , but there is aliasing.
2.94
(a)
X d e j
52
(b)
9
13
2
4
13
4
13
2
9
13
The negative frequency component of X c f shows up as the positive frequency
component of X d e j , and the positive frequency component of X c f shows up as the
negative frequency component of X d e j .
2.95
1 0 n 7
Write x n w n cos n where w n
.
4
0 otherwise
Then
1
W e j
2
4
4
7
1 sin 4 j 2
e
2
4
4
sin
2
sin 4 7
sin 4 7
4 j 2 4 1
4 j 2 4
1
e
e
2
2
1
1
sin
sin
4
4
2
2
X e j
5
4
|X(ej )|
(a)
3
2
1
0
-0.5 -0.375 -0.25 -0.125 0 0.125 0.25 0.375 0.5
/2 (cycles/sample)
(b)
5
|X(ej )|
4
3
2
1
0
-0.5 -0.375 -0.25 -0.125 0 0.125 0.25 0.375 0.5
/2 (cycles/sample)
(c)
5
|X(ej )|
4
3
2
1
0
-0.5 -0.375 -0.25 -0.125 0 0.125 0.25 0.375 0.5
/2 (cycles/sample)
2.96
(a)
4
3.5
|X(ej k)|
3
2.5
2
1.5
1
0.5
0
-0.5 -0.375 -0.25 -0.125 0 0.125 0.25 0.375 0.5
/2 (cycles/sample)
(b)
4
3.5
|Y(ej k)|
3
2.5
2
1.5
1
0.5
0
-0.5 -0.375 -0.25 -0.125 0 0.125 0.25 0.375 0.5
/2 (cycles/sample)
(c)
The DTFT of x n is
sin 4 7
sin 4 7
j
4 2 4 1
4 j 2 4
1
X e j
e
e
2
2
1
1
sin
sin
4
4
2
2
which has zero-crossings at l 1 for l 0 . The length-8 DFT samples X e j at
k
which corresponds to the zero-crossings. On the other hand, the DTFT of y n is
4
sin 4 7
sin 4 7
3 j 2 3 1
3 j 2 3
1
Y e j
e
e
2
2
1
1
sin
sin
3
3
2
2
3l 4
for l 0 . The length-8 DFT samples X e j at
which has zero-crossings at
12
k
which does not correspond to any of the zero-crossings.
4
2.97
X e j
80, 000
3.3
4
3.3
4
2.98
(a)
S d e j
96000
5
1 12
2
5
12
1
2
normalized frequency
(cycles/sample)
(b)
S d e j
88200
1
2
200
441
200
441
1
2
normalized frequency
(cycles/sample)
(c)
The DTFT in part (b) is wider than the DTFT in part (a). This is due to the fact that the sample
rate used in part (b) is closer to the minimum sample rate defined in the sampling theorem than
the sample rate used in part (a).
2.99
S d e j
500
400
1
4
3
2
2
10 10 10
2
10
3
10
4
10
1
2
normalized frequency
(cycles/sample)
2.100
(a)
S d e j
280
224
1
2
24
56
14
56
4
56
4
56
24
56
14
56
1
2
normalized frequency
(cycles/sample)
(b)
S d e j
200
160
1
2
1
4
1
4
1
2
normalized frequency
(cycles/sample)
(c)
There are two key differences between the spectra in parts (a) and (b).The first difference is the
bandwidth: the spectrum in part (a) is the spectrum of an oversampled signal whereas the
spectrum in part (b) is the spectrum of a critically sampled signal (that is, it is sampled at the
minimum sample rate defined by the sampling theorem). The second difference is the fact that
the two spectra are “flipped” relative to each other. In part (a) the positive frequency component
of the continuous-time signal shows up on the positive frequency axis in the DTFT domain. In
part (b) the negative frequency component of the continuous-time signals shows up on the
positive frequency axis in the DTFT domain.
2.101
(a)
Y f
12
10
1
f
1
Y e j
120
100
1
2
1
10
1
10
normalized frequency
(cycles/sample)
1
2
(b)
Z p f
120
100
20
10
f
1
1
10
1
10
1
2
20
Z e j
120
100
1
2
1
10
normalized frequency
(cycles/sample)
(c)
Zp f
120
100
20
10
f
1
positive- and negative-frequency
components of the continuous-time
signal overlap when sampled
1
10
20
Z e j
120
100
1
2
1
10
1
10
1
2
normalized frequency
(cycles/sample)
The fundamental problem with this approach is that the positive- and negative-frequency
components of the z(t) overlap when sampled. This was not a problem when using the complexvalued version of the signal. This problem illustrates one of the advantages of signal-processing
using complex-valued signals: one does not have to worry about negative- and positivefrequency components aliasing on top of each other. The disadvantage is complexity: a complexvalued signal is a “two-dimensional” signal. Consequently, addition and multiplication require
more resources.
2.102
The DTFT of the sampled sequence is
S e j
500
400
0.8 0.6 0.4 0.2
0.2
0.4
0.6
0.8
Neglecting scaling constants, the sampled sequence may be written as (see Exericse 2.48)
s nT I nT cos 0.6 n Q nT sin 0.6 n .
To produce I nT and Q nT from s nT , we need 0 0.6 .
As it is written the system produces
1
I nT
2
1
y nT Q nT
2
x nT
Hence the system must be modified either by changing the sign on y nT or by changing the
sign on sin 0 n -- that is, use sin 0 n in place of sin 0 n on the lower mixer.
2.103
Hf
(a)
H j
1
−5
1
f
5
10
(b)
H e j
H e j
1
−5T
(c)
f
10
1
/ 2
cycles/sample
5T
10 T
10 T
rads/sample
Construct the diagram of the impulse sampled waveform shown below
Xp f
Xf
1
Xf
T
1
T
20
5
0
Hf
5
1
20
T
1
Xf
T
20
1
T
From this diagram, we see that the minimum sampling rate is determined from relation
1
1
20 5 from which we obtain 25 .
T
T
2.104
(a)
H c j e jT
(b)
In the interval we have
Hd e
j
j T T
e
0
(c)
hd (n)
(d)
W
1
2
W
e
W
T
sin n
W
T
W
j n
e d
T
n
W
T
T
sin n
T
hd (n)
T
n
T
(e)
W ,
j T
T
W W
otherwise
T 1
T
2
1
sin n
2
hd (n)
1
n
2
3.1
(a)
X 2 e j
5
4
5
2
5
2
5
4
5
(b)
X 4 e j
3.75
2.5
4
5
2
5
2
5
4
5
3.2
(a)
X 2 e j
5
3
4
3
5
5
3
5
4
5
(b)
X 4 e j
2.5
1.5
4
5
2
5
2
5
4
5
3.3
(a)
X 2 e j
5
3
4
3
5
5
(b)
3
5
4
5
X 4 e j
2.5
1.5
4
5
2
5
2
5
4
5
3.4
(a)
X 2 e j
5
3
4
3
5
5
3
5
4
5
3
5
4
5
(b)
X 2 e j
5
3
4
3
5
5
(c)
The spectra are reversed about
4
5
3.5
(a)
X 4 e j
2.5
1.5
4
5
2
5
2
5
4
5
(b)
X 4 e j
2.5
1.5
(c)
4
5
2
5
The spectra are reversed about
2
5
2
5
4
5
3.6
(a)
X 2 e j
5
3
3
5
4
5
(b)
X 4 e j
2.5
1.5
4
5
2
5
3.7
(a)
X 2 e j
5
3
4
3
5
5
(b)
X 4 e j
2.5
1.5
2
5
4
5
3.8
(a)
X 2 e j
5
3
3
5
(b)
4
5
X 2 e j
5
3
4
3
5
5
(c)
The signal in part (a) is centered at
4
whereas the signal in part (b) is centered at
5
4
. The two are almost frequency-domain reversed versions of each other.
5
3.9
(a)
X 4 e j
2.5
1.5
4
5
2
5
(b)
X 4 e j
2.5
1.5
(c)
2
5
The signal in part (a) is centered at
4
5
2
whereas the signal in part (b) is centered at
5
2
. The two are almost frequency-domain reversed versions of each other.
5
3.10
(a)
H e j
10
X e
j
2
5
5
5
2
5
2
2
X 2 e j
5
4
5
4
5
(b)
H e j
10
X e j
10
3
2
5
5
5
3
2
5
3
X 3 e j
10
3
10
9
3
5
3
5
(c)
H e j
10
7.5
X e j
2
5
5
5
4
2
5
4
X 4 e j
2.5
1.875
4
5
4
5
(d)
H e j
10
X e j
2
5
5
5
2
5
X 5 e j
2
3.11
(a)
X 2 e j
8
4
5
4
5
(b)
X 4 e j
4
2
5
2
5
3.12
(a)
X 2 e j
5
3
4
5
4
5
(b)
X 4 e j
2.5
1.5
2
5
2
5
3.13
(a)
X 2 e j
8
4
5
4
5
(b)
X 2 e j
5
3
(c)
4
5
4
5
The spectrum in part (a) exhibits aliasing due to overlap by the positive and negative
frequency components of the signal. The spectrum in part (b) does not show this aliasing
because it does not have a negative-frequency component.
3.14
(a)
X 4 e j
4
2
5
2
5
(b)
X 4 e j
2.5
1.5
(c)
2
5
2
5
The spectrum in part (a) exhibits aliasing due to overlap by the positive and negative
frequency components of the signal. The spectrum in part (b) does not show this aliasing
because it does not have a negative-frequency component.
3.15
(a)
X 4 e j
2.5
1.5
(b)
2
5
2
5
The spectrum of the signal after multiplication by the complex exponential is
8
6
10
10
After downsampling by 4, the spectrum is
X 4 e j
2.5
1.5
(c)
2
5
2
5
The answers to (a) and (b) are identical. This means these two systems perform equivalent
operations. In general, resampling a bandpass signal produces frequency translations for
free.
3.16
The system illustrated in the block diagram translates the spectrum of x(n) directly to baseband
prior to downsampling. The DTFT of y m is
Y e j
10
N
6
N
N
10
N
10
Clearly, N 10 for distortion-free resampling.
The interpretation of (3.9) given in Section 3.2.2 starts with making N shifted copies of the
2
spectrum, each centered at
k for k 0,1,, N 1 . For the downsample-by-N operation alone
N
to produce the desired result, we must have
2
2
k
2
N
5
for some combination of k and N. This condition reduces to
k 1
1.
N 5
The two distortion-free possibilities are k 4, N 5 and k 8, N 10 .
Downsampling x n by 5 and 10 produces to two spectra below:
X 5 e j
2
6
5
2
2
X 10 e j
1
3
5
3.17
(a)
Label the points in the system as follows:
x n
y n
s n
H e j
e
j
z m
N
n
2
The DTFT of x n is
X e j
8
6
10
10
From this, it is seen that the filter requirements are
H e j
1
10
10
In this ideal case Y e j X e j so that the lowest possible sample rate for z m
corresponds to the maximum downsample factor N. The maximum downsample factor is
N 10 . The DTFT of the output of the downsample block is
Y10 e j Z e j
1
3
5
(b)
Label the points in the system as follows:
x1 m
s n
2
y1 m
x2 m
y2 m
H1 e j
H 2 e j
z k
N
e j1m
After downsampling by 2, the spectrum is
X 1 e j
5
3
4
5
4
5
From this, it is seen the first filter H1 e j must be an ideal high-pass filter:
H 1 e j
1
4
5
4
5
Now, the spectrum should be translated to baseband. This requires 1 . (Note that
e j1n e j n 1, 1,1, 1, ). The resulting spectrum is
X 2 e j
5
3
5
5
Because of the filtering performed by H1 e j , the second filter, H 2 e j , is not needed
and may be removed from the block diagram. The lowest possible sample rate for z m
corresponds to the maximum downsample factor N. The maximum downsample factor at
this point is N 5 . The DTFT of the output of the downsample block is
Z e j
1
3
5
(c)
Label the points in the system as follows:
x1 m
s n
4
y1 m
x2 m
y2 m
H1 e j
H 2 e j
e j1m
After downsampling by 4, the spectrum is
N
z k
X 1 e j
2.5
1.5
2
5
2
5
From this, it is seen the first filter H1 e j must be an ideal low-pass filter:
H 1 e j
1
2
5
2
5
Note that no frequency translation is needed here. Hence the multiplier involving the
complex exponential is not needed. Consequently the second filter H 2 e j , is not
needed and may be removed from the block diagram. The lowest possible sample rate for
z m corresponds to the maximum downsample factor N. The maximum downsample
factor at this point is N 2.5 . (Fractional downsample factors are covered in Chapter 9. It
is performed using a polyphase partition of H1 ( z ) .) The DTFT of the output of the
downsample block is
Z e j
1
3
5
3.18
(a)
X 2 e j
4
5
5
5
4
5
(b)
X 4 e j
9
10
6
5
4
10
10
10
10
10
4
10
5
10
6
10
9
10
3.19
(a)
X 2 e j
4
5
5
5
4
5
(b)
X 4 e j
9
10
6
10
4
10
10
10
4
10
6
10
9
10
3.20
(a)
X 2 e j
7
10
3
10
3
10
7
10
(b)
X 4 e j
17
20
13
20
7
20
3
20
3
20
7
20
13
20
17
20
3.21
(a)
X 2 e j
4
5
5
5
4
5
(b)
X 2 e j
(c)
7
10
3
10
3
10
7
10
Both have two spectral copies in the first Nyquist zone. The copies are located at slightly
different frequencies, as expected. But the copies closest to baseband are frequencyreversed versions.
3.22
(a)
X 4 e j
9
10
6
10
4
10
10
10
4
10
6
10
9
10
(b)
X 4 e j
(c)
17
20
13
20
7
20
3
20
3
20
7
20
13
20
17
20
Both have four spectral copies in the first Nyquist zone. The copies are located at slightly
different frequencies, as expected. But the copies closest to baseband are identical (other
than the center frequency).
3.23
(a)
X 2 e j
4
5
5
5
4
5
(b)
X 4 e j
9
10
6
10
4
10
10
10
4
10
6
10
9
10
3.24
(a)
X 2 e j
7
10
3
10
3
10
7
10
(b)
X 4 e j
17
20
13
20
7
20
3
20
3
20
7
20
13
20
17
20
3.25
(a)
X 2 e j
4
5
5
5
4
5
(b)
X 2 e j
(c)
7
10
3
10
3
10
7
10
The two are identical except for the center frequencies of the spectral copies.
3.26
(a)
X 4 e j
9
10
6
10
4
10
10
10
4
10
6
10
9
10
(b)
X 4 e j
(c)
17
20
13
20
7
20
3
20
3
20
7
20
13
20
The two are identical except for the center frequencies of the spectral copies.
17
20
3.27
(a)
The result of the upsample-by-10 operation is illustrated by the DTFT below.
X 10 e j
4
5
3
5
2
5
5
25
25
5
2
5
3
5
4
5
The goal of the low-pass filter is to remove all the spectral copies except the one at
baseband. The generic filter requirements are
H e j
10
10
But, because we know more about the spectral properties of the signal, we can generate
more specific filter requirements:
This filter is defined by the following intervals on
the frequency axis (these intervals are usually
called frequency bands):
H e j
passband: 0,
25
4
transition band: ,
25 25
4
stop band: ,
25
4
25
25
25
4
25
The DTFT of the filter output is shown below.
DTFT at H output
4
5
3
5
2
5
5
25
25
5
2
5
After frequency translation, the DTFT of the output y m is
3
5
4
5
Y e j
(b)
4
5
3
5
2
5
5
25
25
5
2
5
3
5
4
5
A system such as the one illustrated can produce the desired signal. This fact follows from
the observation that after upsampling, X 10 e j has a spectral copy centered at the
3
. There simply is no need to eliminate the spectral copy at 0 ,
5
keep the copy at 0 , then consume resources to translate the copy at 0 up to the
spot on the frequency axis where there originally was a spectral copy.
desired frequency 0
The filter requirements are frequency-translated versions of the low-pass filter H e j .
The generic requirements are
G e j
5
10
3
5
7
10
Note that G e j is a complex-valued filter. The more specific requirements are
G e j
11
25
3
5
14
25
19
25
16
25
Note that after filtering X 10 e j (shown above) with G e j , the desired spectrum
results.
3.28
H e j
5
5
Y e j
2
2
3.29
H e j
5
5
Y e j
2
3
3
3
2
3
3.30
(a)
The highest frequency is 2
20000 5
. Thus the DTFT of x nT is
48000 6
X e j
5
6
(b)
5
6
44100 147
times the input rate. Hence U 147 and D 160 .
48000 160
The filter specifications are determined by the bandwidth of the signal to be resampled
and the requirement to eliminate spectral energy aliased by the upsample/downsample
operation. The thought process is illustrated below.
The output rate is
X e j
spectral copy resulting
from the upsample-by147 operation
transition band
X e j
spectral copy resulting from the
multiplication by the discretetime impulse train associated
with the downsample-by-160
operation.
transition band
5
6 147
2
147
7
6 147
5
2
6 147
160
241
35280
The transition band required by the upsample operation is
2
5
5
7
.
147 6 147 6 147 6 147
The transition band required by the downsample operation is
241
5
41
35280 6 147 35280
The transition band requirement for the downsample operation is more strict and hence is
the one that must be used. From this requirement, we get
41
F
.
70560
Now the passband and stopband ripples are
0.1 20 log 1 p p 100.1/20 1 0.0116
96 20 log s s 1096/20 1.5849 105
The filter length estimates based on the Kaiser and Harris formulae are
Kaiser Estimate: 6407
Harris Estimate: 7510
clock rate = 7056 kHz
(c)
The starting point is to factor the sample rate conversion as follows
147
3·7·7
160 2·2·2·2·2·5
and perform the resampling operation in two steps. There are a number of options here.
The thing to keep in mind is that the first resampling ratio must be greater than 5/6.
(Otherwise the resampling filter will distort the signal.)
Proceeding as outlined in part (b), except doing it twice, the following are some
representative examples:
U1 = 21 D1 = 20
Kaiser Estimate: 469
Harris Estimate: 550
clock rate = 1008 kHz
U2 = 7 D2 = 8
Kaiser Estimate: 320
Harris Estimate: 375
clock rate = 352.8 kHz
U1 = 7 D1 = 4
Kaiser Estimate: 156
Harris Estimate: 183
clock rate = 336 kHz
U2 = 21 D2 = 40
Kaiser Estimate: 1602
Harris Estimate: 1877
clock rate = 1764 kHz
U1 = 7 D1 = 8
Kaiser Estimate: 625
Harris Estimate: 733
clock rate = 336 kHz
U2 = 21 D2 = 20
Kaiser Estimate: 1642
Harris Estimate: 1924
clock rate = 882 kHz
U1 = 3 D1 = 2
Kaiser Estimate: 67
Harris Estimate: 79
clock rate = 144 kHz
U2 = 49 D2 = 80
Kaiser Estimate: 3204
Harris Estimate: 3755
clock rate = 3528 kHz
U1 = 49 D1 = 40
Kaiser Estimate: 1095
Harris Estimate: 1283
clock rate = 2352 kHz
U2 = 3 D2 = 4
Kaiser Estimate: 160
Harris Estimate: 188
clock rate = 176.4 kHz
(d)
The starting point here is to factor the downsample rate: 160 2·2·2·2·2·5 In this part, the
downsample operation can be performed in three steps where D1 D2 D3 160 . As an
example, suppose D1 40, D2 2, and D3 2 . The first filter has a transition band as
determined below. Note that as a result of the downsample-by-40 operation, the signal
100
now has a bandwidth of
as shown.
441
X e j
X e j
transition band
↓ 40
5
2
6 147 7 147
6 147
100
441
The transition band for the second filter, and the corresponding spectrum after
downsampling by D2 2 is shown below.
X e j
X e j
transition band
100
441
341
441
↓2
200
441
The transition band for the third filter, and the corresponding spectrum after
downsampling by D3 2 is shown below.
X e j
X e j
transition band
200
441
241
441
↓2
400
441
The lengths of the three filters and their clock rates is summarized as follows:
U = 147 D1 = 40 D2 = 2 D3 = 2
Kaiser Estimate for filter 1: 3284
Harris Estimate for filter 1: 3849
clock rate for filter 1 = 7056 kHz
Kaiser Estimate for filter 2: 14
Harris Estimate for filter 2: 16
clock rate for filter 2 = 176.4 kHz
Kaiser Estimate for filter 3: 80
Harris Estimate for filter 3: 94
clock rate for filter 3 = 88.2 kHz
Some other options are summarized as follows:
(f)
U = 147 D1 = 2 D2 = 2 D3 = 40
Kaiser Estimate for filter 1: 3284
Harris Estimate for filter 1: 3849
clock rate for filter 1 = 7056 kHz
Kaiser Estimate for filter 2: 8
Harris Estimate for filter 2: 9
clock rate for filter 2 = 3528 kHz
Kaiser Estimate for filter 3: 1602
Harris Estimate for filter 3: 1877
clock rate for filter 3 = 1764 kHz
U = 147 D1 = 20 D2 = 4 D3 = 2
Kaiser Estimate for filter 1: 3284
Harris Estimate for filter 1: 3849
clock rate for filter 1 = 7056 kHz
Kaiser Estimate for filter 2: 27
Harris Estimate for filter 2: 32
clock rate for filter 2 = 352.8 kHz
Kaiser Estimate for filter 3: 80
Harris Estimate for filter 3: 94
clock rate for filter 3 = 88.2 kHz
U = 147 D1 = 20 D2 = 2 D3 = 4
Kaiser Estimate for filter 1: 3284
Harris Estimate for filter 1: 3849
clock rate for filter 1 = 7056 kHz
Kaiser Estimate for filter 2: 10
Harris Estimate for filter 2: 11
clock rate for filter 2 = 352.8 kHz
Kaiser Estimate for filter 3: 160
Harris Estimate for filter 3: 188
clock rate for filter 3 = 176.4 kHz
U = 147 D1 = 10 D2 = 4 D3 = 4
Kaiser Estimate for filter 1: 3284
Harris Estimate for filter 1: 3849
clock rate for filter 1 = 7056 kHz
Kaiser Estimate for filter 2: 19
Harris Estimate for filter 2: 23
clock rate for filter 2 = 705.6 kHz
Kaiser Estimate for filter 3: 160
Harris Estimate for filter 3: 188
clock rate for filter 3 = 176.4 kHz
A polyphase partition of the filter designed in part (b) may be used. The polyphase
filterbank consists of 147 subfilters. (Note: using the estimates from the first problem,
each subfilter requires approximately 100 coefficients.) The filterbank presents an
upsampled and filtered version of the input signal in parallel as shown in the block
diagram below. The output commutator strides through the available output samples by
160 (the downsample rate). As only one filter in the filter bank has be active to produce
each output sample, this approach is an efficient method for resamping the signal.
H0(z)
H1(z)
stride = 160
H146(z)
3.31
(a)
hc t A1e s1t A2e s2t AN e sN t u t
(b)
hd n Thc nT
s1Tn
AT
A2Te s2Tn AN Te sN Tn u n
1 e
AT
e s1T
1
(c)
(d)
Hd z
n
AT
1
1 e
s1T
z
A2T e s2T
1
n
AN T e sN T
A2T
1 e
s2T
z
1
n
u n
AN T
1 e s N T z 1
The poles are e s1T , e s2T , , e s N T Thus a pole at s p maps to a pole at z e pT in the
discrete-time system generated using the impulse invariance technique.
3.32
(a)
The transform pair xc (t ) X c j implies xc t T e jT X c j . This sugguests
the block diagram blow:
H c j e jT
xc t
xc t T
The interpretation is that a continuous-time LTI system with transfer function
H c j e jT produces a delay of T .
(b)
Applying (2.86) produces H d e
(c)
hd (n)
1
2
W
e
j
T
T
e jn d
W
T
sin n
W
T
W
T
n
W
T
(d)
W
T
sin n
T
hd (n)
T
n
T
(e)
T 1
T
2
1
sin n
2
hd (n)
1
n 2
j
1
2
j TT
e
0
| | W
W | |
T
j n TT W
j n
W
e T
e
T
jn
T
1
3.33
(a)
(b)
(c)
hc (t ) ae at u (t )
hd (n) Thc (nT )
aT
1 e aT e j
a
. The technique outlined in Section 2.6.2 evaluates H c j at .
a j
T
This clearly produces a result that is different from that obtained in (b).
H c j
3.34
(a)
1
a 1 1
1
Sc ( s ) H c s
s sa s s sa
at
sc (t ) 1 e u (t )
(b)
sd (n) sc (nT ) 1 e aTn u (n) 1 e aT
(c)
n
u(n)
1 1e
S d e j H d e j
j
e jn e aT
n 0
n 0
n
e jn
e jn n e jn
n 0
Hd e
(d)
j
n 0
1
1
j
1 e
1 e j
1 e j
1 e 1 e
j
j
1 e j
1 e j
a
. The technique outlined in Section 2.6.2 evaluates H c j at .
a j
T
This clearly produces a result that is different from that obtained in (c).
H c j
3.35
(a)
(b)
(c)
1 z 1
a
aT
1
Hd z
where
1
1
2 1 z
2
z 1
a 1
1
1
T 1 z
0
n
n 1
1
1
hd (n)
u (n)
u (n 1)
1 1
1 1
1
2 1 n
2
1 1
n0
n0
n0
a
. The technique outlined in Section 2.6.2 evaluates H c j at .
a j
T
This clearly produces a result that is different from that obtained in (b).
H c j
3.36
1
hd (n)
2
WcT
WcT
j
j e jn d
T
2 T
WcT
e jn d
WcT
WcT
The integral
e jn d may be evaluated using integration by parts:
WcT
WcT
e
j n
WcT
d e jn
jn
WcT
WcT
WcT
WcT
1 jn
1
j n
e d e jn
e
2
jn
jn
W T
jn
c
W jWcTn jWcTn
1
e
e
2 e jWcTn e jWcTn
jn
n
2W
cos WcTn
The desired result is
hd (n)
WcT
jn
j2
sin WcTn
n2
j
times the integral. Putting this together gives
2 T
W cos WcTn 1 sin WcTn
T
n
T
n2
3.37
The length-3 differentiator is
1
T
hd (0) 0
hd (1)
hd (1)
1
T
Applying the window gives
0.5
T
hd ,FIR (0) 0
hd ,FIR (1)
hd ,FIR (1)
0.5
T
The DTFT is
H d ,FIR e j
0.5 j 0.5 j j
e
e sin
T
T
T
For small , sin so that H d ,FIR j
.
T
3.38
From Table 3.3.3 we have H I e j T
e
j
2
j 2 sin
2
.
The series expansions for the exponential and the sine are
e
j
2
2
3
1 1
1 j j j
2 2! 2 3! 2
3
5
1 1
sin
2 2 3! 2 5! 2
For small , e
j
2
1
1
so that H I T
1 and sin
2 2
j2
j
2
T
3.39
From Table 3.3.3 we have H I e j
cos
T
2 .
j2
sin
2
The series expansions for the cosine and the sine are
2
4
1
1
cos 1
2! 2 4! 4
2
3
5
1 1
sin
2 2 3! 2 5! 2
T 1
1
For small , cos 1 and sin
so that H I
j2 j
2
2 2
2
T
3.40
(a)
Use the identity cos( A B) cos( A) cos( B) sin( A) sin( B) with A c n and B nT .
(b)
Suppose S e j looks like this
S e j
2W
c W
c
c W
c
c W
c W
The input to the upper filter is
s nT cos c n cos nT cos c n sin nT sin c n cos c n
cos nT cos 2 c n sin nT sin c n cos c n
1 1
cos 2 c n
2 2
1
sin 2 c n
2
1
1
1
cos nT cos nT cos 2 c n sin nT sin 2 c n
2
2
2
double frequency term
double frequency term
If the filter H e j is a low-pass filter with a pass-band gain of 2 as follows:
H e j
W
W
then the double-frequency terms are eliminated and x nT cos nT .
The input to the lower filter is
s nT sin c n cos nT cos c n sin nT sin c n sin c n
cos nT cos c n sin c n sin nT sin 2 c n
1
sin 2 c n
2
1 1
cos 2 c n
2 2
1
1
1
cos nT sin 2c n sin nT sin nT cos 2 c n
2
2
2
double frequency term
double frequency term
Using the same low-pass filter described above, the double frequency terms are eliminated
so that y nT sin nT .
(c)
(d)
y nT
y nT sin nT
tan nT . Thus we have nT tan 1
.
x nT
x nT cos nT
nT
d y nT
dt x nT
y 2 nT
1 2
x nT
x nT y ' nT x ' nT y nT
x 2 nT
x nT y ' nT x ' nT y nT
2
y nT
x 2 nT y 2 nT
1 2
x nT
(e)
x nT
D z
x ' nT
zL
x ' nT y nT
z
y nT
scale to a
point on the
unit circle
' nT
L
D z
y ' nT
x nT y ' nT
Note: the length of the derivative filter is assumed to be 2L+1. The delay by L samples is
required to align the input samples with the outputs of the derivative filters.
(f)
0.2
0.004 . Figure 3.3.18 shows that a length-3 (L = 1)
50
derivative filter is more than adequate.
The bandwidth of x nT is
4.1
(a) Q A
(b) Q A
(c) 1 Q A
(d) 1 2Q A
4.2
1
x
0
1
1
t 2
x e dt x e dt
t2
x
e
t 2
dt
0
substitute u x
1
2
x
u
e du
2
0
x
e
0
t 2
dt
1
x
e
0
t 2
dt
4.3
x
2
1
1
et dt
e
u2
du
x
substitute u x
thus
1
x
1
e dt
t 2
1
x e dt
t 2
e
x
u2
1
du
2
x e dt
t 2
e
x
t 2
dt
4.4
1 t 2 /2
1 u2
1
2 u 2
1
x
Q x
e dt
e du
e du erfc
2 x/ 2
2
2
2
x
x/ 2
substitute u t / 2
4.5
(a) 1 2Q 1 1 2 0.1587 0.6827
(b) 1 2Q 2 1 2 0.0228 0.9545
(c) 1 2Q 3 1 2 0.0013 0.9973
(d) 1 2Q 4 1 2 3.1533 105 0.9999
4.6
(a) 1 2Q 1 1 2 0.1587 0.6827
(b) 1 2Q 2 1 2 0.0228 0.9545
(c) 1 2Q 3 1 2 0.0013 0.9973
(d) 1 2Q 4 1 2 3.1533 105 0.9999
4.7
6
1 2Q 1 2Q 2 1 2 0.0228 0.9545
3
4.8 Let be a random variable with pdf f X ( x) and let Z u ( X ) X 2 . The pdf of Z may be
1
1
fX z
fX z .
expressed in terms of f X ( x) using the relationship f Z ( z )
2 z
2 z
For this problem, X ~ N (0, 2 ) so that the pdf of Z1 X 2 is
f Z1 ( z )
1
1
2 z
2 2
e
z
2 2
1
1
2 z
2 2
e
z
2 2
1
2 2
z 1/2 e
z
2 2
Similarly, the pdf of Z 2 Y 2 is
1
f Z2 ( z )
1/2
z
2 2
z e
2 2
Because X and Y are independent, Z1 and Z2 are independent. Hence the pdf of Z Z1 Z 2 is
f Z ( z ) f Z1 ( z ) f Z2 ( z )
f Z1 ( z x) f Z2 ( x)dx
z
2 2
0
2
1
e 2
zx
z
2
e 2
2 2
zx
1
z
dx
( z x) x
0
z
2
2
e
2 2
z
1 2 2
e
2 2
z0
1
2 2
x
1 2 2
e
dx
x
4.9 Let be a random variable with pdf f X ( x) and let Z u ( X ) X 2 . The pdf of Z may be
1
1
fX z
fX z .
expressed in terms of f X ( x) using the relationship f Z ( z )
2 z
2 z
For this problem, X ~ N ( X , 2 ) so that the pdf of Z1 X 2 is
f Z1 ( z )
1
1
2 z
2
1
2
1
2 2 2 z
e
e
z X
2
2 2
z X2
2 2
e
1
1
2 z
2
X z
2
e
z
X2
e
2
z X
2
2 2
Similarly, if Y ~ N ( Y , 2 ) , the pdf of Z 2 Y 2 is
f Z2 ( z )
1
1
2 z
2 2
1
1
2 2 2 z
e
e
z Y
2 2
2
2 2
z Y2
1
1
2 z
2 2
e
z Y
2
2 2
z
Y 2 z
Y2
e e
Because X and Y are independent, Z1 and Z2 are independent. Hence the pdf of Z Z1 Z 2 is
f Z ( z ) f Z1 ( z ) f Z2 ( z )
f Z1 ( z x) f Z 2 ( x)dx
1
0
e
z X2 Y2
2 2
8
e
z X2
2
2
z
0
Y2
2
8 2
e
z x X2
e 2
e
2 2 2 z x
z
z X2 Y2
2 2
8 2
2
X z x
2
1
x( z x)
X
e
e
zx
X 2
z x Y
2
x Y2
2
x
x
1
Y 2
e 2 Y 2
e dx
e
2
2
x
2
x
e
X z x Y x
2
e
X z x Y x
2
e
X z x Y x
2
dx
z X2 Y2
z X2 Y2
z X2 Y2
z X2 Y2
cos( )
cos( )
cos( )
cos( )
1
1
1
1
2
2
2
2
e
d e
d e
d e
d
0
0
0
0
z 2 2
X
Y
4I0
2
e
z X2 Y2
2 2
2 2
Using the substation (4.38) gives the desired answer.
z 2 2
X
Y
I0
2
z 0
4.10
Let Y be a random variable with pdf fY ( y ) and let Z u (Y ) Y . The pdf of Z may be
expressed in terms of fY ( y ) using the relationship f Z ( z ) 2 zfY z 2 .
In this case, Y is a central chi-square random variable whose pdf is (4.36):
y
1 2 2
fY ( y )
e
2 2
Substituting produces
z2
z2
2 z 2 2
z 2
fZ ( z)
e
2 e 2
2
2
4.11
Let Y be a random variable with pdf fY ( y ) and let Z u (Y ) Y . The pdf of Z may be
expressed in terms of fY ( y ) using the relationship f Z ( z ) 2 zfY z 2 .
In this case, Y is a non-central chi-square random variable whose pdf is (4.37):
y s2
1 2 2
s
fY ( y )
e
I0 y 2
2
2
Substituting produces
2z
fZ ( z)
e
2 2
z 2 s2
2 2
s z
I0 z 2 2 2 e
z 2 s2
2 2
s
I0 z 2
4.12
This transformation is defined by two functions
R u1 ( X , Y ) X 2 Y 2
Y
u2 ( X , Y ) tan 1
X
and their inverses
X 1 v1 ( R, ) R cos
X 2 v2 ( R, ) R sin
The joint pdf f R , (r , ) may be expressed in terms of the joint pdf f X ,Y ( x, y ) using the
relationship
f R , (r , ) J f X ,Y v1 r , , v2 r ,
where J is the Jacobian matrix
v1 r ,
r
J
v2 r ,
r
v1 r ,
cos
v2 r , sin
r
r sin
r cos
Starting with
1
f X ,Y ( x, y )
e
2 2
and substituting produces
r
f R , ( r , )
e
2
2
r
e
2 2
r
e
2 2
r cos X 2 r sin Y 2
2 2
r 2 cos2 2 r X cos X2 r 2 sin 2 2 r Y sin Y2
2 2
r 2 X2 Y2 2 r X cos 2 r Y sin
2 2
x X 2 y Y 2
2 2
4.13
Because the covariance matrix M is diagonal, X1 and X2 are uncorrelated and therefore
independent random variables. To see that this is so, observe that the joint pdf can be factored:
x
fX 1
x2
1
1
2
5 0
1/2
e
5 0 x 1
1
x1 1 x2 20 5 x 1 2
2
2
0 5
1
e
2 5
1
1
x1 1 x2 2 5
2
0
0
x 1
1
1 x2 2
5
1 ( x1 1)2 ( x2 2)2
5
5
1 2
e
2 5
( x 1)2
( x 2)2
1
2
1
1
e 25
e 25
2 5
2 5
Thus we see that X1 ~ N(1,5) and X2 ~ N(2,5).
3 1 3 2
(a) P X 1 3 and X 2 3 P X 1 3 P X 2 3 Q
Q
0.0607
5 5
11
2 2 1
(b) P X 1 1 and X 2 2 P X 1 1 P X 2 2 Q
1 Q
5
5 4
4.14
P( X 1 X 2 ) P( X 1 X 2 0) Let Z X 1 X 2 . Then the desired probability is P ( Z 0) .
Z is a Gaussian random variable because it was obtained through a linear operation on the
T
Gaussian random vector X X 1 X 2 . The pdf of X is (4.44) with
2
10 0
μ and M
5
0 10
The random variable Z is related X through Z = AX where A 1 1 . The mean and variance
of Z are
Z Aμ 3
10 0 1
Z2 AMAT 1 1
20
0 10 1
Thus,
03
P( Z 0) Q
0.2512
20
4.15
P( X 1 X 2 ) P( X 1 X 2 0) Let Z X 1 X 2 . Then the desired probability is P ( Z 0) .
Z is a Gaussian random variable because it was obtained through a linear operation on the
T
Gaussian random vector X X 1 X 2 . The pdf of X is (4.44) with
2
10 2
μ and M
5
2 10
The random variable Z is related X through Z = AX where A 1 1 . The mean and variance
of Z are
Z Aμ 3
10 2 1
Z2 AMAT 1 1
16
2 10 1
Thus,
03
P( Z 0) Q
0.2266
4
4.16
(a)
0.25
f X 5
2
1
1
1
2
3/2
0.5 0.1
0.5
1
1/2
e
1 0.5 0.1 0.25 0
1
0.250 5 4 2 20.5 2 0.5 5 4
2
0.1 0.5 1 2 2
0.0403
0.5
0.1 0.5
1
(b)
Let Z X 1 X 2 X 3 . Then the desired probability is P ( Z 5) . Z is a Gaussian random
variable because it was obtained through a linear operation on the Gaussian random vector
T
X X 1 X 2 X 3 . The pdf of X is (4.44) with
0
1 0.5 0.1
μ 4 and M 0.5 2 0.5
2
0.1 0.5 1
The random variable Z is related X through Z = AX where A 1 1 1 . The mean and variance
of Z are
Z Aμ 6
1 0.5 0.1 1
AMA 1 1 1 0.5 2 0.5 1 6.1
0.1 0.5 1 1
56
Thus, P( Z 5) Q
0.6572 .
6.1
2
Z
T
4.17
Because the covariance matrix M is diagonal, X1, X2, and X3 are uncorrelated and therefore
independent random variables. To see that this is so, observe that the joint pdf can be factored:
x1
f X x2
x
3
1
1
2
3/2
1 0 0
0 1 0
1/2
e
1 0 0 x1 1
1
x1 1 x2 1 x3 1 0 1 0 x2 1
2
0 0 1 x3 1
0 0 1
1 2 ( x1 1)2 ( x2 1)2 ( x3 1)2
e
2
1
1
( x1 1)2
2
1
( x2 1)2
2
1
( x3 1)2
2
e
e
e
2
2
2
Thus we see that X1 ~ N(1,1), X2~N(1,1), and X3 ~ N(1,1).
(a)
0.25
f X 5
2
1
2
e
(0.25 1)2
2
1
2
e
(5 1)2
2
1
2
e
(2 1)2
2
9.7517 10 6
(b)
P( X 1 2 and X 2 2 and X 3 2) P( X 1 2) P( X 2 2) P( X 3 2)
2 1 2 1 2 1
Q
Q
Q
1 1 1
0.004
4.18
(a) The vector Y consists of 2 jointly Gaussian random variables because Y was obtained from X
via a linear operation. Consequently, the pdf of Y is
f Y (y )
1
T
y μ Y M Y1 y μ Y
1
2
e
2 | M y |1/2
and all that remains is to compute the mean vector and the covariance matrix:
2
1
5 1 2 4 3 24
μ Y E Y E AX AE X
1
2 4 6 1 2 18
2
1
M Y E YYT E AX RA
T
1
0.3
5 1 2 4 3
0.4
2 4 6 1 2
0.2
0.1
AE XX A
0.3
1
0.3
0.4
0.2
T
0.4
0.3
1
0.3
0.4
0.2
0.4
0.3
1
0.3
AMAT
T
0.1 5
0.2 1
0.4 2
0.3 4
1 3
2
4
99.4 94.1
6
94.1 112.2
1
2
Thus, the pdf of Y is
T
1
24 0.0488 0.0410 24
y
0.0410 0.0433 18
y
1
18
2
f Y (y )
e
2 47.9361
0.5
The value of the pdf at the point Y is
0.5
T
1 0.5 24 0.0488 0.0410 0.5 24
0.0410 0.0433 0.5 18
0.5
1
2 0.5 18
fY
e
0.5 2 47.9361
1.2678 104
(b) P(Y1 Y2 ) P(Y1 Y2 0)
Let Z Y1 Y2 , then the desired probability is P ( Z 0) .
The random variable Z is a Gaussian random variable because it was obtained from Y via a
linear transformation. That is Z Y1 Y2 looks like Z BY where B 1 1 . Consequently,
the pdf of Z is f Z ( z )
1
2 Z2
e
z Z 2
2 Z2
, where the mean and variance of Z are
24
Z E Z E BY BE{Y} Bμ Y 1 1 6
18
Z2 E ZZ T E BY BY
T
BE{YY }B
T
T
99.4 94.1 1
1 1
94.1 112.2 1
23.4
Now, the desired probability is
0 Z
P( Z 0) Q
Z
6
Q
0.8926
23.4
BM Y BT
4.19
xT x
2
1
The pdf of the vector X is f X (x)
e 2
2
2
The vector Y consists of 2 jointly-Gaussian random variables because Y is obtained from X via a
linear operation. Consequently, the pdf of Y is
f Y (y )
1
T
y μ Y M Y1 y μ Y
1
2
e
2 | M y |1/2
and all that remains is to compute the mean vector and covariance matrix:
0 0
μ Y E Y E RX RE X R
0 0
M Y E YYT E RX RX
T
RE XX R
T
T
R 2 IRT 2 RRT
cos 2 ( ) sin 2 ( )
cos( ) sin( ) cos( )sin( )
2
cos 2 ( ) sin 2 ( )
cos( ) sin( ) cos( ) sin( )
2
I
Because the pdf of Y is the form of two uncorrelated Gaussian random variables with zero mean
and common variance 2 .
4.20
The pdf of the vector X is f X (x)
1
xT x
2 2 N
e
N
The vector Y consists of N jointly-Gaussian random variables because Y is obtained from X via
a linear operation. Consequently, the pdf of Y is
2
f Y (y )
N /2
1
2
N /2
| M y |1/2
e
1
y μY T M Y1 y μ Y
2
and all that remains is to compute the mean vector and covariance matrix:
0 0
0 0
μ Y E Y E UX UE X U
0 0
M Y E YYT E UX UX
T
UE XX U
T
T
U 2IUT 2 UUT 2I
Because the pdf of Y is the form of N uncorrelated Gaussian random variables with zero mean
and common variance 2 .
4.21
C XX (n, k ) E X (n) (n) X (k ) (k )
E X (n) X (k ) (k ) X (n) (n) X (k ) (n) (k )
E X (n) X (k ) (k ) E X (n) (n) E X (k ) (n) (k )
RXX (n, k ) (k ) (n) (n) (k ) (n) (k )
RXX (n, k ) (n) (k )
4.22
From the covariance matrix, we obtain the following:
RXX (0) 1
RXX (1) RXX (1) 0.35
RXX (2) RXX (2) 0.13
RXX (3) RXX (3) 0.04
R
S X e j
k
XX
( k ) e j k
0.04e j 3 0.13e j 2 0.35e j 1 0.35e j 0.13e j 2 0.04e j 3
1 0.7 cos 0.26 cos 2 0.08cos 3
4.23
The autocorrelation function is obtained by computing the inverse DTFT of S X e j :
1
1
1
1
1
1
RXX (k ) (k 3) (k 2) (k 1) (k ) (k 1) (k 2) (k 3)
8
4
2
2
4
8
1
1
2
1
8
RXX (k )
1
2
1
4
-4 -3 -2
1
4
-1
0
1
2
1
8
3 4
k
(a) The vector X consists of four jointly Gaussian random variables. The form of the pdf is given
by (4.44). Consequently, all the remains to be found are the mean vector and the covariance
matrix M.
X 1 0
X 0
μ E X E 2
X 3 0
X 4 0
X 1
X 1 X 1 X 1 X 2 X1 X 3 X 1 X 4
X 2
X 2 X1 X 2 X 2 X 2 X 3 X 2 X 4
T
M E XX E
X 1 X 2 X 3 X 4 E X X X X X X X X
3 2
3 3
3 4
X 3
3 1
X 4
X 4 X 1 X 4 X 2 X 4 X 3 X 4 X 4
1 1 1
1 2 4 8
RXX (0) RXX (1) RXX (2) RXX (3) 1
1 1
1
R (1) R (0) R (1) R (2)
2
2 4
XX
XX
XX
XX
RXX (2) RXX (1)
RXX (0) RXX (1) 1 1
1
1
2
RXX (1)
RXX (0) 4 2
RXX (3) RXX (2)
1 1 1
1
8 4 2
Note that M 0.4219
M
1/2
0.6495 and that M 1
4
3
2
3
0
0
2
3
4
3
2
3
0
0
2
3
4
3
2
3
0
0
.
2
3
4
3
(b) Let Z X 1 X 2 X 3 X 4 . Then the desired probability is P ( Z 1) . The random variable Z
is related to the random vector X through the transformation
X1
X
Z 1 1 1 1 2
X3
X4
which we write as Z = AX (where A = [1 1 1 1].) Z is a Gaussian random variable because it is
the result of a linear operation on a sequence of jointly Gaussian random variables. The mean
and variance of Z are
Z E Z AE X 0
Z2 E ZZ T AMAT
1
1
1 1 1 1 2
1
4
1
8
33
4
1
2
1
1
2
1
4
1
4
1
2
1
1
2
1
8
1 1
4 1
1 1
2 1
1
Hence
1 Z
1
P( Z 1) Q
Q
0.3639
33
/
4
Z
4.24
(a)
M E XXT
X1
X
E 2 X1
X 3
X 4
X 1 X 1
X X
E 2 1
X 3 X1
X 4 X 1
X1 X 2
X2X2
X3 X2
X4X2
X1 X 3
X2 X3
X3X3
X4 X3
X2
X3
X 4
X1 X 4
X 2 X 4
X3 X 4
X 4 X 4
1
RXX (0) RXX (1) RXX (2) RXX (3) 1
R (1) R (0) R (1) R (2)
XX
XX
XX
2
XX
RXX (2) RXX (1)
RXX (0) RXX (1) 1
RXX (1)
RXX (0) 4
RXX (3) RXX (2)
1
8
1
2
1
1
2
1
4
1
4
1
2
1
1
2
1
8
1
4
1
2
1
(b)
1
0.5
0
fX
0.25
1
1
2
2
1
0.5 0.25 0.125
0.5
1
0.5 0.25
0.25 0.5
1
0.5
0.125 0.25 0.5
1
1/2
e
0.5 0.25 0.125 0.5
1
0.5
1
0.5 0.25 0
1
0.5 0 0.25 1
0.25 0.5
1
2
0.5 0.25
1 1
0.125 0.25 0.5
0.0136
(c) Let Z X 1 X 2 X 3 X 4 . Then the desired probability is P ( Z 1) . The random variable Z
is related to the random vector X through the transformation
X1
X
Z 1 1 1 1 2
X3
X4
which we write as Z = AX (where A = [1 1 1 1].) Z is a Gaussian random variable because it is
the result of a linear operation on a sequence of jointly Gaussian random variables. The mean
and variance of Z are
Z E Z AE X 0
Z2 E ZZ T AMAT
1
1
1 1 1 1 2
1
4
1
8
1.75
1
2
1
1
2
1
4
1
4
1
2
1
1
2
1
8
1 1
4 1
1 1
2 1
1
Hence
1 Z
1
P( Z 1) Q
Q
0.2248
1.75
Z
(d)
SX e
j
m
1
m
1
e jm
2
m
m
1
1
e jm e jm
2
2
m
m 0
1
1
1
1
1
1 e j
1 e j
2
2
3
4
5
cos
4
4.25
(a)
1
1 1
H e j e j e j
3
3 3
1
1 2 cos
3
2
1
H e j 3 4 cos 2 cos 2
9
SY e j 2 H e j
2
2
3 4 cos 2 cos 2
9
2
2 2
2
2 2
2
(k 2)
(k 1) (k )
(k 1) (k 2)
9
9
3
9
9
The autocorrelation function is plotted below.
(b) RYY (k )
2
9
-3 -2
2 2
9
-1
3 2
9
0
2 2
9 2
9
1
2
RYY (k )
3
k
(c) The averages are correlated one with another and the span of the correlation is determined by
the length of the filter. This makes sense, as the averages computed 1 or 2 samples before the
current sample are based, in part, on the same data.
4.26
(a)
1 W W
H e j
otherwise
0
2
1 W W
H e j
otherwise
0
SY e
j
2
0
W W
otherwise
(b)
RYY (k ) 2
sin Wk
k
4.27
(a)
H e j (n 1)a n e jn
H e j
n0
2
SY e j
1
·
1
1 ae
j 2
1
1 ae 1 ae
j
2
j
2
1
1 a 4a(a 1) cos() 2a 2 cos(2)
4
2
2
1 a 4 4a(a 2 1) cos() 2a 2 cos(2)
(b) RYY (k ) 2 h(k ) h( k )
For k 0
RYY (k ) 2 (m 1)a m (m k 1)a m k
m 0
2a 4
a2
1
2
(2 k )
(1 k ) a k
2 3
2 2
2
(1 a )
1 a
(1 a )
For k 0
RYY (k ) 2 (m 1)a m (m k 1)a m k
m k
1
2a 2
2a 4
1
2k
2k
a k (k 1)
a k
a 2 k k (2 k )
2
2 2
2 3
2
1 a
(1 a )
(1 a ) 1 a
ak
2
2
a
1
2k
2k
k
a
a
k
(
2
)
(1
)
2 2
1 a2
(1 a )
4.28
(a)
H e j
H e j
SY e j
1
sin N1
2
e j n
n N1
sin
2
1
sin 2 N1
2
2
sin 2
2
N1
1
sin 2 N1
2
2
sin 2
2
2 2 N1 1 | n | 2 N1 | n |
(b) RYY (n) h(n) h(n)
0
2 N1 | n |
2
4.29
(a)
H e 1
S e
H e j e jn0
j
j
2
2
Y
(b)
RYY (k ) 2 (k )
4.30
hf k T
hf
1
kT
Thus,
hf
2
3
hf 1 hf 1 hf
kT 2 kT 6 kT
hf
1
kT
e kT 1
Substituting gives
hf
SV ( f )
e
hf
kT
hf
hf
kT
hf
hf
1
1 1
kT
kT
5.1
E=
T2
∫ [r (t ) − rˆ(t )]
2
T1
2
K −1
⎡
⎤
dt = ∫ ⎢r (t ) − ∑ xl φl (t )⎥ dt
l =0
⎦
T1 ⎣
T2
2
K −1
∂
⎡
⎤
E = −2 ∫ ⎢r (t ) − ∑ xl φ l (t )⎥φ k (t )dt
∂x k
l =0
⎦
T1 ⎣
T
T2
K −1
T2
T1
l =0
T1
= −2 ∫ r (t )φ k (t )dt + 2∑ xl ∫ φ l (t )φ k (t )dt
δl −k
T2
= −2 ∫ r (t )φ k (t )dt + 2 x k
T1
2
∂
E = 0 ⇒ x k = ∫ r (t )φ k (t )dt
∂x k
T1
T
5.2
E=
T2
∫ [r (t ) − rˆ(t )]
2
T1
2
K −1
⎡
⎤
dt = ∫ ⎢r (t ) − ∑ xl φl (t )⎥ dt
l =0
⎦
T1 ⎣
T2
The optimum xk satisfy
2
K −1
∂
⎡
⎤
0=
E = −2 ∫ ⎢r (t ) − ∑ xl φl (t )⎥φ k (t )dt
∂x k
l =0
⎦
T1 ⎣
T
e (t )
T2
0 = −2 ∫ e(t )φ k (t )dt ⇒ φ k (t ) is orthogonal to e(t )
T1
5.3
The set is orthogonal (but not orthonormal) means
l=k
l≠k
⎧E
∫ φ (t )φ (t )dt = ⎨⎩ 0
T2
k
l
k
T1
The optimum coefficients are derived as follows:
E=
T2
∫ [r (t ) − rˆ(t )]
2
T1
2
K −1
⎡
⎤
dt = ∫ ⎢r (t ) − ∑ xl φl (t )⎥ dt
l =0
⎦
T1 ⎣
T2
2
K −1
∂
⎡
⎤
(
)
E = −2 ∫ ⎢r t − ∑ xl φl (t )⎥φ k (t )dt
∂x k
l =0
⎦
T1 ⎣
T
T2
K −1
T2
T1
l =0
T1
= −2 ∫ r (t )φ k (t )dt + 2∑ xl ∫ φl (t )φ k (t )dt
Ek when l = k
T2
= −2 ∫ r (t )φ k (t )dt + 2 x k E k
T1
T2
∂
1
E = 0 ⇒ xk =
∂x k
Ek
T2
∫ r (t )φ (t )dt =
k
T1
∫ r (t )φ (t )dt
k
T1
T2
∫ φ (t )dt
2
k
T1
5.4 The set is neither orthogonal nor normalized.
E=
T2
∫ [r (t ) − rˆ(t )]
2
T1
2
K −1
⎡
⎤
dt = ∫ ⎢r (t ) − ∑ xl φl (t )⎥ dt
l =0
⎦
T1 ⎣
T2
2
K −1
∂
⎡
⎤
E = −2 ∫ ⎢r (t ) − ∑ xl φl (t )⎥φ k (t )dt
∂x k
l =0
⎦
T1 ⎣
T
T2
K −1
T2
T1
l =0
T1
= −2 ∫ r (t )φ k (t )dt + 2∑ xl ∫ φl (t )φ k (t )dt
2
K −1
∂
E = 0 ⇒ ∫ r (t )φ k (t )dt = ∑ xl ∫ φl (t )φ k (t )dt
∂x k
l =0
T1
T1
T2
T
k = 0,1,… , K − 1
This defines a set of K equations in the K unknowns xk (for k = 0,1, …, K − 1 ). To see this, Let
T2
T2
T1
T1
Pk = ∫ r (t )φ k (t )dt and Rk ,l = ∫ φl (t )φ k (t )dt for k = 0,1, …, K − 1 .
With these definitions, the system of equations is
K −1
∂
E = 0 ⇒ Pk = ∑ x k Rl ,k for k = 0,1, … , K − 1 .
∂x k
l =0
As is customary with systems of equations, this system of equations may also be expressed in
matrix form as follows:
⎡ P0 ⎤ ⎡ R0,0
⎢ P ⎥ ⎢ R
⎢ 1 ⎥ = ⎢ 1,0
⎢
⎥ ⎢
⎢
⎥ ⎢
⎣ PK −1 ⎦ ⎣ RK −1, 0
P
R0, K −1 ⎤ ⎡ x0 ⎤
R1, K −1 ⎥⎥ ⎢ x1 ⎥
⎢
⎥
⎥⎢
⎥
⎥⎢
⎥
RK −1, K −1 ⎦ ⎣ x K −1 ⎦
R0,1
R1,1
RK −1,1
R
x
The matrix equation is P = Rx. The solution for the optimum x’s is
x = R −1 P .
Note that if the set of basis functions is orthogonal, then R is the diagonal matrix
⎡ E0
⎢
R=⎢
⎢
⎢
⎣
E1
⎤
⎥
⎥.
⎥
⎥
E K −1 ⎦
In this case, the solution for the optimum x’s is trival because
R
−1
⎡1
⎢E
⎢ 0
⎢
=⎢
⎢
⎢
⎢
⎢⎣
1
E1
⎤
⎥
⎥
⎥
⎥
⎥
⎥
1 ⎥
E K −1 ⎥⎦
T2
where E k = ∫ φ k2 (t )dt for k = 0,1, …, K − 1 . The solution for xk is
T1
1
Pk
Ek
which is the solution to Exercise 5.3. Furthermore, when the set of basis functions is
orthonormal, Ek = 1 for k = 0,1, …, K − 1 . As a consequence, R = I, so that R −1 = I . The solution
for xk is
xk =
x k = Pk
which is the solution to Exercise 5.1.
T2
5.5 Let E k = ∫ φ k2 (t )dt for k = 0,1, …, K − 1 and note that
T1
⎧E
∫ φ (t )φ (t )dt = ⎨⎩ 0
T2
k
l
T1
k
l=k
.
l≠k
Now the energy in s (t ) is
K −1
⎡ K −1
⎤
E s = ∫ s (t )dt = ∫ ⎢∑ a k φ k (t )∑ a l φ l (t )⎥ dt
l =0
⎦
T1
T1 ⎣ k = 0
T2
T2
2
K −1 K −1
T2
k =0 l =0
T1
= ∑∑ a k al ∫ φ k (t )φl (t )dt
K −1
= ∑ a k2 E k
k =0
The interpretation is that Es is a scaled squared Euclidean distance from the origin of the signal
space. Alternatively, the expression can be interpreted as the squared Euclidean distance in a
Cartesian coordinate system whose axes do not have the same length.
5.6
+A
1
t
1
−A
−A
+A
t
5.7
+A
1
t
(0,+ A)
−A
+A
1
(+ A,0)
t
5.8
+2A
½
+2A
1
t
t
−2A
−2A
(− A,+ A)
(− A,− A)
+2A
−2A
½
1
½
1
(+ A,+ A)
(+ A,− A)
+2A
½
t
1
t
−2A
5.9
+A
1
t
−A
(0,+ A)
+A
1
t
-A
(− A,0)
(+ A,0)
(0,− A)
+A
t
−A
1
1
t
5.10
½
−
3 −1
A
2
−
3 +1
A
2
1
t
3 +1
A
2
+
3 −1
A
2
t
½
⎛ A 3 A⎞
⎜−
,+ ⎟⎟
⎜
2
2⎠
⎝
⎛ A 3 A⎞
⎜+
,+ ⎟⎟
⎜
2
2⎠
⎝
(0,0)
+2A
(0,− A)
t
−2A
+
½
1
+A
t
−A
1
1
−4A
t
−2A
−4A
−4A
1
t
−2A
½
1
−2A
−2A
−4A
½
½
t
(− A,+ A)
(− 3 A,+ A)
t
(− A,+ A)
1
1
(− 3 A,+ A)
½
+4A
+2A
½
(+ A,+ A) (+ 3 A,+ A)
(+ A,+ A) (+ 3 A,+ A)
½
1
1
t
+4A
+2A
+2A
+2A
t
+4A
+4A
½
½
1
1
t
t
5.11
5.12
1
x 0 = ∫ r (t )φ 0 (t )dt =
0
1
2
1
x1 = ∫ r (t )φ1 (t )dt = 0
0
1
rˆ(t ) = x 0φ 0 (t ) + x1φ1 (t ) = φ 0 (t )
2
r (t )
r̂ (t )
+1
+½
t
½
1
5.13
1
x 0 = ∫ r (t )φ 0 (t )dt =
0
1
2
1
x1 = ∫ r (t )φ1 (t )dt = 0
0
1
rˆ(t ) = x 0φ 0 (t ) + x1φ1 (t ) = φ 0 (t )
2
r (t )
+1
r̂ (t )
+½
t
½
1
5.14
1
x 0 = ∫ r (t )φ 0 (t )dt =0
0
1
x1 = ∫ r (t )φ1 (t )dt = −
0
1
2
1
rˆ(t ) = x 0φ 0 (t ) + x1φ1 (t ) = − φ1 (t )
2
r (t )
+1
+½
−½
−1
r̂ (t )
½
t
1
5.15
1
x 0 = ∫ r (t )φ 0 (t )dt =0
0
1
x1 = ∫ r (t )φ1 (t )dt =
0
1
2
1
rˆ(t ) = x0φ 0 (t ) + x1φ1 (t ) = φ1 (t )
2
r (t )
+1
+½
−½
−1
1
½
t
r̂ (t )
5.16
(a)
3φ0 (t ) + φ1 (t ) →
+4
+2
t
½
1
(b)
1
x 0 = ∫ r (t )φ 0 (t )dt =3
0
1
x1 = ∫ r (t )φ1 (t )dt =1
0
rˆ(t ) = x 0φ 0 (t ) + x1φ1 (t ) = 3φ 0 (t ) + φ1 (t )
r̂ (t )
r (t )
+4
+2
t
½
1
(c) The best approximation for r (t ) in Span{φ 0 (t ), φ1 (t )} is 3φ 0 (t ) + φ1 (t ) .
5.17
½
+A 2
1
t
½
−A 2
−A
+A
1
t
5.18
+A 2
t
½
1
(0,+ A)
+A 2
(+ A,0)
½
1
t
5.19
+A 2
−A 2
+A 2
½
1
t
t
(− A,+ A)
(− A,− A)
1
−A 2
½
1
½
1
(+ A,+ A)
(+ A,− A)
+A 2
t
½
−A 2
t
5.20
+A 2
t
−A 2
½
1
(0,+ A)
+A 2
−A 2
+A 2
½
1
t
(− A,0)
(+ A,0)
(0,− A)
+A 2
½
1
t
−A 2
t
−A 2
½
1
5.21
3
A
2
1
+
A
2
+
+
−
1
A
2
½
t
1
3
A
2
⎛ A 3 A⎞
⎜+
,+ ⎟⎟
⎜
2
2⎠
⎝
⎛ A 3 A⎞
⎜−
,+ ⎟⎟
⎜
2
2⎠
⎝
(0,0)
+2A
t
−2A
½
t
½
1
(0,− A)
½
−A 2
1
t
1
− A3 2
−A 2
− A3 2
−A 2
A 2
½
½
1
1
t
t
−A 2
−A 2
A 2
½
t
(− A,+ A)
(− 3 A,+ A)
t
(− A,+ A)
1
1
(− 3 A,+ A)
½
½
−A 2
A 2
½
(+ A,+ A) (+ 3 A,+ A)
(+ A,+ A) (+ 3 A,+ A)
A 2
A3 2
1
1
t
t
−A 2
A 2
A3 2
A 2
A3 2
½
½
1
1
t
t
5.22
5.23
(a)
φ 0(t)
1
0
-1
0
0.1
0.2
0.3
0.4
0.5
t
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
φ 1(t)
1
0
-1
dimension K = 2
(b)
2
1.5
s1
1
φ1
0.5
0
s2
s0
-0.5
-1
s3
-1.5
-2
-2
-1.5
-1
-0.5
0
φ0
0.5
1
1.5
2
5.24
(a)
φ 0(t)
1
0
-1
0
0.1
0.2
0.3
0.4
0.5
t
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
φ 1(t)
1
0
-1
dimension: K =2
(b)
2
1.5
s2
1
φ1
0.5
0
s1
s3
-0.5
-1
s0
-1.5
-2
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
φ0
The basis functions here are the same as those for Exercise 5.23, but the constellation is a reordered version of the constellation in Exercise 5.23.
5.25
(a)
1.5
1
φ 0(t)
0.5
0
-0.5
-1
-1.5
0
0.1
0.2
0.3
0.4
0.5
t
0.6
0.7
0.8
0.9
dimension: K = 1
(b)
8
6
4
2
0
s3
s2
s1
s0
-2
-4
-6
-8
-8
-6
-4
-2
0
φ0
2
4
6
8
1
5.26
(a)
φ 0(t)
1
0
-1
0
0.1
0.2
0.3
0.4
0.5
t
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
φ 1(t)
1
0
-1
dimension: K = 2
(b)
s2
2
1.5
s3
1
φ1
0.5
s0
0
s1
-0.5
-1
-1.5
-2
-2
-1.5
-1
-0.5
0
φ0
0.5
1
1.5
2
5.27
(a)
φ 0(t)
1
0
-1
0
0.5
1
1.5
t
2
2.5
3
0
0.5
1
1.5
t
2
2.5
3
0
0.5
1
1.5
t
2
2.5
3
φ 1(t)
1
0
-1
φ 2(t)
1
0
-1
dimension K = 3
(b)
5
s3
φ2
s2
0
s0
s1
-5
5
5
0
φ1
0
-5
-5
φ0
5.28
(a)
φ 0(t)
1
0
-1
0
0.5
1
1.5
t
2
2.5
3
0
0.5
1
1.5
t
2
2.5
3
φ 1(t)
1
0
-1
dimension: K = 2
(b)
2.5
2
1.5
s1
s2
1
φ1
0.5
s0
0
-0.5
-1
-1.5
-2
-2.5
-2
-1
0
φ0
1
2
5.29
The average energy may be expressed as
E avg =
M
−1
2
2
A
M
∑ (2m + 1)
m=−
2
=
M
2
2A
M
M
−1
2 2
∑ (2m + 1)
m =0
We will need the following identities:
n
∑i = n
i =0
n
∑i =
i =0
n
∑i
2
i =0
=
n(n + 1)
2
n(n + 1)(2n + 1)
6
Using these identities, we have
M
−1
2
∑ (2m + 1)
m =0
2
M
−1
2
M
−1
2
M
−1
2
m =0
m =0
m =0
= 4 ∑ m2 + 4∑ m +
∑1
⎛M
⎞⎛ M ⎞
⎛M
⎞⎛ M
⎜ − 1⎟⎜ ⎟(M − 2 + 1)
⎜ − 1⎟⎜
2
2
⎠⎝ 2 ⎠
⎠⎝ 2
= 4⎝
+4⎝
6
2
3
M M −1
=
2
3
Putting this all together we have
E avg =
2 A2 M M 3 − 1 M 3 − 1 2
A
=
M 2
3
3
⎞
⎟
⎠+M
2
2
.
5.30
(a) A = 0.05.
LUT
0
1
bits
-0.05
+0.05
↑N
p(nT)
ADC
(b)
-3
2
x 10
1.5
1
s(t)
0.5
0
-0.5
-1
-1.5
-2
0
1
2
3
4
t/Ts
5
6
7
8
(c)
x(t)
r(t)
DAC
N samples/bit
x(kT)
p(−nT)
decision
n = kN
(d)
k
0
1
2
3
----------------------------------x(k) +0.00
-0.01
+0.33
+0.07
----------------------------------a(k) +0.05
-0.05
+0.05
+0.05
----------------------------------bits
1
0
1
1
s(t)
5.31
(a) A = 0.03.
LUT
00
01
10
11
bits
-0.09
-0.03
+0.09
+0.03
↑N
p(nT)
ADC
(b)
-3
2.5
x 10
2
1.5
s(t)
1
0.5
0
-0.5
-1
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
(c)
r(t)
DAC
N samples/symbol
p(−nT)
decision
n = kN
(d)
k
0
1
2
3
----------------------------------x(k) +0.07
+0.06
-0.03
-0.10
----------------------------------a(k) +0.09
+0.03
-0.03
-0.09
----------------------------------bits
10
11
01
00
s(t)
5.32
(a) A = 0.125
000
001
010
011
100
101
110
111
bits
LUT
-0.875
-0.625
-0.125
-0.375
+0.875
+0.625
+0.125
+0.375
↑N
p(nT)
ADC
(b)
0.02
0.018
0.016
0.014
s(t)
0.012
0.01
0.008
0.006
0.004
0.002
0
0.5
1
1.5
t/Ts
2
2.5
3
(c)
r(t)
DAC
N samples/symbol
p(−nT)
decision
n = kN
(d)
k
0
1
2
3
--------------------------------------x(k) +0.601
-0.101
+0.355
-0.777
--------------------------------------a(k) +0.625
-0.125
+0.375
-0.875
--------------------------------------bits
101
010
111
000
s(t)
5.33
2
1
0
-1
-2
-0.5
-0.4
-0.3
-0.2
-0.1
0
t/Ts
0.1
0.2
0.3
0.4
0.5
-0.4
-0.3
-0.2
-0.1
0
t/Ts
0.1
0.2
0.3
0.4
0.5
-0.4
-0.3
-0.2
-0.1
0
t/Ts
0.1
0.2
0.3
0.4
0.5
2
1
0
-1
-2
-0.5
2
1
0
-1
-2
-0.5
(d) All eye diagrams exhibit no ISI at the optimum sampling instant. The peak overshoot
increases as the excess bandwidth decreases. The width of the eye opening (and hence, the
immunity to timing offset) decreases as the excess bandwidth decreases.
5.34
4
2
0
-2
-4
-0.5
-0.4
-0.3
-0.2
-0.1
0
t/Ts
0.1
0.2
0.3
0.4
0.5
-0.4
-0.3
-0.2
-0.1
0
t/Ts
0.1
0.2
0.3
0.4
0.5
-0.4
-0.3
-0.2
-0.1
0
t/Ts
0.1
0.2
0.3
0.4
0.5
5
0
-5
-0.5
5
0
-5
-0.5
(d) All eye diagrams exhibit no ISI at the optimum sampling instant. The peak overshoot
increases as the excess bandwidth decreases. The width of the eye opening (and hence, the
immunity to timing offset) decreases as the excess bandwidth decreases.
5.35
magnitude (dB)
part (a)
0
signal
filter
-20
-40
-5
0
frequency (cycles/symbol)
5
2
0
-2
-1
-0.5
0
t/Ts
0.5
1
magnitude (dB)
part (b)
0
signal
filter
-20
-40
-5
0
frequency (cycles/symbol)
5
2
0
-2
-1
-0.5
0
t/Ts
0.5
1
magnitude (dB)
part (c)
0
signal
filter
-20
-40
-5
0
frequency (cycles/symbol)
5
2
0
-2
-1
-0.5
0
t/Ts
0.5
1
(d) As the bandwidth of the channel decreases, signal distortion increases. The signal distortion
manifests itself in the eye diagram as a narrowing of the eye opening and closure of the eye due
to ISI.
5.36
magnitude (dB)
part (a)
0
signal
filter
-20
-40
-5
0
frequency (cycles/symbol)
5
1
0
-1
-1
-0.5
0
t/Ts
0.5
1
magnitude (dB)
part (b)
0
signal
filter
-20
-40
-5
0
frequency (cycles/symbol)
5
2
0
-2
-1
-0.5
0
t/Ts
0.5
1
magnitude (dB)
part (c)
0
signal
filter
-20
-40
-5
0
frequency (cycles/symbol)
5
2
0
-2
-1
-0.5
0
t/Ts
0.5
1
(d) As the order of the filter increases, an increasing amount of energy is removed from the
signal. This increases distortion seen as eye narrowing and eye closure due to ISI.
5.37
(a) ω 0τ =
(b)
π
2
π
2ω 0
⇒ τ =
π
, then the output of the phase shifter is
2ω0
cos(ω x (t − τ )) = cos(ω x t − ω xτ ) = cos(ω x t − θ )
If the LO is operating at ω x rad/s with τ =
Using ω x = ω 0 ± ∆ω , the phase shift may be expressed as
θ = ω xτ =
Now, 1 =
π ω 0 ± ∆ω π ⎛ ∆ω ⎞
⎟
= ⎜⎜1 ±
ω 0 ⎟⎠
2 ω0
2⎝
π
180
rad. So we want
π
2
−
π
180
≤θ ≤
π
2
+
π
180
which implies
π⎛
π⎛
1 ⎞
1 ⎞
⎜ 1 − ⎟ ≤ θ ≤ ⎜1 + ⎟
2 ⎝ 90 ⎠
2 ⎝ 90 ⎠
Case 1: θ =
∆ω ⎞
⎟
⎜⎜1 +
2⎝
ω 0 ⎟⎠
π⎛
ω
ω
1 ⎞ π ⎛ ∆ω ⎞ π ⎛
1
⎟⎟ ≤ ⎜1 + ⎞⎟ ⇒ − 0 ≤ ∆ω ≤ 0
⎜1 − ⎟ ≤ ⎜⎜1 +
ω 0 ⎠ 2 ⎝ 90 ⎠
2 ⎝ 90 ⎠ 2 ⎝
90
90
π⎛
Case 2: θ =
∆ω ⎞
⎟
⎜⎜1 −
2⎝
ω 0 ⎟⎠
π⎛
ω
ω
1 ⎞ π ⎛ ∆ω ⎞ π ⎛
1 ⎞
⎟⎟ ≤ ⎜1 + ⎟ ⇒ − 0 ≤ ∆ω ≤ 0
⎜1 − ⎟ ≤ ⎜⎜1 −
ω 0 ⎠ 2 ⎝ 90 ⎠
2 ⎝ 90 ⎠ 2 ⎝
90
90
π⎛
5.38
(a)
ω 0τ =
π
2
⇒ τ =
π
2ω 0
π⎞
⎛
The output of the phase shifter is sin ⎜ ω 0 t − ⎟ = − cos(ω 0 t )
2⎠
⎝
(b)
The system outputs are
− cos(ω0t )
sin (ω0t )
but − cos(ω 0 t ) = cos(ω 0 t + π ) and sin (ω 0 t ) = − sin (ω 0 t + π ) . So the system output may be
expressed as
cos(ω0t + θ )
− sin (ω0t + θ )
with θ = π. This system does indeed have the desired relationship.
5.39
(a) θ =
3π
2
(b) ω 0τ =
3π
2
⇒ τ =
3π
2ω 0
(c)
The delay is 3 times the delay from Exercise 5.37. Furthermore, the tunable range over
which the phase error does not exceed 1° is less than that for Exercise 5.37.
5.40
(a) A = √2
ILUT
00
01
10
11
I(t)
-1
-1
+1
+1
↑N
p(nT)
2 cos(Ω 0 n )
bits
s(t)
ADC
− 2 sin (Ω 0 n )
QLUT
00
01
10
11
-1
+1
-1
+1
↑N
p(nT)
Q(t)
(b)
I(t)
0.05
0
-0.05
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
Q(t)
0.05
0
-0.05
s(t)
0.05
0
-0.05
(c)
x(t)
x(kT)
p(−nT)
n = kN
2 cos(Ω 0 n )
r(t)
decision
DAC
N samples/symbol
− 2 sin (Ω 0 n )
y(t)
y(kT)
p(−nT)
n = kN
(d)
1.5
1
0.5
0
-0.5
-1
-1.5
-1.5
-1
-0.5
0
0.5
1
1.5
k
0
1
2
3
----------------------------------x(k) -1.01
+1.02
+1.11
-0.03
y(k) +1.07
-0.99
+1.00
+0.07
----------------------------------a0(k) -1.00
+1.00
+1.00
-1.00
a1(k) +1.00
-1.00
+1.00
+1.00
----------------------------------bits
01
10
11
01
5.41
(a) A = 0.05 = 0.2236
LUT
00
01
10
11
bits
-0.6708
-0.2236
+0.2708
+0.2236
↑N
p(nT)
s(t)
ADC
2 cos(Ω 0 n )
(b)
0.02
I(t)
0.01
0
-0.01
-0.02
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
0.04
s(t)
0.02
0
-0.02
-0.04
(c)
x(t)
r(t)
DAC
N samples/symbol 2 cos(Ω n )
0
x(kT)
p(−nT)
decision
n = kN
(d)
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-3
-2
-1
0
1
2
k
0
1
2
3
----------------------------------x(k) -0.03
+0.55
+1.53
-2.10
----------------------------------a(k) -0.22
+0.67
+0.67
-0.67
----------------------------------bits
01
10
10
00
5.42
(a) A = 2
ILUT
00
01
10
11
I(t)
0
-√3
+√3
0
↑N
p(nT)
2 cos(Ω 0 n )
bits
s(t)
ADC
− 2 sin (Ω 0 n )
QLUT
00
01
10
11
0
+1
+1
−2
↑N
p(nT)
Q(t)
(b)
I(t)
0.05
0
-0.05
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
Q(t)
0.05
0
-0.05
s(t)
0.1
0
-0.1
(c)
x(t)
x(kT)
p(−nT)
n = kN
2 cos(Ω 0 n )
r(t)
decision
DAC
N samples/symbol
− 2 sin (Ω 0 n )
y(t)
y(kT)
p(−nT)
n = kN
(d)
2.5
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-2.5
-2
-1
0
1
2
k
0
1
2
3
----------------------------------x(k) -0.03
-0.01
+0.01
+0.03
y(k) +1.97
+1.97
+1.97
+1.97
----------------------------------a0(k) -1.73
+0.00
+0.00
+1.73
a1(k) +1.00
+0.00
+0.00
+1.00
----------------------------------bits
01
00
00
10
5.43
(a) A = 2
ILUT
000
001
010
011
100
101
110
111
+√2
+1
−1
0
+1
0
-√2
−1
I(t)
↑N
p(nT)
2 cos(Ω 0 n )
bits
s(t)
ADC
QLUT
000
001
010
011
100
101
110
111
− 2 sin (Ω 0 n )
0
+1
+1
+√2
−1
-√2
0
−1
↑N
p(nT)
Q(t)
(b)
I(t)
0.05
0
-0.05
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
Q(t)
0.05
0
-0.05
s(t)
0.05
0
-0.05
(c)
x(t)
x(kT)
p(−nT)
n = kN
2 cos(Ω 0 n )
r(t)
decision
DAC
N samples/symbol
− 2 sin (Ω 0 n )
y(t)
y(kT)
p(−nT)
n = kN
(d)
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-2
-1
0
1
2
k
0
1
2
3
----------------------------------x(k) -1.30
+1.51
-0.09
+0.65
y(k) +1.64
-1.49
-0.91
+0.07
----------------------------------a0(k) -1.00
+1.00
+0.00
+1.41
a1(k) +1.00
-1.00
-1.41
+0.00
----------------------------------bits
010
100
101
000
5.44
(a) A = 2
ILUT
000
001
010
011
100
101
110
111
+2√2
0
+2
+2√2
−2
-2√2
0
-2√2
I(t)
↑N
p(nT)
2 cos(Ω 0 n )
bits
s(t)
ADC
QLUT
000
001
010
011
100
101
110
111
− 2 sin (Ω 0 n )
+2√2
+2
0
−2√2
0
+2√2
−2
−2√2
↑N
p(nT)
Q(t)
(b)
I(t)
0.1
0
-0.1
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
Q(t)
0.1
0
-0.1
s(t)
0.2
0
-0.2
(c)
x(t)
x(kT)
p(−nT)
n = kN
2 cos(Ω 0 n )
r(t)
decision
DAC
N samples/symbol
− 2 sin (Ω 0 n )
y(t)
y(kT)
p(−nT)
n = kN
(d)
8
6
4
2
0
-2
-4
-6
-8
-5
0
5
k
0
1
2
3
----------------------------------x(k) -2.00
+1.00
+7.20
-1.64
y(k) -2.01
+1.64
+0.00
+1.25
----------------------------------a0(k) -2.83
+0.00
+2.00
-2.00
a1(k) -2.83
+2.00
+0.00
+0.00
----------------------------------bits
111
001
010
100
5.45
(a) A = 2
ILUT
000
001
010
011
100
101
110
111
+2
-1
+1
+2
-1
-2
+1
−2
I(t)
↑N
p(nT)
2 cos(Ω 0 n )
bits
s(t)
ADC
QLUT
000
001
010
011
100
101
110
111
− 2 sin (Ω 0 n )
+2
+1
+1
-2
−1
+2
-1
−2
↑N
p(nT)
Q(t)
(b)
I(t)
0.05
0
-0.05
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
Q(t)
0.05
0
-0.05
s(t)
0.1
0
-0.1
(c)
x(t)
x(kT)
p(−nT)
n = kN
2 cos(Ω 0 n )
r(t)
decision
DAC
− 2 sin (Ω 0 n )
N samples/symbol
y(t)
y(kT)
p(−nT)
n = kN
(d)
3
2
1
0
-1
-2
-3
-3
-2
-1
0
1
2
3
k
0
1
2
3
----------------------------------x(k) -1.32
+1.51
+0.71
-0.05
y(k) -1.32
-1.51
+1.32
+0.05
----------------------------------a0(k) -1.00
+2.00
+1.00
-1.00
a1(k) -1.00
-2.00
+1.00
+1.00
----------------------------------bits
100
011
010
001
5.46
(a) A = 0.3
ILUT
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
-0.9
-0.9
-0.9
-0.9
-0.3
-0.3
-0.3
−0.3
+0.9
+0.9
+0.9
+0.9
+0.3
+0.3
+0.3
+0.3
I(t)
↑N
p(nT)
2 cos(Ω 0 n )
bits
ADC
QLUT
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
+0.9
+0.3
-0.9
-0.3
+0.9
+0.3
-0.9
−0.3
+0.9
+0.3
-0.9
-0.3
+0.9
+0.3
-0.9
-0.3
− 2 sin (Ω 0 n )
↑N
p(nT)
Q(t)
s(t)
(b)
I(t)
0.02
0
-0.02
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
Q(t)
0.02
0
-0.02
s(t)
0.05
0
-0.05
(c)
x(t)
x(kT)
p(−nT)
n = kN
2 cos(Ω 0 n )
r(t)
decision
DAC
N samples/symbol
− 2 sin (Ω 0 n )
y(t)
y(kT)
p(−nT)
n = kN
(d)
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-1
-0.5
0
0.5
1
k
0
1
2
3
----------------------------------x(k) -0.03
+0.51
-0.09
+0.65
y(k) +0.64
-0.49
-0.91
+0.07
----------------------------------a0(k) -0.30
+0.30
-0.30
+0.90
a1(k) +0.90
-0.30
-0.90
+0.30
----------------------------------bits
0100
1111
0110
1001
5.47
(a) A = 0.3
ILUT
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
0
-0.9
0
+0.9
-0.3
+0.3
-1.5
-0.9
+0.9
+1.5
-0.3
+0.3
0
-0.9
+0.9
0
I(t)
↑N
p(nT)
2 cos(Ω 0 n )
bits
ADC
QLUT
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
+1.5
+0.9
+0.9
+0.9
+0.3
+0.3
0
0
0
0
-0.3
-0.3
-0.9
-0.9
-0.9
-1.5
− 2 sin (Ω 0 n )
↑N
p(nT)
Q(t)
s(t)
(b)
I(t)
0.05
0
-0.05
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
0
0.5
1
1.5
2
t/Ts
2.5
3
3.5
4
Q(t)
0.05
0
-0.05
s(t)
0.1
0
-0.1
(c)
x(t)
x(kT)
p(−nT)
n = kN
2 cos(Ω 0 n )
r(t)
decision
DAC
N samples/symbol
− 2 sin (Ω 0 n )
y(t)
y(kT)
p(−nT)
n = kN
(d)
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-2
-1
0
1
2
k
0
1
2
3
----------------------------------x(k) +0.90
-1.95
+1.90
+0.05
y(k) +0.54
+0.01
-1.91
-0.94
----------------------------------a0(k) +0.90
-1.50
+0.90
+0.00
a1(k) +0.90
+0.00
-0.90
-0.90
----------------------------------bits
0011
0110
1110
1100
5.48
5.49
5.50
5.51
5.52
5.53
E avg
E avg
4r12 + 12r22 r12 + 3r22
=
=
16
4
2
2
2
4r + 12r2 + 16r3
r 2 + 3r22 + 4r32
= 1
= 1
32
8
5.54
r2
= 2.7236
r1
∆θ = 0
maximum d min = 0.1462
3
2
1
0
-1
-2
-3
-3
-2
-1
0
1
2
3
5.55
r2
= 2.8242
r1
r3
= 4.2143
r1
∆θ 1 = 0
∆θ 2 = 0
maximum d min = 0.073658
5
4
3
2
1
0
-1
-2
-3
-4
-5
-5
0
5
5.56
E avg
E avg
(
) (
) (
(
) (
)
) (
)
4 2 A 2 + 4 9 A 2 + 4 25 A 2 + 4 18 A 2
27 2
=
=
A
16
2
4 2 A2 + 4 9 A2
11
=
= A2
8
2
5.57
r1 = 2 A
θ1 =
π
4
θ2 = 0
π
r3 = 3 2 A θ 3 =
r2 = 3 A
r1 = 5 A
4
θ2 = 0
r1 = 2 A θ1 =
r2 = 3 A
π
4
θ2 = 0
5.58
5.59
5.60
5.61
(a)
1.5
(1,1)
1
(1,1)
0.5
0
-0.5
-1
(-1,-1)
-1.5
-1.5
-1
(1,-1)
-0.5
0
0.5
1
1.5
(b)
k
0
1
2
3
-------------------------------------------x(kTs)
+1.0
-1.0
+1.0
+1.0
y((k+0.5)Ts) -1.0
-1.0
+1.0
+1.0
-------------------------------------------a0(k)
+1.0
-1.0
+1.0
+1.0
a1(k)
-1.0
-1.0
+1.0
+1.0
-------------------------------------------bits
10
00
11
11
5.62
For each case, the bit rate is Rb =
1
N M k where k is the number of data bits per subcarrier.
TM
(a) QPSK (k = 2)
Rb = 100000
symbols
subcarriers
bits
× 64
×2
= 12.8 Mbits/s
sec
symbol
subcarrier
(b) 16-QAM (k = 4)
Rb = 100000
symbols
subcarriers
bits
× 64
×4
= 25.6 Mbits/s
sec
symbol
subcarrier
(c) 64-QAM with rate-2/3 code: k = 6 ×
Rb = 100000
2
=4
3
symbols
subcarriers
bits
× 64
×4
= 25.6 Mbits/s
sec
symbol
subcarrier
5.63
Rb =
NM
1
NM k
TM
Rb
20 × 10 6
=
=
= 25
1
200000 × 4
k
TM
5.64
Assign a point from a constellation with a small number of points (e.g., BPSK, QPSK) to those
subcarriers that are attenuated a lot by the channel. Assign a point from a constellation with a
large number of points (e.g., 16-QAM, 64-QAM) to those subcarriers that are not attenuated a lot
by the channel. The selection criterion would be the maximum allowed bit error rate.
10.1
(a)
I (t ) 2 cos(ω 0 t ) − Q(t ) 2 (1 + ε )sin (ω 0 t + φ )
= I (t ) 2 cos(ω 0 t ) − Q(t ) 2 (1 + ε )[sin (ω 0 t ) cos(φ ) + cos(ω 0 t )sin (φ )]
= I (t ) 2 cos(ω 0 t ) − Q(t ) 2 (1 + ε ) cos(φ )sin (ω 0 t ) − Q(t ) 2 (1 + ε )sin (φ ) cos(ω 0 t )
= [I (t ) − Q(t )(1 + ε )sin (φ )] 2 cos(ω 0 t ) − Q(t )(1 + ε ) cos(φ ) 2 sin (ω 0 t )
~
~
I (t )
Q (t )
Putting the relationships in matrix form gives the desired result.
(b)
0.5
0.4
0.3
0.2
Q(nT)
0.1
0
-0.1
-0.2
-0.3
-0.4
-0.5
-0.5
0
I(nT)
0.5
(c)
0.5
0.4
0.3
0.2
Q(nT)
0.1
0
-0.1
-0.2
-0.3
-0.4
-0.5
-0.5
0
I(nT)
0.5
(d) The phase trajectory is skewed, the same way a circle is skewed to create an ellipse.
(e)
ε = 0 φ = 20 deg
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
Q(nT)
Q(nT)
ε = 0.1 φ = 0 deg
0
0
-0.1
-0.1
-0.2
-0.2
-0.3
-0.3
-0.4
-0.4
-0.5
-0.5
-0.5
0
I(nT)
0.5
-0.5
0
I(nT)
0.5
The two plots above illustrate the effects: The amplitude imbalance stretches the axis. The phase
imbalance rotates the Q-axis but not the I-axis creating the pronounced skewing effect observed
in part (c).
10.2
(a)
I (t ) 2 cos(ω 0 t ) − Q(t ) 2 (1 + ε )sin (ω 0 t + φ )
= I (t ) 2 cos(ω 0 t ) − Q(t ) 2 (1 + ε )[sin (ω 0 t ) cos(φ ) + cos(ω 0 t )sin (φ )]
= I (t ) 2 cos(ω 0 t ) − Q(t ) 2 (1 + ε ) cos(φ )sin (ω 0 t ) − Q(t ) 2 (1 + ε )sin (φ ) cos(ω 0 t )
= [I (t ) − Q(t )(1 + ε )sin (φ )] 2 cos(ω 0 t ) − Q(t )(1 + ε ) cos(φ ) 2 sin (ω 0 t )
~
~
I (t )
Q (t )
Putting the relationships in matrix form gives the desired result.
(b)
2
1.5
1
Q(nT)
0.5
0
-0.5
-1
-1.5
-2
-2
-1
0
I(nT)
1
2
(c)
2
1.5
1
Q(nT)
0.5
0
-0.5
-1
-1.5
-2
-2
-1
0
I(nT)
1
2
(d) The phase trajectory is skewed, the same way a circle is skewed to create an ellipse.
(e)
ε = 0 φ = 20 deg
2
1.5
1.5
1
1
0.5
0.5
Q(nT)
Q(nT)
ε = 0.1 φ = 0 deg
2
0
0
-0.5
-0.5
-1
-1
-1.5
-1.5
-2
-2
-1
0
I(nT)
1
2
-2
-2
-1
0
I(nT)
1
2
The two plots above illustrate the effects: The amplitude imbalance stretches the axis. The phase
imbalance rotates the Q-axis but not the I-axis creating the pronounced skewing effect observed
in part (c).
10.3
(a)
I (t ) 2 cos(ω 0 t ) − Q(t ) 2 (1 + ε )sin (ω 0 t + φ )
= I (t ) 2 cos(ω 0 t ) − Q(t ) 2 (1 + ε )[sin (ω 0 t ) cos(φ ) + cos(ω 0 t )sin (φ )]
= I (t ) 2 cos(ω 0 t ) − Q(t ) 2 (1 + ε ) cos(φ )sin (ω 0 t ) − Q(t ) 2 (1 + ε )sin (φ ) cos(ω 0 t )
= [I (t ) − Q(t )(1 + ε )sin (φ )] 2 cos(ω 0 t ) − Q(t )(1 + ε ) cos(φ ) 2 sin (ω 0 t )
~
~
I (t )
~
~
= I (t ) 2 cos(ω 0 t ) − Q (t ) 2 sin (ω 0 t )
Q (t )
~
~
The outputs of the matched filters are based on I (t ) and Q (t ) . Thus we have
~
~
x (t ) = I (t ) ∗ p(− t )
= [I (t ) − Q(t )(1 + ε )sin (φ )] ∗ p(− t )
= I (t ) ∗ p(− t ) − (1 + ε )sin (φ )Q(t ) ∗ p(− t )
= x(t ) − (1 + ε )sin (φ ) y (t )
~
~
y (t ) = Q (t ) ∗ p(− t )
= (1 + ε ) cos(φ )Q(t ) ∗ p(− t )
= (1 + ε ) cos(φ ) y (t )
Organizing these expressions into a matrix equation produces the desired result.
(b) Using x(t ) = ∑ a0 (m )rp (t − mT ) and y (t ) = ∑ a1 (m )rp (t − mT ) as substitutions in the
m
m
equations from part (a), we have
~
x (t ) = ∑ [a 0 (m ) − (1 + ε )sin (φ )a1 (m )]rp (t − mT )
m
~
y (t ) = ∑ (1 + ε ) cos(φ )a1 (m )rp (t − mT )
m
from which we obtain
~
x (kT ) = ∑ [a 0 (m ) − (1 + ε )sin (φ )a1 (m )]rp (kT − mT ) = a0 (k ) − (1 + ε )sin (φ )a1 (k )
m
~
y (kT ) = ∑ (1 + ε ) cos(φ )a1 (m )rp (kT − mT ) = (1 + ε ) cos(φ )a1 (k )
m
Organizing these expressions into a matrix produces
x (kT )⎤ ⎡1 − (1 + ε )sin (φ )⎤ ⎡a 0 (k )⎤
⎡~
⎥
⎢~
⎥=⎢
⎥⎢
⎣ y (kT )⎦ ⎣0 (1 + ε ) cos(φ ) ⎦ ⎣ a1 (k )⎦
(c)
1
y(kT)
0.5
0
-0.5
-1
-1
-0.5
0
x(kT)
0.5
1
-1
-0.5
0
x(kT)
0.5
1
(d)
1
y(kT)
0.5
0
-0.5
-1
(e) The phase trajectory is skewed, the same way a circle is skewed to create an ellipse.
(f)
ε = 0.1 φ = 0 deg
1
y(kT)
0.5
0
-0.5
-1
-1
-0.5
0
x(kT)
0.5
1
0.5
1
ε = 0 φ = 20 deg
1
y(kT)
0.5
0
-0.5
-1
-1
-0.5
0
x(kT)
The two plots above illustrate the effects: The amplitude imbalance stretches the axis. The phase
imbalance rotates the Q-axis but not the I-axis creating the pronounced skewing effect observed
in part (d).
10.4
(c)
4
3
2
y(kT)
1
0
-1
-2
-3
-4
-4
-2
0
x(kT)
2
4
-4
-2
0
x(kT)
2
4
(d)
4
3
2
y(kT)
1
0
-1
-2
-3
-4
(e) The phase trajectory is skewed, the same way a circle is skewed to create an ellipse.
(f)
ε = 0 φ = 20 deg
4
4
3
3
2
2
1
1
y(kT)
y(kT)
ε = 0.1 φ = 0 deg
0
0
-1
-1
-2
-2
-3
-3
-4
-4
-4
-2
0
x(kT)
2
4
-4
-2
0
x(kT)
2
4
The two plots above illustrate the effects: The amplitude imbalance stretches the axis. The phase
imbalance rotates the Q-axis but not the I-axis creating the pronounced skewing effect observed
in part (d).
10.5
⎡ ⎛m⎞
⎛ m ⎞⎤
~
s (nT ) = ∑ ⎢a0 ⎜ ⎟ + ja1 ⎜ ⎟⎥δ (n − Nm ) ∗ [ p c (nT ) + jp s (nT )]
⎝ N ⎠⎦
⎝N⎠
m ⎣
⎡ ⎛m⎞
⎤
⎛m⎞
= ∑ ⎢a0 ⎜ ⎟δ (n − Nm ) ∗ p c (nT ) − a1 ⎜ ⎟δ (n − Nm ) ∗ p s (nT )⎥
⎝N⎠
⎝N⎠
m ⎣
⎦
⎡ ⎛m⎞
⎤
⎛m⎞
+ j ∑ ⎢a 0 ⎜ ⎟δ (n − Nm ) ∗ p s (nT ) − a1 ⎜ ⎟δ (n − Nm ) ∗ p c (nT )⎥
⎝N⎠
⎝N⎠
m ⎣
⎦
Retaining the real part produces the desired result.
10.6
Start with the diagram in Figure 10.1.2 (d) and consider first the top branch:
a0 (k )
pc (nT )
↑N
Let Pc ( z ) be the z-transform of the filter coefficients p c (nT ) and let
( )
( )
( )
( )
Pc ( z ) = Pc ,0 z N + z −1 Pc ,1 z N + z −2 Pc , 2 z N + " + z − ( N −1) Pc , N −1 z N
be the polyphase partition of Pc ( z ) . A direct application of the discussion in Section 3.2
surrounding Figure 3.2.9 produces the equivalent system
Pc ,0 ( z )
a0 (k )
Pc ,1 ( z )
Pc , 2 ( z )
#
Pc, N −1 ( z )
The same process is applied to the branch involving a1 (k ) and p s (nT ) . Putting it all together
produces the desired block diagram.
10.7
Consider first the processing involving subfilter r:
a0 (k ) + ja1 (k )
Pr ( z )
e jΩ 0 r
Because the coefficients p (nT ) are purely real, the filtering process has the equivalent form
Pr ( z )
a0 (k )
Pr ( z )
a1 (k )
b0 (k )
b1 (k )
e jΩ 0 r
j
The desired output is
{
}
Re [b0 (k ) + jb1 (k )]e jΩ0 r = Re{[b0 (k ) + jb1 (k )][cos(Ω 0 r ) + j sin (Ω 0 r )]}
= b0 (k ) cos(Ω 0 r ) − b1 (k )sin (Ω 0 r )
which is produced by the following block diagram
a0 (k )
Pr (z )
cos(Ω 0 r )
+
−
a1 (k )
Pr (z )
sin (Ω 0 r )
Applying the same procedure to all subfilters generates the desired result.
10.8
We need to show that Pc ,r ( z ) = Pr ( z ) cos(rΩ 0 ) and Ps ,r (z ) = Pr ( z )sin (rΩ 0 ) . From Equation
(10.14) we have
Gr ( z ) = 2 p(r )e jΩ0 r z − r + 2 p( N + r )e jΩ 0 ( N + r ) z − ( N + r ) + 2 p(2 N + r )e jΩ0 (2 N + r ) z − (2 N + r ) + "
Because Ω 0 =
2π
× integer , e jΩ0 ( N + r ) = e jΩ0 so that
N
Gr ( z ) = 2 p (r )e jΩ0 r z − r + 2 p( N + r )e jΩ 0 r z − ( N + r ) + 2 p (2 N + r )e jΩ0 r z − (2 N + r ) + "
[
]
= 2e jΩ0 r p (r )z − r + p ( N + r )z −( N + r ) + p (2 N + r )z −(2 N + r ) + "
= 2e jΩ0 r Pr ( z )
which implies that
g r (k ) = 2 p r (k )e jΩ 0 r .
Now, because pr(k) is purely real, we have
{
}
(k ) = Im{g (k )} = Im{ 2 p (k )e } =
p c ,r (k ) = Re{g r (k )} = Re 2 p r (k )e jΩ0 r = 2 p r (k ) cos(Ω 0 r )
p s ,r
jΩ 0 r
r
which proves the result.
r
2 p r (k )sin (Ω 0 r )
10.9
[
]
r (nT ) ∗ g (nT ) = I r (nT ) 2 cos(Ω 0 n ) − Qr (nT ) 2 sin (Ω 0 n ) ∗ g (nT )
{
}
) ⎤⎥ ∗ g (nT )
+ ([I (nT ) + jQ (nT )]e
= 2 Re [I r (nT ) + jQr (nT )]e jΩ0 n ∗ g (nT )
jΩ 0 n ∗
⎡ [I (nT ) + jQ (nT )]e jΩ0 n
r
r
r
r
= 2⎢
2
⎢⎣
⎥⎦
1
[I r (nT ) + jQr (nT )]e jΩ0n ∗ g (nT ) + 1 [I r (nT ) − jQr (nT )]e − jΩ0n ∗ g (nT )
=
2
2
= [I r (nT ) + jQr (nT )]e jΩ0 n ∗ p(− nT )e jΩ0 n + [I r (nT ) − jQr (nT )]e − jΩ0 n ∗ p (− nT )e jΩ0 n
first term = ∑ [I r (kT ) + jQr (kT )]e jΩ0 k p ((n + k )T )e jΩ0 (n − k )
k
= e jΩ0 n ∑ [I r (kT ) + jQr (kT )] p((n + k )T )
k
⎡
⎤
⎢
⎥
= e jΩ0 n ⎢∑ I r (kT ) p ((n + k )T ) + j ∑ Qr (kT ) p ((n + k )T )⎥
k
k
⎥
⎢
x ( nT )
y ( nT )
⎣⎢
⎦⎥
= e jΩ0 n [x(nT ) + jy (nT )]
second term = ∑ [I r (kT ) − jQr (kT )]e − jΩ 0 k p((n + k )T )e jΩ0 (n − k )
k
= e jΩ0 n ∑ [I r (kT ) + jQr (kT )]e − j 2 Ω0 k p((n + k )T )
k
=0
The second term is zero because Ir(nT), Qr(nT), p(nT) are low-pass sequences.
10.10
[
]
r (nT ) ∗ g (nT ) = I r (nT ) 2 cos(Ω 0 n ) − Qr (nT ) 2 sin (Ω 0 n ) ∗ g (nT )
{
}
) ⎤⎥ ∗ g (nT )
+ ([I (nT ) + jQ (nT )]e
= 2 Re [I r (nT ) + jQr (nT )]e jΩ0 n ∗ g (nT )
jΩ 0 n ∗
⎡ [I (nT ) + jQ (nT )]e jΩ0 n
r
r
r
r
= 2⎢
2
⎢⎣
⎥⎦
1
[I r (nT ) + jQr (nT )]e jΩ0n ∗ g (nT ) + 1 [I r (nT ) − jQr (nT )]e − jΩ0n ∗ g (nT )
=
2
2
jΩ 0 n
jΩ 0 n
= [I r (nT ) + jQr (nT )]e
∗ p(− nT )e
+ [I r (nT ) − jQr (nT )]e − jΩ 0 n ∗ p(− nT )e − jΩ 0 n
first term = ∑ [I r (kT ) + jQr (kT )]e jΩ0 k p((n + k )T )e − jΩ 0 (n − k )
k
= e − jΩ0 n ∑ [I r (kT ) + jQr (kT )]e j 2 Ω0 k p((n + k )T )
k
=0
second term = ∑ [I r (kT ) − jQr (kT )]e − jΩ 0 k p((n + k )T )e − jΩ0 (n − k )
k
= e − jΩ0 n ∑ [I r (kT ) − jQr (kT )] p ((n + k )T )
k
⎡
⎤
⎢
⎥
= e − jΩ0 n ⎢∑ I r (kT ) p ((n + k )T ) − j ∑ Qr (kT ) p ((n + k )T )⎥
k
k
⎢
⎥
⎢⎣
⎥⎦
x ( nT )
y ( nT )
= e − jΩ0 n [x(nT ) − jy (nT )]
The first term is zero because Ir(nT), Qr(nT), p(nT) are low-pass sequences.
Note that the answer
r (nT ) ∗ g (nT ) = e − jΩ0 n [x(nT ) − jy (nT )]
is the complex-conjugate of the result of Exercise 10.9.
10.11
(a)
H ( jω ) =
=
1
1
1
+
+ jωC
R jω L
ωRL
ωRL[ωL − j (ω 2 RLC − R )]
=
2
ωL + j (ω 2 RLC − R )
ω 2 L2 + (ω 2 RLC − R )
=
ω 2 RL2 − jωRL(ω 2 RLC − R )
ω 2 L2 + (ω 2 RLC − R )
The imaginary part is zero when ωRL(ω 2 RLC − R ) = 0 . Thus we have
ω 02 RLC − R = 0
⇒ ω 02 =
2
1
LC
(b)
H ( jω ) =
2
ω 2 R 2 L2
ω 2 L2 + (ω 2 RLC − R )
2
from which we have
H ( jω 0 )
2
1 2 2
R L
LC
=
= R2
2
1 2 ⎛ 1
⎞
L +⎜
RLC − R ⎟
LC
⎝ LC
⎠
Now the 3-dB frequency ω3 satisfies
ω 32 R 2 L2
ω 32 L2 + (ω 32 RLC − R )
2
R2
=
2
(
⇒ 2ω 32 L2 = ω 32 L2 + ω 32 RLC − R
(
)
)
2
⇒ R 2 L2 C 2ω 34 − 2 R 2 LC + L2 ω 32 + R 2 = 0
(2R
)
2
LC + L2 − 4 R 4 L2 C 2
⇒ω =
2 R 2 L2 C 2
1
1
1
1
⇒ ω 32 =
+
±
+ 2 3
2 2
4 4
LC 2 R C
4R C
R LC
2
3
2 R 2 LC + L2 ±
(this is a quadratic equation in ω 32 )
2
1 ⎡ 1
1 ⎤
+
2 ⎢
2 2
LC ⎥⎦
R C ⎣ 4R C
⇒ ω 32 =
1
1
+
±
LC 2 R 2 C 2
⇒ ω 32 =
1
1
1
+
±
2 2
LC 2 R C
RC
2
1
1
+
2 2
LC
4R C
⇒ ω 32 =
1
1
±
2 2
RC
4R C
X2
1
1
1
1
+
+
+
2 2
2 2
LC LC
4R C
4R C
Y2
2 XY
⎡ 1
1
1 ⎤
⇒ω = ⎢
±
+
⎥
2 2
LC ⎦
4R C
⎣ 2 RC
2
2
3
⎡ 1
1
1 ⎤
⇒ ω3 = ± ⎢
±
+
⎥
2 2
LC ⎦
4R C
⎣ 2 RC
There are four solutions. Retaining only the positive roots produces the desired result
(c)
W3−dB = ω 2 − ω1
=
⎡
1
1
1
1
1
1 ⎤
+
+
−
−
+
+
⎢
⎥
2 RC
4 R 2 C 2 LC ⎣ 2 RC
4 R 2 C 2 LC ⎦
1
1
1
1
1
1
+
+
+
−
+
2 2
2 2
LC 2 RC
LC
2 RC
4R C
4R C
1
=
RC
=
(d)
maximum C: ω 0 = 2π × 540 × 10 3
C max =
minimum C: ω 0 = 2π × 1600 × 10 3 C min =
1
(2π × 540 × 10 ) (0.25 × 10 )
3 2
−3
1
= 0.3475nF
(2π × 1600 × 10 ) (0.25 × 10 )
3 2
−3
= 0.0396nF
(e)
30
3-dB bandwidth (kHz)
25
20
15
10
5
0
400
600
800
1000
1200
carrier frequency (kHz)
1400
1600
The relationship between the tuned frequency (the carrier frequency) and the 3-dB bandwidth
follows the equation (see parts (a) and (c))
W3−dB =
L 2
ω0
R
This shows that the 3-dB bandwidth increases as the square of the center frequency. The
component values selected for this problem produce the desired 3-dB bandwidth at 1000 kHz
(approximately the center of the band). For carrier frequencies below 1000 kHz, the 3-dB
bandwidth is too narrow; for carrier frequencies above 1000 kHz, the 3-dB bandwidth is too
high.
10.12
For high side mixing, f LO = f c + f 0 .
The image frequency on the positive frequency axis is the frequency resulting from the
difference. That is, the image frequency satisfies
f i − f LO = f 0
Substituting, we have
f i = f 0 + f LO = f 0 + ( f c + f 0 ) = f c + 2 f 0
10.13
(a) Construct the diagram below where, for low-side mixing, f i = f c − 2 f IF .
image rejection BPF
BW
f
fi
fi +
fc
B
2
Let BW be the bandwidth of the image rejection filter. Then, from the diagram, we have
BW
B⎞
B
B
⎛
= f c − ⎜ f c − 2 f IF + ⎟ = f c − f c + 2 f IF − = 2 f IF −
2
2⎠
2
2
⎝
⇒ BW = 4 f IF − B
(b) Construct the diagram below where, for high-side mixing, f i = f c + 2 f IF .
image rejection BPF
BW
f
fc
fi
B
2
Let BW be the bandwidth of the image rejection filter. Then, from the diagram, we have
fi −
BW ⎛
B⎞
B
B
= ⎜ f c + 2 f IF − ⎟ − f c = f c − f c + 2 f IF − = 2 f IF −
2
2⎠
2
2
⎝
⇒ BW = 4 f IF − B
(c) Both expressions are the same. This issue is not the deciding factor in selecting between highside mixing and low-side mixing.
10.14
(a) For low-side mixing, the image frequency is below the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the highest channel, the image
frequency is in the band. When selecting the highest channel ( f c = 1600 kHz), the image
frequency is
f i = f c − 2 f IF = 1600 − 2 × 455 = 690 kHz
which is within the AM band. Thus, YES: the image rejection filter must be tunable.
(b) For high-side mixing, the image frequency is above the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the lowest channel, the image
frequency is in the band. When selecting the lowest channel ( f c = 540 kHz), the image
frequency is
f i = f c + 2 f IF = 540 + 2 × 455 = 1450 kHz
which is within the AM band. Thus, YES: the image rejection filter must be tunable.
10.15
(a) For low-side mixing, the image frequency is below the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the highest channel, the image
frequency is in the band. When selecting the highest channel ( f c = 107.9 MHz), the image
frequency is
f i = f c − 2 f IF = 107.9 − 2 × 10.7 = 86.5 MHz
which is not within the FM band. Thus, NO: the image rejection filter does not need to be
tunable.
(b) For high-side mixing, the image frequency is above the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the lowest channel, the image
frequency is in the band. When selecting the lowest channel ( f c = 88.1 MHz), the image
frequency is
f i = f c + 2 f IF = 88.1 + 2 × 10.7 = 109.5 MHz
which is not within the FM band. Thus, NO: the image rejection filter does not need to be
tunable.
10.16
(a) For low-side mixing, the image frequency is below the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the highest channel, the image
frequency is in the band. When selecting the highest channel ( f c = 848.97 MHz), the image
frequency is
f i = f c − 2 f IF = 848.97 − 2 × 10.7 = 827.57 MHz
which is within the AMPS band. Thus, YES: the image rejection filter must be tunable.
(b) For high-side mixing, the image frequency is above the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the lowest channel, the image
frequency is in the band. When selecting the lowest channel ( f c = 824.04 MHz), the image
frequency is
f i = f c + 2 f IF = 824.04 + 2 × 10.7 = 845.44 MHz
which is within the AMPS band. Thus, YES: the image rejection filter must be tunable.
10.17
(a) For low-side mixing, the image frequency is below the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the highest channel, the image
frequency is in the band. When selecting the highest channel ( f c = 848.97 MHz), the image
frequency is
f i = f c − 2 f IF = 848.97 − 2 × 70 = 708.97 MHz
which is not within the AMPS band. Thus, NO: the image rejection filter does not need to be
tunable.
(b) For high-side mixing, the image frequency is above the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the lowest channel, the image
frequency is in the band. When selecting the lowest channel ( f c = 824.04 MHz), the image
frequency is
f i = f c + 2 f IF = 824.04 + 2 × 70 = 964.04 MHz
which is not within the AMPS band. Thus, NO: the image rejection filter does not need to be
tunable.
10.18
(a) For low-side mixing, the image frequency is below the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the highest channel, the image
frequency is in the band. When selecting the highest channel ( f c = 893.97 MHz), the image
frequency is
f i = f c − 2 f IF = 893.97 − 2 × 10.7 = 872.57 MHz
which is within the AMPS band. Thus, YES: the image rejection filter must be tunable.
(b) For high-side mixing, the image frequency is above the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the lowest channel, the image
frequency is in the band. When selecting the lowest channel ( f c = 869.04 MHz), the image
frequency is
f i = f c + 2 f IF = 869.04 + 2 × 10.7 = 890.44 MHz
which is within the AMPS band. Thus, YES: the image rejection filter must be tunable.
10.19
(a) For low-side mixing, the image frequency is below the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the highest channel, the image
frequency is in the band. When selecting the highest channel ( f c = 893.97 MHz), the image
frequency is
f i = f c − 2 f IF = 893.97 − 2 × 70 = 753.97 MHz
which is not within the AMPS band. Thus, NO: the image rejection filter does not need to be
tunable.
(b) For high-side mixing, the image frequency is above the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the lowest channel, the image
frequency is in the band. When selecting the lowest channel ( f c = 869.04 MHz), the image
frequency is
f i = f c + 2 f IF = 869.04 + 2 × 70 = 1009.04 MHz
which is not within the AMPS band. Thus, NO: the image rejection filter does not need to be
tunable.
10.20
(a) For low-side mixing, the image frequency is below the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the highest channel, the image
frequency is in the band. When selecting the highest channel ( f c = 914.8 MHz), the image
frequency is
f i = f c − 2 f IF = 914.8 − 2 × 10.7 = 893.4 MHz
which is within the GSM band. Thus, YES: the image rejection filter must be tunable.
(b) For high-side mixing, the image frequency is above the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the lowest channel, the image
frequency is in the band. When selecting the lowest channel ( f c = 890.2 MHz), the image
frequency is
f i = f c + 2 f IF = 890.2 + 2 × 10.7 = 911.6 MHz
which is within the GSM band. Thus, YES: the image rejection filter must be tunable.
10.21
(a) For low-side mixing, the image frequency is below the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the highest channel, the image
frequency is in the band. When selecting the highest channel ( f c = 914.8 MHz), the image
frequency is
f i = f c − 2 f IF = 914.8 − 2 × 70 = 774.8 MHz
which is not within the GSM band. Thus, NO: the image rejection filter does not need to be
tunable.
(b) For high-side mixing, the image frequency is above the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the lowest channel, the image
frequency is in the band. When selecting the lowest channel ( f c = 890.2 MHz), the image
frequency is
f i = f c + 2 f IF = 890.2 + 2 × 70 = 1030.2 MHz
which is not within the GSM band. Thus, NO: the image rejection filter does not need to be
tunable.
10.22
(a) For low-side mixing, the image frequency is below the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the highest channel, the image
frequency is in the band. When selecting the highest channel ( f c = 959.8 MHz), the image
frequency is
f i = f c − 2 f IF = 959.8 − 2 × 10.7 = 838.4 MHz
which is within the GSM band. Thus, YES: the image rejection filter must be tunable.
(b) For high-side mixing, the image frequency is above the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the lowest channel, the image
frequency is in the band. When selecting the lowest channel ( f c = 935.2 MHz), the image
frequency is
f i = f c + 2 f IF = 935.2 + 2 × 10.7 = 956.6 MHz
which is within the GSM band. Thus, YES: the image rejection filter must be tunable.
10.23
(a) For low-side mixing, the image frequency is below the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the highest channel, the image
frequency is in the band. When selecting the highest channel ( f c = 959.8 MHz), the image
frequency is
f i = f c − 2 f IF = 959.8 − 2 × 70 = 819.8 MHz
which is not within the GSM band. Thus, NO: the image rejection filter does not need to be
tunable.
(b) For high-side mixing, the image frequency is above the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the lowest channel, the image
frequency is in the band. When selecting the lowest channel ( f c = 935.2 MHz), the image
frequency is
f i = f c + 2 f IF = 935.2 + 2 × 70 = 1075.2 MHz
which is not within the GSM band. Thus, NO: the image rejection filter does not need to be
tunable.
10.24
(a) For low-side mixing, the image frequency is below the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the highest channel, the image
frequency is in the band. When selecting the highest channel ( f c = 136.975 MHz), the image
frequency is
f i = f c − 2 f IF = 136.975 − 2 × 10.7 = 115.575 MHz
which is not within the ATC band. Thus, NO: the image rejection filter does not need to be
tunable.
(b) For high-side mixing, the image frequency is above the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the lowest channel, the image
frequency is in the band. When selecting the lowest channel ( f c = 118 MHz), the image
frequency is
f i = f c + 2 f IF = 118 + 2 × 10.7 = 139.4 MHz
which is not within the ATC band. Thus, NO: the image rejection filter does not need to be
tunable.
10.25
(a) For low-side mixing, the image frequency is below the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the highest channel, the image
frequency is in the band. When selecting the highest channel ( f c = 136.975 MHz), the image
frequency is
f i = f c − 2 f IF = 136.975 − 2 × 21.4 = MHz
which is not within the ATC band. Thus, NO: the image rejection filter does not need to be
tunable.
(b) For high-side mixing, the image frequency is above the carrier frequency. Consequently the
image rejection filter needs to be tunable if, when selecting the lowest channel, the image
frequency is in the band. When selecting the lowest channel ( f c = 118 MHz), the image
frequency is
f i = f c + 2 f IF = 118 + 2 × 21.4 = MHz
which is not within the ATC band. Thus, NO: the image rejection filter does not need to be
tunable.
10.26
(a)
c
f
− fc
fc
d
f c > f1
− ( f c + f1 )
− ( f c − f1 )
f c − f1
f c + f1
f
e
− ( f c − f1 )
f
f c − f1
f
f 2 < f c − f1
f
− ( f c − f1 − f 2 ) f c − f1 − f 2
− ( f c − f1 + f 2 )
f c − f1 + f 2
g
f
− ( f c − f1 − f 2 ) f c − f1 − f 2
(b) From the diagrams in part (a) and the assumed relationships, we have
f IF = f c − f 1 − f 2
(c)
To prevent spectral overlap at f = 0, we require
B
2
B
f c − f1 − f 2 >
2
f c − f1 >
(d)
BPF 1
f
− fc
B
f c − 2 f1 −
2
fc
f c + 2 f1 +
BPF 2
− ( f c − f1 )
f
f c − f1
f c − f1 − 2 f 2 −
B
2
f c − f1 + 2 f 2 +
B
2
BPF 3
B
f
− ( f c − f1 − f 2 ) f c − f1 − f 2
B
2
10.27
There are many correct answers to this question. To illustrate the approach, assume that all
frequency translations use low-side mixing. With these assumptions, the first and second IF
frequencies are
first IF = fc – f1
second IF = fc – f1 – f2
Using these assumptions and the constraint on the center frequency to bandwidth ratios, we have
12000
≤ 10
4 × (12000 − f1 ) + 6
(1)
12000 − f1
≤ 10
4 × (12000 − f 1 − f 2 ) + 6
(2)
In addition, we require the second IF to be 70 MHz. This creates a third condition:
12000 − f 1 − f 2 = 70
(3)
f1 ≤ 11701.5
(4)
39 f 1 + 40 f 2 ≤ 468060
(5)
Solving (1) for f1 gives
Condition (2) may be re-expressed as
Solving (3) for f2 and substituting into condition (5) produces
f1 ≥ 9140
(6)
9140 ≤ f 1 ≤ 11701.5
(7)
Putting (4) and (6) together gives
The design procedure is to select f1 that satisfies condition (7). Given the choice for f1, f2 is given
by
f 2 = 11970 − f1
(8)
As an example, choose f1 = 10000 MHz. This forces f2 = 1930 MHz. The bandwidths (and center
frequency to bandwidth ratios) of the three bandpass filters are
BW 1 = 4 × (12000 − 10000 ) + 6 = 8006
12000
Q1 =
= 1.5
8006
BW 2 = 4 × (2000 − 1930 ) + 6 = 286
2000
Q2 =
=7
286
BW 3 = 6
70
Q3 =
= 11.7
6
The block diagram below illustrates the complete design.
2000
BPF
70
fixed
BPF1
10000
fixed
BPF2
1930
LO1
LO2
tuning control
12000
7997
1857
2000
2143
67
70
73
tuning control
16003
12000
from
antenna
to
demod
10.28
(a)
r (t ) 2 cos(ω c t ) = 2 I r (t ) cos 2 (ω c t ) − 2Qr (t )sin (ω c t ) cos(ω c t )
= I r (t ) + I r (t ) cos(2ω c t ) − Qr (t )sin (2ω c t )
(b) The low-pass filter output is I r (t ) . It is impossible to recover both I r (t ) and Qr (t ) from the
LPF output.
10.29
(a)
r (t ) 2 cos(ω c t ) = 2 I r (t ) cos(ω c t + θ ) cos(ω c t ) − 2Qr (t )sin (ω c t + θ ) cos(ω c t )
= I r (t ) cos(θ ) + I r (t ) cos(2ω c t + θ ) − Qr (t )sin (θ ) − Qr (t )sin (2ω c t + θ )
(b) The low-pass filter output is I r (t ) cos(θ ) − Qr (t )sin (θ ) . It is impossible to recover both I r (t )
and Qr (t ) from the LPF output.
10.30
(a) Channel k requires M complex-valued multiplications
(b) The number of complex-valued multiplications required to produce K channels in parallel is
M×K.
(c) The condition is
M ( p1 + p 2 + " + pν − ν ) < MK
which requires
p1 + p 2 + " + pν − ν < K
10.31
(a) 4 Mbits/s QPSK → 2 Msymbols/s: the sample rate must be a multiple of 2.
The bandwidth of 4 Mbit/s QPSK with the SRRC pulse shape with 50% excess bandwidth is
1 + α Rb
1.5 4
2×
×
= 2×
× = 3 MHz
2
2
2 2
Thus the sample rate must be greater than 6 Msamples/s
From Table 10.1.1, the lowest sample rate that satisfies both requirements is 8 Msamples/s.
(b) From Table 10.1.1, the sample rates that meet both requirements are 280 Msamples/s, 56
Msamples/s, 40 Msamples/s, and 8 Msamples/s.
10.32
(a)
Maximum
|
Maximum
Sample Rate
Bandwidth
| Sample Rate
Bandwidth
(Msamples/s)
(MHz)
| (Msamples/s)
(MHz)
--------------------------------------------------------180
90.0000
|
4
16/41
2.1951
60
30.0000
|
4
8/43
2.0930
36
18.0000
|
4
2.0000
25
5/7
12.8571
|
3
39/47
1.9149
20
10.0000
|
3
33/49
1.8367
16
4/11
8.1818
|
3
9/17
1.7647
13 11/13
6.9231
|
3
21/53
1.6981
12
6.0000
|
3
3/11
1.6364
10 10/17
5.2941
|
3
3/19
1.5789
9
9/19
4.7368
|
3
3/59
1.5254
8
4/7
4.2857
|
2
58/61
1.4754
7 19/23
3.9130
|
2
6/7
1.4286
7
1/5
3.6000
|
2
10/13
1.3846
6
2/3
3.3333
|
2
46/67
1.3433
6
6/29
3.1034
|
2
14/23
1.3043
5 25/31
2.9032
|
2
38/71
1.2676
5
5/11
2.7273
|
2
34/73
1.2329
5
1/7
2.5714
|
2
2/5
1.2000
4 32/37
2.4324
|
2
26/77
1.1688
(b) The lowest sample rate that can accommodate 8 MHz of bandwidth is 16 4/11 Msamples/s.
10.33
(a)
Maximum
|
Maximum
Sample Rate
Bandwidth
| Sample Rate
Bandwidth
(ksamples/s)
(kHz)
| (ksamples/s)
(kHz)
--------------------------------------------------------------42800
21400.0000
|
701
23/36
350.8194
14266
2/3
7133.3333
|
679
23/63
339.6825
8560
4280.0000
|
658
6/13
329.2308
6114
2/7
3057.1429
|
638
25/31
319.4032
4755
5/9
2377.7778
|
620
11/38
310.1447
3890 10/11
1945.4545
|
602
58/71
301.4085
3292
4/13
1646.1538
|
586
22/73
293.1507
2853
1/3
1426.6667
|
570
2/3
285.3333
2517 11/17
1258.8235
|
555
27/32
277.9219
2252 12/19
1126.3158
|
541
61/79
270.8861
2038
2/21
1019.0476
|
528
15/38
264.1974
1860 20/23
930.4348
|
515
55/83
257.8313
1712
856.0000
|
503
9/17
251.7647
1585
5/27
792.5926
|
491
83/87
245.9770
1475 25/29
737.9310
|
480
80/89
240.4494
1380 20/31
690.3226
|
470
30/91
235.1648
1296 32/33
648.4848
|
460
20/93
230.1075
1222
6/7
611.4286
|
450
10/19
225.2632
1156 28/37
578.3784
|
441
14/59
220.6186
1097 17/39
548.7179
|
432
11/34
216.1618
1043 37/41
521.9512
|
423
77/101
211.8812
995 15/43
497.6744
|
415
55/103
207.7670
951
1/9
475.5556
|
407
13/21
203.8095
910 30/47
455.3191
|
400
200.0000
873 23/49
436.7347
|
392
35/53
196.3302
839 11/51
419.6078
|
385
24/41
192.7927
807 29/53
403.7736
|
378
51/67
189.3806
778
2/11
389.0909
|
372
4/23
186.0870
750 50/57
375.4386
|
365
95/117
182.9060
(b) The lowest sample rate that can accommodate 200 kHz of bandwidth is 400 ksamples/s.
10.34
(a)
Maximum
|
Maximum
Sample Rate
Bandwidth
| Sample Rate
Bandwidth
(ksamples/s)
(kHz)
| (ksamples/s)
(kHz)
--------------------------------------------------------------1820
910.0000
|
29
51/61
14.9180
606
2/3
303.3333
|
28
8/9
14.4444
364
182.0000
|
28
14.0000
260
130.0000
|
27
11/67
13.5821
202
2/9
101.1111
|
26
26/69
13.1884
165
5/11
82.7273
|
25
45/71
12.8169
140
70.0000
|
24
68/73
12.4658
121
1/3
60.6667
|
24
4/15
12.1333
107
1/17
53.5294
|
23
7/11
11.8182
95 15/19
47.8947
|
23
3/79
11.5190
86
2/3
43.3333
|
22
38/81
11.2346
79
3/23
39.5652
|
21
77/83
10.9639
72
4/5
36.4000
|
21
7/17
10.7059
67 11/27
33.7037
|
20
80/87
10.4598
62 22/29
31.3793
|
20
40/89
10.2247
58 22/31
29.3548
|
20
10.0000
55
5/33
27.5758
|
19
53/93
9.7849
52
26.0000
|
19
3/19
9.5789
49
7/37
24.5946
|
18
74/97
9.3814
46
2/3
23.3333
|
18
38/99
9.1919
44 16/41
22.1951
|
18
2/101
9.0099
42 14/43
21.1628
|
17
69/103
8.8350
40
4/9
20.2222
|
17
1/3
8.6667
38 34/47
19.3617
|
17
1/107
8.5047
37
1/7
18.5714
|
16
76/109
8.3486
35 35/51
17.8431
|
16
44/111
8.1982
34 18/53
17.1698
|
16
12/113
8.0531
33
1/11
16.5455
|
15
19/23
7.9130
31 53/57
15.9649
|
15
5/9
7.7778
(b) The lowest sample rate that can accommodate 10 kHz of bandwidth is 20 ksamples/s.
10.35
Maximum
|
Maximum
Sample Rate
Bandwidth
| Sample Rate
Bandwidth
(ksamples/s)
(kHz)
| (ksamples/s)
(kHz)
--------------------------------------------------------------85600
42800.0000
|
1403
5/18
701.6389
28533
1/3
14266.6667
|
1358
19/26
679.3654
17120
8560.0000
|
1316
12/13
658.4615
12228
4/7
6114.2857
|
1277
11/18
638.8056
9511
1/9
4755.5556
|
1240
11/19
620.2895
7781
9/11
3890.9091
|
1205
19/30
602.8167
6584
8/13
3292.3077
|
1172
44/73
586.3014
5706
2/3
2853.3333
|
1141
1/3
570.6667
5035
5/17
2517.6471
|
1111
11/16
555.8438
4505
5/19
2252.6316
|
1083
43/79
541.7722
4076
4/21
2038.0952
|
1056
15/19
528.3947
3721 17/23
1860.8696
|
1031
13/40
515.6625
3424
1712.0000
|
1007
1/17
503.5294
3170 10/27
1585.1852
|
983
79/87
491.9540
2951 21/29
1475.8621
|
961
71/89
480.8989
2761
9/31
1380.6452
|
940
29/44
470.3295
2593 16/17
1296.9706
|
920
40/93
460.2151
2445
5/7
1222.8571
|
901
1/19
450.5263
2313 19/37
1156.7568
|
882
9/19
441.2368
2194 34/39
1097.4359
|
864
11/17
432.3235
2087 33/41
1043.9024
|
847
32/61
423.7623
1990 30/43
995.3488
|
831
3/44
415.5341
1902
2/9
951.1111
|
815
5/21
407.6190
1821
5/18
910.6389
|
800
400.0000
1746 15/16
873.4688
|
785
9/28
392.6607
1678 22/51
839.2157
|
771
6/35
385.5857
1615
3/32
807.5469
|
757
12/23
378.7609
1556
4/11
778.1818
|
744
8/23
372.1739
1501 43/57
750.8772
|
731
73/117
365.8120
A.1
NRZ
RZ
0.5
0.5
rp(τ)
1
rp(τ)
1
0
0
-0.5
-0.5
-1
-0.5
0
/T
τ s
0.5
1
-1
-0.5
MAN
0
/T
τ s
0.5
1
0.5
1
HS
0.5
0.5
rp(τ)
1
rp(τ)
1
0
0
-0.5
-0.5
-1
-0.5
0
0.5
1
-1
-0.5
τ/Ts
0
τ/Ts
For all pulse shapes, the autocorrelation function is zero for τ > Ts . Hence the pulse shape
autocorrelation function is zero for all non-zero multiples of Ts .
A.2
(a) E =
Ts / 2
∫
0
2
⎛
t
4 A2 2
+
t
dt
4 A 2 ⎜⎜1 −
2
∫
Ts
⎝ Ts
T2 / 2
T
2
⎞
A 2Ts
⎟⎟ dt =
3
⎠
Ts
2
Ts / 2
τ +Ts / 2
⎛
2A 2A
(t − τ )dt + ∫ 2 A⎜⎜1 − t
rp (τ ) = ∫
t
Ts
⎝ Ts
τ Ts
Ts / 2
⇒
3
Ts
A=
(b) For 0 ≤ τ ≤
=
T
2 A2 3 2 A2 2 1 2
τ −
τ + A Ts
Ts
3
Ts2
⎛τ
= 6⎜⎜
⎝ Ts
For
s
⎞ 2A
⎛
⎟⎟
(t − τ )dt + ∫ 2 A⎜⎜1 − t
⎠ Ts
⎝ Ts
τ +Ts / 2
3
2
⎞
⎛τ ⎞
⎟⎟ − 6⎜⎜ ⎟⎟ + 1
⎠
⎝ Ts ⎠
Ts
≤ τ ≤ Ts
2
⎛
t
rp (τ ) = ∫ 2 A⎜⎜1 −
⎝ Ts
τ
2 A2
=− 2 τ3 +
3Ts
Ts
⎛τ
= −2⎜⎜
⎝ Ts
3
⎞ ⎛ t −τ ⎞
⎟dt
⎟⎟2 A⎜⎜1 −
Ts ⎟⎠
⎠ ⎝
2 A2 2
2
τ + −2 A 2τ + A 2Ts
Ts
3
2
⎞
⎛τ ⎞
⎛τ ⎞
⎟⎟ + 6⎜⎜ ⎟⎟ − 6⎜⎜ ⎟⎟ + 2
⎠
⎝ Ts ⎠
⎝ Ts ⎠
Using the property rp (− τ ) = rp (τ ) we have
⎧0
3
2
⎪
⎪2⎛⎜ τ ⎞⎟ + 6⎛⎜ τ ⎞⎟ + 6⎛⎜ τ ⎞⎟ + 2
⎜T ⎟
⎜T ⎟
⎪ ⎜⎝ T3 ⎟⎠
⎝ 3⎠
⎝ 3⎠
⎪
3
2
⎪− 6⎛⎜ τ ⎞⎟ − 6⎛⎜ τ ⎞⎟ + 1
⎜T ⎟
⎪⎪ ⎜ T ⎟
⎝ 3⎠
rp (τ ) = ⎨ ⎝ 33⎠
2
⎛τ ⎞
⎪ ⎛⎜ τ ⎞⎟
⎪6⎜ T ⎟ − 6⎜⎜ T ⎟⎟ + 1
⎪ ⎝ 3⎠ 3 ⎝ 3⎠ 2
⎪ ⎛τ ⎞
⎛τ ⎞
⎛τ ⎞
⎪− 2⎜⎜ ⎟⎟ + 6⎜⎜ ⎟⎟ − 6⎜⎜ ⎟⎟ + 2
⎝ T3 ⎠
⎝ T3 ⎠
⎪ ⎝ T3 ⎠
⎪⎩0
τ ≤ −Ts
− Ts ≤ τ ≤ −
−
Ts
≤τ ≤ 0
2
0 ≤τ ≤
Ts
2
Ts
≤ τ ≤ Ts
2
Ts ≤ τ
Ts
2
⎞ ⎛ t −τ
⎟⎟2 A⎜⎜1 −
Ts
⎠ ⎝
⎞
⎟⎟dt
⎠
(c)
P( f ) =
Ts / 2
∫
0
=−
=
P( f )
2
2
⎛
2 A − j 2πft
t ⎞
te
dt + ∫ 2 A⎜⎜1 − ⎟⎟e − j 2πft dt
Ts
⎝ Ts ⎠
Ts / 2
T
[
A
1 − e jπfTs
2
2π f Ts
2
]
2
2A
⎛ πfT ⎞
sin 2 ⎜ s ⎟e jπfTs
2
π f Ts
⎝ 2 ⎠
2
4 A2
⎛ πfT ⎞
= 4 4 2 sin 4 ⎜ s ⎟
π f Ts
⎝ 2 ⎠
⎛ πfT ⎞
sin 4 ⎜ s ⎟
AT
⎝ 2 ⎠
=
4
4
⎛ πfTs ⎞
⎟
⎜
⎝ 2 ⎠
⎛ πfT ⎞
sin 4 ⎜ s ⎟
3T
⎝ 2 ⎠
= s
4 ⎛ πfTs ⎞ 4
⎟
⎜
⎝ 2 ⎠
2
2
s
(d)
Babs = ∞
B90 =
0.85
Ts
B99 =
1.30
Ts
B−60 dB =
20
Ts
(I used the closest null in the plot below for B-60dB)
0
-10
-20
|P(f)|2 (dB)
-30
-40
-50
-60
-70
-80
-90
0
5
10
15
fTs (cycles/symbol)
20
25
A.3
2
⎡
⎛ 2πt ⎞ ⎤
3 A 2Ts
2
⎜
⎟
=
E
A
1
cos
dt
=
−
(a)
∫0 ⎢
⎜ T ⎟⎥
2
⎝ s ⎠⎦
⎣
Ts
⇒
A=
2
3Ts
(b) For 0 ≤ τ ≤ Ts
⎡
⎛ 2πt ⎞⎤ ⎡
⎛ 2π (t − τ ) ⎞⎤
⎟⎟⎥ ⎢1 − cos⎜⎜
⎟⎟⎥ dt
rp (τ ) = ∫ A2 ⎢1 − cos⎜⎜
⎝ Ts ⎠⎦ ⎣
⎝ Ts
⎠⎦
τ
⎣
Ts
= A2 (Ts − τ ) +
⎛
A2
(Ts − τ )cos⎜⎜ 2πτ
2
⎝ Ts
⎞ 3 A2Ts ⎛ 2πτ ⎞
⎟⎟ +
⎟⎟
sin ⎜⎜
4π
⎠
⎝ Ts ⎠
⎛
τ ⎞
τ ⎞ 1
2⎛ τ ⎞ 1⎛ τ ⎞ ⎛
= ⎜⎜1 − ⎟⎟ + ⎜⎜1 − ⎟⎟ cos⎜⎜ 2π ⎟⎟ +
sin ⎜⎜ 2π ⎟⎟
3 ⎝ Ts ⎠ 3 ⎝ Ts ⎠ ⎝ Ts ⎠ 2π
⎝ Ts ⎠
Using the property rp (− τ ) = rp (τ ) we have
⎧0
⎪2 ⎛
⎛ τ ⎞
τ ⎞ 1⎛ τ ⎞ ⎛ τ ⎞ 1
⎪ ⎜⎜1 + ⎟⎟ + ⎜⎜1 + ⎟⎟ cos⎜⎜ 2π ⎟⎟ −
sin ⎜⎜ 2π ⎟⎟
Ts ⎠ 3 ⎝ Ts ⎠ ⎝ Ts ⎠ 2π
⎪3
⎝ Ts ⎠
rp (τ ) = ⎨ ⎝
⎪ 2 ⎛⎜1 − τ ⎞⎟ + 1 ⎛⎜1 − τ ⎞⎟ cos⎛⎜ 2π τ ⎞⎟ + 1 sin ⎛⎜ 2π τ ⎞⎟
⎜ T ⎟
⎪ 3 ⎜⎝ Ts ⎟⎠ 3 ⎜⎝ Ts ⎟⎠ ⎜⎝ Ts ⎟⎠ 2π
s ⎠
⎝
⎪0
⎩
(c)
⎡
⎛ 2πt ⎞⎤ − j 2πft
⎟⎟⎥ e
P( f ) = ∫ A⎢1 − cos⎜⎜
dt
T
⎝ s ⎠⎦
0
⎣
1
jA
1 − e − j 2πfTs
=
2
2π f ( fTs ) − 1
Ts
)[
(
=−
P( f ) =
2
=
e − jπfTs
sin (πfTs )
π f ( fTs )2 − 1
A
(
)
2
sin 2 (πfTs )
AT
3 ( fT )2 − 1 2 (πfTs )2
s
2
(
2
s
)
2
sin 2 (πfTs )
Ts
3 ( fT )2 − 1 2 (πfTs )2
s
(
)
]
τ < Ts
− Ts ≤ τ ≤ 0
0 ≤ τ ≤ Ts
Ts < τ
Babs = ∞
B90 =
0.95
Ts
B99 =
1.41
Ts
B−60 dB =
7
Ts
(I used the closest null in the plot below for B-60dB)
0
-10
-20
-30
|P(f)|2 (dB)
(d)
-40
-50
-60
-70
-80
-90
0
2
4
6
fTs (cycles/symbol)
8
10
A.4
∞
Let p (t ) ↔ P( f ) be a Fourier transform pair. Also note that
∑ δ (t − kTs ) ↔
k = −∞
(a)
G( f ) = Rp ( f ) ∗
1
Ts
∞
∑
m = −∞
⎛
δ ⎜⎜ f −
⎝
m⎞ 1
⎟=
Ts ⎟⎠ Ts
(b)
G( f ) =
∞
∫ g (t )e
− j 2πft
dt
−∞
∞
=
=
∞
⎤
⎡
(
)
r
t
δ (t − kTs )⎥ e − j 2πft dt
∫− ∞ ⎢⎣ p k∑
= −∞
⎦
∞
∞
∑ ∫ r (t )e
k = −∞ − ∞
=
p
∞
∑ r (kT )e
k = −∞
p
s
− j 2πft
δ (t − kTs )dt
− j 2πfTs k
⎛
∞
m⎞
⎟⎟
s ⎠
∑ R ⎜⎜ f − T
m = −∞
p
⎝
1
Ts
∞
⎛
m = −∞
⎝
m⎞
⎟⎟
s ⎠
∑ δ ⎜⎜ f − T
A.5
Rp ( f ) =
∞
r (τ )e
∫
τ
− j 2πfτ
p
= −∞
dτ
⎡ ∞
⎤ − j 2πfτ
dτ
⎢ ∫ p(t ) p (t − τ )dt ⎥ e
∫
τ = −∞ ⎣t = −∞
⎦
∞
⎡ ∞
⎤
= ∫ p(t )⎢ ∫ p (t − τ )e − j 2πfτ dτ ⎥ dt
t = −∞
⎣τ = −∞
⎦
∞
⎡ ∞
⎤
= ∫ p(t )⎢ ∫ p( x )e − j 2πf (t − x )dx ⎥ dt
t = −∞
⎣ x = −∞
⎦
∞
⎡ ∞
⎤
= ∫ p(t )e − j 2πft dt ⎢ ∫ p( x )e j 2πfx dx ⎥
t = −∞
⎣ x = −∞
⎦
∞
=
∞
=
∫ p(t )e
− j 2πft
t = −∞
= P( f )P∗ ( f )
= P( f )
2
⎡ ∞
⎤
dt ⎢ ∫ p( x )e − j 2πfx dx ⎥
⎣ x = −∞
⎦
∗
A.6
p (t ) =
⎛ π (1 − α ) ⎞ 4α
⎛ π (1 + α ) ⎞
sin ⎜⎜
t ⎟⎟ +
t cos ⎜⎜
t ⎟⎟
1
⎝ Ts
⎠ Ts
⎝ Ts
⎠=
2
Ts
π t ⎡ ⎛ 4α ⎞ 2 ⎤
⎟ t ⎥
⎢1 − ⎜
T s ⎢ ⎜⎝ T s ⎟⎠ ⎥
⎦
⎣
N (t )
T s D (t )
1
The derivatives of the numerator and denominators are
⎛ π (1 + α ) ⎞ 4α
⎛ π (1 + α ) ⎞ 4αt π (1 + α ) ⎛ π (1 + α ) ⎞
− cos⎜⎜
t ⎟⎟ +
t ⎟⎟ −
t ⎟⎟
cos⎜⎜
sin ⎜⎜
Ts
Ts
⎝ Ts
⎠ Ts
⎝ Ts
⎠ Ts
⎝ Ts
⎠
2
2
π ⎡ ⎛ 4α ⎞ 2 ⎤
πt ⎛ 4α ⎞
⎟⎟ t ⎥ − 2 ⎜⎜
⎟ t
D′(t ) = ⎢1 − ⎜⎜
Ts ⎢ ⎝ Ts ⎠ ⎥
Ts ⎝ Ts ⎟⎠
⎣
⎦
N ′(t ) =
π (1 − α )
(a)
lim p(t ) =
t →0
N (t )
N ′(t )
1
1
1 N ′(0)
1 ⎡
4α ⎤
=
=
=
lim
lim
1−α +
⎢
t
→
0
t
→
0
D(t )
D′(t )
π ⎥⎦
Ts
Ts
Ts D′(0)
Ts ⎣
(b)
lim p(t ) =
1
Ts
=
1
Ts
=
1
Ts
T
t→ s
4α
=
⎛T ⎞
N ′⎜ s ⎟
N (t )
N ′(t )
1
1
⎝ 4α ⎠
lim
lim
=
=
Ts D (t )
Ts D′(t )
Ts t →
Ts D′⎛ Ts ⎞
t→
4α
4α
⎜
⎟
⎝ 4α ⎠
π (1 − α ) ⎛ π 1 − α ⎞ 4α
⎛π 1+ α ⎞ π 1+α ⎛π 1+ α ⎞
cos⎜
cos⎜
sin ⎜
⎟
⎟−
⎟+
Ts
⎝4 α ⎠
⎝4 α ⎠ 4 α
⎝ 4 α ⎠ Ts
2
2
2
π ⎡ ⎛ 4α ⎞ ⎛ Ts ⎞ ⎤ 2π ⎛ Ts ⎞⎛ 4α ⎞ ⎛ Ts ⎞
⎟ ⎜
⎟ ⎜
⎢1 − ⎜
⎟ ⎥−
⎜
⎟⎜
⎟
Ts ⎢ ⎜⎝ Ts ⎟⎠ ⎝ 4α ⎠ ⎥ Ts ⎝ 4α ⎠⎜⎝ Ts ⎟⎠ ⎝ 4α ⎠
⎣
⎦
2πα ⎡⎛
2⎞ ⎛ π ⎞ ⎛
2 ⎞ ⎛ π ⎞⎤
−
⎜1 + ⎟ sin ⎜
⎟ + ⎜1 − ⎟ cos⎜
⎟
⎢
2Ts ⎣⎝ π ⎠ ⎝ 4α ⎠ ⎝ π ⎠ ⎝ 4α ⎠⎥⎦
2π
−
Ts
α ⎡⎛
2⎞ ⎛ π ⎞ ⎛
2 ⎞ ⎛ π ⎞⎤
⎟
⎜1 + ⎟ sin ⎜
⎟ + ⎜1 − ⎟ cos⎜
⎢
2Ts ⎣⎝ π ⎠ ⎝ 4α ⎠ ⎝ π ⎠ ⎝ 4α ⎠⎥⎦
In a similar way it can be shown that the limit as t → −
T2
is the same.
4α
A.7 Let f3 be the 3-dB bandwidth. Then
2
Ts2
= Rp ( f3 )
2
T
T
⇒ s = Rp ( f3 ) = s
2
2
⎡
⎛ πf 3Ts π (1 − α ) ⎞⎤
−
⎟⎥
⎢1 + cos⎜
2α ⎠⎦
⎝ α
⎣
⎛ πf T π (1 − α ) ⎞
⇒ 2 = 1 + cos⎜ 3 s −
⎟
2α ⎠
⎝ α
πf T π (1 − α )
⇒ cos −1 2 − 1 = 3 s −
2α
α
(1 − α ) = f T
α
⇒ cos −1 2 − 1 +
3 s
2
π
⇒ 0.5 − 0.368α = f 3Ts
(
)
(
)
A.8 Let f3 be the 3-dB bandwidth. Then
2
Ts2
= Rp ( f3 ) = Rp ( f3 ) ⇒
2
f 3Ts =
1
by construction.
2
B.1
e − jω 0 t + e jω 0 t
cos(− ω 0 t ) =
2
B.2
e jω0t + e − jω0t
− cos(ω 0 t ) = −
2
B.3
e − jω0t − e jω0t
sin (− ω 0 t ) =
j2
B.4
e jω 0 t − e − jω 0 t e − jω 0 t − e jω 0 t
− sin (ω 0 t ) = −
=
j2
j2
B.5
e − jω0t = cos(ω 0 t ) − j sin (ω 0 t )
B.6
e j (ω0t +θ ) + e − j (ω0t +θ )
cos(ω 0 t + θ ) =
2
jθ
e
e − jθ − jω 0 t
=
e jω0t +
e
2
2
B.7
e j (ω0t +θ ) − e − j (ω0t +θ )
sin (ω 0 t + θ ) =
j2
=
e jθ jω 0 t e − jθ − jω 0 t
e
−
e
j2
j2
B.8
e j (ω0t +θ ) = cos(ω 0 t + θ ) + j sin (ω 0 t + θ )
B.9
Given e jω0t ↔ 2πδ (ω − ω 0 ) and
x(t ) ↔ X ( jω ) ⇒ x ∗ (t ) ↔ X ∗ (− jω ) ,
we have
(
e − jω 0 t = e jω 0 t
)
∗
↔ 2πδ (− ω − ω 0 ) = 2πδ (ω + ω 0 )
if we assume the Dirac delta function is symmetric about 0.
B.10
cos(ω 0 t ) =
1 jω0t 1 − jω0t
e
+ e
2
2
1
1
↔ 2πδ (ω − ω 0 ) + 2πδ (ω + ω 0 )
2
2
= πδ (ω − ω 0 ) + πδ (ω + ω 0 )
B.11
sin (ω 0 t ) =
1 jω 0 t 1 − jω 0 t
e
e
−
j2
j2
1
1
↔
2πδ (ω − ω 0 ) −
2πδ (ω + ω 0 )
j2
j2
=
π
j
δ (ω − ω 0 ) −
π
j
δ (ω + ω 0 )
B.12
cos(ω 0 t + θ ) =
1 j (ω0t +θ ) 1 − j (ω0t +θ )
e
+ e
2
2
jθ
e
e − jθ − jω 0 t
jω 0 t
=
e
+
e
2
2
e jθ
e − jθ
↔
2πδ (ω − ω 0 ) +
2πδ (ω + ω 0 )
2
2
= πe jθ δ (ω − ω 0 ) + πe − jθ δ (ω + ω 0 )
B.13
sin (ω 0 t + θ ) =
1 j (ω0t +θ ) 1 − j (ω0t +θ )
e
−
e
j2
j2
=
e jθ jω 0 t e − jθ − jω 0 t
e
−
e
j2
j2
↔
e jθ
e − jθ
2πδ (ω − ω 0 ) −
2πδ (ω + ω 0 )
2
2
=
π
j
e jθ δ (ω − ω 0 ) −
π
j
e − jθ δ (ω + ω 0 )
B.14
Given e jω0t ↔ 2πδ (ω − ω 0 ) and
x(t ) ↔ X ( jω ) ⇒ x ∗ (t ) ↔ X ∗ (− jω ) ,
we have
(
e − j (ω0t +θ ) = e − jθ e jω0t
)
∗
↔ e − jθ 2πδ (− ω − ω 0 ) = e − jθ 2πδ (ω + ω 0 )
B.15
x(t ) ↔ X ( jω ) ⇒ x(t )e − jω0t ↔ X ( j (ω + ω 0 ))
B.16
I (t ) cos(ω 0 t ) − Q(t )sin (ω 0 t ) = A(t ) cos(ω 0 t + θ (t ))
where
A(t ) = I 2 (t ) + Q 2 (t )
⎛ Q(t ) ⎞
⎟⎟
⎝ I (t ) ⎠
θ (t ) = tan −1 ⎜⎜
B.17
(a) A straightforward application of the frequency shift property in Table 2.4.3 gives
~
~
u~ (t ) ↔ U ( jω ) ⇒ u~ (t )e jω0t ↔ U ( j (ω − ω 0 )) .
(b) A straight forward application of the conjugation property in Table 2.4.3 gives
~
u~ (t )e jω0t ↔ U ( j (ω − ω 0 ))
∗
~
u~(t )e jω0t ↔ U ∗ ( j (− ω − ω 0 ))
~
u~ ∗ (t )e − jω0t ↔ U ∗ (− j (− ω + ω ))
[
]
0
(c) Let F.T.{x(t )} be the Fourier transform of x(t ) . Then
{ {
}}
1
⎧1
⎫
F.T. Re u~ (t )e jω0t = F.T.⎨ u~ (t )e jω0t + u~ ∗ (t )e − jω0t ⎬
2
⎩2
⎭
1~
1 ~
= U ( j (ω − ω 0 )) + U ∗ (− j (ω + ω 0 ))
2
2
B.18
(a)
r (t ) =
1
2
[
]
I r (t ) e jω0t + e − jω0t + j
[
]
[
1
2
[
Qr (t ) e jω0t − e − jω0t
r (t ) 2e − jω0t = I r (t ) 1 + e − j 2ω0t + jQr (t ) 1 − e − j 2ω0t
= I r (t ) + jQr (t ) + [I r (t ) − jQr (t )]e
(b)
r (t ) =
1
2
([I (t ) + jQ (t )]e
r
r
jω 0 t
]
− j 2ω 0 t
+ [I r (t ) − jQr (t )]e − jω0t
r (t ) 2e − jω0t = [I r (t ) + jQr (t )] + [I r (t ) − jQr (t )]e − j 2ω0t
)
]
B.19
MLTED
e(k ) = Re{[a 0 (k ) − ja1 (k )][x (k ) + jy (k )]}
= Re{a 0 (k )x (k ) + a1 (k ) y (k ) + j[a0 (k ) y (k ) − a1 (k )x (k )]}
= a 0 (k )x (k ) + a1 (k ) y (k )
ELTED
e(k ) = Re{[a 0 (k ) − ja1 (k )][x(k + 1 / 2 ) + jy (k + 1 / 2 ) − x(k − 1 / 2 ) − jy (k − 1 / 2 )]}
⎫
⎧a 0 (k )x(k + 1 / 2 ) − a 0 (k )x(k − 1 / 2 ) + a1 (k ) y (k + 1 / 2 ) − a1 (k ) y (k − 1 / 2 )
= Re⎨
⎬
⎩+ j[a 0 (k ) y (k + 1 / 2 ) − a 0 (k ) y (k − 1 / 2 ) − a1 (k ) y (k + 1 / 2 ) + a1 (k ) y (k − 1 / 2 )]⎭
= a 0 (k )[x(k + 1 / 2 ) − x(k − 1 / 2 )] + a1 (k )[ y (k + 1 / 2 ) − y (k − 1 / 2 )]
ZCTED
e(k ) = Re{[x(k ) + jy (k )][a 0 (k − 1) − ja1 (k − 1) − a 0 (k ) + ja1 (k )]}
⎫
⎧ x(k )a 0 (k − 1) − x(k )a 0 (k ) + y (k )a1 (k − 1) − y (k )a1 (k )
= Re ⎨
⎬
⎩+ j[ y (k )a 0 (k − 1) − y (k )a 0 (k ) − x(k )a1 (k − 1) + x(k )a1 (k )]⎭
= x(k )[a 0 (k − 1) − a 0 (k )] + y (k )[a1 (k − 1) − a1 (k )]
GTED
e(k ) = Re{[x(k − 1 / 2) + jy (k − 1 / 2)][x(k − 1) − jy (k − 1) − x(k ) + jy (k )]}
⎧ x(k − 1 / 2)x(k − 1) − x(k − 1 / 2 )x(k ) + y (k − 1 / 2) y (k − 1) − y (k − 1 / 2 ) y (k )
⎫
= Re⎨
⎬
⎩+ j[ y (k − 1 / 2)x(k − 1) − y (k − 1 / 2)x(k ) − x(k − 1 / 2) y (k − 1) + x(k − 1 / 2) y (k )]⎭
= x(k − 1 / 2)[x(k − 1) − x(k )] + y (k − 1 / 2 )[ y (k − 1) − y (k )]
MMTED e(k ) = Re{[a 0 (k − 1) − ja1 (k − 1)][x(k ) + jy (k )] − [a 0 (k ) − ja1 (k )][x(k − 1) + jy (k − 1)]}
⎫
⎧a 0 (k − 1)x(k ) + a1 (k − 1) y (k ) − a 0 (k )x(k − 1) − a1 (k ) y (k − 1)
= Re ⎨
⎬
⎩+ j[a 0 (k − 1) y (k ) − a1 (k − 1)x(k ) − a 0 (k ) y (k − 1) + a1 (k )x(k − 1)]⎭
= [a 0 (k − 1)x(k ) − a 0 (k )x(k − 1)] + [a1 (k − 1) y (k ) − a1 (k ) y (k − 1)]
B.20
{
}
e(k ) = Im a~ ∗ (k )~
x ′(k )
= Im{[a 0 (k ) − ja1 (k )][x ′(k ) + jy ′(k )]}
= Im{a 0 (k )x ′(k ) + a1 (k ) y ′(k ) + j[a 0 (k ) y ′(k ) − a1 (k )x ′(k )]}
= a 0 (k ) y ′(k ) − a1 (k )x ′(k )
B.21
~
x′ = ~
x e − jθ
x ′ + jy ′ = ( x + jy )e − jθ
= ( x + jy )(cos θ − j sin θ )
= x cos θ + y sin θ + j (− x sin θ + y cos θ )
x′
y′
Organizing the above into a matrix equation produces the desired result.
B.22
(a)
The upper output:
[
]
r (t ) 2 cos(ω 0 t ) = I r (t ) 2 cos(ω 0 t ) − Qr (t ) 2 sin (ω 0 t ) 2 cos(ω 0 t )
= I r (t ) + I r (t ) cos(2ω 0 t ) − Qr (t )sin (2ω 0 t )
The output of the low-pass filter is I r (t ) .
The lower output:
[
]
r (t ) 2 sin (ω 0 t ) = I r (t ) 2 cos(ω 0 t ) − Qr (t ) 2 sin (ω 0 t ) 2 sin (ω 0 t )
= I r (t )sin (2ω 0 t ) − Qr (t ) + Qr (t ) cos(2ω 0 t )
The output of the low-pass filter is − Qr (t ) .
(b)
r (t )
LPF
I r (t ) − jQr (t )
e jω 0 t
The output of this system is the complex conjugate of the system that uses − 2 sin (ω 0 t ) .
(d)
− ω0
ω
ω0
LPF
0
2ω0
ω
ω
0
C.1
H a (s ) =
k0 k p k
s + k0 k p k
ˆ (s ) = H (s )Θ(s ) =
and Θ
a
∆θ
s
k 0 k p k∆θ
k0 k p k
s + k0 k p k
(a) Θ(s ) =
ˆ (s ) =
Θ
s (s + k 0 k p k )
[
θˆ(t ) = ∆θ 1 − e
(b) Θ(s ) =
ˆ (s ) =
Θ
− k 0 k p kt
=
∆θ
∆θ
−
s
s + k0 k p k
]u(t )
∆ω
s2
k 0 k p k∆ω
s (s + k 0 k p k )
2
=−
∆ω
k0 k p k
s
+
⎡
∆ω
k0k p k
∆ω
−
s 2 s + k0k p k
1
1
− k k kt ⎤
+t +
e 0 p ⎥u (t )
k0k p k
⎢⎣ k 0 k p k
⎥⎦
⎡
∆ω
− k k kt ⎤
= ⎢∆ωt −
1 − e 0 p ⎥u (t )
k0k p k
⎣⎢
⎦⎥
θˆ(t ) = ∆ω ⎢−
(
)
Θ(s )
C.2
ω n2
ω n2
ˆ
H a (s ) = 2
and Θ(s ) = H a (s )Θ(s ) = 2
Θ (s )
s + 2ζω n s + ω n2
s + 2ζω n s + ω n2
The roots of s 2 + 2ζω n s + ω n2 are − ζω n ± ω n ζ 2 − 1 . These roots are real and distinct for ζ > 1 ,
real and repeated for ζ = 1 , and complex conjugates when 0 < ζ < 1 . (When ζ = 0 , the roots
are purely imaginary. But we are not interested in this solution for this problem.)
(a) Θ(s ) =
∆θ
s
Case 1, ζ > 1 :
ˆ (s ) =
Θ
∆θ
∆θωn2
ωn2
=
s 2 + 2ζω n s + ωn2 s
s s + ζω n − ωn ζ 2 − 1 s + ζω n + ωn ζ 2 − 1
)(
(
)
∆θ
2
⎡
∆θ ⎡
ζ ⎤
ζ ⎤
1+ 2 ⎥
1− 2 ⎥
⎢
⎢
2 ⎣ ζ − 1⎦
∆θ
⎣ ζ − 1⎦ −
=
−
2
s
s + ζω n − ωn ζ − 1 s + ζω n + ωn ζ 2 − 1
⎡ 1 ⎛⎡
ζ ⎤ − ⎛⎜ ζω
θˆ(t ) = ∆θ ⎢1 − ⎜⎜ ⎢1 + 2 ⎥ e ⎝
2
ζ −1
n
− ω n ζ 2 −1 ⎞⎟ t
⎠
⎡
ζ ⎤ − ⎛⎜ ζω n +ω n
+ ⎢1 − 2 ⎥ e ⎝
⎣ ζ − 1⎦
ζ 2 −1 ⎞⎟ t
⎠
⎢⎣
⎦
⎝⎣
⎛
⎡
⎤⎞
ζ
= ∆θ ⎜⎜1 − e −ζω n t ⎢cosh ωn ζ 2 − 1 t + 2
sinh ωn ζ 2 − 1 t ⎥ ⎟⎟u (t )
ζ −1
⎣
⎦⎠
⎝
))
((
Case 2, ζ = 1 :
ˆ (s ) =
Θ
ω n2
∆θω n
∆θ ∆θ
∆θ
=
−
−
2
2
s
(s + ω n ) s + ω n
(s + ω n ) s
[
]
θˆ(t ) = ∆θ 1 − ω n te −ω t − e −ω t u (t )
n
n
((
))
⎞⎤
⎟⎥u (t )
⎟
⎠⎥⎦
Case 3, 0 < ζ < 1 :
ˆ (s ) =
Θ
ω n2
∆θω n2
∆θ
=
s 2 + 2ζω n s + ω n2 s
s s + ζω n − jω n 1 − ζ 2 s + ζω n + jω n 1 − ζ 2
)(
(
)
ζ ⎤
ζ ⎤
∆θ ⎡
∆θ ⎡
⎢1 + j
⎥
⎢1 − j
⎥
2
2
2
2
⎢
⎥⎦
⎢
⎥
ζ
ζ
−
1
1
−
∆θ
⎣
⎣
⎦ −
=
−
s
s + ζω n − jω n 1 − ζ 2 s + ζω n + jω n 1 − ζ 2
⎡
⎛
1
θˆ(t ) = ∆θ ⎢1 − ⎜1 − j
⎞ −⎛⎜ ζω n − jωn
⎟e ⎝
2 ⎟
1−ζ ⎠
ζ
1−ζ 2 ⎞⎟ t
⎠
ζ
1⎛
− ⎜1 + j
2⎜
1−ζ 2
⎝
⎢ 2 ⎜⎝
⎣
⎡
⎛
ζ
sin ω n ζ 2 − 1t
= ∆θ ⎢1 − e −ζω nt ⎜ cos ω n ζ 2 − 1t +
2
⎜
⎢
1−ζ
⎝
⎣
)
(
(
⎞ −⎛⎜ ζω n + jωn
⎟e ⎝
⎟
⎠
1−ζ 2 ⎞⎟ t
⎠
⎤
⎥u (t )
⎥
⎦
)
⎞⎤
⎟⎥u (t )
⎟⎥
⎠⎦
∆ω
s2
Case 1, ζ > 1 :
(b) Θ(s ) =
ˆ (s ) =
Θ
∆ωω n2
ω n2
∆ω
=
s 2 + 2ζω n s + ω n2 s 2
s 2 s + ζω n − ω n ζ 2 − 1 s + ζω n + ω n ζ 2 − 1
2∆ωζ
=
)(
(
∆ω
−
s2
ωn
s
⎡
2ζ
⎢
⎣
ωn
θˆ(t ) = ∆ω ⎢t −
+
)
∆ω ⎡
2ζ 2 − 1 ⎤
2ζ 2 − 1 ⎤
∆ω ⎡
⎢2ζ −
⎥
⎢2ζ +
⎥
2
2
2ω n ⎢
2ω n ⎢
⎥⎦
⎥
−
−
1
1
ζ
ζ
⎣
⎣
⎦+
+
s + ζω n − ω n ζ 2 − 1 s + ζω n + ω n ζ 2 − 1
1⎡
2ζ 2 − 1 ⎤ −⎛⎜⎝ ζω n −ω n
⎢2ζ +
⎥e
2
2⎢
⎥⎦
−
1
ζ
⎣
⎡ 2ζ
1 −ζω nt ⎛⎜ 2ζ 2 − 1 ωn
= ∆ω ⎢t −
+
e
e
⎜ ζ 2 −1
⎢ ω n 2ω n
⎝
⎣
(
ζ 2 −1 ⎞⎟ t
ζ 2 −1t
⎠
+
2ζ 2 − 1 ⎤ −⎛⎜⎝ ζω n +ωn
1⎡
⎢2ζ −
⎥e
2
2⎢
⎥⎦
−
1
ζ
⎣
+ 2ζe ω n
)
ζ 2 −1t
−
2ζ 2 − 1
ζ 2 −1
(
e −ω n
ζ 2 −1t
⎡ 2ζ
1 −ζω nt ⎛⎜
2ζ 2 − 1
= ∆ω ⎢t −
+
e
2ζ cosh ω n ζ 2 − 1t +
sinh ω n ζ 2 − 1t
2
⎜
ω
ω
⎢
ζ −1
n
n
⎝
⎣
ζ 2 −1 ⎞⎟ t
⎠
+ 2ζe −ω n
)⎞⎟⎟⎥⎥u(t )
⎤
⎠⎦
⎤
⎥u (t )
⎥
⎦
ζ 2 −1t
⎞⎤
⎟⎥u (t )
⎟⎥
⎠⎦
Case 2, ζ = 1 :
2∆ω
2∆ω
ω
ωn
∆ω
∆ω
ˆ (s ) = ∆ω ω
Θ
= 2 − n +
+
2
2
2
s
s (s + ω n )
s
(s + ω n ) s + ω n
2
n
⎡
2
⎣
ωn
θˆ(t ) = ∆ω ⎢t −
e −ωnt + te −ωnt +
⎤
e −ωnt ⎥u (t )
ωn
⎦
2
Case 3, 0 < ζ < 1 :
ˆ (s ) =
Θ
=
∆ωω n2
ω n2
∆ω
=
s 2 + 2ζω n s + ω n2 s 2
s 2 s + ζω n − jω n 1 − ζ 2 s + ζω n + jω n 1 − ζ 2
(
∆ω
−
s2
2ζ
∆ω
ωn
s
⎡
2ζ
⎢
⎣
ωn
θˆ(t ) = ∆ω ⎢t −
+
+
(
)
2ζ 2 − 1 ⎤
∆ω ⎡
⎢2ζ − j
⎥
2
2ω n ⎢
⎥⎦
1
−
ζ
⎣
s + ζω n − jω n 1 − ζ
(
)(
2
+
s + ζω n + jω n 1 − ζ
)
(
)
)
2ζ 2 − 1 ⎤
∆ω ⎡
⎢2ζ + j
⎥
2
2ω n ⎢
⎥⎦
1
−
ζ
⎣
2ζ 2 − 1 ⎤ −⎛⎜⎝ ζω n − jω n
1⎡
⎢2ζ − j
⎥e
2
2⎢
⎥⎦
1
−
ζ
⎣
⎡ 2ζ e −ζω n t
= ∆ω ⎢t −
+
ωn
⎢ ωn
⎣
(
)
1−ζ 2 ⎞⎟ t
⎠
(
+
2
(
)
2ζ 2 − 1 ⎤ −⎛⎜⎝ ζω n + jω n
1⎡
⎢2ζ − j
⎥e
2
2⎢
⎥⎦
1
−
ζ
⎣
)
(
)
2
⎛
⎞⎤
⎜ 2ζ cos ω 1 − ζ 2 t + 2ζ − 1 sin ω 1 − ζ 2 t ⎟⎥u (t )
n
n
⎜
⎟⎥
1− ζ 2
⎝
⎠⎦
1−ζ 2 ⎞⎟ t
⎠
⎤
⎥u (t )
⎥
⎦
C.3
2ζω n s + ω n2
2ζω n s + ω n2
ˆ
H a (s ) = 2
Θ (s )
and Θ(s ) = H a (s )Θ(s ) = 2
s + 2ζω n s + ω n2
s + 2ζω n s + ω n2
The roots of s 2 + 2ζω n s + ω n2 are − ζω n ± ω n ζ 2 − 1 . These roots are real and distinct for ζ > 1 ,
real and repeated for ζ = 1 , and complex conjugates when 0 < ζ < 1 . (When ζ = 0 , the roots
are purely imaginary. But we are not interested in this solution for this problem.)
∆θ
s
Case 1, ζ > 1 :
(a) Θ(s ) =
ˆ (s ) =
Θ
(
)
∆θ 2ζω n s + ω n2
2ζω n s + ω n2 ∆θ
=
s 2 + 2ζω n s + ω n2 s
s + ζω n − ω n ζ 2 − 1 s + ζω n + ω n ζ 2 − 1 s
(
)(
)
(ζ − ζ − 1) ∆θ (ζ + ζ − 1) ∆θ
1 − (ζ − ζ − 1 )
1 − (ζ + ζ − 1 )
∆θ
=
+
+
2
2
2
2
s + ζω n − ω n ζ 2 − 1
(
)
2
2
s + ζω n + ω n ζ 2 − 1
⎡ ζ − ζ 2 − 1 2 −⎛⎜ ζω −ω
n
n
θˆ(t ) = ⎢
e ⎝
2
⎢1 − ζ − ζ 2 − 1
⎣
(
2
2
s
⎤
(
ζ + ζ − 1)
+
e
+ 1⎥ ∆θu (t )
⎥
1 − (ζ + ζ − 1 )
⎦
ζ − 1)
1 − (ζ − ζ − 1 )
and the second term by
and
ζ − 1)
1 − (ζ − ζ − 1 )
ζ 2 −1 ⎞⎟ t
2
2
−⎛⎜ ζω n +ω n ζ 2 −1 ⎞⎟ t
⎠
⎝
⎠
)
2
2
(
Multiplying the first term by
1 − (ζ +
1− ζ +
2
2
2
2
2
2
2
2
simplifying produces
⎡⎛ ⎡ 1
⎤ ω
ζ
ˆ
⎥e n
θ (t ) = ⎢⎜ ⎢− +
⎢⎜ ⎢ 2 2 ζ 2 − 1 ⎥
⎦
⎣⎝ ⎣
⎡⎛ 1
= ⎢⎜ − e ωn
⎢⎜⎝ 2
⎣
ζ 2 −1t
Now using cosh (x ) =
⎡
⎛
1
− e −ω n
2
⎡ 1
⎤ −ω
ζ
+ ⎢− −
⎥e n
2
2
⎢⎣
2 ζ − 1 ⎥⎦
ζ 2 −1t
ζ 2 −1t
+
⎢
⎣
⎜
⎝
2 ζ 2 −1
e ωn
ζ 2 −1t
−
⎤
⎞
⎟e −ζω nt + 1⎥ ∆θu (t )
⎟
⎥
⎠
⎦
ζ
2 ζ 2 −1
e −ω n
e x + e−x
e x − e−x
and sinh (x ) =
we have
2
2
(
)
θˆ(t ) = ⎢1 − e −ζω t ⎜ cosh ω n ζ 2 − 1t −
n
ζ
ζ 2 −1t
ζ
ζ 2 −1
(
sinh ω n ζ
2
⎞
− 1t )⎟⎥ ∆θu (t )
⎟⎥
⎤
⎠⎦
ζ 2 −1t
⎞ −ζω t ⎤
⎟e n + 1⎥ ∆θu (t )
⎟
⎥
⎠
⎦
Case 2, ζ = 1 :
ˆ (s ) =
Θ
=
2ω n s + ω n2 ∆θ ∆θ (2ω n s + ω n2 )
=
s 2 + 2ω n s + ω n2 s
(s + ω n ) 2 s
ω n ∆θ
∆θ
∆θ
−
+
2
(s + ω n ) s + ω n s
[
]
= [1 − (1 − ω t )e ]∆θu (t )
θˆ(t ) = ω n te −ω t − e −ω t + 1 ∆θu (t )
n
n
−ω n t
n
Case 3, 0 < ζ < 1 :
ˆ (s ) =
Θ
2ζω n s + ω n2 ∆θ
∆θ (2ζω n s + ω n2 )
=
s 2 + 2ζω n s + ω n2 s
s + ζω n − jω n 1 − ζ 2 s + ζω n + jω n 1 − ζ 2 s
)(
(
)
(ζ − j 1 − ζ ) ∆θ (ζ + j 1 − ζ ) ∆θ
1 − (ζ + j 1 − ζ )
1 − (ζ − j 1 − ζ )
∆θ
+
+
=
2
2
2
2
2
2
)
⎡ ζ − j 1 − ζ 2 2 −⎛⎜ ζω − jω
n
n
ˆ
θ (t ) = ⎢
e ⎝
2
⎢1 − ζ − j 1 − ζ 2
⎣
(
2
s + ζω n + jω n ζ 2 − 1
s + ζω n − jω n ζ 2 − 1
(
2
)
ζ 2 −1 ⎞⎟ t
⎠
s
(ζ + j 1 − ζ ) e
+
1 − (ζ + j 1 − ζ )
2
2
2
(
Multiplying the first term by
1 − (ζ + j
1− ζ + j 1−ζ 2
1−ζ 2
)
)
2
2
−⎛⎜ ζω n + jω n ζ 2 −1 ⎞⎟ t
⎠
⎝
2
⎤
+ 1⎥ ∆θu (t )
⎥
⎦
(
and the second term by
1 − (ζ − j
1− ζ − j 1−ζ 2
1−ζ 2
)
)
2
2
and
simplifying produces
⎡⎛ ⎡ 1
⎢⎜ ⎢ 2
⎣⎝ ⎣
θˆ(t ) = ⎢⎜ ⎢− − j
⎡⎛ 1
= ⎢⎜ − e jωn
⎢⎜⎝ 2
⎣
⎤ jω
⎥e n
2
2 1 − ζ ⎥⎦
ζ
Now using cos( x ) =
⎡
⎛
1
− e − jω n
2
ζ 2 −1t
ζ 2 −1t
ζ 2 −1t
⎢
⎣
⎜
⎝
−j
ζ
2 ζ 2 −1
e jω n
ζ 2 −1t
+ j
ζ 2 −1t
(
)
2 ζ 2 −1
(
)
⎞⎤
sin ω n ζ 2 − 1t ⎟⎥ ∆θu (t )
⎟⎥
ζ 2 −1
⎠⎦
ζ
⎤
⎞
⎟e −ζω nt + 1⎥ ∆θu (t )
⎟
⎥
⎠
⎦
ζ
e jx − e − jx
e jx + e − jx
and sin ( x ) =
we have
2
j2
θˆ(t ) = ⎢1 − e −ζω t ⎜ cos ω n ζ 2 − 1t −
n
⎡ 1
⎤ − jω
ζ
+ ⎢− + j
⎥e n
2
⎢⎣ 2
2 1 − ζ ⎥⎦
e − jω n
ζ 2 −1t
⎞ −ζω t ⎤
⎟e n + 1⎥ ∆θu (t )
⎟
⎥
⎠
⎦
∆ω
s2
Case 1, ζ > 1 :
(b) Θ(s ) =
ˆ (s ) =
Θ
(
)
∆ω 2ζω n s + ω n2
2ζω n s + ω n2 ∆ω
=
s 2 + 2ζω n s + ω n2 s 2
s + ζω n − ω n ζ 2 − 1 s + ζω n + ω n ζ 2 − 1 s 2
(
∆ω
∆ω
1
ωn 2 ζ − 1
s + ζω n − ω n ζ − 1
2
⎡ ∆ω
θˆ(t ) = ⎢
1
⎢⎣ ω n 2 ζ 2 − 1
e
+
+
s + ζω n + ω n ζ − 1
2
− ⎛⎜ ζω n +ω n ζ 2 −1 ⎞⎟ t
⎠
⎝
−
)
1
ωn 2 ζ 2 − 1
2
=−
)(
∆ω
1
ωn 2 ζ 2 − 1
e
∆ω
s2
−⎛⎜ ζω n −ω n ζ 2 −1 ⎞⎟ t
⎠
⎝
⎤
+ ∆ωt ⎥u (t )
⎥⎦
)
(
⎡
⎤
∆ω
1
= ⎢∆ωt −
sinh ω n ζ 2 − 1t e −ζω nt ⎥u (t )
ωn ζ 2 − 1
⎢⎣
⎥⎦
Case 2, ζ = 1 :
ˆ (s ) =
Θ
2ω n s + ω n2 ∆ω ∆ω (2ω n s + ω n2 )
=
s 2 + 2ω n s + ω n2 s 2
(s + ω n )2 s 2
=−
[
∆ω
(s + ω n )
2
+
∆ω
s2
]
θˆ(t ) = − ∆ωte −ω t + ∆ωt u (t )
n
[
]
= ∆ωt 1 − e −ωnt u (t )
Case 3, 0 < ζ < 1 :
ˆ (s ) =
Θ
(
)
∆ω 2ζω n s + ω n2
2ζω n s + ω n2 ∆ω
=
s 2 + 2ζω n s + ω n2 s 2
s + ζω n − jω n 1 − ζ 2 s + ζω n + jω n 1 − ζ 2 s 2
(
∆ω
=−
∆ω
1
ωn j2 1 − ζ 2
s + ζω n − jω n 1 − ζ 2
⎡ ∆ω
θˆ(t ) = ⎢−
1
⎢⎣ ω n j 2 1 − ζ 2
e
+
s + ζω n + jω n 1 − ζ 2
+
)
1
ωn j2 1 − ζ 2
−⎛⎜ ζω n − jω n 1−ζ 2 ⎞⎟ t
⎠
⎝
(
)(
∆ω
+
1
ωn j2 1 − ζ 2
)
⎡
⎤
∆ω
1
= ⎢∆ωt −
sin ω n 1 − ζ 2 t e −ζω nt ⎥u (t )
ωn 1 − ζ 2
⎢⎣
⎥⎦
∆ω
s2
e
−⎛⎜ ζω n − jω n 1−ζ 2 ⎞⎟ t
⎠
⎝
⎤
+ ∆ωt ⎥u (t )
⎥⎦
C.4
Kω n2
Kω n2
ˆ
Θ (s )
H a (s ) = 2
and Θ(s ) = H a (s )Θ(s ) = 2
s + 2ζω n s + ω n2
s + 2ζω n s + ω n2
2ζ k 2
where K = 1 +
− 2
ωn
ωn
The roots of s 2 + 2ζω n s + ω n2 are − ζω n ± ω n ζ 2 − 1 . These roots are real and distinct for ζ > 1 ,
real and repeated for ζ = 1 , and complex conjugates when 0 < ζ < 1 . (When ζ = 0 , the roots
are purely imaginary. But we are not interested in this solution for this problem.)
(a) Θ(s ) =
∆θ
s
Case 1, ζ > 1 :
ˆ (s ) =
Θ
Kω n2
K∆θω n2
∆θ
=
s 2 + 2ζω n s + ω n2 s
s s + ζω n − ω n ζ 2 − 1 s + ζω n + ω n ζ 2 − 1
)(
(
)
⎤
K∆θ ⎡
ζ
K∆θ ⎡
ζ ⎤
⎢1 +
⎥
⎢1 −
⎥
2 ⎢
2 ⎢
ζ 2 − 1 ⎥⎦
ζ 2 − 1 ⎥⎦
K∆θ
⎣
⎣
=
−
−
s
s + ζω n − ω n ζ 2 − 1 s + ζω n + ω n ζ 2 − 1
⎡
⎛⎡
⎢
⎣
2 ⎜⎢
⎝⎣
1
θˆ(t ) = K∆θ ⎢1 − ⎜ ⎢1 +
⎤ −⎛⎜ ζω n −ωn
⎥e ⎝
2
ζ − 1 ⎥⎦
ζ
ζ 2 −1 ⎞⎟ t
⎠
⎡
ζ ⎤ −⎛⎜⎝ ζωn +ωn
+ ⎢1 −
⎥e
⎢⎣
ζ 2 − 1 ⎥⎦
)
(
(
)
ζ 2 −1 ⎞⎟ t
⎛
⎡
⎤⎞
ζ
= K∆θ ⎜1 − e −ζω nt ⎢cosh ω n 1 − ζ 2 t +
sinh ω n 1 − ζ 2 t ⎥ ⎟u (t )
⎜
⎢⎣
⎥⎦ ⎟⎠
ζ 2 −1
⎝
Case 2, ζ = 1 :
ˆ (s ) =
Θ
Kω n2
(s + ω n )
[
2
K∆θω n
∆θ K∆θ
K∆θ
=
−
−
2
s
s
(s + ω n ) s + ω n
]
θˆ(t ) = K∆θ 1 − ω n te −ω t − e −ω t u (t )
n
n
⎠
⎞⎤
⎟⎥u (t )
⎟⎥
⎠⎦
Case 3, 0 < ζ < 1 :
ˆ (s ) =
Θ
Kω n2
K∆θω n2
∆θ
=
s 2 + 2ζω n s + ω n2 s
s s + ζω n − jω n 1 − ζ 2 s + ζω n + jω n 1 − ζ 2
)(
(
)
ζ ⎤
ζ ⎤
K∆θ ⎡
K∆θ ⎡
⎢1 − j
⎥
⎢1 + j
⎥
2
2
2
2
⎢
⎥
⎢
⎥⎦
ζ
ζ
1
1
−
−
K∆θ
⎣
⎦ −
⎣
=
−
s
s + ζω n − jω n 1 − ζ 2 s + ζω n + jω n 1 − ζ 2
⎡
⎛
1
θˆ(t ) = K∆θ ⎢1 − ⎜1 − j
⎞ −⎛⎜ ζω n − jωn
⎟e ⎝
2 ⎟
1−ζ ⎠
ζ
1−ζ 2 ⎞⎟ t
⎠
ζ
1⎛
− ⎜1 + j
2⎜
1−ζ 2
⎝
⎢ 2 ⎜⎝
⎣
⎡
⎛
ζ
sin ω n ζ 2 − 1t
= K∆θ ⎢1 − e −ζω nt ⎜ cos ω n ζ 2 − 1t +
2
⎜
⎢
1− ζ
⎝
⎣
)
(
(b)
Θ(s ) =
(
⎞ −⎛⎜ ζω n + jωn
⎟e ⎝
⎟
⎠
1−ζ 2 ⎞⎟ t
⎠
⎤
⎥u (t )
⎥
⎦
)
⎞⎤
⎟⎥u (t )
⎟⎥
⎠⎦
∆ω
s2
Case 1, ζ > 1 :
ˆ (s ) =
Θ
Kω n2
K∆ωω n2
∆ω
=
s 2 + 2ζω n s + ω n2 s 2
s 2 s + ζω n − ω n ζ 2 − 1 s + ζω n + ω n ζ 2 − 1
2 K∆ωζ
=
)(
(
K∆ω
−
s2
ωn
s
)
2ζ 2 − 1 ⎤ K∆ω ⎡
2ζ 2 − 1 ⎤
K∆ω ⎡
⎥
⎢2ζ +
⎥
⎢2ζ −
2
2
2ω n ⎢
2ω n ⎢
⎥
⎥⎦
ζ
1
ζ
1
−
−
⎦+
⎣
⎣
+
s + ζω n − ω n ζ 2 − 1
s + ζω n + ω n ζ 2 − 1
⎡ 2ζ 1 ⎡
2ζ 2 − 1 ⎤ −⎛⎜⎝ ζω n −ωn ζ 2 −1 ⎞⎟⎠ t 1 ⎡
2ζ 2 − 1 ⎤ −⎛⎜⎝ ζω n +ωn ζ 2 −1 ⎞⎟⎠ t ⎤
ˆ
⎥u (t )
⎢
θ (t ) = K∆ω t −
+ ⎢2ζ −
+ ⎢2ζ +
⎥e
⎥e
2
2
2
⎥
⎢ ω n 2 ⎢⎣
⎢⎣
ζ − 1 ⎥⎦
ζ − 1 ⎥⎦
⎦
⎣
⎡ 2ζ
⎞⎤
2
2
1 −ζω nt ⎛⎜ 2ζ 2 − 1 ωn ζ 2 −1t
2ζ 2 − 1 −ωn ζ 2 −1t
e
e
e
= K∆ω ⎢t −
+
+ 2ζe ωn ζ −1t −
+ 2ζe −ωn ζ −1t ⎟⎥u (t )
⎜ ζ 2 −1
⎟⎥
⎢ ω n 2ω n
ζ 2 −1
⎝
⎠⎦
⎣
⎡ 2ζ
⎞⎤
1 −ζω nt ⎛⎜
2ζ 2 − 1
e
2ζ cosh ω n ζ 2 − 1t +
sinh ω n ζ 2 − 1t ⎟⎥u (t )
= K∆ω ⎢t −
+
⎜
⎟⎥
⎢ ωn ωn
ζ 2 −1
⎝
⎠⎦
⎣
(
)
(
)
Case 2, ζ = 1 :
2 K∆ω
ωn
K∆ω
ˆ (s ) = ∆ω Kω
Θ
= 2 −
2
2
s
s (s + ω n )
2
n
⎡
2
⎣
ωn
θˆ(t ) = K∆ω ⎢t −
+
s
e −ωnt + te −ωnt +
2 K∆ω
K∆ω
(s + ω n )
2
+
ωn
s + ωn
⎤
e −ωnt ⎥u (t )
ωn
⎦
2
Case 3, 0 < ζ < 1 :
ˆ (s ) =
Θ
=
Kω n2
K∆ωω n2
∆ω
=
s 2 + 2ζω n s + ω n2 s 2
s 2 s + ζω n − jω n 1 − ζ 2 s + ζω n + jω n 1 − ζ
K∆ω
−
s2
2ζ
(
K∆ω
ωn
s
⎡
2ζ
⎢
⎣
ωn
θˆ(t ) = K∆ω ⎢t −
+
+
(
)
K∆ω ⎡
2ζ 2 − 1 ⎤
⎢2ζ − j
⎥
2
2ω n ⎢
⎥⎦
−
1
ζ
⎣
s + ζω n − jω n 1 − ζ
(
)(
2
+
)
1⎡
2ζ 2 − 1 ⎤ −⎛⎜⎝ ζω n − jωn
⎢2ζ − j
⎥e
2
2⎢
⎥⎦
−
1
ζ
⎣
⎡ 2ζ e −ζω nt
= K∆ω ⎢t −
+
ωn
⎢ ωn
⎣
(
)
)
)
K∆ω ⎡
2ζ 2 − 1 ⎤
⎢2ζ + j
⎥
2
2ω n ⎢
⎥⎦
−
1
ζ
⎣
s + ζω n + jω n 1 − ζ
1−ζ 2 ⎞⎟ t
⎠
(
(
2
+
2
(
)
1⎡
2ζ 2 − 1 ⎤ −⎛⎜⎝ ζω n + jω n
⎢2ζ − j
⎥e
2
2⎢
⎥⎦
−
1
ζ
⎣
)
(
)
2
⎛
⎞⎤
⎜ 2ζ cos ω 1 − ζ 2 t + 2ζ − 1 sin ω 1 − ζ 2 t ⎟⎥u (t )
n
n
⎜
⎟⎥
1−ζ 2
⎝
⎠⎦
1−ζ 2 ⎞⎟ t
⎠
⎤
⎥u (t )
⎥
⎦
k p k 0 F (s )
In general, H (s ) =
C.5
(a)
s + k p k 0 F (s )
F (s ) = k
H (s ) =
H ( jω ) =
H ( jω ) =
2
(k
k0 k )
2
p
ω 2 + (k p k 0 k )2
=
k p k0 k
s + k p k0 k
k p k0 k
jω + k p k 0 k
(k
k0 k )
2
p
ω 2 + (k p k 0 k )2
k p k0 k
1
⇒ ω 3 = k p k 0 k ⇒ B3 =
2
2π
B3 k p k 0 k
4
2
=
×
= <1
Bn
k p k0 k π
2π
So, B3 < Bn
(b)
F (s ) =
k
s+k
H (s ) =
ω n2
s 2 + 2ζω n s + ω n2
H ( jω ) =
H ( jω ) =
2
(ω
ω n4
2
n
−ω
) + (2ζω ω )
2 2
2
=
n
B3 =
B3 ω n
=
1 − 2ζ 2 +
Bn 2π
So, B3 < Bn
ω n2
ω n2 − ω 2 + j 2ζω nω
(ω
ω n4
2
n
−ω2
) + (2ζω ω )
2
2
n
1
⇒ ω 3 = ω n 1 − 2ζ 2 +
2
(1 − 2ζ )
2 2
+1
ωn
2
1 − 2ζ 2 + (1 − 2ζ 2 ) + 1
2π
(1 − 2ζ )
2 2
+1 ×
8ζ
ωn
=
8ζ
1 − 2ζ 2 +
2π
(1 − 2ζ )
2 2
+ 1 < 1 for all ζ > 0
(c)
F (s ) = k 1 +
k2
s
2ζω n s + ω n2
H (s ) = 2
s + 2ζω n s + ω n2
H ( jω ) =
H ( jω ) =
2
(ω
ω n4 + (2ζω nω )2
2
n
−ω
) + (2ζω ω )
2 2
2
=
n
j 2ζω nω + ω n2
ω n2 − ω 2 + j 2ζω n ω
(ω
ω n4 + (2ζω nω )2
2
n
−ω2
) + (2ζω ω )
2
2
n
1
⇒ ω 3 = ω n 1 + 2ζ 2 +
2
(1 + 2ζ )
2 2
ωn
2
1 + 2ζ 2 + (1 + 2ζ 2 ) + 1
2π
B3 ω n
2
=
1 + 2ζ 2 + (1 + 2ζ 2 ) + 1 ×
Bn 2π
+1
B3 =
2
⎛
ω n ⎜⎜ ζ +
⎝
=
2
1 1 + 2ζ +
π
(1 + 2ζ )
1
ζ+
4ζ
2 2
+1
1
4ζ
⎞
⎟⎟
⎠
< 1 for all ζ > 0
(d)
F (s ) =
H (s ) =
H ( jω ) =
H ( jω ) =
2
(ω
C2
2
n
−ω
) + (2ζω ω )
2 2
2
=
n
k1 + s
k2 + s
⎛
k
where C = ω n2 + ω n ⎜⎜ 2ζ − 2
ωn
⎝
C
s + 2ζω n s + ω
2
2
n
⎞
⎟⎟
⎠
C
ω − ω + j 2ζω nω
2
n
(ω
2
C2
2
n
) + (2ζω ω )
⇒ ω = ω (2ζ
−ω2
C2
2
2
2
n
2
n
3
2
)
(
) (
− 1 + ω n4 2ζ 2 − 1 − ω n4 − 2
2
(
)
(
) (
)
(
)
(
) (
)
2
1
ω n2 2ζ 2 − 1 + ω n4 2ζ 2 − 1 − ω n4 − 2
2π
B3
2
1
ω n2 2ζ 2 − 1 + ω n4 2ζ 2 − 1 − ω n4 − 2 ×
=
Bn 2π
B3 =
8ζ
=
2πω n
)
8ζ
2
⎡ ⎛
k2 ⎞ ⎤
ω n ⎢1 + ⎜⎜ 2ζ − ⎟⎟ ⎥
ωn ⎠ ⎥
⎢⎣ ⎝
⎦
ω n2 (2ζ 2 − 1) + ω n4 (2ζ 2 − 1) − (ω n4 − 2 )
2
⎛
k ⎞
1 + ⎜⎜ 2ζ − 2 ⎟⎟
ωn ⎠
⎝
2
This is a mess. It appears to be less than 1 for all values of ζ, but it also depends on k2 and ωn.
C.6
H a (s ) =
2ζω n s + ω n2
k
where ζ = 1
2
2
2
s + 2ζω n s + ω n
k0k p
k2
and ω n = k 0 k p k 2
If k1 = 0, then ζ = 0 and we have
H a (s ) =
which has two poles at s = ± jω n .
This is an oscillator, which we do not want!
ω n2
s 2 + ω n2
C.7
k0 k p k
(a) H a (s ) =
s + k0 k p k
=
τ
s +τ
τT
2
τT
τT
2
+
z −1
τT
1+
1+
⎛ 2 1 − z −1 ⎞
τ
2
2
⎟=
=
(b) H a ⎜⎜
−1 ⎟
−1
τ
T
T
1
2
1
+
−
z
z
⎝
⎠
1−
+τ
2 z −1
T 1 + z −1
1−
τT
1+
2
(c) H d ( z ) =
1−
(d)
K p K 0 Kz −1
1 − (1 − K p K 0 K )z −1
τT
2 = 1− K K K
p
0
τT
1+
2
(e) Bn =
k0 k p k
4
⇒ K p K0 K =
⇒ k 0 k p k = 4 Bn
τT
=
τT
1+
2
k 0 k p kT
1
1 + k 0 k p kT
2
⇒ K0 K p K =
4 Bn T
1 + 2 Bn T
C.8
ω n2
(a) H a (s ) = 2
s + 2ζω n s + ω n2
(b)
⎛ 2 1 − z −1 ⎞
ω n2
⎟=
H a ⎜⎜
−1 ⎟
2
⎛ 2 1 − z −1 ⎞
⎝ T 1 + z ⎠ ⎛ 2 1 − z −1 ⎞
⎟ + ω n2
⎟
⎜
⎜⎜
2
ζω
+
n⎜
−1 ⎟
−1 ⎟
⎝T 1+ z ⎠
⎝T 1+ z ⎠
=
z −2
where θ n =
ω nT
2
K p K0
(c) H d ( z ) =
z −2 +
(d) K =
θ n2
(
z − 2 + z −1 + 1)
2
1 − 2ζθ n + θ n
θ n2 − 1
1 + 2ζθ n + θ n2
−1
+2
+
z
1 − 2ζθ n + θ n2
1 − 2ζθ n + θ n2
z −1
K
K p K0 − K −1
K
z −1 +
1
K
1 − 2ζθ n + θ n2
1 + 2ζθ n + θ n2
ωn
ωT
1
we have Bn T = n =
8ζ
8ζ
4ζ
⎛ ω nT ⎞ θ n
⎜
⎟=
⎝ 2 ⎠ 4ζ
Making this substitution into the answer from part (d) gives
(e) Using Bn =
K=
⇒ θ n = 4ζ (BnT )
1 − 8ζ 2 (BnT ) + 16ζ 2 (BnT )
2
1 + 8ζ 2 (BnT ) + 16ζ 2 (BnT )
2
C.9
⎡
(a) H a (s ) =
k
1 ⎛
⎜⎜ 2ζ − 2
ωn
⎣ ωn ⎝
s 2 + 2ζω n s + ω n2
ω n2 ⎢1 +
⎞⎤
⎟⎟⎥
⎠⎦
=
ω n2 C
s 2 + 2ζω n s + ω n2
(b)
⎛ 2 1 − z −1 ⎞
ω n2 C
⎟
=
H a ⎜⎜
−1 ⎟
2
⎝ T 1 + z ⎠ ⎛ 2 1 − z −1 ⎞
⎛ 2 1 − z −1 ⎞
⎜⎜
⎟
⎟ + ω n2
+ 2ζω n ⎜⎜
−1 ⎟
−1 ⎟
⎝T 1+ z ⎠
⎝T 1+ z ⎠
θ n2 C
(z −2 + z −1 + 1)
1 − 2ζθ n + θ n2
=
θ n2 − 1
1 + 2ζθ n + θ n2
−1
+
z −2 + 2
z
1 − 2ζθ n + θ n2
1 − 2ζθ n + θ n2
K p K0
(c) H d ( z ) =
z −2 +
K 2 − K p K 0 K1
K p K0 − K2 −1
K 2 − K p K 0 K1
(
z −1 1 − K 1 z −1
z −1 +
)
1
K 2 − K p K 0 K1
(d)
K p K0 − K2 −1
K 2 − K p K 0 K1
=
(
)
2 θ n2 − 1
1 − 2ζθ n + θ n2
1 − 2ζθ n + θ n2
K 2 − K p K 0 K1 =
1 + 2ζθ n + θ n2
K1 = 1 −
⇒
4θ n2
1
K p K 0 1 + 2ζθ n + θ n2
1 − 2ζθ n − 3θ n2
K2 = K p K0 +
1 + 2ζθ n + θ n2
Digital Communications: A Discrete-Time Approach
M. Rice
Errata
Foreword
Page xiii, first paragraph, “bare witness” should be “bear witness”
Page xxi, last paragraph, “You know who you.” should be “You know who you are.”
Chapter 1
Page 3, second new paragraph, “Pittsburg” should be “Pittsburgh”
Page 9, The end of the second lines reads, “… signal sideband AM.” This should be “… single
sideband AM.”
Page 10, the second line of Section 1.2, “information baring” should be “information bearing”
Page 12, Equation (1.1) should read
Page 14, Second new paragraph, “The same power/bandwidth exists with digitally modulated
carriers.” should read, “The same power/bandwidth trade off exists with digitally modulated
carriers.”
Chapter 2
Page 27, The sentence after Equation (2.14) should read
“An energy signal is a signal with finite nonzero energy whereas a power signal is
a signal with finite nonzero power.”
Page 28, Equation (2.19) should read
Page 28, Equation (2.20) should read
Page 33, Equation (2.30) should read
Page 36, Equation (2.36) should read
Page 37: Equation (2.37) should read
Page 40, the sixth row in Table 2.4.4 should be
Page 41, second line of text, “complex plain” should be “complex plane”
Page 50, Equations (2.57) and (2.58) should read
Page 56, Equation (2.72) should be
Page 65, the third paragraph of Section 2.6.2. The two occurrences of
should be
Page 75, Exercise 2.24, the equation for
should be
Page 76, Exercise 2.27, the line after the equation should read
“where the integral on the left is the power contained in the interval
Page 83, Exercise 2.46, the plot for
should be
Page 83, Exercise 2.47, the plot for
should be
Page 84, the second figure of Exercise 2.48 should be
Page 85: The first figure of Exercise 2.50 should be
…”
Page 86: The figure at the top of the page (Exercise 2.50) should be
Page 86: The first Figure in Exercise 2.51 should be
Page 87, The figure at the top of the page (Exercise 2.51) should be
Page 95, The figure of Exercise 2.63 should be
Page 97, Exercise 2.67. The Fibonacci sequence is
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 377, 610, 987, …
Page 98, Exercise 2.67. The initial conditions should be
.
Page 98, Exercise 2.69 should begin “Consider an LTI system with input …”
Page 99, Exercise 2.70 should begin “Consider an LTI system with input …”
Page 99, the figure of Exercise 2.71 should be
Page 102, the figure of Exercise 2.77 should be
Page 103, Exercise 2.81 (c) should read
Page 103, Exercise 2.81 (f) should read
Page 103, Exercise 2.82 (a) should read
Page 106, Exercise 2.90. The plot for
should be
and the statement after the figures should read
Show that the length-4 DFT of both series have the same magnitude. (Hint: do the
analysis in the time domain.)
Chapter 3
Page 125, Forth line after equation (3.16):
should be
Page 125, Fifth line after equation (3.16):
should be
Page 132, Equation (3.23) should read
Page 133, Equation (3.24) should read
Page 138, second line “… Figure 3.3.8 requires three multipliers,…” should read “… Figure
3.3.8 requires four multipliers,…”
Page 154, Equation (3.66) should read,
Page 165, the figure for Exercise 3.15 (b) should be
Page 174, Exercise 3.33 (b) should read
Sample the impulse response at instants
impulse response
.
and scale by
to produce a discrete-time
Page 175, Exercise 3.37, The last sentence should read “Show that for small
approximately
.
, the DTFT is
Chapter 4
Page 195, Equation (4.45) should be
Page 198, Equation (4.52): µ should be boldface so that Equation (4.52) reads as follows:
Page 198, Equation (4.54): µ should be boldface so that Equation (4.54) reads as follows:
Page 198, Equation (4.55): AMA should be boldface so that Equation (4.55) reads as follows:
Page 200, the last equation on the page.
should be
.
Page 201, figure 4.4.1 should be
Page 201, the first line of the equation for
should be
Page 201, Example 4.4.1: In the properties of the Gaussian sequence,
Page 202, Figure 4.4.2 should be
should be
.
Page 211, Exercise 4.19: µ should be boldface.
Page 211, Exercise 4.20: µ should be boldface.
Page 212, Exercise 4.23: part (b) should read
(b) Compute the probability that
.
Page 212, Exercise 4.24: The autocorrelation function should be
Page 212, Exercise 4.24: part (c) should read
(c) Compute the probability that
.
Page 212, Exercise 4.26: the impulse response should be written as follows:
Page 213, Exercise 4.28: the impulse response should be written as follows:
Chapter 5
Page 225, Equation (5.17) should read
Page 229, First line after equation (5.29): 2 should be
.
Page 234, Four lines below equation (5.40), ADC should be DAC.
Page 239, Equation (5.53) should be
Page 239, Equation (5.54) should be
Page 239, Equation (5.56) should be
Page 241, Equation (5.62) should be
Page 249, Figure 5.3.11 should be
Page 253, Figure 5.3.13 should be
Page 254, The table at the end of Example 5.3.9 should be
k
0
+1.00
−0.95
+1
−1
1
+1.02
+1.02
+1
+1
2
−0.97
+0.98
−1
+1
3
−0.97
−1.00
−1
−1
Page 270, Figure 5.5.2 (b) should be
Page 273, the sentence just above Equation (5.107) should read, “Using Bayes’ rule, …”
Page 286, The constellation plot in Exercise 5.17 should be
Page 286, The constellation plot of Exercise 5.18 should be
Page 286, The constellation of Exercise 5.19 should be
Page 286, The constellation of Exercise 5.20 should be
Page 287, The constellation of Exercise 5.21 should be
Page 291, The constellation of Exercise 5.30 should be
Page 291, The constellation of Exercise 5.31 should be
Page 291, The constellation of Exercise 5.32 should be
Page 292, Exercise 5.33 should read
Consider a binary PAM system using the SRRC pulse shape.
(a) Produce an eye diagram corresponding to 200 randomly generated symbols for
a pulse shape with 100% excess bandwidth and
. Use a sampling rate
equivalent to 16 samples/symbol.
(b) Repeat part (a) for a pulse shape with 50% excess bandwidth and
.
(c) Repeat part (a) for a pulse shape with 25% excess bandwidth and
.
(d) Compare and contrast the eye diagrams from parts (a) – (c).
Page 292, Exercise 5.34 should read
Consider a 4-ary PAM system using the SRRC pulse shape.
(a) Produce an eye diagram corresponding to 200 randomly generated symbols for
a pulse shape with 100% excess bandwidth and
. Use a sampling rate
equivalent to 16 samples/symbol.
(b) Repeat part (a) for a pulse shape with 50% excess bandwidth and
(c) Repeat part (a) for a pulse shape with 25% excess bandwidth and
.
.
(d) Compare and contrast the eye diagrams from parts (a) – (c).
Page 300, The figure for Exercise 5.46 should be
Page 301, The first line after the figure should read “… the average energy is 1.215 J …”
Chapter 6
Page 306, Equation (6.2) should be
Page 306, Equation (6.3) should be
Page 313, The legend of Figure 6.1.3 is incorrect. Figure 6.1.3 should be as follows:
Page 314, Equation (6.30) should be
Page 314, Equation (6.31) should be
Page 325, the sentence just above Equation (6.87) should read, “The union bound for the
conditional probabilities are”
Page 343, Footnote 5. The last line should read “… utility of these added features.”
Page 349, Exercise 6.10 should read as follows:
Use the union bound to upper bound the probability of bit error for the 4+12+16 APSK
constellation shown in Figure 5.3.5 using
,
,
,
,
,
and
.
Page 357-358, Exercise 6.38. The paragraph (on page 357) that begins “The other kind of
repeater is the regenerative repeater” should be part (c). Part (d) is
Suppose a 1 Mbit/s QPSK link has
dB W/Hz and
dB W/Hz.
Which type of repeater provides the lowest composite bit-error rate: the bent-pipe
repeater or the regenerative repeater?
Chapter 7
Page 424, Exercise 7.7: (7.45) should be (7.44).
Chapter 8
Page 445, Equation (8.20) should be as follows:
Page 488, In the caption of Figure 8.4.26, “Figure 8.4.26” should be “Figure 8.4.25”.
Page 471, The lower diagram of Figure 8.4.16 should be
Page 507, Equation (8.130) should be
Page 507, The exponential term of the first line of equation (8.131) should be
Pages 517-518, Exercises 8.12-8.15 and 8.18 should be deleted. The remaining exercises should
be renumbered in the obvious way.
Chapter 9
Page 560, the first line under Figure 9.3.3. Equation (9.60) should be (9.59).
Page 560, the first new paragraph the text “… a filter’s transfer frequency response …” should
be “ … a filter’s transfer function …”
Page 562: Equation (9.61) should be
Page 579: Figure 9.4.1 should be
Page 580: Table 9.4.1 should be
k
0
1
2
3
4
5
6
7
(degrees)
45.00
26.57
14.04
7.13
3.58
1.79
0.90
0.45
(degrees)
45.00
71.57
57.53
50.40
46.83
48.62
49.51
49.96
(degrees)
+50.00
+5.00
-21.57
-7.53
-0.40
+3.17
+1.38
+0.49
+1
+1
-1
-1
-1
+1
+1
+1
+1.00
+1.00
+0.50
+0.88
+1.05
+1.13
+1.09
+1.07
+0.00
+1.00
+1.50
+1.38
+1.27
+1.20
+1.24
+1.25
Page 581: Figure 9.4.2 should be
90
k
!
80
θ
70
angles (degrees)
δn−1 θn
n=0
60
50
40
30
20
10
0
0
1
2
3
4
iteration index (k)
5
6
7
Page 582: The first line of equation (9.107) should be
Page 587: The first line of equation (9.117) should be
Page 596: The first word of the second sentence of the third new paragraph on the page (10 lines
from the bottom of the page) should be CoRDiC.
Page 602: The first equation of Exercise 9.25 should be
Page 602: The first equation of Exercise 9.26 should be
Page 603: The equation of Exercise 9.26 (c) should be
Chapter 10
Page 660: Figure 10.2.18 should be
Page 661: line 15, the parenthetical statement should read
(ratio of RF power out to DC power in)
Page 662: The matrix equation for Exercise 10.1 (a) should be
Page 663: The matrix equation for Exercise 10.3 (a) should be
Page 664: The second line of Exercise 10.6 should reference Figure 10.1.2 (d).
Page 665: Exercise 10.8 should read
The QAM modulator of Exercise 10.6 was derived from the QAM modulator of Figure
10.1.2 (d) by using polyphase partitions of
and
. The QAM modulator of
Exercise 10.7 was derived from the QAM modulator of Figure 10.1.4 by using a
polyphase partition of
. Show that the filterbanks in these two modulators are
exactly identical.
Page 668: The second line of Exercise 10.20 should begin “consists of 124 channels …”
Page 669: The second line of Exercise 10.22 should begin “consists of 124 channels…”
Page 669, Exercise 10.26: the last sentence above the figure should read
Assume
and
.
Page 669, Exercise 10.26 (b) should read
Derive an expression for
in terms of
,
, and
.
Page 670, Exercise 10.28 (a) should read
Express the product of
and the LO in terms of baseband signals and double
frequency signals (centered at
rads/s).
Appendix A
Page 677, Equation (A.9) should read
Page 679, Equation (A.11) should read
Page 684, Figure A.2.4 should be
Appendix B
Page 715, The equation below the first line of Exercise B.18 should be
Page 715, The equation below the first line of Exercise B.18 (b) should be
Bibliography
Page 752. Ref. 23 should be
T. Cover and J. Thomas, Elements of Information Theory, John Wiley & Sons, New York, 1991.
Page 758: Ref.182 should be
C. Dick and f. harris, “On the structure, performance, and applications of recursive all-pass
filters with adjustable and linear group delay,” Proceedings of the IEEE International
Conference on Acoustics, Speech, and Signal Processing, Orlando, FL, May 14-17, 2002, pp.
1517-1520.
