Solutions Manual
to accompany
Communication
Systems
An Introduction to Signals and Noise in
Electrical Communication
Fourth Edition
A. Bruce Carlson
Rensselaer Polytechnic Institute
Paul B. Crilly
University of Tennessee
Janet C. Rutledge
University of Maryland at Baltimore
Solutions Manual to accompany
COMMUNICATION SYSTEMS: AN INTRODUCTION TO SIGNALS AND NOISE IN ELECTRICAL COMMUNICATION,
FOURTH EDITION
A. BRUCE CARLSON, PAUL B. CRILLY, AND JANET C. RUTLEDGE
Published by McGraw-Hill Higher Education, an imprint of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas,
New York, NY 10020. Copyright © The McGraw-Hill Companies, Inc., 2002, 1986, 1975, 1968. All rights reserved.
The contents, or parts thereof, may be reproduced in print form solely for classroom use with COMMUNICATION SYSTEMS: AN
INTRODUCTION TO SIGNALS AND NOISE IN ELECTRICAL COMMUNICATION, provided such reproductions bear copyright
notice, but may not be reproduced in any other form or for any other purpose without the prior written consent of The McGraw-Hill
Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance
learning.
www.mhhe.com
Chapter 2
2.1-1
Ae jφ
T0
cn =
 Ae jφ n = m
j 2 π ( m−n )f 0t
jφ
e
dt
=
Ae
sinc(
m
−
n
)
=

∫− T0 / 2
0 otherwise
T0 / 2
2.1-2
c0 v (t ) = 0
cn =
2
T0
n
cn
∫
T0 / 4
0
0
0
A cos
T0 / 2
2π nt
2π nt
2A
πn
dt + ∫ ( − A)cos
dt =
sin
T
/
4
0
T0
T0
πn
2
1
2A/π
2
0
4
0
5
2 A / 5π
±180°
0
arg cn
3
2 A / 3π
6
0
7
2 A / 7π
±180°
0
2.1-3
c0 = v (t ) = A / 2
cn =
2
T0
n
∫
T0 /2
0

2 At 
2π nt
A
A
dt =
sin π n −
(cos π n − 1)
A−
 cos
T0 
T0
πn
(π n) 2

0
1
2
3
4
5
6
0.5A 0.2A 0 0.02A 0 0.01A 0
cn
0
arg cn
0
0
0
2.1-4
c0 =
2
T0
∫
T0 / 2
0
A cos
2π t
=0
T0
(cont.)
2-1
2
cn =
T0
2π t
2π nt
2 A  sin (π − π n ) 2t / T0 sin (π + π n ) 2t / T0 
A cos
cos
dt =
+


T0
T0
T0  4(π − π n) / T0
4(π + π n ) / T0  0
T0 / 2
∫
T0 / 2
0
n = ±1
A/2
A
[ sinc(1 − n) + sinc(1 + n )] = 
2
otherwise
 0
=
2.1-5
c0 = v (t ) = 0
cn = − j
2
T0
∫
T0 / 2
0
n
cn
1
2A/π
arg cn
−90°
A sin
2
0
2π nt
A
dt = − j
(1 − cos π n )
T0
πn
3
2 A / 3π
4
5
2 A / 5π
−90°
−90°
2.1-6
c0 = v(t ) = 0
2
cn = − j
T0
= −j
2π t
2π nt
2 A  sin (π − π n ) 2t / T0 sin ( π + π n ) 2t / T0 
A sin
sin
dt = − j
−


T0
T0
T0  4(π − π n ) / T0
4(π + π n )/ T0  0
T0 / 2
∫
T0 / 2
0
n = ±1
m jA / 2
A
[sinc(1 − n ) − sinc(1 + n ) ] = 
2
otherwise
 0
2.1-7
cn =
T0
1  T0 / 2
v ( t) e− jnω0 t dt + ∫ v(t )e − jnω 0t dt ]
∫
T0 / 2
T0  0
where
∫
T0
T0 / 2
v(t )e − jnω0 t dt = ∫
T0 / 2
0
v (λ + T0 /2) e− jnω 0λ e− jnω 0T0 / 2 d λ
= −e jnπ ∫
T0 / 2
0
v (t )e − jnω0 t dt
since e jnπ = 1 for even n, cn = 0 for even n
2-2
2.1-8
∞
P = c0 + 2 ∑ cn = Af 0τ + 2 Af 0τ sinc f 0τ + 2 Af 0τ sinc2 f 0τ + 2 Af 0τ sinc3 f 0τ + L
2
2
2
2
2
2
n =1
1
= 4 f0
τ
1
A2
f >
P=
τ
16
where
1 + 2sinc2 1 + 2sinc2 1 + 2sinc2 3  = 0.23 A2

4
2
4 
2
A2 
1
1
3
5
3
7
f >
P=
1 + 2sinc2 + 2sinc2 + 2sinc2 + 2sinc2 + 2sinc 2 + 2sinc 2  = 0.24 A2

τ
16 
4
2
4
4
2
4
2
1
A 
1
1
f >
P=
1 + 2sinc2 + 2sinc 2  = 0.21A2

2τ
16 
4
2
2.1-9
 0

cn =  2 2
 π n 


1
a) P =
T0
n even
n odd
2
 4t 
2
∫−T0 / 2  1 − T0  dt = T0
T0 / 2
2
2
2
∫
T0 / 2
0
 4t 
1
 1 −  dt =
3
 T0 
2
 4 
 4 
 4 
P′ = 2  2  + 2  2  + 2 
= 0.332 so P′ / P = 99.6%
2 
π 
 9π 
 25π 
8
8
8
b) v′(t ) = 2 cos ω 0t + 2 cos3ω 0t +
cos5ω 0t
π
9π
25π 2
2.1-10
 0

cn =  − j 2
 π n
1
a) P =
T0
n even
n odd
 2 2  2  2  2 2 
′
∫−T0 / 2 (1) dt = 1 P = 2  π  +  3π  +  5π   = 0.933 so P′ / P = 93.3%


T0 / 2
2
(cont.)
2-3
4
4
4
cos ( ω0t − 90° ) +
cos ( 3ω0t − 90° ) +
cos ( 5ω 0t − 90°)
π
3π
5π
4
4
4
= sin (ω 0t ) +
sin ( 3ω 0t ) +
sin ( 5ω0t )
π
3π
5π
b) v′(t ) =
2.1-11
1
P=
T0
2
∫
T0
0
n=0
1 / 2
cn = 
1 / 2π n n ≠ 0
 t 
1
  dt =
3
 T0 
∞
 2 
2  1 1 1
 1
P =2∑ 
 = 2    4 + 4 + 4 +L =
π  1 3 5
 3
n odd  π n 
4
4
1 1 1
4π 2
+
+
+
L
=
12 22 32
2
Thus,
1 1  π2
3− 4= 6


2.1-12
2
P=
T0
2
∫
T0 / 2
0
 4t 
1
 1 −  dt =
3
 T0 
n even
0
cn = 
2
 (2/ π n) n odd
∞
1 2
1
 1 
P =   + 2∑ 
 = + 2
4 4π
2
n =1  2π n 
2
2
1 1 1
 1
 2 + 2 + 2 +L =
2
3
1
 3
1 1 1
π4 1 π 4
Thus, 4 + 4 + 4 + L =
=
1 3 5
2 ⋅ 24 3 96
2.2-1
πt
cos2π ftdt
τ
τ

π
 sin τ −2π f 2 sin
= 2A 
+
π −2π f
2
2

τ

V ( f ) = 2∫
τ /2
0
A cos
(
(
)
)
( πτ 2π f ) τ2  Aτ
[ sinc( f τ −1/2) + sinc( f τ + 1/2) ]
=
(πτ 2π f )  2
+
+

(cont.)
2-4
2.2-2
2π t
cos2π ftdt
0
τ

2π −2π f τ sin
sin

τ
2−
= − j2 A 
2
π
2
 2 τ −2π f

V ( f ) = − j 2∫
τ /2
A sin
(
(
)
)
( 2τπ 2π f ) τ2  = − j Aτ sinc( f τ −1) − sinc( f τ + 1)
[
]
( 2τπ 2π f )  2
+
+
2.2-3
τ
t
2 Aτ 
 ωτ
V ( f ) = 2∫  A − A  cos ω tdt =
2sin 2 
2

0
τ
(ωτ ) 

 2


2
 − 1 + 1 = Aτ sinc f τ


2.2-4
τ
t
2 Aτ
V ( f ) = − j 2∫ A sin ω tdt = − j
(sin ωτ − ωτ cos ωτ )
0
τ
(ωτ ) 2
A
= −j
(sinc2 f τ − cos2π f τ )
πf
2.2-5
v (t ) = sinc2Wt ↔
∫
∞
−∞
1
 f 
Π

2W  2W 
sinc2Wt dt = ∫
2
∞
−∞
2
∞
1
1
1
 f 
Π
df = ∫
df =

2
−∞ 4W
2W  2W 
2W
2-5
2.2-6
W
A2
A2
A2
2πW
E′ = 2∫ 2
df
=
arctan
2
0
0 b + (2π f )
2b
πb
b
E′ 2
2πW 50% W = b / 2π
= arctan
=
E π
b
84% W = 2b / π
E=∫
∞
( Ae − bt ) dt =
2
2.2-7
∞
 ∞ W ( f ) e jω t df dt
v
(
t
)
w
(
t
)
dt
=
v
(
t
)
∫
−∞
−∞
 ∫−∞

∞
∞
∞
= ∫ W ( f )  ∫ v (t) e− j (− ω ) t dt  df = ∫ W ( f )V (− f ) df
−∞
 −∞

−∞
∫
∞
V ( − f ) = V * ( f ) when v (t) is real, so
∫
∞
−∞
∞
∞
−∞
−∞
v 2 ( t ) dt =∫ V ( f )V * ( f )df = ∫ V ( f ) df
2
2.2-8
∗
∗
∞
∞
∫−∞ w∗ (t)e− j 2π ft dt =  ∫−∞ w(t )e j 2π ft dt  =  ∫−∞ w( t)e− j 2π ( − f )t dt  = W ∗ ( f )
Let z ( t ) = w∗ ( t ) so Z ( f ) = W ∗ ( − f ) and W ∗ ( f ) = Z ( − f )
∞
Hence
∫
∞
−∞
∞
v( t )z( t) dt = ∫ V ( f ) Z ( − f ) df
−∞
2.2-9
1  f 
t 
Π   ↔ A sinc Af so sinc At ↔ Π  
A  A
 A
2t
τ  fτ 
2
v (t ) = sinc ↔ V ( f ) = Π 
 for A =
τ
2  2 
τ
2.2-10
πt  t 
Bτ
B cos Π   ↔
[ sinc( f τ − 1/2) + sinc( f τ + 1/2) ]
τ τ 
2
Bτ
π (− f )  − f 
πf  f 
so
Π
Π 
[sinc(tτ − 1/2) + sinc(tτ + 1/2) ] ↔ B cos
 = B cos
2
τ
τ
 τ 
τ 
Let B = A and τ = 2W ⇒ z (t ) = AW [sinc(2Wt − 1/2) + sinc(2Wt + 1/2) ]
2.2-11
2π t  t 
Bτ
B sin
Π  ↔− j
[sinc( f τ − 1) + sinc( f τ + 1) ]
τ
2
τ 
Bτ
2π (− f )  − f 
2π f  f 
so − j
Π
Π 
[ sinc(tτ − 1) + sinc( tτ + 1) ] ↔ B sin
 = − B sin
2
τ
τ
 τ 
τ 
Let B = − jA and τ = 2W ⇒ z ( t ) = AW [ sinc(2Wt − 1) + sinc(2Wt + 1) ]
2-6
2.2-12
e −b t ↔
∫
∞
−∞
2b
4π a
a /π
⇒ e−2π a t ↔
= 2
2
2
2
b + (2π f )
(2π a ) + (2π f )
a + f2
2
(e
−2 π a t
Thus,
∫
)
2
∞
0
2
∞
1
a /π
a
dt =
=∫
df = 2  
2
2
2π a −∞ a + f
π 
(a
∞
0
(a
df
2
+ f2)
2
2
dx
2
∫
+ x2 )
2
1 π  1
π
=  
= 3
2  a  2π a 4 a
2.3-1
z (t ) = v (t − T ) + v( t + T ) where v( t ) = AΠ (t /τ ) ↔ Aτ sinc f τ
so Z( f ) = V ( f ) e− jω T + V ( f )e jωT = 2 Aτ sinc f τ cos2π fT
2.3-2
z (t ) = v (t − 2T ) + 2v( t ) + v( t + 2T ) where v(t ) = aΠ (t /τ ) ↔ Aτ sinc f τ
Z ( f ) = V ( f )e − j 2ωT + V ( f ) + V ( f )e j2ωT = 2 Aτ (sinc f τ )(1 + cos4π fT )
2.3-3
z (t ) = v (t − 2T ) − 2v( t ) + v( t + 2T ) where v (t ) = aΠ (t /τ ) ↔ Aτ sinc f τ
Z ( f ) = V ( f )e − j 2ωT − 2V ( f ) + V ( f )e j2ωT = 2 Aτ (sinc f τ )(cos4π fT − 1)
2.3-4
 t −T 
 t −T / 2 
v (t ) = AΠ 
 + ( B − A)Π 

 2T 
 T

− j ωT
V ( f ) = 2 AT sinc2 fTe
+ ( B − A)T sinc fTe− jω T / 2
2-7
2.3-5
 t − 2T 
 t − 2T 
v (t ) = AΠ 
 + ( B − A)Π 

 4T 
 2T 
V ( f ) = 4 AT sinc4 fTe − j 2ωT + 2( B − A)T sinc2 fTe− j 2ωT
2.3-6
Let w(t ) = v( at ) ↔ W ( f ) =
1
V ( f / a)
a
Then z (t ) = v[a(t − td / a)] = w(t − td / a) so Z ( f ) = W ( f ) e− jω td / a =
1
V ( f / a) e− jω td / a
a
2.3-7
∞
∞
−∞
−∞
F  v (t) e jω ct  = ∫ v(t )e jω ct e− jω t dt =∫ v (t) e− j 2π ( f − fc ) t dt =V ( f − f c )
2.3-8
v (t ) = AΠ (t /τ )cos ωc t with ω c = 2π f c = π / τ
V( f ) =
Aτ
Aτ
Aτ
sinc( f − f c )τ +
sinc( f + f c )τ =
[ sinc( fτ − 1/2) + sinc( f τ + 1/2) ]
2
2
2
2.3-9
v( t ) = AΠ (t /τ )cos(ωc t − π /2) with ω c = 2π f c = 2π / τ
e− jπ / 2
e jπ / 2
Aτ sinc( f − f c )τ +
Aτ sinc( f + f c )τ
2
2
Aτ
= −j
[sinc( f τ − 1) − sinc( f τ + 1) ]
2
V( f ) =
2.3-10
2A
1 + (2π f ) 2
1
1
A
A
Z ( f ) = V ( f − fc ) + V ( f + fc ) =
+
2
2
2
2
2
1 + 4π ( f − f c ) 1 + 4π ( f + f c )2
z (t ) = v (t) c o s ω ct
v(t ) = Ae − t ↔
2.3-11
z (t ) = v ()cos(
t
ωc t − π /2)
v (t ) = Ae −t for t ≥ 0 ↔
A
1 + j 2π f
e − jπ / 2
e jπ / 2
− jA / 2
jA / 2
V ( f − fc) +
V ( f + fc ) =
+
2
2
1 + j 2π ( f − f c ) 1 + j 2π ( f + f c )
A/2
A/2
=
−
j − 2π ( f − f c ) j − 2π ( f + f c )
Z( f ) =
2-8
2.3-12

 ↔ 2 A sinc2 f τ

d
d  sin2π f τ 
2A
 (2πτ )2 f cos2π f τ − 2πτ sin2π f τ 
Z ( f ) = 2A 
=

2 
df
df  2π f τ  (2π f τ )
v (t ) = t z (t ) z (t ) =
A t
Π
τ  τ
1 d
− jA
Z( f ) =
( sinc2 f τ − cos2π f τ )
− j 2π df
πf
V( f ) =
2.3-13
z (t ) = tv (t ) v( t ) = Ae
−b t
↔
2 Ab
2
b + (2π f )
2
1 d 
2 Ab

j 2 Abf
=
 2
2
2
2
− j 2π df  b + (2π f )   b + (2π f ) 2 


Z( f ) =
2.3-14
z (t ) = t 2v( t ) v (t ) = Ae −t for t ≥ 0 ↔
1
Z( f ) =
( − j 2π f )
2
d
df
A
b + j2π f

A

2A
 b + j 2π f  =
3

 [ b + j 2π f ]
2.3-15
2
2
1
v (t ) = e−π (bt) ↔ V ( f ) = e −π ( f / b)
b
2
d
j 2π f −π ( f / b)2
( a) v (t ) = −2π b 2te −π ( bt ) ↔
e
dt
b
2
1 d
f −π ( f / b )2
(b ) te− π (bt) ↔
V( f ) =
e
− j 2π df
jb
Both results are equivalent to bte −π ( bt ) ↔ − jf e−π ( f / b)
2
2.4-1
y (t ) = 0
t<0
t
= ∫ Aλ d λ =
0
2
At
2
2
= ∫0 Aλ d λ = 2A
0<t <2
t >2
2-9
2
2.4-2
y (t ) = 0
t < 0, t > 5
t
= ∫ Aλ d λ =
0
At
2
2
0 <t< 2
= ∫0 Aλ d λ = 2 A
2
2<t <3
2
A
 4 − ( t − 3) 2  3 < t < 5
2
= ∫ Aλ d λ =
t− 3
2.4-3
y (t ) = 0
t < 0, t > 3
At 2
0 < t <1
2
t
A
= ∫ Aλ d λ = (2t − 1)
1< t < 2
t −1
2
2
A
= ∫ Aλ d λ =  4 − (t − 1) 2  2 < t < 3
t −1
2
t
= ∫0 Aλ d λ =
2.4-4
y (t ) = 0
t <4
t
= ∫ 2 Ad λ = 2 At − 8 A 4 ≤ t ≤ 6
4
8
= ∫ 2A dλ = 4 A
6
t >6
2-10
2.4-5
y (t ) = 0
t<0
t
= ∫ 2e−2λ d λ = 1 − e −2t
0 ≤t≤ 2
0
t
= ∫ 2e −2λ d λ = e −2t  e4 − 1
t− 2
t>2
2.4-6
y (t ) = 0
t < −1, t ≥ 3
t
= ∫ 2(1 + λ ) d λ = t2 + 2t + 1
−1 ≤ t < 0
−1
0
t
−1
0
= ∫ 2(1 + λ ) d λ + ∫ 2(1 − λ )d λ = −t 2 + 2t + 1 0 ≤ t < 1
0
1
t− 2
0
= ∫ 2(1 + λ )d λ + ∫ 2(1 − λ ) d λ = −t 2 + 2t + 1 1 ≤ t < 2
1
= ∫ 2(1 − λ ) d λ = t2 − 6t + 9
2≤t <3
t− 2
2.4-7
y (t ) = 0
t≤0
= ∫ Ae− aλ Be− b( t− λ )d λ =[ AB /( a − b)][ e− bt −e − at ] t > 0
t
0
2.4-8
j jπ t j − jπ t
e − e
= B1e −b1t + B2e− b2 t
2
2
y (t ) = v ∗ w1 (t ) + v ∗ w2 (t ) = [ AB1 /(a − b1 )][ e −b1t − e − at ] + [ AB2 /( a −b2 )][ e− b2 t − e − at ]
Let B1 = j /2, b1 = − jπ , B2 = − j /2, b2 = jπ and simplify
v (t ) = Ae −at
w(t ) = sin π t =
2.4-9
∞
v ∗ w(t ) = ∫ v (λ )w(t − λ ) d λ
−∞
let µ = t − λ
−∞
∞
∞
−∞
= − ∫ v (t − µ ) w( µ ) d µ = ∫ w( µ) v (t − µ )d µ = w ∗ v (t )
2-11
2.4-10
∞
Let y (t ) = ∫ v( λ ) w(t − λ )d λ where v( −t ) = v(t ), w(−t ) = w(t )
−∞
∞
∞
y (−t ) = ∫ v (λ )w( −t − λ ) d λ = ∫ v (λ )w(t + λ ) d λ
−∞
−∞
∞
∞
−∞
−∞
= − ∫ v( −µ ) w( t − µ ) d µ = ∫ v( µ ) w(t − µ )d µ = y (t )
2.4-11
∞
Let y (t ) = ∫ v (λ )w(t − λ ) d λ where v ( −t ) = −v (t ), w(−t ) = −w(t)
−∞
∞
∞
y (−t ) = ∫ v (λ )w(−t − λ ) d λ = − ∫ v (λ )w(t + λ ) d λ
−∞
−∞
∞
∞
−∞
−∞
= ∫ v (− µ ) w(t − µ ) d µ = ∫ v (µ ) w( t − µ )d µ = y (t )
2.4-12
Let w(t ) = v ∗ v (t ) = τΛ (t /τ )
t +τ / 2
3
(τ − λ) d λ = τ 2 − t 2 0 ≤ t < τ / 2
4
2
τ
1 3 
= ∫t −τ / 2 (τ − λ) d λ =  t − τ 
τ / 2 ≤ t < 3τ / 2
2 2 
 3 2 2
t <τ / 2
 4τ −t

2
1 
3 
Thus v ∗ v ∗ v (t ) =   t − τ  τ / 2 ≤ t < 3τ / 2
2 
2

0
t ≥ 3τ / 2


v ∗ w(t ) = ∫
0
t −τ / 2
(τ + λ) d λ + ∫
0
2.4-13
F {v (t ) ∗ [ w(t ) ∗ z (t )]} = V ( f ) [W (f )Z ( f )] = [V ( f )W( f ) ] Z ( f )
so v (t ) ∗[w(t ) ∗ z (t) ] = F −1 {[V ( f )W( f )]Z ( f )} = [v( t ) ∗ w(t) ] ∗ z (t )
2-12
2.4-14
1 f 
Π   W ( f ) = 4Π (2 f )
4 4
Y ( f ) = V ( f )W( f ) = Π (2 f ) ↔ y (t ) = (1/2)sinc( t /2)
V( f ) =
2.5-1
Aτ
Aτ
t
z (t ) = AΠ   cos ωc t Z ( f ) =
sinc( f − f c )τ +
sinc( f + f c )τ
2
2
τ 
As τ → 0 the cosine pulse z (t) gets narrower and narrower while maintaining height A.
This is not the same as an impulse since the area under the curve is also getting smaller.
As τ → 0 the main lobe and side lobes of the spectrum Z ( f ) get wider and wider, however
the height gets smaller and smaller. Eventually the spectrum will cover all frequenci es with
almost zero energy at each frequency. Again this is different from what happens in the case
of an impulse.
2.5-2


W ( f ) = v ( f )e − j 2π ftd =  ∑ cv ( nf 0 )δ ( f − nf 0 )  e− j 2π ftd
 n

− j 2π n f0 t d
δ ( f − nf 0 ) ⇒ cw ( nf 0 ) = cv (nf 0 )e − j2π n f0 td
= ∑ cv ( nf 0 ) e
n
2.5-3


W ( f ) = j 2π fV ( f ) = j 2π f  ∑ cv (nf 0 )δ ( f − nf 0 )  = ∑ [ j 2π nf 0 cv ( nf 0 ) ]δ ( f − nf 0 )
 n
 n
⇒ cw ( nf 0 ) = j 2π nf 0 cv ( nf 0 )
2.5-4
1
1

[V ( f − mf0 ) ] = ∑ cv ( nf0 )δ ( f − kf0 − mf0 ) + ∑ cv ( nf0 )δ ( f − kf0 + mf0 ) 
2
2 n
n

1

= ∑ cv [(k − m) f 0 ]δ ( f − kf 0 ) + ∑ cv [(k + m) f 0 ]δ ( f − kf 0 ) 
2 k
k

1
= ∑ {cv [(n − m) f 0 ] + cv [(n + m) f 0 ]}δ ( f − nf 0 )
n 2
1
so cw ( nf 0 ) = {cv [(n − m) f 0 ] + cv [(n + m) f 0 ]}
2
W( f ) =
2-13
2.5-5
v (t ) = Au (t ) − Au (t − 2τ )
 1
1
 1
1
 − j 4π f τ 
V( f ) = A
+ δ ( f ) −
+ δ ( f ) e

 j 2π f 2

 j 2π f 2

− j 4π ft
− j0
But δ ( f ) e
= e δ ( f ), so
V( f ) =
A
1 − e − j 4π fτ ) = 2 Aτ sinc2 f τ e − j 2π f τ
(
j 2π f
 t −τ 
− j 2π f τ
Agrees with v( t ) = Π 
 ↔ 2 Aτ sinc2 f τ e
 2τ 
2.5-6
v (t ) = A − Au (t + τ ) + Au (t − τ )


 1

 1

1
1
V ( f ) = A δ ( f ) − 
+ δ ( f )  e j 2π f τ − 
+ δ ( f )  e− j 2π fτ 
 j 2π f 2

 j 2π f 2



j 2π f τ
− j 2π f τ
j0
But δ ( f ) e
= δ ( f )e
= e δ ( f ), so


1
V ( f ) = A δ ( f ) −
e j 2π f τ − e− j2π f τ )  = Aδ ( f ) − 2 Aτ sinc2 f τ
(
j 2π f


Agrees with v( t ) = A − AΠ (t / 2τ ) ↔ Aδ ( f ) − 2 Aτ sinc2 f τ
2.5-7
v (t ) = A − Au (t + T ) − Au( t − T )

 1
1
 j 2π fT  1
1
 − j 2π fT 
V ( f ) = A δ ( f ) − 
+ δ ( f ) e
−
+ δ ( f ) e

 j 2π f 2

 j 2π f 2



But δ ( f ) e j 2π fT = δ ( f ) e− j2π fT = e j 0δ ( f ) = δ ( f ), so
−A
−A
e j 2π fT + e − j2π fT ) =
cos2π fT
(
j 2π f
jπ f
−A
If T → 0, v(t ) = − A sgn t ↔ V ( f ) =
, which agrees with Eq. (17)
jπ f
V( f ) =
2.5-8
V ( f ) = sinc f ε and V (0) = 1, so
sinc f ε 1
W(f ) =
+ δ(f)
j 2π f
2
If ε → 0, w( t ) = u (t ) and W ( f ) =
1
1
+ δ ( f ), which agrees with Eq. (18)
j 2π f 2
2-14
2.5-9
V( f ) =
1/ ε
and V (0) = 1, so
1/ ε + j 2π f
W(f ) =
1/ ε
1
+ δ(f )
( j 2π f )(1/ ε + j 2π f ) 2
If ε → 0, w( t ) = u (t ) and W ( f ) =
1
1
+ δ ( f ), which agrees with Eq. (18)
j 2π f 2
2.5-10
z (t ) = AΠ ( t / τ ) ∗ [δ (t − T ) +δ (t + T ) ]
so Z( f ) = ( Aτ sinc f τ ) ( e− jω T + e jωT ) = 2 Aτ sinc f τ cos2π fT
2.5-11
z (t ) = AΠ ( t / τ ) ∗ [δ (t − 2T ) + 2δ (t ) + δ (t + 2T ) ]
so Z ( f ) = ( Aτ sinc f τ ) ( e − jω 2T + 2 + e jω 2 T ) = 2 Aτ sinc f τ (1 + cos4π fT )
2.5-12
z (t ) = AΠ ( t / τ ) ∗ [δ (t − 2T ) − 2 δ ( t ) + δ (t + 2T ) ]
so Z ( f ) = ( Aτ sinc f τ ) ( e − jω 2T − 2 + e jω 2T ) = 2 Aτ sinc fτ (cos4π fT − 1)
2.5-13
n
sin(π t) δ (t − 0.5n)
v (t )
0 1 2 3 4 5 6 7 8
0 1 0 1 0 1 0 1 0
0 1 1 2 2 3 3 4 4
2.5-14
n
cos(2π t )δ (t − 0.1n)
v (t )
-10
1
-9
0.81
-8
0.31
-7
-0.31
-6
-0.81
-5
-1
-4
-0.81
-3
-0.31
-2
0.31
-1
0.81
0
1
1
1.81
2.12
1.81
1
0
-0.81
-1.12
-0.81
0
1
v ()t for n = 1,10
1.81
2.12
1.81
1
0
-0.81
-1.12
-0.81
0
1
2-15
Chapter 3
3.1-1
y (t ) = h(t ) ∗ A[δ (t + t d ) −δ (t −t d ) ] = A[ h (t + t d ) − h(t − t d ) ]
(
)
Y ( f ) = H ( f )A e jω td − e − jω td = j 2 AH ( f )sin2π ft d
3.1-2
y (t ) = h( t ) ∗ A δ ( t + t d ) + δ ( t )  = A h ( t + td ) + h ( t )
(
)
Y ( f ) = H ( f )A e jω td + 1 = 2AH ( f )cos π ft d e jπ ftd
3.1-3
y (t ) = h( t ) ∗ Ah (t − t d ) = Ah(t ) ∗ h(t − t d )
Y ( f ) = H ( f ) AH ( f )e − jω td = AH 2 ( f ) e− jω ftd
3.1-4
y (t ) = h( t ) ∗ Au (t − t d ) = A ∫
t −t d
∞
h (λ ) d λ
 1
1

A
A
Y ( f ) = H( f )A 
+ δ ( f )  e − jω td =
H ( f )e − j2 π ftd + H (0)δ ( f )
j 2π f
2
 j 2π f 2

3.1-5
 1
1

1
1
F [ g (t )] = H ( f ) 
+ δ ( f ) =
H ( f ) + H(0)δ ( f )
2
 j 2π f 2
 j 2π f
F [ dg (t) / dt ] = j 2π f F [ g (t ) ] = H ( f ) = F [ h( t )] Thus h( t ) = dg (t) / dt
3.1-6
j 2π fY ( f ) + 4πY ( f ) = j 2π fX ( f ) + 16π X ( f )
Y( f )
j 2π f + 2π 8 8 + jf
H(f ) =
=
=
X ( f ) j 2π f + 2π 2 2 + jf
H( f ) =
64 + f 2
4+ f2
arg H ( f ) = arctan
f
f
− arctan
8
2
3-1
3.1-7
j 2π fY ( f ) + 16π Y ( f ) = j 2π fX ( f ) + 4π X ( f )
Y ( f ) j 2π f + 2π 2 2 + jf
H(f ) =
=
=
X ( f ) j 2π f + 2π 8 8 + jf
4+ f2
H( f ) =
64 + f 2
arg H ( f ) = arctan
f
f
− arctan
2
8
3.1-8
j 2π fY ( f ) − 4π Y ( f ) = − j2 π fX ( f ) + 4π X ( f )
Y ( f ) − j2 π f + 2π 2 2 − jf
H(f ) =
=
=
X( f )
j 2π f − 2π 2 2 + jf
2+ f2
H( f ) =
= 1 for all f
2+ f2
arg H ( f ) = −2arctan
f
2
3.1-9
H(f ) ≈
B
jf
Thus Y ( f ) ≈
f ≥W? B
for
B
1
X ( f ) = 2π B
X ( f ) for
jf
j 2π f
t
and y (t ) ≈ 2π B∫−∞ x (λ ) d λ
f ≥W
since X (0) ≈ 0
3.1-10
H(f ) ≈
jf
B
for f ≤ W = B
jf
1
X(f)=
j 2π fX ( f ) for
B
2π B
1 dx (t )
so y(t) ≈
2π B dt
Thus Y ( f ) ≈
3-2
f ≤W
3.1-11
2
 f
Π
4W  4W
∞
2W
2
1
1
Ex = ∫ X ( f ) df = ∫
df =
−∞
−2W 4W 2
W
1
1
f 
Y( f ) =
Π 
1 + j ( f / B ) 2W  4W 
x (t ) = 2sinc4Wt ↔ X ( f ) =
2W
E y = 2∫0
1

 f 
Π
=

 2W  4W 
1 / 4W 2
B
2W
df =
arctan
2
2
2W
B
1+ ( f / B )
Ey
B
2W
=
arctan
E x 2W
B
3.1-12
h (t ) = F −1 [ H1 ( f ) H 2 ( f ) ] = h1 (t ) ∗ h2 (t )
where h1( t ) = u (t ) − u (t − T1 )
h2 ( t ) = u (t ) − u (t − T2 )
3.1-13
−1
h (t ) = F [ H 1 ( f ) H 2 ( f ) ] = h1 (t ) ∗ h2 (t )
where h1( t ) = 2π Be−2π Bt u(t )
h2 ( t ) = u (t ) − u (t − T )
3.1-14
j 2π f
1
1
=
j 2π f
1 + jK 2π f K
1/ K + j 2π f
1 d −t / K
1
1
e u (t )  = δ (t ) − 2 e −t / K u (t )
h (t ) =
K dt
K
K

1/ K
 1 −t / K
g ( t ) = F −1 
 = e u (t )
1/ K + j2π f  K
H(f ) =
3-3
3-1-15
K
1
=
1
1 + Kj 2π f
+ j 2π f
K
so h (t ) = e− t / K u (t )
H( f ) =
g ( t ) = ∫ h( λ ) dλ = K(1 − e−t / K ) u (t )
t
−∞
3.1-16
Since h( t ) is real, H r ( f ) = H e ( f ) and Hi ( f ) = H o ( f ), so
∞
[ He ( f ) + jHo ( f ) ] e jω t df
−∞
h (t ) = ∫
∞
∞
∞
= 2∫ H r ( f )cos ω t df + j 2 ∫ jH i ( f )sin ωt df
0
0
∞
= 2  ∫ H r ( f )cos ωt df − ∫ Hi ( f )sin ω t df 
0
 0

∞
∞
h (t ) = 0 for t < 0 ⇒ ∫ H i ( f )sin ωt df = ∫ Hr ( f )cos ω t df
0
0
∞
∞
Hence, for t > 0, − ∫ Hi ( f )sin ω t df = ∫ H r ( f )cos ω t df
0
0
∞
∞
0
0
so h( t ) = ∫ Hi ( f )sin ω t df = ∫ H r ( f )cos ω t df
3.2-1
−1 / 2
1
( f / B )2 + L ≈ 1
2
3
f
f 1 f 
f
arg H ( f ) = − arctan = − +   + L ≈ −
B
B 3 B 
B
2
H ( f ) = 1 + ( f / B ) 


=1 −
f ≤W= B
for
3.2-2
−1 / 2
2
H ( nf 0 ) = 1 + ( n /3 ) 
arg H ( nf 0 ) = − arctan(n /3)
y (t ) = (0.95)(4)cos (ω 0t − 18° ) + (0.71)(4/9)cos ( 3ω 0t − 45° ) + (0.5)(4/25)cos ( 5ω 0t − 59°)
= 3.79cos (ω 0t −18° ) + 0.31cos ( 3 ω 0t − 45° ) + 0.08cos ( 5ω0t − 59°)
3-4
3.2-3
H ( nf 0 ) =
n /3
1 + (n / 3)
2
arg H ( nf 0 ) = 90° − arctan( n /3)
y (t ) = (0.32)(4)cos (ω 0t − 72° ) + (0.71)(4/9)cos ( 3ω 0t − 45° ) + (0.86)(4/25)cos ( 5ω0t − 31°)
= 1.28cos ( ω0t − 72° ) + 0.31cos ( 3ω0t − 45° ) + 0.14cos ( 5ω 0t − 31°)
3.2-4
2  f  1  f 
Π
=
Π
40  40  20  40 
20  f  − jω 5
Y( f ) =
Π
e
40  40 
1  f  − jω 5
Π
e
Y ( f ) 2  40 
H(f ) =
=
= 10e − jω 5
1  f 
X( f )
Π
20  40 
X( f ) =
Note that 300π = 2π and the phase actually wrapped around several times. Under
normal plotting conventions we would go from −π to π and repeat this pattern 300
times.
3.2-5
− arctan( f / B )
2π f
f kHz t d ( f ), ms
0
-0.08
0.5
-0.078
1
-0.074
2
-0.062
td ( f ) =
B = 2kHz
lim t d ( f ) =
f→0
3-5
1
2π B
3.2-6
 πf
− 30
arg H ( f ) = 
 −π
 2
f ≤ 15
f > 15
 1  π  1
1 d
−
−  =
tg ( f ) = −
arg H ( f ) =  2π  30  60
2π df

0

 π f /30 1
=
f ≤ 15

60
− arg H ( f )  2π f
td ( f ) =
=
2π f
 π /2 = 1
f > 15
 2π f 4 f
so t d ( f ) = t g ( f ) for
f ≤ 15
f > 15
f ≤ 15
3.2-7
1


(a) H c ( f ) = 1 + 2α ( e jωT + e − jωT )  e− jω T = α + e − jω T + α e − jω 2T
2


Thus, y (t ) = α x(t ) + x (t − T ) + α x (t − 2T )
(b) τ =
2T
3
τ=
4T
3
3-6
3.2-8


α2
exp  − j (ωT − α sin ωT )  = e− jω T e jα sinωT = e − jωT  1 + jα sin ωT −
sin2 ωT + L 
2


α
If α = π /2, H c ( f ) ≈ e − jωT + jα sin ωT e − jωT = e − jωT + ( e jω T − e − jωT ) e − jω T
2
α
α
≈ + e − jωT − e − jω 2 T
2
2
α
α
Thus, y (t ) ≈ x( t ) + x (t − T ) − x (t − 2T )
2
2
{
14243
leading echo
inverted trailing echo
3.2-9
− j ω ( t d −T ) j 0.4sin ω T
H eq ( f ) = Ke
e
e j0.4sin ωT = 1 + j 0.4sin ωT − 0.8sin 2 ωT + L ≈ 1 + 0.2 ( e jω T − e − jω T )
so H eq ( f ) ≈ Ke
− jω ( td −T )
(1 + 0.2e
jω T
− jω T
− 0.2e
)
Take K = 1 and t d = 2T , so H eq ( f ) ≈ ( 0.2e jω T + 1 − 0.2e − jω T ) e − jω T
Hence, ∆ = T , M = 1, c−1 = 0.02, c0 = 1, c1 == −0.2
3.2-10
H eq ( f ) = Ke − jω ( td − T) (1 + 0.8cos ωT )
−1
Expanding using the first 3 terms
(1 + 0.8cos ω T )
−1
= 1 − 0.8cos ωT + 0.64cos 2 ωT − 0.51cos3 ωT
1 jω T
1 1
1 1
e + e − jω T ) , cos2 ωT = + cos2ωT = + ( e j 2ω T + e − j 2ω T )
(
2
2 2
2 4
1
3
1
cos 3 ωT = ( 3cos ω T + cos3ωT ) = ( e jω T + e − jωT ) + ( e j 3ω T + e− j3ωT )
4
8
8
Take K = 1 and t d = 4T , so
where cos ωT =
 0.13 j3ω T 0.64 j2ω T  0.8 0.38  jωT
0.64  0.8 0.38  − jω T
H eq ( f ) =  −
e
+
e
−
+
e + 1+
−
+
e

4
2 
2
2 
 2
 2
 2
0.64 − j 2ωT 0.13 − j3ωT  − j3ω T
+
e
−
e
 e
4
2
Hence, ∆ = T , M = 3, c−3 = c3 = −0.065, c−2 = c2 = 0.16, c−1 = c1 = −0.59, c0 = 1.32
3-7
3.2-11
y (t ) = 2 A cos ω 0t − 3 A3 cos3 ω 0t
3 A3 cos 3 ω0t =
9 A3
3 A3
cos ω 0t +
cos3ω 0t
4
4

9 A3 
3 A3
so y (t ) =  2 A −
cos
ω
t
−
cos3ω 0t
0
4 
4

2 nd harmonic distortion = 0
3 A3
300% A = 1
4
3rd harmonic distortion =
×
100
=

9 A3
 42% A = 2
2A −
4
3.2-12
y (t ) = 5 A cos ω 0t − 2 A2 cos 2 ω0 t + 4 A3 cos 3 ω0 t
2 A2 cos 2 ω 0t = A2 + A2 cos2ω0t
4 A3 cos 3 ω 0t = 3 A3 cosω 0 t + A3 cos3ω 0 t
so y (t ) = − A2 + ( 5 A + 3 A3 ) cos ω 0t − A 2 cos2ω0 t + A3 cos3ω0 t
2 nd harmonic distortion =
12.5%
A2
×100 = 
3
5 A + 3A
11.8%
12.5%
A3
3 harmonic distortion =
×
100
=

5 A + 3 A3
23.5%
rd
A =1
A=2
A =1
A=2
3.3-1
Pin = 0.5W = 27dBm l = 50km α = 2dB/km
Pout = 50mW = 17dBm 20µW = −17dBm
27dBm − 2l1 = −17dBm ⇒ l1 = 22km ⇒ l 3 = 50 − 22 = 28km
−17dBm + g2 − 2 × 28 = −17dBm ⇒ g2 = 56dB
−17dBm + g4 = 17dBm ⇒ g 4 = 34dB
3.3-2
Pin = 100mW = 20dBm l = 50km α = 2dB/km
Pout = 0.1W = 20dBm 20µW = −17dBm
20dBm − 2l1 = −17dBm ⇒ l 1 = 18.5km ⇒ l 3 = 40 − 18.5 = 21.5km
−17dBm + g2 − 2 × 21.5 = −17dBm ⇒ g2 = 43dB
−17dBm + g4 = 20dBm ⇒ g 4 = 37dB
3-8
3.3-3
Pout 50 ×10 −3
=
= −16 dB, mLi = 0.4 × 400 = 160 dB, gi ≤ 30 dB
Pin
2
m × 30 dB − 160 ≥ −16 ⇒ m ≥ 4.8 so m = 5
g = (160 − 16)/5 = 28.8 dB
3.3-4
Li = 0.5 × 3000/ m = 1500/ m dB
Pin = 5mW = 7dBm
Pin
≥ 67µW = −11.75dBm
L1
1500
7dBm −
≥ −11.75 ⇒ m ≥ 80
m
Pout mgi
1500
=
= 1 ⇒ g i = Li =
= 18.75dB
Pin mLi
80
3.3-5
Li = 2.5 × 3000/ m = 7500/ m dB
Pin = 5mW = 7dBm
Pin
≥ 67µW = −11.75dBm
L1
7500
7dBm −
≥ −11.75 ⇒ m ≥ 400
m
Pout mgi
7500
=
= 1 ⇒ gi = Li =
= 18.75dB
Pin mLi
400
3.3-6
Pout
= 2 ×10 −6 /5 = −64 dB, L = 92.4-6 + 26 = 112.4 dB
Pin
g 2 Pout
=
L
Pin
so
⇒ g = (112.4 − 64)/2 = 24.2 dB = 263
4π (π r 2 )( 0.5 ×109 )
(3 × 10 )
5 2
2
= 263 ⇒ r = 1.55 × 10−3 km = 1.55 m
3-9
3.3-7
Pout 2 ×10−6
=
= −64dB
Pin
5
L = 92.4 − 14 + 20 = 98.4
g 2 Pout
98.4 − 64
=
⇒g=
= 17.2dB = 52.5
L
Pin
2
so
4π (π r 2 )( 0.2 ×109 )
(3 × 10 )
5 2
2
= 52.5 ⇒ r = 1.7 ×10−3 km = 1.7m
3.3-8
With repeater Pout =
gT gR grpt
Pin
L1L2
Without repeater Pout =
gT g R
P
L in
g T g R g rpt
g g
LL
= 1.2 T R ⇒ grpt = 1.2 1 2
L1L2
L
L
L1d B = 92.4 − 20log f GHz + 20log25km = 120 + 20log f
L2 = L1
L = 92.4 − 20log f + 20log50 = 126 + 2 − log f
Let f = 1GHz
L1 = L2 = 120dB ⇒ 1012
g rpt = 1.2
L = 126dB ⇒ 3.98 ×1012
1012 ×1012
= 0.3 ×1012 = 115dB
3.98 × 1012
3.3-9
Lu = 92.4 + 20log17 + 20log3.6 × 104 = 208
Ld = 92.4 + 20log12 + 20log3.6 ×10 4 = 205
Pin = 30dBW so Psatin = 30 + 55 − 208 + 20 = −103dBW
based on parameters from Example 3.3-1 g amp = 18 + 144 = 162dB
Psatout = −103 +162 = 59dBW so
Pout = 59 + 16 − 205 + 51 = −79dBW ⇒ 1.26 ×10 -8W
3.4-1
 f  − jω td
H ( f ) = Ke− jω td − K Π 
e
 2 fl 
h (t ) = K δ (t − t d ) − 2 Kf l sinc2 f l ( t − td )
3-10
3.4-2
H ( f ) = Ke− jω td − H BP ( f ) where H BP ( f ) = Eq. (1)
Thus, from Exercise 3.4-1, h( t ) = Kδ (t − t d ) − 2 BK sinc B( t − td ) cos ω c (t − t d )
3.4-3
−1
H (0.7 B) = 1 + (0.7)2 n  ≥ 10−1/10 = 1/1.259
so 1 + (0.7) 2n ≤ 1.259 or (0.7) 2n ≤ 0.259
2
n (0.7) 2n
1
2
⇒ select n = 2
0.49
0.24
H (3B) = 1 + 32n 
−1/2
= 1 + 34 
−1 / 2
= 0.11 = −19dB
3.4-4
−1
H (0.9B ) = 1 + (0.9) 2n  ≥ 10 −1/10 = 1/1.259
so 1 + (0.9) 2n ≤ 1.259 or (0.9) 2n ≤ 0.259
2
n (0.9) 2n
6
7
0.282
0.229
⇒ select n = 7
H (3B) = 1 + 32n 
−1/2
= 1 + 314 
−1 / 2
= 4.6 ×10− 4 = −66.8dB
3.4-5
2

f  f  
H ( f ) = 1 + j 2 −   
B  B  

−1
from Table 3.4-1
−1
2

s
 s  
H ( s) = H ( f ) | f =s / j 2π = 1 + j 2
−

2π B  2π B  

2b2
=
b = 2π B / 2
2
( s + b ) + b2
so h( t ) = 2be − bt sin b t u (t )
3-11
3.4-6
(a) H ( f ) =
Z RC
R / jω C
LC
where Z RC =
=
Z RC + jω L
R + 1/ jω C 1 + jω LC
Thus, H ( f ) =
1
1 + jω LC − ω LC
2
−1
2
2
1
2
2
4
so H ( f ) = (1 − ω 2 LC ) + (ω 2 LC )  = 1 − ( f / f 0 ) + ( f / f 0) with f 0 =


2π LC
(b) H ( B) = 1 / 2 ⇒ 1 − ( B / f 0 ) + ( B / f0 ) =2
2
2
2
B 1
so   = 1 + 5
2
 f0 
(
)
B=
4
(
)
1
1 + 5 f0 = 1.27 f 0
2
3.4-7
−2π Bt1
1− e
= 0.1 ⇒ t1 = 0.11/2π B 
2.30 − 0.11
1
=
 tr = t 2 − t1 =
−2π Bt2
2π B
2.87 B
1− e
= 0.9 ⇒ t 2 = 2.30/2π B 
3.4-8
g ( t ) = ∫ h( λ )d λ = 2b ∫ e − bλ sin bλ d λ = 1 − e − bt ( sin bt + cos bt ) for t ≥ 0
t
t
−∞
0
bt1 / π ≈ 0.1, bt2 / π ≈ 0.6
tr ≈
0.5π 0.5π 2
1
=
=
b
2π B
2.8 B
3-12
3.4-9
1
A  f 
, X( f ) =
Π
W
2W  2W 
 A  f 
 2W Π  2W  for B > W



 f 
Y( f ) = Π
X(f )=

 2B 
 A Π  f  for B < W
 2W  2 B 
 A sinc2Wt ⇒ τ y = 1/ W for B > W

y (t ) =  B
W A sinc2 Bt ⇒ τ y = 1/ B for B < W
x (t ) = A sinc2Wt ⇒ τ x =
3.4-10
∞
H (0) = ∫ h(t )e − jω t dt
−∞
h( t ) =
∫
∞
−∞
∞
f =0
= ∫ h ( t) dt
−∞
H ( f ) e jω t df ≤ ∫
Thus τ eff =
∞
−∞
H (0)
≥
h( t ) max
H ( f )e jω t df = ∫
H (0)
∫
∞
−∞
H ( f ) df
∞
−∞
=
H ( f ) df
1
2Beff
3.4-11
(a) H ( f ) = 2 KB∫
2 td
0
sinc2 B ( t − td ) e − jω t dt = 2 KBe − jωtd ∫ sinc2Bλ e− jωλ d λ
td
−t d
1
sin2π ( f + B ) λ − sin2π ( f − B ) λ 
2
t d sin2π ( f ± B ) λ
1 2π ( f ± B)td sin α
1
and ∫
dλ =
dα =
Si  2π ( f ± B ) td 
∫
0
0
2π Bλ
2π B
α
2π B 
K
Thus H ( f ) = e− jω td Si  2π ( f + B ) td  − Si 2π ( f − B ) td 
π
where sin2π Bλ cos2π f λ =
{
}
(cont.)
3-13
(b) td ?
1
B
td =
1
2B
3.5-1
1 ∞ δ (λ )
1 1
1
(a) δˆ ( t ) = ∫
dλ =
=
−∞
π
t −λ
π t − λ λ=0 π t
F δˆ (t )  = ( − j sgn f ) F [δ (t ) ] = − j sgn f
1
Thus, F - 1 [ − j sgn f ] = δˆ (t ) =
πt
1
1 1
 −1 
·1 
(b) δˆ ( t ) ∗ = δ( t) and δˆ ( t ) ∗   = − ∗ = − 
πt
πt πt
πt 
 πt 
·1 
Thus,   = −δ ( t )
πt 
3.5-2
t
AΠ   = x ( t + τ / 2 ) where x (t ) = A [u ( t ) − u (t − τ )]
τ 
ˆ  t  = xˆ  t + τ  = A ln t + τ / 2 = A ln 2t + τ
so AΠ
 


t +τ / 2 −τ π
2t − τ
τ 
 2 π
A 2t + τ
A
t
Now let v(t ) = lim AΠ   so vˆ (t ) = lim ln
= ln1 = 0
τ →∞
τ
→∞
π
2t − τ π
τ 
3.5-3
1
 f 
Π
2W  2W 
j
j
 f +W / 2 
 f −W / 2 
=
Π
−
Π


2W  W
 2W  W

j
Thus, xˆ (t ) = sinc Wt ( e − jπWt − e jπ Wt ) = sincWt sinc π Wt = π Wt sinc2 Wt
2
F [ xˆ (t )] = ( − j sgn f )
3-14
3.5-4
1
1
x (t ) = cos ω 0t − cos3ω 0t + cos5ω 0t
3
5
1
1
xˆ (t ) = sin ω0t − sin3ω0t + sin5ω0t
3
5
3.5-5
4
4
x (t ) = 4cos ω 0 t + cos3ω 0t + cos5ω 0t
9
25
4
4
xˆ (t ) = 4sin ω0 t + sin3ω0t + sin5ω 0t
9
25
3.5-6
x (t ) = sinc2Wt ↔ X ( f ) =
1
 f 
Π

2W  2W 
X(f) =
1
 f 
Π

2W  2W 
f −
1
xˆ (t ) = πWt sinc Wt = sinc Wt sin π Wt ↔ Xˆ ( f ) =
Π
2W  W
2
W
2
 − jπ / 2 1
 f + W2
+
Π
 e
2W  W

 jπ / 2
 e

 f − W2  − jπ /2
 f + W2  + jπ / 2
1
1
Xˆ ( f ) =
Π
e
+
Π


e
2W  W 
2W  W 
Note that the cross term is zero since there is no overlap. From the graph we see that the
two rectangle functions form one larger function so
1
 f
Xˆ ( f ) =
Π
2W  2W
3.5-7
x (t ) = A cos ω0t
∫
∞
−∞

 = X(f )

xˆ (t ) = A sin ω0 t
∞
x ( t ) xˆ( t) dt = A2 ∫ cos ω 0t sinω 0t dt
 A2
= lim 
T →∞
 2
−∞
A2
sin
ω
−
ω
t
dt
+
(
)
0
0
∫−T
2
T
T



A2 1
sin
ω
+
ω
t
dt
=
lim
0
+
cos2ω 0t 
(
)
0
0
 T →∞ 
∫− T
2 2ωo


−T
T
 A2

= lim 
( cos2ω0T − cos ( −2ω 0T ) )  = 0
T →∞ 4ω
 0

3-15
3.5-8
F [ he ( t ) ] = ∫
∞
−∞
∞1
1
h ( t ) e− jω t dt = 2∫ h ( t ) cos ω t dt
0
2
2
∞
∞
= ∫0 h (t) c o s ω t dt = ∫−∞ h (t) c o s ωt dt = H e ( f )
H ( f ) = F (1 + sgn t ) he (t )  = H e ( f ) +
)
Thus, H o ( f ) = − He ( f )
1
∗ H e ( f ) = He ( f ) −
j 2π f
)

1 
j H e ( f ) ∗
= He ( f ) − jH e ( f )

πf

3.6-1
Rwv (τ ) = w(t) v∗ ( t − τ ) = w∗ (t )v(t −τ )
∗
∗
*
= v [ t + ( −τ ) ] w∗ ( t ) = Rvw
( −τ )
3.6-2
Rv (τ ± mT0 ) = v ( t + τ ± mT0 ) v ∗ (t )
but v ( t + τ ± mT0 ) = v ( t + τ ) so Rv (τ ± mT0 ) = v ( t + τ ) v∗ ( t ) = Rv (τ )
3.6-3
Pw = v (t +τ )
2
= v (t )
Rv (τ ) = v( t) w∗ (t )
2
2
2
= Pv
≤ Pv Pv = Rv2 (0) so Rv (τ ) ≤ Rv (0)
3.6-4
1
cos2ω 0τ
2
1
y (t ) = sin2ω 0t = cos ( 2ω 0t − 90° ) ⇒ R y (τ ) = cos2ω0τ
2
(Note that the phase delay does not appear in the autocorrelation)
Since Ry (τ ) = Rx (τ ) we conclude that y (t ) is similar to x (t ). This is the expected conclusion
x (t ) = cos2ω 0t
From Eq. (12) Rx (τ ) =
since y( t ) is just a phase shifted version of x (t) .
3.6-5
V ( f ) = AD sinc f D e− jωtd
Gv ( f ) = ( AD ) 2 sinc2 fD ⇒ Rv (τ ) = A2 D Λ (τ / D ) , Ev = Rv (0) = A2 D
3-16
3.6-6
 A
V( f ) = 
 4W
  f  − jω td
 Π  4W  e
 

2
A2
A2
 A   f 
Gv ( f ) = 
Π
⇒
R
(
τ
)
=
sinc4
W
τ
,
E
=
R
(0)
=
 

v
v
v
4W
4W
 4W   4W 
3.6-7
V( f ) =
A
b + j 2π f
Gv ( f ) =
A2
A2 −b τ
⇒
R
(
τ
)
=
e ,
v
b 2 + (2π f ) 2
2b
Ev = Rv (0) =
A2
2b
3.6-8
A1 jφ jω 0 t A1 − jφ − jω 0t
e e + e e
2
2
2
A
Gv ( f ) = A02δ ( f ) + 1  δ ( f − f 0 ) + δ ( f + f 0 )
4
2
A
A2
Rv (τ ) = A02 + 1 cos ω 0τ , Pv = Rv (0) = A02 + 1
2
2
v (t ) = A0 +
3.6-9
(
)
(
)
A1 jφ1 jω0 t
A
e e + e − jφ1 e − jω0 t + 2 e− jπ / 2e j 2ω 0 t e jφ1 + e jπ / 2e − j 2ω0 t e− jφ1
2
2
2
A
A2
Gv ( f ) = 1 δ ( f − f 0 ) + δ ( f + f 0 ) + 2 δ ( f − 2 f 0 ) + δ ( f + 2 f 0 ) 
4
4
2
2
A
A
A2 A2
Rv (τ ) = 1 cos ω oτ + 2 cos2ω0τ
Pv = Rv (0) = 1 + 2
2
2
2
2
v (t ) =
3.6-10
0 t < τ
A2 u(t )u (t − τ ) dt where u(t )u (t − τ ) = 
−T / 2
1 t > τ
T/2
T /2
T
Take T / 2 > τ > 0, so ∫ u (t )u (t − τ ) dt = ∫ dt = − τ
-T/2
τ
2
2
2
A T
 A
Thus Rv (τ ) = lim
 2 −τ  = 2 for all τ
T →∞ T


1
T →∞ T
Rv (τ ) = lim
∫
A2
Pv = Rv (0) =
2
T /2
A2
Gv ( f ) =
δ(f)
2
3-17
3.6-11
t 
1
f
x (t ) = Π (10t ) = Π 
↔ X ( f ) = sinc

10
10
 1/10 
 f 
 f 
H ( f ) = Ke − jω td Π 
= 3e− jω 0.05Π  

 2B 
 40 
2
2
Gy ( f ) = H ( f ) Gx ( f ) = H ( f ) X ( f )
2
2
since x( t ) is an energy signal
2
f
  f   1
f 
 f  1
= 3e
Π 
sinc
=  9Π   
sinc 2 
10
10 
 40  10
  40   100
20 9
f
so Ry (τ ) = ∫
sinc 2 e j2π f τ df
−20 100
10
− jω 0.05
3-18
Chapter 4
4.1-1
vi (t ) = v1 (t ) + v2 (t) c o s α
vq (t ) = v2 (t) s i n α
A(t ) = v12 ( t ) + 2 v1(t ) v2 (t) c o s α + v22 (t ) ≈ v1 (t ) + v2 ()cos
t
α
φ (t ) = arctan
v2 (t) s i n α
v (t) s i n α
≈ 2
v1(t ) + v2 (t) c o s α
v1 (t )
4.1-2
vi (t ) = [ v1( t ) + v2 (t )] cos ω 0t
vq (t ) = [ v2 ( t ) − v1 (t )] sinω 0t
A(t ) = v12 ( t ) + 2 v1 (t ) v2 ( t) c o s 2ω0t + v 22 (t ) ≈ v1 (t ) + v2 (t) c o s 2ω 0t
φ (t ) = arctan
[ v2 (t) − v1(t )] sinω 0t
≈ −ω 0t
[ v1 (t) + v2 (t )] cos ω0t
4.1-3
(a)
∫
∞
∫
∞
∫
∞
−∞
∞
∞
vbp (t )dt = ∫ vi (t) c o s ω ct dt − ∫ vq ()sin
t
ωc t dt
−∞
−∞
∞
∗
1
1
vi (t) c o s ω ct dt = ∫ Vi ( f ) δ ( f − f c ) + δ ( f + f c )  df = [Vi ( f c ) +Vi ( − fc) ] = 0
−∞
2
2
since f c > W and Vi ( f ) = 0 for f > W
−∞
−∞
∞
∗
1
vq (t) s i n ω ct dt = ∫ Vq ( f ) e − jπ / 2δ ( f − fc ) + e jπ / 2δ ( f + f c )  df
−∞
2
1
= Vq ( f c ) e− jπ / 2 + Vq (− fc )e jπ / 2  = 0
2
Thus,
∫
∞
v (t ) dt = 0
−∞ bp
(cont).
4-1
(b) Ebp = ∫
∞
−∞
( v (t) c o s ω t − v
i
c
sinω ct ) dt
2
q
∞
∞
∞
∞
1 ∞ 2
vi dt + ∫ vq2dt + ∫ vi2 cos2ωc t dt + ∫ vq2 sin2ω ct dt + ∫ vivq sin2ωc t dt 
∫
−∞
−∞
−∞
−∞

2  −∞
2
2
but vi , vq , and vi vq are bandlimited in 2W < 2 fc so, from the analysis in part (a)
=
∫
∞
−∞
∞
∞
−∞
−∞
vi2 cos2ω ct dt = ∫ vq2 sin2ω ct dt = ∫ vivq sin2ωc t dt = 0
Hence, Ebp =
∞
1 ∞ 2
1
vi dt + ∫ vq2 dt  = ( Ei + Eq )
∫
−∞
 2
2  −∞
4.1-4
f + 100 
Vlp ( f ) = Π 

 400 
vlp (t ) = 400sinc400t e − j 2π 100t
= 400sinc400t ( cos2π 100t + j sin2π 100t )
vi (t ) = 800sinc400t cos2π 100t
vq (t ) = −800sinc400t sin2π 100t
4.1-5
1  f − 75 
 f + 50 
Π
 +Π 

2  100 
 150 
150
j 2 π 75 t
− j 2π 50 t
vlp (t ) =
sinc150t e
+ 100sinc100t e
2
vi (t ) = 2Re  vlp (t )  = 150sinc150t cos2π 75t + 200sinc100t cos2π 50t
Vlp ( f ) =
vq (t ) = 2Im  vl p (t )  = 150sinc150t sin2π 75t − 200sinc100t sin2π 50t
4.1-6
vbp (t ) = 2 z( t ) cos ( ±ω 0t + α ) cos ω ct − sin ( ±ω 0t + α ) sinω c t 
so vi (t ) = 2 z( t) cos ( ±ω0t + α )
vlp (t ) =
vq (t ) = 2 z( t )sin ( ±ω 0t + α )
1
2 z ( t )  cos ( ±ω 0t + α ) + j sin ( ±ω 0t + α )  = z (t) e j( ±ω 0 t+α )
2
4.1-7
−1
2

 f
f0  
1
f 
2 f
H ( f ) = 1 + Q  −   =
⇒ Q  − 0  = ±1
2

 f 0 f  
 f0 f 
Q 2
f
so
f ± f = Qf 0 = 0 ⇒ f l , f u = 0 1 + 4Q 2 ± 1
f0
2Q
f
f
f
B = f l − f u = 0 1 + 4Q2 + 1 − 0 1 + 4Q2 − 1 = 0
2Q
2Q
Q
2
(
(
)
(
)
)
4-2
4.1-8
f
= 1+ δ ,
f0
f0
−1
= (1 + δ ) ≈ 1 − δ , so
f
{
}
H ( f ) ≈ 1 + jQ  1+ δ − ( 1− δ ) 
But δ =
H(f ) ≈
−1
=
f
f − f0
−1 =
so
f0
f0
1
1 + j 2 Q( f − f 0 ) / f 0
for
1
1 + j 2Qδ
f = f 0 (1 + δ ) > 0
f − f0 = δ f0 = f 0
4.1-9
H(f ) ≈
≈
1
1 + ( f − fc + b) / b 2 1 + ( f − fc − b) / b2
2
1
1 + ( f − f c ) / 2b 2
2
2
stagger-tuned
single tuned
4.1-10
H lp ( f ) =
1
πB
=
1 + j 2 f / B π B + j2π f
⇒ hlp (t ) = π Be −π Bt u(t )
A
A
jω t 
xbp (t ) = 2Re  u (t )e c  ⇒ xlp (t ) = u (t )
2
2

π BA t −π B(t −λ )
A
ylp (t ) = hlp ∗ xlp (t ) =
e
d λ = (1 − e− π Bt ) u (t )
∫
0
2
2
j ωc t
−π Bt
ybp (t ) = 2Re  ylp (t ) e  = A (1 − e ) cos ωc t u(t )
4-3
4.1-11
f
H lp ( f ) = Π   e− j(ω +ω c )td ⇒ hlp ( t ) = Be − jω ctd sinc B ( t − t d )
B
A
A

xbp (t ) = 2Re  u (t )e jω ct  ⇒ xlp (t ) = u (t )
2
2

BA − jωc td t
ylp (t ) = hlp ∗ xlp (t ) =
e
∫−∞ sinc B ( λ − td ) d λ
2
0
B ( t− td )
A
= e − jω ctd  ∫ sinc µ d µ + ∫
sinc µ d µ 

−∞
0


2
A
1 1
= e − jω ctd  + Siπ B ( t − t d ) 
2
2 π

1 1

ybp (t ) = 2Re  ylp (t )e jωc t  = A  + Si π B ( t − t d )  cos ω c ( t − t d )
2 π

4.1-12
jα
± jω 0t
xlp (t ) = 2e u( t) e
⇒


1
jα
Xlp ( f ) = e 
+ δ ( f m f0 ) 
 jπ ( f m f 0 )

B
 f 
H lp ( f ) = Π   with
= f 0 so δ ( f m f 0 ) falls outside passband.
2
B
e jα
eα
B
f 
f 
Thus, Ylp ( f ) =
Π ≈
Π   since f 0 ? f for f <
jπ ( f m f 0 )  B  m jπ f 0  B 
2
 e jα 
ylp (t ) ≈ ± j 
 B sinc Bt
π
f
0


2B
2B
ybp (t ) ≈
sinc Bt Re  ± je jα e jω ct  = m
sinc Bt sin (ω ct + α )
π f0
π f0
4-4
4.1-13
1
B
f 
X lp ( f ) = Z ( f ) = 0 f ≤ W ≤
B
2
2
 
2
1
1
f 2
f2
Ylp ( f ) = e jf /b Z ( f ) ≈ 1 + j  Z ( f ) since
= 1 for f ≤ W
2
2
b 
b

1
j
2
1
j d2

≈  Z ( f ) − 2 ( j 2π f ) Z ( f )  ⇒ ylp (t ) ≈  z ( t ) − 2
z (t)
2
2
4π b
2
4π b dt


H lp ( f ) = e jf
2
/b
Thus, ybp (t ) ≈ z ()cos
t
ω ct −

1  d2
z (t )  sin ω ct
2 
2
4π b  dt

4.2-1
4.2-2
4.2-3
AM: BT = 400Hz
DSB: BT = 400Hz
1 2
100
Ac (1 + µ 2 Sx ) =
1 + 0.6 2 ) = 68W
(
2
2
1
100
ST = Ac2S x =
= 50W
2
2
ST =
4-5
4.2-4
1  f 
Λ 
40  40 
BT = 2W = 80 Hz
sinc2 40t
↔
4.2-5
2
Amax
= ( 2 Ac ) = 32kW ⇒ A c2 = 8kW
2
µ = 1, S x =
1
⇒
2
ST =
1 2
A (1+ µ 2 S x ) = 6kW
2 c
4.2-6
Sx =
A
2
max
1
1  µ2 
4
2
, ST = Ac2 1 +
kW
 = 1kW ⇒ Ac =
2
2 
2 
2+ µ2
= (1 + µ ) A = 4
2
2
c
(1 + µ )2
so 1 + 2 µ + µ 2 ≤ 2 + µ 2
2 + µ2
⇒
kW ≤ 4kW
µ ≤ 0.5
4.2-7
x max = x(0) =3 K (1 + 2) ≤1 ⇒
K ≤1 / 9
2
13
45 2 2 45 2
 1
2
Psb =  KAc  + ( 3KAc ) =
K Ac =
K Pc
2 2
8
4
 2
45 2
K Pc
2Psb
45K 2
45
2
=
=
≤
≈ 22%
2
45
ST
2 + 45 K
207
Pc + K 2 Pc
2
4-6
4.2-8
x (t ) = 2 K cos20π t + K cos12π t + K cos28π t
x max = x(0) = K (2 +1 + 1) ≤ 1 ⇒ K ≤ 1 / 4
2
1
11
3
2
 3
Psb = ( KAc ) + 2 ×  KAc  = K 2 Ac2 = K 2 Pc
2
22
2
 4
2Psb
3K 2 Pc
3
=
=≤
≈ 16%
2
ST
Pc + 3K Pc
19
4.2-9
x (t ) = 4sin
π
1
t = 4sin2π t
2
4
B T = 2W =
1
kHz
2
BT
< 0.1 ⇒ 10BT < f c < 100 BT
fc
5 kHz < f c < 50 kHz
0.01 <
4.2-10
xc (t ) = Ac [1 + x (t )] cos ω ct
⇒
A(t ) = Ac [1 + x (t )] ≥ 0 for no phase reversals to occur
Since x(t ) min = −4 there is no value of Ac that can keep A( t) from going negative.
Therefore phase reversals will occur whenever x (t ) goes negative.
4.2-11
1


xc (t ) = Ac  cos2π 40t + cos2π 90t  cos2π f ct
2


4-7
4.3-1
 2a

(a) vout = a1 x(t ) + a2 x 2 (t ) + a2 cos 2 ω c t + a1 1 + 2 x (t ) cos ω c t
a1


1444
424444
3
desired term
Select a filter centered at f c =10 kHz with a bandwidth of 2W = 2 ×120 = 240 Hz.
 2a

 1

(b) a1 1 + 2 x(t ) cos ωc t = Ac [1 + µ x( t )] cos ω ct = 10 1 + x(t ) cos ωc t
a1
 2



2a2 1
5
⇒ a1 = 10
=
⇒ a2 =
a1
2
2
4.3-2
xc (t ) = aK 2 ( x + A cos ω c t ) − b ( x − A cos ω ct )
2
2
= ( aK 2 − b )( x2 + A2 cos 2 ω c t ) + 2 A ( aK 2 + b ) x cos ω c t
= 4 Abx(t) c o s ωc t
if K =
b
a
4.3-3
xc (t ) = aK 2 ( v + A cos ω ct ) − b ( v − A cos ω c t )
2
2
= ( aK 2 − b )( v 2 + A2 cos 2 ω ct ) + 2 A ( aK 2 + b ) v cos ω ct
= 4 Ab [1 + µ x (t )] cos ω ct
if K =
b
a
4-8
and
v(t ) = 1 + µ x( t )
4.3-4
Take vin = x + cos ω 0t so
vout = a1 ( x + cos ω 0t ) + a3 ( x3 + 3x 2 cos ω0t + 3x cos 3 ω0t + cos3 ω 0t )
3 
3
3
1



=  a1 + a3  x + a 3x3 +  a1 + a 3 + 3a3 x 2  cos ω 0t + a3 x cos2ω 0t + a3 cos3ω0t
2 
4
2
4



Take f c = 2 f 0 where f 0 + 2W < 2 f 0 − W so f c > 6W
4.3-5
Take vin = y + cos ω 0t ,
where y = Kx (t) , so
vout = a1 ( y + cos ω 0t ) + a3 ( y 3 + 3 y 2 cos ω0t + 3 y cos 2 ω 0t + cos 3 ω0t )
3 
3
3
1



=  a1 + a3  y + a 3 y 3 +  a1 + a 3 + 3a3 y 2  cos ω 0t + a3 y cos2ω 0t + a 3 cos3ω 0t
2 
4
2
4



Take f c = 2 f 0
where f 0 + 2W < 2 f 0 − W
so f c > 6W
 3a K

3

xc (t ) =  a3Kx( t ) + Ac  cos ω ct = Ac 1 + 3 x (t )  cos ω ct
2 Ac
2



4-9
4.3-6
2
Let vout+
3
1 
1 
1 



= a1  Ac cos ω c t + x  +a 2  Ac cos ω c t + x  + a3  Ac cos ω c t + x 
2 
2 
2 



2
3
1 
1 
1 



vout− = b1  Ac cos ω c t − x  + b2  Ac cos ωc t − x  + b3  Ac cos ω c t − x 
2 
2 
2 



1 1
3
1
Expanding using cos 2 ω ct = + cos2ω c t , cos3 ωc t = cos ωc t + cos3ω c t
2 2
4
4
Since BPFs reject components outside f c − W < f < f c + W ,
xc (t ) = vout+
BPF
− vout−
BPF
3
3 

2
=  a1 + a3 − b1 − b3  cos ω c t + 2 ( a2 + b2 ) x (t) c o s ω ct + 3 ( a3 − b3 ) x ()cos
t
ω ct
4
4 

so there's unsuppressed carrier and 2nd harmonic distor tion
4.3-7
20  f  1  f 
Λ
=
Λ
400  400  20  400 
4
4
4
vout (t ) = x(t) c o s ω c t −
x(t) c o s 3ω ct +
x (t) c o s 5ωc t −L
π
3π
5π
x (t ) = 20sinc 2 400t ↔
X(f)=
need f c + 200 < 3 f c − 200 ⇒ f c > 100 Hz
But f c must meet fractional bandwidth requirements as well
so 400 < 0.1 f c ⇒ f c > 4000 Hz which meets the earlier requirements as well.
4.4-1
1
xc (t ) = 2Re  Ac [ x( t ) ± jxˆ (t ) ] e jω ct 
4

A
= c Re {[ x( t) c o s ωct ± ( −1) xˆ( t )sin ω c t ] + j [x (t) s i n ω ct ± xˆ ()cos
t
ωc t ]}
2
A
= c [ x (t) c o s ω ct m xˆ ()sin
t
ωc t ]
2
4-10
4.4-2
x ()cos
t
ω ct ↔
1
1
X ( f − fc ) + X ( f + fc )
2
2
(
)
1 jω c t − jωc t
e −e
and Xˆ ( f ) = ( − j sgn f ) X ( f ) so
j2
1
1
xˆ (t) s i n ωc t ↔ − sgn ( f − f c ) X ( f − f c ) + sgn ( f + f c ) X ( f + f c )
2
2
Ac
Thus, X c ( f ) =
1 ± sgn ( f − fc )  X ( f − f c ) + 1 m sgn ( f + fc )  X ( f + f c )
4 
sin ω ct =
{
}
4.4-3
Upper signs for USSB, so
2
1 + sgn ( f − f c ) = 
0
 Ac
 2 X(f −

Xc ( f ) = 
0
A
 c X(f +
2
f > fc
0
, 1 − sgn ( f + f c ) = 
f < fc
2
fc )
f > fc
fc )
f < − fc
f > − fc
f < − fc
,
f < fc
4.4-4
1
Let θ = ω mt so xˆ (t ) = sin θ + sin3θ
9
2
2
1
1

 

x + xˆ =  cos θ + cos3θ  +  sinθ + sin3θ 
9
9

 

1 2
2
82 2
= 1 + + cos θ cos3θ + sin θ sin3θ = + cos2θ
81 9
9
81 9
1
1
1
9
A(t ) = Ac x2 (t ) + xˆ 2 ( t ) = × 81×
82 + 18cos2θ =
82 + 18cos2θ
2
2
9
2
2
2
4-11
4.4-5
For LSSB, upper cutoffs of BPFs should be f 1 and f 2 , respectively.
4.4-6
2 β = 400 ≥ 0.01 f1 ⇒ f1 ≤ 40kHz
0.01 f 2 ≤ 2 f1 + 400 ≤ 80.4kHz
f 2 ≤ 8.04MHz and f c = f 1 + f 2 ≤ 8.08MHz
4.4-7
1 2
1 1 2 1 2 1
2
7
Ac S x = Ac2  (1) + ( 3) + ( 2 )  = Ac2
4
4 2
2
2
 4
BT = W = 400 Hz
ST =
4-12
or calculate directly from
the line spectrum
4.4-8
Check to make sure BPF meets requirements:
B
B 10,000 − 9600
0.01 <
< 0.1 ⇒
=
⇒ 0.01 < 0.04 < 0.1 P
fc
fc
10 4
Also f c < 200 β = 200× 100 = 20 kHz P
Note that a LPF at 10 kHz would have violated the fractional bandwidth
requirements so a BPF must be used.
4.4-9
cos ( ωc t − 90° + δ ) = sin (ω ct + δ ) = cos δ sin ωc t + sin δ cos ω c t ≈ sin ω ct + δ cos ωc t
Thus, xc ( t ) ≈
A(t ) ≈
Ac
2
t
ω t}
{[ x (t) m δ xˆ(t )] cos ω t m xˆ()sin
c
c
1/2
A 2
 x (t ) + xˆ 2 ( t ) m 2δ xˆ ( t )x( t ) 
2
4.4-10
(1 − ε ) cos (ω mt − 90° + δ ) = (1 − ε ) [ cos δ sin ω mt + sin δ cos ω mt ] ≈ (1 − ε ) sinω mt +δ
cosω mt
Ac
 cos ω m cos ω c t − (1 − ε ) sin ω mt sin ω c t − δ cos ω mt sinω ct 
2 
A
= c 2cos ( ωc + ωm ) t + ε cos ( ωc − ω m ) t − cos (ω c + ω m ) t 
4
−δ sin (ω c − ω m ) t + sin ( ωc + ω m ) t 
xc (t ) ≈
{
}
But ε cos θ − δ sin θ = ε 2 + δ 2 cos (θ + arctan ( δ / ε ) )
( 2 − ε ) cosθ − δ sin θ
=
( 2 − ε ) 2 + δ 2 cos  θ + arctan

δ 
2 − ε 
≈ 2 1 − ε /2cos (θ + δ / 2)
Thus xc ( t ) ≈
Ac
A
δ

1 − ε / 2 c o s  (ω c + ω m ) t + δ / 2 + c ε 2 + δ 2 cos  (ω c − ωm ) t + arctan 
2
4
ε

4-13
4.4-11
The easiest way to find the quadrature component is graphically from the phasor diagram.
xcq (t ) =
1
1
1

aAm Ac sin2π f mt − (1 − a ) Am Ac sin2π f mt =  a −  Am Ac sin2π f mt
2
2
2

4.4-12
Ac
( 0.5 + a ) cos (ω c + ω m ) t + ( 0.5 − a ) cos (ω c − ω m ) t 
2 
A 1
1

= c   cos (ω c + ω m ) t + cos (ω c − ωm ) t  + 2a  cos (ω c + ω m ) t − cos ( ωc − ω m ) t  
2 2
2

xc (t ) =
Ac
[ cos ω mt cos ωc t − 2a sin ω mt sinω ct ]
2
A
a = 0 ⇒ xc (t ) = c cos ωmt cos ω ct DSB
2
A
A
a = ±0.5 ⇒ xc (t ) = c [ cos ω mt cos ωc t m sin ω mt sin ω ct ] = c cos (ω c ± ω m ) t SSB
2
2
=
4.4-13
µ


xc (t ) = Ac  cos ω ct + cos (ω c + ω m ) t 
2


1/2
2
2
 µ
 µ
 
A(t ) = Ac  1 + cos ω mt  +  sinω mt  
 2
 
  2
1/2

µ2 
= Ac 1 + µ cos ω mt + 
4

4-14
4.5-1
f1 ± 199.25 = 66 MHz ⇒
f 2 ± 66 = 67.25 MHz ⇒
f1 = 265.25 or 133.25
f 2 = 133.25 or 1.25
Take f LO = 133.25 MHz
4.5-2
f1 ± 651.25 = 66 MHz ⇒
f 1 = 717.25 or 585.25
f 2 ± 66 = 519.25 MHz ⇒
f 2 = 585.25 or 453.25
Take f LO = 585.25 MHz
4.5-3
Output is unintelligible because spectrum is reversed, so low- frequency components
become high frequencies, and vice versa.
Output signal can be unscrambled by passing it through a second, identical scrambler
which again reverses the spectrum.
4.5-4
LPF input = ( Kc + K µ x ) cos ω ct − K µ xq sin ω ct  cos (ω ct + φ )
= ( Kc + K µ x ) cos φ + ( Kc + K µ x ) cos ( 2ω ct + φ ) + Kµ xq sin φ − K µ xq sin ( 2ω ct + φ )
yD (t ) =  Kc + K µx (t ) cos φ + K µ xq ( t) s i n φ
Modulation
Kc
Kµ
xq (t )
y D (t )
AM
Ac
µ Ac
0
Ac [1 + µ x(t )] cos φ
DSB
SSB
0
0
Ac
Ac / 2
0
m xˆ (t )
Ac x()cos
t
φ
VSB
0
Ac / 2
xˆ (t ) + xβ ( t )
Ac / 2 [ x(t) c o s φ m xˆ (t) s i n φ ]
{
Ac / 2 x( t) c o s φ +  xˆ ( t ) + xβ (t ) sinφ
4-15
}
4.5-5
From equation for xc (t ) we see that
1
a = will produce standard AM with no distortion at the output.
2
a = 1 will produce USSB + C 
 maximum distortion from envelope detector.
a = 0 will produce LSSB + C
4.5-6
Envelope detector follows the shape of the positive amplitude portions of xc ( t ) .
Envelope detector output is proportional to x(t ) .
4.5-7
A square wave, like any other periodic signal, can be written as a Fourier series of
harmonically spaced sinusoids. If the square wave has even symmetry and a fundamental
of f c , it will have terms like a1 cos ω c t + a3 cos ω 3t + a5 cos ω ct +L . This will cause
signals at f c , 3 f c , 5 f c K to be shifted to the origin. If f c is large enough, and our desired
signal can be isolated, our synchronous detector will work fine. Otherwise there may be
noise or intelligible crosstalk. Note that any phase shift will cause amplitude distortion.
For any periodic signal in general, as long as the Fourier series has a term at f c and our
signal can be isolated, this can also serve as our local oscillator signal.
4-16
4.5-8
Between peaks v (t ) ≈ Ac [1 + cos2π Wt1 ] e −( t− t1 ) /τ , t1 < t < t1 +1/ f c
1
and we want
4W




1
1 
2π W 
v  t1 +  ≈ Ac e−1/τ fc < Ac 1 + cos2π W  t1 +   = Ac 1− sin

fc 
fc  
fc 




1
2π W
1
so 1 −
<1 −
if τ ?
and f c ? W
τ fc
fc
fc
1
We also want τ ?
for linear decay between peaks.
fc
Thus 2π W ≤ R1C1 = f c and f c / W ≥ 2π ×10 ≈ 60
Maximum negative envelope slope occurs at t1 =
4-17
Chapter 5
5.1-1
PM
FM
5.1-2
PM
FM
5.1-3
dx(t )
t 2 + 16
= −4 A
2
dt
(t 2 − 16 )
PM
t > 4,
4A
∫ x ( λ ) d λ = 2 log ( t
t
FM
5-1
2
− 16) t > 4
5.1-4
f (t ) = a + bt for 0 < t < T
f (0) = a = f1 ,
f (T ) = a + bT = f 2
f2 − f1
T
⇒ b=
t
t
f −f
f −f
θ c (t ) = 2π ∫ f (λ ) d λ = 2π ∫  f1 + 2 1 λ  d λ = 2π  f1t + 2 1 t 2 
0
0
T
T




5.1-5
Type
φ (t )
Phase-integral
K
φ max
f (t )
K d 2x (t )
2π dt 2
φ dx (t )
fc + ∆
2π dt
dx(t )
dt
fc +
PM
φ ∆ x(t )
FM
2π f ∆ ∫ x (λ ) d λ
fc + f∆ x (t)
Phase-accel.
t
µ
2π K ∫  ∫ x (λ )d λ d µ


f c + K ∫ x( λ )d λ
t
K 2π f m
t
φ∆
f∆
fm
K
2π f m2
f max
f c + K 2π f m2
fc + φ∆ f m
fc + f∆
fc +
K
2π f m
5.1-6
xc (t ) = Ac  cos ( β sin ω mt ) cos ω c t − sin ( β sin ω mt ) sin ωc t 


= Ac  J 0 ( β ) cos ωc t + ∑ 2 J n ( β ) cos nω mt cos ω c t − ∑ 2 J n ( β ) sin nω mt sin ωc t 
n even
n odd


1
where cos nω mt cos ωc t =  cos (ω c − nω m ) t + cos (ω c + nω m ) t 
2
1
sin nω mt sin ω c t = cos ( ωc − nω m ) t − cos ( ωc + nωm ) t 
2
so xc (t ) = Ac J 0 ( β ) cos ω ct + ∑ J n ( β )  cos (ω c + nω m ) t + cos ( ω c − nω m ) t 
n even
+
∑ J ( β ) cos ( ω
n
c
n odd
+ nω m ) t − cos (ω c − nωm ) t 
5.1-7
∞
e jβ sinω mt = ∑ cne jnω mt with period Tm = 2π / ω m
−∞
so cn =
1
Tm
∫
Tm
e j β sinω mt e− jnωm t dt =
1
2π
∫
π
−π
e
j ( β sin λ −nλ )
dλ = Jn ( β )
 ∞

Thus, cos ( β sin ω mt ) = Re e jβ sinω mt  = Re  ∑ J n ( β ) e jnω mt 
 n=−∞

(cont.)
5-2
=
∞
∑ J ( β ) cos nω
n
n =−∞
∞
m
t = J 0 ( β ) + ∑  J n ( β ) + J −n ( β )  cos nω mt
n =1
 ∞

sin ( β sin ω mt ) = Im  e jβ sinω mt  = Im  ∑ J n ( β ) e jnω mt 
 n=−∞

=
∞
∑ J ( β ) sin nω
n
n =−∞
∞
m
t = 0 + ∑  J n ( β ) − J − n ( β )  sin nω mt
n=1
But J − n ( β ) = ( −1) J n ( β ) so
n
2J
Jn + J −n =  n
 0
 0
J n − J−n = 
2 J n
n even
n odd
Hence, cos ( β sin ω mt ) = J 0 ( β ) +
n even
n odd
∞
∑  2 J ( β )  cos nω t
n
m
n even
sin ( β sin ω mt ) =
∞
∑  2 J ( β )  sin nω t
n
m
n odd
5.1-8
β = φ∆ Am for PM, β = Am f ∆ / f m for FM
(a) Line spacing remains fixed, while line amplitudes change in the same way since β is
proportional to Am .
(b) Line spacing changes in the same way but FM line amplitudes also change while PM
line amplitudes remain fixed.
(c) Line spacing changes in the same way but PM line amplitudes also change while FM
line amplitudes remain fixed.
5.1-9
(a) f (t ) = f c + f ∆ x (t ) = f c +
Assuming Am = 1
β fm
cos ω mt
Am
f (t ) = 30 + 20cos ω mt kHz
(b) "Folded" component at f c − 4 f m = 10 kHz
1
1002
2
2
2
2
2
2
2
ST = ( −13) + ( 35 − 3) + ( −58 ) + 22 + 58 + 35 + 13 + 3 = 4988.5 <

2
2
2
5-3
5.1-10
(a) f (t ) = f c + f ∆ x (t ) = f c +
Assuming Am = 1
β fm
cos ω mt
Am
f ( t ) = 40 + 40cos ω mt kHz
(b) "Folded" components at f c − 3 f m = 20 kHz and
f c − 4 f m = 40 kHz
1
1002
2
2
ST = 352 + ( −58 −13 ) + ( 22 + 3) + 582 + 352 +132 + 32  = 6441.5 >

2
2
5.1-11
ω mt = 0
ω mt =
π
4
A = 9.4 + 2 ×.3 = 10 φ = 0
(
A = 9.4 2 + 2.4 2
0.5 rad × sin
)
2
= 9.99 φ = arctan
2.4 2
= 0.347 rad
9.4
π
= 0.356 rad
4
(cont.)
5-4
ω mt =
π
2
A=
0.5 rad × sin
5.1-12
ω mt = 0
ω mt =
π
4
1 rad × sin
ω mt =
π
2
1 rad × sin
( 9.4 − .6 ) + ( 2 × 2.4 )
2
2
= 10.02 φ = arctan
2 × 2.4
= 0.499 rad
9.4 − .6
π
= 0.5 rad
2
A = 7.7 + 2 ×1.1 = 9.9 φ = 0
(
A = 7.7 2 + 4.6 2
)
2
= 10.08 φ = arctan
4.6 2
= 0.702 rad
7.7
π
= 0.707 rad
4
A=
( 7.7 − 2.2 ) + ( 2 × 4.4)
2
2
= 10.02 φ = arctan
π
= 1 rad
2
5-5
2 × 2.4
= 1.012 rad
7.7 − 2.2
5.1-13
Amin = 0.77 Ac
Amax = .77 2 + .88 2 Ac = 1.17 Ac
5.1-14
Amin = 0.18 Ac
Amax = .182 + .662 Ac = .68 Ac
5.1-15
Want f 0 plus 3 harmonics ⇒ select β = 1.0
Generate FM signal with 24,300 < f c < 243,000 to meet fractional bandwidth
requirements since BT = 6 × 405 = 2,430 Hz. Apply BPF to select carrier plus
3 sidebands. Use frequency converter at f LO = fc − f 0 .
5-6
5.1-16
(a) For 0 < t < T / 2 φ (t ) = 2π f ∆ t + φ (0) = 2π f ∆ t
T
2
For t <
cn =
=
=
1
T0
∫
φ (t ) = 2π f ∆ t =
− j 2 π ( β + n ) t / T0
0
−T0 / 2
e
sin π ( β + n ) / 2
π ( β + n)
1 jπβ
e
2
1
T0
jπ ( β + n ) / 2
+
∫
T0 / 2
0
e
j 2π ( β − n ) t/ T0
dt
sin π ( β − n ) / 2
π ( β − n)
e
jπ ( β − n) / 2

 n + β  jπ n / 2
 n − β  − jπ n / 2 
+ sinc 
e
 sinc  2  e



 2 


(b) β f 0 = f ∆
5.2-1
f ∆ , kHz
0.1
0.5
1
5
10
50
100
500
e
dt +
2πβ
t
T0
cn ≈
1
n+β 
n−β 
 sinc 
 + sinc 
  when β ? 1
2
 2 
 2 
Eq.
D
BT ,kHz
0.01
5
(0.01+1)(30) = 30.3
0.03
5
(0.03+1)(30)= 31
0.07
5
(0.07+1)(30) = 32
0.33
4
(1.8)(30) = 54
0.67
4
(2.2)(30) = 66
3.33
6
(3.33+2)(30) ? 160
6.67
6
(6.67+2)(30) ? 260
33 5 or 6 (33+1)(30) = 1030
5.2-2
5-7
f ∆ , kHz
0.1
0.5
1
5
10
50
100
500
Eq.
D
BT ,kHz
0.02
5
(0.02+1)(10) = 10.2
0.1
5
(0.1+1)(10)= 11
0.2
5
(0.2+1)(10) = 12
1
4
(2.7)(10) = 27
2
4
(3.8)(10) = 38
10
6
(10+2)(10) = 120
20 5 or 6 (20+1)(10) = 210
100
5
(100+1)(10) = 1010
5.2-3
W
D M(D)
BT
(a) 2 kHz
5
6
(lower curve)
2? 6? 2=24 kHz
(b) 3.2 kHz 3.1 3.1+2 (middle curve) 2? 5.1? 3.2? 33 kHz
(c) 20 kHz 0.5 2.5 (upper curve)
2? 2.5? 20=100kHz
5.2-4
FM:
f ∆ 25
=
=5
BT = 2(4 + 2)5 = 60 MHz
W
5
DSB:
BT = 2W = 10 MHz
f c > 10 BT = 100 MHz
D=
5.2-5
W = 15
0.01 <
f c > 10 BT = 600 MHz
f c = 5 ×1014
BT
< 0.1 ⇒
fc
( 0.01) (5 ×1014 ) < BT < ( 0.1) (5 ×1014 )
5 ×1012 < BT < 5 × 1013
BT ≈ 2 ( D + 1) W = 2 ( f ∆ + W ) ≈ 2 f ∆
⇒ 2.5 ×10 < f ∆ < 2.5 ×10
12
13
5.2-6
For CD: BT = 2(5 + 2)15 = 210 kHz
For talk show: BT = 2(5 + 2)5 = 70 kHz
Since station must broadcast at CD bandwidth, the fraction of the available
bandwidth used during the talk show is
BTused
BT
70
= talk =
× 100 = 33.3%
BTavailable BTCD 210
5.2-7
D = φ∆ = 30/10 = 3 ⇒ BT = 2 M (3)W ≈ 100 kHz
5-8
60%
62%
70%
PM
B = 2 M ( β ) fm
1
10
50
5.2-8
Take x (t ) = Am cos ω mt , β = φ max , and
B ≈ 2 ( β + 1) f m
f m , kHz
β
0.1
1.0
5.0
300
30
6
FM
B = 2 M ( β ) fm
600×0.1
62×1.0
14×5.0
B / BT
B / BT
1%
10%
50%
Phase-integral modulation
Phase-acceleration modulation
φ (t )
−2π KAm f m sin ωm t
− ( KAm / 2π f m2 ) cos ω mt
β
B
2π KAm fm
KAm / 2π fm2
2 ( KAm / 2π f m + f m )
2 ( 2π KAm f m2 + f m )
4π KW 2 2 π KW ? 1
 K / π fmmin K / 2π ? f mmin W

 2W
K / 2π = f mmin W
2π KW = 1
 2W

In both cases, spectral lines are spaced by f m and B increases with Am . However, in
phase- integral modulation, tones at f m = W occupy much less than BT if 2π KW ? 1 .
In phase-acceleration modulation, mid- frequency tones may occupy the most bandwidth
and will determine BT when K / 2π ≈ f mmin W .
BT
5.2-9
xlp (t ) =
1
1
Ac e jφ (t ) ≈ Ac [1 + jφ ( t )] , φ (t ) = φ ∆ x (t )
2
2




1
1
1
πf
1
Ylp ( f ) =
Ac [δ ( f ) + jφ ∆ X ( f )] = Ac δ ( f ) + jφ∆ c
X ( f )
1 + j 2Qf / f c 2
2 
Q π f c + j 2π f



Q
ylp (t ) =
1 
πf

Ac 1 + jφ ∆ c x% ( t ) where x% ( t ) =  e−π fc t / Qu (t )  ∗ x(t )
2 
Q

 jω t

πf
yc (t ) = Ac Re e c + jφ∆ c x% (t )[ cos ω ct + j sinω ct ] 
Q



πf

= Ac cos ωc t − φ ∆ c x% ( t) s i n ω c t 
Q


 πf

so φ (( t ) = arctan φ∆ c x% (t ) 
Q


5.2-10
5-9
Ylp ( f ) =  K0 − K 2 f 2  X lp ( f ) = K0 X lp ( f ) −
ylp (t ) = K0 xlp (t ) +
K2
( j2π f )
2
( j2π f )2 X lp ( f )
K2
&&
x (t )
4π 2 lp
1
j
Ac e jφ ( t) and &&
x( t ) = Ac φ&&(t )e jφ ( t ) + jφ& 2( t ) e jφ ( t ) 
2
2
with φ&(t ) = 2π f ∆x (t) , φ&&( t) = 2π f ∆x& (t )
where xlp (t ) =
K
yc (t ) = Ac Re  K0e j[ω c t+φ ( t )] + 22  jφ&&(t ) − φ& 2 (t )  e j[ωc t+φ ( t )] 
4π



K
K


= Ac  K0 − 22 4π 2 f ∆2 x2 (t ) cos [ωc t + φ ( t ) ] − 22 2π f ∆ x& ( t) s i n [ω c t + φ (t ) ]
4π
4π



2

2
K f
 
so A(t ) = Ac K 0 − K2 f∆2 x2 (t )  +  2 ∆ x& (t )  
 2π
 

5.2-11
yc (t ) = Ac cos ω ct + φ y (t )  with φ y (t ) = φ (t ) + arg H [ f (t ) ]
f (t ) − fc = f ∆ x ( t ) ⇒ arg H [ f (t ) ] = α1 f ∆ x (t ) + α3 f ∆3 x3 ( t)
f y (t ) = f c +
1 &
α f
3α f 2
φy (t ) = f c + f ∆x (t ) + 1 ∆ x& (t ) + 3 ∆ x (t )x&( t)
2π
2π
2π
3
5.2-12
−1

2Qf ∆x (t ) 
−1
H [ f (t ) ] = 1 + j
= [1 + jα x( t ) ] , α 2 = 1

fc


−1 / 2
1
H [ f (t ) ] = 1 + α 2 x 2 (t ) 
≈ 1 − α 2 x 2 (t ), arg H [ f (t ) ] = − arctan α x(t ) ≈ −α x (t )
2
t
 1

Thus, y c (t ) ≈ Ac 1 − α 2 x 2 ( t ) cos ω ct + 2π f ∆ ∫ x ( λ ) d λ − α x( t ) 


 2

f y (t ) = f c +
1 &
α
φ y (t ) = f c + f ∆x (t ) −
x& (t )
2π
2π
5.2-13
5-10
B1 ≈ 2 ( D + 1)W , B3 ≈ 2 ( 3D +1 ) W since 3φ( t ) = 2π ( 3 f ∆ ) ∫ x ( λ ) d λ
t
B3
B
> fc + 1
2
2
( f −W )
hence f ∆ < c
2
We want 3 f c −
⇒ 2 f c > ( D + 1) W + ( 3D + 1) W = 4 f∆ + 2W
5.2-14
nφ (t ) = 2π ( nf ∆ ) ∫ x ( λ ) d λ ⇒
t
We want 2 f c +
B2
B
< 3 fc + 3
2
2
Bn ≈ 2(nD + 1)W
and 3 f c +
B3
B
< 4 fc + 4
2
2
1
( B3 + B4 ) = ( 3D + 1 + 4D + 1) W = 7 f∆ + 2W
2
f − 2W
and f ∆ < c
7
so f c >
5.2-15
nφ (t ) = 2π ( nf ∆ ) ∫ x ( λ ) d λ ⇒
t
We want 3 f c +
Bn ≈ 2 ( nD + 1) W
B3
B
B
B
< 4 f c − 4 and 4 f c + 4 < 5 f c − 5
2
2
2
2
1
( B4 + B5 ) = ( 4D + 1 + 5 D + 1) W = 9 f∆ + 2W
2
( f − 2W )
and f ∆ < c
9
so f c >
5-11
5.3-1
Let α = 1/ NVB = 1
c (t ) = c1 +
c2
−1 / 2
c
(1 +α x ) = c1 + 2
VB
VB
3 2 2
 1

1 − α x + α x + L 
8
 2

3 2
1 α
α ≤
⇒
8
100 2
Then c (t ) ≈ c0 − cx(t ) with
c
c α
c2
c0 = c1 + 2
c= 2
=
VB
V B 2 2 NVB VB
Since x ≤ 1, we want
Thus, c ≤
so
NVB ≥
300
4
c2
c
, c0 > 2
150 VB
VB
f∆
c
1
=
<
f c 2c0 300
5.3-2
f ∆ = 150 kHz
0.01 <
BT ≈ 2 ( f ∆ + 2W )
BT
< 0.1 ⇒ 0.01 f c < 2 f ∆ + 4W < 0.1 f c
fc
If f c = 10 MHz
( 0.01× 10 MHz ) - ( 2×150 kHz )
4
Since W cannot be negative, W < 175 kHz
<W <
( 0.1× 10 MHz ) - ( 2×150 kHz )
4
5.3-3
x&c (t ) = − Acθ&c (t) s i n θ c (t ),
∫ θ& (t ) x (t) dt = A ∫ cosθ
c
c
c
c
(t )
dθ c
dt = Ac sin θ c (t)
dt
Thus, −θ&c (t ) ∫θ&c ( t ) xc (t) dt = −θ&c (t ) Ac sin θ c (t ) = x&c (t )
For FM, we want θ&c ( t ) = 2π f (t ) = 2π [ f c + f ∆x (t ) ]
5.3-4
The frequency modulation index is proportional to the message amplitude and inversely
proportional to the message frequency, whereas the phase modulation index is
proportional to amplitude only. Therefore the output of an FM detector tends to boost
higher frequencies, resulting in the higher frequencies in the output message signal being
boosted relative to the original message signal.
5-12
5.3-5
The lower frequencies would have much more phase deviation than a PM modulator
would have given them. Since the output from a PM demodulator is proportional to the
phase deviation, the lower frequencies in the output message signal would be boosted
relative to the original message signal.
5.3-6
f∆
⇒ f ∆ = DW = 5 ×15 = 75 kHz
W
 φ 
f ∆ = n  ∆  so 75,000 < n × 20 ⇒ n > 3750
 2π T 
Since we are using triplers we need n = 3m > 3750
D=
7 37 = 2187
For m = 
therefore m = 8 triplers are needed.
8
8 3 = 6561
If the local oscillator is placed at the end, 6581f c1 − f LO = 915 ×106
Thus, f LO = 6581× ( 500× 103 ) − 915 × 10 6 = 2.37 ×10 9 Hz
5.3-7
25 kHz
= 1250
20 Hz
One doubler and 6 triplers yield n = 2 × 36 = 1458
φ∆
25 kHz
so
=
= 17.1 Hz
2π T
1458
200 kHz × 1458 = 291.6 MHz > 100 MHz
f ∆ = DW = 25 kHz,
n>
Use down-converter before last tripler, where 291.6/3 = 97.2 MHz
so f LO = 97.2 − ( 4.5/3) = 95.7 MHz
5.3-8
5-13
φ∆
2π T
f∆
> 120
φ∆
2π T
Using doublers only 2 7 = 128 > 120 ⇒ 7 doublers
nf c1 = 128 ×10 kHz = 1.28 MHz
f∆ = n
⇒ n=
Since this doesn't exceed 10 MHz, the down converter can be located at any point.
⇒ Choose to place it after the last doubler.
φ
f n ( t ) = nfc1 + n ∆ x (t ) at the end of the last doubler
2π T
f c = nf c1 ± f LO ⇒ 1 MHz = 128 ×10 kHz ± f LO ⇒ f LO = 280 kHz
5.3-9
φ
(a) ∆ ∫ Am cos ω mt dt = β sinω mt
T
NBFM output = Ac cos ω c1 t − Ac β sin ω mt sin ω c1 t = A( t) c o s ωc1 t + arctan ( β sinω mt ) 
f 1( t ) = f c1 +
1 d
cos ωm t
arctan ( β sinω mt )  = f c1 + β f m
2π dt
1 + β 2 sin 2 ωm t
cos ωm t
= cos ω mt 1 − β 2 sin 2 ωmt + β 4 sin4 ω mt + L
2
2
1 + β sin ω mt

1 1

≈ cos ω mt 1 − β 2  − cos2ωmt  , β = 1
2 2


 β2 
β2
= 1−
cos
ω
t
+
( cos ω mt + cos3ωmt )

m
2 
4

 β 2 

β2
Thus, f 1( t ) ≈ f c1 + β f m  1 −
cos3ω mt 
 cos ω mt +
4 
4


2


β 
≈ f c1 + β f m  cos ω mt +   cos3ωmt 
 2


(cont.)
5-14
2
2
β
 φ  A
(b) 3 harmonic distortion =   ×100 = 25  ∆  m2 ≤ 1%
2
 2π T  f m
Worst case occurs with Am maximum and fm minimum, so
2
rd
 φ  1
5 ∆ 
≤1 ⇒
 2π T  30 Hz
φ∆
≤ 6Hz
2π T
5.3-10
f (t ) = f c + f ∆x (t ) = f 0 −b + f ∆x (t )
  2Q 2

2
H [ f (t ) ] = 1 + 
 (b − f ∆ x) 
  f 0 

−1 / 2
2

f∆ x  
2
= 1 + α 1 −

b  


2
−1 / 2
where α =
2Qb
f0
2
1 
f x
f x

≈ 1 − α 2  1− ∆  for α 2  1 − ∆  = 1
2 
b 
b 

2
2
 α 
α f∆ 
A(t ) = Ac H [ f ( t ) ] ≈ Ac 1−
+ Ac 

 x (t )
2 

 b 
2
 2Q 
α2
so y D ( t ) ≈ K D f ∆ x (t ) where K D = Ac
= Ac 
 b
b
 f0 
5.3-11

f (t ) = f c + f ∆x ()t and H ( f ) = 1 +

Let α = f ∆ / f c so α x = 1
2
H [ f (t ) ] = 1 + (1 + α x ) 


−1 / 2
 f 
j  
 f c 
−1
1 
α2 2 
=
x 
1 + α x +
2
2

1  1
α 2 x2 
=
1 −  α x +
+
2 
2  2 
1  1
1 2 2
≈
1− α x + α x 
8
2 2

−1 / 2
2

3
α 2 x2 
α
x
+
+
L



8
2 

2
Ac 
f∆
1  f∆  2 
A(t ) = Ac H [ f (t ) ] ≈
1 −
x( t ) +   x ( t ) 
8  fc 
2  2 f c

Ac
Ac
so y D ( t ) ≈ −K1 f ∆x (t ) + K2 f∆2 x2 (t ) with K1 =
, K2 =
2
2 2 fc
8 2 fc
(cont.)
5-15
1 1
+ cos2ω mt
2 2
K2 f ∆2 / 2
K
harmonic distortion =
×100 ≈ 100 2 f ∆
2
K1 f ∆ − K 2 f ∆ / 2
2 K1
If x(t ) = cos ω mt , then x 2 (t ) =
So 2
nd
= 100
f∆
< 1% ⇒
8 fc
5.3-12
yD (t ) = Ac Hu [ f ( t ) ] − H l [ f ( t ) ]
{
}
f∆
< 0.08
fc
f (t ) = f c + f ∆x (t )
−1/2
−1 / 2
 α 2

 α2
2
2
= Ac 1 + 2 ( f c + f ∆ x − fc − b )  − 1 + 2 ( f c + f ∆ x − fc + b)  

 b
 
 b
2
y D (t) 
f∆x  
2
= 1 + α 1 −

Ac
b  


 α2
= 1 −
2

−1/2
2

f∆ x  
2
− 1 + α  1 +

b  


−1 / 2
2
4
2
4
  α2 

f∆ x  3 4 
f∆ x 
f∆ x  3 4 
f∆x 

1
−
+
α
1
−
+
L
−
1
−
1
+
+
α
1
+
+ L
 








b  8 
b 
2 
b  8 
b 

 

3
2
1 α 2 4 f∆ x 3 4   f∆ x   f ∆ x  
f∆ x 
2
≈
− α 8 
+8
= 1
 when α  1 ±
2 b
8   b   b  
b 

3

f x
 f x 
so y D ( t ) ≈ Ac ( 2α 2 − 3α 4 ) ∆ − 3α 4  ∆   = K1 x(t ) − K3 x3 (t )
b
 b  

2α 2 − 3α 4 ) f ∆
(
2α 2
3α 4
where K1 = Ac
≈ Ac
f ∆ and K3 = Ac 3 f ∆3
b
b
b
K3 3  α f ∆  2 2  α f ∆ 
=
= α
= 1 since α 2 = 1 and b ≤ f ∆
K1 2  b  3  b 
2
2
5.4-1
v (t ) = Ac [1 + x( t )] cos ωc t + Ai [1 + xi ( t )] cos ( ω c + ωi )t + φi  where
Ai = Ac
Av (t ) ≈ Ac [1 + x ] + Ai [1 + xi ] cos (ωi t + φ i ) = Ac {1 + x (t ) + ρ [1 + xi (t )] cos (ω it + φi )}
yD (t ) ≈ K D  x (t ) + ρ cos (ω it + φi ) + ρ xi ()cos
t
(ω it + φ i ) 
xi ()t will be unintelligible if ω i ≠ 0
5-16
5.4-2
v (t ) = Ac cos [ω ct + φ (t ) ] + Ai cos ( ωc + ω i ) t + φi (t ) where φ (t ) = φ∆ x(t ) = 1, Ai = Ac
φ v (t ) = arctan
Ac sin φ + Ai sin ( ωi t + φi )


A
≈ arctan φ + i sin (ω it + φ i ) 
Ac cos φ + Ai cos (ω it + φ i )
Ac


≈ φ∆ x( t ) + ρ sin [ωi t + φi (t ) ]
φ i ( t ) will be unintelligible if ω i ≠ 0.
5.4-3
v (t ) = Ac [1 + µ x (t )] cos ω ct + α Ac 1 + µ x ( t − td )  cos (ω c t − ωc t d )
Envelope detection: yD (t ) = K D  Av (t ) − Av (t ) 
({
where Av (t ) = Ac 1 + µ x( t ) + α 1 + µ x ( t − td )  cos ω c td
Synchronous detection: y D (t ) = K D  vi (t ) − vi (t ) 
{
where vi (t ) = Ac 1 + µ x(t ) + α 1 + µ x ( t − t d )  cos ω ct d
} {
2
+ α 1+ µ x ( t − t c )  sin ω c td
})
2 1/2
}
Thus, Av (t ) always has the same or more distortion than vi ( t ). If ω c td ≈ π / 2 then
cosω c td = 0 and vi ( t ) is distortionless. If ω ct d ≈ π then sinω c td = 0 and Av ( t ) ≈ vi (t) .
5.4-4
Motorized electric appliances generate electromagnetic waves that can interfere with the
amplitude of the AM signals. FM signals do not suffer in quality when the amplitude of
the transmitted signal is corrupted. In addition, most FM cordless phones are above the
frequencies of these interfering signals and other household remote-controlled devices
such as garage door openers.
5-17
5.4-5
Preemphasis increases the energy above 500 Hz so S x will increase.
ST = Pc + 2 Psb
for AM
ST = 2 Psb
for DSB
 Sx
 4 DSB
Psb
but 2 = 
Amax  S x
AM
16
Assuming the peak envelope power allowed by the system is the same
for both AM and DSB
S 2
S 2
ST = Pc + 2 x Amax
for AM
ST = 2 x Amax
for DSB
16
4
Thus, the transmitted power for DSB is increased much more than it is for AM.
5.4-6
Transmitted power is the same in both cases since it depends only on the carrier
amplitude.
Transmitted bandwidth is greater if preemphasis is done prior to transmission since the
frequency deviation is increased by a factor of W / Bde . However, since speech has very
little energy at high frequencies, the bandwidth is driven by the higher amplitude lower
frequencies that are not affected by the preemphasis.
Preemphasis after transmission will amplify any noise or interference signals along with
the signal of interest. Therefore preemphasis prior to transmission is less susceptible to
interference.
Overall, the greater difference is in susceptibility to interference since BT is not much
larger with preemphasis before transmission. Therefore preemphasis at the microphone
end is better than at the receiver end.
5-18
5.4-7
Gxp e ( f ) = H pe ( f )
Thus, for
2

Gx ( f )

2
G x ( f ) ≈  f 

 Gx ( f )
 Bde 
f < Bde
f > Bde
f < Bde , G x pe ( f ) ≤ Gx ( f ) max = G max
2
2
 f 
 Bde 
while for f > Bde , G x pe ( f ) = 
 Gx ( f ) ≤ Gmax if Gx ( f ) ≤ 
 Gmax
 Bde 
 f 
Since BT is essentially determined by the combinati on of maximum amplitude
and maximum-frequency sinusoidal components in the modulating signal, BT is
not increased if Gx pe ( f ) ≤ Gmax .
5.4-8
yD (t ) = α ( ρ ,ω it ) ω i
where
 ρ
 1+ ρ

 ρ2
α=
2
1 + ρ
 −ρ

 1− ρ
ωi t = 0
ω it =
π
2
ωi t = π
5.4-9
(1 ± ε ) − (1 ± ε ) = ε 2 ± ε
α (1 ± ε , π ) =
2
ε2
1 + (1 ± ε ) − 2 (1 ± ε )
Thus, α (1 ± ε ,π ) → ± ∞ as ε → 0
2
= 1±
5-19
1
ε
5.4-10
v (t ) = Ac cos [ω ct + φ (t ) ] + ρ Ac [ωc + θ i (t )]
so φ v (t ) = arctan
sin φ + ρ sin θi
cos φ + ρ cos θi
−1
2
1 &
1   sin φ + ρ sin θ i   d  sin φ + ρ sin θi 
1 + 
yD (t ) =
φ (t ) =
 
2π v
2 π   cos φ + ρ cos θ i   dt  cos φ + ρ cosθi 


&
&
1 ( cos φ + ρ cos θ i ) φ cos φ + ρθ i cos θ i − ( sin φ + ρ sin θ i ) −φ& sin φ − ρθ&i sinθ i
=
2π
( cos φ + ρ cos θ i ) 2 + ( sin φ + ρ sin θi ) 2
{1 + ρ cos [ φ(t ) − θi (t)]} φ&(t) / 2π + {ρ + cos [φ (t ) −θi (t )]} ρ f i
=
1 + ρ 2 + 2ρ cos [φ (t ) −θ i (t ) ]
(
)
5-20
(
)
Chapter 6
6.1-1
1
n
sinc
2
2
cn =
y (t ) = 1 +
⇒ c0 = 1/2, 2 c1 = 2 / π, 2c2 = 0, 2c3 = −2 / 3π
1
2
cos2π20t + cos2π30t + cos2π70t
π
π
6.1-2
1
n
sinc
2
2
cn =
⇒ c0 = 1/2, 2 c1 = 2 / π, 2c2 = 0, 2c3 = −2 / 3π
y (t ) = 1 + cos2π30t +
1
2
cos2π40t + cos2π70t
π
π
6.1-3
Take f s = f l + W + ε. Amplifier then passes xs (t ) since f l > f s − W and f l + B >> f s .
Second chopper with synchronization yields Kxs (t )s (t ) = Kx( t) s 2 (t ) = Kx(t ) since s 2 (t ) = 1.
6-1
6.1-3 continued
6.1-4
xs ( t ) = xL ( t )s (t ) + xR ()[1
t − s( t )]
1
n
1
πn
1 2
2
sinc =
sin
⇒ s (t ) =
+ cos ωs t − cos3ωst + ...
2
2
πn
2
2 π
3π
2
2
1 2

1 2

xs ( t ) = xL ( t ) +  + cos ωs t − cos3ωst + ... + xR (t )  − cos ωs t + cos3ωs t + ...
3π
3π
2 π

2 π

Since LPF rejects f ≥ 99 kHz and 3 f s − W = 3 x 38 - 15 = 99 kHz,
cn =
K1
2K
ω
[ xL (t ) + xR ( t) ] + 2 [ xL (t) − xR (t ) ] + A cos s t
2
π
2
so we want K1 = 2 and K2 = π / 2
xb (t ) =
6.1-5
1
n
1
πn
1 2
2
sinc =
sin
⇒ s (t ) =
+ cos ωs t − cos3ωst + ...
2
2
πn
2
2 π
3π
1
1
2
x1 ( t ) = [ xL (t ) + xR (t ) ] + x [ xL (t ) − xR (t ) ] + high-frequency terms
2
2 π
1
1  2
x2 (t ) = [ xL (t ) + xR ( t ) ] + x  −  [ xL (t ) − xR (t ) ] + high-frequency terms
2
2  π
cn =
6-2
6.1-5 continued
 1 1 
 1 1 
 1 1 
 1 1 
(a) v L ( t ) =  +  − K  −   xL (t ) +  −  − K  +   x R (t ) + ...
 2 π 
 2 π 
 2 π 
 2 π 
 1 1 
 1 1 
 1 1 
 1 1 
v R ( t ) =   −  − K  +   xL (t ) +   +  − K  −  xR ( t ) + ..
 2 π 
 2 π 
 2 π 
 2 π 
π−2
1 1
1 1
we want  −  − K  +  = 0 ⇒ K =
= 0.222
π+2
 2 π
 2 π
so lowpass filtering yields vL' (t ) = 0.778 xL ( t ), v'R (t ) = 0.778 xR (t )
(b) If K = 0, then lowpass filtering yields
v'L (t ) = 0.818x L (t ) + 0.182xR (t) , vR' ( t ) = 0.182 x L ( t ) + 0.818x R (t )
So there's incomplete separation of left and right channels at output.
6.1-6
Let v (t ) = s δ ( t ) = ∑ δ( t − kTs ) with period Ts = 1/ fs so
k
c (nf 0 ) =
1
Ts
Ts / 2
∫
sδ (t ) dt = f s
−Ts / 2
Ts / 2
∫
δ(t )dt = f s
−Ts / 2
Thus S δ ( f ) = V ( f ) = ∑ f s δ( f − nf s ) = f s ∑ δ( f − nf s )
n
n
6.1-7
f s = 60 kHz Recover using LPF 25 ≤ B ≤ 35 kHz
f s = 45 kHz Can't recover by filtering
f s = 25 kHz Recover using BPF over f l ≤ f ≤ 25 kHz with 10 < f l < 15 kHz
6-3
6.1-8
(a)
fs
1
=
2( fu / W )
W M
(b) m = 2, f s =
1
x 2 x 2.5 W = 2.5 W and f s = 4W
2
6.1-9
x (kTs) = sinc 2 5(0.1k ) = sinc 2 0.5k
since sinc2 0.5k = 1 for k ≥ 2,
y (t ) ≈ 0.405 sinc10(t + 0.1) + sinc10t + 0.405sinc10( t -0.1) ≈ sinc2 5t
6-4
6.1-9 continued
6.1-10
 kT  1, kTs ≤ 1
x (kTs ) = Π  s  = 
 2  0, kTs > 1
M
1
1
1
Take K =
and t d = 0 so y (t ) = ∑ sinc f s ( t − kTs ) where
−1 < M ≤
fs
Ts
Ts
k =− M
6.1-11
(a) h(t ) = u( t ) − u (t − Ts )
y( t ) = h(t) * xδ (t ) = ∑ x( kTs )[u (t − kT s) −u (t − kTs − Ts )]
k
6-5
6.1-11 continued
(b)
H ( f ) = Ts sinc fTs
since W = f s , Y ( f ) ≈ Ts X ( f ) for f ≤ W
so x ()t can be recovered using a simple LPF to remove f ≥ f s − W
6.1-12
(a) Let hz (t ) = impulse response of a ZOH = u (t ) − u( t − Ts )
Then h(t ) =
1
hz (t) * h z (t ) + h z (t ) − hz ( t − Ts )
Ts
 t − Ts 
1
hz (t) * hz (t ) = Λ 

Ts
 Ts 
y( t ) is a linear piecewise approximation obtained by extrapolating forward from the
where
two previoius values.
6-6
6.1-12 continued
(b) Let H Z ( f ) = ℑ[ hz (t ) ] = Tssinc f Ts e− jωTs / 2
H( f ) =
1 2
H z ( f ) + Hz ( f ) − H z ( f )e − jω Ts
Ts
(
)
= Ts sinc 2 f Ts e − jωTs + Ts sinc fTs e jωTs / 2 − e− jωTs e − jωTs


1
= Ts 1 +
j 2sin πfTs  sinc 2 f Ts e− jωTs
 sinc fTs

2
− j ωTs
= Ts (1 + j 2πfTs ) sinc fTs e
H ( f ) = Ts
(1 + ( j 2πfTs )
2
sinc 2 fTs
Note that high frequency components of X ( f ) are accentuated.
6-7
6.1-13
∞

k 
f − jωk / 2 W
 k 

 k  1
X ( f ) = ℑ∑ x 
sinc
2
Wt
t
−
=
x
Π
e
∑





 2W   k =−∞  2W  2W 2W
 k  2W 
∞
+W

 k  1 − jωk / 2W  
 m  1 + jω m/ 2 W 
E = ∫ X ( f ) X * ( f )df = ∫ ∑ x 
e
x* 
e
∑




 df
 2W  2W
  m  2W  2W

−∞
−W  k
1
=
2W
=
1
2W
1
=
2W
 k
x
∑∑
 2W
k
m
  m  1
 x  2W  2W
 

W
∫e
+ j π (m −k ) f / W
df
−W
∑∑ x  2W  x  2W  sinc(m − k )
k
k
m
m
∞
∑
k =−∞
 k 
x

 2W 
2
1 m = k
since sinc( m - k ) = 
0 m ≠ k
6.1-14
∞
∑
v (t ) =
n =−∞
∞
cv ( nf 0 ) e jnω0 t ⇒ V ( f ) = ∑ cv (nf 0 ) δ( f − nf 0 )
n =−∞
where
1
cv ( nf 0 ) =
T0
T0 / 2
∫
v (t )e− jnω0 t dt =
−T0 / 2
1
T0
T
∫
x(t )e − jnω0 t dt =
−T
1
T0
∞
∫ x (t )e
− jnω0 t
dt = f 0 X ( nf 0 )
−∞
But x (t ) = v (t ) Π (t / 2T ) ⇒ X ( f ) = V ( f )*(2T sinc 2Tf ) so


X ( f ) =  ∑ f 0 X (nf 0 )δ( f − nf 0 )  *(2T sinc 2Tf )
 n

∞
= 2Tf 0
∑ X (nf
n =−∞
0
) sinc 2T ( f − n f 0 )
Hence, X ( f ) is completely determined by the sample values of X (nf 0 ).
6.1-15
B = f s / 2 ⇒ f s ≤ 2B = 12 MHz
f x = 1/ Tx = 12.5 MHz, 1 − α = f s / f x =0.96
⇒ α = 0.04
2 m + 1 < 1/ α = 25 ⇒ mmax = 11
Presampling bandwidth ≤ 11 x 12.5 = 137.5 MHz
6-8
6.1-16
B = f s / 2 ⇒ f s ≤ 2B
1 − α = f s / f x ≤ 2 BTx < 2/3 ⇒ α =1/3
2 m + 1 < 1/ α < 3 ⇒ mmax = 0
so only the dc component could be displayed
6.1-17
W = 15 kHz, f s = 150 kHz
with
HZOH ( f ) = Ts sinc(fTs ) and H FOH ( f ) = Ts 1 + (2 πfTs ) 2 sinc 2 (fTs )
(a) For a ZOH, the maximum aperature error in the signal passband occurs at
f = 15 kHz and thus:
HZOH ( f )
f =15 kHz
= 0.9836 and H ZOH ( f ) f =0 kHz = 1
⇒ % aperatureerror =
1 − 0.9836
x 100%=1.640%
1
(b) For a FOH, the maximum aperature error in the signal passband occurs at
f = 15 kHz and thus:
H FOH ( f ) f =15kHz = 1.1427 and H FOH ( f ) f = 0kHz =1
⇒ % aperatureerror =
1 − 1.1427
x 100%= − 14.27%
1
6.1-18
W = 15 kHz, f s = 150 kHz , Error % =
⇒ f a = 150-15=135 kHz, ⇒ Error % =
1/0.707
1 + ( f a / B )2
x 100% and B = W
1/0.707
1 + (135/15) 2
6-9
x 100% = 15.61%
6.1-19
If x(t ) is a sinusoid with period 2T0 with its zero crossings occuring at t = T0 and
the sampling function has period Ts = 2T0 . It is possible for the sampler to sample x (t ) at t = T0 .
Therefore, the output of the sampler is always = 0.
6.1-20
1
f
Π(
) ⇒ to sample, f s ≥ 100
100 100
1
f
(b) sinc 2 (100t ) = sinc2 (2 x 50t ) ⇔
Λ(
) ⇒ to sample, f s ≥ 200
100 100
10
(c) 10cos3 2πx105t = (3cos2πx105 t + cos2π x3x105t)
4
Its bandwidth = (3 -1) x 105 = 2 x 105 Hz ⇒ f s > 4 x 105 Hz.
(a) sinc(100t ) = sinc(2 x 50t ) ⇔
6.1-21
At f =159 kHz the signal level is down -3 dB and we want aliased components down -40 dB.
⇒ at f =159 kHz, aliased components should be down -43 dB = 5 x 10-5 .
1
1
H( f ) =
⇒ 5 x 10-5 =
⇒ f = 3172 MHz.
2
1+ ( f / B )
1 + ( f /159) 2
6.2-1
P( f ) = τsinc f τ =
Ts
f
sinc
2
2 fs
K
K
=
for f ≤ W
sinc (f / 2 f s ) sinc ( f / 5 fW )
H eq (0) = K , H eq (W ) = 1.07 K , so equalization is not essential
H eq ( f ) =
6-10
6.2-2
τ /2
P( f ) = 2 ∫ cos
0
=
πt
cos2πft dt
τ
 f − fs
τ
1
1  T 


sinc  f τ −  + sinc  f τ +  = s  sinc 

2
2
2  4 


 2 fs

 f + fs
 + sinc 

 2 fs
−1

 f − 2.5W 
 f + 2.5W  
H eq ( f ) = K sinc 
+ sinc 
for f ≤ W


 5W

 5W


H eq (0) = 0.785K , Heq (W ) = 0.816K so equalization is not essential.
6.2-3
1 t
(a) Let x ( t ) = ∫ x (λ )d λ = x( t) * h(t )
τ t −τ
where
t
h (t ) =
1
1
δ(λ) d λ = [u (t ) − u( t − τ) ] ⇒ H ( f ) = sinc f τ e− jωτ / 2
∫
τ t −τ
τ
Averaging filter
X( f )
Xδ(f )
Xp (f )
X ( f ) → H ( f ) = sinc f τ e− jωτ / 2 → Ideal sampler → P( f ) →
X p ( f ) = P( f ) X δ ( f )
where X δ ( f ) = f s ∑ X ( f − nf s ) = f s ∑ H ( f − nf s )X ( f − nf s )
n
n
(b) X p ( f ) = P ( f ) f s H ( f ) X ( f ) for f ≤ W , where P( f ) = τ sin c f τ
Thus, H eq ( f ) = Ke − jω (td − τ /2) /sinc2 f τ,
f ≤W
6-11



6.2-4
(a) Let v(t ) = A0 [1 + µx(t )] ↔ V ( f ) = A0 [δ ( f ) + µX ( f )]
x p ( t ) = ∑ v( kTs ) p( t − kTs ) = p( t) * vδ ( t )
k


X p ( f ) = P( f )Vδ ( f ) = A0 fs P( f ) ∑ [ δ ( f − nf s ) + µX ( f − nf s ) ]
n

(b) P( f ) = τsinc f τ =
µX ( f ) =
1
f
sinc
2 fs
2 fs
1
[δ ( f − fm ) + δ( f + fm ) ]
2
6.2-5
(a) P( f ) = τ sinc f τ ( e − jωτ / 2 − e jωτ / 2 ) = τ sinc f τ ( −2 j sin πf τ) = −
(b) X p ( f ) = − A0
H eq ( f ) =
j πf
f
2
sinc
µX ( f ) for
8 fs
4 fs
Ke − jωtd
− jf sinc 2 ( f / 4 f s )
f ≤W
fl ≤ f ≤ W
If f l → 0 then H eq (0) → ∞ and equalization is not possible.
6-12
j πf
f
sinc 2
2
8 fs
4 fs
6.2-6
The spectrum of a PAM signal is like that of the chopper sampled signal of Fig. 6.1-4
and can be written as X s ( f ) = c0 X ( f ) + c1[ X ( f − f s ) + X ( f + f s ) ] + ...
With product detection ⇒ xs ( t ) x cos2πf st giving a frequency domain expression of
c0
c
X ( f − f s ) + 0 X ( f + fs )
2
2
c
c
c
c
+ 1 X ( f − fs + f s ) + 1 X ( f − fs − f s ) + 1 X ( f + fs + f s ) + 1 X ( f + fs − f s )
2
2
2
2
Combining terms and using a LPF the output spectra from the product detector gives
c1
c
X ( f ) + 1 X ( f ) = c1 X ( f )
2
2
6.3-1
τ min =
Ts
1
(1− 0.8) =
5
25 f s
⇒ tr ≤
1
100 f s
so BT ≥ 1 / 2t r ≥ 50 f s = 400 kHz
6.3-2
τ min = τ0 (1 − 0.8) ≥ 3t r and t r ≥ 1 / 2 BT ⇒ 0.2 τ0 ≥ 3 / 2 BT
τ max = τ 0 (1 + 0.8) ≤ Ts /3
⇒ τ0 ≤
1
= 23.1 µs
1.8 x 3 f s
Thus, 15 ≤ τ0 ≤ 23.1 µs
6.3-3

 2πk  
(a) τk = 0.4Ts 1 + 0.8cos 

 3 


 2 πk  
(b) t k = Ts  k + 0.5 + 0.2cos 

 3 

6-13
⇒ τ 0 ≥ 15 µs
6.3-3 continued
6.3-4

 πk 
(a) τk = 0.4Ts 1 + 0.8cos  
 3 


 πk  
(b) t k = Ts  k + 0.5 + 0.2cos   
 3 

6.3-5
Take t 0 = Ts so that t 0 x& (t ) < 1. Apply -x(t ) to PPM generator to get


x p ( t ) = A( t ) 1 + ∑ 2cos [ nωs t + nω s t0 x (t ) ]

n

where A(t ) = Af s [1 + t 0 x& (t ) ] > 0
6-14
6.3-5 continued
− x (t)
x p (t)
BPF
→ PPM →
v(t )
f 0 = mf s
> Lim → BPF → xc (t)
f 0 = mf s
First BPF yields v (t ) = 2 A(t) c o s [ωc t + φ∆ x (t )]
with f c = mf s , φ∆ = 2πmf s t 0 = π if m =
Ts
2t0
Limiter and second BPF give xc (t ) = Ac cos [ ωc t + φ∆ x (t ) ]
6.3-6
(a) sδ (t ) = ℑ−1 [ S δ ( f ) ] = f s ∑ e − j 2πn fs t and sδ (t ) = ∑ δ( t − kTs )
n
∞
Thus,
∑e
± j 2 πnt / L
=L
n=-∞
∞
∑ δ(t − kL)
k=-∞
k
where L = 1/ f s = Ts
(b) Sδ ( f ) = ℑ [ sδ ( t ) ] = ∑ e− j2 πfkTs and Sδ ( f ) = f s ∑ δ ( f − nf s )
k
∞
∑e
Thus,
± j 2 πkf / L
k =−∞
n
∞
= L ∑ δ ( f − nL ) where L = 1/ Ts = f s
n =−∞
6.3-7
g ( t ) = v ⇒ t = g −1 (v ) and λ = g −1 (0)
dv
1
1
g& ( t ) =
⇒ dt =
dv =
dv
−1
dt
g& (t )
g&  g ( v) 
b
Thus, ∫ d [g( t )]dt =
a
v =g (b )
δ( v)
dv
&  g −1 (v ) 
v= g ( a ) g
∫
If g& (λ ) > 0, then g (b) > 0 > g ( a) so
g ( b)
δ(v )
1
1
dv =
=
∫
−1
−1
&  g (v ) 
g&  g (0)  g& ( λ)
g ( a) g
6-15
6.3-7 continued
If g& (λ ) < 0, then g (b ) < 0 < g ( a ) so
g ( b)
∫
g ( a)
δ(v )
δ(v )
1
dv = − ∫
dv = −
−1
−1
g& (λ)
&  g (v ) 
g&  g (v ) 
g ( b) g
g ( a)
b
b
1
1
Hence ∫ δ[ g (t )] dt =
=∫
δ(t − λ ) dt
g& ( λ) a g& (t )
a
so δ[ g (t )] = δ( t − λ) / g& (t )
6-16
Chapter 7
7.1-1
10
kHz ⇒ f IF ≥ (1605 − 540)/2 = 532.5 kHz
2
= 1072.5 to 2132.5 kHz, BT = 10 kHz < BRF < 2 f IF = 1065 kHz
f c' = f c + 2 f IF ≥ 1600 +
f LO = fc + f IF
7.1-2
250
kHz ⇒ f IF ≥ (107.9 − 87.975)/2 = 9.9625 MHz
2
= 78.1375 to 97.9375 MHz, BT = 250 kHz < BRF < 2 fIF = 19.925 kHz
f c' = f c − 2 f IF ≤ 88,100 −
f LO = fc − f IF
7.1-3
C = 1 / 4π 2 Lf lo2 = 2.533 x 10 4 / f lo2
f lo = fc + f IF = 995 − 2055 kHz ⇒ C = 6.0 - 25.6 nF
f lo = fc − f IF = 85 − 1145 kHz ⇒ C = 19.3 - 3,506 nF
7.1-4
f c = 1 / 2π LC ⇒ C = 1 / 4π2 Lfc2 = 9.9 - 86.9 nF
Q=R
C
f
1
= c =
L BRF 2π LCBRF
BRF > BT
BRF > 2 f IF
⇒ R=
1
2πBRF C
1
= 1.6 kΩ ,
2π x 10 kHz x 9.9 nF
1
⇒ R>
= 2.0 Ω
2π x 910 kHz x 86.9 nF
⇒ R<
7.1-5
f IF ≈ BT /0.02 = 200 kHz since BT = W
f LO = fc + f IF = 3.77 - 3.83 MHz, f c' = f LO + f IF = 3.97 - 4.03 MHz
Take BRF ≈ 0.02 x 3.6 MHz = 72 kHz centered at 3.6 MHz
IF must pass f IF − W ≤ f ≤ fIF
7-1
7.1-5 continued
7.1-6
f IF ≈ BT /0.02 = 300 kHz since BT = W
f LO = fc + f IF = 7.34 - 7.46 MHz, f c' = f LO + f IF = 7.54 - 7.66 MHz
Take BRF ≈ 0.02 x 7.2 MHz = 144 kHz centered at 7.2 MHz
IF must pass f IF ≤ f ≤ fIF + W
7.1-7
7-2
7.1-8
θ IF = ωc t + φ − θ where φ = 2πf ∆ x(t ), θ& = 2π [ f c − fIF + kv (t ) + ε(t ) ]
(
)
KD &
θIF − 2 πf IF = K D [ f c + f ∆ x − f c + fIF − Kv − ε − f IF ]
2π
= KD [ f ∆ x ( t ) − Kv( t ) − ε (t ) ]
yD =
so v (t ) = KD [ − Kv( t ) − ε (t ) ]
Thus,
v (t ) = − K D ε(t) / ( 1 + K D K )




− KD K ε (t )
1
and y D (t ) = K D  f ∆x (t ) −
− ε(t )  = K D  f ∆ x (t ) −
ε (t ) 
1 + KD K
1+ K D K




≈ K D f ∆ x ()t if K D K ? 1
7.1-9
(a) With f c = 50 → 54 MHz and f IF = 455 kHz ⇒ f L O = 50.455 → 54.455 MHz.
⇒ f c' = f c + 2 f IF = 50.910 → 54.910 MHz.
(b) With f c = 50 → 54 MHz and f IF = 7 MHz ⇒ f LO = 57 → 61 MHz.
⇒ f c' = f c + 2 f IF = 64 → 68 MHz.
7.1-10
If W = signal bandwidth, then the incomming signal is 50 + W → 54 + W MHz.
With f IF =100 MHz, to avoid sideband reversal use f LO = 150 → 154 MHz.
At the product detector stage, use an oscillator frequency of 100 MHz.
The image frequency is f c' = f c + 2 f IF and its range is thus 250 → 254 MHz.
7-3
7.1-11
Image frequency = f c' = f c + 2 f IF = 2 + 2 x 455 = 2.91 kHz.
1
For a BPF with center frequency of f 0 =f c ⇒ H ( f ) =
1+Q 2 (
f
f
− 0 )2
f0
f
and Q = f 0 /B. We use the BPF to reject images and thus we have
H ( f ) f =2.9 MHz =
1
2.91
2 2
1+4 (
−
)
2
2.91
= 0.3123 ⇒ 20log(0.3123) =-10 dB.
2
Images are rejected by -10 dB.
7.1-12
(a) With f LO = 2.455 MHz and f IF = 455 KHz, then f c = 2 MHz, and f c' = 2.910 MHz (image).
With f LO = 2.455 x 2 = 4.910 MHz ⇒ Input frequencies accepted are:
f c'' = 4.910 − 0.455 = 4.455 MHz, and f c''' = 4.455 + 2x 0.455=5.365 MHz.
Given the RCL BPF with B = 0.5 MHz ⇒ Q = 2/0.5 = 4
1
H ( f ) f =2.9 MHz =
= 0.3123 ⇒ 20log(0.3123) =-10 dB
2 2
2 2.91
1+4 (
−
)
2
2.91
We repeat the above calculation for the spurious frequencies of 4.455 and 5.360 MHz.
But because the LO oscillator harmonic is 1/2 that of the fundamental we multiply the result
by 1/2. Hence,
1
H ( f ) f =4.455 MHz =
= 0.139 x 1/2 = 0.070 ⇒ 20log(0.070) = - 23.1 dB
2 2
2 4.455
1+4 (
−
)
2
4.455
and
1
H ( f ) f =5.365 MHz =
= 0.108 x 1/2=0.054 ⇒ 20log(0.054) =-25.4 dB
2 2
2 5.365
1+4 (
−
)
2
5.365
7-4
7.1-12 continued
(b) To reduce spurious inputs: (1) use a more selective BPF, (2) Use filter to reject the
LO second harmonic, (3) use a higher f IF .
7.1-13
If f c1 = 50 → 51 MHz and f IF1 = 7 → 8 MHz.
We could choose a fixed frequency output LO with f LO = 43 MHz.
(a) With f c1 = 50 MHz and fIF1 = 7 MHz, and fL O =43 MHz, the image frequency is
f c'1 = f c1 − 2 x f IF1 = 50 − 2 x 7 = 36 MHz
But, the original 7 MHz receiver also suffers from images, so if the incomming signal is
supposed to be 7.0 MHz, it could also be 7 + 2 x 0.455 = 7.910 MHz ⇒ f IF' 1 = 7.910 MHz.
⇒ f c1 = 43 + 7.910 = 50.910 MHz will also be heard.
'
(b) Use a more selective BPF at the output of the first mixer and/or at the input of the 7 MHz
receiver.
7.1-14
With f c = 50 → 54 MHz, let's use fc = f 0 = 52 MHz. Assume f LO = fc + f IF
(a) With f I F = 20 MHz ⇒ f LO = 72 MHz and ⇒ fc' = 52 + 2 x 20 =92 MHz.
Q = f 0 / B = 52/4 = 13
H ( f ) f =92 MHz =
1
92 52
1+132 ( − )2
52 92
= 0.064 ⇒ 20log(0.064) =-23.9 dB
(b) With f I F = 100 MHz ⇒ f LO = 152 MHz and ⇒ fc' = 52 + 2 x 100 =252 MHz.
H ( f ) f =152 MHz =
1
252 52 2
1+13 (
−
)
52 252
= 0.017 ⇒ 20log(0.017) =-35.6 dB
2
7-5
7.1-15
Given f c1 = 850 MHz and f c2 = 1950 MHz, let's pick a common 500 MHz IF ⇒ f IF = 500 MHz.
For f c1 = 850 MHz, select f LO1 = f c1 + f IF ⇒ f LO = 1350 MHz
and
for f c2 = 1950 MHz, select f LO2 = fc 2 − f IF ⇒ f LO = 1450 MHz.
⇒ f L O = 1350 → 1450 MHz.
Image frequencies:
f c = 850 MHz ⇒ f c' = 850 + 2 x 500 = 1850 MHz
and
f c = 1950 MHz ⇒ f c' = 1950 - 2 x 500 = 950 MHz.
7.1-16
BIF2 = 2W , f IF2 ≈ 2W /0.02 = 1 MHz
From Exercise 7.1-2, f IF1 ≈ 9.5 f c = 38 MHz so BIF1 ≈ 0.02 x 38 = 760 kHz
f LO 1 = f c + f IF1 = 42 MHz,
f LO 2 = f IF1 ± f IF2 = 37 or 39 MHz
7.1-17
f LO 1 = f c + f IF1 = 330 MHz ⇒
fc' = 330 + 30 = 360 MHz
f LO 2 = f IF1 + f IF2 = 33 MHz, so image frequency at input of 2nd mixer is
f LO2 + f IF2 = 36 MHz produced by
f c' − f LO1 = 36 MHz ⇒ fc' = 294 and 366 MHz
7.1-18
f LO 1 = f c − f IF1 = 270 MHz ⇒
fc' = 270 − 30 = 240 MHz
f LO 2 = f IF1 − f IF2 = 27 MHz, so image frequency at input of 2nd mixer is
f LO2 − f IF2 = 24 MHz produced by
f c' − f LO1 = 24 MHz ⇒ fc' = 246 and 294 MHz
7-6
7.1-19
1/ T0 = 20 Hz, so take B < 20 Hz to resolve lines
f 1 = 0, f 2 = 10/ T0 = 200 Hz, T ≥
f 2 − f1
B
2
>
200 Hz
= 0.5 sec
2
(20 Hz)
7.1-20
Take B < fm = 1 kHz to resolve lines, β = 5 ⇒ 8 pairs of sideband lines.
f 1 = f c − 8 f m = 92 kHz, f 2 = f c + 8 f m = 108 kHz
f 2 − f1 16 kHz
>
= 16 ms
B2
(1 kHz) 2
T≥
7.1-21
hbp (t ) = cos α t 2 cos ωct − sin αt 2 sin ωct so
hlp (t ) =
1
1 2
cos αt 2 + j sin α t 2 ) = e jαt
(
2
2
xbp (t ) = v(t) c o s α t 2 cos ωc t −  −v( t) s i n α t 2  sin ωc t so
2
1
1
v (t) c o s αt 2 − jv(t) s i n αt 2  = v( t) e− j αt
2
2
∞
1
1 j αt 2 ∞
− j αλ 2 j α( t −λ )2
ylp (t ) = x lp * hlp = ∫ v( λ) e
e
d λ = e ∫ v (λ )e − j 2αtλ d λ
4 −∞
4
−∞
xlp (t ) =
Ay (t ) = ylp (t ) =
1 ∞
1
v( λ) e− j 2π( αt / π) λ d λ = V ( f )
∫
4 −∞
4
f =αt / π
7.2-1
7-7
7.2-2
7.2-3
DSB
SSB
7.2-4
We want H ( f ) ≤ 0.1 for
W / 2 + Bg
f − f0 ≥
W
+ Bg
2
1
ln(1/0.1) ≈ 1.26 ⇒ Bg ≥ 0.76W
W
1.2
Thus, BT = 10W + 9Bg ≥ 17W
so
≥
7-8
7.2-4 continued
7.2-5
Let f i = i th subcarrier, take B = 3 kHz
f 0 = f i − 0.2 kHz -
B
= f i − 1.7kHz
2
2
We want H ( f ) ≤ 0.01 for f − f 0 ≥ B /2 + 1 kHz = 2.5 kHz
 2 x 2.5 
Thus, 1 + 

 3 
2n
≥ 100 ⇒ n ≥
1 ln99
≈ 4.5 ⇒ n = 5
2 ln(5/3)
7.2-6
(a) Bi = 2M ( D)Wi = 2α M ( D) f i . fi + Bi / 2 + Bg = fi +1 − Bi+1 / 2
Thus f i +1 =
[1 + α M ( D)] f i + Bg
1 − α M (D)
7-9
7.2-6 continued
1
B1 / f1 = 0.2 ⇒ f i+1 = (1.2 f i + 400)/0.8
2
so f 2 = 3.5 kHz, f 3 = 5.75 kHz, f 4 = 9.125 kHz
(b) αM ( D) =
7.2-7
xc (t ) = x1 (t) c o s ωc t + x2 ()cos(
t
ωc t ± 900 ) taking Ac = 1
2 xc ()cos(
t
ωc t + φ' ) = x1 (t )  cos φ ' + cos(2ωc t + φ ' ) 
+ x2 (t ) cos(φ' ± 900 ) + cos(2ωc t + φ' ± 900 ) 
2 xc ()cos(
t
ωc t + φ' ± 900 ) = x1 ( t )  cos(φ ' ± 900 ) + cos(2ωc t + φ' ± 1800 ) 
+ x2 (t ) cos φ ' + cos(2ωc t + φ' ± 1800 ) 
Thus, LPF outputs are
y1 ( t ) = K  x1 (t) c o s φ' m x2 (t) s i n φ ' 
y2 (t ) = K m x1 (t) s i n φ' + x2 ()cos
t
φ' 
7.2-8
We want
x0 + x1 = 2( LF + LR ) 
 ⇒ x1 (t ) = LF + LR − ( RF + RR )
x0 − x1 = 2( RF + RR ) 
x0 + x1 + x3 = 3LF + L R + RF − RR
x0 + x1 − x3 = LF + 3LR − RF + RR
x0 − x1 + x3 = LF − LR + 3RF + RR
x0 − x1 − x3 = − LF + LR + RF + 3 RR
Take x2 ( t ) = LF − L R − RF + RR so that
x0 + x1 + x2 + x3 = 4 LF
x0 + x1 − x2 − x3 = 4 LR
x0 − x1 − x2 + x3 = 4 RF
x0 − x1 + x2 − x 3 = 4 RR
7-10
7.2-8 continued
7.2-9
ℑ [ x2 (t) s i n ωc t ] = −
j
[ X 2 ( f − fc ) − X 2 ( f + fc )] so
2
Ac
[ X ( f − fc ) + X1 ( f + fc ) m jX 2 ( f − f c ) ± jX 2 ( f + fc )]
2 1
Yc ( f ) = HC ( f ) X c ( f )
1
ℑ [ yc (t) c o s ωc t ] = [ H C ( f − f c ) Xc ( f − f c ) + H C ( f + f c ) Xc ( f + f c ) ]
2
A
= c {HC ( f − f c ) [ X 1 ( f − 2 f c ) + X1 ( f ) m jX 2 ( f − 2 f c ) ± jX 2 ( f ) ]
4
+ H C ( f + fc ) [ X1 ( f ) + X1 ( f + 2 f c ) m jX 2 ( f ) ± jX 2 ( f + 2 f c ) ]}
Xc ( f ) =
The output of the lower LPF is
A
Y1 ( f ) = c {[ H C ( f − f c ) + HC ( f + f c )] X1 ( f ) ± j[ HC ( f − f c ) − H C ( f + f c )]X 2 ( f )}
4
To remove cross t alk from X 2 ( f ), we must have HC ( f − f c ) − H C ( f + f c ) = 0 for f < W
Ac
[ H C ( f − fc )] X1 ( f ) so Y1 ( f ) H eq (f) = KX 1 ( f ) e− jωtd
4
4K / Ac
where the equalizer has H eq ( f ) =
e − jω td
H C ( f − fc )
Then, Y1 ( f ) =
7.2-10
r = (24 + 1) x 8 kHz = 200 kHz, τ = 0.5
Ts
1
= 2.5 µs, BT ≥
= 200 kHz
24 + 1
2τ
7-11
7.2-11
r = (24 + 1) x 6 kHz = 150 kHz, τ = 0.3
Ts
1
= 2 µs, BT ≥
= 250 kHz
24 + 1
2τ
7.2-12
f s = 2W + Bg = 10 kHz
Ts
= 1.25 µs ⇒ BT ≥ 1/ τ = 800 kHz
20
1
(b) BT = Bb =
x 20 fs = 100 kHz
2
(a) τ = 0.25
7.2-13
f s = 2W + Bg = 5 kHz
Ts
= 4 µ s ⇒ BT ≥ 1/ τ = 250 kHz
10
1
(b) Bb =
x 10 fs = 25 kHz, D = f ∆ / Bb = 3, BT ≈ 2(3 + 2) Bb = 250 kHz
2
(a) τ = 0.2
7.2-14
Sampling rate (kHz)
Minimum
16
7
4
3.6
3
2.4
Actual
2x8
8
4
4
4
4
FDM - SSB: BT ≥ ∑ Wi = 18 kHz
i
7-12
7.2-15
Sampling rate (kHz) Minimum
24
8
2
1.8
1.6
1.0
0.6
Actual
3x8
8
2
2
2
1
1
FDM - SSB: BT ≥ ∑ Wi = 19.5 kHz
i
7.2-16
Sampling rate (kHz)
Sampling rate (kHz) Minimum
24
7
4
1
0.8
0.6
0.4
0.2
Actual
3x8
8
4
1
1
1
0.5
0.5
7-13
7.2-16 continued
FDM - SSB: BT ≥ ∑ Wi = 19 kHz
i
7.2-17
−54.5 BTg ≤ −40 ⇒ Tg ≥ 0.734/ B
Ts
0.734
0.2
= Tg + 2t 0 + τ ≥
+ 3t 0 , t0 = τ =
= 1 µs
M
B
25 x 8 kHz
0.734
1
≤
− 3t 0 = 2 µs ⇒ B ≥ 367 kHz
B
Mf s
7.2-18
−54.5 BTg ≤ −30 ⇒ Tg ≥ 0.55/ B
Ts
0.25Ts 0.55 3 Ts
= Tg + 2t 0 + τ = Tg + 3 x
>
+
M
M
B
4M
1 Ts 0.55
1
B
Thus,
>
⇒ M <
= 28.4 ⇒ M = 28
4M
B
4 x 0.55 fs
7.2-19
7-14
7.3-1
(a)
(b)
φV
similarly, we get
450
1350
1800
2250
3150
__________________________________________________________
y / A -6/8
0
2/8
6/8
-2/8
7.3-2
similarly, we get
(b)
φV
450
1350
1800
2250
3150
__________________________________________________________
0
y / A 2/8
-4/8
-2/8
2/8
7-15
7.3-3
t < 0, ε ss = ∆f / K
∆f + f1
1 so assume ε 1 and sinε ≈ ε
K
Thus, ε& + 2πK ε = 2π ( ∆f + f1 ) ⇒ trial solution ε = A + Be st
t > 0, φ' =2πf1 and
Then Bse st + 2πKA + 2πKBe st = 2π ( ∆f + f1 )
∆f + f1
⎧
2πKA = 2π ( ∆f + f1 ) ⎫⎪
⎪A =
so
K
⎬ ⇒ ⎨
( s + 2πK ) Be st = 0 ⎪⎭
⎪⎩ s = −2πK
∆f + f1
and ε(t ) =
+ Be −2 πKt , t > 0,
K
Since ε(t ) can make a step change at t = 0,
ε(0+ ) =
∆f + f1
∆f
+ B = ε(0− ) =
K
K
⇒ B=−
f1
K
Hence,
⎧ ∆f
⎪⎪ K
ε(t ) = ⎨
⎪ ∆f + f1 (1 − e −2 πKt )
⎪⎩ K K
t<0
t>0
7.3-4
1
Ac ⎡ x(t ) cos ωc t − xq (t ) sin ωc t ⎤⎦ where xq (t ) = ± x% (t ) for SSB
2 ⎣
=A(t ) cos [ ωc t + φ(t ) ]
xc (t ) =
xq (t )
1
Ac x 2 (t ) + xq2 (t ), φ(t ) = arctan
x(t )
2
If loop locks to φ(t ) and ε ss ≈ 0, then the output is proportional to A(t).
with A(t ) =
Otherwise, φ(t ), may be too rapid for loop to lock.
7-16
7.3-5
1
cos [ θv (t ) − (ω1t + φ1 )] + high frequency term
2
Thus, cos [ θv (t ) − (ω1t + φ1 )] = cos ( ωc t + φ0 + 900 − ε ss )
cos θv (t ) x cos(ω1t + φ1 ) =
so cos θv (t ) = cos ⎡⎣(ωc + ω1 )t + φ0 + φ1 + 900 − ε ss ⎤⎦
7.3-6
cos [ θv (t ) / n ] = cos(ωc t + φ0 + 900 − ε ss )
so cos θv (t ) = cos(nωc t + nφ0 + n900 − nε ss )
7.3-7
Let subcarrier be cos(ωsc t + φsc ) so pilot signal is cos [ (ωsc t + φsc ) / 2]
and output of PLL doubler will be cos φv (t ) = cos ⎡⎣ 2(ωsc t + φsc ) / 2 + 2 x 900 ⎤⎦
7.3-8
7-17
7.3-9
f LO = f c + f IF = 98.8 to 118.6 MHz in steps of 0.2 MHz = 120.0 MHz ÷ 600
120.0 - 98.8 = 106 x 0.2 MHz,
120.0 - 118.6 = 7 x 0.2 MHz
7.3-10
f LO = f c + f IF = 955 to 2055 kHz in steps of 10 kHz = 2 x 2105 kHz ÷ 421
2105 - 955 = 115 x 10 kHz,
2105 - 2055 = 5 x 10 kHz
7.3-11
1
Y ( f ) and Φ ( f ) = φ∆ X ( f ) for PM, so
j 2πf
φ
φ
1 1 jfKH ( f )
Z( f )
KH ( f )
=
φ∆ = ∆
= ∆ HL ( f )
2πK v jf + KH ( f ) 2πK v
X ( f ) j 2πf K v jf + KH ( f )
Z( f ) =
7.4-1
(a) The frame should have an odd number of lines so that each field has a half-line to
fill the small wedge at the top and bottom of the raster.
7-18
7.4-1 continued
(b) A linear sweep (sawtooth or triangular) is needed to give the same exposure time
to each horizontal element. A triangular sweep would result in excessive retrace time,
equal to the line time.
7.4-2
(a) No vertical dependence. Video signal is rectangular pulse train with
τ = ( H / 4) / sh = 1/ 4 f h and T0 = 2τ.
Thus, f 0 = 2 f h
c(nf 0 ) = Ksinc
n
2
(b) No horizontal dependence. Video signal is rectangular pulse train with
τ = (V / 4) / sv = 1/ 4 f v and T0 = 2τ.
Thus, f 0 = 2 f v
n
2
Same spectrum as (a) with f h replaced by f v
c(nf 0 ) = Ksinc
7-19
f h , so much smaller bandwidth.
7.4-3
(a)
⎧
⎪1
I (h, v) = ⎨
⎪⎩0
cmn
1
=
HV
H αH
H αH V β V
V βV
−
<h<
+
−
<v< +
,
2
2
2
2
2
2
2
2
otherwise
(1+α ) H / 2
∫
(1−α ) H / 2
(1+β )V / 2
e
− j 2 πmh / H
dh
∫
e− j 2 πmv / V dv
(1−β )V / 2
1 ⎛ e− jπmα − e jπmα − jπm ⎞ ⎛ e− jπmβ − e jπmβ − jπn ⎞
e
e
⎟
⎜
⎟⎜
HV ⎝ − j 2πm / H
⎠
⎠ ⎝ − j 2πn / V
sin πmα sin πnβ
= αβ sinc mα sinc nβ
Thus, cmn =
πm
πn
=
(b) cmn =
1
m
n
n ⎞
⎛
sinc
sinc
, f mn = mf h + nf v = ⎜ m +
⎟ fh
8
2
4
100 ⎠
⎝
7.4-4
nv = 0.7 x 230, n p = 1 x nv2 = 25,921
B = 0.35 x 1 x 230/100 µs = 805 kHz
7.4-5
7-20
nv = 0.7 (1125 − N vr ) ≈ 787, n p = 5/3 x 787 2 = 1.03 x 106
Tline =
(2 / 60) sec
= 29.6 µs,
1125
B = 0.35 x
5
1125
x
= 27.7 MHz
3 (1 - 0.2)29.6 µs
7.4-6
nv = 0.7 (625 − 48) = 404, n p = 4/3 x 4042 = 2.18 x 105
Tline =
1
4
625 - 48
= 64 µs, B = 0.35 x x
= 4.99 MHz
15.625 kHz
3 (64 - 10) µs
7.4-7
(a) Since x% (t ) is proportional to x(t ) averaged over the previous τ seconds, the picture
will be smeared int he horizontal direction and five vertical lines will be lost.
(b) x% (t ) =
t
t −τ
t
t
−∞
−∞
−∞
−∞
∫ x(λ)d λ − ∫ x(λ)d λ = ∫ x(λ)d λ − ∫ x(λ − τ)d λ
1
1
1 − e − j 2 πf τ
X( f )−
X ( f )e − j 2 πf τ =
X(f )
j 2πf
j 2πf
j 2πf
Y ( f ) = H eq ( f ) X% ( f ) = KX ( f )e − jωtd
so X% ( f ) =
j 2πfKe− jωtd
K
Thus, H eq ( f ) =
=
e − jω(td −τ / 2)
− j 2 πf τ
τ sinc f τ
1− e
which can only hold for f < 1/ τ since H eq ( f ) → ∞ at f = 1/ τ, 2/τ,...
7.4-8
(a) If gain of the chrominance amp is too high, then xc will be too large and all colors
will be saturated and pastel colors will be too bright. If the gain of the chrominance
amp is too low, then xc will be too small and all colors will be unsaturated and
appear as "washed-out" pastels.
(b) If +900 error, then red → blue, blue → green, green → red.
If -900 error, then red → green, blue → red, green → blue.
If 1800 error, then red → blue-green, blue → yellow (red-green),
green → purple (red-blue).
7.4-9
7-21
Let xb' (t ) be the BPF output in Fig. 7.4-11 so, from Eq. (15),
xb' (t ) = xYH (t ) + xQ (t ) sin ωcc t + xI (t ) cos ωcc t + xˆIH (t ) sin ωcc t
where xYH (t ) is the high-frequency portion of xY (t ).
7.4-9 continued
Thus,
vI (t ) = xb' (t ) x 2 cos ωcc t = 2 xYH cos ωcc t + xQ (t ) sin 2ωcc t + xI (t )(1 + cos 2ωcc t ) + xˆIH (t ) sin 2ωcc t
= xI (t ) + 2 xYH cos ωcc t + xI (t ) cos 2ωcc t + [ xQ (t ) + xˆ IH (t )]sin 2ωcc t
vQ (t ) = xb' (t ) x 2sin ωcc t = 2 xYH sin ωcc t + xQ (t )(1 − cos 2ωcc t ) + xI sin 2ωcc t + xˆIH (1 − cos 2ωcc t )
= xQ (t ) + xˆIH (t ) + 2 xYH (t ) sin ωcc t + xI (t ) sin 2ωcc t − [ xQ (t ) + xˆ IH (t )]cos 2ωcc t
7.4-10
To modify Eq. (15) to account for asymmetric sidebands in Q channel, let
xQH (t ) be the high-frequency portion of xQ (t ). Then
xb (t ) = xY (t ) + [ xI (t ) cos ωcc t + xˆIH (t ) sin ωcc t ] + [ xQ (t ) cos(ωcc t − 900 )
1442443
sin ωcc t
+xˆQH (t ) sin(ωcc t − 90 )
144244
3
0
− cos ωcc t
Let xb' (t ) be the BPF output at the receiver, so
xb' (t ) =xYH + [ xI (t ) cos ωcc t + xˆIH (t ) sin ωcc t ] + [ xQ sin ωcc t − xˆQH cos ωcc t ]
7-22
Thus
vI (t ) = xb' x 2 cos ωcc t = xI (t ) − xˆQH (t ) + 2 xYH (t ) cos ωcc t + [ xI (t ) − xˆQH (t )]cos 2ωcc t
+[xQ (t ) + xˆIH (t )]sin 2ωcc t
vQ (t ) = xb' x 2sinωcc t = xQ (t ) + xˆIH (t ) + 2 xYH (t ) sin ωcc t + [ xI (t ) − xˆQH (t )]sin 2ωcc t
-[xQ (t ) + xˆIH (t )]cos 2ωcc t
and lowpass filtering with B = 1.5 MHz yields
vI' (t ) = xI (t ) − xˆQH (t ) + 2 xYH (t ) cos ωcc t
vQ' (t ) = xQ (t ) + xˆIH (t ) + 2 xYH (t ) sin ωcc t
Now we have cross talk between I and Q channels since both xˆQH (t ) and xˆIH (t )
have components in 0.5 MHz < f < 1.5 MHz. This quadrature color cross talk is
eliminated by reducing the bandwidth of xQ (t ) to 0.5 MHz so xˆQH (t ) = 0 and the
Q-channel LPF removes xˆ IH (t ).
7-23
Chapter 8
8.1-1
M = 12 equally likely outcomes
P(A) = 6/12, P(B) = 4/12, P(C) = 3/12
A 1
2
3
4
5
6
7
8
9
10
11
12
2,1
B
P(AB) = 2/12, P(AC) = 0, P(BC) = 1/12
P(AcB) = 2/12
B
C
8.1-2
M = 16 equally likely outcomes
A
P(A) = 4/16, P(B) = 6/16, P(C) = 6/16
1,2
P(AB) = 0, P(AC) = 2/16, P(BC) = 2/16
P(AcB) = 6/16
1,1
1,3
1,4
2,2
2,3
2,4
3,2
3,3
3,4
C
4,4
P ( A B c ) = N A B c / N = (N A - N A B )/ N = P ( A ) - P (A B )
8.1-4
P( A + B ) =
N B = N A B + N Ac B
N AB + NABc + NAc B
N
=
N A + N B − N AB
= P ( A) + P ( B) − P( AB)
N
8.1-5
N A = N AB + N ABc
P(C ) =
N ABc + N Ac B
N
N B = NAB + NAc B
=
N A + N B − 2 N AB
= P( A) + P( B ) − 2 P( AB)
N
8-1
4,1
4,2
4,3
8.1-3
N A = N AB + N ABc
3,1
8.1-6
P(match) = P( HH + TT ) = P( HH ) + P(TT ) = P( H ) P (H ) + P (T )P (T )
1+ ∈ 1−∈
P(T ) = P( H ) = 1 −
=
2
2
c
2
 1+ ∈   1− ∈  1+ ∈ 1
P(match) = 
+
=
>
 

2
2
 2   2 
2
2
8.1-7
Let A = “A fails,” B = “B fails,” C = “computer inoperable”
P(A) = 0.01, P(B) = 0.005, P(B|A) = 4 × 0.005 = 0.02
P(C) = P(AB) = P(B|A)P(A) = 0.0002, P(A|B) = P(AB)/P(B) = 0.04
8.1-8
Let M = “match,” H1 = “heads on first toss,” etc.
(a)
P(H1 ) = ½, P(MH1 ) = P(H1 H2 ) = (½)2 , P(M|H1 ) = P(MH1 )/P(H1 ) = ½
(b)
Let A = “H1 or H2 ,” P(A) = P(H1 T2 + T1 H2 + H1 H2 ) = ¾
P(MA) = ¼, P(M|A) = P(MA)/P(A) = 1/3
(c)
P(M) = P(H1 H2 + T1 T2 ) = ½, P(A|M) = P(A)P(M|A)/P(M) = ½
8.1-9
Let M = “match,” H1 = “heads on first toss,” etc.
(a)
P(H1 ) = ¼, P(MH1 ) = P(H1 H2 ) = (¼)2 , P(M|H1 ) = P(MH1 )/P(H1 ) = ¼
(b)
Let A = “H1 or H2 ,” P(A) = P(H1 T2 + T1 H2 + H1 H2 ) = 2 × ¼ × ¾ + (¼)2 =7/16
P(MA) = P(H1 H2 ) = (¼)2 , P(M|A) = P(MA)/P(A) = 1/7
(c)
P(M) = P(H1 H2 ) + P(T1 T2 ) = (¼)2 + (¾)2 = 10/16, P(A|M) = P(A)P(M|A)/P(M) = 1/10
8.1-10
Since P(AB) = P(A|B)P(B) = P(B|A)P(A), P(XYZ) = P(X)P(YZ|X) where
P(YZ|X) =
P ( XYZ ) P ( XY ) P ( XYZ )
=
= P(Y|X)P(Z|XY) so P(XYZ) = P(X)P(Y|X)P(Z|XY)
P( X )
P( X ) P( XY )
8.1-11
Let F = “fair coin,” L = “loaded coin,” A = “all tails,” P(F) = 1/3, P(L) = 2/3, P(A|F) = (½)2 , P(A|L) =
(¾)2
(a)
P(A) = P(A|F)P(F) + P(A|L)P(L) = 11/24
(b)
P(L|A) = P(L)P(A|L)/P(A) = 9/11
8-2
8.1-12
Let F = “fair coin,” L = “loaded coin,” A = “all tails;” P(F) = 1/3, P(L) = 2/3, P(A|F) = (½)2 , P(A|L) =
(¾)2
(a)
P(A) = P(A|F)P(F) + P(A|L)P(L) = 31/96
(b)
P(L|A) = P(L)P(A|L)/P(A) = 27/31
8.1-13
Let R1 = “first marble is red,” etc., M = “match;” P(R1 ) = 5/10, P(W1 ) = 3/10, P(G1 ) =2/10,
P(M|R1 ) = P(R2 |R1 ) = (5 – 1)/(10 – 1) = 4/9, P(M|W1 ) = 2/9, P(M|G1 ) = 1/9
(a)
P(M) = P(M|R1 ) × P(R1 ) + P(M|W1 ) × P(W1 ) + P(M|G1 ) × P(G1 )
=
(b)
4 5 2 3 1 2 14
× + × + × =
9 10 9 10 9 10 45
P(W1 |M) = P(W1 )P(M|W1 )/P(M) = 3/14
8.1-14
Let R1 = “first marble is red,” etc., M = “match;” P(R1 ) = 5/10, P(W1 ) = 3/10,
P(G1 ) =2/10, P(M|R1 ) = P(R3 R2 |R1 ) = P(R3 |R2 R1 )P(R2 |R1 ) = 5 − 2 × 5 − 1 = 3 × 4 , P(M|W1 ) = 1 × 2 ,
10 − 2 10 − 1
P(M|G1 ) = 0 × 1
8
(a)
9
P(M) = P(M|R1 ) × P(R1 ) + P(M|W1 ) × P(W1 ) + P(M|G1 ) × P(G1 )
= 12 × 5 + 2 × 3 + 0 × 2 = 11
72 10
(b)
72 10
10
120
P(W1 |M) = P(W1 )P(M|W1 )/P(M) = 1/11
8-3
8
9
8 9
8.2-1
P(Ni) = 1/5
N
xi
PX(x i)
FX(x i)
0
0.2
0.2
0.5
0.4
0.6
2
2
0.2
0.8
3
4.5
0.2
1.0
0
-1,1
FX
1.0
0.8
0.6
0.2
0
0.5
2.0
4.5
x
(cont.)
P(X W 0) = FX(0) = 0.2, P(2 < X W 3) = FX(3) – FX(2) = 0, P(X < 2) = FX(2 - ∈) = 0.6,
P(X D 2) = 1 – 0.6 = 0.4
8.2-2
P(Ni) = 1/5
N
xi
PX(x i)
FX(x i)
3
-4
0.2
0.2
2
-2
0.2
0.4
-1,1 2
0.4
0.8
0
0.2
1.0
4
FX
1.0
0.8
0.4
0.2
-4
-2
0
2
4
8-4
x
P(X W 0) = FX(0) = 0.4, P(2 < X W 3) = FX(3) – FX(2) = 0, P(X < 2) = FX(2 - ∈) = 0.4,
P(X D 2) = 1 – 0.4 = 0.6
8.2-3
0

FX ( x) = ∫ pX (λ) d λ =  x − λ
−x
−∞
 ∫ λe d λ = 1 − ( x + 1)e
0
x
x≤0
x>0
P(X W 1) = FX(1) = 0.264, P(X > 2) = 1 – FX(2) = 0.406, P(1 < X W 2) = FX(2) – FX(1) = 0.330
8.2-4
x 1 λ
1 x
x≤ 0
 ∫ e dλ = e
x
2
 −∞ 2
FX ( x) = ∫ pX (λ) d λ = 
x
−∞
 1 + 1 e−λ dλ = 1 − 1 e−x
x>0
2 ∫ 2
2

0
P(X W 0) = FX(0) = 1/2, P(X > 1) = 1 – FX(1) = 0.184, P(0 < X W 1) = FX(1) – FX(0) = 0.316
8.2-5
FX(∞) = 100K = 1 ⇒ K = 0.01 so pX(x) = dFX(x)/dx = 0.2x[u(x) – u(x – 10)]
P(X W 5) = FX(5) = K × 52 = 0.25, P(5 < X W 7) = FX(7) – FX(5) = 0.49 – 0.25 = 0.24
8.2-6
FX(∞) = K/ 2 = 1 ⇒ K = 2 so pX(x) = dFX(x)/dx =
2π
πx
cos [u(x) –
40
40
u(x – 10)]
P(X W 5) = FX(5) = K sin π = 0.541, P(5 < X W 7) = FX(7) – FX(5) = K sin 7π – 0.541 = 0.198
8
40
8.2-7
P(Z < 0) = 0, P(Z W 0) = P(X W 0) = ½, P(Z W z) = P(X W z) for z > 0
z<0
0

FZ (z ) =  1 1
 2 + π arctan z z ≥ 0
0
d

p Z (z) =
FZ ( z ) =  1
1
dz
 2 δ( z ) + π(1 + z 2 )

8-5
z<0
z≥0
8.2-8
P(Z < -1) = 0, P(Z W -1) = P(X W 0) = ½, P(Z W z) = P(X W z) for z > 0

0
z < −1

FZ ( z ) = 1 / 2
−1 ≤ z ≤ 0
1 1
 + arctan z
z≥0
2 π
1
 2 δ( z + 1)
d
p Z (z) =
F ( z) = 
1
dz Z

 π(1 + z 2 )
z≤0
z>0
8.2-9
p Z (z) =
1 −2( z +5)/2  z + 5  − ( z+ 5)
2e
u
u ( z + 5)
=e
2
 2 
2
1
-5
pZ
pX
-4
0
0.5
x, z
8.2-10
p Z (z) =
1
 z − 1  ( z −1)
2e −2( z −1)/( −2) u 
 = e u [− (z − 1)]
−2
 −2 
2
pZ
1
-2
-1
pX
0
0.5
x, z
8.2-11
Monotonic transformation with g-1 (z) = z2 – 1, dg-1 /dz = 2z, pX(x) = ¼ for –1 W x W 3, so
p Z (z) =
1
z
2 z [ u ( z ) − u ( z − 2)] = [u ( z ) − u ( z − 2)]
4
2
8-6
8.2-12
g1 (x) = -x[u(x + 1) – u(x)], gl-1 (z) = -z[u(z) – u(z – 1)], dg1 -1 /dz = -1
g2 (x) = x[u(x) – u(x – 3)], g2 -1 (z) = z[u(z) – u(z – 3)], dg2 -1 /dz = 1, pX(x) = ¼ for –1 W x W 3, so
1
1
1
 4 −1 + 4 1 = 2
p Z (z ) = 
1 1 = 1
 4
4
0 ≤ z ≤1
1< z ≤ 3
8.2-13
g1 (x) =
− x [u(x + 1) – u(x)], gl-1 (z) = -z2 [u(z) – u(z – 1)], dg1 -1 /dz = -2z
g2 (x) =
x [u(x) – u(x – 3)], g2 -1 (z) = z2 [u(z) – u(z – 3 )], dg2 -1 /dz = 2z, pX(x) = ¼ for –1 W x W
3, so
1
1
 4 −2 z + 4 2 z = z
p Z (z ) = 
1 2z = z
 4
2
0 ≤ z ≤1
1< z ≤ 3
8.2-14
g1 (x) = x 2 u(-x), gl-1 (z) = - z u(z), dg1 -1 /dz = -1/2 z , g2 (x) = x 2 u(x),
g2 -1 (z) = + z u(z), dg2 -1 /dz = +1/2 z
pZ(z) = 0 for z < 0, p Z ( z ) = pX ( z )
p Z (z) =
(cont.)
1
2 z
+ pX ( − z ) −
1
2 z
for z > 0, so
1 
p X ( z ) + pX ( − z )  u ( z )
2 z
8.2-15
∞
∞
pY ( y ) = ∫ pXY ( x , y) dx = ye− yu ( y ) ∫ e− yx dx = e − yu ( y ) ,
−∞
0
p XY ( x , y ) = ye− yx u( x) e − yu ( y ) ≠ pX ( x) pY ( y) , pX(x|y) = p XY ( x , y ) / pY ( y ) = ye− yx u( x)
8-7
8.2-16
∞
pY ( y ) = ∫ pXY ( x , y) dx =
−∞
p XY ( x , y ) =
1 y 1 2
1
 y
2
2
Π   ∫ ( x + 2 xy + y ) dx =
1+ 3 y )Π   ,
(
40  6  −1
60
6
3 ( x + y )2  x  1
y
Π   (1 + 3y 2 ) Π   ≠ p X ( x) pY ( y)
2
2 (1 + 3y )  2  60
6
3( x + y) 2  x 
pX(x|y) = p XY ( x , y ) / p Y ( y ) =
Π
2(1 + 3y 2 )  2 
8.2-17
∫
∞
−∞
pX(x|y) dx =
∫
∞
−∞
∞
p XY ( x , y)
1
dx =
pXY ( x , y ) dx = 1
∫
pY ( y )
pY ( y) −∞
For any given Y = y, X must be somewhere in the range -∞ < x < ∞.
8.2-18
∞
∞
−∞
−∞
pXY(x,y) = pX(x|y)pY(y), p X ( x) = ∫ pXY ( x , y) dy = ∫ pX (x|y)pY(y) dy
∞
Thus, pY(y|x) = pXY(x,y)/pX(x) = pX(x|y)pY(y)/ ∫ pX (x|y)pY(y) dy
−∞
8.3-1
∞
∞
mX = a ∫ xe − ax dx = 1/ a, X 2 = a ∫ x2 e− ax dx = 2 / a 2 , so σ X =
0
0
2
2 1
1
−  =
2
a  a
a
8.3-2
2
∫
mX =
2
π
∫
X2 =
2
π
∞
∞
x2
2  ∞ λ2
2λ
a2

2
dx
=
d
λ
+
d
λ
+
d
λ

 = 1 + a , so
∫−∞ 1 + ( x − a )4
∫
4
∫
4
∫
4
−∞ 1 + λ
−∞ 1 + λ
π  −∞ 1 + λ

σX =
(1+ a ) − a
mX = a
2 − ax
xe
0
dx = 2 / a , X = a
2
2
∫
∞
6 2
2
dx = 6 / a , so σ X =
−  =
2
a a
a
∞
2
0
3 − ax
xe
2
8.3-3
∞
x
2 ∞ λ
a

dx =
dλ + ∫
d λ  = a,
4
∫
4
4

−∞ 1 + ( x − a)
−∞ 1 + λ
π  −∞ 1 + λ

∞
∞
2
2
=1
8-8
8.3-4
P(X = b) = 1 – p, mX = ap + b(1 – p), X 2 = a2 p + b2 (1 – p),
σX2 = a2 p + b2 (1 – p) – [ap + b(1 – p)]2 = (a – b)2 p(1 – p), σX = |a – b| p (1 − p)
8.3-5
m X = ∑ i =0 ai
K −1
X = ∑ i= 0
K −1
2
1
a ( K − 1) K K − 1
=
=
a,
K K
2
2
1 a 2 ( K −1) K [2( K − 1) + 1] ( K −1)(2 K − 1) 2
( ai)
=
=
a
K K
6
6
2
σ X = X 2 − mX =
2
2
( K − 1) a 2 2(2 K − 1) − 3( K − 1) K 2 − 1 2
=
a , σX =
2
6
12
K 2 −1 a
3 2
8.3-6
∞
mY = ∫ a cos x pX ( x) dx =
−∞
a θ +2 π
cos xdx = 0 ,
2π ∫θ
∞
Y 2 = ∫ a2 cos 2 x pX ( x) dx =
−∞
a 2 θ +2 π 2
a2
cos
xdx
=
2π ∫θ
2
σY =
a2
2
−0 =
a
2
8.3-7
∞
mY = ∫ a cos x pX ( x) dx =
−∞
∞
a θ+π
2a
cos xdx = − sin θ ,
∫
π θ
π
Y 2 = ∫ a2 cos 2 x pX ( x) dx =
−∞
a2
π
∫
θ+π
θ
cos 2 xdx =
a2
2
2
σY =
a 2  2a
1 4

−  − sin θ  = a
− 2 sin 2 θ
2  π
2 π

8.3-8
mY = αmX + β, Y 2 = E ( αX + β) 2  = E α 2 X 2 + 2αβX + β2  = α 2 X 2 + 2αβ mX + β2
(
σY = Y 2 − mY = α 2 X 2 − mX
2
2
2
)=ασ
2
2
X
, σY = α σ X
8-9
8.3-9
Y 2 = E ( X + β) 2  = E  X 2 + 2β X + β2  = X 2 + 2βmX + β 2
d 2
Y = 2 mX + 2β = 0 ⇒ β = − mX
dβ
8.3-10
∞
P( X ≥ a ) = ∫ p X ( x ) dx and
a
pX ( x ) = 0 for x < 0
∞
∞
∞
0
a
a
E [ X ] = ∫ p X ( x ) dx ≥ ∫ xp X ( x ) dx ≥ a ∫ p X ( x) dx = aP ( X ≥ a) , so P( X ≥ a) ≤ m X / a
8.3-11
2
E ( X ± Y )  = E  X 2 ± 2 XY + Y 2  = X 2 ± 2 XY + Y 2 ≥ 0


(
so 2 XY ≥ − X + Y
2
2
) and 2 XY ≤ ( X
2
+Y
2
)
X 2 +Y2
X2 +Y2
, ⇒ −
≤ XY ≤
2
2
8.3-12
CXY = E [ XY − mX Y − mY X + mX mY ] = XY − mX mY
(a) XY = XY = m X mY
⇒ C XY = 0
(
)
(b) XY = E [ X (αX + β) ] = α X 2 + βmX and mY = αmX + β so C XY = α X 2 − mX =ασ X
8.3-13
2
∈2 = E Y 2 − 2 (α X + β) Y + ( αX + β)  = Y 2 − 2α XY − 2 βY + α 2 X 2 + 2αβ X + β2


∂ ∈2 ∂ α = −2 XY + 2α X 2 + 2 βY = 0 and ∂ ∈2 ∂ β = −2Y − 2α X + 2β = 0 so
(
)
α = XY − X Y / σX 2 and β = Y − α X
8.3-14
dn
dn
n
Φ
(
ν
)
=
E e jνX  = E  ( jX ) e jνX  = j n E  X ne jνX  , so
X
n
n


dν
dν
n
dn
n
n
n
−n d




Φ X (0) = j E  X  ⇒ E  X  = j
Φ X (ν)
ν=0
d νn
d νn
8-10
2
8.3-15
a
F  ae− atu (t )  =
=
a + j 2 πf
Φ X (2πt ) =
⇒ F
−1
a
 aae− af u ( f )  =
aso
a + j 2π( −t )
−1
a
a − j 2π t
and Φ X ( ν ) =
−2
dΦ X
ν  j

= − 1 − j   −  ⇒
dν
a  a

−3
d 2Φ X
ν  j

= 2 1 − j   − 
2
dν
a  a

−4
a
ν

= 1− j 
a − jν 
a
 j 1
X = j −1   =
a  a
2
2
2
 j
X 2 = j −2 2   = 2
a
a
⇒
d 3Φ X
ν   j

= −6  1 − j   − 
3
dν
a  a

3
3
6
 j
X 3 = j −3 ( −6 )   = 3
a
a
⇒
8.3-16
∞
∞
Φ Y (ν ) = E  e jνx  = ∫0 e jνx 2axe − ax dx = ∫0 e jνλ ae −aλ d λ ⇒
2
2
2
pY ( y ) = ae − ay u ( y )
8.3-17
Φ Y (ν ) = E  e jνsin x  = ∫
π/ 2
−π /2
e jν sin x
1
dx
π
Let λ = sin x, dλ = cos x dx where cos x = 1 − sin 2 x = 1 − λ2 , so
Φ Y (ν ) = ∫
sin π / 2
sin ( −π / 2 )
e jνλ
1
1 dλ
1
jνλ
=
e
dλ ⇒
∫
−1
π 1− λ2
π 1− λ2
pY ( y ) =
 y
Π
 
π 1− y2  2 
1
8.4-1
Binomial distribution with α = (1 - α)= ½, so m = 10 × ½ = 5, σ2 = 5 × ½ = 2.5, m ± 2σ ≈ 2 to 8
10  10 
P(i < 3) = FI (2) =   +   +
 0   1 
10
10   1 
1 + 10 + 45
 2   2  = 1024 = 0.0547
   
8.4-2
Binomial distribution with α = 3/5 and (1 - α)= 2/5, so m = 10 × 3/5 = 6, σ2 = 5 × 2/5 = 2.4,
m ± 2σ ≈ 3 to 9
10 
 28 (1× 4 + 10 × 6 + 45 × 9)256
10 
 10 
P(i < 3) = FI (2) =   3022 +   3121 +   3220  10 =
= 0.0122
9.87 × 106
1
2
 0 
5
8-11
8.4-3
Let I = number of forward steps, binomial distribution with mI = 100 × ¾ = 75, σI2 =75 × ¼,
I 2 = 75 4 + 752 , X = Il – (100 – I)l = (2I – 100)l so mX = (2mI - 100)l = 50l and
X 2 = (2 I − 100) 2l 2 = (4 I 2 − 400m I + 10 4 )l 2 = 2575l 2 , σ X = 2575l 2 − (50l ) 2 = 75l
8.4-4
Binomial distribution with 1 - α = 0.99 so
 10 
0
10
P( I > 1) = 1 − PI (0) − PI (1) = 1 −   0.010.99
−
0
 
 10 
1
9
 1  0.010.99 = 0.0042
 
Poisson approximation with m = 10 × 0.01 = 0.1
P( I > 1) ≈ 1 − e −0.1
(0.1)0
(0.1) 1
− e −0.1
= 0.0047
0!
1!
8.4-5
µ = 0.5 particles/sec, T = 2 sec, µT = 1, so
11
(a) PI (1) = e
= 0.368
1!
−1
1
10
−1 1
(b) P( I > 1) = 1 − PI (0) − PI (1) = 1 − e
−e
= 0.264
0!
1!
−1
8.4-6
E [ I ] = ∑ i=0 ie − m
∞
em = 1+ m +
mi
∞
mi
= e − m ∑ i =0 i ,
i!
i!
E  I 2  =e− m ∑ i= 0 i 2
∞
mi
, where
i!
i
i
1 2
∞ m
d m d
∞ m
mi−1 1 ∞ mi
m + L = ∑ i= 0
so
e =
=
i
∑
∑ i ! = m ∑ i= 0 i i! and
2!
i!
dm
dm i =0 i !
d2 m d
∞
mi −1
mi− 2
1
∞
mi
2
e
=
i
=
i
(
i
−
1)
=
(
i
−
i
)
∑
∑
∑
dm2
dm i= 0 i !
i!
m2 i= 0
i!
(cont.)
But
d m
d2 m
e =
e = em so
2
dm
dm
∑ i =0 i
∞
mi
= mem and
i!
∑ i =0 (i2 − i)
∞
mi
∞
mi
= ∑ i= 0 i 2
− mem = m2 e m
i!
i!
Thus, E[I] = e-m(mem) = m and E[I2 ] = e-m(m2 em + mem) = m2 + m
8-12
8.4-7
X = m = 100, σ2 = X 2 − X
2
⇒ X 2 = σ 2 + m 2 = 10,004
P( X < m − σ or X > m + σ) = P ( X ≤ m − σ ) + P ( X > m + σ ) = 2 Q (1) ≈ 0.32
8.4-8
2
m = X = 2, σ = X 2 − X = 3
P( X > 5) = P( X > m + σ) = Q (1) ≈ 0.16 ,
P(2 < X ≤ 5) = P ( X > m) − P( X > m + σ) =
1
− Q (1) ≈ 0.34
2
8.4-9
m = 10, σ = 500 −100 = 20,
P( X > 20) = P( X > m + σ / 2 ) = Q (0.5) ≈ 0.31
P(10 < X ≤ 20) = P( X > m) − P( X > m + σ /2) = 1 / 2 − Q (0.5) ≈ 0.19
P(0 < X ≤ 20) = P( X − m < σ / 2 ) =1 − 2 Q(0.5) ≈ 0.38
P( X > 0) = 1 − P ( X ≤ m − σ /2) = 1 − Q (0.5) ≈ 0.69
8.4-10
m = 100 × ½ = 50, σ2 = 50 × ½ = 25, σ = 5
(a) P( X > 70) = P( X > m + 4σ ) = Q(4) ≈ 3.5 × 10−5
(b) P(40 < X ≤ 60) = P( X − m ≤ 2σ ) =1 − 2 Q(2) = 0.95
8.4-11
Let a = m – k 1 σ and b = m + k 2 σ so
 m−a
 m−b 
P( a < X ≤ b) = 1 − Q ( k1) − Q ( k2 ) = 1 − Q 
 − Q

 σ 
 σ 
Q(k 1 )
Q(k 2 )
a
m
b
8-13
8.4-12
c
c
m = 0, σ = 3, P( X ≤ c) = P  X − m ≤ σ  = 1 − 2Q  
σ 

σ
(a) 1 – 2Q(c/3) = 0.9 ⇒ Q(c/3) = 0.05, c ≈ 3 × 1.65 = 4.95
(b) 1 – 2Q(c/3) = 0.99 ⇒ Q(c/3) = 0.005, c ≈ 3 × 2.57 = 7.71
8.4-13
1
2π
Q( k ) =
∫
∞
e−λ
k
1
2π
∫
∞
k
dλ =
/2
1
=
2π k 2
1
so Q ( k ) <
2
2 πk
2
e− k
2
e−k
/2
2
/2
∫
−
1
1
2π k
e−k
2
k
(
)
 1
−λ 2 / 2
=
− λ d e


∫
2π
∞
k
1  1 −λ 2 / 2 ∞ ∞ −λ2 / 2  1  
− e
d  − 
− e
k ∫k
2π  λ
 λ 
1 − λ2 / 2
e
dλ
2
λ
and
1 −λ 2 / 2
1 1
e
dλ <
2
λ
2π k 2
Thus, Q ( k ) ≈
∞
1
2π
2
/2
∫
∞
k
e−λ
2
/2
dλ =
1
Q( k ) = Q( k ) i f k ? 1
k2
for k ? 1
8.4-14
E  ( X − m)  =
n
1
2πσ 2
∫
∞
−∞
(2σ )
dx =
2 n/2
(x − m) e
n
−( x − m ) 2 / 2 σ 2
π
∫
∞
−∞
λ ne −λ d λ
2
(a)
E  ( X − m) n  = 0 for odd n since λn e−λ has odd symmetry
(b)
E  ( X − m)  = K n 2 ∫ λ e
0
2
∞
n
(2σ )
=
2 n/2
n −λ 2
d λ for even n, where Kn
π
=
2 n / 2 σn
π
( ) so
But e −λ d λ = − 12λ d e−λ
2
2
∞
∞
2

2 ∞
2

∞
2
E  ( X − m) n  = −K n ∫0 λn −1 d e−λ = − Kn λ n−1e −λ
− ∫0 e−λ d ( λn−1 ) = K n (n − 1) ∫ λn −2e −λ d λ
0
0


( )
= (n − 1)
∞
2
Kn
K n −2 2∫ λ n− 2e −λ d λ = ( n − 1) σ2 E ( X − m) n−2 
0
2K n − 2
(cont.)
8-14
Thus, E  ( X − m ) 4  = (4 − 1)σ2 E  (X − m ) 2  = 3σ4 , E  ( X − m ) 6  = (6 − 1)σ2 ( 3σ4 ) = 3⋅ 5σ 6 , and
E  ( X − m ) n  = 1⋅ 3 ⋅ 5L (n − 1)σn , n = 2,4,6, K
8.4-15
pX ( f ) =
1 − π[( f − m)/ b]2
e
b
where b = 2πσ 2
If m = 0, Φ X (2πt ) = e −π( bt) = e−σ
2
2
(2 πt )2/ 2
, so Φ X ( ν ) = e− σ ν / 2
2 2
For m ≠ 0, use frequency-translation theorem with ωc = 2πm, so
Φ X (2πt ) = e −π( bt) e jωct = e −σ
2
2
(2 πt) 2 / 2 jm(2 πt )
and Φ X (ν ) = e − σ ν / 2e jmν
2 2
e
8.4-16
Φ Z ( ν ) = Φ X (ν ) ΦY ( ν ) = e−σX
2 2
ν /2
e jmX ν e−σY
2 2
ν / 2 jmY ν
e
2 2
= e−σZ ν / 2 e jmZ ν
where σ Z = σX + σY , mZ = mX + mY . Hence, p Z ( z ) =
2
If Z =
2
2
2
2 2
Then Φ Z ( ν ) = Φ Y1 ( ν) Φ Y2 ( ν )L Φ Yn ( ν) = e− σ Z ν / 2e jmZ ν where
σZ = ∑ σYi =
2
1
2
n
∑
n
i =1
σX i , mZ = ∑ mYi =
2
1 n
∑ mX
n i=1 i
8.4-17
X = ln Y ⇒ Y = eX
E [Y ] = E  e X  = E  e jνX 
jν = 1
= Φ X ( − j ) = eσX
2
/ 2 mX
e
E Y 2  = E  e 2 X  = E  e jνX 
= Φ X ( − j 2) = e 2σX e2 mX
jν = 2
2
8-15
2
−[ z − ( mX + mY ) ] /2( σ X 2 +σ Y 2 )
2
2π(σ X + σY )
1 n
X
X i = Y1 + Y2 + L + Yn where Yi = i is gaussian with
∑
i =1
n
n
σ Xi
X i 2 X i2
X i2 − X i
2
Yi =
, Yi = 2 , σYi =
=
n
n
n2
n2
2
1
2
2
e
8.4-18
∞
(a) Φ Z ( ν ) = ∫ e
jνx 2
−∞
(b)
pX ( x ) dx =
−3 / 2
dΦZ
= jσ 2 (1 − j 2σ2ν )
dν
2
2πσ 2
∫
∞
0
 1

−  2 − jν x2
 2σ

e
dx = (1 − j 2σ2 ν )
−1 / 2
⇒ E [ Z ] = j −1( j σ2 ) = σ2
−5 / 2
d 2Φ Z
4
2
=
−
3
σ
1
−
j
2
σ
ν
(
)
d ν2
⇒ E  Z 2  = j −2 ( −3σ4 ) = 3σ4
−7 / 2
d 3Φ Z
6
2
=
−
j
15
σ
1
−
j
2
σ
ν
(
)
d ν3
⇒ E  Z 3  = j −3 (− j15σ6 ) =15 σ6
Thus, E  X 2  = E [ Z ] = σ2 , E  X 4  = E  Z 2  = 3σ 4 , E  X 6  = E  Z 3  = 15σ6
8.4-19
R 2 = 2σ 2 = 32 ⇒ σ2 = 16 so p R ( r ) =
Thus, P( R > 6) = 1 − P (R ≤ 6) = e−6
2
/32
(
)
r − r2 /32
− r 2 /32
e
u ( r ) and P( R ≤ r ) = FR ( r ) = 1 − e
u (r )
16
= 0.325 and
P(4.5 < R < 5.5) = P( R ≤ 5.5) − P (R ≤ 4.5) = e −4.5 / 32 − e−5.5 / 3 2 = 0.143
2
2
8.4-20
X 2 = 2σ 2 = 18 ⇒ σ2 = 9 so p X ( x) =
Thus, P( X < 3) = P ( X ≤ 3) = 1 − e −3
2
/18
(
)
x − x2 /18
− x 2 /18
e
u ( x ) and P( X ≤ x ) = 1− e
u( x)
9
= 0.393 , P( X > 4) = 1 − P( X ≤ 4) = e−4
P(3 < X ≤ 4) = P (X > 3) − P( X > 4) = [ 1− P( X ≤ 3) ] − [1− P ( X ≤ 4) ] = e −3
2
/18
2
/18
= 0.411 , and
− e−4
2
/18
= 0.195
8.4-21
Since Z D 0 and X D 0, monotonic transformation with g(x) = x 2 , g-1 (z) = + z , dg −1 / dz =
p X ( x) =
p Z (z) =
x − x 2 / 2σ 2
e
u ( x ), m = Z = E  X 2  = 2σ2 . Thus
2
σ
z
σ
2
− z
e( )
2
P( Z ≤ km) = ∫
km
0
/ 2σ 2
(
u + z
) 2 1 z = m1 e
−z / m
u( z)
 0.632 k = 1
1 − z/m
e
dz = 1 − e − k = 
m
 0.095 k = 0.1
8-16
1
2 z
8.4-22
(a) A = R12 = X 2 +Y 2 where X and Y are gaussian with X = Y = 0, σ X 2 = σY 2 = σ2
2
Φ A ( ν ) = Φ X 2 ( ν )Φ Y 2 ( ν ) =  (1− j 2σ2 ν) −1 / 2  = (1 − j 2σ2 ν )−1
Φ A (2πt ) =
2
1
b
1
1
=
, b = 2 so p A ( a ) = be− bau (a ) = 2 e − a / 2σ u( a )
2
1 + j 2σ (−2πt ) b + j 2π( −t )
2σ
2σ
(b) pW ( w) = p R 2 (w ) * pR 2 ( w), pR 2 = pR 2 = p A
1
2
1
2
w<0
0

2
=  w 1 −λ / 2σ2 1 − ( w−λ) / 2σ2
 1  − w / 2 σ2 w
e
dλ =  2  e
 ∫0 2σ2 e
∫0 d λ w > 0
2σ2
 2σ 

so pW ( w) =
w − w / 2 σ2
e
u ( w)
4
4σ
8.4-23
Let a2 =
1
a
− a 2 ( x2 − 2 ρxy +y 2 )
so
p
(
x
,
y
)
=
e
XY
2σ2 (1 − ρ2 )
2 π2 σ2
∞
pY ( y ) = ∫ pXY ( x , y) dx =
−∞
pX(x|y) =
a
2 π2 σ 2
e
−a 2 y 2
∫
∞
−∞
e
− a 2 ( x 2 −2 ρxy)
e− a
dx =
2
y2
2π 2 σ 2
a 2 ρ2 y2
e
∞
2∫ e
0
2
2
2
pXY ( x , y)
1
=
e − ( x −ρy ) / 2σ (1−ρ )
pY ( y )
2 πσ2 (1 − ρ2 )
8.4-24
Since Z is a linear combination of gaussian RVs, pZ(z) is a gaussian PDF with
mZ = E [ X + 3Y ] = mX + 3mY = 0
σZ = E  X 2 + 6 XY + 9Y 2  = (σ X + m X ) + 6 E [ XY ] + 9(σY + mY ) = 100
2
2
2
2
2
8.4-25
∞
E  X n  = ∫ x n p X ( x ) dx = 0 for n odd, E [Y ] = E  X 2  = σX 2
−∞
E [ ( X − mX )(Y − mY )] = E  X ( X 2 − σ X 2 )  = E  X 3  − σX 2 E [ X ] = 0 ⇒ ρ = 0
8-17
e− y
2
− λ2
dλ =
/ 2σ 2
2 πσ2
Chapter 9
9.1-1
E  e Xt  =
1 2 xt
1
3
e dx = ( e 2t − 1) , v( t ) = E  6e Xt  = ( e2t − 1)
∫
2 0
2t
t
Rv (t1 ,t 2 ) = E  36e X (t1 +t2 )  =
18
9
e 2( t1 + t2 ) − 1 , v 2 (t ) = ( e 4t −1)


t1 + t 2
t
9.1-2
E [ cos Xt ] =
1 2
1
cos xtdx = s i n 2t ,
∫
2 0
2t
3
v(t ) = E [6 cos Xt ] = s i n 2t
t
Rv (t1 ,t 2 ) = E [ 36cos Xt1 cos Xt2 ] = 18 E [ cos X ( t1 − t2 ) + cos X (t1 + t 2 ) ]
 sin2( t1 − t2 ) sin2(t1 + t2 ) 
= 18 
+

2(t1 + t 2 ) 
 2(t1 − t 2 )
 sin4t 
2
v (t ) = 18  1 +

4t 

9.1-3
X = 0,
X 2 = 1/3, v (t ) = E [Y + 3 Xt ] t = Yt + 3 Xt 2 = 2t
2
Rv (t1 ,t 2 ) = E Y 2 + 3YX ( t1 + t 2 ) + 9 X 2t1t 2  t1t 2 =  Y 2 + 3 X Y (t1 + t 2 ) + 9 X 2t1t 2  t1t 2 = 6t1t2 + 3 ( t1t 2 )


2
2
4
v (t ) = 6t + 3t
9.1-4
E  e Xt  =
1 1 xt
1
1
e dx = ( et − e− t ) , v (t ) = E Ye Xt  = Y E e Xt  = ( et − e− t )
∫
−
1
2
2t
t
Rv (t1 ,t 2 ) = E Y 2e X (t1 +t2 )  = Y 2 E e X ( t1 +t2 )  =
3
3
 e (t1 + t2 ) − e − (t1 +t2 )  , v 2 (t ) = ( e2t − e −2t )


t1 + t2
2t
9.1-5
E [ cos Xt ] =
1 1
sin t
sin t
cos xtdx =
, v( t ) = E [Y cos Xt ] = YE [ cos Xt ] = 2
∫
−
1
2
t
t
(cont.)
9-1
1
Rv (t1 ,t 2 ) = E Y 2 cos Xt1 cos Xt 2  = Y 2 E [ cos X (t1 − t 2 ) + cos X (t1 + t 2 ) ]
2
 sin(t1 − t 2 ) sin(t1 + t 2 ) 
= 3
+

t1 + t2 
 t1 − t 2
 sin2t 
2
v (t ) = 3  1 +

2t 

9.1-6
pFΦ (f,ϕ) =
1
pF ( f ), 0 ≤ ϕ ≤ 2 π,
2π
∞
Rv (t1, t 2 ) = A 2 ∫ g ( f ) pF ( f ) df where
−∞
1 2π
cos(2πft1 + ϕ)cos(2πft 2 + ϕ) d ϕ
2 π ∫0
1 2π
1 2π
=
cos2πf (t1 −t 2 ) d ϕ +
cos[2 πf ( t1 + t 2 ) + 2ϕ] d ϕ = 12 cos2πf (t1 − t 2 )
∫
0
4π
4π ∫0
g( f ) =
Thus, with f = λ, Rv (t1 ,t 2 ) =
A2
2
∫
∞
−∞
cos2 πλ( t1 − t2 ) p F ( λ) d λ
∞  1
2π
A2

v (t ) = A∫  ∫ cos(2πft + ϕ) d ϕ pF ( f ) df = 0, v 2 ( t ) = Rv ( t ,t ) =
−∞ 2π 0
2


∫
∞
−∞
pF ( f ) df =
A2
2
9.1-7
v(t 1 )w(t 2 ) = XY(cos ω0 t 1 cos ω0 t 2 – sin ω0 t 1 sin ω0 t 2 ) – X2 cos ω0 t 1 sin ω0 t 2
+ Y2 sin ω0 t 1 cos ω0 t 2
E [ XY ] = X Y = 0, E  X 2  = E Y 2  = σ2 , so
Rvw (t1, t 2 ) = E [v (t1 )w(t2 ) ] = σ2 (sin ω0t1 cos ω0t 2 − cos ω0t1 sin ω0t 2 ) = σ2 sin ω0 ( t1 − t 2 )
9.1-8
(a) v (t ) = E [ X cos ω0t + Y sin ω0t ] = X cos ω0t + Y sin ω0 t = 0
Rv (t1 ,t 2 ) = E  X 2 cos ω0t1 cos ω0t 2 + XY (cos ω0t 1 sin ω0t2 + sin ω0t1 cos ω0t2 ) + Y 2 sin ω0t 1 sin ω0t 2 
= X 2 cos ω0t1 cos ω0t 2 + Y2 sin ω0t 1 sin ω0t 2 = σ2 cos ω0 ( t1 − t2 )
so Rv (τ) = σ2 cos ω0τ
(b) v 2 ( t ) = Rv (0) = σ2
(cont.)
9-2
<v i2(t)> = <(Xi cos ω0t + Yi sin ω0t)2> = Xi2<cos2 ω0t> + 2XiYi <cos ω0t sin ω0t>
+ Yi2<sin2 ω0t>
1 2 1 2
X i + Yi ≠ σ 2
2
2
=
9.1-9
∞
2π
−∞
0
(a ) v (t ) = ∫ ap A ( a) da ∫ cos(ω0 t + ϕ)
∞
2π
−∞
0
dϕ
= A× 0 = 0
2π
Rv (t1 ,t 2 ) = ∫ a 2 p A ( a ) da ∫ cos(ω0t1 + ϕ)cos( ω0t2 + ϕ)
so Rv(τ) =
dϕ
1
= A2 × cos ω0 (t1 − t 2 )
2π
2
A2
cos ω0 τ
2
(b ) v 2 ( t ) = A2 / 2, < vi 2 (t ) >=< Ai 2 cos 2 ( ω0t + Φ i ) >= Ai 2 < cos 2 (ω0t + Φ i ) >= Ai2 / 2 ≠ A2 / 2
9.1-10
mZ = z( t ) = E [v ( t ) − v (t + T ) ] = v( t )− v (t+ T ) = 0
σZ = z 2 (t ) = v 2 ( t ) − 2v( t )v (t + T ) + v 2 (t + T ) = Rv (0) − 2Rv (T ) + Rv (0) = 2 [ Rv (0) − Rv (T ) ]
2
9.1-11
mZ = z (t ) = E [v ( t ) + v( t −T ) ] = v (t ) + v (t − T ) = 2 Rv ( ±∞)
z 2 (t ) = v 2 ( t ) + 2v( t )v (t − T ) + v 2 (t − T ) = Rv (0) + 2Rv (T ) + Rv (0) ,
σZ = 2 [ Rv (0) + Rv (T ) − 2Rv ( ±∞) ]
2
9.2-1
F e −(

πbt ) 2
<v(t)> =
 = 1 e −(
 b
π f / b )2
2
so Gv (f) = 2 πe − ( πf /8) + 9δ ( f )
Rv ( ±∞) = ±3, < v 2 ( t ) >= Rv (0) = 25, vrms = 25 − 9 = 4
9.2-2
Gv (f) =
32  f  4
Λ   + [δ ( f − 8) + δ ( f + 8) ]
8 8 2
< v (t ) >= mV = 0, < v 2 (t ) >= Rv (0) = 36, vrms = 36 − 0 = 6
9-3
9.2-3
A2
Rv ( τ) =
2
∫
∞
−∞
cos2 πλτ pF (λ ) d λ ,
Gv ( f ) = F τ [ Rv ( τ) ] =
A2
2
∫
∞
−∞
F τ [ cos2πλτ] pF ( λ) d λ =
=
p F ( f ) = δ( f − f 0 ) ⇒ Gv ( f ) =
A2
2
∫
∞
−∞
1
[ δ( f − λ) + δ( f + λ )] pF (λ) d λ
2
A2
[ pF ( f ) + pF ( − f )]
4
A2
[ δ( f − f0 ) + δ( f + f0 ) ] since δ( − f − f0 ) = δ( f + f0 )
4
9.2-4
T 
∞
τ
τ
(a ) E G% v ( f )  = ∫− T 1 −  Rv (τ ) e− jωτ d τ = ∫−∞ Λ   Rv ( τ) e− jωτ dτ
T 
 T
 τ

= F τ  Λ   Rv ( τ)  = (T sinc 2 fT ) *G v ( f )
 T 

τ
(b ) lim Λ 
T →∞
T

 = 1 so

2
lim T sinc fT = F τ [1] = δ( f ) and
T →∞
lim E G% v ( f )  = δ( f ) * Gv ( f ) = Gv ( f )
T →∞
9.2-5
t
vT (t ) = A cos(ω0 t + Φ) Π 
T
AT

sinc(f − f 0 )T e jΦ + sinc(f + f 0 )T e− jΦ 
 , VT ( f , s ) =
2

{
2
A2T 2
E  VT ( f , s)  =
sinc 2 ( f − f 0 )T + sinc 2 ( f − f 0 )T + E e j 2Φ + e− j 2Φ  sinc( f − f 0 )T sinc( f + f 0 )T


4
j 2Φ
− j 2Φ
 = E [ 2cos2Φ ] = 0 and lim T sinc 2 fT = δ( f ) , so
But E  e + e
T →∞
Gv ( f ) = lim
T →∞
A2
A2
T sinc 2 ( f − f 0 )T + T sinc2 ( f + f 0 )T  =
[ δ( f − f 0 ) + δ( f + f 0 ) ]
4
4
9.2-6
Rvw (t1, t 2 ) = E [v (t1 )w(t2 ) ] = mV mW
so Rvw ( τ) = Rwv ( τ) = mV mW
Rz(τ) = Rv (τ) + Rw(τ) ± 2mVmW, Gz(f) = Gv (f) + Gw(f) ± 2mVmWδ(f)
(cont.)
9-4
}
Rz(±∞) = Rv (±∞) + Rw(±∞) ± 2mVmW = mV2 + mW2 ± 2mVmW = (mV ± mW)2
z 2 = Rz (0) = Rv (0) + Rw (0) ± 2mV mW = v 2 + w2 ± 2mV mW
= σV + mV + σW + mW ± 2mV mW = σV + σW + ( mV ± mW ) 2 > 0
2
2
2
2
2
2
9.2-7
Rwv (t1, t2 ) = E [w( t1 )v (t 2 )] = E [v (t 2 ) w( t1) ] = Rvw (t 2 − t1 ) so Rwv ( τ) = Rvw (−τ)
Gwv ( f ) = F τ [ Rvw (−τ) ] =
1
G ( − f ) = Gvw ( − f )
−1 vw
9.2-8
Rz (t1 ,t 2 ) = E [v (t1 )v (t2 ) + v (t1 + T )v (t 2 + T ) − v(t1 + T )v (t 2 ) − v( t1)v (t 2 + T )]
= Rv (t1 − t 2) + Rv (t1 + T − t2 − T ) − Rv (t1 + T − t 2) − Rv (t1 − t2 − T ) so
Rz ( τ) = 2Rv (τ) − Rv (τ + T ) − Rv (τ − T ) and
G z ( f ) = 2Gv ( f ) − Gv ( f ) ( e jωT + e − jω T ) = 2Gv ( f ) (1 − cos2πfT )
9.2-9
Rz (t1 ,t 2 ) = E [v (t1 )v (t2 ) + v (t1 − T )v (t 2 − T ) + v(t1) v (t 2 − T ) + v( t1 − T )v (t 2 )]
= Rv (t1 − t 2) + Rv (t1 − T − t2 + T ) + Rv (t1 − t2 + T) + Rv (t1 −T − t 2 ) so
Rz ( τ) = 2Rv (τ) + Rv (τ + T ) + Rv (τ − T ) and
G z ( f ) = 2Gv ( f ) + Gv ( f ) ( e jωT + e − jω T ) = 2Gv ( f ) (1 + cos2πfT )
9.2-10
z(t) = v(t) cos (2πf 2t + Φ 2) with v(t) = A cos (2πf 1t + Φ 1) so
Gv (f) = (A2/2)[δ(f – f 1) + δ(f + f 1)]
2
Thus, Gz ( f ) = A [ δ( f − f1 − f2 ) + δ( f + f1 − f2 ) + δ( f − f1 + f2 ) + δ( f + f1 + f2 ) ]
16
2
For f 1 = f 2, Gz ( f ) = A [ 2δ ( f ) + δ ( f − 2 f2 ) + δ ( f + 2 f2 )]
16
9-5
9.2-11
∞
Ry (t1 ,t 2 ) = E [ y( t1 ) y(t 2) ] , y (t 2 ) = ∫ h( λ) x (t 2 − λ) d λ so
−∞
∞
Ry (t1 ,t 2 ) = ∫ h (λ) E [ y (t1) x( t2 − λ) ] d λ
(cont.)
−∞
But E [y (t1 ) x (t 2 − λ) ] = Ryx (t1, t2 − λ) = Ryx ( t1 − t 2 + λ ) = Ryx ( τ + λ) so
∞
∞
−∞
−∞
Ry ( τ) = ∫ h( λ) Ryx ( τ + λ) d λ = ∫ h( −µ )Ryx (τ − µ) d µ = h (−τ) * Ryx ( τ)
9.2-12
−1
−1
Ry ( τ) = F τ (2πf ) 2 G x ( f )  = −F τ ( j 2πf ) 2G x ( f )  = − d 2 Rx ( τ) / d τ2
Gyx ( f ) = F τ [ h(τ )* Rx ( τ) ] = H ( f )Gx ( f ) where H ( f ) = j 2πf ,
so R yx (τ) = F τ
−1
[ ( j 2 πf )Gx ( f ) ] = dRx ( τ) / d τ
9.2-13
If x(t) is deterministic, then Y(f) = X ( f ) − α X ( f ) e− jωT
2
H ( f ) = 1 + α 2 − α (e jω T + e − jωT ) = 1 + α 2 − 2α cos ωT
⇒ H ( f ) = 1− α e− jωT
so
Gy ( f ) = (1 + α 2 − 2α cos ωT )Gx ( f ), Ry ( τ) = (1 + α 2 ) Rx ( τ) − α [ Rx ( τ + T ) + Rx (τ − T )]
9.2-14
∞
1
Rx ( λ) d λ
−∞ π( τ − λ)
Rxx
ˆ ( τ) = hQ ( τ) * Rx (τ) = ∫
∞ 1
∞ 1
1
Rx ( λ) d λ = −∫
Rx (µ − τ) d µ = −∫
Rx ( τ − µ ) d µ
−∞ π( − τ − λ)
−∞ πµ
−∞ πµ
Rxxˆ (τ) = Rxxˆ ( −τ) = ∫
∞
= −  hQ ( τ) * Rx (τ)  = − Rˆ x ( τ)
9-6
9.2-15
1
Thus, h (t ) = 
0
t −T / 2 < 0< t + T / 2
otherwise
1/ T
1 t +T / 2
δ
(
λ
)
d
λ
=

T ∫t −T / 2
0
Let x(t) = δ(t) so y(t) = h(t) =
t <T / 2
⇒ H(f) = sinc fT, and
t >T / 2
1 τ
−1
Ry ( τ) = F τ sinc 2 f T Gx ( f )  = Λ 
T T
1

 * Rx (τ) = T


λ
1
−

 Rx ( τ − λ) d λ
−T
 T 
∫
T
9.3-1
1 2
x +L ≈
2
Let x = h f / kT , e x −1 = x +
(e x − 1)
−1
Gv ( f ) ≈
1 1 
≈  1 + x
x 2 
2 Rh f
h f / kT
−1
=
 1 
x 1 + x  for
 2 
x = 1 . Then
1 1
1 2
 1 1 
 1 − x + x + L  ≈  1 − x  , so
x 2
2
 x 2 
 1h f 
1 −
 = 2 RkT
 2 kT 
 h f 
 1
 2 kT 
9.3-2
Ryx ( τ) = h (τ) *
N0
N
δ( τ) = 0 h( τ)
2
2
Ry ( τ) = h( −τ) *
N0
N
h ( τ) = 0
2
2
∫
∞
−∞
h( −λ) h( τ − λ ) d λ =
N0
2
∫
∞
−∞
h( t )h (t + τ) dt
9.3-3
N0 T 2
N T  τ
NT
Gy ( f ) =
sinc 2 fT , Ry ( τ) = 0 Λ   , y 2 = Ry (0) = 0
2
2
2
T 
9.3-4
Gy ( f ) =
Ry ( τ) =
N0 K 2 −2 (af ) 2 N0 K 2 −π( f /
e
=
e
2
2
N0 K 2
a
π −(π τ/
e
8
2 a )2
π / 2 a2 ) 2
, y 2 = Ry (0) =
,
N0 K 2
a
9-7
π
8
9.3-5
N0 K 2
Gy ( f ) =
2
  f − f0 
 f + f0
Π  B  + Π  B


 
Ry ( τ) = N 0 K 2 B sinc 2 Bτ cos2πf 0τ,

 ,

y 2 = R y (0) = N0 K 2 B
9.3-6
N0 K 2
2
Gy ( f ) =

 f 
N0 K 2
,
R
(
τ
)
=
1
−
Π
[ δ( τ) − 2 f0 sinc2 f0τ ] ,



y
2
f
2

0



y 2 = Ry (0) = ∞
9.3-7
H(f ) =
R
1
R
N 0v / 2
2
=
,B=
, so Gy ( f ) = H ( f ) Gx ( f ) =
R + jωL 1 + j ( f / B)
2πL
1 + ( f / B)2
R y ( τ) =
N0v
N R −R τ /L
−2 πB τ
πBe
= 0v e
,
2
4L
y 2 = Ry (0) =
N 0v R
4L
9.3-8
H(f ) =
j ωL
j 2 πf
R
=
, b = , so
R + j ωL b + j 2 πf
L
2
Gy ( f ) = H ( f ) Gx ( f ) =
Ry ( τ) = −
=−
=
N 0v (2 πf ) 2
N
2b
= − 0v ( j 2 πf ) 2 2
.
2
2
2 b + (2πf )
4b
b + (2 πf ) 2
N0v d 2  − b τ 
N d
 −be− bτu ( τ) + bebτu ( −τ) 
e  = − 0v
2 
4b d τ
4b d τ
N0 v 2 − bτ
 b e u( τ) − bδ (τ) + b2e bτu (−τ) − bδ ( −τ) 
4b 
N0 v
 2δ( τ) − be −b τ  ,

4 
y 2 = R y (0) = ∞
9.3-9
∞
i 2 = ∫ Gi ( f ) df = N0 v ∫
−∞
∞
0
df
N
1 N
1
= 0v . Thus, L 0v = k T
2
R + (2πfL )
4LR
2 4 LR 2
2
9-8
⇒
N 0v = 4 Rk T
9.3-10
y is gaussian with y = 0, y 2 = σ2 = 4 RkT 0 B = 4 ×10 −10
z =∫
∞
−∞
y pY ( y) dy = 2∫
∞
y
2
2πσ2
0
e− y
/ 2σ2
2 ∞ − λ2
2
σ ∫ e d ( λ2 ) =
σ ≈ 16 µV
π 0
π
dy =
2
 2 
z2 =y2 so z = y = σ , σZ = σ −  σ  ≈ 12 µV
 π 
2
2
2
2
9.3-11
y is gaussian with y = 0, y 2 = σ2 = 4 RkT 0 B = 4 ×10 −10 and z = y u(y)
(cont.)
∞
z = ∫ ypY ( y ) dy = ∫
0
0
z =∫
2
∞
∞
0
y2
2πσ2
e
− y2 / 2 σ 2
y
2 πσ2
e−y
2σ2
dy =
π
2
/ 2σ2
∫
∞
0
σ
∫
2π
dy =
∞
0
e− λ d ( λ2 ) =
2
σ
2π
≈ 8 µV
2
2 −λ 2
λe
σ2
σ2  σ 
dλ =
, σZ =
−
≈ 12 µV
2  2π 
2
9.3-12
y = z = 0, y 2 = z 2 = σ 2 = 4 RkT 0 B = 4 ×10 −10
E [ ( Y − mY )(Z − mZ ) ] = E [ y (t ) y ( t − T ) ] = Ry (T ) = σ2sinc2 BT = 0 since 2BT = 5
Thus, ρ = 0 and pYZ ( y , z ) =
2
2
2
1
e ( y + z ) / 2σ
2
2πσ
9.3-13
y = z = 0, y 2 = z 2 = σ 2 = 4 RkT 0 B = 4 ×10 −10
E [ ( Y − mY )(Z − mZ ) ] = E [ y (t ) y ( t − T ) ] = Ry (T ) = σ2sinc2BT so ρ = σ sinc 0.5 = 0.637
2
σσ
Thus, pYZ ( y , z ) =
2
2
2
1
e −( y + z −1.274 yz )/1.19σ
2
2π0.77σ
9-9
9.3-14
∞
g = H (0) = K 2 , BN = ∫ e−2 a
2
2 2
f
0
2
2 −2 a 2 B 2
At f = B, H ( B) = K e
df =
π
2 2a
⇒
B=
K2
=
2
ln2 1
B
π
so N =
= 1.06
2 a
B 2 ln2
9.3-15
K2
 K2
 − jω t
A
δ
(
t
−
T
)
e
dt
=
∑ k
k 
∑ Ake− jωTk
−T / 2
k =− K1
 k = − K1

X T ( f , s) = ∫
T /2
where T− K1 > −
T
T
and TK 2 <
2
2
2
2
X T ( f , s ) = ∑∑ Ak Am e− jω(Tk −Tm ) , E  XT ( f , s)  = ∑∑ E [Ak Am ] E  e− jω(Tk −Tm ) 

 k m
k
m
σ 2
where E [A k Am ] = 
0
m= k
2
. So E  XT ( f , s)  = ∑ σ2 E e − jω (Tk −Tk )  = σ2 ( K1 + K 2 )

 k
m≠ k
with K1 + K2 = expected number of impulses in T seconds = µT
1 2
σ µT =µσ 2
T →∞ T
Thus, G x ( f ) = lim
9.4-1
T

10log10  N W  = 10log10 ( 4 ×106 ) ≈ 66 dB, S R = 2 × 10-5 mW ≈ -47 dBm
 T0 
( S / N ) DdB ≈ −47 + 174 − 66 = 61 dB
9.4-2
T

10log10  N W  = 10log10 ( 5 × 2 × 106 ) = 70 dB, S R = 4 × 10-6 mW ≈ -54 dBm
 T0 
( S / N ) DdB ≈ −54 + 174 − 70 = 50 dB
9-10
(cont.)
9.4-3
ST/LN0W = 46 dB = 4 × 104 ⇒ LN0 = 5 × 10-10
(a ) W = 20 kHz, (S/N)D = 55-65 dB, STdBm = ( S / N ) D − 10log10 ( 5 ×10 −10 × 20 ×103 ) + 30 =
35 to 45 dBm, S T = 3.2-32 W
(b ) W = 3.2 kHz, (S/N)D = 25-35 dB, STdBm = ( S / N ) D − 10log10 ( 5 ×10 −10 × 3.2 ×10 3 ) + 30 =
-3 to +7 dBm, S T = 0.5-5 mW
9.4-4
(S/N)D = SR/N0BN = (W/BN)(SR/N0W)
(a ) BN =
π
B = 23.6 kHz = 2.36W ⇒ (S/N)D = 0.424(SR/N0W)
2
(b ) BN =
πB
= 13.4 kHz = 1.34W ⇒ (S/N)D = 0.746(SR/N0W)
4sin π / 4
9.4-5
SD = ∫
∞
−∞
N
ND = 0
2
∞
HC ( f ) H R ( f ) Gx ( f ) df = K 2 ∫ Gx ( f ) df = K 2 ST
2
2
−∞
∫
∞
−∞
H R ( f ) df = N0 K L ∫
2
2
W
0
  f 2 
2 ST
S
2 3W
so   =
1 +    df = N0 K L
2
 N  D 3 LN0W
  W  
9.4-6
SD = ∫
∞
−∞
N
ND = 0
2
∞
HC ( f ) H R ( f ) Gx ( f ) df = K 2 ∫ Gx ( f ) df = K 2 ST
2
2
−∞
∫
∞
−∞
H R ( f ) df = N0 K L∫
2
2
W
0
  2 f 4 
5 ST
S
2 21W
so   =
1 + 
  df = N 0 K L
5
 N  D 21 LN0W
  W  
9.4-7
(a ) STdBm − L + 174 −10log10 (10 × 5 × 103 ) = 60 dB,
L = 3 × 40 = 120 dB, ST = 53 dBm = 200 W
2L
(b) L1 = 60 dB = 106, L = 120 dB = 1012, ST = 1 × 200 W = 0.4 mW
L
9-11
9.4-8
L1 =
240
1
ST
S
= 40 dB = 104 ,   =
= 103
4
6
 N  D 6 ×10 N0W
⇒
ST
= 6 ×107
N 0W
1
S
(a ) L1 = 20 dB = 100,   =
× 6 ×107 = 5 ×104 = 47 dB
N
12
×
100
 D
1
S
(b ) L1 = 60 dB = 106 ,   =
× 6 ×10 7 = 15 ≈ 12 dB
6
 N  D 4 ×10
9.4-9
L=
0.5 × 400 = 200 dB, L1 = 200/m dB,
 ST

200
20
S
≤5
 N  = 10log10  mL N W  = 80 − 10log10 m − m ≥ 30 dB so log10 m +
m
 D
 1 0 
m
log m + 20/m
10
1.0 + 2 = 3
5
0.7 + 4 = 4.7 ⇒ mmin = 5
4
0.6 + 5 = 5.6
9.4-10
L1
dB
=
LdB
m
⇒
−1
S
S
L1 = L1/m , so   = Km−1L− m where K = T
N0W
 N D
−1
−1
d S
= K  ( −m)−2 L− m + m−1L− m (ln L)( m−2 )  = 0 so


dm  N  D
m = ln L =
ln10
(10log10 L ) = 0.23LdB
10
9-12
(cont.)
9.4-11
2RkTN
R
A cos 2πf 0t
+
C
N = kTN/C (from Example 9.3-1)
y
S = (A2/2)[1 + (2πf 0RC)2]-1
–
S
A2
C
=
N 2kT N 1 + (2 πf 0 RC ) 2
and C =
d S
A2 1 + (2 πf 0 RC) 2 − (2 πf 0 RC) 2 2C × C
=
=0
 
2
dC  N  2kT N
1 + (2πf 0 RC ) 2 
so
1
2π f 0 R
9.5-1
N0 BN k T N BN τ 4 ×10−21 × 1
 σA 
=
=
=
= 0.4
 A
A2
Ep
10 −20
 
2
9.5-2
σt 2 =
t r 2 N 0 BN t r 2 N 0 BN τ
=
⇒
A2
A2 τ
σ A2 = N 0 BN
⇒
σA
=
A
σt
=
tr
N 0 BN
=
A2
N0 B N τ
Ep
N0 BN τ
Ep
9.5-3
Take BN ≈ 1/2τ = 100 kHz « BT, so σ A
2
N
 A 
≈ 0 A2 ≤ 

2 Ep
 100 
2
104 N0
⇒ Ep ≥
= 5 ×10−9
2
Then σt / τ ≈ N 0 / 4BN E p = 0.01
9.5-4
2
N 0τ
 τ 
Take BN ≈ BT = 1 MHz, so σt ≈
≤
4BT E p  100 
2
Then σ A / A = N 0 BN τ / E p = 1
9-13
⇒
104 N 0
Ep ≥
= 5 × 10−11
4BT τ
9.5-5
σA
2
2
E /τ
E
 A 
= N 0 BN ≤ 
= p4
⇒ BN ≤ 4 p = 1011 E p

10
10 N0 τ
 100 
2
N 0τ
 τ 
σt =
≤
4BN Ep  1000 
⇒
2
1011 E p ≥
1
4 × 103 Ep
106 N 0
1
BN ≥
=
. Thus,
4τE p
4 ×10 3 E p
⇒ E p ≥ 5 ×10−8
and BN = 1011 E p min = 5 kHz so
1
< BN < BT
2τ
9.5-6
B » 1/τ ⇒ y(t) ≈ x R(t), Ep = Ap2τ, so A2 ≈ Ap2 = Ep/τ
2
2
Ap
2Ep
2Ep
 A
2
BN = πB/2 ⇒ σ = N0BN = N0πB/2, so   =
=
=
N 0 π B / 2 π N 0 Bτ
N0
σ
9.5-7
Assuming pulse arrives at t = 0, y (t ) = Ap (1 − e−2 πBt ) , 0 < t < τ , so A = y(τ) = Ap (1 − e −2 πB τ )
2
Ap2 (1 − e−2 πBτ )
1 − e−2 πBτ ) 2E p
(
 A
2
σ = N0BN = N0πB/2, so   =
=
N 0πB / 2
π Bτ
N0
σ
2
2
9.5-8
P(f) = τ sinc2 fτ ⇒ H opt ( f ) =
2K τ
2 K τ  t − td 
− j ωt
sinc2 f τ e d and hopt (t ) =
Λ
N0  τ 
N0
Want hopt(t) = 0 for t < 0 for realizability, so t d D τ.
9.5-9
P( f ) =
1
b + j 2πf
⇒
H opt ( f ) =
2K
1
2 K − b( td −t )
− j ωt
e d and hopt (t ) =
e
u (t d − t )
N0 b − j 2πf
N0
Want hopt(t) ≈ 0 for t < 0 for approximate realizability, so take t d D 5/b which yields
hopt(t) « 2K/N 0 for t < 0.
9-14
Chapter 10
10.1-1

b
b 2b 
2b 
n 2 = 2 × 5  + 2b +  = 35b , ni 2 = 2 × 5 2b + +  = 35b

 2
2 
2
2 
10
5
0 b 2b
4b f
10.1-2
b

b 2b 
2b 
n 2 = 2 × 5  + 2b +  = 35b , ni 2 = 2 × 5 2b + +  = 35b
2 
2
2 

 2
10
5
0
2b 3b
f
10.1-3
Gn ( f ) =
N0 /2
4  f
f 
1 +   − c 
f 
 3  fc
2
f/fc
0
±0.5
±1
±1.5
±2
Gn/N0
0
0.1
0.5
0.22
0.1
10.1-4
(a) Glp ( f ) =
N0
2
2
H R ( f + fc )u( f + fc ) = 0 for f < -fc
For f > 0, Gn ( f ) =
N0
2
2
H R ( f )u( f ) = Glp ( f − fc )
For f < 0, Gn ( f ) = Gn (−f ) =
N0
2
2
H lp (−f − fc ) =
N0
2
2
H lp ( f + fc ) = Glp ( f + fc )
But Glp ( f − fc ) = 0 for f < 0 and Glp ( f + fc ) = 0 for f > 0, so
(cont.)
10-1
Glp ( f − fc )
Gn ( f ) = Glp ( f − fc ) + Glp ( f + fc ) = 
Glp ( f + fc )

(b) Gn ( f )u( f ) = Glp ( f − fc )
⇒
Gn ( f )[1 − u( f )] = Glp (f + fc )
⇒
f >0
f <0
Gn ( f + fc )u( f + fc ) = Glp ( f − fc + fc ) = Glp ( f )
Gn (f − fc )[1 − u( f − fc )] = Glp ( f − fc + fc ) = Glp ( f )
Thus, Gni ( f ) = Glp (f ) + Glp (f ) = 2Glp ( f )
10.1-5
(a) H lp ( f ) = H R ( f + fc )u( f + fc ) ≈ 1/(1 + j 2 f / BT ) for f > -fc. Thus,
N0
1
Glp ( f ) =
2 1 + j 2 f / BT
2
=
N0 /2
for f > -fc, which looks like the output of a
1 + (2 f / BT )2
1st-order LPF with B = BT/2.
(b) Gni ( f ) = 2Glp ( f )
ni 2 =
∫
∞
−∞
NB
df
= 0 T
2
− fc 1 + (2 f / B )
2
T
2Glp ( f )df ≈ N 0 ∫
∞
π

 + arctan 2 fc 
2
BT 

≈ N0 πBT/2 since2fc/BT = 2Q » 1
10.1-6
y(t ) = 2 ni (t )cos ωct − nq (t )sin ωct  cos(ωct + θ)
= ni (t )cos θ + nq (t )sin θ + ni (t )cos(2ωct + θ) − nq (t )sin(2ωct + θ)
ylp (t )
ybp (t )
Since ni and nq are independent and ni = nq = 0
ylp = 0, ylp 2 = ni 2 cos2 θ + nq 2 sin2 θ = n 2
ybp = 0, ybp 2 = ni 2 cos2 (2ωct + θ) + nq 2 sin2 (2ωct + θ) = n 2 [cos2 (2ωct + θ) + sin2 (2ωct + θ)]
= n2
10.1-7
y(t ) = An (t ) − An with An 2 = 2σn 2 = 8 and An = πσn 2 / 2 = 2π
pY (y ) = pA (y + An ) = 14 (y + 2π )e −(y +
n
2 π )2 / 8
u(y + 2π )
10-2
(cont.)
y = 0, σy 2 = y 2 = (An − An ) = An 2 − An = 8 − 2π
2
2
⇒
σy = 8 − 2π = 1.3
10.1-8
y = An 2 = 2σn 2
y 2 = An 4 =
∫
0
∞
An 4
∞
An −An2 / 2σn2
e
dAn = 4σn 4 ∫ λ 2e −λ d λ = 8σn 4
0
σn 2
Since An ≥ 0 and y ≥ 0, transformation with g-1(An) = y1/2 yields
pY (y ) = pA (y 1/ 2 )
n
dy 1/ 2
y 1/ 2 −y / 2σn2 1/ 2 1 −1/ 2
1 −y / 2σn2
e
u(y ) 2 y
e
u(y )
=
=
dy
σn 2
2σn 2
10.1-9
F [ jvˆ(t )] = j (−j sgn f )V ( f ) = (sgn f )V ( f ) so
Vlp ( f ) = 12 V ( f + fc ) + 12 sgn( f + fc )V ( f + fc ) = 12 [1 + sgn( f + fc )]V ( f + fc )
0
= u( f + fc )V ( f + fc ) since 1 + sgn( f + fc ) = 
2

f < −fc
f > −fc
Vlp(f)
0
b
3b f
10.1-10
Gn ( f ) =
Rn (τ) =


N 0   f − fc 
 + Π  f + fc  ,

Π


2   BT 
 BT 

N0
2
BT sinc BT τ (e j ωc τ + e − j ωc τ ) = N 0BT sinc BT τ cos ωc τ
(−j sgn f )Gn ( f ) =


−jN 0   f − fc 
 − Π  f + fc  ,

Π


2   BT 
 BT 

−jN 0
Rˆn (τ) =
BT sinc BT τ (e j ωc τ − e − j ωc τ ) = N 0BT sinc BT τ sin ωc τ
2
(cont.)
10-3
Rni (τ) = (N 0BT sinc BT τ cos ωc τ) cos ωc τ + (N 0BT sinc BT τ sin ωc τ) sin ωc τ
= N 0BT sinc BT τ (cos2 ωc τ + sin2 ωc τ) = N 0BT sinc BT τ
Gni ( f ) =
 f  N
 f 
 f 
Π   + 0 Π   = N 0 Π   so Rni (τ) = Fτ−1 Gni ( f )


2  BT 
2  BT 
 BT 
N0
10.1-11
Let f0 = fc + BT/2 so ω0 = ωc + πBT. Then Gn ( f ) =
Rn (τ) =
N0
 f + f 
N 0   f − f0 
0 
 + Π 

Π
 B  and
2   BT 
 T 

BT sinc BT τ (e j ω0τ + e − j ω0τ ) = N 0BT sinc BT τ cos ω0 τ
2
(−j sgn f )Gn ( f ) =
 f + f 
−jN 0   f − f0 
0 
 − Π 

Π
 B 
2   BT 
 T 

−jN 0
Rˆn (τ) =
BT sinc BT τ (e j ω0τ − e − j ω0τ ) = N 0BT sinc BT τ sin ω0 τ
2
Rni (τ) = N 0BT sinc BT τ cos (ωc + πBT ) τ cos ωc τ + N 0BT sinc BT τ sin (ωc + πBT ) τ sin ωc τ




= N 0BT sinc BT τ cos πBT τ =
Gni ( f ) =
N0
πτ
sin πBT τ cos πBT τ =
N0
2πτ
sin 2πBT τ = N 0BT sinc 2BT τ
 f + B / 2  N


N 0   f − BT / 2 
T
 + Π 
 = 0 Π  f  so Rn (τ) = Fτ−1 Gn ( f )

Π


i
 i 
BT
BT
2  
2  2BT 




10.1-12
∞
Rni (τ) = Fτ−1 2Glp ( f ) = 2∫ Glp ( f )e j ωτ df


−∞
Rn (τ) = Fτ−1 Glp ( f − fc ) + Glp ( f + fc ) =


=
∫
∞
−∞
= 2∫
Glp (λ1 )e j 2π(λ1 + fc ) d λ1 + ∫
∞
−∞
∞
−∞
∫
∞
−∞
Glp ( f − fc )e j ωτ df + ∫
∞
−∞
Glp ( f + fc )e j ωτ df
Glp (λ 2 )e j 2π(λ2 −fc ) d λ 2
Glp (λ)e j 2πλτ d λ 21 (e j ωc τ + e − j ωc τ ) = Rni (τ)cos ωc τ
Rˆn (τ)sin ωc τ = Rni (τ) − Rni (τ)cos ωc τ cos ωc τ = Rni (τ)sin2 ωc τ so Rˆn (τ) = Rni (τ)sin ωc τ


10-4
10.1-13
E [nq (t )nq (t − τ)] = E1 − E2 − E 3 + E 4 where
E1 = E [nˆ(t )cos ωct × nˆ(t − τ)cos ωc (t − τ)] = 12 Rnˆ (τ)[cos ωc τ + cos ωc (2t − τ)]
E 2 = E [nˆ(t )cos ωct × n(t − τ)sin ωc (t − τ)] = 12 Rnn
(τ)[− sin ωc τ + sin ωc (2t − τ)]
ˆ
E 3 = E [n(t )sin ωct × nˆ(t − τ)cos ωc (t − τ)] = 21 Rnnˆ (τ)[sin ωc τ + sin ωc (2t − τ)]
E 4 = E [n(t )sin ωct × n(t − τ)sin ωc (t − τ)] = 12 Rn (τ)[cos ωc τ − cos ωc (2t − τ)]
Thus, Rnq (t, t − τ) = 12 [Rnˆ (τ) + Rn (τ)]cos ωc τ + 12 [Rnˆn (τ) − Rnnˆ (τ)]sin ωc τ
+ 21 [Rnˆ (τ) − Rn (τ)]cos ωc (2t − τ) − 21 [Rnn
(τ) + Rnnˆ (τ)]sin ωc (2t − τ)
ˆ
But Rnˆ = Rn so Rnˆ + Rn = 2Rn and Rnˆ − Rn = 0 and
−Rnnˆ = Rnn
= Rˆn so Rnn
− Rnnˆ = 2Rˆn and Rnn
+ Rnnˆ = 0 . Hence,
ˆ
ˆ
ˆ
Rnq (t, t − τ) = Rnq (τ) = Rn (τ)cos ωc τ + Rˆn (τ)sin ωc τ
10.1-14
E [ni (t )nq (t − τ)] = E1 − E 2 + E 3 − E 4 where
E1 = E [n(t )cos ωct × nˆ(t − τ)cos ωc (t − τ)] = 21 Rn nˆ (τ)[cos ωc τ + cos ωc (2t − τ)]
E2 = E [n(t )cos ωct × n(t − τ)sin ωc (t − τ)] = 12 Rn (τ)[− sin ωc τ + sin ωc (2t − τ)]
E 3 = E [nˆ(t )sin ωct × nˆ(t − τ)cos ωc (t − τ)] = 12 Rnˆ (τ)[sin ωc τ + sin ωc (2t − τ)]
E 4 = E [nˆ(t )sin ωct × n(t − τ)sin ωc (t − τ)] = 12 Rnn
(τ)[cos ωc τ − cos ωc (2t − τ)]
ˆ
Thus, Rninq (t, t − τ) = 21 [Rn nˆ (τ) + Rnˆn (τ)]cos ωc τ + 21 [Rn (τ) − Rnˆ (τ)]sin ωc τ
+ 21 [Rn nˆ (τ) − Rnn
(τ)]cos ωc (2t − τ) − 21 [Rn (τ) + Rnˆ (τ)]sin ωc (2t − τ)
ˆ
But Rnˆ = Rn so Rnˆ + Rn = 2Rn and Rnˆ − Rn = 0 and
Rnnˆ = −Rnn
= −Rˆn so Rn nˆ − Rnn
= −2Rˆn and Rnnˆ + Rnn
= 0 . Hence,
ˆ
ˆ
ˆ
Rninq (t, t − τ) = Rninq (τ) = Rn (τ)sin ωc τ + Rˆn (τ)cos ωc τ
10.1-15
(a) Rninq (τ) = [Rni (τ)cos ωc τ ]sin ωc τ − [Rni (τ)sin ωc τ ]cos ωc τ = 0
10-5
(cont.)
(b) Gn ( f )u( f ) = Glp ( f − fc )
⇒
Gn ( f )[1 − u( f )] = Glp ( f + fc )
⇒
Gn ( f + fc )u( f + fc ) = Glp ( f + fc − fc ) = Glp ( f )
Gn ( f − fc )[1 − u( f − fc )] = Glp ( f − fc + fc ) = Glp ( f )
Thus, Rninq (τ) = Fτ−1 { j[Glp ( f ) − Glp ( f )]} = 0
10.1-16
Gn ( f + fc )u( f + fc ) =

 N
Rninq (τ) = Fτ−1  j 0
 2

 f − B / 2 
 f + B / 2 
N
T
T
, Gn ( f − fc )[1 − u( f − fc )] = 0 Π 
Π 


BT
BT
2 
2 

N0
  f − B / 2
 f + B / 2  
N


T
T
 Π 
  = j 0 B sincB τ (e j πBT τ − e − j πBT τ )
−
Π

T
 


BT
BT
2 T


 
 
= −N 0BT sincBT τ sin πBT τ = −πN 0BT 2 τ sinc2BT τ
10.1-17
BT/4
3BT/4
f
-jN0/2
Gn ( f + fc )u( f + fc ) − Gn ( f − fc )[1 − u( f − fc )] =
 N
Rninq (τ) = Fτ−1  j 0
 2

=−
N 0BT
2
 f + B / 2 
N 0   f − BT / 2 
T
 − Π 


Π
 B / 2 
2   BT / 2 


T

  f − B / 2
 f + B / 2  
BT τ j πB τ
N 0 BT


T
T
 Π 
T
 
−
Π
− e − j πBT τ )
  B / 2 
 B / 2   = j 2 2 sinc 2 (e


 
 
T
T
sinc
BT τ
2
sin πBT τ = −
πN 0BT 2 τ
2
sinc
BT τ
2
sincBT τ
10.2-1
N 0 = k TN = k T 0 (T N / T 0 ) = 4 × 10−21 × 10 = 4 × 10−20
(S / N )D = S R / N 0W = 20 × 10−9 /(4 × 10−20 × 5 × 106 ) = 105 = 50 dB
10.2-2
N 0 = k TN = k T 0 (T N / T 0 ) = 4 × 10−21 × 10 = 4 × 10−20
S 
0.4
20 × 10−9
  =
= 2.86 × 104 = 44.6 dB
 N D 1 + 0.4 4 × 10−20 × 5 × 106
10-6
10.2-3
y(t ) = {Ac x (t ) + ni (t ) cos ωct − nq (t )sin ωct } 2 cos(ωct + φ ')
= Ac x (t )cos φ '+ ni (t )cos φ '+ nq (t )sin φ '+ high-frequency terms
yD (t ) = Ac x (t )cos φ '+ ni (t )cos φ '+ nq (t )sin φ ' so S D = Ac 2 x 2 cos2 φ '
N D = E ni 2 cos2 φ '+ 2ninq cos φ ' sin φ '+ nq 2 sin2 φ '
= ni 2 cos2 φ '+ nq 2 sin2 φ ' = n 2 (cos2 φ '+ sin2 φ ') = n 2 = 2N 0W
Thus, (S / N )D = Ac 2 x 2 cos2 φ '/ 2N 0W = S R / N 0W × cos2 φ ' = γ cos2 φ '
10.2-4
DSB: S p = xc 2 max = Ac 2
⇒
S D = Ac 2 x 2 = S pS x so (S / N )D = S x S p / 2N 0W = 21 S x γ p
AM: S p = Ac 2 (1 + x 2 ) max = 4Ac 2
⇒
S D = Ac 2 x 2 = 14 S pS x so
(S / N )D = 14 S x S p / 2N 0W = 18 S x γ p
10.2-5
v(t ) = Ac x 1(t ) + ni (t ) cos ωct ± Ac x 2 (t ) ∓ nq (t ) sin ωct so yD (t ) = Ac x 1(t ) + ni (t ) and
1
yD (t ) = Ac x 2 (t ) ∓ nq (t ) where x 12 = x 22 = S x , S R = 21 Ac 2 x 12 + 21 Ac 2 x 22 = Ac 2S x , and
2
SR
S 
A 2S
ni 2 = (∓nq )2 = n 2 = 2N 0W . Thus, both outputs have   = c x =
= 12 γ
 N D
2N 0W
2N 0W
10.2-6
For USSB, any noise component in fc – W < f< fc will be translated to f < W and cannot be
removed by LPF; similarly, for LSSB, noise components in fc < f< fc + W cannot be removed
by LPF. For DSB, noise components outside fc – W < f< fc + W are translated to f > W and
can be removed by LPF.
10.2-7
Gni ( f )
0
W 1.5W
(cont.)
10-7
With ideal LPF at output, N D =
SR
 S 
3

G
f
df
=
N
W
so
=
= 2γ
(
)

2
n

0
∫−W i
 N D ( 23 N 0W ) 3
W
10.2-8
USSB: yD (t ) = 21 Ac x (t ) + ni (t ), S D = S R
N D = 2∫
N 0 fc
W
2( f + fc )
0
df = N 0 fc  ln (W + fc ) − ln fc  = N 0 fc ln (1 + W / fc )
1.10 γ W / fc = 1/ 5
SR
 S 
W / fc

  =
=
γ
=

 N D
1.01γ W / fc = 1/ 50
N 0 fc ln (1 + W / fc ) ln (1 + W / fc )

DSB: yD (t ) = Ac x (t ) + ni (t ), S D = 2S R , Gni ( f ) =
N D = 2∫
fc 2 − f
0
 S 

 N D =
N 0 fc 2
W
df = 2N 0 fc
2
N 0 fc  1
N 0 fc 2
1 

−
=

f − fc  fc 2 − f 2
2  f + fc
 f + W 
1 + W / fc 
1
 = N 0 fc ln 

ln  c
2 fc  fc −W 
 1 −W / fc 
0.99 γ W / fc = 1/ 5
2S R
2W / fc
=
γ = 
1 + W / fc 
1 + W / fc 
1.00 γ W / fc = 1/ 50

 ln 

N 0 fc ln 
 1 −W / fc 
 1 −W / fc 
Note: no significant difference between LSSB and DSB when W/fc « 1.
10.2-9
v(t ) =  21 Ac x (t ) + ni (t ) cos ωct −  21 Ac xˆ(t ) + nq (t ) sin ωct
y(t ) = v(t )2 cos [ωct + φ(t )] =  21 Ac x (t ) + ni (t ) cos φ(t ) +  21 Ac xˆ(t ) + nq (t ) sin φ(t ) +
high-frequency terms. Since φ(t) has slow variations compared to x(t),
yD (t ) ≈  12 Ac x (t ) + ni (t ) cos φ(t ) +  12 Ac xˆ(t ) + nq (t ) sin φ(t ) = 12 Ac x (t ) when φ = ni = nq = 0
which implies that K = 2 / Ac . Then
{
E x (t ) − KyD (t )
2
}
2
2




2 
2 
2 

2
2


= E x + x + ni  cos φ + xˆ + nq  sin2 φ − 2x x + ni  cos φ


Ac 
Ac 
Ac 






2 
2 
2 
−2x xˆ + nq  sin φ + 2 x + ni  xˆ + nq  cos φ sin φ



Ac 
Ac 
Ac 


(cont.)
10-8
where x 2 = xˆ2 = S x , xxˆ = 0,
S
4
= x , ni 2 = nq 2 = N 0W , ni = nq = 0, ninq = 0 . Thus,
Ac 2
SR
∈2 = S x + (S x + S x N 0W / S R ) cos2 φ + (S x + S x N 0W / S R ) sin2 φ − 2S x cos φ / S x


= 1 + cos2 φ + sin2 φ − 2cos φ +
1
cos2 φ + sin2 φ
S R / N 0W
(
)
But cos2 φ + sin2 φ = cos2 φ + sin2 φ = 1 so ∈2 = 2 (1 − cos φ) + 1/ γ
10.2-10
Envelope detection without mutilation requires SR » NR = N0BN, where BN is the noise equivalent
bandwidth of HR(f), so BN should be as small as possible, namely BN = BT = 2W for an ideal BPF.
With synchronous detection, there is no mutilation and noise components outside
fc – W < f < fc + W are translated to f > W and can be removed by the LPF.
10.2-11
1
1 S 
2
With S x = ,   =
γ = 104
1


2 N D 1+ 2
⇒
γ = 3 × 104 , whereas γth ≈ 20 .
Thus, gT = γ / γth ≈ 1500 = 32 dB
10.2-12
1
 S 
2
  =
γ = 103
 N 
1
1+ 2
D
 S 
γth ≈ 20 =  R 
 N 0W min
⇒
⇒
γ=
SR
N 0W
Wmax =
= 3 × 103
⇒
SR
N0
= γW = 24 × 106
1 SR
= 1.2 MHz
20 N 0
10.2-13
yD (t ) = y(t ) = An (t ) + Ac x (t )cos φn (t ) − An
= Ac x (t ) if n(t ) = 0, so K = 1/ Ac
2

 
1
1



∈ = E  x − An − x cos φn + An  / S x , S x = 1

Ac
Ac  


2
(cont.)
10-9

A2
A2
2
2
∈2 = E x 2 + n + x 2 cos2 φn + n − An x − 2x 2 cos φn + An x

Ac 2
Ac 2 Ac
Ac
−

2
2
2
xAn cos φn −
An An − An x cos φn 

Ac
Ac 2
Ac
where x 2 = 1, x = 0, An 2 = 2n 2 = 4N 0W , An 2 = πN 0W , An cos φn = ni = 0, cos φn = 0, .
cos2 φn =
∈2 = 1 +
1 π
1
cos2 φn d φn = , Ac 2 = 2S R /(1 + S x ) = S R . Thus,
∫
2π −π
2
4N 0W
SR
+
1 πN 0W
2
3 4−π
+
−
πN 0W = +
with γ < γth ≈ 20
SR
SR
γ
2
2
10.3-1
N 0 = k TN = k T 0 (T N / T 0 ) = 4 × 10−21 × 10 = 4 × 10−20
PM: N D = N 0W / S R = 4 × 10−20 × 500 × 103 /10 × 10−9 = 2 × 10−6
FM: N D = N 0W 3 / 3S R = 4 × 10−20 (500 × 103 )3 / 3 × 10 × 10−9 = 1.67 × 105
Deemphasized FM: N D ≈ N 0Bde 2W / S R = 4 × 10−20 (5 × 103 )2 500 × 103 /10 × 10−9 = 50
10.3-2
ND =
≤
∫
∞
−∞
N0f 2
N 0W 3
1
df
=
2n
−∞ 1 + ( f / 2W )
SR
2S R
2
H D ( f ) G ξ ( f )df ≤ ∫
N 0W 3
SR
∞
∫
0
∞
λ2
dλ
1 + λ 2n
NW3
π / 2n
≈ 0
if n » 1
sin(3π / 2n )
3S R
10.3-3
Gnq ( f ) = 2
ND = 2
N0
N0
2S R
2
H R ( f + fc ) =
2
∫
W
0
N0
1 + (2 f / BT )2
N 0BT 3
f2
df
=
1 + (2 f / BT )2
8S R
3

N 0BT 3  2W  2W 1  2W 
=
−
− 

 +
8S R  BT
3  BT 
 BT


, Gξ (f ) =
N0
f2
2S R 1 + (2 f / BT )2
 2W
 2W 



 B − arctan  B 


T 
 T
3
  N B 3 
3
  ≈ 0 T 1  2W  = N 0W if B
 

T
8S R 3  BT 
3S R

 
10-10
W
10.3-4
N 0 = k TN = k T 0 (T N / T 0 ) = 4 × 10−21 × 10 = 4 × 10−20 , D = f∆ /W = 2 × 106 / 500 × 103 = 4
S
S 
10−9
FM:   = 3D 2S x R = 3 × 42 × 0.1
= 240 × 103 = 53.8 dB
 N D
N 0W
4 × 10−20 × 500 × 103
Deemphasized FM:
2
2
 f 
 2 × 106 
SR
 S 
10−9

∆

  ≈   S x
0.1
=
= 800 × 106 = 89.0 dB

3
3
−20

 N D  B 
N 0W
4 × 10 × 500 × 10
 5 × 10 
de
10.3-5
ND =
NB
N0
W
1
df = 0 de arctan
2
−W 1 + ( f / B ) 2S
SR
Bde
de
R
∫
W
 2 W 
φ∆2S x S R
W / Bde
S 
2
 φ 2S γ

 N  = N B arctan(W / B ) = arctan(W / B ) φ∆ S x γ ≈ π
 ∆ x
B

D


0 de
de
de
de
10.3-6
S D = f∆ x = f∆ / 2 and N D = 2∫
2
2
2
300
100
N0f 2
2S R
df =
N0
SR
×
26
× 106 so
3
 S 
1
3 × 10−6 S R
  = f∆2
= 290 = 24.6 dB
 N D
2
26 N 0
10.3-7
W
2
N D = 2∫ e −( f / Bde )
0
N0f 2
2S R
N 0Bde 3
df =
SR
3
4 f 2S S
 S 
4  W 
  > ∆ x R =
 
 N D
πN 0Bde 3
π  Bde 
∫
W / Bde
0
2
λ 2e −λ d λ <
N 0Bde 3
SR
∫
∞
0
2
λ 2e −λ d λ =
2
 f∆ 
  S x γ so
W 
Improvement factor > (4 / π )(W / Bde )3 ≈ 770 when Bde = W / 7
10.3-8
PM: (S / N )D = φ∆2S x γ = 103
FM: D = φ∆ for same BT, so (S / N )D = (W / Bde )2 φ∆2S x γ = 102 × 103 = 50 dB
10-11
N 0Bde 3
SR
π
4
10.3-9
(S / N )D = (W / Bde )2 D 2S x γth ≈ 20(W / Bde )2 D 2 (D + 2)S x , D > 2
th
20 × 52 D 2 (D + 2)/ 2 = 105
⇒
D 3 + 2D 2 = 400
⇒
D ≈ 6.7 (by trial and error)
BT ≈ 2(6.7 + 2) × 10 = 174 kHz, S R ≥ 20(6.7 + 2) × 10−8 × 104 = 17.4 mW
10.3-10
γth = 20M (φ∆ ), (S / N )D = φ∆2S x γth = 20M (φ∆ )φ∆2S x
th
φ∆ ≤ π, M (π) ≈ 2(π + 2), S x ≤ 1, so (S / N )D ≤ 40(π + 2)π2 ≈ 2030 ≈ 33 dB
th
10.3-11
IF input = Ac cos [ωct + φ(t )] × 2 cos (ωc − ωIF )t + K φD (t ) so
vIF (t ) = Ac cos ωIF t + φIF (t )
Thus, yD (t ) =
and DIF =
where
φIF (t ) = φ(t ) − K φD (t )
f
1
1
φIF (t ) = f∆x (t ) − KyD (t ) so yD (t ) = ∆ x (t ) =
φ (t )
2π
1+K
2π IF
f∆
(1 + K )W
=
D
1+K
10.4-1
S x = 1/ 2, (S / N )D = 104 , γ = ST / LN 0W = 10ST
(a) (S / N )D = 10ST
⇒
ST = 1 kW
S 
1/ 2
(b) µ = 1,   =
10ST
 N D 1 + 1/ 2
S 
1/ 8
10ST
µ = 21 ,   =
 N D 1 + 1/ 8
(c) (S / N )D = π2 × 12 × 10ST
⇒
⇒
⇒
ST = 3 kW
ST = 9 kW
ST ≈ 200 W
(d) (S / N )D = 3D 2 × 12 × 10ST provided that γ ≥ γth = 20M (D ) so ST ≥ 2M (D )
(cont.)
10-12
D
104/15D2
2M(D)
ST
1
667
5
667 W
5
26.7
14
26.7 W
10
6.7
24
24 W
Threshold limited
10.4-2
S x = 1, (S / N )D = 104 , γ = ST / LN 0W = 5ST
(a) (S / N )D = 5ST
⇒
ST = 2 kW
S 
1
(b) µ = 1,   =
5S
 N D 1 + 1 T
⇒
S 
1/ 4
µ = 21 ,   =
5S
 N D 1 + 1/ 4 T
(c) (S / N )D = π2 × 1 × 5ST
⇒
ST = 4 kW
⇒
ST = 10 kW
ST ≈ 200 W
(d) (S / N )D = 3D 2 × 1 × 5ST provided that γ ≥ γth = 20M (D ) so ST ≥ 4M (D )
D
104/15D2
4M(D)
ST
1
667
10
667 W
5
26.7
28
28 W
Threshold limited
10
6.7
48
48 W
Threshold limited
10.4-3
L = 10α
/10
= 10
/10
, γ = ST / LN 0W = 1010 × 10−
(a) (S / N )D = γ = 1010 × 10−
/10
= 104
S 
1/ 2
1
(b)   =
γ = × 1010 × 10−
 N D 1 + 1/ 2
3
⇒
/10
(c) (S / N )D = 3 × 22 × 21 γ = 6 × 1010 × 10−
(S / N )D = 3 × 82 × 12 γ = 96 × 1010 × 10−
/10
, S x = 1/ 2
= 10(10 − 4) = 60 km
= 104
/10
/10
⇒
= 104 ⇒
= 104 ⇒
10-13
= 10(10 − 4 − log10 3) = 55.2 km
= 10(10 − 4 + log10 6) = 67.8 km
= 10(10 − 4 + log10 96) = 79.8 km
10.4-4
 4π × 3 × 108
L = 
 3 × 105
γ=
gT g RST
≈
LN 0W
2

 = 1.58 × 108

2
and gT × gR = 52 dB = 1.58 × 105 so
107
2
(a) (S / N )D = γ = 107 /
2
= 104
S 
1/ 2
γ = 107 / 3
(b)   =
 N D 1 + 1/ 2
⇒
2
= 1000 = 31.6 km
= 104
⇒
(c) (S / N )D = 3 × 22 × 21 γ = 6 × 107 /
2
(S / N )D = 3 × 82 × 12 γ = 96 × 107 /
= 104
2
= 104
= 1000 / 3 = 18.3 km
⇒
= 6000 = 77.5 km
⇒
= 96, 000 = 310 km
10.4-5
S 
1/ 2
AM:   =
γ = 20
 N D 1 + 1/ 2
⇒
FM: γ ≥ γth = 20M (D )
M (Dmax ) = 60 / 20
⇒
γ = 60
⇒
Dmax ≈ 1
(S / N )D = 3D 2 × 21 γ = 90 = 19.5 dB
10.4-6
At output of the kth BPF,
kW0
2
Sk = ( f∆α k ) x k 2 = f∆2α k 2 , N k = 2
∫
(k −1)W0
N0f 2
2S R
df =
N0
k 3W 3 − (k − 1)3W 3  . Thus,
0
0 
3S R 
f∆2α k 2
 S 
  =
 N 
(3k 2 − 3k + 1)(N 0W0 3 / 3S R )
k
10.4-7
3S R f∆2
α2
 S 
S
  = 2 k
=
 N k
N
3k − 3k + 1 N 0W0 3
⇒
α k 2 = C (3k 2 − 3k + 1) with C =
x 2 = 1
K
K
K K

k
xb 2 = E  ∑ ∑ α k x k α i x i  = ∑ ∑ α k α i x k x i where x k x i = 
 k =1 i =1


k =1 i =1
x k x i = 0
10-14
N 0W0 3 S
3S R f∆2 N
i =k
i ≠k
(cont.)
so
K
K
K
k =1
k =1
k =1
xb 2 = ∑ α k 2 = 1 where ∑ α k 2 = C ∑ 3k 2 − 3k + 1
 K (K + 1)(2K + 1)

K (K + 1)
1
= C 3
−3
+ K  = CK 3 = 1 ⇒ C = 3


K
6
2
2
 f  SR
3S f 2 1
S 
S
= R∆
= 3  ∆ 
Thus,   =
, where W = KW0
3
 N k
W  N 0W
N
N 0W0 3 K
10.4-8
2
(a) Gn1 ( f ) = H de ( f ) G ξ ( f ), n12 =
1
N 0Bde 3  W
W 
10 N 0
−
≈
×
arctan
5.3
10
S R  Bde
Bde 
SR
2
Gn2 ( f ) = H de (f ) G ξ ( f − f0 ) + G ξ ( f + f0 )
2
 2

= H de ( f )
n2 =
2
∫
W
−W
N0
2
N0
2 N
 f 
0
( f − f )2 + ( f + f )2  = H ( f )
( f 2 + f02 )Π 
0
0 
de

 2W 
2S R
2S R
f 2 + f02
2N 0
df =
2
S R 1 + ( f / Bde )
SR
W / Bde
W / Bde

λ2
dλ 
2
Bde 2 ∫

d
λ
+
B
f
de 0 ∫0
0
1 + λ2
1 + λ 2 



 f 2

2N 0Bde 3  W
N

W



0 
=
+   − 1 arctan
≈ 8.8 × 1012 0


S R  Bde  Bde 
Bde 
SR





n12
(b) y1 − y2 = f∆ (x L + x R − x L + x R ) + n1 − n2 ≈ 2 f∆x R − n2 , since n22
n12
y1 + y2 = f∆ (x L + x R + x L − x R ) + n1 + n2 ≈ 2 f∆x L + n2 . Thus,
(S / N )D = (S / N )D ≈ (2 f∆ )2 x R 2 / n22 = 1.5 × 10−13 f∆2S R / N 0
L
R
For mono signal with x 2 = 1, (S / N )D = f∆2 x 2 / n12 = 1.9 × 10−11 f∆2S R / N 0 . Thus,
(S / N )D (stereo)/(S / N )D (mono) ≈ 8 × 10−3 = −21 dB
10.6-1
N 0 = k T N = k T 0 (T N / T 0 ) = 4 × 10−21 × 10 = 4 × 10−20 , µ p = t0 = 0.1/ fs = 1/12 × 106
(cont.)
10-15
2
 W  S x S
 0.4 × 10 × 10−9
 S 
 1  7 
500 × 103
R

  = 4µ p 2BT 


=
4
10


12 × 106 
 N D
1.2 × 106 /12 × 106  4 × 10−20 × 500 × 103
 fs τ 0  N 0W
= 2.78 × 105 = 54.4 dB
10.6-2
N 0 = k T N = k T 0 (T N / T 0 ) = 4 × 10−21 × 10 = 4 × 10−20 , µ p = µτ 0 = 0.2 / 50 × 250 = 16 × 10−6
 W  S x S


0.1S R
S 
100
3
−6 2
R

  = 4µ p 2BT 

=
×
×
4
16
10
3
10
(
)

−20



 N D
 fs τ 0  N 0W
 250 / 50 × 250  4 × 10 × 100
= 384 × 1012 S R ≥ 104
⇒
S R ≥ 26 pW
10.6-3
S 
0.3
0.2
2
1
 N  ≤ 8 × 20 S x γ = 50S x γ . But with µ p = t0 = f , τ 0 = τ = f , and BT = 20W
D
s
s
4µ p 2BTW
fs τ 0
2
W 
= 36  
 f 
⇒
s
2
 S 
 1 
  = 36   S x γ ≈ 5.8S x γ
 2.5 
 N D
10.6-4
S R = 9 fs A2 τ + fs A2 3τ ⇒ A2 =
SR
12 fs τ
, µ p = t0 = 3.6 µs, τ 0 = τ = 2.5 µs,
S
4µ 2B A2
4t 2B W
S 
γ
and BT = 400 kHz, so   = p T S x = 0 T S x R = 2.49  
9
 N D
N0
N 0W
12 fs τ
10.6-5
S R = Mfs A2 τ ≥ Mfs A2 / BT

1 T
µ p = t0 =  s − τ − Tg  ≤

2 M
2
⇒
A2 ≤ BT S R / Mfs
1  1
2 
1 BT − 2Mfs

−
=
2  Mfs BT  20 Mfs BT
2
 B − 2Mfs  B 2S
 B − 2Mfs 
 S 
 T R S x =  T

  ≤  T

 N D  Mf B  Mf


 Mfs BT 
s T
s
2
W 
SR
γ
1  BT 
  S
Sx
< 
, fs ≥ 2W

x


 f  MN W
M
8  MW 
s
0
10-16
Chapter 11
11.1-1
11.1-2
11.1-3
11.1-4
11.1-5
11-1
11.1-6
ak
Nat. code
Gray codes
____________________________________
7 A /2
111
100 010 001
5 A /2
110
101 110 011
3 A /2
101
111 100 111
A /2
100
110 101 101
- A /2
011
010 111 100
-3 A /2
010
011 011 110
-5 A /2
001
001 001 010
-7 A /2
000
000 000 000
11.1-7
1
r = 160 kHz
2 b
(b) r = 320 kpbs /log2 M ≤ 2 B = 120 kbaud
(a) rb = 16 x 20,000 = 320 kpbs, B ≥
log 2 M ≥ 320/120 = 2.67 ⇒ M ≥ 23 = 8
11.1-8
(a) 128 = 27 ⇒ 7 bits/character
rb = 7 x 3000 = 21 kbps, B ≥
1
rb = 10.5 kHz
2
(b) r = 21 kbps/log 2 M ≤ 2 B = 6 kHz
log 2 M ≥ 21/6 = 3.5 ⇒ M ≥ 2 4 = 16
11.1-9
(a) p% (t ) = g (t + Tb ) − g (t )
p% (0) = K 0 (1 − e−2 ) = 1
⇒ K0 =
1
= 1.157
1 − e −2
11-2
11.9 continued
(b)
tk
y (t k )
ISI
_________________________________
0
1
1
1e = 0.135
2
0
-2
0.135
1 + 1e = 1.018
-4
−4
0.018
−6
3
K 0 (1 − e ) + 1e
= 1.138
4
K 0 (1 − e −4 )e −2 + 1e −8 = 0.154 0.138
11.1-10
(a) p% (t ) = g ( t + Tb ) − g (t )
p% (0) = K0 (1 − e −1 ) = 1 ⇒ K0 =
1
= 1.582,
1 − e −1
(b)
11-3
0.138
11.1-10 (b) continued
tk
y (t k )
ISI
__________________________________________________
0
1
1
-1 + 1e = -0.632
-1
2
3
0
-1
1 - 1e
+ 1e = 0.767
-2
K 0 (1 - e ) - 1e
4
+0.368
-2
-2
+ 1e
-0.233
-3
= 1.282
-1 + K0 (1- e-2 ) e -1 - 1 e-3 + 1e-4 = -0.528 +0.472
11.1-11
p ( D) = e −π ( B / b ) ≤ 0.01
2
2
P ( B)
= e−π ( B / b ) ≤ 0.01
P(0)
⇒
π(bD )2 ≥ ln100 where D = 1/ r
⇒ π( B / b) 2 ≥ ln100
thus π(b / r) 2 x π( B / b) 2 ≥ (ln100) 2 ⇒ r ≤
π
B ≈ 0.7 B
ln100
11.1-12
m a = ak = 0, σ2a = ak2 = ( A /2) 2 , P( f ) =
⇒ Gx ( f ) =
+0.282
1
f
sinc
2rb
2rb
A2
f
sinc2
16 rb
2rb
11-4
11.1-12 continued
∞
∫ Gx ( f )df = 2
x2 =
−∞
∞
A2
A2
2 f
sinc
df
=
16 rb ∫0
2rb
8
A waveform with polar RZ format, amplitude = ± A / 2, period = Tb and 50% duty cycle
2
T /2 A
A2
⇒x = b   =
Tb  2 
8
2
11.1-13
1 2
A , σ a2 = A2 /4, P ( f ) = ( τ sinc2f τ)/[1 − (2 f τ) 2 ]
2
(a) τ = 1 / 2rb , P(0) = 1 / 2rb , P (± rb ) = 1 / 4 rb , P( nrb ) = 0, n ≥ 2
ma = A /2, ak2 =
2
2
A2 2
A2 2  1 
A2 2  1 
2
Gx ( f ) =
rb P( f ) +
rb  2  δ( f ) +
rb  2  [ δ( f − rb ) + δ( f + rb )]
4
4
2
r
4
 b 
 4rb 
τ = 1/ rb , P (0) = 1/ rb , P( nrb ) = 0, n ≠ 0
2
A2
A2 2  1 
2
Gx ( f ) =
rb P ( f ) +
rb   δ( f )
4
4
 rb 
11-5
11.1-13 continued
Note larger dc component and smoother waveform.
11.1-14
A

+ (1 − α )( − A /2) = (2α − 1) A / 2 2
2
2
2
2
2
 σa = [1 − (2α − 1) ] A / 4 = (α − α ) A

ak2 = α( A /2) 2 + (1 − α)(− A /2) 2 = A4 / 4

ma = ak = α
2
P( f ) =
x2 =
∞
1
(α − α 2 )
 2α − 1 
Π ( f / rb ), Gx ( f ) =
Π ( f / rb ) + 
 δ( f )
rb
rb
 4 
∫ Gx ( f )df =
−∞
(α − α 2 ) 2
A x 2 rb /2 + (2α − 1) 2 A2 / 4 = A2 / 4
rb
11.1-15
 t + Tb / 4 
 t − Tb / 4 
p (t ) = Π 
 −Π

 Tb / 2 
 Tb / 2 
11-6
11.1-15 continued
P( f ) =
Tb
f
πf
 fT 
sinc  b  ( e j 2πTb / 4 − e − j 2πTb / 4 ) = jTbsinc
sin
2
2rb
2rb
 2 
ak = ± A / 2 ⇒ ma = 0, σa2 = ak2 = A 2 / 4
Gx ( f ) =
A2
f
πf
sinc 2
sin2
4rb
2rb
2rb
11.1-16
Ra ( n) = E[ ak − na k ]
n=0
a P (a )
k
k
_________
0
1/2
± A 1/4
1
1
⇒ Ra (0) = x 0 + 2 x A2 = A2 / 2
2
4
11-7
11.1-16 continued
n =1
ak −1
ak
P( ak −1ak )
_____________________
0
0
1/2 x 1/2
0
A
1/2 x 1/4
0
-A
1/2 x 1/4
A
0
1/4 x 1/2
-A
0
1/4 x 1/2
A
- A
-A
A
1/2 x (1/2) 2  1
 = P(1,1)
1/2 x (1/2) 2  2

 not allowed

1
1
1
Ra ( ±1) = x 0 + 4 x x 0 + 2 x ( A)(− A) = − A2 / 4
4
8
8
n≥2
ak −n a k
P( ak − na k )
A
-A
A
-A
0
0
_____________________
0
0
1/2 x 1/2
0
A
1/2 x 1/4
0
-A
1/2 x 1/4
A
0
1/4 x 1/2
-A
0
1/4 x 1/2
A
- A
-A
A
A
A
−A
-A
Ra ( n) =
(1/4) 2 

(1/4) 2 
 = P(1) P(1)
(1/4) 2 
(1/4) 2 
1
1
1
1
x0+4x x0+2x
( A)(− A) + 2 x
(± A )2 = 0
4
8
16
16
11-8
n ≥2
11.1-17
K
K
K
∑ ∑ g (k − i) = ∑ [ g (k − K ) + g (k − K − 1) + .... + g (k + K )
k =− K i =− K
k =− K
[ g( −2 K ) + g( −2 K + 1) + ... + g (0)]


+[g (−2K + 1) + ....g (0) + g (1)]
 2 K + 1 sums
+g (0) + g (1) + ...g (2 K )] 
= g (−2 K ) + 2 g ( −2 K + 1) + ..... + (2 K + 1) g (0) + 2 Kg (1) + ...2 g (2K − 1) + g (2 K )
=
=
0
2K
n=-2K
n =1
∑ (2 K + 1 + n) g(n ) + ∑ (2 K + 1 − n )g( n)
2K
∑
(2 K + 1 − n )g ( n) = (2 K + 1)
n =−2 K

n 
1 −
 g ( n)
2K + 1 
n =−2 K 
2K
∑
E[ ak ai ] = E[ ak ak −(k − i) ] = Ra ( k − i)
K
= (2 K + 1)
K
∑ ∑ g (k − i)
thus, ρ k ( f ) =
k =− K i = − K
where g ( k − i ) = Ra ( k − i ) e− j ω(k −i) D

n 
− jω nD
1 −
 Ra ( n)e
2K +1 
n =−2 K 
2K
∑
11.1-18
1
1
1
1
1
1
x 1 + x 0 = , ak2 = x 1 2 + x 0 2 =
2
2
2
2
2
2
1
1
bk = 1 − ak = , bk2 = 1 − 2 ak + ak2 = , a k bk = ak − ak2 = 0
2
2
1
1
1
for i ≠ k , a k ai = ak ai = , bk bi = bk bi = , ak bi = ak bi =
4
4
4
(a) ak =
(b) x T ( t ) =
K
∑ [a ρ (t − kT ) + b ρ (t − kT )]
k
k =− K
XT ( f ) =
1
b
k
K
0
b
∑ [a P ( f ) + b P ( f )]e
− j ωkTb
k =− K
k 1
k
0
X T ( f ) = ∑∑ [ ak P1 + bk P0 ]e − jωkTb [ ai P1* + bi P0* ]e
2
k
=
i
∑ ∑ [a a P P
k
k
+ jω iTb
*
i 1 1
+ akbi P1 P0* + ai bk P1* P0 + bk bi P0 P0* ]e
i
11-9
+ j ω( k −i ) Tb
11.1-18 (b) continued
let A = P1 P0* + P0 P0* and B = P1 P1* + P1P0* + P1* P0 + P0 P0* = P1 + P0 , so
2


K
K
K
K
2
1 − j ω(k −i) Tb 
1
1
1 − j ω(k −i)Tb 
1


E XT ( f ) = ∑  A + ∑ Be
=
A
−
B
+
Be



∑
∑

 k =− K 2
4
i =− K 4
k =− K  2
i =− K 4



i ≠k


1
1
1
1
2
where A − B = [ P1 P1* + P1 P0* + P1* P0 + P0 P0* = P1 − P0
2
4
4
4
K
K
2
2K + 1
2
1
2
thus, E[ X T ( f ) ] =
P1 − P0 + P1 + P0 ∑ ∑ e− j ω(k −i) Tb
4
4
k =− K i =− k
rb
rb2
2
Hence, Gx ( f ) = P1 ( f ) − P0 ( f ) +
4
4
K
(c)
K
∑∑
e − jω (k −i )TB = (2K + 1)
k=-K i =− K
so E[ X T ( f ) ] =
2
Gx ( f ) = lim
K →∞
∞
∑
P1 (nrb ) + P0 ( nrb ) δ( f − nrb )
2
n =−∞

n  − jωnTb
1 −
e
2K + 1 
n =−2 K 
2K
∑
2K + 1 
2
 P1 − P0 + P1 + P0
4 
2K

n =−2 K

2
n

∑  1 − 2 K + 1 e
− jω nTb




1
2
E[ X T ( f )
(2 K + 1)Tb
2K
1
2
1
2 
 2K − jω nTb
1

=
P1 − P0 +
P1 + P0 lim  ∑ e
−
n e − jωnTb  
∑
4Tb
4Tb
2 K + 1 n=−2 K

 K →∞  n=−2K
∞
1
2
1
2
=
P1 − P0 +
P1 + P0 ∑ e− jωnTb
4Tb
4Tb
n =−∞
where
1
= rb and
Tb
Hence, Gx ( f ) =
∞
∑
n =−∞
e − jω nTb =
1
Tb
∞

n =−∞

n

b 
∑ δ f − T
rb
r2
2
P1 ( f ) − P0 ( f ) + b
4
4
∞
∑
n =−∞
P1 ( nrb ) + P0 ( nrb ) δ( f − nrb )
11.2-1
 1
Pe = Q 
(S / N )R
 2
Polar: Pe = Q
(

−3
⇒ ( S / N ) R ≈ 2 x 3.12 = 19.2
 = 10

)
( S / N ) R = Q(4.38) ≈ 6 x 10-6
11-10
2
11.2-2
Pe = Q( A / 2σ) ≤ 10 −6 ⇒ A/2σ ≥ 4.76
Polar: 4.76 2 ≤ ( A / 2σ) 2 ≤ S R /( N0 rb /2) ⇒ S R ≥ 4.76 2 x
Unipolar: ( A / 2σ) 2 ≤
1
x 10-8 = 0.113 µW
2
1
S R /( N0 rb /2) ⇒ SR ≥ 0.226 µW
2
11.2-3
(a) From symmetry
Vopt = 0 ⇒ Pe = Pe0 =
∫
∞
0
pn ( y + A /2) dy =
1
2σ 2
∞
−
∫e
2 (y+ A / 2 )
0
1
⇒ A ≥ 8.8σ
-3
2 x 10
Gaussian noise: Pe = Q ( A / 2σ ) ≤ 10-3 ⇒ A ≥ 6.2σ
(b) Pe ≤ 10 −3 ⇒ A / 2σ 2
≥ ln
Note that impulse noise requires more signal power for same Pe
11.2-4
(a) From symmetry, Pe = Pe1
= P( y < 0 | ε = −∞ ) + P (ε = −∞ ) + P ( y < 0 | ε = ∞) + P(ε = ∞)
1  A − 2α  1  A + 2α 
= Q
+ Q
2  2σ  2  2σ 
pY ( y | H1 , ε = −α ) pY ( y | H1 , ε = 0)
pY ( y | H1 , ε = α )
11-11
dy =
1 −A
e
2
2σ 2
11.2-4 continued
(b) ( A ± 2α ) / 2σ = (1 ± 0.2)4,
1
1
Q(3.2) + Q(4.8)
2
2
1
= (7.4 x 10-4 + 8.5 x 10 -7 ) ≈ 3.7 x 10-4
2
If ε = 0, then Pe = Q(4.0) = 3.4 x 10-5
Pe =
11.2-5
(a) From symmetry, Pe = Pe1
=P( y < 0 | ε = − ∞ ) P(ε = − ∞ ) + P ( y < 0 | ε = 0) P(ε = 0)+P ( y < 0 | ε = α ) P(ε = α )
1  A − 2α  1  A  1  A + 2α 
= Q
+ Q
+ Q
4  2σ  2  2σ  4  2σ 
pY ( y | H1 , ε = −α ) pY ( y | H1 , ε = 0) pY ( y | H1 , ε = α )
(b) ( A ± 2α ) / 2σ = (1 ± 0.2)4,
1
1
1
Pe = Q(3.2) + Q(4) + Q(4.8)
4
2
2
1
1
1
= (7.4 x 10-4 + x 3.4 x 10 -5 + x 8.5 x 10 -7 ) ≈ 2 x 10-4
4
2
4
If ε = 0, then Pe = Q(4.0) = 3.4 x 10 -5
11-12
11.2-6
p y ( y | H 0 ) = pn ( y + A /2), p y ( y | H1 ) = pn ( y − A /2), pn (n)=
so P0
1
2πσ 2
2
e-(V + A / 2 )
P0 / P1 = e[(V + A / 2 )
2
/ 2σ
2
= P1
−(V − A /2)]2 / 2 σ 2
1
2
2πσ 2
= eVA / σ
e -(V − A / 2 ) / 2 σ
1
2πσ 2
e-n / 2 σ
2
2
2
Hence, VA / σ 2 = ln( P0 / P1 ) ⇒ Vopt =
σ2
ln( P0 / P1 )
A
11.2-7
dPe
dP
dP
= P0 e0 + P1 e1 = 0 when V = Vopt
dV
dV
dV
b (V )
dPe 0
d
=
g ( v , y ) dy where a(V ) = V , b(V ) = ∞ , g (V , y ) = p y ( y | H 0 )
dV
dV a(∫V )
= 0 − p y (V | H0 ) + 0 since
d b(V )
= 0,
dV
∂
[ p ( y | H 0 )] = 0
∂V y
dPe1
= 0 + p y (V | H1 ) + 0
dV
hence, − P0 p y (V | H 0 ) + P1 p y (V | H1 ) = 0 ⇒ P0 p y (V | H 0 ) = P1 p y (V | H1 )
similarly,
11.2-8
( S / N )1 = 20 dB=100
1
Regenerative: Pe ≈ 20Q  100  ≈ 20
e −100/2 ≈ 1.5 x 10 -22
2π 100
 1

Nonregenerative: Pe ≈ Q 
x 100  = Q(2.24) ≈ 1.2 x 10-2
 20

11.2-9
Regenerative: Pe = 50Q  ( S / N )1  = 10−4 ⇒


( S / N )1 ≈ 4.62
so ( S / N )1 = 21.3 = 13.3 dB
 1

Nonregenerative: Pe = Q 
( S / N )1  = 10−4 ⇒
 50

2
so ( S / N )1 = 50 x 3.73 = 696 = 28.4 dB
11-13
1
( S / N )1 ≈ 3.73
50
2
11.2-10
11.2-11
t
(a) 0 ≤ t ≤ Tb / 2 : Ap (t) * h (t ) = AK0 ∫ e− bλ d λ =
0
t
t ≥ Tb / 2 : Ap(t) * h(t ) = AK 0
∫
t −Tb / 2
e− bλ d λ =
AK0
A
(1 − e− bt ) =
(1 − e −bt )
b
(1 − e −bTb / 2 )
− b ( t− Tb / 2 )
AK 0 −b ( t− Tb /2)
[e
− e −bt ] = Ae
b
11-14
11.2-11 continued
K02 N0 ∞ −2bt
K 02 N0
∫−∞ h (t) dt = 2 ∫0 e dt = 4b
∞
A2 Tb
A2
Eb = ak2 ∫ p( t ) 2 dt =
=
−∞
2 2 4rb
N
(b) σ = 0
2
2
∞
2
2
4Eb rb
4br
4r
 A
− bT / 2 2
thus,   =
= 2b γb = b (1 − e b ) γb ≤ 0.812γ b if b ≥ 2.3rb
2
4( K 0 N0 / 4b)
K0
b
 2σ 
11.2-12
(a) Q( 2 γb ) = 10 −4 ⇒ γb =
(b) 2
γb =
 6x3
7
Q 
γb
8 x 3  63
1
x 3.72 , S R ≥ N0 rb γb = 34.2 pW
2

−4
⇒ Q( 0.286 γb ) = 1.7 x 10 -4
 = 10

1
x 3.62 , S R ≥ N 0 rb γ b = 227 pW
0.286
11.2-13
500 x 103
50
r=
≤ 2 B = 160 x 103 ⇒ log 2 M ≥
= 3.125 so M min = 2 4 = 16
log 2 M
16
2
 6x4
15
Q 
γ
16 x 4  255 b
so γ b =

−4
⇒ Q( 0.0941γb ) = 2.13 x 10 -4
 = 10

1
x 3.552 , and S R ≥ N0 rb γb = 670 pW
0.0941
11-15
11.2-14
With matched filtering; Pbe ≈ 2
 600log 2 M
M −1
Q
M log 2 M 
M2 −1



M −1
600log 2 M
Q
Pbe
M log 2 M
M2 −1
__________________________________________________________
M
2
4
0.75
8.94
<10-12
<10-12
8
0.583
5.35
4.5 x 10-8
2.6 x 10 -8 ⇒ take M = 8
16
0.469
3.07
>10 -3
>10-4
11.2-15
1
a =
M
2
k
2
2
2
2
2
  ( M − 1) A 
 −3 A   A   3 A 
 ( M − 1) A  
−
 + ... +  2  +  − 2  +  2  + ... + 
 
2
2
 

2
1 2
A
1 + 32 + ... + ( M − 1) 2   
M
2
2 M /2
2 M /2
A
A
=
(2i − 1) 2 =
(4i 2 − 4i + 1)
∑
∑
2M i=1
2 M i =1
=2x
=
A2
2M
( M /2)(1 + M /2) M 
A2
 ( M /2)(1 + M / 2)( M + 1)
2
4
−
4
+
=
(
M
−
1)

6
2
2 
12
11.2-16
Let m% = regenerated m and consider m = 00
P( m% = 01) = Q ( A / 2σ) − Q (3 A / 2σ) 1 bit error
P( m% = 11) = Q (3 A / 2σ) − Q (5 A / 2σ) 2 bit errors
P( m% = 10) = Q (5 A / 2σ)
1 bit error
Similarly when m = 10.
11-16
11.2-16 continued
Now consider m = 01, and similarly m = 11.
P( m% = 00) = Q( A / 2σ)
P( m% = 11) = Q( A / 2σ) − Q(3 A / 2σ)
1 bit error
P( m% = 10) = Q(3 A / 2σ)
2 bit errors
1 bit error
1
{[ Q(k ) − Q(3k )] + 2 [Q(3k ) − Q(5k )] + Q(5k)}
4
1
+ 2 x {Q( k ) + [Q( k ) − Q(3k ) ] + 2Q(3k )}
4
3
1
3
= Q(k ) + Q(3k ) − Q(5k ) ≈
Q(k ) when k > 1 since Q(5k ) << Q(3k ) << Q( k )
2
2
2
Thus, Pbe = 2 x
11.2-17
Let m% = regenerated m and consider m = 00 (similarly for m = 11)
P( m% = 01) = Q ( A / 2σ) − Q (3 A / 2σ) 1 bit error
P( m% = 10) = Q (3 A / 2σ) − Q (5 A / 2σ) 1 bit error
P( m% = 11) = Q (5 A / 2σ)
2 bit errors
11-17
11.2-17 continued
Now consider m = 01 (similarly m = 10)
P (m% = 00) = Q( A / 2σ)
1 bit error
P (m% = 10) = Q ( A / 2σ) − Q(3 A / 2σ ) 2 bit errors
P (m% = 11) = Q(3 A / 2σ)
1 bit error
1
{[Q( k ) − Q (3k ) ] + 2 [ Q(3k ) − Q(5k ) ] + 2Q (5k )}
4
1
+ 2 x {Q (k ) + 2 [Q (k ) − Q (3k )] + 2Q(3k )}
4
1
1
= 2Q( k ) − Q(3k ) + Q (5k ) ≈ 2Q( k ) when k > 1 since Q(5k ) << Q(3k ) << Q (k )
2
2
Thus, Pbe = 2 x
11.3-1
pβ ( t ) = sinc rt /2
rt
1
sinc rt ,
= D
2
r
No additional zero crossings.
p (t ) = sinc
11-18
11.3-2
pβ ( t ) = sinc (2rt /3)
2 rt
1
sinc rt ,
= D
3
r
Additional zero crossings at t = ±3D /2, ± 9 D / 2, ± 15D /2,...
p (t ) = sinc
11.3-3
Given B = 3 kHz
r
r r
+ β, 100% ⇒ β = r / 2 ⇒ B = + = r ⇒ r = 3 kbps.
2
2 2
r
r r 3
(b) 50% ⇒ β = ⇒ B = + = r ⇒ r = 4 kpbs.
4
2 4 4
r
r r 5
(c) 25% ⇒ β = ⇒ B = + = r ⇒ r = 4.8 kpbs
8
2 8 8
11.3-4
(a) B =
Figure P11.3-4 are baseband waveforms for 10110100 using Nyquist pulses with
β=r/2 (dotted plot), β=r/4 (solid plot). Note that the plot with β=r/2 is the same as the plot of
Figure 11.3-2.
In comparing the two waveforms, the signal with β=r/4 exhibits higher intersymbol
interference (ISI) than the signal with β=r/4.
11-19
11.3-4 continued
1.5
1
p(t)
0.5
0
-0.5
-1
-1.5
0
2
4
6
time, D-units
8
10
Figure P11.3-5
11.3-5
(a) Given B = r / 2 + β where β represents excess bandwidth.
With a data rate of r = 56 kpbs ⇒ theoretical bandwidth of B = r / 2 = 28 kHz.
But with 100 rolloff ⇒ 100% excess bandwidth ⇒ β = r /2 ⇒ B = r / 2 + r /2 = r = 56 kHz.
(b) 50% excessive bandwidth ⇒ β = r / 4 ⇒ B = r / 2 + r / 4 = 3 / 4 r ⇒ B = 42 kHz.
(c) 25% excessive bandwidth ⇒ β = r /8 ⇒ B = r / 2 + r /8 = 5 / 8 r ⇒ B = 35 kHz.
11.3-6
p (t ) =
cos πrt
1 − (2rt ) 2
1
sin2πrt
sin πrt
sinc 2rt
2
=
=
2
πrt
[1 − (2rt ) ]πrt 1-(2rt ) 2
At t = ± D /2 = ±1 / 2 r , sinc 2rt = 0, and 1 − (2rt ) 2 = 0
sin2πrt
So use L'Hopital's rule with p (t ) =
2πrt [1 − (2rt ) 2 ]
d
d
sin2πt = 2πr cos2πrt ,
{2πrt[1 − (2rt )2 ]} = 2π r − 24πr 3t 2
dt
dt
 1  2 π cos( ±π) 1
thus p  ±  =
=
2
 2 r  2πr − 6πr
11-20
11.3-7
Pβ ( f ) =
f <
π
πf  f 
cos Π   ,
4β
2β  2β 
r
− β,
2
P( f ) =
β
P( f ) =
r
r
− β < f < + β,
2
2
π
πλ 1
1 π/2
cos
d
λ
=
cos θ d θ = 1/ r
∫
2β r
2r −π∫/ 2
−β 4β
β
P( f ) =
π
πλ 1
1
cos
dλ=
∫
2β r
2r
f −r / 2 4β
π/ 2
π
2β
=
∫
cos θ d θ
( f −r /2)
 1 

1 
π
π
1 − sin ( f − r /2)  =
1 + cos ( f − r / 2 + β) 


2r 
2β
2β
 2r 

1
π
= cos 2
( f − r / 2 + β)
r
4β
f >
r
+ β,
2
P( f ) = 0
Since P( f ) has even symmetry,
1/ r

π
1
P( f ) =  cos 2
( f − r / 2 + β)
4β
r
0

β
f < r /2+β
r / 2 − β < f < r / 2+ β
f > r /2+β
β
π
πf
π
πf
pβ ( t ) = ℑ  Pβ ( f )  =
cos e jωt df =
cos cos2πft df
∫
∫
4β − β
2β
2β 0
2β
−1
=
π  sin(2 πrt − π / 2β)β sin(2πrt + π / 2β)β  1  − cos2πβt cos2πβ t 
+
+

= 

2β  2(2 πrt − π / 2β)β
2(2πrt + π / 2β)β  2  4βt − 1
4β t + 1 
Thus, p (t ) = pβ (t )sinc rt =
cos2πβt
sinc rt
1 − (4βt ) 2
11-21
11.3-8




1

(a) ℑ  p (t )∑ δ( t − kD)  = P ( f ) *  ∑e − j2 πfkD  = P( f ) *  ∑ δ( f − n / D) 

k

k

D n

= r ∑ P( f − nr ) = 1
n


ℑ  ∑ p (kD)δ(t − kD )  = ∑ p ( kD) e− j 2πfkD
 k
 k
Thus,
∑ p(kD) e
− j 2 πfkD
=1
k
(b)
1
for all f so p( kD ) = 
0
k =0
k≠0
1
=D
r
∞
Clearly,
∑ P( f − nr ) = D
n=-∞
11.3-9
(a) Conditions: P ( f ) = 1/ r
for r − B <
f < r − B and P( f ) + P ( f − r ) = 1/ r
f <B
11-22
11.3-9 continued
(b) Condition: P( f ) = 1/ r
f ≤ B = r / 2 so P( f ) =
1  f 
Π 
r r
(c) Can't satisfy the theorem when B < r /2.
11.3-10
We want smallest possible M to minimize ST
r=
rb
≤ 2B
log 2 M
⇒ log 2 M ≥
rb
= 1.5 ⇒ take M = 2 2 = 4
2B
r = rb / 2 = 300 kbaud, β = B − r / 2 = 50 kHz
Pbe = 2
3
 A
Q   ≤ 10−5
4 x 2  2σ 
A
 A 4
-5
⇒ Q
≥ 4.23
 ≤ x 10 ⇒
2σ
 2σ  3
2
6 x 2 SR
15
 A
so ST = LS R ≥ 4.232 LN 0 rb = 1.34 W
  =
15 N 0 rb
12
 2σ 
11-23
11.3-11
We want smallest possible M to minimize ST
r=
rb
≤ 2B
log 2 M
⇒ log 2 M ≥
rb
= 2.5 ⇒ take M = 23 = 8
2B
r = rb / 3 = 200 kbaud, β = B − r / 2 = 20 kHz
Pbe = 2
7
 A
Q   ≤ 10 −5
8 x 3  2σ 
A
 A  24
⇒ Q  ≤
x 10 -5 ⇒
≥ 4.15
2σ
 2σ  14
2
6 x 3 SR
63
 A
so ST = LSR ≥
4.152 LN0 rb = 3.61 W
  =
63 N0 rb
18
 2σ 
11.3-12
We want smallest possible M to minimize ST
r=
rb
≤ 2B
log 2 M
⇒ log 2 M ≥
rb
= 3.75 ⇒ take M = 24 = 16
2B
r = rb / 4 = 150 kbaud, β = B − r / 2 = 5 kHz
Pbe = 2
15
 A
Q   ≤ 10 −5
16 x 4  2σ 
A
 A  64
⇒ Q
x 10 -5 ⇒
≥ 4.10
≤
2σ
 2σ  30
2
6 x 4 SR
255
 A
so ST = LS R ≥
4.10 2 LN 0 rb = 10.7 W
  =
255 N 0 rb
24
 2σ 
11.3-13
(a) P( f ) = 1/ r cos 2 ( πf / 2r )Π ( f / 2r), Px ( f ) =
2
H R ( f ) = 500 g
1
sinc(f / 2r ), r = 20,000
2r
cos 2 ( πf / 2 r )
Π( f / 2r )
1 + 3 x 10-4 f
-4
2
80 (1 + 3 x 10 f )cos ( πf / 2 r )
HT ( f ) =
Π ( f / 2r )
g
sinc2 (f / 2r )
2
11-24
11.3-13 continued
 A
−6
(b) Pe = Q 
 = 10 ⇒
2
σ


∞
where I =
∫
P Gn
HC
−∞
2
A
3ST
 A
= 4.75, 
I −2
 =
2σ
2
σ
(4
−
1)
r

max
10−5
2  πf 
-4
∫ 0.01r cos  2r  (1 + 3 x 10 f ) df
−r
r
df =
10 −3
12 
 πf 

(1 + 3 x 10-4 f )cos 2   df = 10−3  4 − 2  ≈ 2.78 x 10-3
∫
r 0
π 
 2r 

r
=2
2
 A
Thus, ST ≥ rI   = 2 x 104 x (2.78 x 10-3 ) 2 x 4.752 ≈ 3.49 W
 2σ 
2
11.3-14
(a) P( f ) =
1
Π ( f / r ),
r
Px ( f ) =
1
f 
sinc 
,
10r
 10r 
r = 100
2
 f 
H R ( f ) = 10 6 g 1 + 32 x 10-4 f 2 Π  
r
2
100 1 + 32 x 10-4 f 2  f 
Π 
g sinc 2 ( f /10 r )
r 
1
2
≈
H R ( f ) since sinc2 ( f /10r ) ≈ 1 for f ≤ r / 2
4 2
10 g
HT ( f ) =
11-25
11.3-14 continued
 1  A 
(b) Pe = 2 1 −  Q   = 10−6 ⇒
 4   2σ 
where I =
∞
∫
−∞
P
Gn
HC
2 x 10-2
=
r
2
3ST
A
 A
= 4.85, 
=
I −2

2σ
 2σ max (16 − 1)r
10−5 1
df = ∫
1 + 32 x 10-4 f 2 df
-3
r 10
−r /2
r /2
∫
0
r /2

 32

100
1 + 32 x 10 -4 f 2 df ≈ 2 x 10-4  75 +
ln 
+ 3   ≈ 1.81 x 10-2
2 32  2
 

2
 A
Thus, ST ≥ 5rI 2   = 500(1.81 x 10-2 ) 2 (4.85) 2 = 3.85 W
 2σ 
11.3-15
(a)
K
HC ( f )
Px ( f )
Gx ( f )
ST
Px* ( f ) H c* ( f )e − jω td
+
SR
N0 / 2
P( f ) e− jωtd = KH C Px* H c* e− jωtd Px ⇒ P( f ) = K Px H C
2
since P( f ) is real, even, and non-negative, p( t ) is even an maximum at t = 0.
∞
p (0) =
∫ P( f )df
∞
=K
−∞
σ2 =
N0
2
∞
∫
∫
Px H C
−∞
2
Px H C df =
−∞
2
∞

2
df = 1 ⇒ K =  ∫ Px HC df 
 −∞

N0
2K
2
Gx ( f ) = σ2a r Px HC , σa2 = ( M 2 − 1) A2 /12
11-26
−1
P( f ) e − jωtd
σ2
11.3-15 continued
∞
M 2 −1 2 ∞
2
ST = K ∫ Gx ( f ) df = K
A r ∫ Px ( f ) df
12
−∞
−∞
2
2
2
1
 A
Thus,   =
4
 2σ 
12ST
∞
2K
3ST
−1
=
I HR
2
N0 ( M − 1)r
( M 2 − 1) K 2r ∫ Px ( f ) df
2
−∞
∞
where I HR =
∫
N0
∞
N0
2
K ∫ Px ( f ) df =
2 −∞
2
Px ( f ) df
−∞
∞
2 ∫ Px ( f ) H C ( f ) df
2
−∞
∞
(b) SR =
∞
∫
ST
=
SR
⇒
2
HC ( f ) K 2 Gx ( f ) df
−∞
∞
∫ Gx df
=
−∞
∞
∫
2
H C ( f ) Gx df
−∞
∞
∫
2
3
 A
Thus,   =
2
( M − 1) r
 2σ 
∞
∫
2 ∫ Px H C df
2
−∞
HC
−∞
SR
2
2
Px df
−∞
∞
∫
2
Px df
HC
2
2
Px df
−∞
∞
Px df
∫
N0
−∞
2
∞
∫
2
Px df
=
6S R
( M − 1) N 0 r
2
−∞
11.3-16
(a)
x (t )
HT
HC
ST
+
HR
H eq = 1/ L H C
SR
N0 / 2
HT
2
=
P
N0 L
2 g Px
2
HR
2
=
2L
gP
N0
∞
∞
M 2 −1 2
2
M 2 − 1 2 N0 L 1
ST =
A r ∫ HT Px df =
Ar
∫ P df where
12
12
2 g −∞
−∞
11-27
∞
∫
−∞
P df =1
y( t )
11.3-16 continued
N
σ = 0
2
2
∞
∫
H R H eq
−∞
2
2L g ∞ P
df
N0 L −∫∞ H C 2
N
df = 0
2
2
1 12 gST
 A
Thus,   =
4 ( M 2 − 1)r
 2σ 
2
2
N0L N0
N0 L  ∞ P
∫
2 L g  −∞
 HC
2

df 

−1
=
6ST / L
K ( M 2 − 1) N0 r
∞
P( f )
1
where K = ∫
df
L −∞ H C ( f ) 2
r
r
  2 f 2 
  2 f 2
1 1
2
2  πf 
2  πf 
(b) K = ∫ cos  L 1 + 
  df = ∫ cos   1 + 
 df
L −r r
r 0
 2 r    r  
 2r    r  
4
=
π
= 1+
π /2
∫
0
64
cos λd λ + 3
π
2
π /2
∫λ
2
cos 2 λd λ
0
64  π
π
−  = 1.52 = 1.83 dB
3 
π  48 8 
3
11.3-17
1.0 0.4 0.0  c−1   0
0.2 1.0 0.4  c  = 1  ⇒ c = −10/21,
−1

 0   
0.0 0.2 1.0  c1   0
1.0

10
25
5
 −0.19
peq (t k ) = −
p% k +1 +
p% k −
p% k −1 = 
21
21
21
 −0.15
0
c0 = 25/21, c1 = −5/21
k =0
k = −2
k=2
otherwise
11-28
11.3-18
1 ε 0
1 ε 0  c−1   0 
δ 1 ε  c  = 1  Let ∆ = δ 1 ε = 1 − 2εδ

 0   
0 δ 1  c1   0 
0 δ 1
0 ε 0
1 0 0
1
−ε
1
1
c−1 = 1 1 ε =
, c0 = δ 1 ε =
,
∆
1 − 2εδ
∆
1 − 2εδ
0 δ 1
0 0 1
1 ε 0
1
−δ
c1 = δ 1 1 =
∆
1 − 2εδ
0 δ 0
11.3-19
1 0.1 0.0 0.0 0.0   c−2   0
1.0 0.1 0.0 0.0 0.0   c   0

  −1   
-0.2 1.0 0.1 0.0 0.0   c0  = 1 

   
0.1 -0.2 1.0 0.1 0.0   c1   0
0.0 0.0 0.1 -0.2 1.0   c2   0
Solving yields c−2 = 0.0094, c− 1 = −0.0941, c0 = 0.9596, c1 = 0.2068, c2 = −0.0549
peq (t k ) = c−2 p% k+ 2 + c−1 p% k + 1 + c0 p% k + c1 p% k −1 + c2 p% k −2
0
 0.00094

0

1
=
0
 −0.0468

 −0.0055
0

k ≤ −4
k = −3
k = −2, −1
k =0
k = 1,2
k=3
k=4
k≥5
t k = k + 2D
worst-case ISI now occurs at k = 3 (t = 5D) but has not been reduced significantly in magnitude.
11-29
11.3-20
(a) h(t ) = sinc rb t + 2 sinc rb (t − Tb ) + sinc rb (t − 2Tb )
1 f 
1  f 
Π   1 + 2e − jω Tb + e − jω 2Tb = Π   ( 2 + cos ωTb ) e− jωTb
rb  rb 
rb  rb 
1 f  
πf 
4
πf  f  − jωTb
= Π   2 1 + cos2  e − jωTb = cos 2
Π  e
rb  rb  
rb 
rb
rb  rb 
H(f ) =
(
)
(b) a'k = ak + 2 ak −1 + ak− 2
1
1

=  mk' − + 2 mk' −1 − 1 + mk' −2 −  A = ( mk' + 2 mk' −1 + mk' − 2 − 2 ) A
2
2

'
'
'
'
mk
m k−1 m k − 2 ak / A
_______________________
0
0
0
-2
0
0
1
-1
0
1
0
0
0
1
1
1
1
0
0
-1
1
0
1
0
1
1
0
1
1
1
1
2
Thus,
2A

A

y (t k ) =  0

−A
 −2 A

mk' = mk' −1 = mk' −2 = 1
mk' −1 = 1, m'k ≠ mk' −2
mk' −1 ≠ mk' , mk' = m'k − 2
mk' −1 = 0, m'k ≠ mk' −2
mk = mk −1 = mk − 2 = 0
'
'
'
11-30
11.3-21
(a) h(t ) = sinc rbt + 2 sinc rb ( t − Tb ) + sinc rb (t − 2Tb )
1 f 
1  f 
Π   1 − 2e − jωTb + e− jω4Tb = Π   ( 2cos2ωTb − 2 ) e − jω 2Tb
rb  rb 
rb  rb 

1 f 
πf 
4
2πf  f  − j ω 2Tb
= Π   (−2)  1 − cos2  e− jω2Tb = − sin2
Π  e
rb  rb 
rb 
rb
rb

 rb 
(
H(f ) =
)
(b) a'k = ak − 2 ak− 2 + ak −4
1
1

=  mk' − − 2mk' −2 + 1 + mk' − 4 −  A = ( mk' − 2 mk' − 2 + mk' −4 ) A
2
2

'
'
'
'
mk
mk −1 mk −4 ak / A
_______________________
0
0
0
0
1
1
1
1
Thus,
0
0
1
1
0
0
1
1
2A

A

y (t k' ) = 0
−A

 −2 A

0
1
0
1
0
1
0
1
0
1
-2
-1
1
2
-1
0
mk' −2 = 0, mk' = m'k −4 = 1
mk' − 2 = 0, mk' ≠ m'k −4
mk' = mk' −2 = mk' −4
mk' −2 = 1, mk' ≠ mk' − 4
mk' − 2 = 1, mk' = m'k −4 = 0
11-31
11.3-22
1
L
HT
2
πf  f 
cos Π  
rb
rb  rb 
HT ( f ) H R ( f ) = H ( f ) =
2
ST =
=L
A2 rb
4
2
H
2
HR
, Px ( f ) = 1, Gn ( f ) = N 0 / 2
∞
∫
σ2 =
2
HT Px df
−∞
N0 ∞
2
H R df
∫
2 −∞
2
2
∞

A
2 ST  ∞ H
2

Thus,   =
∫
df
H
df
∫ R 
N0rb L  −∞ H R 2 −∞
 2σ 

2 ST
2SR
where
=
= 2 γb
N 0rb L N0 rb
∞
H
−∞
HR
∫
∞
2
2
df
∫
∞
2
H R df ≥
−∞
with equality when
2
So take H R ( f ) =
∞
Then, since
∫
−∞
∫
−∞
H
HR
−1
2
H
H R df
HR
= g HR
1
H( f )
g
4
H ( f ) df =
rb
2
and HT ( f ) = gL H ( f )
rb / 2
∫
0
cos
πf
4
df =
rb
π
2
2 γb
 A
 2σ  = (4/ π) 2
  max
11-32
11.2-22 continued
 A
Pe1 = 2Q 
 and
 2σ 
1  A 1  A
1
'
Pe 0 = Q 
+ Q
 since P ( ak = ± A | H 0 ) =
2  2σ  2  2σ 
2
Hence, Pe =
1
3  A  3 π

( Pe1 + Pe 0 ) = Q   = Q 
2 γb 
2
2  2σ  2  4

11.3-23
(a) Assume m−1 = mˆ −1 = 0.
Use y (t k ) = (mk + mk −1 − 1) A to calculate y (t k ) given mk , and mk −1
and to calculate mˆ k given y (t k )and mˆ −1 .
k
0
1
2
3
4
5
6
7
8
mk
1
0
1
0
1
1
1
0
1
mk −1
0
1
0
1
0
1
1
1
0
y (t k )
0
0
0
0
0
2
2
0
0
mˆ k −1
mˆ k
0
1
0
1
0
1
1
1
0
1
0
1
0
1
1
1
0
1
(b) dc value: y(tk ) = (2 + 2)/9 = 0.44
ˆ 2 is received in error such that
(c) As the table below indicates, if bit m
mˆ 2 = 0 ⇒ y (t 2 ) = −2 instead of 0.
Because mk = f ( mk −1 ) ⇒ errors in mˆ k =3 will affect all subsequent values of mˆ k as indicated
in t he table below.
k
y (t k )
mˆ k −1
mˆ k
0
0
1
0
2
−2
3
0
4
0
5
2
6
2
7
0
8
0
0
1
1
0
0
0
0
1
1
0
0
2
2
0
0
1
1
0
x
x
x
x
x
x
11-33
x = errors
11.3-24
(a) Use the precoder of Fig. 11.3-9 to convert mk → mk' , Eq. (23) for y (t k ), Eq. (24)
to determine mˆ k from the received value of y( tk ). Note that with precoding mˆ k is
not a function of mˆ 'k −1. Also, assume m'−1 = 0.
k
mk
0
1
1
0
2
1
3
0
4
1
5
1
6
1
7
0
8
1
m'k −1
0
1
1
0
0
1
0
1
1
m'k
1
y (t k ) 0
mˆ k
1
1
2
0
0
0
−2
1
0
0
0
1
0
1
2
0
0
0
1
0
1
1
1
0
1
(b) dc value: y (t k ) = (2 − 2 + 2 ) / 9 = 0.22
(c) If bit mˆ 2 is received in error, only that bit is affected since with
precoding mˆ k is not a function of mˆ k' −1 .
11.3-25
(a) Use the precoder of Fig. 11.3-9 to convert mk → mk' except use two stages of delay
⇒ mk' = mk ⊕ mk' −2 .
Then use Eq. (27) to determine y( tk ) from mk' and mk' − 2 and mˆk from y (tk ).
Assume m'−1 = m−' 2 = 0.
k
mk
0
1
1
0
2
1
3
0
4
1
5
1
6
1
7
0
8
1
m'k − 2
0
0
1
0
0
0
1
1
0
m
1
y (t k ) 2
mˆ k
1
0
0
0
−2
0
0
1
2
1
2
0
−2
1
0
1
2
0
1
0
1
1
1
0
1
'
k
(b) dc value: y(tk ) = (2 − 2 + 2 + 2 − 2 + 2 ) / 9 = 0.44
(c) If bit mˆ 2 is received in error, only that bit is affected since
we can obtain mˆ k directly from y (t k ).
11-34
11.4-1
(a) Using structure of Fig. 11.4-6a to scramble the input sequence we get:
mk
0
1
1
1
1
1
1
0
1
1
1
0
1
1
1
m'k −1 0
0
1
0
0
1
1
1 0
1
1
0
1
1
0
0
0
0
1
0
0
1
1
1
0
1
1
0
1
1
m'k −3 0
0
0
0
1
0
0
1
1
1
0
1
1
0
1
m'k − 4 0
0
0
0
0
1
0
0
1
1
1
0
1
1
0
m'k −5 0
0
0
0
0
0
1
0
0
1
1
1
0
1
1
m''k
0
0
1
1
0
0
0
0
0
0
1
1
0
1
1
'
0
1
0
0
1
1
1
0
1
1
0
1
1
0
0
m
'
k−2
mk
dc values of unscrambled and scrambled sequences:
mk = (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1)/15 = 12/15 = 0.80
m'k = (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1)/15 = 8/15 = 0.53
(b) Using the structure of Fig 11.4-6b to unscramble mk' we get:
m'k
0
1
0
0
1
1
1
0
1
1
0
1
1
0
0
m'k −1 0
0
1
0
0
1
1
1 0
1
1
0
1
1
0
m'k − 2 0
0
0
1
0
0
1
1
1
0
1
1
0
1
1
m'k −3 0
0
0
0
1
0
0
1
1
1
0
1
1
0
1
m'k − 4 0
0
0
0
0
1
0
0
1
1
1
0
1
1
0
m'k −5 0
0
0
0
0
0
1
0
0
1
1
1
0
1
1
m''k
0
0
1
1
0
0
0
0
0
0
1
1
0
1
1
mk
0
1
1
1
1
1
1
0
1
1
1
0
1
1
1
11.4-2
Using the results from Exercise 11.4-1, we get the output sequence and its shifted versions to
generate the following table used to calculate the autocorrelation function.
11-35
11.4-2 continued
τ original/shifted
0 1111100100110000101101010001110
1111100100110000101101010001110
1 1111100100110000101101010001110
0111110010011000010110101000111
2 1111100100110000101101010001110
1011111001001100001011010100011
3 1111100100110000101101010001110
1101111100100110000101101010001
4 1111100100110000101101010001110
1110111110010011000010110101000
5 1111100100110000101101010001110
0111011111001001100001011010100
6 1111100100110000101101010001110
0011101111100100110000101101010
7 1111100100110000101101010001110
0001110111110010011000010110101
8 1111100100110000101101010001110
1000111011111001001100001011010
9 1111100100110000101101010001110
0100011101111100100110000101101
10 1111100100110000101101010001110
1010001110111110010011000010110
11 1111100100110000101101010001110
0101000111011111001001100001011
12 1111100100110000101101010001110
1010100011101111100100110000101
13 1111100100110000101101010001110
1101010001110111110010011000010
14 1111100100110000101101010001110
0110101000111011111001001100001
v (τ) R (τ) = v( τ) / N
31 1
-1
-0.032
-1
-0.032
-1
-0.032
-1
-0.32
-1
-0.032
-1
-0.032
-1
-0.032
-1
-0.032
-1
-0.032
-1
-0.032
-1
-0.032
-1
-0.032
-1
-0.032
-1
-0.032
11-36
11.4-2 continued
τ original/shifted
v( τ )
R( τ) = v( τ) / N
15 1111100100110000101101010001110 -1
1011010100011101111100100110000
-0.032
16 1111100100110000101101010001110 -1
0101101010001110111110010011000
-0.032
17 1111100100110000101101010001110 -1
0010110101000111011111001001100
-0.032
28 1111100100110000101101010001110 -1
0001011010100011101111100100110
-0.032
19 1111100100110000101101010001110 -1
0000101101010001110111110010011
20 1111100100110000101101010001110 -1
-0.032
-0.032
1000010110101000111011111001001
21 1111100100110000101101010001110 -1
-0.032
1100001011010100011101111100100
22 1111100100110000101101010001110 -1
-0.032
0110000101101010001110111110010
23 1111100100110000101101010001110 -1
-0.032
0011000010110101000111011111001
24 1111100100110000101101010001110 -1
-0.032
1001100001011010100011101111100
25 1111100100110000101101010001110 -1
-0.032
0100110000101101010001110111110
26 1111100100110000101101010001110 -1
0010011000010110101000111011111
-0.032
27 1111100100110000101101010001110 -1
1001001100001011010100011101111
-0.032
28 1111100100110000101101010001110 -1
1100100110000101101010001110111
-0.032
29 1111100100110000101101010001110 -1
1110010011000010110101000111011
-0.032
30 1111100100110000101101010001110 -1
1111001001100001011010100011101
-0.032
31 1111100100110000101101010001110 31
1111100100110000101101010001110
1
11-37
11.4-2 continued
1.20
1.00
0.80
0.60
R ( τ)
0.40
0.20
0.00
-0.200.00
τ
10.00
20.00
30.00
40.00
R( τ) is periodic with period = 31.
Is the output a ml sequence? Apply ML rules: (1) #1s =16, #0s = 15⇒ obeys balance
property; (2) obeys the run property; (3) has a single autocorrelation peak; (4) obeys Mod-2
property, and (5) all 32 states exist during sequence generation.
11.4-3
m1 = m2 + m4 and the output sequence = m4
shift m1 m2m3m 4
_______________
0
1 1 1 1
1
0 1 1 1
2
0 0 1 1
3
1 0 0 1
4
1 1 0 0
5
1 1 1 0
6
1 1 1 1
11-38
11.4-3 continued
The autocorrelation function is calculated as follows:
τ original/shifted
0 1 1 1 1 0 0
v ( τ)
R (τ) = v( τ) / N
6
6/6=1
2
0.333
-2
-0.333
-2
-0.333
4 1 1 1 10 0
1 1 0 01 1
-2
-0.333
5 1 1 1 1 0 0
2
0.333
2
1.
1 1 1 1 0 0
1 1 1 1 1 0 0
01 1 1 1 0
2 1 1 1 1 0 0
0 0 1 1 1 1
3 1 1 1 1 0 0
1 0 01 1 1
1 1 1 0 0 1
6 1 1 1 1 0 0
1 1 1 1 0 0
R( τ) is periodic with period = 6.
1.50
R( τ)
1.00
0.50
0.00
0.00
-0.50
τ
2.00
4.00
6.00
8.00
The [4,2] register does not produce a ml sequence since there are only 5/16 possible states
exist in the output sequence and the period is not 24 -1.
11-39
Chapter 12
12.1-1
1

vf s ≤ BT 
2 BT
2 BT
= 3.33 ⇒ v = 3, f s ≤
= 33.3 kHz
2
 v≤
2
W
v
f s ≥ 2W 
q = M 3 ≥ 200 ⇒ M = 2n ≥ 2001 / 3 = 5.85 ⇒ n = 3
12.1-2
1

vf s ≤ BT 
2 BT
2BT
= 5.33 ⇒ v = 5, f s ≤
= 32 kHz
2
 v≤
2
W
v
f s ≥ 2W 
q = M 5 ≥ 200 ⇒ M = 2n ≥ 2001 / 5 = 2.88 ⇒ n = 2
12.1-3
Transmit one quantized pulse, having q N possible values, for every N successive
1
quantized samples. The output pulse rate is f s / N so BT ≥ ( f s / N ) ≥ W / N ,
2
allowing BT < W .
12.1-4
εk ≤ 1/ q  1/ q
P
≤

2
100
2 x( t ) = 2 
Thus, v ≥ log M (50/ P)
⇒ q=Mv ≥
50
P
12.1-5
( S / N ) D = 3q 2 x 0.25 ≥ 104 ⇒ q ≥ 116, v ≤ BT / W = 5.3, q = M v
M
v
q
comment
__________________________
2
5
32
q too small, try M > 2
3
5
243
ok
3
4
81
q too small
⇒ Take M = 3, v = 5, f s ≤ 2 BT / v = 6.4 kHz
12-1
12.1-6
( S / N ) D = 3q 2 x 0.25 ≥ 4000 ⇒ q ≥ 73, v ≤ BT / W = 6.7, q = M v
M
v
q
comment
__________________________
2
6
64
q too small, try M > 2
3
6
729
q excessively large
3
5
81
ok
⇒ Take M = 3, v = 5, f s ≤ 2 BT / v = 10 kHz
12.1-7
W = 5 kHz, ( S / N ) D = 40 − 50 dB = 10 − 10
4
5
(a) ( S / N ) D = 0.9q 2 ≥ 104 − 105 ⇒ q = 2v = 105 − 333 so v = 7 or 8
BT ≥ vW = 35 − 40 kHz
(b) v ≤ 4, q = M 4 = 105 − 333 ⇒ M ≥ 4
12.1-8
W = 20 kHz, ( S / N ) D = 55 − 65 dB = 3.2 x 105 − 3.2 x 106
(a) ( S / N ) D = 0.9q 2 ≥ 3.2 x 105 − 3.2 x 10 6 ⇒ q = 2 v = 596 − 1886 so v = 10 ⇒ BT ≥ vW = 200 kHz
(b) v ≤ 4, q = M 4 = 596 − 1886 ⇒ M ≥ 5
12.1-9
xmin = 5 x 10 V and x max = 200 x 10 V ⇒ normalize ⇒ xmin = 25 x 10 V and x max = 1 V
-6
-3
-6
Assume a sinusoidal input, the power of the smallest signal is ⇒ S x = (25 x 10 -6 )2 / 2 = 3.12 x 10 -10
Using Eq. (7) ⇒ 40 dB = 4.8 + 10l og(2 2v x 3.12 x 10-10 ) ⇒ v = 21.7 ⇒ v = 22.
12.1-10
Scale +/- 10V input by a factor of 10 to make input +/- 1V, then because its sinusoidal ⇒ S x = 0.5
q = 212 ⇒ quantum size = 2/ q = 2/4096 = 0.488 mV.
(S/N) D = 3q2 S x = 3(212 ) 2 x 0.5=2.52 x 107 ⇒ 74 dB.
12-2
12.1-11
Scale +/- 10V input by a factor of 10 to make input +/- 1V, then because its sinusoidal ⇒ S x = 0.5
q = 216 ⇒ quantum size = 2/ q = 2/65,536 = 30.5 uV.
(S/N) D = 3q2 S x = 3(216 ) 2 x 0.5=6.44 x 109 ⇒ 98 dB.
12.1-12
Let v max , v min = maximum and minimum input voltages
⇒ Dynamic range = 20log(v max /v min )
Assuming the largest signal = 1 volt ⇒ the smallest signal = 1/ q volts
⇒ Dynamic range = 20log(q )=20log(2 v )
Dynamic range = 20log(2v ) = 120 dB ⇒ v = 20 bits.
12.1-13
If ( S / N ) D = 35 dB and assuming Sx = 1
⇒ 35 = 4.8 + 6v ⇒ v = 5.03
Memory must hold:
10 min x 60 secs/min x 8000 samples/sec = 4.8 Msamples @ 5 bits/sample = 24 Mbits
12.1-14
(a) v = 12 bits ⇒ q = 212 =4096. With x max = 10 V ⇒ each step size = 10 x 2/ q = 4.88 x 10-3 V
Maximum input = x( kTs ) max = ( q − 1)/ q = 4095/4096 x 10 = 9.9976 V
For positive inputs from 0 to 10 V, xq ( kTs ) = 0.00244, 0.00732, 0.01221, 0.01709, 0.02197......
⇒ x(t ) = 0.02 V ⇒ xq ( kTs ) = 0.02197 ⇒ ε k = 0.02 − 0.02197 = 0.00197
Quantization error% = 0.00197/0.02 x 100% = 9.85%
(b) For x(t ) = 0.2 V ⇒ xq ( kTs) = 0.19775 ⇒ εk = 0.2 − 0.19775 = 0.00225
Quantization error% = 0.00225/0.2 x 100% =1.125%
12.1-15
Let v = 2 + n, n ≥ 1 then hi = 2 / q =
2
1
1 1
v= n x ≤
2
2 4
2
so p x (x) is constant over each band.
ai = xi − 1/ q , bi = xi + 1/ q
12-3
12.1-15 continued
n
2 1
=
⇒ n = q /8
q 4
q / 2 − n = 3q / 8
ε = px ( xi )
xi +1/ q
2
i
∫
( xi − x) 2 dx =
xi −1/ q
σq = 2
2
2
3q 3
2 px ( xi )
3q 3

1
q
 1
 n x 1 +  2 − n  x 3  = 3q2




 14

1
1  22
 2
2
S x = 2  ∫ x dx +∫ x x dx  =
= 0.229
3
96
1
0



4

2
Thus, ( S / N ) D = 3q x 0.229 ≈ 0.7 q2
12.1-16
hi = 2 / q, ai = xi − 1/ q , bi = xi + 1/ q, p x ( x ) ≈ px ( xi ) for ai < x < bi
Thus, εi2 ≈ px ( xi )
xi +1/ q
∫
( xi − x ) 2 dx =
xi −1/ q
so σ2q ≈
2 q/ 2
∑ px (xi )
3q 3 i =− q / 2
1
But
∫ px (x )dx =
-1
q /2
Hence,
2 px ( xi )
3q 3
∑
i = −q / 2
q /2
2
∑ q p x ( xi ) and
i =− q / 2
p x ( xi ) ≈
1
∫ p (x )dx = P  x < 1 ≈ 1
x
-1
q
2 q
1
so σq2 ≈ 3 = 2
2
3q 2 3q
12-4
12.1-17

1

 Sx ≤
1
2
2.6 ≈ 0.148
P  x > 1 = 2Q (1/ σ x ) ≤ 0.01 ⇒
≥ 2.6 
σx

Thus, ( S/N ) D ≤ 10log10 (3 x 22v x 0.148) = −3.5 + 6v dB
x = 0 ⇒ σx = Sx
12.1-18
∞
α
P  x > 1 = 2∫ e −αx dx = e − α ≤ 0.01 ⇒ α ≥ − ln0.01
2
1
so S x = 2/ α 2 ≤ 2 / ( ln100 ) = 0.0943
2
and ( S/N )D ≤ 10log10 (3 x 2 2v x 0.0943) = -5.5 + 6v dB
12.1-19
2
 z
(a) x =  2
- z
(b) z ' ( x ) =
z>0
z<0
⇒ x( z ) = (sgn z ) z
2
1 −3 / 2
x
,
2
1
1 / 4

1 1
K z = 2∫ 4 x3 px ( x )dx = 8  ∫ x3 dx + ∫ x 3dx  = 0.672
31 / 4
0
0

12.1-20
(a) z > 0 : ln(1 + µx) = z ln(1 + µ) = ln(1 + µ ) 2 ⇒ x = [(1 + µ) 2 − 1]/ µ
z < 0 : ln(1 − µx) = − z ln(1 + µ ) = ln(1 + µ) −2 ⇒ x = −[(1 + µ )−2 −1]/ µ
(1 + µ ) − 1
µ
z
Thus, x ( z ) = (sgn z )
12-5
12.1-20 continued
2
 ln(1 + µ)  1
(b) K z = 2 
(1 + µx) 2 p x ( x) dx

∫
 µ  0
1
1

ln 2 (1 + µ )  1
2
=2
p
(
x
)
dx
+
2
µ
xp
(
x
)
dx
+
µ
x2 p x ( x) dx 
∫ x
x
2
∫
∫
µ
0
0
0

where
1
1
0
0
∫ px (x )dx = 1 / 2 , 2∫ xpx ( x)dx =
1
∫ x 2 px (x )dx =
0
1
∫ x p ( x)dx =
x
x
−1
1
1
1
1
x 2 px ( x )dx = x 2 = S x
∫
2 −1
2
2
(
ln 2 (1 + µ )
Thus, K z =
1 + 2µ x + µ2 S x
2
µ
)
12.1-21
(a) With x(t ) = 0.02 V ⇒ with companding using Eq. (12) we have
 ln(1 + 255 x 0.02/9.9976) 
z ( x) = 9.9976 
 = 0.7432
ln(256)


z ( x) is fed to the quantizer giving z ( kTs) = 0.74463
Using Eq. (13) with xq ( kTs ) = z (kTs ) gives
xˆ =
9.9976
 (1 + 255) 0.74463/9.9976 − 1 = 0.02005 ⇒ ε k = 0.02 − 0.02005 ≈ 0 ⇒ 0% quantization error.
255 
(b) With x(t ) = 0.2V ⇒ with companding using Eq. (12) we have
 ln(1 + 255 x 0.2/9.9976) 
z ( x) = 9.9976 
 = 3.26059
ln(256)


z ( x) is fed to the quantizer giving z ( kTs ) = 3.25928
Using Eq. (13) with xq = z (kTs ) gives
xˆ =
9.9976
 (1 + 255) 3.25928/9.9976 − 1 = 0.19983 ⇒ εk = 0.2 − 0.19983 ≈ 0 ⇒ 0% quantization error.
255
12.1-22
(a) z ' ( x ) = 3e −3x , x > 0
1
α
α
α
1
K z = 2∫ e6 x e −αx dx =
e6− α − 1) ≈
(0 − 1) = for α ? 1
(
9
2
9(6 − α)
9( −α)
9
0
1
( S / N ) D ≈ 10log10 (9 x 3 x 22v S x ) = 14.3 + 6.0v + S x
12-6
dB
12.1-22 continued
(b)
( S / N ) D = 52.9 + S x − K z
α
Sx (dB)
dB
K z (dB)
( S / N )D
____________________________________
4
-9
1.5
42.4
8
-15
-4.2
42.1
16
-21
-7.5
39.4
? 1
-9.5
62.4 + S x
12.1-23
 A
1 + ln A
'
(a) z ( x ) = 
 A 1
1 + ln A x
0 ≤ x ≤ 1/ A
1/ A ≤ x ≤ 1
1/ A
1

1
2 
2
K z = (1 + ln A )  2 ∫ 2 p x ( x) dx + 2 ∫ x px ( x ) dx 
 0 A
1/ A

1/ A
 1

 1

=(1 + ln A )2 2 ∫ x 2 px ( x )dx + 2 ∫  2 − x 2  px ( x )dx 

 0
0  A

1
where 2 ∫ x 2 p x ( x) dx =
0
1
∫x
2
px ( x) dx = S x
−1
12-7
12.1-23 continued
1/ A
(b) 2 ∫
0
=
α
 1
2
 2 − x  p x ( x) dx = 2
A
A

1/ A
∫e
−α
dx − α
1/ A
0
∫
x 2e − αx dx
0
2  α  −α / A 1
2
2
+1 e
+ 2 − 2 , Sx = 2
2 
α A 
A
α
α
2  2  α
 −α / A 1 
Thus, K z = (1 + ln A)  2  + 1 e
+ 2
A 
α  A 
 1 + ln A 
=

 A 
(c)
2

A  A  − α / A   1 + ln A 
1 + 2  + 1  e
 ≈

αα 

  A 
2
( S / N ) D = 52.9 + S x − K z
α
S x (dB)
if α ? A
dB
K z (dB)
( S / N )D
____________________________________
4
-9
5.9
38
16
- 21
-6.1
38
64
-33
-17.9
37.8
? 100
-25
77.9 + S x
12.2-1
( S / N )D
= 3q 2 x 1/2 ≥ 4 x 103 ⇒ q > 51, v ≤ BT / W = 2.5
so q ≤ M > 51 ⇒ M > 7. Thus, take M = 8, v = 2, q = 64
S
γ = R ≥ γth = 6 ( BT / W ) ( M 2 − 1) = 945 ⇒ S R ≥ 945 N0 W = 56.7 mW
N0 W
2
12-8
12.2-2
( S / N )D
= 3q 2 x 1/2 ≥ 4 x 103 ⇒ q > 51, v ≤ BT / W = 3.33
so q ≤ M > 51 ⇒ M > 3.7 Thus, take M = 4, v = 3, q = 64
S
γ = R ≥ γth = 6 ( BT / W ) ( M 2 − 1) = 300 ⇒ SR ≥ 300 N0 W = 18 mW
N0 W
3
12.2-3
( S / N )D
= 3q 2 x 1/2 ≥ 4 x 103 ⇒ q > 51, v ≤ BT / W = 8.33
so q ≤ M > 51 ⇒ M > 1.2 Thus, take M = 2, v = 6, q = 64
S
γ = R ≥ γth = 6 ( BT / W ) ( M 2 − 1) = 150 ⇒ SR ≥ 150 N0W = 9 mW
N0 W
8
12.2-4
PCM: Pe = 20Q  ( S / N )1  ≤ 10−5 ⇒ ( S / N )1 ≥ 4.92
BT /W ≥ v = 8, γ = ( BT /W ) (S / N )1 ≥ 192 ≈ 22.8 dB
1
( S / N )1 = 37 dB ≈ 5000, γ = ( S / N )1 = 105 = 50 dB
20
PCM advantage: 50-22.8 = 27.2 dB
Analog: ( S / N ) R =
12.2-5
PCM: Pe = 100Q  ( S / N )1  ≤ 10 −5 ⇒ ( S / N )1 ≥ 5.22
BT /W ≥ v = 8, γ = ( BT /W ) (S / N )1 ≥ 216 = 23.4 dB
1
( S / N )1 = 37 dB ≈ 5000, γ = ( S / N )1 = 5 x 105 = 57 dB
100
PCM advantage: 57 - 23.4 = 33.6 dB
Analog: ( S / N ) R =
12-9
12.2-6
10log10 (1 + 4q 2 Pe ) = 1 dB ⇒ 1 + 4 q2 Pe =100.1 =1.259
so Pe = 0.259/4q 2 ≈ 1/15q 2
Pe = Q  ( S / N ) R  ≈
1
15 x 22v
W
1
γ ≤ γ ⇒ γth ≈ v ( S / N )R
BT
v
( S / N ) D = 4.8 + 6.0v dB
(S / N ) R =
v
Pe
(S / N )R
γ th (dB)
( S / N ) D , dB
________________________________________________
4 2.6 x 10-4
3.5
16.9
28.8
-6
4.8
22.7
52.8
-9
5.8
26.1
76.8
8 1.0 x 10
12 4.0 x 10
12.2-7
Errors in magnitude bits have same effect as before, and there are q / 2
equiprobable values for i. Thus
2
2


1  v=2  2 m 
1 q/ 2  2 
2
4  v− 2 m 2 q / 2 2
ε = ∑  2  +
2
i
−
1

=
4
+
(4
i
−
4
i
+
1)
)
∑
  (
∑
∑


2
v  m= 0  q  q / 2 i= 0  q 
q i= 0
 vq  m=0

2
m
4  4v −1 − 1 2  q /2( q / 2 + 1)( q + 1)
q /2( q / 2 + 1) q  
+ 4
−4
+   , 4v = q 2
2 
vq  3
q
6
2
2 
4  5q 2 2  5q 2 − 8 5
= 2
− =
≈
if 5q 2 ? 8
2
vq  12 3 
3vq
3v
=
12-10
12.2-8
γ ≥ γth ≈ 6b ( M 2 − 1) ⇒ M 2 ≤
γ
+ 1, v ≤ b
6b
v /2
Thus, qmax = M
vmax
max
 γ

=  +1
 6b

12.2-9
M b = 256,
γth = 6b( M 2 − 1)
M
b
γth
dB
___________________________
2
8
144
21.6
4
4
360
25.6
16
2
3060
34.9
256
1
3.93 x 105
55.9
12.3-1
2 πf m Am
fs
2π3
If W = 3 kHz and normalized input with Am = 1 ⇒ f m = 3 kHz ⇒ ∆ ≥
= 0.628
30
Using Eq. (5) and a sine wave ⇒ f s ∆ ≥ x& (t ) max = 2πf m Am ⇒ ∆ ≥
12-11
12.3-2
Using Eq. (5) and a sine wave ⇒ f s ∆ ≥ x& (t ) max = 2πf m Am ⇒ Am ≤
If W = 1 kHz and Nyquist sampling ⇒ f s = 20 kHz
⇒ Am ≤
20 x 103 x 0.117
= 0.372
2π x 1000
12.3-3
Note slope overload when ∆ =1.
12.3-4
DM: ( S / N ) D = 5.8b3 /(ln2b )2 = 7.6 + 10log10
PCM: ( S / N ) D = 3 x 2 2b x
b3
dB
(ln2b) 2
1
= 6.0b − 10 dB
30
b
DM
PCM
________________
4
19.3 14
8
25.8 38
16
32.9 86
12-12
fs ∆
2 πf m
12.3-4 continued
12.3-5
f s = 2Wb ≈ 8b kHz, σ = S x = 1 / 3
sopt =
∆ opt
f s ∆opt
≈ l n 2b
2πσWrms
1.3π l n 2b
≈
12 b
b
f s (kHz)
∆opt
________________________
4
32
0.177
8
64
0.118
16
128
0.0737
12.3-6
f s ∆ ≥ 2πf 0 ⇒ ∆ ≥ 2πf 0 / fs , f s = 2Wb
( S / N )D
2
3 fs
3 f s2
3 8W 3b 3
6 W 
= 2 Sx ≤ 2 2 Sx = 2 2
S x = 2   b 3 Sx
∆W
4 π f0 W
4π f 0 W
π  f0 
12-13
12.3-7
Sx =
∞
∫ G ( f )df
x
−∞
2
Wrms
=
=
1
Sx
∞
∫
W
= 2K ∫
0
df
2K
W
=
arctan
2
f + f
f0
f0
⇒ K=
2
0
f 0 Sx
2arctan(W / f 0 )
2K W
f2
df
Sx ∫0 f 02 + f 2
f 2 G x ( f ) df =
−∞
W
 2K
f 02
2K  W f 02 + f 2
df
−
df  =
∫ 2
2
∫
2
2
Sx  0 f 0 + f
0 f0 + f
 Sx

W
W − f 0 arctan 
f0 


f0
W  

Thus, Wrms = 
W − f 0 arctan  
f 0  
 arctan(W / f 0 ) 
when W = 4 kHz, f 0 = 0.8 kHz
1/2
= 1.3 kHz
12.3-8
∆ = 2πσWrms s / f s , b = f s / 2W , σ = S x
2
W ∆2
π2  W 
Ng =
= 3  rms  s 2 S x
fs 3
6b  W 
2
N g + Nso = K  s + a(3s + 1) e
2
−3 s
π2  W 
16
 where K = 3  rms  S x , a = b3
9
6b  W 
d
N g + N so ) = K  2s + 3ae−3s − 3a(3s + 1)e −3s  = 0 ⇒ 2 − 9ae −3s = 0
(
ds
1 9a 1
Thus, s opt = ln
= ln8b3 = l n 2b
3 2
3
12.3-9
ρ0 = Rx (0)/ S x = 1
n = 1 : ρ0 c1 = ρ1 ⇒ c1 = ρ1 , Gρ = [1 − ρ12 ] −1 = 2.77 = 4.4 dB
1
n = 2: 
 0.8
0.8  c1  0.8 
=
⇒ c1 = 8 / 9 , c2 = −1 / 9
1  c2  0.6 
-1
8
1
Gp = 1 − x 0.8 + x 0.6  = 2.81 = 4.5 dB
9
 9

12-14
12.3-10
ρ0 = Rx (0)/ S x = 1
n = 1 : ρ0 c1 = ρ1 ⇒ c1 = ρ1 , Gρ = [1 − ρ12] −1 = 10.26 = 10.1 dB
0.95  c1  0.95 
1
n = 2: 
  = 
 ⇒ c1 = 0.9744, c2 = −0.0256
 0.95 1  c2  0.90 
Gp = [1 − 0.9744 x 0.95 + 0.0256 x 0.9]-1 = 10.27 = 10.1 dB
12.3-11
x[( k − 1) Ts ] ≈ xq ( k − 1)
dx( t ) 1
dx (t )
≈ [ x( t ) − x (t − Ts ) ] ⇒ Ts
≈ xq (k − 1) − xq ( k − 2)
dt
Ts
dt ( k −1) Ts
Thus, take °
xq ( k ) = xq (k − 1) + [ xq ( k − 1) − xq ( k − 2)] = 2 xq ( k − 1) − xq ( k − 2)
so c1 = 2, c2 = −1
12.3-12
x%q ( k ) = cxq ( k − 1) ≈ cx( k − 1) so εq ( k ) ≈ x( k ) − cx( k − 1)
Then ε 2 = E[ x 2 ( k ) − 2 cx( k ) x (k − 1) + c 2 x 2 (k − 1)]
where E  x2 ( k )  = E  x2 ( k − 1)  = E  x2 (t )  = Sx
E [ x (k )x ( k − 1) ] = E [x (t )x (t − Ts ) ] = Rx (Ts )
so ε 2 = S x − 2cRx (Ts ) + c 2 S x = (1 + c 2 ) S x − 2cRx (Ts )
d ε2
= 2cS x − 2 Rx ( Ts ) = 0 ⇒ c = Rx (Ts ) / S x = ρ1
dc
12.3-13
x%q ( k ) = c1 xq ( k − 1) + c 2xq ( k − 2) ≈ c1 x( k − 1) + c2 x (k − 2)
so ε q ( k ) ≈ x ( k ) − c1 (k − 1) − c2 ( k − 2)
 x2 ( k ) + c12 x2 ( k − 1) + c22 x2 ( k − 2) − 2c1 x( k) x (k − 1) 
ε2 = E 

 −2c2 x( k) x ( k − 2) + 2c1 c2 x( k − 1) x( k − 2)

2
2
where E  x ( k − n)  = E  x (t )  = S x
E [ x (k − n) x( k − m) ] = E [x ( t − nTs ) x( t − mTs ) ] = Rx [(m − n)Ts ]
Hence, ε2 = (1 + c12 + c22 ) S x + 2c1 ( c2 − 1) Rx (Ts ) − 2c2 Rx (2Ts )
12-15
12.3-13 continued
We want
∂ ε 2 / ∂ c1 = 2c1S x + 2(c 2 − 1) Rx ( Ts ) = 0
∂ ε 2 / ∂c2 = 2c2 Sx + 2c1 Rx (Ts ) − 2Rx (2Ts ) = 0
so
Rx (Ts )
R (T ) 
c2 = x s 
Sx
Sx

 ⇒
Rx (Ts )
Rx (2Ts ) 
c1 + c2 =
Sx
S x 
c1 +
1 ρ1  c1  ρ1 
ρ 1  c  = ρ 
 1
 2   2
Same result as Eq. (16b) with n = 2 since ρ0 = Rx (0)/ S x = 1
12.4-1
Assume just music samples and no parity or control information
70 min/CD x 1.4112 Mbits/sec x 60 sec/min = 5.927 Gbits.
12.4-2
981 pages x 2 columns/page x 57 lines/column x 45 characters/line x 7 bits/character =35 Mbits.
Based on problem 12.4-1, a CD can store 5.9 Gbits=5900 Mbytes ⇒ 35/5900 x 100% = 0.59%
12.4-3
Assume with the 2 Gbyte hard drive there is no need to store extra control or parity bits.
Music ⇒ 1.4112 Mbits/sec x 1 byte/8bits
1 sec
x 8 bits/byte x 1 min/60secs x 2 x 109 bytes/hard drive = 189 minutes
1.4112 x 106 bits
If we do incorporate the same error control used on the CD, the recording time is:
1 sec
x 1 min/60 secs x 1 frame/561 bits x 8 bits/byte x 2 x 109 bytes/hard drive =65 minutes.
7350 frames
12-16
12.5-1
rb = vf s ≥ 12 x 2 x 15 kHz = 360 kbps
N≤
1.544 Mbps
= 4.2 ⇒ N = 4
rb
1

x 1.544 Mbps = 772 kHz
60
= 7.8%
2
 Eff =
772

Analog BT ≥ NW = 60 kHz

Digital BT ≥
12.5-2
rb = vf s ≥ 12 x 2 x 15 kHz = 360 kbps
N≤
2.048 Mbps
= 5.6 ⇒ N = 5
rb
1

x 2.048 Mbps = 1.024 MHz
75
= 7.3%
2
 Eff =
1024

Analog BT ≥ NW = 75 kHz

Digital BT ≥
12.5-3
From Fig. 12.5-8, if we subtract Transport and Path overhead,
a SONET frame has 9 rows x 86 bytes/row =774 bytes of user data.
Thus a STS-1 has a capacity of 774 bytes/frame x 8 bits/byte x 8000 bits/frame = 49.536 Mbps.
A DS0 line is 64 kpbs ⇒ and STS-1 can handle 49536/64 = 774 DS0 lines.
However, in practice, a VT is used to interface DS0 and DS1 lines to a STS-1.
The additional overhead of the VT reduces the number of DS0 inputs to 672 and the
number of DS1 inputs to 28. See Bellamy (1991) for more information.
12.5-4
(600 dots/inch) 2 x (8 inches x 11 inches)/page = 31,680 kbits/page.
2 BRI channel ⇒ 128 kpbs ⇒ 31,680 kbits/128 kbps = 247 seconds/page.
Obviously, some image compression is needed for this to be practical!
12.5-5
(600 dots/inch) 2 x (8 inches x 11 inches)/page = 31,680 kbits/page.
1-56 kbps channel ⇒ 31,680 kbits/56 kbps = 566 seconds/page.
12-17
Chapter 13
13.1-1
P( no errors ) = P( 0,4) = (1 − 0.1)4 = 0.6561
P( detected errors ) = P(1,4) = 4 × 0.1× 0.93 = 0.2916
P( undetedcte d errors ) = 1 − P( 0,4) − P(1,4) = 0.0523
13.1-2
P( no errors ) = P( 0,9) = (1 − 0.05 )9 = 0.5971
P( detected errors ) = P(1,9) = 9 × 0.05 × 0.95 8 = 0.2985
P( undetedcte d errors ) = 1 − P(0,9) − P(1,9) = 0.0712
13.1-3
(a) Two errors not in the same row or column yields 4
intersections as possible error locations. Two errors in
the same row (or column) yields two columns (or
rows) as possible error locations.
(b) L shaped error pattern yields no parity failures and is
undetectable. Other patterns yield 4 or 6 parity
failures and are detectable.
13.1-4

(31,26) t = 1 : Q


(31,21) t = 2 : Q


(31,16) t = 3 : Q

(
1/ 2
52   1
31
−4 

γ b  ≤  ×10  ⇒ γ b ≥ × 2.9 2 = 7.0dB
31   30
52

1/ 3
42   2
31
−4 

γb ≤ 
× 10  ⇒ γ b ≥
× 2.532 = 6.7 dB
31   30 × 29
42

1/ 4
32  
3× 2
31
−4 

γb ≤ 
×10  ⇒ γ b ≥
× 2.252 = 6.9dB
31   30 × 29 × 28
32

)
1
× 3.732 = 8.4dB
2
Thus, use (31, 21) code to save 1.7 dB.
Uncoded
: Q 2γ b ≤ 10 −4 ⇒ γ b ≥
13-1
13.1-5
1/ 2
 52   1
31
(31,26) t = 1 : Q
γ b  ≤  ×10 −6  ⇒ γ b ≥ × 3.62 = 8.9dB
52

 31   30
1/ 3
 42   2
31
(31,21) t = 2 : Q
γ b  ≤ 
×10− 6  ⇒ γ b ≥
× 3.02 = 8.2dB
31
30
×
29
42


 
1/ 4
 32  
3× 2
31
−6 
(31,16) t = 3 : Q
γ b  ≤ 
× 10  ⇒ γ b ≥ × 2.652 = 8.3dB
32

 31   30 × 29 × 28
1
Uncoded : Q 2γ b ≤ 10 −6 ⇒ γ b ≥ × 4.762 = 10.5dB
2
Thus, use (31, 21) code to save 2.3 dB.
(
)
13.1-6
(31,26) t = 1 :
(
)
 52 
Pube = Q 2γ b ,α = Q
γ b , Pbe = 30α 2
 31 
γb
2
5
10
dB
3
7
10
Pube
2.3×10-2
8.5×10-4
4×10-6
α
3.5×10-2
2×10-3
2.2×10-5
Pbe
3.7×10-2
1.2×10-4
1.5×10-8
13.1-7
(31,21) t = 2 :
(
)
 42 
30 × 29 3
Pube = Q 2γ b , α = Q
γ b , Pbe =
α
2
 31 
γb
2
5
10
DB
3
7
10
Pube
2.3×10-2
8.5×10-4
4×10-6
α
5×10-2
4.7×10-3
1.2×10-4
Pbe
5.4×10-2
4.5×10-5
7.5×10-10
13-2
13.1-8
Coded transmission has rb / r = Rc ' and Q( 2Rc 'γ b ) = α
(12,11) l = 1 :
1/ 2
1

α =  ×10 −5 
 11

(15,11) l = 2 :
= 9.5 × 10− 4 , Rc ' =
11
(1 − 12α ) = 0.906, γ b = 3.122 / 2 Rc ' = 7.3dB
12
1/3
 2

α =
×10 −5 
 14 ×13

(16,11) l = 3 :
= 4.8 ×10 −3 , Rc ' =
 3×2

α =
× 10−5 
 15 × 14 × 13

11
(1 − 15α ) = 0.681, γ b = 2.6 2 / 2Rc ' = 7.0dB
15
1/4
= 1.22 × 10−2 , Rc ' =
11
(1 − 16α ) = 0.554, γb = 2.252 / 2Rc ' = 6.6dB
16
Uncoded transmission rb / r = 1
Q
(
)
2γ b = 10 −5 ⇒ γ b ≈
1
× 4 .27 2 = 9.6 dB
2
13.1-9
Coded transmission has rb / r = Rc ' and Q( 2Rc 'γ b ) = α
(12,11) l = 1 :
1/ 2
1

α =  ×10− 6 
 11

(15,11) l = 2 :
= 3.0 ×10− 4 , Rc ' =
11
(1 − 12α ) = 0.913, γ b = 3.452 / 2 Rc ' = 8.1dB
12
1/ 3
 2

α =
×10 − 6 
 14 ×13

(16,12) l = 3 :
= 2.2 ×10 − 3 , Rc ' =
11
(1 − 15α ) = 0.709, γ b = 2.852 / 2 Rc ' = 7.6dB
15
11
 3× 2

α =
×10− 6  = 6.9 ×10 −3 , Rc ' = (1 − 16α ) = 0.612, γ b = 2.452 / 2 Rc ' = 6.9dB
16
 15 ×14 ×13

Uncoded transmission rb / r = 1
1/ 4
Q
(
)
2γ b = 10 −6 ⇒ γ b ≈
1
× 4 .77 2 = 10 .6dB
2
13-3
13.1-10
14 × 13 3
α ≈ 10 −6
2
Tw = n / r = 30 µs,

2t d
= 10
⇒ N ≥
5
td ≥ 45km/3 ×10 km/s = 150 µs 
Tw
11 1 − 0.035
p = nα ≈ 0.035, Rc ' =
= 0.538 ⇒ rb = 269kbps
151 + 9 × 0.035
α = Q ( 2 × 4 ) = 2 .3 × 10 −3 , l = 2 , Pbe =
13.1-11
15 × 14 × 13 4
α ≈ 10 − 8
3×2
Tw = n / r = 32 µs,

2t d
= 9.38 ⇒ N = 10
⇒ N ≥
5
td ≥ 45km/3 ×10 km/s = 150 µs 
Tw
11 1 − 0.037
p = nα ≈ 0.037, R c ' =
= 0.497 ⇒ rb = 248kbps
161 + 9 × 0.037
α = Q ( 2 × 4 ) = 2 .3 × 10 −3 , l = 3, Pbe =
13.1-12
Pbe = kα 2 ≤ 10−6 ⇒ α < 10−3 and p = (k + 1 )α<< 1 if k < 100
D ≥ 2 × 18km/3 ×10 5 km/s = 120 µ s
1.2
 ⇒ D / Tw ≥
k +1
Tw = ( k + 1)/ r = 100( k + 1) µ s 
Rc ' ≤
k
1− p
k
7200
×
≈
≥
⇒ k ≥ 5.66 ⇒ k = 6
k + 1 1 + D / Tw k + 1 + 1.2 10,000
Then,
1
Rc ' ≈ 0.732 , α = Q ( 2 Rc 'γ b ) ≤ ( × 10 −6 )1 / 2 = 4 .1 × 10 −4
6
So, γ b ≥ 3.35 2 / 2 Rc ' = 8.8dB
13.1-13
Pbe = kα 2 ≤ 10−6 ⇒ α < 10−3 and p = (k + 1 )α<< 1 if k < 100
D ≥ 2 × 60km/3 × 105 km/s = 400 µ s
4
 ⇒ D / Tw ≥
k +1
Tw = ( k + 1) / r = 100( k + 1) µs 
Rc ' ≤
k
1− p
k
7200
×
≈
≥
⇒ k ≥ 12.9 ⇒ k = 13
k + 1 1 + D / Tw k + 1 + 4 10,000
Then, Rc ' ≈ 0.737, α = Q( 2 Rc ' γ b ) ≤ (
1
× 10 −6 )1 / 2 = 2.8 × 10 −4
13
So, γ b ≥ 3.45 2 / 2 Rc ' = 9.1dB
13-4
13.1-14
m = 1(1 − p) + (1 + N ) p(1 − p) + (1 + 2 N ) p 2 (1 − p) + (1 + 3N ) p 3 (1 − p) + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
[
= (1 − p)[(1 + p + p
]
+ ⋅ ⋅ ⋅) ]
= (1 − p) 1 + (1 + N ) p + (1 + 2 N ) p 2 + (1 + 3N ) p 3 + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
+ p + ⋅ ⋅ ⋅) + Np(1 + 2 p + 3 p
But (1 + p + p + p + ⋅ ⋅ ⋅) = (1 − p) −1 ,
and (1 + 2 p + 3 p 2 + ⋅ ⋅ ⋅) = (1 − p) −2
2
2
3
2
3
 1
Np
+
2
1 − p (1 − p )
Thus, m = (1 − p )

Np
 =1+1− p

13.1-15
A word has a detected error and must be retransmitted if the number of bit error is
t < i ≤ l , so
l
 n  t +1
 n  t +1
p = ∑ P(i , n) ≈ P(t + 1, n) = 
α (1 − α ) n −t −1 ≈ 
α
i =t +1
 t + 1
 t + 1
With d min = 4,l = 2 and t = 1 so p ≈
n ( n − 1) 2
α
2
13.1-16
 n  t +1
 n  t +1
(a) p = P(t + 1, n) = 
α (1 − α ) n −t −1 ≈ 
α
 t + 1
 t + 1
n
 n  t+ 2
 n  t+ 2
Pwe = ∑ P(i, n) ≈ P(t + 2, n) = 
α (1 − α) n − t −2 ≈ 

α
i =t +2
t + 2
t + 2
t+2
t + 2 n(n − 1) ⋅⋅⋅ (n − t − 1) t +2  n − 1 t +2
P&be =
Pwe ≈
α =
α
n
n
(t + 2)!
 t +1 
(b) d min = 2 t + 2 = 8 ⇒ t = 3, R c ' ≈
12 1
= , assuming p<<1
24 2
23 × 22 × 21 × 20 5
α = 10 −5 ⇒ α = 1.62 × 10 −2
4 × 3× 2
24 × 23 × 22 × 21 4
p≈
α = 7.3 × 10− 4 << 1 as assumed
4× 3 × 2
Pbe ≈
1
α = Q ( 2 × γ b ) = 1.62 × 10 − 2 ⇒ γ b = 2 .15 2 = 4 .62
2
Uncoded: Pbe = Q( 2 γ b ) ≈ 1.3 × 10 −3
13-5
13.2-1
(a) Let nu = number of 1s in U, nv = number of 1s in V, and
nuv = number of 1s position in U and V. Then,
W (U ) + W (V ) = nu + nv
d (U , V ) = W (U + V ) = nu + nv − nuv ≤ W (U ) + W (V )
(b) U + V = X + Y + Y + Z
= ( x1 ⊕ y1 ⊕ y1 ⊕ z1 ,⋅ ⋅ ⋅), where y1 ⊕ y1 = 0 , etc.
= ( x1 ⊕ z1 ,⋅ ⋅ ⋅)
= X +Z
Thus,
d (U , V ) = W (U + V ) = W ( X + Z ) = d ( X , Z )
and W(U) + W(V) = W(X + Y) + W(Y + Z) = d(X,Y) + d(Y,Z)
so
d(X,Z) ≤ W(U) + W(V) = d(X,Y) + d(Y,Z)
13.2-2
d ( X , Y ) = i ≤ l,
d ( X , Y ) + d (Y , Z ) ≥ d ( X , Z ) ≥ d min ≥ l + 1
so,
l + 1 ≤ d min ≤ i + d (Y , Z ) ≤ l + d (Y , Z ) ⇒ d (Y , Z ) ≥ 1
Thus, Y cannot be a vector in the code, and the errors are detectable.
13.2-3
d ( X , Y ) = i ≤ t,
d ( X , Y ) + d (Y , Z ) ≥ d ( X , Z ) ≥ d min ≥ 2t + 1
so,
2t + 1 ≤ d min ≤ i + d (Y , Z ) ≤ t + d (Y , Z ) and d (Y , Z ) ≥ t + 1 > d ( X , Y )
Thus, Y is closer to X than to any other valid code vector, and the errors can be corrected.
13.2-4
n = 3, k = 1, q = 2
1 1
T
H = 1 0, S = EH T ⇒
0 1
13-6
13.2-4 continued
13.2-5
1
0

1
T
H = 
1
0

0
0
1
1
,
0
0
1
1
1
0
0
1
0
S = EH T ⇒
13.2-6
1
0

H T = 1

1
0
13.2-7
0
0

0

0
1

P = 1
1

1
1

1
1

1
1
0,

0
1
S = EH T ⇒
1
0
1
1
0
1
1
1
0
1
1
1
0
0
1
c1 = m5 ⊕ m6 ⊕ m7 ⊕ m8 ⊕ m9 ⊕ m10 ⊕ m11
0
1
0 ,

c2 = m2 ⊕ m3 ⊕ m4 ⊕ m8 ⊕ m9 ⊕ m10 ⊕ m11
0
1
1
c3 = m1 ⊕ m3 ⊕ m4 ⊕ m6 ⊕m7 ⊕m10 ⊕ m11
1
0
0
1
0
1
1
1
0
1
1
1 



c4 = m1 ⊕ m2 ⊕ m4 ⊕ m5 ⊕ m7 ⊕ m9 ⊕ m11


13-7
13.2-8
(a)
z q − 1 ≥ n

q = n −k 
zn
≥ z k = 64
n +1
⇒ n ≥ 10 (by trial and error)
Take n = 10, so q = 4 and ( q + n ) × z q = 224 bits
(b) Any 6×4 matrix whose rows are all different and contain at least 2 ones.
Example:
0 0 1 1
0 1 0 1


0 1 1 0
P=

1 0 0 1
1 0 1 0


1 1 0 0
13.2-9
zq − 1 ≥ n
(a)

q = n −k 
zn
≥ z k = 256
n +1
⇒ n ≥ 12 (by trial and error)
Take n = 12, so q = 4 and ( q + n ) × z q = 256 bits
(b) Any 8×4 matrix whose rows are all different and contain at least 2 ones.
Example:
0 0 1 1
0 1 0 1


0 1 1 0


0 1 1 1
P=
1 0 0 1


1 0 1 0
1 0 1 1 


1 1 0 0
13.2-10
n
a ij = ∑ g im hmj T (mod 2), where
m =1
g im = m th element of i th row of G
 0
 1

=
 0
 pi ( m −k )
1 ≤ m ≤ i −1
m=i
i +1 ≤ m ≤ k
k +1 ≤ m ≤ n
hmj T = mth element of j th row of HT
 pmj
 0

=
1
 0
13-8
1≤ m ≤n −k
n − k + 1 ≤ m≤ j + k − 1
m = j +k
j + k + 1≤ m ≤ n
13.2-10 continued
Thus, a ij = 1 ⋅ pij ⊕ pi ( j + k − k ) ⋅1 = pij ⊕ pij = 0
13.2-11
(a) The binary number
location,
i.e.,
000 ⇒ no error
s1 s 2 s3 equals the error
001 ⇒ 1st bit
010 ⇒ 2
etc.
nd
bit
(b)
s1 =
x4 ⊕ x5 ⊕ x 6 ⊕ x 7 = 0
s2 =
x 2 ⊕ x3
⊕ x6 ⊕ x7 = 0
s 3 = x1
⊕ x3
⊕ x5
⊕ x7 = 0
Since x1 , x 2 , and x 4 appear only once, they must be the check bits.
Thus,
X = (c1 c 2 m1 c3 m2 m3 m4 )
m2 ⊕ m3 ⊕ m4
c3 =

⇒ c2 = m1
⊕ m3 ⊕ m4
c = m ⊕ m
⊕ m4
 1
1
2
13.2-12
(a)
(b) Note from check-bit equations in Example 13.2-1 that
c1 ⊕ c2 ⊕ c3 = m2 , so m1 ⊕ m 2 ⊕ m3 ⊕ m 4 ⊕ c1 ⊕ c2 ⊕ c3 = m1 ⊕ m3 ⊕ m4
Thus, c 4 = m1 ⊕ m 3 ⊕ m 4
n
d = 0 ⇒ no errors or even number of errors
(c) Form d = ∑ y i (mod 2) so 
i =1
 d = 1 ⇒ odd number of errors
13-9
13.2-12 continued
If S = (000) and d = 0, assume no errors so X = Y ,
∧
If S ≠ (000) and d = 1, assume single error so X = Y + E,
If S ≠ (000) and d = 0, assume detectedbut uncor rectable errors.
13.2-13
G
111
}
11101 1010000
11101
10010
11101
11110
11101
0011
⇒ QM ( p ) = p 2 + p + 1, C ( p) = 0 + 0 + p + 1 ⇒ X = (101 M 0011)
X ' = (0 1 0 0 1 1 1)
Y ( p) = 0 + p5 + 0 + 0 + p 2 + p + 1
011
11101 0100111
11101
11101
11101
0000
⇒ S ( p) = 0
13.2-14
G
100
67
8
10111 1010000
10111
1100
⇒ QM ( p ) = p 2 + 0 + 0, C ( p) = p 3 + p 2 + 0 + 0 ⇒ X = (101 M1100)
X ' = (0 1 1 1 0 0 1)
Y( p) = 0 + p5 + p4 + p3 + 0 + 0 + 1
13-10
13.2-14 continued
011
10111 0111001
10111
10111
10111
0000
⇒ S ( p) = 0
13.2-15
G
1011
}
1011 1000000 = p 6
1011
1100
1011
1110
1011
101 = R1
101
1011 0100000 = p 5
1011
1100
1011
111 = R2
Similarly, R3 = 110 and R4 = 011
 R1  1 0 1
 R  1 1 1
 , which is the P matrix in Example 13.2-1.
P =  2 = 
 R3  1 1 0
  

 R4  0 1 1
13-11
13.2-16
G
1110
}
1101 1000000 = p 6
1101
1010
1101
1110
1101
110 = R1
0111
1101 0100000 = p 5
1101
1010
1101
1110
1101
011 = R2
Similarly, R3 = 111 and R4 = 101
 R1  1 1 0
 R  0 1 1
 , same rows as P matrix in Example 13.2-1, but different order.
P= 2=
 R3  1 1 1
  

 R4  1 0 1
13.2-17
p 3 M ( p ) = 1100000
13-12
13.2-18
Initialize register to (00……0), close
feedback, open output switch, and shift y into
register. After 7 shift cycles, open feedback
switch and close output shift, shift out S(p)
from register in 3 shift cycles.
Input bit
y
r01 =
y ⊕ r2
Register before shift
r0
r1
r2
r1' =
r0 ⊕ r2
r2' =
r1
____________________________________________________________
1
0
0
0
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
1
1
1
0
1
1
1
1
0
1
1
1
0
1
1
0
1
0
0
0
0
0
0
0 S ( p) = 0 + 0 + 0
13.2-19
Pbe = 1 x 10-5 = Q( 2 γ b ) and d min = 3
From Table T.6 ⇒ 2 γb =4.3 ⇒ γb =9.2
(a) For (7,4) code ⇒ n = 7, k = 4, Rc = 4/7, t = 1
( ) = i !( nn−! i)! ⇒ ( ) = 1!(66!− 1)! = 6
n
i
n-1
t
t +1
Pbe = ( n-1
= 6[Q( 8/7 x 9.68)]2 = 2.2 x 10-6
t ) [Q ( 2 Rc γ b )]
α = Q( 2Rc γ b ) = Q( 8/7 x 9.2) = 6 x 10 -4
(b) (15,11) code ⇒ d min = 3 ≥ 2t + 1 ⇒ t = 1 and Rc = 11/15
( ) = 1!(1414!− 1)! = 14 ⇒ P
n-1
t
be
= 14Q( 22/15 x 9.2)]2 = 3.2 x 10-7
α = Q( 2Rc γ b ) = Q( 22/15 x 9.68) = 1.5 x 10-4
13-13
13.2-19 continued
(c) (31,26) code ⇒ Rc = 26/31, t = 1
( ) = 1!(3030!− 1)! = 30 ⇒ P
n-1
t
be
= 30Q( 52/31 x 9.2)] 2 = 6.1 x 10 -8
α = Q( 2Rc γ b ) = Q( 52/31 x 9.2) = 4.5 x 10-5
13.2-20
With γc = Rc γb , α =Q( 2γc ) =Q( 2Rc γb ) ⇒ if γc fixed ⇒ Rc γb fixed ⇒ α fixed
Let's assume α=1 x 10 -2
With a (7,4) code and t = 1 ⇒ Pbe = 6Q 2 = 6 x 10-4
The percent reduncancy of the (7,4) code is 3/7 x 100%=43%
With a (31,26) code and t = 1 ⇒ Pbe = 30Q2 = 3 x 10-3
The percent reduncancy of the (31,26) code is 5/31 x 100%=16%
⇒ Pbe (7,4) < Pbe (31,26)
Because the (31,26) code has less redundancy than the (7,4) code, we woul d therefore
expect Pbe(7,4) < Pbe (31,26) .
13.2-21
Uncoded system: Pube = 1 x 10-5 = Q ( 2 γb ) ⇒ γb = 9.2
If we transmit at 3 times data rate ⇒ Rc = 1/3 ⇒ Pe = Q( 2 x (1/3) x 9.2) = 7 x 10 -3
3
An error occurs if 2 or 3 transmissions are received in error ⇒ Pbe = ∑ (3j ) p j (1 − p )3 − j
j =2
with p = 7 x 10-3
( 3j ) =
3!
⇒ ( 3j ) j = 2 = 3 and ( 3j ) j =3 = 1
j !(3 − j )!
⇒ Pe = 3(7 x 10-3 ) 2 (1 − 7 x 10−3 )1 + (7 x 10-3 ) 3 (1 − 7 x 10 -3 ) 0
= 1.46 x 10-4 = Pe (triple reduncancy)
versus the case from problem 13.2-19 where Pe (7,4) = 2.2 x 10-6 and Pe (15,11) = 3.2 x 10-7
13-14
13.2-22
101 
111 


110 
1110100 


Y = [0100101] and with a (7,4) code H= 0111010 ⇒ S = YH T = [0100101]  011 = [ 010]
100 
1101001 


010 
001 


Using Table 13.2-2 S = [010] ⇒ error in 6th bit ⇒ X= [ 0100111]
101 
111 


110 


(b) Y= [0111111] ⇒ S = YH T = [0111111]  011 = [101] ⇒ X = [1111111]
100 


 010
 001


101 
111 


110 


(c) Y= [1010111] ⇒ S = YH T = [1010111]  011 = [100] ⇒ X = [1010011]
100 


 010
 001


101 
111 


110 


(d) Y= [1101000] ⇒ S = YH T = [1101000] 011  = [ 001] ⇒ X = [1101001]
100 


 010
 001


13-15
13.2-23
G( p) = p12 + p 11 + p 3 + p 2 + p + 1
ASCII character "E " = 1000101 ⇒ M ( p) = p6 + p2 + 1 ⇒ p12 M ( p) = p18 + p14 + p12
 p q M( p) 
C ( p ) = rem 
⇒
 G ( p) 
p 6 + p 5 + p4 + p3 + 1
p12 + p 11 + p 3 + p 2 + p + 1 p 18 + p14 + p12
p18 + p 17 + p 9 + p 8 + p 7 + p 6
_____________________________
p17 + p 14 + p 12 + p 9 + p 8 + p 7 + p 6
p17 + p16 +
+p8 + p7 + p 6 + p 5
______________________________________
p16 + p14 + p12 + p 9 +
p5
p16 + p15 +
p7 + p6 + p5 + p4
_______________________________________
p15 + p 14 + p 12 + p 9 + p 7 + p6 +
p4
p15 + p14 +
p6 + p5 + p4 + p3
________________________________________________
p12 + p 9 + p 7 + p 5 + p 3
p12 + p11 +
p3 + p 2 + p +1
__________________________________________________
p11 + p 9 + p 7 + p 5 + p 2 + p + 1 = C( p )
X ( p) = p12 M ( p) + C ( p) = p18 + p14 + p 12+ p 11+ p 9+ p 7 + p 5 + p 2 + p + 1
X = [1000101101010100111]
13.2-24
X is transmitted ⇒ X ⇒ Y .
 Y ( p) 
rem 
 = 0 ⇒ no errors
G ( p ) 
From problem 13.2-23, errors in received vector ⇒ Y=1000101101010100100
⇒ Y ( p) = p18 + p 14 + p 12+ p 11+ p 9+ p 7 + p 5 + p 2
13-16
13.2-24 continued
Y ( p) / G( p) =
p12 + p 11 + p 3 + p 2 + p + 1
p6 + p5 + p4 + p3 + 1
p18 + p 14 + p12 + p 11 + p 7 + p 5 + p 2
p18 + p 17 + p 9 + p 8 + p 7 + p 6
_____________________________
p17 + p 14 + p 12 + p 11+ p 8+ p 6 + p 5 + p 2
p17 + p16 +
+p8 + p7 + p 6 + p 5
______________________________________
p16 + p 14 + p12 + + p11 + p 7 +
p2
p16 + p15 +
p7 + p6 + p5 + p4
_______________________________________
p15 + p14 + p12 + p11 + p 6 + p 5 + p 4 + p 2
p15 + p14 +
p6 + p5 + p4 + p3
________________________________________________
p12 + p 11 + p 3 + p 2
p12 + p11 +
p 3 + p2 + p + 1
__________________________________________________
p +1 ⇒
⇒ remainder ≠ 0 ⇒ an error has occured.
13.2-25
9600 bps ⇒ 1/9600 ⇒ 0.1 ms/bit ⇒ 125 ms noise burst ⇒ 125 ms x 10 bit/ms=1250 bits in error
for each burst.
A (63,45) code can correct for 3 errors ⇒ distribute 1200 errors over 400 blocks ⇒ 3 errors/block.
We int erleave the bits so we have 63+3=66 bits between errors, ⇒ each block=66 bits long
latency delay between interleaving and deinterleaving ⇒
2 x (400 blocks x 66 bits/block x 0.1 ms/bit)=2 x 2.64 seconds = 5.28 seconds
13-17
13.3-1
13.3-2
13.3-3
(a)
Input
1
0
1
1
0
0
1
1
1
1
State
a
b
c
b
d
c
a
b
d
d
d
Output
111
011
000
100
100
111
111
100
011
011
(b)
Minimum-weight paths: abce = D8 I 1 and abcde = D8 I 2
Thus, d f = 8, M (d f ) = 1 + 2 = 3
13-18
13.3-4
(a)
Input
State
Output
1
a
0
b
11
1
c
00
1
f
10
0
d
11
0
g
10
1
e
10
1
b
00
d
00
(b)
Minimum-weight paths
abdgei = D6 I2
abdce = D8 I2
Thus,
df = 6, M(df) = 2
13.3-5
M = 110101110111000000 ⇒ 1 + D + D3 + D5 + D6 + D7 + D9 + D10 + D11
G1 = 1 + D + D3 and G2 = D + D 2 and G3 = 1
X ''j = G1 M = (1 + D + D3 )(1 + D + D3 + D5 + D6 + D7 + D9 + D10 + D11 )
= 1+D2 + D 5 + D 6 + D 10 + D 13 + D 14
X 'j = G2 M = ( D + D2 )(1 + D + D3 + D5 + D6 + D7 + D9 + D10 + D11 )
= D + D3 + D4 + D5 + D6 + D9 + D10 + D13
X '''j = M = (1 + D + D3 + D5 + D6 + D7 + D9 + D10 + D11 )
j = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
x j = 011 101 010 101 100 111 111 001 000 101 111 001 000 110 010 000 000
13-19
1
1
h
01
h
10
13.3-6
Redrawing Figure 13.3-5 to eliminate the input distributor, we have the following
equivalent convolutional encoder:
m j−3
m j −2
m j −1
mj
Inpu
x 'j
Outpu
x ''j
x '''j
With M = 110101110111000000 ⇒ 1 + D + D3 + D5 + D6 + D7 + D9 + D10 + D11
and G1 = 1 + D 2 + D 3 , G2 = 1 + D + D 3 and G3 = 1 + D 2
X 'j = G1 M = (1 + D2 + D3 )(1 + D + D3 + D5 + D6 + D7 + D9 + D10 + D11 )
= 1 + D + +D 2 + D 3 + D4 + D8 + D9 + D12 + D14 + 0
Because we eliminated the input distributor ⇒ we will partition the output bits in groups of 2
and select the second bit for the ouput ⇒
11 11 10 00 11 00 10 10
⇒ x'j =
1
1 0
0
0
0
0
0
X ''j = G2 M = (1 + D + D3 )(1 + D + D3 + D5 + D6 + D7 + D9 + D10 + D11 )
= 1 + D5 + D6 + D10 + D12 + D13 + D14
⇒
10 00 01 10 00 10 11 10
⇒ x''j =
0
0
1
0
0
0
1
0
13-20
13.3-6 continued
X '''j = G3 M = (1 + D2 )(1 + D + D3 + D5 + D6 + D7 + D9 + D10 + D11 )
= 1 + D + D2 + D6 + D8 + D10 + D12 + D 13
⇒
11 10 00 10 10 10 11 00
⇒ x'''j =
1
0
0
0
0
0
1
0
Interleaving x'j , x''j and x'''j we get
Input message of
⇒ xj =
11 01 01 11 01 11 00 00
101 100 010 000 100 000 011 000
13.3-7
Redrawing Figure 13.3-5 to eliminate the input distributor, we have the following
equivalent convolutional encoder:
m j− 3
m j −2
Input
m j −1
mj
x 'j
Output
x ''j
x '''j
13-21
13.3-7 continued
3
5
6
7
9
10
11
With M = 110101110111000000 ⇒ 1 + D + D + D + D + D + D + D + D
and G1 = 1, G2 = 1 + D2 and G3 = D2 + D3
X j = G1 M = (1)(1 + D + D + D + D + D + D + D + D )
'
3
5
6
7
9
10
11
= 1 + D + +D 3 +D5 + D6 + D7 + D9 + D10 + D11 + 0 + 0
Because we eliminated the input distributor ⇒ we will partition the output bits in groups of 2
and select the second bit for the ouput
⇒
11 01 01 11 01 11 00 00
⇒ x'j =
1
1
1
1
1
1
0
0
X ''j = G2 M = (1 + D2 )(1 + D + D3 + D5 + D6 + D7 + D9 + D10 + D11 )
= 1 + D + D2 + D6 + D8 + D10 + D12 + D13
⇒
11 10 00 10 10 10 10 11
⇒ x''j =
1
0
0
0
0
0
0
1
X '''j = G3 M = (D 2 + D3 )(1 + D + D3 + D5 + D6 + D7 + D9 + D10 + D11 )
= D2 + D4 + D5 + D6 + + D7 + D10 + D11 + D12 + D14 + 0
⇒
00 10 11 11 00 11 10 10
⇒ x'''j =
0
0
1
1
0
1
0
0
Interleaving x'j , x''j and x'''j we get
Input mess age of
⇒ xj =
11 01 01 11 01 11 00 00
110 100 101 101 100 101 010 000
13.3-8
(a) The state transition diagram does not have a transition with a nonzero input that has a
zero output ⇒ noncatastrophic.
Alternatively, factoring G2 ( D ) = 1 + D2 = (1 + D)(1 + D) and dividing
G1 ( D) = D2 + D + 1 by (D + 1) we get
13-22
13.3-8 continued
D
D +1 D + D +1
2
D2 + D
________
1
⇒ there are no common factors to G1 ( D) and G2 ( D) ⇒ noncatastrophic
(b) The state transition diagram of Fig 13.3-5 does not have a transition with a nonzero input that has a zero
output ⇒ noncatastrophic.
Alternatively, with G1 ( D) = D 3 + D2 + 1, G2 ( D) = D 3 + D + 1, and G3 ( D) = D 2 + 1 = ( D + 1)(D + 1)
Dividing G1 ( D) and G2 ( D) by D + 1 we get
D2
D2 + D
D + 1 D3 + D2 + 1 and D + 1 D3 + D + 1
D3 + D2
________
1
D3 + D2
_____________
D2 + D + 1
D2 + D
________
1
⇒ there are no common factors to G1 ( D), G2 (D) and G3 ( D) ⇒ noncatastrophic
(c) G1 ( D) = D + D 2 = D( D + 1) and G2 ( D) = D 2 + 1 = ( D + 1)( D + 1)
⇒ ( D + 1) is common to both G1 ( D) and G2 ( D) ⇒ catastrophic
13-23
13.3-9
First, we consider W b → W c section where
1
T1 =
,
1 − DI
D2 I
T2 = DI × T1 × D =
,
1 − DI
D
T3 = D + T2 =
1 − DI
Then,
T4 =
T3
D
,
=
1 − IT3 1 − 2 DI
and T ( D,I ) = D 2 I × T4 × D 2 =
D5I
1 − 2 DI
13.3-10
∂ T ( D, I )
=
∂I
Thus,
∞
∑
∞
∑ A(d , i ) D d i I i−1 ⇒
d =d f i =1
[
]
d
1 ∞
M ( d ) 2 α (1 − α )
∑
k d =d
1 
d
d
≤ M ( d f )2 [α (1 − α ) ]
k 
Pbe ≤
∂T ( D , I )
∂I
=
I =1
∞
∑
d =d f
∞
where M (d ) = ∑ iA(d , i )
i =1
f
f
f
/2
+
∞
∑ M (d )2 d [α (1 − α )]
d = d f +1
If α << 1 , then 1 − α ≈ 1 and M ( d )2d [ α (1 − α )]
for d ≥ d f + 1, so
Pbe ≈
1
d
d
M (d f ) 2 α
K
f
f
∞
 d
∑ iA( d , i )  D
 i =1

/2
13.3-11
(a)
13-24
df /2
d /2



<< M ( d f )2
df
[ α (1 − α ) ]d / 2
13.3-11 continued
Minimum-weight paths: abc = D3 I 1 so, d f = 3, M ( d f ) = 1
(b)
T ( D, I ) = D 2 I ×
1
D3 I
×D=
1 − DI
1 − DI
(c)
∂ T ( D, I )
D3
=
∂I
(1 − DI ) 2
Eq.(9) yields
1 2 3 [α (1 − α )]
Pbe ≤ ×
1 1 − 2 α (1 − α )
3/ 2
[
]
2
≈ 8α 3 / 2 , α << 1
Eq.(10) yields
Pbe ≈ 2 3 α 3 / 2 = 8α 3 / 2
13.3-12
(a)
Minimum-weight paths: abce = D 6 I 1 so, d f = 6, M (d f ) = 1
13-25
13.3-12 continued
(b)
T1 =
1
1− D 2 I
T2 = D 2 I × T1 × D =
T3 = D + T2 =
D
1− D 2 I
T ( D, I ) = D 3 I ×
(c)
∂ T ( D, I )
D6
=
∂I
(1 − 2 D 2 I ) 2
Eq.(9) yields
1 4 3 α 3 (1 − α 3 )
Pbe ≤ ×
≈ 64α 3 ,α << 1
2
1 [1 − 8α (1 − α )]
Eq.(10) yields
Pbe ≈ 2 6 α 6 / 2 = 64α 3
13.3-13
13-26
D3I
1 − D2I
T3
D6I
× D2 =
1 − DIT3
1 − 2 D2 I
13.3-14
13.3-15
From g 0 = g 3 = g 4 = 1 and g 1 = g 2 = 0 , We can get:
s j = e' j − 4 ⊕ e' j −3 ⊕ e' j ⊕ e'' j
Thus, s j −4 , s j −1 and s j are orthogonal on e' j-4
13.3-16
From g 0 = g 1 = g 4 = 1 and g 2 = g 3 = 0 , We can get:
s j = e' j −4 ⊕ e' j −1 ⊕ e' j ⊕ e'' j
Thus, s j −4 , s j −3 and s j are orthogonal on e' j-4
13-27
13.3-16 continued
13.3-17
(a) (37,21) ⇒ (358 , 218 ) ⇒ (11 1012 , 10 0012 ) where the 1s and 0s define the feedback
and output connections respectively.
(b) (34,23) ⇒ (348 , 238 ) ⇒ (11 1002 , 10 0112 )
The corresponding RSC block diagrams are shown as follows:
xk
+
D
D
D
D
yk
+
(35,21) RSC encoder
xk
+
D
D
D
(34,23) RSC encoder
13-28
D
+
yk
13.4-1
(a) p = 5, q = 11 ⇒ pq = 55 and φ( n)=4 x 10 = 40=2 x 2 x 2 x 5 ⇒ pick e = 7 (since 7 is relatively
prime to 40.
7d = 40Q + 1 ⇒ if Q = 4 ⇒ 7d = 160 + 1 ⇒ d = 23
y = xe mod n and x = y d mod n
x = 8 ⇒ y = 87 mod55 = 2
To check ⇒ x = 223 mod55 = 8
x = 27 ⇒ y = 277 mod55 = 3
To check ⇒ x = 323 mod55 = 27
x = 51 ⇒ y = 517 mod55 = 6
To check ⇒ x = 6 23 mod55 = 51
(b) p = 11, q = 37 ⇒ pq = 407 and φ( n)=10 x 36 = 360=2 x 2 x 2x 3 x 3 x 5 ⇒ pick e = 7
(since 7 is relatively prime to 360
7d = 360Q +1 ⇒ if Q = 2 ⇒ 7 d = 720 + 1⇒ d = 103
y = xe mod n and x = y d mod n
x = 8 ⇒ y = 87 mod407 = 288
To check ⇒ x = 288103 mod407 = 8
x = 27 ⇒ y = 277 mod407 = 212
To check ⇒ x = 212103 mod407 = 27
x = 51 ⇒ y = 517 mod407 = 171
To check ⇒ x = 17123 mod407 = 51
(c) p = 13, q = 37 ⇒ pq = 481 and φ( n)=12 x 36 = 432=2 4 x 33 ⇒ pick e =5
(since 5 is relatively prime to 432)
5d = 432Q + 1 ⇒ if Q = 2 ⇒ 5d = 864 + 1⇒ d = 173
y = x e mod n and x = y d mod n
x = 8 ⇒ y = 85 mod481 = 60
To check ⇒ x = 60173 mod481 = 8
x = 27 ⇒ y = 275 mod481 = 196
To check ⇒ x = 196173 mod481 = 27
x = 51 ⇒ y = 515 mod481 = 103
To check ⇒ x = 103173 mod481
= {[((1032 mod481)6 mod481)14 mod481][1035 mod481]mod481} = 51
13-29
13.4-2
p = 11, q=31 ⇒ n=341 and φ( n) = 10 x 30 = 300 = 2 2 x3x5
⇒ select d to be any prime number greater than 5
de = Q φ(n ) + 1 ⇒ if e = 7 and Q = 1 ⇒ d = 43
There are as many values of e as there are prime numbers greater than 5.
if e = 7 and Q=8 ⇒ de = Q φ ( n) + 1 ⇒ d = 343
⇒ pattern is such that for a given value of e and d ,
successive values of d can be found by adding φ( n) to previous values of d
It is also observed that the values of p and q must be such that the product of p, q
must be greater than the number you are trying to encrypt.
13-30
Chapter 14
14.1-1
∞
xc2 = ∫ Gc ( f ) df =
−∞
=
A
2 ∫ Glp (λ) df
4 −∞
Ac2
2
∫ [G
lp
( f − fc) +Glp ( f + fc )]df
−∞
∞
2
c
since Glp ( f − f c) and Glp ( f + f c ) don't overlap if f c >> r
 M 2 −1 ∞
( M − 1) 2
2

∫−∞ sinc ( f / r )df + 4
 12r
Ac2
=
2
=
2 ∞
Ac
4
∞

δ
(
f
)
df

∫
−∞

2
 M 2 −1
( M − 1)2  Ac
r
+
=
( M − 1)(2 M − 1)


4
 12r
 12
Pc = 2 x
Ac2 ( M − 1) 2
,
4
4
Pc
2
c
x
=
3M − 3 1/2
=
4M − 2 3/4
M =2
M ? 1
14.1-2
(a) xi (t ) = ∑ ak p(t ) where p(t)= u(t)-u(t-Tb / 2 ), where xq (t ) = 0
k
ak = 0, 1 ⇒ ma = 1/2, σa2 = 1 / 4
2
T 
 fT
| P( f ) |2 =  b  sinc 2  b
 2
 2
⇒
( ma rb )
2
| P( nrb ) |2
Glp ( f ) = Gi ( f ) =
1
16rb

1
2 f 
 = 4r 2 sinc  2r 

b
 b
1/16
n=0

= 0
n = ±2, ± 4,...

2
1/ ( 2πn ) n odd
f
1
1
sinc 2
+ δ( f ) + ∑
δ( f − nrb )
2
2rb 16
± n odd ( 2 πn )
1 Ac2
1 Ac2
(b) x =
from waveform, Pc = 2 x
from spectrum, Pc / xc2 = 1 / 4
4 2
16 4
2
c
14-1
14.1-3
ma = 1/2,
Glp ( f ) =
σ2a = 1/4,
r = rb
rb
r2
| P( f ) |2 + b
4
4
∑| P (nr ) |
2
b
δ( f − nrb ),
n

τ
f =0

P( f ) =  τ /2
f = ±1 / 2T

m
0
f = ± , m ≥0
2τ

(a) 2τ = Tb = 1/ rb
Glp ( f ) =
1 sinc 2 ( f / rb ) 1
1
+ δ( f ) + [ δ( f − rb ) + δ( f + rb ) ]
2
16rb 1 − ( f / rb )
16
64
(b) 2τ = 2Tb = 2 / rb
Glp ( f ) =
1 sinc 2 (2 f / rb ) 1
+ δ( f )
4rb 1 − (2 f / rb ) 2 4
14-2
14.1-4
(a) For kD < t < (k + 1 )D, xi (t) = a2k and xq (t) = a2k +1 with ak = ±1
Thus xi2 ( t ) + xq2 ( t ) = 2 for all t so A(t) = 2 Ac for all t
 (a ) 
whereas φ(t ) = ∑ arctan  2 k +1  pD (t − kD)
k
 a2k 
(b) For kD < t < (k + 1 )D, x(t)
= a 2 k p(t-kD)
i
and x q (t ) = a 2k +1 p(t-kD), ak = ±1
Thus xi2 (t ) + xq2 ( t ) = 2 p 2 (t-kD) so A( t ) = 2 Ac ∑ p( t − kD)
k
 xq (t ) 
 a2k +1 
and φ (t ) = ∑ arctan 
 pD (t − kD)
 p D (t − kD) = ∑ arctan 
k
k
 xi (t ) 
 a2 k 
14.1-5
ma = ak =0, σ2a =ak2 = (m2 -1)/12 from Eq. (21), Sect 11.2, with A = 1
P(f) =
1
 πf   f 
cos 2   Π   so P(f) = 0 for | f |≥ r
r
 2 r   2r 
Glp (f) = G(f)
=
i
M 2 −1
 πf   f 
cos 4   Π  
12r
 2r   2r 
Ac2
 G ( f − f c ) + Gl p ( f + fc ) 
Gc ( f ) =| H VSB ( f ) |
4  lp
f > f c + βv
1

where | HVSB ( f ) |= 1/2
f = fc
0
0 < f < f c − βv

2
14-3
14.1-6
(a)
Since xq changes from ± 1 to m 1 while xi stays constant at ± 1,
and vice versa, it follows that ∆ φ = ± π / 2
(b) a2k and a2 k +1 , are independent sequences with ma = 0,
σ2a = 1, r = rb /2.
Time delay of xq (t) affects only the phase of P ( f ), so
Gq ( f ) = Gi ( f ) with | P( f ) | =| PD ( f )|where D = 2Tb
Hence, Glp (f) = 2 x r| PD ( f ) |2 =
4
sinc2 (2 f / rb ), same as QPSK.
rb
14.1-7
Ak
0
0
0
0
1
1
1
1
Bk
0
0
1
1
0
0
1
1
Ck
0
1
0
1
0
1
0
1
Ik
α
β
-β
-α
α
β
-β
-α
Qk
β
α
α
β
-β
-α
-α
-β
14-4
14.1-7 continued
Let I ko = (1 − Ck )α + Ck β and Qko = Ck α + (1 − Ck )β
and construct reduced table
Ak
Bk
Ik
Qk
0
0
1
0
1
0
Ik 0
-I k 0
Ik 0
Qko
Qko
-Qko
1
1
-I k 0
-Qko
Thus, I k = (1 − 2Bk )I ko and Qk = (1 − 2Ak )Qko
where (1 − 2 AK ) = ak , (1 − 2Bk ) = bk , and Ck = (1 − ck )/2, so
 bk α ck = +1
β = 
ck = −1
 bk β
1 + ck  ak β ck = +1
1 − ck
Qk = ak 
α+
β = 
2
 2
 ak α ck = −1
1 − ck
1 + ck
I k = bk 
α+
2
 2
14.1-8
xc (t ) = Ac [x 1 (t ) + x0 (t )] where, with ak = 0,1


x1 ( t ) =  ∑ ak pTb (t − kTb )  cos( ω1 t + θ1 ),
 k



x0 (t ) =  ∑ (1 − ak ) pTb (t − kTb )  cos(ω0 t + θ0 )
 k

14-5
14.1-8 continued
Then, from Eq. (7) with M = 2 and r = rb ,
1
f 1
G1lp ( f ) = G 0lp ( f ) =
sinc 2 + δ( f )
4rb
rb 4
Ac2
 G1lp ( f − f1 ) + G0lp ( f − f 0 ) + G1lp ( f + f1 ) +G 0lp (f + f 0 )
4 
Since f c ? rb , for f > 0 we have
and Gc ( f ) =
Gc ( f ) =
Ac2
16
1
f − f0

1
2 f − f1
+ si nc2
+ δ( f − f1 ) + δ( f − f 0 ) 
 r sinc
rb
rb
rb
 b

14.1-9
 r
π   t − Tb 
p (t ) = cos  2 π b t −  Π 
 so modulation theorem yields
2   Tb 
 2
P(t) =
=
Tb
2

rb  − jπ ( f −rb / 2 )Tb − jπ / 2
r 



e
+ sinc  f + b  Tb e − jπ (f + rb / 2 )Tb e+ jπ / 2 
 sinc  f − 2  Tb e
2





 ( f − ( rb /2)) 
 ( f + (rb /2))   − jπf Tb
1 
sinc 
 + sinc 
 e
2rb 
rb
rb




 ( f − ( rb /2)) 
 ( f + ( rb /2))  
1 
Thus, | P( f ) | = 2 sinc 
 + sinc 

4 rb 
rb
rb




 f 1
 πf π 
1
2 cos( πf / rb )
But sinc  ±  =
sin 
± =±
π (2 f / rb ) 2 ± 1
 rb 2  π  f ± 1   rb 2 


 rb 2 
2
2
2
1  2   −(cos( πf / rb ) (cos( πf / rb ) 
4
+

 = 2 2

2 
4rb  π   (2 f / rb ) − 1
(2 f / rb ) + 1 
π rb
2
so | P( f ) | =
2
14-6
 (cos( πf / rb ) 


2
 (2 f / rb ) − 1 
2
14.1-10
xc (t ) = A ∑ [cos( ωd ak t )cos( ωc t + θ ) − sin( ωd ak t )sin( ωc t + θ )] pT b (t − kTb )
c k
with ak = ±1 and ωd = πN / Tb = πNrb
xi (t ) = ∑ cos( ωd ak t) pTb (t − kTb ) = ∑ cos ωd tpTb ( t − kTb )
k
k
Nr
=cos ωd t ∑ pTb (t − kT ) = cos ωd t = cos2π b t for all t
b
2
k
1
Thus, Gi ( f ) = [δ ( f − Nrb /2) + δ( f + Nrb /2) ]
4
xq ( t ) = ∑ sin( ωd ak t) pTb (t − kTb ) = ∑ ak sin( ωd akt pTb (t − kTb ) = ∑ a k sin ωd t pTb (t − kTb )
k
k
k
 πN

where sin ωd ak t = sin 
(t − kTb ) + πNk 
 Tb

 πN

Nk
= cosπNk sin 
(t − kTb )  = ( −1) sin [ πNrb (t − kTb )]
 Tb

so xq ( t ) = ∑ Qk p(t − kTb ) with Qk = ( −1)
Nk
and
k
 Nr
p (t ) = sin  2π b
2


t  [u (t ) − u( t − Tb ) ]

Qk = ( −1) Nk ak = 0 and Qk2 = ak2 = 1 so Gq ( f ) = rb P ( f )
2
  Nr
π   t − Tb / 2 
P( f ) = ℑ[ p (t )] = ℑ cos  2π b t −  Π 

2
2   Tb

 
T 
Nr 
Nr 



= b sinc  f - b  Tb e− jπ( f − Nrb / 2 )Tb e− jπ / 2 + sinc  f + b  Tb e − jπ ( f + Nrb /2) Tb e + jπ / 2 
2
2 
2 



But e
− j π( f m rb /2) Tb
e m jπ / 2 = e− jπfTb e ± j( N −1) π / 2 = ( ± j ) N −1 e − jπfTb
 f - Nrb / 2 
 f + Nrb / 2 
1 N −1
N −1
Thus Gq ( f )
j sinc 
 + ( − j ) sinc 

4r
rb
rb




b
2
14.1-11
f − rb
f + rb 
1
1 
N = 2: Gi ( f ) = [ δ( f − rb ) + δ( f + rb ) ] , G q ( f ) =
− sinc
sinc

4
4rb 
rb
rb 
14-7
2
14.1-11 continued
f − 3rb / 2
f + 3rb / 2 
1 
3 
3 
1 

N = 3: Gi ( f ) = δ  f − rb  + δ  f + rb   , Gq ( f ) =
+sinc
 sinc

4 
2 
2 
4rb 
rb
rb


14.1-12
(a)
 π

(b) For 2kTb < t < ( 2 k + 1 )Tb , xi ( t ) = a2k cos 
(t − 2kTb )  and
 2Tb

 π
 π

π
xq ( t ) = a2k +1 = cos 
(t − 2kTb ) −  = a2k +1 sin 
(t − 2kTb )
2
 2Tb
 2Tb

 π

 π

so xi2 (t ) + xq2 (t ) = a22k cos2 
(t − 2kTb )  + a22k +1 sin 2 
(t − 2kTb ) = 1 since ak2 = 1
 2Tb

 2Tb

Thus, A(t ) = Ac for all t
14-8
2
14.1-12 continued
(c) Let xi (t ) =
∑I
k
p' (t − kTb ) and xq (t) =
k even
∑Q
k
p' ( t − kTb )
k odd
where I k = ak for k even and Qk = ak for k odd so
I k2 = Q2k = 1 and I k = Qk = 0
 π
  t − kTb − Tb 
p ' (t − kTb ) = cos 
(t − 2kTb )  Π 

2Tb
 2Tb
 

p ' (t ) = cos ( πrb t / 2 ) [u (t + Tb ) − u (t − Tb )]
These expressions are identical to MSK, so Glp ( f ) is as in Eq. (22).
14.1-13
Consider (k -1)Tb < t < (k + 1)Tb with k odd, so
xq ( t ) = sin(φ k −1 + ak −1ck −1 ) pTb [t − (k − 1)Tb ] + sin( φk + ak ck ) pTb [ t − kTb ]
since cos φ k = 0, sin(φ k + ak ck ) = cos( ak ck )sin φ k = cos ck sin φk
also sin φ k -1 = 0, φ k −1 = φ k − ak −1π /2, and ck −1 = ck + π /2 so

π 
π


sin(φ k −1 + ak −1ck −1 ) = sin  ak −1  ck +   cos  φ k − ak −1 
2 
2



π

= ak −1 sin  ck +  sin φk sin ak −1 π / 2 = (ak −1 cos ck )( ak −1 sin φk )
2

=cosck sin φk
Thus xq (t ) = cos ck sin φk for ( k -1) Tb < t < ( k + 1)Tb with k odd,
and xq ( t ) =
∑Q
k
p(t − kTb ) where Qk = sin φk and
k odd
 πr

p (t − kTb ) = cos  b (t − kTb )  {u [ t − ( k − 1)Tb ] − u [t − (k + 1)Tb ]}
 2

so p (t ) = cos ( πrbt /2)[u (t + Tb ) − u (t − Tb )]
14.1-14
1
sinc 2 ( f / r ) and r = rb ⇒ Glp max = Glp ( f = 0)
r
Because we are at baseband ⇒ use f = BT /2 ⇒ f = 1500 and Glp ( f = 0) = 1/ rb
Glp ( f ) =
 Glp ( f = 1500) 
 Glp ( f = 1500) 
10log 
≤ −30 dB ⇒ log 
=log(sinc 2 (1500/ rb )) ≤ −3

 G ( f = 0) 
 G ( f = 0) 
lp
lp




⇒ sinc 2 (1500/ rb ) ≤ 0.001 ⇒ from sinc tables ⇒ 3.9=1500/rb ⇒ rb < 385 bps.
14-9
14.1-15
(a) Assume Sunde's FSK ⇒ f d = rb /2; and f = BT /2 =1500.
Using Eq. 18 and neglecting impulses we have
4
Glp ( f = 0) = 2 2 =Glp max
π rb
 Glp ( f = 1500) 
 cos(1500π /rb ) 
⇒ 10log 
 ≤ −30 dB ⇒ log 
 = ≤ −3
2
 (2 x 1500/rb ) − 1) 
 Glp ( f = 0) 
We get rb ≤ 656 ⇒ since f d = rb /2 ⇒ f d = 328 Hz.
2
(b) Using Eqs. (21) and (22) in a similar way as (a), we get rb = 1312 bps and f d = 328 Hz.
14.2-1
Since - s 0 (t) = s1 (t ), Vopt = 0 and hopt (t) = 2 KAc pTb (Tb − t ) cos ω c (Tb − t)
= 2 KAc cos(ωc t − 2π Nc ) pTb (t ) = 2 KAc cos(ω c t )
0 < t < Tb
14.2-2
π

(a) hopt (t ) = KAc sin 2  (Tb − t )  pTb (Tb − t )cos ωc (Tb − t )
 Tb


πt 
= KAc sin 2  π −  pTb (t )cos(ω c t − 2π Nc )
Tb 

πt 
= KAc sin 2   cos ω c t 0 < t < Tb
 Tb 
14-10
14.2-2 continued
(b)
14.2-3
Tb
T
Ac2 b
E1 = ∫ A cos 2π ( f c + f d ) tdt =
[1 + cos4π ( f c + f d )t ]dt
2 ∫0
0
2
c
=
2
Ac2Tb Ac2 sin4π ( f c + f d )Tb Ac2Tb
+
=
[1 + sin4( Nc + f d Tb )]
2
2
4π ( f c + f d )
2
Tb
Ac2Tb
similarly, E0 = ∫ A cos 2π ( f c − f d )tdt =
[1 + sinc4( Nc − f d Tb )]
2
0
2
c
2
2
AT
1
( E0 + E1 ) = c b [2 + sinc4( Nc + f d Tb ) + sinc4( Nc − f d Tb )]
2
2
If Nc − f d Tb >> 1 then N c − f d Tb >> 1 so |sinc4( Nc ± f d Tb ) |<< 1
and Eb =
and Eb ≈
Ac2Tb
2
14.2-4
−
E10
= -sinc(4f d / rb )
Eb
−sincλ |max ≈ .216 at λ ≈ 1.4
so take 4 f d / rb ≈ 0.216 ⇒ f d ≈ .35rb
Then E b − E10 = 1.216 Eb and Pe = Q
(
1.216γ b
)
14.2-5
Vopt = 0
Take K = 1/Ac so
14-11
14.2-5 continued
 2π 

1
Ks1,0 (t − kTb ) = cos   N c ±  ( t − kTb ) kTb < t < ( k + 1)Tb
2
 Tb 

= cos[2π ( Nc ± 1/2) rb t m π k ]
cos2π ( N c rb ± rb /2)t ]
=
− cos2π ( N c rb ± rb /2)t ]
k even 

k odd 
14.2-6
Pe = Q
(
)
2rb = 10 −5 ⇒
2rb = 4.27
Q(4.27cos θ ε ) < 10−4 ⇒ θ ε < arccos
3.74
≈ 290
4.27
14.2-7
± Ac cos ωc t + noise
Tb
X
∫
S/H
0
± Ac cos( ωc t + θε )
ωc = 2πNc / Tb
14-12
y (Tb ) = z (tb ) + N0
14.2-7 continued
Tb
Z ( tb ) = ∫ KAc cos(ωc t + θ ε )( ±Ac cos(ω c t ) dt
0
Ac2
= ±K
2
Tb
∫ [ cos(θ
ε
) + cos(2ωc t + θ ε ) dt ]
0

Ac2 
1
Tb  cosθ ε +
[sin(4π Nc + θε ) − sinθε ]
2 
2ω c Tb

= ± KEb cos θ ε
= ±K
14.2-8
s( λ ) = Ac cos ωc λ
0 < λ < Tb
h (t − λ ) = KAc cos ω c ( t − λ ) t - Tb < λ < t
for Tb < t < 0,
Z ( t) = 0
for 0 < t < Tb
t
Z (t ) = KAc2 ∫ cos ωc λ cos ω c (t − λ )d λ
0
KA 
2sin ω c t 
t cos ωc t +

2 
2ω c 
t
=KE [ cos ωc t + sinc(2 N c t / Tb )]
Tb
=
for Tb < t < 2Tb ,
2
c
Z (t ) = KAc2
t
∫
cos ω c λ cos ω c ( t − λ ) d λ
t −Tb

sin ω c (2Tb − t ) − sin ω c (t − Tb ) 
 (2Tb − t )cos ω c t +

2ω c



2sin ω c t 
t 
= KE  2 −  cos ωc t −

Tb 
2ω cTb 


 2 N t   
t 
= KE  2cos ωc t − cos ωc t + sinc  c   
Tb 
 Tb   

=
KAc2
2
14-13
14.2-8 continued
If Nc >> 1, then |sinc(2 Nc t / Tb ) |<< 1 for t ≠ 0, so
t

 KE T cos ω c t
b

z (t ) ≈ 
 KE  2 − t  cos ω t


c

Tb 


 t − Tb
≈ KE Λ 
 Tb
0 < t < Tb
Tb < t < 2 Tb

 cos ωc t

14.2-9
E1 = A (1 + α )
2
c
Tb
2
∫ cos
2
(ω c t ) dt = (1 + α ) 2 Ac2Tb /2
since ω cTb = 2π Nc
0
Tb
E0 = Ac2 (1 − α ) 2 ∫ cos 2 (ω c t )dt = (1 − α )2 Ac2Tb / 2
0
Tb
E10 = − A (1 − α )(1 + α ) ∫ cos (ωc t ) dt = −(1 − α )Ac Tb / 2
2
c
2
2
2
0
2
(1 + α ) + (1 − α ) Ac Tb
= (1 + α 2 )Ac2Tb / 2
2
2
2
Eb − E10 =  (1 + α ) + (1 − α 2 )  Ac2Tb / 2 = 2Eb /(1 + α 2 )
2
Eb =
⇒
2
Pe = Q  2γ b /(1 + α 2 ) 


14.2-10
cos(ω c t − π /2) = sin ω c t and w cTb = 2π Nc so
E1 = A
2
c
Tb
∫ (cos
2
ω c t + 2α sin ωc t cos ω c t + α 2 sin 2 ω c t) dt
0
= (1 + α ) Ac2Tb / 2
2
Similarly, E0 = E1 and Eb =
E10 = − A
2
c
Tb
∫ (cos
2
1
( E0 + E1 ) = (1 + α 2 ) Ac2Tb / 2
2
ω c t − α 2 sin 2 ω c t ) dt = −(1 − α 2 )Ac2Tb / 2
0
Thus,
Eb − E10 = Ac2Tb / 2 =
Eb
1+ α 2
⇒
Pe = Q  2γ b /(1 + α 2 ) 


14-14
14.2-11
s%m (t )
sm (t )
+
Hw( f )
S/H
hopt (t )
z m + N0
Gn ( f )
| H w ( f ) |2 = N0 / 2Gn ( f )
S° m ( f ) = ℑ[ s%m (t )] = S m ( f ) H w ( f ) where Sm ( f ) = ℑ[ sm (t )], m = 0,1
If hopt (t ) = K [ %s1 (Tb − t) − s%0 (Tb − t)] then
2
 z1 − z0 
1 %
%
%
 2σ  = 2 N ( E1 + E0 − 2 E10 ) where

 max
0
∞
Tb
E% m = ∫ s%m2 (t ) dt =
0
2
∫ | S%m ( f ) | df =
−∞
Tb
∞
0
−∞
E%10 = ∫ s%1 (t ) s%0 (t ) dt =
−∞
∞
∫|S
m
( f ) |2 | H w ( f ) |2 df
−∞
∫ s% (t)s% (t)dt
1
∞
*
∫ S%1 ( f )S%0 ( f )df =
=
∞
2
∫ s%m (t )dt =
0
∞
∫ S ( f )S
1
-∞
−∞
∞
∞
or = ∫ S%1* ( f )S%0 ( f ) df =
-∞
∫S
*
1
*
0
2
( f ) Hw ( f ) df
2
( f )S0( f ) H w ( f ) df
−∞
Thus, we can write
∞
∫S
2 E%10 =
*
1
( f ) S 0 ( f ) | Hw ( f ) | df +
2
−∞
∞
∫ S ( f )S
1
*
0
( f ) | H w ( f ) |2 df
−∞
so E%1 + E% 0 − 2E10
=
∞
∫  S ( f )
1
−∞
2
+ S0 ( f ) − S1 ( f ) S 0* ( f ) − S1* ( f )S0( f )  | H w ( f ) |2 df

∞
=
∫ |S (f)−S
1
2
0
(f )|
−∞
2
N0
df
2Gn ( f )
∞
| S ( f ) − S0 ( f ) |
z −z 
Hence,  1 0  = ∫ 1
df
4Gn ( f )
 2σ  max −∞
2
14-15
14.2-12
Tb
Tb
∫ s (t ) s (t )dt = A ∫ cos[2 π( f
1
0
2
c
0
c
− f d )t ]cos[2π( f c + f d )t ]dt
0
Tb
Ac2 Tb
Ac2 1
1
=
{∫ (cos4πf c t )dt + ∫ (cos4π f dt )dt } =
[
sin4πf c Tb +
sin4πf d Tb ]
2 0
2 4πf c
4 πf d
0
since f cTb is an integer ⇒ sin4πf cTb = 0, and rb = 1/ Tb
Ac2 Tb sin4πf dTb Ac2Tb
⇒ ∫ s1 ( t ) s0 (t ) dt =
=
sinc4 f d / rb
2 Tb
4πfd
2
0
Tb
Ac2Tb
If E1 = E0 =
⇒ρ=
2
1
E1 E0
Tb
∫ s (t)s (t)dt = sinc4 f
1
0
d
/ rb
0
14.2-13
Eq. (9) ⇒ Pe = Q  ( Eb (1 − ρ) / N 0  ⇒ to minimize Pe ⇒ maximize (1- ρ)
⇒ make ρ as negative as possible. With ρ=sinc(4 f d / rb )
From the sinc Table, the maximum negative value of the sinc function =-0.216
⇒ ρ=sincλ = −0.216 = ρ
with λ = 1.4
14.2-14
(a) Given Pe = 10
−5
and N0 = 10
−11
Ac2Tb
Ac2
, Eb =
=
, Pe = Q
2
2rb
(
2E b / N 0
)
18.3
x 10-11 = 9.16 x 10-11
2
= 1.76 x 10-6
From Table, Pe = 10−5 ⇒ 2 Eb / N0 = 18.3 ⇒ Eb =
⇒ Ac2 = 2rb Eb = 2 x 9600 x 9.16 x 10-11
⇒ Ac = 0.00133 V.
(b) rb = 28.8 kpbs ⇒ Ac2 = 2 rb Eb = 2 x 28,800 x 9.16 x 10-11 = 5.28 x 10-6
⇒ Ac = 0.0023 V
14-16
14.2-15
(a) Given Pe = 10 −5 and N0 = 10−11 , Eb =
Assuming Sunde's FSK, ⇒ Pe = Q
(
Ac2Tb
A2
= c ,
2
2rb
E b /N0
)
−5
From Table, Pe = 10 ⇒ Eb / N0 = 18.3 ⇒ Eb = 18.3 x 1 x 10-11 = 1.83 x 10-10
⇒ Ac2 = 2rb Eb = 2 x 9600 x 1.83 x 10 -10 = 3.51 x 10-6
⇒ Ac = 0.00187V.
(b) rb = 28.8 kpbs ⇒ Ac2 = 2rb Eb = 2 x 28,800 x 1.83 x 10-10 = 1.05 x 10-5
⇒ Ac = 0.00325 V
14.3-1
( γ ) < 10
≈ Q ( γ ) < 3 x 10
1  −rb / 2
e
+Q
2
so Pe1
 1

⇒ γ b > 2ln 
≈ 12.4 ≈ 10.9 dB
-3 
 2 x 10 
−3
b
−4
b
14.3-2
( γ ) < 10
≈ Q ( γ ) < 2 x 10
1  −rb / 2
e
+Q
2
so Pe1
−5
b
1


⇒ γ b > 2ln 
≈ 21.6 ≈ 13.4 dB
-5 
 2 x 10 
−6
b
14.3-3
(a) Given Pe = 10 −5 and N 0 = 10−11 , Eb =
Ac2Tb Ac2
=
,
2
2 rb
1 − Eb / 2 N0 1 − γb / 2
e
= e
= 1 x 10 -5
2
2
-5
ln(2 x 10 ) = − γb / 2 ⇒ γ b = 21.6 = Eb / N 0 ⇒ Eb = 21.6 x 1 x 10-11 = 2.16 x 10-10
Noncoherent FSK, ⇒ Pe =
⇒ Ac2 = 2rb Eb = 2 x 9600 x 2.16 x 10 -10 = 4.14 x 10-6
⇒ Ac = 0.00204V.
(b) rb = 28.8 kpbs ⇒ Ac2 = 2rb Eb = 2 x 28,800 x 2.16 x 10-10 = 1.24 x 10 -5
⇒ Ac = 0.00353 V
14-17
14.3-4
(a) Sunde's FSK ⇒ f d = rb / 2 , BT ≈ rb , and E10 = 0
⇒ f d = 14,400/2 = 7200, S / N =12 dB = 15.9
 S  B 
γb = Eb / N 0 =    T  = 15.9 x 1
 N   rb 
Coherent FSK ⇒ Pe = Q
(
)
E b / N0 = Q
(b) Noncoherent FSK ⇒ Pe =
(
)
15.9 = 3.4 x 10
-5
1 − γb / 2 1 −1 5 . 9 / 2
e
= e
= 1.76 x 10-4
2
2
14.3-5
KAc = Ac2 / E1 = 2 / Tb
N
σ = 0
2
2
=
∞
N0
∫−∞ | H ( f ) | df = 2
2
 4f c
N0 
1 + sinc 
Tb 
 rb
∞
N0 2
∫−∞ | h(t) | dt = 2 Tb
2
Tb
∫ cos
2
(ω c t )dt
0
 N 0
since f c >> rb
 ≈
  Tb
Thus Ac2 / σ 2 = Ac2Tb / N0 = 4 Eb / N0
where Eb = Ac2Tb /4
14.3-6
∞
∞
df
4( f − f c ) 2
0 
1 +
2
B

σ 2 = ( N0 / 2) * 2 ∫ | H ( f ) | 2 df = N0 ∫



≈ ( N 0 B / 2)π since f c / B >> 1 ⇒ arctan(-2f c / B) ≈ −π / 2
0
Pe ≈ 1 / 2 Peo = 1 / 2e − A c /8σ where A2 c / 8σ 2 = A2c / 4πBN 0 and
2
2
A2c = 4 Eb rb so A2 c / 8σ 2 = 4Eb rb / 4πBN 0 = ( 2rb / πB) rb / 2
⇒ increase rb by πB / 2rb ≥ π ≈ 5dB
14.3-7
Take K = Ac / E1 and thresholds at Ac / 2 and 3 Ac / 2
14-18
14.3-7 continued
2
2
 A 
A 
Then Peo = e − Ac / 8σ , Pe1 ≈ 2Q  c  , Pe 2 ≈ Q  c 
 2σ 
 2σ 
2
2
E0 = 0, E1 ≈ Ac D /2, E2 ≈ (2 Ac ) D /2 where D = 1/ r
1
so E = (E0 + E1 + E 2 ) = 5 Ac2 D /6 ⇒ Ac2 D = 6 E / 5
3
2
Ac
Ac2 D
4 Ac2 D 6E
and 2 =
=
from Eq.(9) with Eb =
σ
N0 4
5 N0
4
1
1
Thus, Pe = ( Pe0 + Pe 1 + Pe 2 ) ≈ e −3E /20 N0 + Q 3E /10 N 0
3
3
(
)
14.3-8
ST
ST
2 x 105
= S R = Eb rb = N 0γ b rb ⇒ rb =
=
L
LN 0γ b
γb
1
1
(a) e− γ b / 2 ≤ 10 −4 ⇒ γ b ≥ 2ln
≈ 17 so rb ≤ 2 x 105 /17 = 11.8 kbps
-4
2
2 x 10
1
1
(b) e− γ b ≤ 10−4 ⇒ γ b ≥ ln
≈ 8.5 so rb ≤ 2 x 105 /8.5 = 23.5 kbps
-4
2
2 x 10
1
(c) Q 2γ b ≤ 10−4 ⇒ γ b ≥ x 3.752 ≈ 7 so rb ≤ 2 x 105 /7 = 28.4 kbps
2
(
)
14.3-9
ST
ST
2 x 105
= S R = Eb rb = N 0γ b rb ⇒ rb =
=
L
LN 0γ b
γb
1
1
(a) e− γ b / 2 ≤ 10 −5 ⇒ γ b ≥ 2ln
≈ 21.6 so rb ≤ 2 x 105 /21.6 = 9.2 kbps
2
2 x 10-5
1
1
(b) e− γ b ≤ 10−5 ⇒ γ b ≥ 2ln
≈ 10.8 so rb ≤ 2 x 105 /10.8 = 18.4 kbps
2
2 x 10 -5
1
(c) Q 2γ b ≤ 10− 5 ⇒ γ b ≥ x 4.272 ≈ 9.1 so rb ≤ 2 x 105 /9.1 = 21.9 kbps
2
(
)
14-19
14.3-10
DPSK :
1 −γb
e ≤ 10−4
2
BPSK: Q
(
 1
γ b ≥ ln 
-4
 2 x 10
⇒
)
2
2γ b cos θ ε ≤ 10
2

 = 8.52

−4
1  3.75 
7.03
⇒ γb ≥ 
 =
2
2  cosθ ε 
cos θ ε
BPSK requires less energy if |θ ε |< arccos 7.03/8.52 ≈ 250
14.3-11
DPSK :
1 −γb
e ≤ 10−6
2
BPSK: Q
(
 1
γ b ≥ ln 
-6
 2 x 10
⇒
)
2
2γ b cos θ ε ≤ 10
2

 = 13.12

−6
1  4..75 
11.28
⇒ γb ≥ 
 =
2
2  cosθ ε 
cos θ ε
BPSK requires less energy if |θ ε |< arccos 11.28/13.12 ≈ 220
14.3-12
p y ( y ) = pnq ( y) =
1
2πσ 2
e− y
2
/ 2σ 2
1
p xy ( x, y ) = px ( x) p y ( y) =
1
, px ( x ) = pni (x − Ac ) =
e−[( x− Ac )
2
2πσ 2
e − ( x − Ac )
2
/ 2σ 2
+ y 2 ] / 2σ 2
2πσ 2
For polar transformation: x = A cos φ , y = A sin φ , dxdy = A dA d φ
so p Aφ ( A, φ ) dA dφ = pxy ( x , y )dxdy = p xy ( A cos φ , Asin φ ) A dA d φ
where ( x - Ac ) 2 + y 2 = A2 cos 2 φ − 2 AAc cos φ + + Ac2 + A2 sin 2 φ = A2 − 2 AAc cos φ + Ac2
A − ( A2 −2 AAc cosφ + Ac2 ) / 2σ 2
Thus pAφ ( A, φ ) = Ap xy ( A cos φ , Asin φ ) =
e
2πσ 2
14.4-1
rb / BT ≥ 1.25
⇒ Modulation types with rb / BT = 2
(a) QAM/QPSK: Q
(
)
2γ b ≤ 10-6 ⇒ γ b ≥ 1/2 x 4.752 = 10.5 dB
(b) DPSK with M = 4 so K = 2:
2
Q
2
(
)
4 x 2γ b sin π / 8 ≤ 10
2
2
1  4.75 
γb ≥ 
 = 12.8 dB
8  0.383 
14-20
−6
14.4-2
rb / BT ≥ 2.5
⇒ modulation types with rb / BT = 3
(a) PSK with M = 8 so K = 3:
2
Q
3
(
)
2 x 3γb sin 2 π / 8 ≤ 10−6
2
1  4.7 
γb ≥ 
= 14.0 dB
6  0.383 
(b) DPSK with M = 8 so K = 3:
2 
Q
3 
4 x 3γ b sin 2 (π /16  ≤ 10−6

2
1  4.7 
γb ≥ 
= 16.8 dB
12  0.195 
14.4-3
rb / BT ≥ 3.2
⇒ modulation types with rb / BT = 4
(a) QAM with M = 16 so K = 4:
γb ≥
4 1  3 x 4
γb
1-  Q 
4  4   15
15 2
4.7 = 14.4 dB
12
(b) PSK with M = 16 so K = 4:
2
Q
4
(

−6
 ≤ 10

)
2 x 4γb sin π /16 ≤ 10
2
−6
2
1  4.6 
γb ≥ 
 = 18.4 dB
8  0.195 
14.4-4
rb / BT ≥ 4.8
⇒ modulation types with rb / BT = 5 or 6
(a) QAM with M = 64 so K = 6:
γb ≥
4 1   3 x 6
γb
1 −  Q 
6  8   63

−6
 ≤ 10

63
4.652 = 18.8 dB
18
(b) PSK with M = 32 so K = 5:
2
Q
5
(
)
2 x 5γ b sin 2 (π /32 ≤ 10−6
2
1  4.55 
γb ≥ 
 = 23.3 dB
10  0.098 
14-21
14.4-5
xc (t ) = Ac cos( wc t + φk )
Upper delay output = xc (t − D)2cos[ωc ( t − D) + θε ]
= Ac [cos( θε − φk ) + cos(2ωc t − 2ωc D + θε + φk )]
Lower delay output = xc ( t − D) { −2sin[ ωc (t − D) + θε ]}
= − Ac [sin( θε − φk ) + sin(2ωc t − 2ωc D + θ ε + φk )]
LPF input = Ac sin φˆ k cos( θε − φ k ) − (− cos φˆ k )sin(θ ε − φ k ) + high frequency terms 
1
1
Thus, v (t ) = Ac  sin( φˆ k − θ ε + φ k ) + sin(φˆ k + θ ε − φk )
2
2
1
1

si n(θ ε − φk − φˆ k ) + sin(θ ε − φ k + φˆ k ) 
2
2

= Ac sin(θ ε + φˆ k − φ k ) = Ac sin(θ ε ) when φˆ k = φk
+
14.4-6
 3E 
 4E 2

QAM: Pe ≈ 3Q 
DPSK : Pe ≈ 2Q 
sin (π /32) 

 15 N0 
 N0

Since magnitude of Pe is dominated by argument of Q, we want
4 EQAM
4 EDPSK
x (0.098) 2 ≈
N0
15 N 0
⇒
EDPSK
3
≈
= 5.2
EQAM 15 x 4 x (0.098) 2
14.4-7
 2E  π 2 
PSK: Pe ≈ 2Q 
x  
 No
 M  

 3E

QAM : Pe ≈ 4Q 
γb 
 MN 0 
1
≈ 1, M − 1 ≈ M
M
Magnitude of Pe is dominated by argument of Q , so we want
since sin π / M ≈ π / M
3 EQAM
MN0
since 1-
2
EQAM 2 π2
2 EPSK  π 
≈
⇒
≈
No  M 
EPSK 3 M
14-22
14.4-8
2
2
∞
e − Ac / 2σ ∞ −( A2 − 2 AAc cosφ ) / 2σ 2
p (φ ) = ∫ pAφ ( A, φ ) dA =
dA
∫ Ae
φ
2πσ 2 0
0
let λ = ( A - Ac cos φ ) / σ and λ0 = ( Ac cos φ ) / σ
so A = σ (λ + λ ) and (A2 − 2 AA cos φ ) / 2σ 2 = ( λ 2 / 2 ) − (λ 2 /2)
0
∞
2
2
eλ0 / 2 ∫ σ ( λ + λ0 ) e− λ / 2σ d λ
2πσ 2
−λ0
∞

1 − Ac2 / 2σ 2 − λ 2 / 2  ∞ − λ 2
−λ2
e
e
 ∫ λ e d λ + λ0 ∫ e d λ 
2π
 −λ0

λ0
=
∞
∫
λe − λ
0
e
Then pφ (φ ) =
where
c
− A2c / 2σ 2
2
/2
d λ = e − λ0 / 2
2
−λ0
∞
∫
− λ0
−λ2 / 2
λe
dλ =
∞
∫e
−λ 2 / 2
dλ −
−∞
− λ0
∫
e− λ
2
/2
dλ =
2π [1 − Q( λ0 ) ]
−∞
Ac2 / 2σ 2 − λ02 / 2 = Ac2 (1 − cos 2 φ ) / 2σ 2 = Ac2 sin 2 φ / 2σ 2
1 − Ac2 / 2σ2 λ02 / 2 − λ02 / 2
Thus p (φ ) =
e
e
e
+ λ0 2π [1 − Q (λ0 ) ]
φ
2π
1 − Ac2 / 2σ 2 Ac cos φ − Ac2 sin 2 φ / 2 σ 2 
 A cos φ  
=
e
+
e
1− Q c


2
2π
 σ

2πσ

and
{
}
14.4-9
Use the design of Fig. 14.4-2 and (1) change the 4th law device to a second law device,
(2) change the 4f c BPF to a 2f c BPF, (3) change the ÷ 4 block to a ÷ 2 block, (4) eliminate
the +90 deg block. ⇒ The output of the ÷ 2 block is the reference signal and is cos(2 πf c t + πN ).
The πN term is a phase ambiguity that depends on the lock-in transient and have to be accounted
for. This could be done using a known preamble at the beginning of the transmission.
14.4-10
Use the design of Fig. 14.4-2 and (1) change the 4th law device to a M th-law device,
(2) change the 4f c BPF to a Mf c BPF. The output of the PLL is cos[2M πt + M φ k + 2 πN ].
The 2πN term is a phase ambiguity that depends on the lock-in transient and will have to be
accounted for. This could be done using a known preamble at the beginning of the transmission
(3) At the output of the PLL, change the ÷ 4 bl ock to a ÷ M block, giving an output of
cos[2πt + φ k + 2 πN / M ]. (4) Replace +90 deg block with an M output phase network.
14-23
14.4-11
γb = 13 dB=20, Pe = Pbe K and K = log 2 M
(a) FSK ⇒ Pe =
1 − γ b / 2 1 −2 0 / 2
e
= e
= 2.3 x 10 -5
2
2
(b) BPSK ⇒ Pe = Q
(
)
2γ b = Q
(
)
2 x 20 = 1.8 x 10-10
2 
π  2 
π 
Q  2 K γbsin 2
= Q  2 x 7 x γb x sin 2


K 
M  7 
64 
= 0.0571 ⇒ Pe = 0.031 x 7 = 0.40
(c) 64-PSK ⇒ Pbe =
4
1   3K γb 
1−


 Q
K
M   M − 1 
4
1   3 x 7 x 20 
-3
K = 7, M = 64 ⇒ Pbe =  1 −
Q 
 = 2.4 x 10

7
63
64  

-3
⇒ Pe = Pbe x K =2.4 x 10 x 7=0.017
(d) 64-QAM ⇒ Pe =
14.5-1
Eq. (6) ⇒ Pe = N min Q
(
)
2
d min
/ 2 N 0 = 1 x 10 -5
For uncoded QPSK ⇒ N min = 2 and ( dmin ) uncoded = 2
⇒ 1 x 10-5 = 2Q
(
)
2 / 2 N 0 ⇒ solving gives N 0 = 0.052
TCM with 8 states, m = m% = 2 ⇒ N min = 2 and g =3.6 dB=2.3
From Eq. (5) ⇒ g =
2
2
( d min
) coded
( dmin
) uncoded
2
=
= 2.3 ⇒ (d min
) coded = 4.6
2
( dmi n ) uncoded
2

4.6
⇒ Pe = 2Q 
 2 x 0.052

-11
 = 4.0 x 10

14-24
14.5-2
From Ungerbroeck (1982), we have
•
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o
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o
•
o
o
o
o
o
•
o
o
o
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o
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o
o
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o
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o
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o
o
o
o
o
•
o
o
o
•
o
•
o
o •
• o
o •
• o
d1 =
o
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o
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d0 = 1
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• o
o •
• o
o•
2
o
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o
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• o
oo
• o
oo
d2 = 2
o
o
o
o
o
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o
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• o
oo
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oo
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o
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o
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o
• o
oo
o o
oo
Input:
x2 x1 00 → 01 → 10 → 01 → 11 → 00
Output: y3 y2 y1 000 → 100 → 011 → 010 → 111 → 111
a
b
g
b
h
e
14.5-4
What is the distance between paths (0,2,4,2) and (6,1,3,0)?
Using Figs. 14.5-4 and 14.5-7 we have:
d 02→6 + d 22→1 +d 42→3 + d22→ 0 =
d12 + d02 + d02 + d12
= ( 2) 2 + (2sin π /8)2 + (2sin π /8)2 + ( 2) 2 = 5.2 = 2.3
14.5-5
See error event Trellis diagram of Fig. P14.5-5
2
d min
= d02→6 + d02→1 + d02→7 + d02→2 = d12 + d02 + d02 + d12
= 2 + 4sin 2 π /8 + 4sin 2 π /8+ 2 = 5.17
 5.17 
⇒ coding gain = 10log 
 = 4.13 dB
 2 
14-25
o
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o
o
o
14.5-3
State:
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o
o
o
o
o
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o
o
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o
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o
•
o
o
o
o
o
•
o
o
d2 = 2 2
Chapter 15
15.1-1
(a) With DSS (S/N)D = ( S / N ) R =
SR
Er
E
= b b = b = 60 dB = 106
N R N 0 rb N 0
with N 0 = 10−21 ⇒ Eb = 10−15
SR = Eb × rb = 10-15 × 3000 = 3 × 10-12 ⇒ J = 15 x 10 -12
Since J ? N 0 we can neglect noise and use Eq. (13) and J / SR =5 giving
 2Pg 
Pe = 10-7 = Q 
 ⇒ 5.22 = 2 Pg /5 ⇒ Pg = 67.6
 5 


Eq. (9) with Wx =3000 Wc = 67.6 x 3000 = 203 kcps
(b) BT = 2 x Wc = 2 x 203 x 103 = 0.406 MHz
15.1-2
 2 x 1000 
−7
Pg = 30 dB = 1000 With Eq. (13) we have Pe = Q 
 = 10

J
/
S
R


2000
2
⇒ 5.2 =
⇒ J / S R = 74.0 ⇒ jamming margain = 10log(74) = 18.7 dB
J / SR
Jamming margain = 10 × log(J/Sr )
= 10 × log(Pg ) − 10 × log(Eb /N J )
= 10 × log( 1000 ) − 10 × log( 1352
. ) = 1869
.
15.1-3
(a) ( S / N ) D =20 dB = 100 = Eb / N 0
Wc = 10 x 10 6 and rb = 6000 = Wx ⇒ Pg = 10 × 106 / 6000 = 1.67 x 103


1

With Eq. (19) we have Pe = 10-7 = Q 
 M/ (3 x 1.67 x 103 ) + N / 2 E 
0
b 

1
⇒ 5 .2 2 =
⇒ M = 159 additional users for a tot al of 160 users.
M
1
+
5000 200
15-1
15.1-3 continued
(b) If each user reduces their power by 6 dB ⇒ 6 dB = 4 ⇒ Eb / N 0 = 100/4 = 25
Using the results of part (a) we have
1
5.22 =
⇒ M − 1 = 85 additional users for a total of 86 users.
M −1 1
+
5000 50
15.1-4
Let d = distance between the transmitter and authorized receiver then
d m = d + 500 = distance of the multipath, Tm multipath travel time and c speed of light.
With dm = c × T m ⇒ Tm = d m /c = 500 / 3 x 108 = 167
. µs
To avoid multipath interference Tc < Tm ⇒ Wc > 600 kcps
15.1-5
 3 Pg 
Using Eq. (2) with M − 1 = 9 additional users (10 total users), we have Pe = Q 
 = 10−7
 9 


2
9 x 5.2 / 3 = Pg = 81.1 ⇒ Wc = Pg x rb = 81.6 x 6000 = 487 kcps = Wc
15.1-6
 2 Eb
-9
With Pe = 10 = Q 
 N
0


2 Eb
2
= 6 = 36
 ⇒
N0



1
-7
Eq. (19), M − 1 = 9 additional users and Pe = 10 = Q 
 9
1
 3P + 36
g

Wc
1
⇒ 5.2 2 =
⇒ Pg = 326 =
⇒ Wc = 1.96 Mcps
9
1
6000
+
3Pg 36
15.1-7
rb = 9kbps, J/S R = 30dB = 1000 and Pe < 10-7 ;
 2Pg 
-7
 ⇒ 5.2 2 = 2Pg /1000 ⇒ Pg = 13,520
Eq. (13) 10 = Q 
 1000 


15-2






15.1-8
Pg = 30 dB = 1000 and assuming negligible noise,
 3Pg 
Using Eq. (20) Pe = 10-7 = Q 
 ⇒ 3 x 1000/( M − 1) = 5.22 ⇒ M − 1 = 111
 M −1 


⇒ 112 total users
15.1-9
rb = 6kbps, Wc = 10 Mcps ⇒ Pg = 10 x 106 /6000 = 1667
 2 Eb
For single user Pe = 10-10 , then Eq. (6) ⇒ 10-10 = Q 
 N0

2 Eb
 ⇒ 6.42 = 41 =
N0

If each user reduces their power by 3 dB = 2 ⇒ Eb → Eb / 2 ⇒
2 Eb
→ 41/2 = 20.5
N0




1
1
 ⇒ 4.32 =
Eq. (19) we have 10 −5 = Q 
⇒
M −1
1

M −1
1 
+
+


3 x 1667 20.5
 3 x 1667 20.5 
M − 1 = 26 additional users ⇒ M = 27 total users
15.2-1
(a) Eb /N0 = 60dB = 106 and if N0 = 10−21 ⇒ Eb = 106 /10−21 = 10 −15
SR = Eb × rb = 10-15 × 3000 = 3 × 10-12
J = 5 × S r = 15× 10-12
Eb
10 − 15
− 12
1 −
1 − −21
Pe = e 2( N 0 + N J ) ⇒ 10-7 = e 2(10 +1 5×1 0 ) ⇒ N J = 3.24 x 10 -17
2
2
J
15 × 10-12
NJ =
⇒ Wc =
= 4.62 x 105
-17
Wc
3.2424 × 10
4.62 x 105
Pg =
=
= 154
Wx
3 × 103
Wc
If k = 7 ⇒ Pg = 27 = 128
If k = 8 ⇒ Pg = 28 = 256
⇒ if Pg = 256 ⇒ Wc = 256 × 3103 = 768 kHz
(b) BT = Wc = 768kHz
15-3
15.2-2
10 users ⇒ M = 10
Pe =
 (K - 1) 
(1/2)(K − 1)
+ (1/2) × e- Eb/(2×N 0 ) 1
Pg
Pg 

With Pe = 10−10 for one user ⇒ first term in above Eq. dominates.

 ⇒ Pg = 450000

k
But Pg = 2 ≥ 450,000 ⇒ k = 19 ⇒ Pg = 219 =524,288
2 × 10-5 =
Pg =
Wc
Wx
 9
9
+ 10 - 10 × 1  Pg
Pg

⇒ Wc = 524,288 x 3000 = 1.57 GHz
15.2-3
From problem 15.2.1 S R = 3 × 10-12 , J = 1.5 × 10-11 , Wx =3000
 2 Pg
Using Sect 15.1, Eq. (13) we have 10−7 = Q 
 J / SR

2Wc /3000
⇒
= 5.22 ⇒ Wc = 2 x 105
5
Pg = Wc /W x = 2 x 105 / 3000 = 67

 2Wc /3000 
 = Q



5



BT = 2 × Wc = 400 kHz
15.2-4
With k = 10,
⇒ Pg = 2 = 1024
10
and with rb =6000 ⇒ Wc = Pg Wx = 1024 × 6000 = 6144 kbps
N J = J/Wc = 6 × 10-3 / 1024 = 9.7 6 × 10-10
−
1 − 0.1 −2 x 10-11 /2 x10 −12 0.1 2(2 x 1 0
e
+
e
2
2
Pe = 2.04× 10 -5 + 4.99 × 10 -2 = 0.05
From Eq. (4) Pe =
−2 x 10 -11
-12
+9.766 x 10-10 /0.1
)
15.2-5
d = 5 miles
⇒
d = 5 ×1610 = 8050meters
∆d = 2 × 8050 = 16100
∆d = v × t
⇒
t = ∆d / t = 16100/ ( 2.99 × 108 ) = 5.38 × 10−9 s
f = 1/t = 1/ ( 5.38 × 10−9 ) = 18.6 kHz
15-4
15.3-1
(a) A shift register with [4,1] configuration, and initial state of 0100 has the following
contents after each clock pulse:
Clock shift
0
1
2
3
4
5
6
7
Register contents
0100
0010
0001
1000
1100
1110
1111
0111
Clock shift
8
9
10
11
12
13
14
15
Register contents
1011
0101
1010
1101
0110
0011
1001
0100
Thus, the output sequence= 001000111101011…
(b) PN sequence length = N = 15
(c) f c = 10 MHz ⇒ Tc = 10−7 s. With TPN sequence = NTc = 15 x 10-7 =1500 ns
(d) Autocorrelation function, R[4,1],[4,1] ( τ) versus τ
1
0.8
0.6
0.4
0.2
0
-0.2
5
10
15
15-5
20
25
30
15.3-2
(a) Shift register with [4,2] configuration, and initial state of 0100 has the following
contents after each clock pulse:
Clock shift
0
1
2
3
4
5
6
Register contents
0100
1010
0101
0010
0001
1000
0100
The output sequence: 001010…
(b) PN sequence length = N = 6
(c) f c = 10 MHz ⇒ Tc = 10 −7 s. With TPN sequence = NTc = 6 x 10-7 =600 ns
(d) To calculate the autocorrelation function, we use the method of Example 11.4-1 to
get:
____________________________________________________________
τ
original/shifted
v(τ)
R[4,2][4,2] (τ ) = v(τ) / N
____________________________________________________________
0
001010
6-0=6
6/6=6
001010
1
001010
000101
2-4=-2
-2/6 =-0.33
2
001010
100010
4-2=2
0.33
3
001010
010001
2-4=-2
-0.33
4
001010
101000
4-2=2
0.33
5
001010
010100
2-4=-2
-0.33
6
001010
6-0=6
1
001010
_____________________________________________________________
15-6
15.3-2 continued
The plot of R[4,2][4,2] ( τ) versus τ
R[4,2][4,2] ( τ) R[4,2][4,2] ( τ)
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
1
2
3
4
5
6
7
8
9
10
11
12
15.3-3
Given the results of Problem 15.3-1,
(1) Number of 1s= 8, number of 0s: 7 ⇒ satisfies balance property.
(2) Length of single type of digit: 4/8 of length 1, 2/8 of length 2, 1/8 of length 3, 1/8 of length 4
⇒ satisfies run property.
(3) Single autocorrelation peak ⇒ satisfies autocorrelation property.
(4) Mod 2 addition of the output with a shifted version results in another shifted versi on
(5) All 15 states exist.
⇒ A [4,1] register produces a ml sequence.
15.3-4
A shift register with [5,4] configuration, and initial state of 11111 has the following
contents after each clock pulse:
15-7
15.3-4 continued
Clock pulse
0
1
2
3
4
5
6
7
8
9
10
register contents
11111
01111
00111
00011
00001
10000
01000
00100
00010
10001
11000
clock pulse
register contents
10
11000
11
01100
12
00110
13
10011
14
01001
15
10100
16
01010
17
10101
18
11010
19
11101
20
11110
21
11111
The output sequence is thus: 1111100001000110010101 ⇒ N = 21
To be a ml sequence, the PN length should be N = 2 n − 1 = 25 − 1 = 31
21 ≤ 31 ⇒ the shift register does not produce a ml sequence
15.3-5
(a) From Ex. 11.4-1 and Sect. 11.4 we get the output sequences from [5,4,3,2] and [5,2]
configurations to obtain [5,2] ⊕ [5,4,3,2]=
1111100110100100001010111011000
⊕ 1111100100110000101101010001110
_______________________________
0000000010010100100111101010110 = output sequence
 n2+1

2 +1 

(b) Rst = 
= ( 23 − 1) /31 = 0.29

 N 


15.3-6
0.01 miles x 1610 meters/mile = 16.1 meters
From Eq. (7) we have Tc = ∆d / c = 16.1/3 x 108 = 53.7 ns
⇒ f c = 1/ Tc = 1/53.8 x 10-9 = 18.6 MHz
15-8
15.4-1
With 5 Hz/hour drift ⇒ chip uncertainty/day = 5 chips/hour x 24 hour/day = 120 chips
15.4-2
f c = 900 MHz, f clock = 10 MHz, and v = 500 Mph, and light speed = c = 3 x 108 M/s
vf c 500 Mph x 1610 M/mile x 1 hour/3600 s x 900 x 106 s -1
=
c
3 x 108 M/s
= 671 Hz
vf
500 Mph x 1610 M/mile x 1 hour/3600 s x 10 x 106 s-1
(b) Doppler shift = ∆f clock = ± clock =
c
3 x 108 M/s
= 7.45 chips
(a) Doppler shift = ∆f c = ±
15.4-3
(a) Using Eq. (2) with a preamble of l = 2047, Tc = 1/ f clock =1 x 10 -7 s α =100,
PD = 0.9, PFA = 0.01, and assuming that the average phase error is 2048/2 chips we have
Tacq =
2 − PD
2-0.9
-7
(1 + αPFA ) N c lTc =
(1+100 x 0.01) x 1024 x 2047 x 10
PD
0.9
= 0.51
(b) Using Eq. (3) we have
1
1 
 1
σ2Tacq = (2 x 1024 x 2047 x 10-7 ) 2 x (1+100 x 0.01) 2  +
−
 = 0.15
2
 12 0.9 0.9 
σTacq = 0.38
15.4-4
(a) 12 stage shift register ⇒ l = 4095, then using Eq. (2) with Tc = 1/ f clock =2 x 10 -8 s α =10,
PD = 0.9, PFA = 0.001, and assuming that the average phase error is 4096/2 chips we have
Tacq =
2 − PD
2-0.9
-8
(1 + αPFA ) N c lTc =
(1+10 x 0.001) x 2048 x 4097 x 2 x 10
PD
0.9
= 0.21
(b) Using Eq. (3) we have
1
1 
1
σ2Tacq = (2 x 2048 x 4095 x 2 x 10-8 ) 2 x (1+10 x 0.001)2  +
−
= 0.024
2
0.9 
 12 0.9
σTacq = 0.15
15-9
Chapter 16
16.1-1
P(not F) = 4/5 ⇒ I = log 5/4 = 0.322 bits, P(specific grade) = 1/5 ⇒ I = log 5 = 2.322 bits
so Ineeded = 2.322 – 0.322 = 2 bits
16.1-2
(a) P(heart) = 13/52 = 1/4 ⇒ I = log 4 = 2 bits, P(face card) = 12/52 = 3/13 ⇒
I = log 13/3 = 2.12 bits, Iheart face card = 2 + 2.12 = 4.12 bits
(b) P(red face card) = 6/52 ⇒ Igiven = log 52/6 = 3.12 bits, P(specific card) = 1/52 ⇒
I = log 52, Ineeded = log 52 – 3.12 = 2.58 bits
16.1-3
Including the direction of the first turn, the number of different combinations is
2 × 102 × 102 × 102 , assumed to be equally likely. Thus, I = log (2 × 106 ) = 20.9 bits
16.1-4
1
H ( X ) = 12 × 1 + 41 × 2 + 81× 3 + ln2
( 2 × 201 ln20 + 401 ln40) = 1.94 bits, 6H(X) = 11.64
P(ABABBA) = ( 12 ) × ( 14 ) =
3
3
1
⇒ I = log 29 = 9 bits < 6H(X)
9
2
P(FDDFDF) = ( 401 ) × ( 201 ) =
3
3
1
⇒ I = log (512 × 106 ) = 28.93 bits > 6H(X)
6
512 ×10
16.1-5
H (X ) =
1 
1
1
1
1
1 
+ 0.2ln
+ 0.12ln
+ 2 × 0.1ln
+ 0.08ln
 0.4ln
 = 2.32 bits,
ln2 
0.4
0.2
0.12
0.1
0.08 
6H(X) = 13.90 bits
P(ABABBA) = (0.4)3 (0.2)3 = 5.12 × 10-4 ⇒ I = log
1
= 10.93 bits < 6H(X)
−4
5.12 ×10
P(FDDFDF) = (0.1)3 (0.08)3 = 5.12 × 10-7 ⇒ I = log
16-1
1
= 20.90 bits > 6H(X)
−7
5.12 ×10
16.1-6
Since the first symbol is always the same, there are M = 8 × 8 = 64 different blocks, and
H(X) = log 64 = 6 bits/block. Thus, R = 1000 blocks/sec × 6 bits/block = 6000 bits/sec
16.1-7
There are M = 1615 different blocks, and H(X) = log 1615 = 60 bits/block at the rate
r =1/(15 + 5)ms = 50 blocks/sec, so R = 50 × 60 = 3000 bits/sec
16.1-8
Pdot + Pdash = 32 Pdot = 1 ⇒
Tdash = 2Tdot = 0.4,
1
r
Pdot = 23 , Pdash =
1
3
= T = 32× 0.2 + 13 × 0.4 =
so H ( X ) = 23 log 32 + 31 log3 = 0.920 bits/symbol
0.8
3
sec/symbol
Thus, R = (3/0.8) × 0.920 = 3.45 bits/sec
16.1-9
P3 = 1 – (P1 + P2 ) =
2
3
− p ⇒ H ( X ) = 13 log3 + plog
1
1
+ ( 23 − p ) log 2
p
( 3 − p)
When p = 0 or 2/3, H ( X ) = 13 log3 + 23 log 32 = 0.918 bits
16.1-10
Let P1 = α so Pi = (1 - α)/(M – 1) for i = 2, 3, …, M
H ( X ) = α log
1
1− α
M −1
1
+ (M − 1)
log
= α log + (1 − α) [log( M − 1) − log(1 − α ) ]
α
M −1
1− α
α
1


= log( M −1) + α  log − log( M − 1)  − (1 − α )log(1 − α)
α


But log(1 − α ) =
1
1
ln(1 − α ) =
− α − 12 α 2 − 13 α 3 − L) so
(
ln2
ln2
−(1 − α)log(1 − α) ≈
(1 − α)α
α
ln e
≈
=α
= α log e . Thus,
ln2
ln2
ln2
1


H ( X ) ≈ log( M − 1) + α  log − log( M − 1) + log e 
α


1
1
≈ log( M −1) + α log α if α ? M −1and α1 ? e
16-2
16.1-11
– log Pi W Ni < - log Pi + 1 ⇒ 2 log Pi ≥ 2− N i > 2log Pi 2−1 so Pi ≥ 2 − Ni > 12 Pi
M
Thus,
M
∑ Pi ≥ ∑ 2
i =1
−Ni
>
i =1
M
1
2
∑ Pi where
i =1
M
M
∑ Pi = 1 . Hence, 1 ≥ ∑ 2 − Ni >
i =1
i =1
1
2
⇒
1
2
< K ≤1
16.1-12
xi
Pi
1
2
3
4
5
Codewords
PiNi
A
1/2
0
0
0.5
B
1/4
1
0
10
0.5
C
1/8
1
1
0
110
0.375
D
1/20
1
1
1
0
1110
0.2
E
1/40
1
1
1
1
0
11110
0.25
F
1/40
1
1
1
1
1
11111
0.125
H(X) = 1.940 bits from Prob. 16.1-4, and N = 1.950 , so H ( X ) / N = 1.940/1.950 = 99.5%
16.1-13
Since PA = 0.4 and PA + PB = 0.6, the dividing line at the first coding step can be between A and B
or between B and C.
xi
A
B
C
D
E
F
Pi
0.4
0.2
0.12
0.1
0.1
0.08
0
1
1
1
1
1
0
0
1
1
1
0
1
0
1
1
0
1
Code I
0
100
101
110
1110
1111
PiNi
0.4
0.6
0.36
0.3
0.4
0.32
0
0
1
1
1
1
0
1
0
0
1
1
0
1
0
1
Code II
00
01
100
101
110
111
H(X) = 2.32 bits from Prob. 16.1-5, and N ≈ 2.4 , so H ( X ) / N ≈ 2.32/2.4 ≈ 97%
16.1-14
H(X) = 0.5 +
1 
1
1 
+ 0.1ln
 0.4ln
 = 1.36 bits
ln2 
0.4
0.1 
(a) N = 1.5 so H ( X ) / N = 1.36/1.5 = 90.5%
xi
A
B
Pi
0.5
0.4
0
1
0
Code
0
10
PiNi
0.5
0.8
16-3
PiNi
0.8
0.4
0.36
0.3
0.3
0.24
C
0.1
1
1
11
0.2
(b) 2 N = 2.78 so H ( X ) / N = 1.36/1.39 ≈ 97.8%
(cont.)
x ij
AA
AB
BA
BB
AC
CA
BC
CB
CC
Pij
0.25
0.2
0.2
0.16
0.05
0.05
0.04
0.04
0.01
0
0
1
1
1
1
1
1
1
0
1
0
1
1
1
1
1
1
0
1
1
1
1
1
0
0
1
1
1
0
1
0
1
1
0
1
Codewords
00
01
10
110
11100
11101
11110
111110
111111
16.1-15
H (X ) =
1 æ
1
1 ö÷
çç0.8ln
+ 0.2ln
÷ = 0.7219 bits
ln2è
0.8
0.2 ø
(a) 2 N = 1.56 so H ( X ) / N = 0.7219/ 0.78 = 92.6%
x ij
AA
AB
BA
BB
Pij
0.64
0.16
0.16
0.04
0
1
1
1
0
1
1
0
1
Codeword
0
10
110
111
PijNij
0.64
0.32
0.48
0.12
(b) 3N = 2.184 so H ( X ) / N = 0.7219/0.728 = 99.2%
x ijk
AAA
AAB
ABA
BAA
Pijk
0.512
0.128
0.128
0.128
0
1
1
1
0
0
1
Codeword
0
100
101
110
0
1
0
16-4
Pijk Nijk
0.512
0.384
0.384
0.384
PijNij
0.5
0.4
0.4
0.48
0.25
0.25
0.20
0.24
0.06
ABB
BAB
BBA
BBB
0.032
0.032
0.032
0.008
1
1
1
1
1
1
1
1
1
1
1
1
0
0
1
1
0
1
0
1
11100
11101
11110
11111
0.160
0.160
0.160
0.040
16.1-16
H ( X ) = P0 H ( X | 0) + PH
( X |1) =
1
1
2
[ H ( X | 0) + H ( X | 1) ]
P01 = P(0 | 1) = 3 / 4 ⇒ P11 = 1 − P (0 | 1) = 1 / 4
P10 = P(1| 0) = 3 / 4 ⇒
P00 = 1 − P(1| 0) = 1 / 4
Thus, H ( X | 0) = H ( X | 1) = 34 log 43 + 14 log4 = 0.811 and H ( X ) = 2 × 12 0.811 = 0.811 bits
16.2-1
P(x iyj) = P( xi | y j ) P( y j ) and
1
∑ P(x y ) = P( y ) so H (X , Y ) = ∑ P(x y )log P( x | y
i
j
i
j
x, y
x
=
é
å êêå
ëx
y
ù
1
P (x i y j )úlog
+
úû P ( y j )
j
å
P (x iy j )log
x ,y
i
j
) P( y j )
1
= H (Y ) + H ( X | Y )
P (x i | y j )
16.2-2
P( xi | y j ) = P(x i y j ) / P (y j ) so
I ( X ; Y ) = ∑ P(x i y j )log
x, y
where

1
1
1 
= ∑ P(x i y j )  log
+ log
− log

P( xi )P (y j ) x, y
P( xi )
P( y j )
P (x i y j ) 

P (x i y j )

1

1
∑ P(x y )log P( x ) =∑  ∑ P(x y ) log P( x ) = H ( X )
i
j
i
x, y
i
j
y
14
4244
3
x
i
P ( xi )
and likewise
1
∑ P(x y )log P( y ) = H (Y )
i
x, y
j
j
Thus, I ( X ; Y ) = H ( X ) + H (Y ) − ∑ P (x i y j )log
x,y
1
= H ( X )+ H (Y ) − H (X , Y )
P(x i y j )
16.2-3
 P( x )
P(xi y j ) = P ( y j | xi )P (xi ) =  i
0
M
j=i
⇒ P( y j ) = ∑ P( x i y j ) = P( x j ) .
j≠i
i =1
(cont.)
16-5
Thus, H (Y | X ) = ∑∑ P(x i y j )log
i
j
1
1
= ∑ P( xi )log
= ∑ P( xi )log1 = 0
P( y j | xi ) i
P( yi | xi ) i
I ( X ; Y ) = H (Y ) − H (Y | X ) = H ( X ) − 0 = H ( X )
16.2-4
P(y1 ) = p(1 - α) + (1 – p)β = β + (1 -α - β)p, P(y2 ) = (1 – p)(1 - β) + pα = 1 - β - (1 - α β)p
= 1 – P(y1 ) so H (Y ) = P ( y1 )log
1
1
+ P( y2 )log
= Ω β + (1 − α − β ) p 
P( y1 )
P ( y2 )
1
1

1
1 

H (Y | X ) = p (1 − α)log
+ α log  + (1 − p )  β log + (1 −β)log
1− α
α
β
1 − β 


= pΩ( α ) + (1 − p) Ω( β) so
I ( X ; Y ) = H (Y ) − H (Y | X ) = Ω[β + (1 − α − β) p ] − pΩ( α ) − (1 − p )Ω( β)
16.2-5
If β = 1 - α, then P(y1 |x 1 ) = P(y1 |x 2 ) and P(y2 |x 1 ) = P(y2 |x 2 ) so the occurrence of y1 or y2 gives
no information about the source. Analytically, if β = 1 - α, then Ω(β) = Ω(1 - α) and Ω[β + (1 α - β)p] = Ω(1 - α). Thus, I ( X ; Y ) = Ω( α) − pΩ ( α ) − (1 − p) Ω( α) = 0 so no information is transferred.
16.2-6
Ω (a + bp) = (a + bp)log
1
1
+ (1− a − bp )log
a + bp
1 − a − bp
= − [ ( a + bp)log( a + bp ) + (1− a − bp)log(1 − a − bp )] and
so
d
1 dx
[log 2 x( p )] = ( log 2 e )
dp
x dp
d
b
Ω ( a + bp) = −b log( a + bp ) − ( a + bp )(log e)
− (−b)log(1 − a − bp)
dp
a + bp
−(1 − a − bp )(log e)
−b
1 − a − bp
= −b [ log( a + bp ) − log(1 − a − bp ) ] = b log
16-6
1 − a − bp
a + bp
(cont.)
d
d
1 − α − (1− 2 α) p
I(X ; Y ) =
Ω [α + (1 − 2 α) p ] + 0 = (1− 2 α)log
= 0 so
dp
dp
α + (1 − 2 α) p
1 − α − (1 − 2 α ) p
= 1 ⇒ 1 − 2α = 2(1 − 2 α) p ⇒
α + (1 − 2 α) p
p=
1
2
16.2-7
I ( X ; Y ) = Ω( p /2) − pΩ (1/2) − (1 − p )Ω (0) = Ω ( p / 2 ) − p since Ω (1/2)=1 and Ω(0)=0
and
d
1− 1 p
2− p
Ω ( 0 + 12 p ) = 12 log 1 2 = 12 log
so
dp
p
2 p
d
2− p
I ( X ; Y ) = 12 log
−1 = 0 ⇒
dp
p
2− p
= 22
p
⇒
p=
2
5
Thus, Cs = Ω ( 15 ) − 25 = 15 log5 + 45 log 54 − 25 = 0.322 bits/symbol
16.2-8
I ( X ; Y ) = Ω(3 p /4) − pΩ(1/4) − (1 − p )Ω (0) = Ω(3 p /4) − 0.811 p since Ω(0)=0
and
d
1− 3 p
4−3p
Ω ( 0 + 34 p ) = 34 log 3 4 = 34 log
so
dp
p
3p
4
d
4−3p
I ( X ; Y ) = 34 log
− 0.811 = 0 ⇒
dp
3p
4−3p
×
)
= 2( 4/30.811
= 2.12 ⇒
3p
p = 0.428
Thus, Cs = Ω (0.321) − 0.811 ×0.428 = 0.557 bits/symbol
16.2-9
P(0) = P(1) = 12 (1 − α ), P( E) = 12 α + 12 α = α
H (Y ) = 2
1− α
2
1
1
1
log
+ α log = (1 − α )log2 + (1 − α )log
+ α log = 1 − α + Ω( α)
2
1− α
α
1− α
α
1
1
1

H (Y | X ) = 2 × 12 (1 − α)log
+ α log + 0log  = Ω( α)
1− α
α
0

Thus, Cs = max I (X ; Y ) = H (Y ) − H (Y | X ) = 1 − α
16.2-10
P( n − n ≥ kσ ) = P( n − n ≥ k σ) + P( n − n ≤ −k σ) ≤ 1/ k 2 so
P( n ≥ n + k σ ) ≤
1
k2
− P (n − n ≤ −k σ) ≤ 1/ k 2
16-7
d−n
where d = N β, n = N α , σ2 = Nα (1 − α)
σ
Let d = n + k σ ⇒ k =
σ 
N α (1 − α)
α(1 − α)
Then P( n ≥ d ) ≤ 
=
 =
2
( Nβ − N α)
N (β − α ) 2
d −n
2
16.3-1
p ( x ) = 1 / 2a for
∞
x ≤ a, S = ∫ x 2 p ( x ) dx = a 2 / 3
−∞
1
log2adx = log2 a = log 12 S = 12 log12 S < 12 log2πeS since 2πe =17.08 > 12
− a 2a
H (X ) = ∫
a
16.3-2
p (x ) =
α −α x
2
1
2
e
and S = 2 . For x D 0, log
= log + α x log e , so
2
α
p (x )
α
∞α
2
21
1
 ∞ α − αx


− αx
2
H ( X ) = 2 ∫
e log dx + ∫
e αx log edx  = α  log  + α ( log e ) 2
0 2
α
α α
α
 0 2


= log
2
2e
+ log e = log
= log 2e2 S = 12 log2e 2S < 12 log2πeS since π > e
α
α
16.3-3
∞
S = ∫ ax 2e− ax dx =
0
2
1
, log
=ax log e − log a for x D 0
2
a
p( x)
∞
∞
H ( X ) = ∫ ae − ax ( ax log e) dx − ∫ ae− ax (log a ) dx = a2 (log e)
0
0
1
e
− log a = log
2
a
a
= log e 2S / 2 = 12 log e 2 S / 2 < 12 log2πeS since 2π > e/2
16.3-4
pZ ( z) =
1
 z −b
pX 

a
 a 
(cont.)
16-8
∞
H ( Z ) = ∫−∞




1
1
 z −b 
 dz
pX 
log
a
+
log

a
 z −b 
 a 
pX 


 a  
= log a
= log a
∞
1
dz
 z − b  dz
 z −b 
p
+
X


∫−∞  a  a ∫−∞ pX  a  log  z − b  a
pX 

 a 
∞
∫
∞
−∞
∞
p X (λ ) d λ + ∫ pX (λ)log
−∞
1
d λ = log a + H ( X )
pX ( λ )
16.3-5
∫
∞
0
p( x) dx = 1 ⇒ F1 = p , c1 = 1,
Thus, −
∫
∞
0
ln p + 1
+ λ1 + λ 2 x = 0
ln2
∂F1
= 1 and
∂p
⇒ p = e(
∫
∞
0
xp( x) dx = m ⇒ F2 = xp, c2 = m,
λ1 l n 2 −1)
e(
λ 2 l n 2) x
= Ke −ax
∂F2
=x
∂p
x≥0
∞
Ke − ax dx = K / a = 1, ∫ xKe − ax dx = K / a2 = m ⇒ K = a = 1/ m
0
∞
x
Hence, p( x) = 1 e − x / mu (x) and H ( X ) = ∫ p ( x )  log m + log e dx = log m + log e = log em
0
m
m


16.3-6
∫
∞
0
p( x) dx = 1 ⇒ F1 = p , c1 = 1,
Thus, −
∫
∞
0
∂F1
= 1 and
∂p
ln p + 1
+ λ1 + λ 2 x 2 = 0 ⇒
ln2
Ke − ax dx =
2
Hence, p ( x ) =
H (X ) = ∫
∞
0
K
2
p = e(
∫
∞
0
x 2 p( x) dx = S ⇒ F2 = x 2 p , c2 = S ,
λ1 l n 2−1 )
e(
λ 2 ln2) x 2
∞
2
π
K
π
= 1, ∫ x 2 Ke −ax dx =
=S
0
a
a a 4
= Ke − ax
⇒
2
K=
∂F2
= x2
∂p
x≥0
2
1
,a =
2S
2πS
2
2
e − x / 2 S u ( x ) and
2 πS


πS x 2
πS log e
πS 1
πeS
p ( x )  log
+
log e  dx = log
+
S = 12 log
+ 2 log e = 12 log
2 2S
2
2S
2
2


16-9
16.3-7
∞
pX ( x) pY ( y )
dxdy and
p XY ( x , y )
I ( X ; Y ) = − ∫ ∫ p XY ( x , y)log
−∞
log
p X ( x ) pY ( y)
1
p ( x ) pY ( y )
1  pX ( x) pY ( y ) 
=
ln X
≤
−1 
pXY ( x , y )
ln2
p XY ( x , y )
ln2  p XY ( x , y)

Thus, I ( X ; Y ) ≥
∞
 p X ( x ) pY ( y ) 
1
p
(
x
,
y
)
XY
1 − p ( x, y )  dxdy
ln2 ∫ ∫−∞

XY

=
where
∫∫
∞
−∞
p XY ( x , y) dxdy = 1,
∞
1  ∞
p XY ( x , y) dxdy − ∫ ∫ p X ( x ) pY ( y ) dxdy 
∫
∫
−∞

l n 2  −∞
∫∫
∞
−∞
p X ( x) pY ( y) dxdy =
Hence, I(X;Y) D 0
16.3-8

S  B  10 4 
R ≤ C = B log  1 +
ln  1 +
=

B 
 N0 B  ln2 
B = 10 3 ⇒
B = 10 4
R≤
103
ln11 = 3459 bits/sec
ln2
⇒ R ≤ 10 4 log 2 2 = 10,000 bits/sec
B = 10 5 ⇒
R≤
105
ln1.1 = 13,750 bits/sec
ln2
16.3-9
R ≤ C = 3000log(1 + S / N ) ⇒ S / N ≥ 2 R /3000 −1
R = 2400 ⇒ S / N ≥ 20.8 −1 = 0.741 ≈ −1.3 dB
R = 4800 ⇒ S / N ≥ 21.6 − 1 = 2.03 ≈ 3.1 dB
R = 9600 ⇒ S / N ≥ 23.2 −1 = 8.19 ≈ 9.1 dB
16-10
∫
∞
−∞
∞
p X ( x ) dx ∫ pY ( y ) dy = 1
−∞
16.3-10
S
B
≥ ( 2R / B −1) , N 0 B = 10 −6 ×103 = 10 −3,
N0 R R
S ≥ 10 −3 ( 2R /1000 − 1)
R = 100 ⇒ S ≥ 10−3 (20.1 − 1) = 0.072 mW
R = 1000 ⇒ S ≥ 10−3 (2 −1) = 1 mW
R = 10,000 ⇒ S ≥ 10 −3 (210 −1) = 1023 mW
16.3-11
For an ideal system with the same parameters,

SR 
S

  = 1+
N0 BT 
 N D 
BT / W
− 1 = (1 + 3)10 −1 = 60.2 dB
Since the claimed performance approaches an ideal system, the claim is highly doubtful.
16.3-12
4
S
 γ
b = 4 :   =  1 +  − 1 = 104
 N D  4 
⇒ γ ≈ 36
12
S
 36 
b = 3 × 4 :   =  1 +  −1 ≈ 412 = 72.2 dB
 N  D  12 
16.3-13
4
S
 γ
b = 4 :   =  1 +  − 1 = 104
 N D  4 
⇒ γ ≈ 36
1/2
36 
S

b = 12 :   =  1 +

 N D  1 / 2 
− 1 = 7.54 = 8.8 dB
16.3-14
b = 4,
LN0 = 106 ×100 ×10−12 = 10−4
4
S
 γ
3
 N  = 1 + 4  − 1 = 10
 D 

⇒ γ = 4 (10011/4 − 1) = 18.5 ⇒ ST = γLN 0W = 5.55 W
16-11
16.3-15
LN0 = 106 ×100 ×10−12 = 10−4
b = 2,
2
S
 γ
3
 N  = 1 + 2  − 1 = 10
 D 

⇒ γ =2
(
)
1001 − 1 = 61.3 ⇒ ST = γLN 0W = 36.8 W
16.4-1
2
v = Ev = 4,
2
2
w = Ew = 4,
2
v+w = v + w
2
z
2
2
= v + w = Ez = 8
2
⇒ ( v, w ) = 0, Ex = x = 6 2 + 12 = 37
3v = 6
z =2 2
w =2
− 0.5w = 1
x
v =2
16.4-2
2
v = Ev = 4,
2
2
2
w = Ew = 12,
v+w = v + w
2
z
2
2
= v + w = Ez = 16
⇒ (v, w ) = 0,
Ex = x
2
= 62 +
z =4
( 3)
2
= 39
3v = 6
w =2 3
− 0.5w = 3
x
v =2
16.4-3
v + w = v + 2(v , w) + w and (v , w) ≤ v w so
2
2
2( v, w ) ≤ 2 v w
2
⇒
v+ w ≤ v +2 v w + w =
2
2
2
16-12
(v+
w
)
2
. Thus, v + w ≤ v + w
16.4-4
With v w = αw, (v – v w,w) = (v - αw,w) = (v,w) - α(w,w) so (v – v w,w) = 0
α = (v,w)/(w,w) = (v,w)/||w||2
2
Hence, vw =
( v, w)
w
2
w and vw
2
 (v, w) 
2
(v , w) 2
= (α w, αw) = α2 w = 

w
=
2
2
 w 
w
2
But (v,w)2 W ||v||2 ||w||2 so vw ≤  v

2
1/2
2
2
w / w 

= v
16.4-5
s1
= ∫ 1 dt = 2 ⇒ ϕ1 = s1 / s1 = 1/ 2 t ≤ 1
1
2
−1
( s2 ,ϕ1 ) = ∫
1
−1
( s3 ,ϕ1) = ∫
1
−1
t
dt = 0 ⇒
2
g 2 = s2
1
t2
2
dt =
, ( s3 , ϕ2 ) = ∫ t 2
−1
3
2
2
ϕ1 = t 2 − 13 and g3
3
g 3 = s3 −
⇒
Thus, s1 = 2 ϕ1 ,
s2 =
2
3
ϕ2 ,
2
3
2
1
= ∫ t 2 dt =
2
g2
−1
2
−1
s3 =
8
45
⇒ ϕ2 =
s2
=
2/3
3
t t ≤1
2
tdt = 0 so
= ∫ ( t 2 − 13 ) dt =
1
2
3
ϕ3 + 13 =
2
3
8
45
⇒ ϕ3 =
ϕ1 +
8
45
45
8
(t
2
− 13 )
ϕ3
16.4-6
s1
= ∫ 1 dt = 2 ⇒ ϕ1 = s1 / s1 = 1/ 2 t ≤ 1
1
2
−1
1
( s2 ,ϕ1 ) = ∫ cos
−1
g2
2
πt 1
2 2
dt =
2 2
π
⇒
g 2 = s2 −
2 2 1
π
2
( s2 − 2π )
4
πt 4 
8

2 πt
=
= ∫  cos
− cos + 2  dt = 1 − 2 so ϕ2 =
−1
g2
2 π
2 π 
π

1
( s3 ,ϕ1) = ∫ sin πtdt = 0, ( s3 , ϕ2 ) = ∫ sin πt
g 3 = s3 ,
1
1
−1
−1
g3
2
πt 2 

 cos 2 − π  t ≤ 1

π −8 
π
2
πt 2 

 cos 2 − π  dt = 0 , so

π −8 
2
= ∫ sin 2 πt d t = 1 ⇒ ϕ3 = s3 = sin πt t ≤ 1 . Thus,
s1 = 2 ϕ1 , s2 =
π
1
−1
π2 − 8
2 2 2
π2 − 8
ϕ2+ =
ϕ1 +
ϕ2 , s3 = ϕ3
π
π
π
π
16-13
16.5-1
(a) ϕ2
c1 = c2 = a2 /2 so take
a
s2
z1 = (y,s1 )
z2 = (y,s2 )
ϕ1
s1
a
z1
×
∫
S/H
s1 (t)
Choose
y(t)
larger
m̂
z2
×
∫
S/H
s2 (t)
c1 = c2 = a2 so take
(b)
s2
s1
-a
0
ϕ1
z1 = (y,s1 ) z2 = (y,s2 ) = -(y,s1 )
a
Since z2 = -z1 , z1 > z2 ⇒ z1 – z2 = 2z1 > 0 ⇒ z1 > 0
y(t)
×
∫
S/H
Choose
0
larger
s1 (t)
m̂
(c)
s2
s1
0
a/2
ϕ1
z1 = (y,s1 ) – a2 /2
a
z2 = 0
z1 > z2 ⇒ (y,s1 ) > a2 /2
y(t)
×
∫
s1 (t)
S/H
Choose
a2 /2
larger
16-14
m̂
16.5-2
ϕ2 a
(a)
Since all ci = a2 /2 take
s2
z1 = (y,s1 ), z2 = (y,s2 )
s3
ϕ1
s1
-a
z3 = (y,s3 ) = -(y,s1 ), z4 = (y,s4 ) = -(y,s2 )
s4
×
∫
S/H
s1 (t)
-1
Choose
y(t)
largest
×
∫
m̂
S/H
s2 (t)
-1
(b) ϕ2
a
s2
c1 = c2 = a2 /2, c3 = a2 , c4 = 0
s3
z1 = (y,s1 ) – a2 /2, z2 = (y,s2 ) – a2 /2
z3 = (y,s1 + s2 ) – a2 = z1 + z2
a/2
z4 = 0
s4
0
s1
a/2
ϕ1
a
×
∫
S/H
+
s1 (t)
Choose
-a2 /2
y(t)
×
∫
S/H
+
largest
+
s2 (t)
0
16-15
m̂
16.5-3
ϕ2
s4
a
s1
Ei =
(
2a
)
2
for all i so E = 14 × 4
(
2a
)
2
= 2 a2
ϕ1
-a
a
-a
s3
s2

 a
For all i, P( c | mi ) = ∫− a pβ (β1) dβ1 ∫− a pβ (β 2 ) d β2 = 1 − Q 
 N /2

0

∞
∞
2

 E 
 E 
2
Thus, Pe = 1 − 14 × 4 1 − Q 
 = 2Q 
 − Q
N
N
0 
0 




2

  where a =


 E 


 N0 
16.5-4
Ei = ( 2 a) 2 i = 1,2,3,4
ϕ2
s4
a
ϕ1
s5
-a
s3
E5 = 0
s1
E = 15 × 4( 2 a) 2 = 85 a 2
a
-a
s2
Since s5 is nearest neighbor to all other si, d j = 2a = 5 E / 4
 2a
Thus, Pe ≤ 45 × 5Q 
 2N
0


 5E
 = 4Q 
 8N 0




16-16
j = 1,2,...,5
E/2
16.5-5
ϕ2
2a
Ei = (2a)2
s4
s3
= 2(2a)2 i = 2, 4, 6, 8
s2
a
=0
s5
-2a
s9
i = 1, 3, 5, 7
i=9
s1
-a
a
ϕ1
2a
E = 19  4 × (2 a) 2 + 4 × 2(2a )2  =
48
9
a2
-a
-2a
s6
s7
 a 
 9E
Let q = Q 
= Q 

 N /2 
0
 24 N0


s8



P( c| m9 ) = ∫ pβ (β1 ) dβ1 ∫ pβ (β2 ) d β2 = (1 −2 q) 2
a
a
−a
−a
∞
a
−a
−a
For i = 1, 3, 5, 7 P( c| mi ) = ∫ pβ (β1) dβ1 ∫ pβ (β2 ) d β2 = (1 − q )(1 − 2q)
∞
∞
For i = 2, 4, 6, 8 P( c| mi ) = ∫ pβ ( β1 ) d β1 ∫ pβ ( β 2 ) d β2 = (1 − q ) 2
−a
−a
Thus, Pc = 19  4(1 − q )(1 − 2 q ) + 4(1 − q) + (1 − 2 q )2  = 19 ( 9 − 24q + 16 q 2 ) and
2
Pe = 1 − Pc =
24
9
q − 169 q =
2
24
9
 9E
Q 
 24 N 0
 16 2  9E
 − 9 Q 

 24 N 0



 2a 
 9E
For union bound note that dj = 2a, j = 1, 2, ..., 9 so Pe ≤ 89 × 9Q 
= 8Q 

 2N 
0 
 24 N 0

16-17



Appendix
A-1
4R2 kT2
R1
R1 + R2
R2
⇒
4R1 kT1
vn ( f ) = 4R1kT 1 + 4 R2k T 2 , in ( f ) =
2
2
v n 2(f)
2
vn ( f )
Z( f )
2
=
4k ( R1T1 + R2T 2 )
( R1 + R2 )
2
If T1 = T2 = T, then vn 2 ( f ) = 4( R1 + R2 ) kT , in 2 ( f ) =
2
4 kT
R1 + R2
A-2
4kT1 /R1
4kT2 /R2
R1
R2
⇒
in 2(f)
R1 R2 /(R1 + R2 )
in ( f ) =
2
4kT 1 4 kT 2
+
R1
R2
2
 R R  4 kR2T 1 + 4kR1T 2
4k R1 R2
vn ( f ) = Z ( f ) in ( f ) =  1 2 
=
( R2T 1 + R1T 2 )
R1 R2
( R1 + R2 ) 2
 R1 + R2 
2
2
2
If T1 = T2 = T, then in 2 ( f ) = 4
R1 + R2
RR
2
k T and vn ( f ) = 4 1 2 kT
R1 R2
R1 + R2
A-3
9(1 + jf )
Z( f ) =
10 + jf
10 + f 2
⇒ R ( f )= 9
100 + f 2
10 + f 2
⇒ vn ( f ) = 36kT
100 + f 2
2
A-4
Z( f ) =
R( R − j / ωC )
2 R 2 + (1/ ωC ) 2
⇒ R( f ) = R 2
2 R − j / ωC
4 R + (1/ ωC ) 2
A-1
⇒ vn2 ( f ) = 4 Rk T
1 + 2(2 πRCf ) 2
1 + 4(2 πRCf ) 2
A-5
If qV/kT » 1, then I » Is so in2(f) ≈ 2qI and r ≈ kT/qI. Thus, in2(f) ≈ 2rkT = 4(r/2)kT, which looks
like thermal noise from resistance R = r/2.
A-6
2
 H ( f ) 50  50 1
2
With Rs = ri = ro = 50 Ω, Eq. (8) yields g a ( f ) = 
= H ( f ) . Then,

 100  50 4
∞
since Π 2( ) = Π( ), Eq. (10) becomes gBN = ∫ g a ( f ) df =
0
Then, from Eq. (11), k T e =
1
gBN
∫
∞
0
ηint ( f ) df =
2002
4
∫
∞
0
 f − fc
Π
 B

4
10
 df = 10 B = 10 .

2 × 10−16 B
= 2 ×10 −20 . Finally, with Ts = T0 ,
1010
Eq. (12) becomes
N o = gBN (k T s + kT e ) ≈ 1010 (4 ×10 −21 + 2 ×10−20 ) = 2.4 × 10−10 = 240 pW.
A-7
Ts = T0: N o = 105 k (T 0 + T e ) × 20 ×103 = 8 ×10 −12
T0 + Te
= 80 ×10−12 so (T0 + Te)/T0 = 10 and
T0
Te = 9T0 , F = 1 + 9 = 10
Ts = 2T0 : N o = 105 × 4 ×10−21
2T 0 + 9T 0
× 20 ×10 3 = 88 pW
T0
A-8
Ts = T0: N o = gk (T 0 + T e ) BN ,
Ts = 2T0 : N o = gk (2T 0 + T e ) BN = 43 gk (T 0 + T e ) BN so
2T 0 + T e = 43 (T 0 + T e ) . Thus, Te = 2T0 , F = 1 + 2 = 3.
A-9
Rs, T0
Assume matched impedances (Rs = ri), so
N1 = CNo = Cgk(T0 + Te)BN
N2 = N1 + CgS = 2N1 ⇒ S = k(T0 + Te)BN
Thus, Te =
S
− T0 ⇒
kBN
F=
vs
S
k T 0 BN
ri
(cont.)
A-2
To determine F by this method, we need an absolute power meter to measure S, and we must also
know BN. Both requirements are disadvantages compared to the noise-generator method.
A-10
Rs = 300 Ω
300 Ω
+
Pi, ri ⇒
vs
50 Ω
v L PL ⇒
RL = 50 Ω
–
50 × 50
 325

= 325 Ω Pi = 
vs  / ri
50 + 50
 300 + 325 
2
50 50
 25 
vL =
v PL = 
vs  / RL
300 + 300 + 50 50 s
 625 
2
ri = 300 +
2
P  25vs  1  625 
1
g = L =
×
. Thus, F = L = 1/g = 26.
 × 325 =

Pi  625  50  325vs 
26
2
A-11
Te = Te1 + Te2/g where F2 = 13.2 dB = 21 ⇒ Te2 = (21 – 1)T0, so
Te = 3T0 + 20T0/10 = 5T0 and
Ss
Ss
S
3
−12
 N  = k (T + T ) B = 15kT B = 10 ⇒ Ss = 6 ×10 = 6 pW
 o
s
e
N
0 N
A-12
1S
S
S
Since Ts = T0 ,   =   ≥ 0.05  
 N o F  N  s
 N s
⇒ F ≤ 20
Since F1 = 1/g1 = L and F2 = 7 dB ≈ 5, F = F1 +
F2 − 1
= L + L(5 − 1) = 5L
g1
Thus, we want 5L W 20 ⇒ L W 4 = 6 dB ⇒ l W 6 dB/2 dB/km = 3 km
A-13
Preamp: F = 2 and g = 100; Cable: F = 4 and g = 1/4; Receiver: F = 20
(cont.)
A-3
(a) F = 2 +
(b) F = 4 +
4 −1
20 − 1
+
= 2.79 = 4.5 dB
100 100 ×1 / 4
2 −1
1
+
4
20 − 1
= 8.76 = 9.4 dB
1 ×100
4
A-14
L1 = 100.15 = 1.413, g2 = 100, F3 = 10
T e = 0.413T + 1.413 ×50K+
1.413
(10 −1)290K=0.413T + 107.5K
100
so Te W 150 K ⇒ T W 103 K
A-15
With unit #1 first, cascade has F12 = F1 + (F2 – 1)/g1 . Similarly, interchange the subscripts for unit #2

F −1 
F − 1 
F − 1 
F −1
first. Thus, F12 − F21 =  F1 + 2  −  F2 + 1  = F1 − 1  − F2 − 2 
g1  
g2  
g2  
g1 

If unit #1 first yields F12 < F21 , then F12 – F21 < 0 and
F1 −


F1 −1
1 
F −1
1
= ( F1 − 1) 1 −  + 1 < F2 − 2
= ( F2 −1)  1 −  +1 so
g2
g1
 g2 
 g1 
F1 − 1
F −1
< 2
1 − 1/ g1 1 − 1/ g 2
⇒ M1 < M 2
A-4