Communication Systems [4 ed.] 9780070111271, 0070111278

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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

∫

T0 / 4

0

0 0

n cn

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

∫

T0 / 2

0

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

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  4

4

∞  2  2  1 1 1  1 P =2∑   = 2    4 + 4 + 4 +L = π  1 3 5  3 n odd  π n 

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 

2

n even 0 cn =  2  (2/ π n) n odd

2

∞ 1 2 1  1  P =   + 2∑   = + 2 4 4π 2 n =1  2π n 

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 

2

∞

0

df

(a

2

+ f2)

2

2

dx

(a

∫

+ 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 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

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
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 − 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
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
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 ), vR' (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
0, K Since ε(t ) can make a step change at t = 0, ε(0+ ) =

∆f + f1 ∆f + B = ε(0− ) = K K

Hence,

⎧ ∆f ⎪⎪ K ε(t ) = ⎨ ⎪ ∆f + f1 (1 − e −2 πKt ) ⎪⎩ K K

⇒ B=−

f1 K

t0

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 ) =

with A(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).

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 Z( f ) =

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 )

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

c(nf 0 ) = Ksinc

n 2 Same spectrum as (a) with f h replaced by f v ฀ f h , so much smaller bandwidth.

7-19

7.4-3 (a)

⎧ ⎪1 I (h, v) = ⎨ ⎪⎩0 cmn

1 = HV

H αH H αH V β V V βV − 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 z0

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 )
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 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

= = Xi2 + 2XiYi + Yi2 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

=

 = 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

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 2

E [ ( Y − mY )(Z − mZ ) ] = E [ y (t ) y ( t − T ) ] = Ry (T ) = σ2sinc2BT so ρ = σ sinc 0.5 = 0.637 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 2Rk TN

+

R

A cos 2πf 0t

C

N = k TN/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 -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 ∫

∞

≈ N0 πBT/2 since2fc/BT = 2Q » 1

π   + arctan 2 fc  2 BT  

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.)

2

2

y = 0, σy 2 = y 2 = (An − An ) = An 2 − An = 8 − 2π

⇒

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 ) σn 2 dy 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 T N = k T 0 (TN / 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 T N = k T 0 (TN / T 0 ) = 4 × 10−21 × 10 = 4 × 10−20 S  0.4 20 × 10−9 4   =  N D 1 + 0.4 4 × 10−20 × 5 × 106 = 2.86 × 10 = 44.6 dB 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  N 0 fc ln (1 + W / fc ) ln (1 + W / fc ) 1.01γ W / fc = 1/ 50 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 γ = 104 With S x = ,   = 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 γ SR SR 2 2

10.3-1 N 0 = k T N = k T 0 (TN / 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

H D ( f ) G ξ ( f )df ≤ ∫ 2

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 ) =

f2 2S R 1 + (2 f / BT )2 N0

 2W  2W      B − arctan  B    T   T

3   3   ≈ N 0BT 3 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 (TN / 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  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 2

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   2S γ ≈   φ 2S x γ   = = φ x ∆ π  N D N 0Bde arctan(W / Bde ) arctan(W / Bde )  Bde  ∆ 10.3-6 S D = f∆ x = f∆ / 2 and N D = 2∫ 2

2

300

2

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

N D = 2∫ e −( f / Bde ) 0

2

N0f 2 2S R

N 0Bde 3

df =

SR

4 f 2S S S  4  W    > ∆ x R =    N D πN 0Bde 3 π  Bde 

3

∫

W / Bde

0

λ 2e −λ d λ < 2

N 0Bde 3 SR

∫

0

∞

λ 2e −λ d λ = 2

 f∆    S x γ so W  2

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 γ = × 1010 × 10− (b)   =  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

  = 1.58 × 108  2

2

and gT × g R = 52 dB = 1.58 × 105 so

107 2

(a) (S / N )D = γ = 107 /

2

⇒

= 104

S  1/ 2 (b)   = γ = 107 / 3  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 γ = 20 AM:   =  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

Sk = ( f∆α k ) x k 2 = f∆2α k 2 , N k = 2 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  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 2

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

2

2 N  f  0 ( f − f )2 + ( f + f )2  = H ( f ) ( f 2 + f02 )Π  0 0  de   2W  2S R 2S R

N0

f 2 + f02

2N 0 df = 2 S R 1 + ( f / Bde ) SR

N0

W / Bde W / Bde  λ2 dλ  2 Bde 2 ∫  λ + d 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   f τ  N W  N D 1.2 × 106 /12 × 106  4 × 10−20 × 500 × 103  s 0 0

= 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

W  = 36    f 

2

⇒

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

⇒

A2 ≤ BT S R / Mfs

 1 1 1 T 2  1 BT − 2Mfs µ p = t0 =  s − τ − Tg  ≤  − =   2  Mfs B  20 Mfs B 2 M T T  B − 2Mfs  B 2S  B − 2Mfs   S   T R S x =  T    ≤  T   N D  Mf B  Mf    Mfs BT  s T s 2

2

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 P ( B) = e−π ( B / b ) ≤ 0.01 P(0)

⇒ π( B / b) 2 ≥ ln100

2

π(bD )2 ≥ ln100 where D = 1/ r

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, σ2a = 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 )]

XT ( f ) =

1

k

k =− K

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

10-4

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 ) 0 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

(S / N )R

Pe

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 ' =

11 (1 − 15α ) = 0.681, γ b = 2.6 2 / 2Rc ' = 7.0dB 15

1/4

 3×2  α = × 10−5   15 × 14 × 13 

= 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

1/ 4

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 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 then N c − f d Tb >> 1 so |sinc4( Nc ± f d Tb ) |> 1, then |sinc(2 Nc t / Tb ) |> 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

e

Then pφ (φ ) =

∞  1 − Ac2 / 2σ 2 − λ 2 / 2  ∞ − λ 2 −λ2 e e  ∫ λ e d λ + λ0 ∫ e d λ  2π  −λ0  λ0

= ∞

∫

where

λe − λ

0

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 • • • • o • o •

o o o o

o o o o

o o • o

o o o o

• o o o

• o • o

o o o o

• o • o

o o o o

• o o o

• o • o

o • • o o • • o

d1 =

o • o •

o o o o

o o • o

o o o •

o o o o

o • o o

o o o o

o o o o

• • • •

• • • •

d0 = 1

• • • •

• o o • • o o•

2

o • o •

o o o o

o • o o

• o oo • o oo

d2 = 2

o o o o

o o o •

• o oo oo oo

o o o o

o o • o

o o o o

• o • o

o o • o

o o o o

o o o 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 • o •

o o o o

14.5-3

State:

• o • o

o o o o

o • o o

o o o o

o o o •

o • o •

o o o o

o • o •

o o o o

o o 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 + 32 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

0.032 0.032 0.032 0.008

ABB BAB BBA BBB

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 2

2

2

v + w = v + 2(v , w) + w and (v , w) ≤ v w so

2( v, w ) ≤ 2 v w

⇒

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

= ∫ 1 dt = 2 ⇒ ϕ1 = s1 / s1 = 1/ 2 t ≤ 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 + 31 =

2 3

8 45

⇒ ϕ3 =

ϕ1 +

8 45

45 8

(t

2

− 13 )

ϕ3

16.4-6

s1

1

= ∫ 1 dt = 2 ⇒ ϕ1 = s1 / s1 = 1/ 2 t ≤ 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

1

1

−1

−1

( s3 ,ϕ1) = ∫ sin πtdt = 0, ( s3 , ϕ2 ) = ∫ sin πt

g 3 = s3 ,

g3

2

πt 2    cos 2 − π  t ≤ 1  π −8 

π

2

πt 2    cos 2 − π  dt = 0 , so  π −8  2

1

= ∫ sin 2 πt d t = 1 ⇒ ϕ3 = s3 = sin πt t ≤ 1 . Thus,

s1 = 2 ϕ1 , s2 =

π

−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

z1 = (y,s1 )

s2

z2 = (y,s2 )

ϕ1

s1 a

z1

×

∫

S/H

s1 (t)

Choose larger

y(t)

mˆ

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)

mˆ

(c) s2

s1

0

a/2

ϕ1

z1 = (y,s1 ) – a2 /2 z2 = 0

a

z1 > z2 ⇒ (y,s1 ) > a2 /2 y(t)

×

∫ s1 (t)

S/H

Choose

a2 /2

larger

16-14

mˆ

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 largest

y(t) ×

∫

mˆ

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

mˆ

16.5-3 ϕ2 s4

a

Ei =

s1

(

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

=0

a 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 k T2 R1

R1 + R2

R2 ⇒

4R1 k T1

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 4k T1 /R1

4k T2 /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/k T » 1, then I » Is so in2(f) ≈ 2qI and r ≈ k T/ qI. Thus, in2(f) ≈ 2rk T = 4(r/2)k T, 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 = 10 k (T 0 + T e ) × 20 ×10 = 8 ×10 5

3

−12

T0 + Te T0

= 80 ×10−12 so (T0 + Te)/T0 = 10 and

Te = 9T0 , F = 1 + 9 = 10 Ts = 2T0 : N o = 10 × 4 ×10 5

−21

2T 0 + 9T 0 T0

× 20 ×10 3 = 88 pW

A-8 Ts = T0: N o = gk (T 0 + T e ) BN , Ts = 2T0 : N o = gk (2T 0 + T e ) BN =

4 3

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 Assume matched impedances (Rs = ri), so

Rs, T0

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