Vo tuyen so

7/31/2019 Vo tuyen so

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HNG DN GII BI TP

CHNG 2Bi 1Cho mt dy xung ch nht bin A, chu k T v thi gian xung t (t


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[ ]T

1lim exp(-2T)-1 02T

= =w

c) L hm kiu nng lng

Bi 4Cho hm

1/ 1 t+

a) Tm tng nng lng b) Tm cng sut trung bnha) L hm kiu g?

Gii

a)2

e 0

- -

1E[ ]= s(t) dt dt log (1 t) Joule1 t

= = + = + b)

2

T T- -

1 1 1P[ ]= lim s(t) dt lim dtT T 1 t

= + eT

1lim log (1 T) 0T

w

= + =c) L hm kiu nng lng

Bi 5Tm ACF v PSD ca hm cosin di y

s(t)=Acos(2 f 1t+ )tma) ACF b) PSDc) Cng sut trung bnh

Giia) V y l hm tun hon c gi tr thc nn

0

1 1( ) lim ( ) ( ) ( ) ( )+

= + = +

T T

T

s t s t dt s t s t dt T T

{ }2 2

1 1 10

os(2 f t+ ) os[2 f (t+ )+ ] os(2 f )2

= = T A Ac c dt c

T

b) Ta c th biu din

1 12 f 2 f 1

1 os(2 f )2

= + j jc e e

V th

2


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[ ]2 2

1 1A A(f ) F cos(2 f ) (f f ) (f f )2 4

= = + +

c) Cng sut trung bnh tnh nh sau:2

AP( ) ( 0) (f )df 2

w

= = = =

Bi 6Cho dy xung ch nht bin A , chu k T nh hnh v di y

+A

-A

T/2

T

t0

a)Tm bin i Fourier b) Tm PSDc) Tm ACFd) Tm cng sut trung bnh

Gii

a) Bin i Fourier c dng sau:

S(f)=ASinc(Tf/2)f1(f)-A(f)

trong f 1=1/Tf1 1

k (f ) (f kf ) =

b) (f)=A2 Sinc2(Tf/2)f1(f)- A2(f)

c) T quan h bin i Fourier:1 2 1 2 1 2s (z)s (z )dz s ( ) s ( ) S (f ).S (f )

=

Ta c2

T( ) 2A (2 / T) ( ) = =2

2A (2 / T) pI(-kT)-A2

d) P= (0)=A2

Bi 7Cho dy xung trong l qu trnh ngu nhin c biu din theo cng thc sau:

( ) ( )2

== + K T

k

T X t A p t kT

trong Ak ={+A,-A} vi xc xut xut hin +A v -A bng nhau v bng 1/2. Tm:a) ACF b) PSDc) Cng sut trung bnh

Giia)

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[ ]

2

2

( ) ( ) ( )

1 ,

0 ,( )

x

T

E X t X t

A T T

A

= + =

= nu kh c

b) 2 2( ) ( ) X f A TSinc fT =c) P= (0)=A2

Bi 8Mt ng truyn dn bng gc trong mi k hiu truyn c 2 bit c tha s dc=1. Nu tc

s liu cn truyn l 9600 bps , tm:

a) Tc truyn dn b) Bng thng Nyquist

Giia) R s=2R b=2x9600 kps= 19200 sps b) Bng thng Nyquist

B N=(1+)R s/2=19200 Hz

Bi 9Mt ng truyn dn bng thng c d liu nh bi 10. Tm:

a) Tc truyn dn

b) Bng thng Nyquist

Gii

a) R s=2R b=2x9600 kps= 19200 sps b) Bng thng Nyquist

B N=(1+)R s=2x19200=38400 Hz

Bi 10Mt tn hiu c o ti u ra ca b lc bng thng l lng vi bng thng l B Hz. Khi khng c

tn hiu ti u vo b lc, cng sut o c l 1x10-6W. Khi c tn hiu NRZ lng cc cng sut o c l1,1x10-5W. Tp m c dng tp m trng. Tnh:a) T s tn hiu trn tp m theo dB b) Xc sut my thu nhn bit sai xung NRZ

Giia) SNR= (11-1)/1=10SNRdB]= 10lg10= 10dB b) p dng cng thc (2.54) vi: A2/ 2=SNR =10 v tra bng trong ph lcta c Pe=Q( 10 )=Q(3,16)= 8x10-4

11. Nu bng thng b lc trong bi 10 tng gp i v mc cng sut tn hiu o ti u ra b lc . Hi:a) Khi khng c tn hiu th cng sut o c ti u ra cu b lc bng bao nhiu b) T s tn hiu trn tp m bng bao nhiuc) Xc sut li xung NRZ bng bao nhiu

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Giia) Trong trng hp ny cng sut tp m tng gp i v bng 2x10-6 cn cng sut tn hiu khng i v cngsut o ti u ra cu b lc s bng 1,2x10-6, V th t s tn hiu trn tp m nh sau:

SNR= (12-2)/2=5SNR[dB]= 10lg5= 6,98970 dB b) Xc sut li xung nh sau:

Pe=Q( 5 )=Q(2,24)=0,0125

Bi 11Gi s cc mc cng sut tn hiu v tp m ging nh trong bi11 v tn hiu l mt tn s duy nht

ti tm ca b lc bng thng. Tm bng thng b lc khi t s tn hiu trn tp m o c l SNR=30dB.GiiChuyn SNR vo s ln: SNR=1030/10=1000

T cng thc SNR ta c th vit: SNR=1000=

5

6

1x10 2B10B' B'102B

=

Vy bng thng ca b lc l: B'=2Bx10-2

12. Cho mt chui nh phn di v tn c phn b 1 v 0 ngu nhin i qua knh AWGN. Tm xc sut li xungkhi:Cc xung l NRZ n cc {0,A} vi SNR=10dB.

Gii

a) Do NRZ n chc nhn hai mc tn hiu {0,A}. Nn cc hm mt xc sut trong trng hp ny c dngsau:

y

2 2( ) /(2 )1( |1)2

= yY f y e

2 2( ) /(21( | 0)2

y AY f y e

=

2 2( ) /(2 )1(1)2

= ye

u

P e dy

2 2( ) /(2 )1(0)2

u y A

e P e

=

U=A/2 A (bit 0)0 (bit 1)

Pe = P0Pe(0) + P1Pe(1)= (1/2)[ Pe(0) + Pe(1)]

trong P0 = P1 = 1/2, Pe(0) = Pe(1). V th ta c th vit

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2 2e

U

1P exp( y / 2 )dy2

=

t y/=z ta c

2

e A/2

1P exp( z /2)dz2

= =Q(A/2)Do A2/2=SNR=10, nn xc sut li xung bng: Pe=Q( 10 /2)=Q(1,58)=0,0571

CHNG 3

Bi 1

Cho mt tn hiu bn mc ai={-3a/2,-a/2,a/2,3a/2}, i=1,2,3,4 vi thi gian truyn mi mc l T . Hy biu din tn hiu ny trong khng gian tn hiu.a) Tm vect n v b) Biu din tn hiu trong khng gian tn hiu

Gii

a) Vect n v c xc nh nh sau:

31 T

23

0

a a / 2 1(t)E T

(a /2) dt = = =

T phng trnh (3.4) ta c:

T

i1 i i0

1s s (t) dt a TT

= = b)

3a T2

1 a T2

1 a T2

3a T2

Bi 2

Gi s mi mc ca tn hiu trong bi 1 truyn hai bit tng ng nh sau {00,01,11,10}. Tm xc sutc iu kin thu sai cp bit 00.Gii

Xc sut c iu kin thu sai cp bit 00 nh sau:.

Pe(00)= P(01|00)+ P(10|00)+ P(11|00)T phng trnh (3.32) ta c th biu din cc hm xc sut c iu kin nh trn hnh v sau.

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1 Y 1f (y | 00) 1 Y 1f (y | 01)

1 Y 1f (y | 11) 1 Y 1f (y | 10)

3a T

21y

P(11| 00) eP (10|00)

1 a T2

1a T

23

a T2

eP (01| 00)a T 0 a T

:Phn din tch t cho trn hnh v ta xc sut sai c iu kin nh sau.

Pe(00)=2

1 100a T

1 1 3exp - (y a T) dy N 2 N

+

t z= 1 03

y a T / N2

+ , sau khi bin i ta c:

P e(00)=

0

2

a T2 N

1 2 exp -z dz2

=

0

1 a Terfc2 2 N

t z= 1 032 y a T / N2

+ , ta c

P e(00)=

0

T

2 N

2

a1 exp -z / 2 dz2

=0

T2N

Q a

Bi3(Tip) Tm xc sut c iu kin thu sai cp bit 01.

Gii

Pe(01)= P(00|01)+P(11|01)+ P(10|01)T phng trnh (3.32) ta c th biu din cc hm xc sut c iu kin nh trn hnh v sau.

1 Y 1f (y | 00) 1 Y 1f (y | 01)

1 Y 1f (y | 11) 1 Y 1f (y | 10)

3a T

21y

P(11| 01) eP (10|01)

1 a T2

1a T

23

a T2

eP (00| 01)a T 0 a T

Phn din tch t cho trn hnh v ta xc sut sai c iu kin nh sau.

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Pe(01)=a T

21 1

00

1 1 1exp - (y a T) dy N 2 N

+ +2

1 100 0

1 1 1exp - (y a T) dy N 2 N

+

=22

1 100 0

1 1 1exp - (y a T) dy N 2 N

+

t z= 1 01y a T / N2

+ , sau khi bin i ta c:

P e(01)=

0

2

a T2 N

2 exp -z dz

=0

a Terfc2 N

z= 1 012 y a T / N2

+

P e(01)=

0

2

Ta2N

2exp -z / 2 dz2

=2 0TQ a 2N Bi 4

Tim xc sut c iu kin thu sai cp bit 11

GiiPhn tch nh bi 3 ta c:

Pe(10)=

02

1 100

1 1 1exp - (y a T) dy N 2 N

+

21 1

00a T

1 1 1exp - (y a T) dy N 2 N

=22

1 100a T

1 1 1exp - (y a T) dy N 2 N

t z= 1 01y a T / N2

, sau khi bin i ta c:

P e (01)=

0

2

a T2 N

2 exp -z dz

=0

a Terfc2 N

z= 1 012 y a T / N2

P e(01)=

0

2

Ta2N

2 exp -z / 2 dz2

=2

0

TQ a2N

Bi 5 Tm xc sut c iu kin thu sai cp bit 10.

Gii

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Phn tch nh bi 3 ta c:

Pe(11)=a T

21 1

00

1 1 3exp - (y a T) dy N 2 N

=2

1 1002a T

1 1 3exp - (y a T) dy N 2 N

t z= 1 03y a T / N2

, sau khi bin i ta c:

P e(11)=

0

2

a T2 N

1 2 exp -z dz2

=

0

1 a Terfc2 2 N

z= 1 032 y a T / N2

P e(11)=

0

2

Ta2N

1 exp -z / 2 dz2

=

0

TQ a2N

Bi 6Tm xc sut thu sai k hiu trung bnh vi gi thit xc sut pht cc mc ai bng nhau v bng 1/4.

GiiPe= P(00)Pe(00) + P(01)Pe(01)+ P(11)Pe(11)+ P(10)Pe(10)

trong P(xx) v Pe(xx) l xc sut pht k hiu xx v xc sut c iu kin thu sai k hiu xx tng ng.

T cc kt qu bi 2,3,4,5 v P(xx)=1/4 ta c:

Pe= [Pe(00)+Pe(01)+Pe(11)+Pe10)]/4

=0

3 a Terfc4 2 N

=0

6 TQ a4 2N

Bi 7Cho mt tn hiu iu ch 4-ASK c xc nh nh sau:

si(t) =TE2 aicos(2f ct+)

vi:E l nng lng trn mt k hiu = 2E b, E b l nng lng trn mt bit.T l thi gian ca mt k hiu bng 2T b, T b l thi gian ca mt bit,

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i = 1, 2, 3,4; f c tn s sng mang, l mt gc pha ban u bt k khng nh hng ln qu trnh phn tch nn ta s b qua; ai={-3,-1,1,3}.

a) Tim vect n v b) Biu din tn hiu trong khng gian tn hiu

Gii

a)3 c 3 c

1 c2T3

3 c0

2E 2Ea cos(2 f t) a cos(2 f t) 2T T(t) cos(2 f t)TE 2Ea cos(2 f t) dt

T

= = =

b) T phng trnh (3.4) ta c:

T T

i1 i c i c c0 0

2 2E 2s s (t) cos(2 f t)dt a cos(2 f t) cos(2 f t)dtT T T

= =

S dng bin i lng gic:

i c c2E 2a cos(2 f t) cos(2 f t)T T

= [ ]i cE a 1 cos(4 f t)

T+

ta c

si1= iEa vi biu th trn hnh v nh sau:

1-1 3-3 0E E E E

Bi 8(Tip) Gi s mi mc ca tn hiu trong bi 1 truyn hai bit tng ng nh sau {00,01,11,10}. Tm

xc sut c iu kin thu sai cp bit 00.

GiiXc sut c iu kin thu sai cp bit 00 nh sau:.

Pe(00)= P(01|00)+ P(10|00)+ P(11|00)T phng trnh (3.32) ta c th biu din cc hm xc sut c iu kin nh trn hnh v sau.

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1 Y 1f (y | 00) 1 Y 1f (y | 01)

1 Y 1f (y | 11) 1 Y 1f (y | 10)

1y

P(11| 00) eP (10|00)eP (01| 00)03 E 2 E E E 2 E 3 E

:Phn din tch t cho trn hnh v ta xc sut sai c iu kin nh sau.

Pe(00)=2

1 1002 E

1 1exp - (y 3 E) dy N N

+

t z=( )1 0y 3 E / N

+, sau khi bin i ta c:

P e(00)=

0

2

E N

1 2 exp -z dz2

=

0

1 Eerfc2 N

t z= ( )1 02 y 3 E / N+ , ta c

P e(00)=

0

2E N

21 exp -z / 2 dz2

=

0

2E

NQ

Bi 9(Tip) Tm xc sut c iu kin thu sai cp bit 01.

Gii

Pe(01)= P(00|01)+P(11|01)+ P(10|01)

T phng trnh (3.32) ta c th biu din cc hm xc sut c iu kin nh trn hnh v sau.

1 Y 1f (y | 00) 1 Y 1f (y | 01)

1 Y 1f (y | 11) 1 Y 1f (y | 10)

1y

P(11| 01) eP (10|01)eP (00|01)03 E 2 E E E 2 E 3 E

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Ta xc sut sai c iu kin nh sau.

Pe(01)=2 E

21 1

00

1 1exp - (y E) dy N N

+ +2

1 100 0


+

=22

1 1

00 0


+

t z=( )1 0y E / N+ , sau khi bin i ta c:

P e(01)=0

2

E N

2 exp -z dz

=0

Eerfc N

z= ( )1 02 y E / N+

P e(01)=

0

2

2E N

2 exp -z / 2 dz2

=20

2EQ N

Bi 10

(tip) Tm xc sut li bit trung bnh.Gii

V mi k hiu pht ng thi hai bit nn xc sut li bit trung bnh s bng 1/2 xc sut li k hiu v c tnnh sau:

P b=1/2[P(00)Pe(00)+ P(01)Pe(01)+P(11)Pe(11)+P(10)Pe(10)]/4

Da trn cc bi 8, 9 ta c th vit:

Pe(00)=0

1 Eerfc2 N

=0

2E N

Q

Pe(01)=0

Eerfc

N

=20

2E N

Q

Pe(11)=0

Eerfc N

=20

2E N

Q

Pe(10)=0

1 Eerfc2 N

=0

2E N

Q

V th xc sut li bit tringbinhg nh sau:

Pe=0

3 Eerfc8 N =

0

2E N3Q4

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Bi 11

Tm s bt li xy ra trong mt ngy i vi my thu iu ch BPSK nht qun hot ng lin tc.R b=10.000bps, P=0,1.10-6 W v N0= 10-11W.Hz-1 . Gi thit rng cng sut v nng lng bit c chun ha titr ti bng 1 m.

GiiE b=P/R b= 0,1.10-10; 2E b/N0= 2; tra bng ta c P b =0,0228

Vy s bt li xy ra trong mt ngy bng:

N b= R b.(3600. 24).0,0228= 10.000(3600.24).0,02281970 bit

Bi 12Mt h thng BPSK hat ng lin tc mc li trung bnh 5000 bit li trn mt ngy. R b=10000bps,

N0=10-10WHz-1 .a) Tm xc sut li bit b) Tm cng sut thu tng ng c xc sut li bit nh a)

Gii

a) P b=50000

10000(3600x24)=5,79.10-5

b) Tra bng trong ph lc ta c:

2E b/N0=2P/(R b N0)= 3,2. Vy cng sut thu tng ng s l: P=3,2xR b N0/2= 3,2.10000.10-10/2=1,6.10-4W

Bi 13

Tn hiu thu ca h thng BPSK nht qun c nh ngha nh sau:

r(t) = k b

b

TE2

sin (2f ct) 21 k b

b

TE2

cos(2f ct) +x(t), 0 tT b

trong du cng tng ng vi k hiu '0' v du tr tng ng vi '1', thnh phn th nht th hin sngmang ng b my thu vi my pht, T b l rng bit v E b l nng lng bit, x(t) l tp m Gauss trngcng.a) Vit cng thc lin h xc sut li bit trung bnh P b vi xc sut pht k hiu 1: P(1), xc sut pht k hiu 0:P(0), xc sut c iu kin Pe(0|1): xc sut pht k hiu mt nhng quyt nh thu k hiu 0 v xc sut c iukin Pe(1|0): xc sut pht k hiu 0 nhng quyt nh thu k hiu 1 b) Tm cc biu thc xc nh Pe(0|1) v Pe(1|0)

Giia) Cng thc lin h xut li Pe:

P b= P(1).P(0|1)+ P(0).P(1|0)= 1/2[P(0|1)+P(1|0) b) Sau tch phn gii iu ch BPSK ta c:

Y1=T

c0

2s(t) cos(2 f t)dtT

= bE)k ( 21 + X1 trong Y1 l tn hiu sau mch tch phn, '+" tng ng k hiu '0' c pht, '-' tng ng k hiu '1' c pht, W1 l tp m Gauss trng cng trung bnh khng.

V Y1 l bin ngu nhin Gauss c gi tr trung bnh bE)k (21 nn:

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f Y1(y1|0)= ( )

22

100

111 bE)k (y Nexp

Nv

f Y1(y1|1)= ( )

+

221

00

111 bE)k (y N

exp N

Vy xc sut c iu kin m my thu quyt nh 1 khi k hiu 0 c pht:0

Y1 1 1P(1| 0) f (y | 0)dy

= =20

21 b

00

1 1exp y (1 k )E dt N N

=2

21 b

00 0


v xc sut c iu kin m my thu quyt nh thu 0 khi k hiu 1 c pht:

Y1 1 10

P(0 |1) f (y |1)dy

= =2

21 b

00 01 1exp y (1 k )E dt N N

+ Do tnh i xng nn: P(1|0)=P(0|1)

v v th: P(1|0)=P(0|1)=2

21 b

00 0


t

z= 21 b 02 y (1 k )E / N +

ta c

P(1|0)=P(0|1)= ( )2 b

0

2

2(1 k )E N

1 exp z / 2 dt2

=

2 b

0

2(1 k )EQ N

Bi 14

(Tip) i vi h thng BPSK nh cho trong bi 13.

a) Chng minh rng xc sut li trung bnh bng:

)k ( N

EP be2

012

trong : N0 l mt ph cng sut tp m Gauss trng. b) Gi thit 10% cng sut tn hiu pht c phn b cho thnh phn sng mang chun ng b tm E b/N0 m bo xc sut li bit trung bnh bng 3.10-4.c) So snh gi tr SNR h thng ny i vi h thng BPSK thng thng

Giia) Xc sut li bit trung bnh c xc nh nh sau:

P b= P(1).P(0|1)+ P(0).P(1|0)= 1/2[P(0|1)+P(1|0)

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=P(0|1)=P(1|0)=2

b

0

2(1 k )EQ N

b) Cng sut trung bnh chun ha cho in tr 1m ca hnh phn sng mang ng b c xc nh nhsau;

b

2T b

c c b b0

2E1P k sin(2 f t) dtT T

= =k 2P=0,1Pk 2=0,1, nn cng sut cho tn hiu bn tin s chim 1-

k 2=0,9 tng cng sut . Rt ra:

u=2 b b

0 0

2E 1,8E(1 k ) N N

= 2

b

0

E u N 1,8

=

Tra bng trong ph lc cho P b=3.10-4 ta c u=3,4, V th:

2 b

0

E 3,4 6,42 N 1,8

= =

c) i vi h thng BPSK thng thng m bo xc sut li bit 3.10-4, ta cn t s E b/N0 nh sau:

b

0

2E 3,4 N

= 2

b

0

E 3,4 5,78 N 2

= =

Nh vy h thng BPSK s dng thnh phn ng b sng mang cn c t s SNR ln hn h thng thnthng l:

6,42:5,78=1,1 ln

Bi 15

Mt h thng BPSK c xc sut truyn bit "0" bng xc sut truyn bit "1". Gi thit rng khi h thnng b tt, E b/N0=9,6 dB dn n xc sut li bit bng 10-5. Trong trng hp vng kha pha PLL b mc li pha .a) Xc sut li bit s gim cp nh th vo nu =250 b) Sai pha ln no s dn n xc sut li bit bng 10-3

Giia) Sau tch phn gii iu ch BPSK ta c:

Y1=T

c0

2s(t) cos(2 f t)dtT

=T

bc c

b b0

2E 2cos(2 f t) cos(2 f t )dtT T

+ + X1= bE cos +X1

V Y1 l bin ngu nhin Gauss c gi tr trung bnh bE cos nn:

f Y1(y1|0)= ( )2

1 b00

1 1exp y E cos N N

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f Y1(y1|1)= ( )2

1 b00

1 1exp y E cos N N

+

Ta c th biu din cc hm mt xc sut ni trn theo gi tri y1 nh sau.

y 1

21 0

1

( os ) /1

0

1( |1)2

+= b y E c N Y f y e N

21 0

1

( os ) /1

0

1( | 0)2

= b y E c N Y f y e N

21 0( os ) /

10 0

1(1)2

+= b y E c N e P e dy N

21 0

0( os ) /

10

1(0)2

= b y E c N e P e dy N

osb E c os b E c 0

Do xc sut truyn bit "1" bng xc sut truyn bit "0" nn: P(0) =P(1)=1/2 v tnh i xng ca cc hm mt xc sut c iu kin, nn cc sut li bit trong trng hp ny c tnh nh sau:

P b=P(0)Pe(0)+P(1)Pe(1)= 1/2[Pe(0)+Pe(1)]=Pe(0)=Pe(1)

= ( )2

1 b 100 0

1 1exp y E cos dy N N

+

t

z= 1 b

0

y E cos2

N+

ta c:

P b= ( ) b

0

21

2E cos N

1 exp z / 2 dy2

= b

0

2EQ cos N

Thay E b/N0= 109,6/10=9,12; =250cos250=0,8; b0

2E cos 2x9,12x0,8 3,417 N

= = .

Tra bng trong ph lc ta c: P b=3.10-4

b) Tra bng trong ph lc cho xc sut li bit 10-3 ta c:

b

02E 2x9,12xcos 3,1 N = = cos =0,726 =43,46

0

16


17/52

Bi 16

Cho hai my thu h thng truyn dn nht qun 16-QAM vi cc thng s sau: cng sut thu trung bnh Pavr =10-5W, R b=5000bps, N0=10-10WHz-1.a) Tm xc sut li bit trong hai h thng b) Tm v bng thng Nyquist ca hai h thng khi cho h s dc=0,2

Gii

a) Tc k hiu cho 16-QAM v QPSK c xc nh nh sau:

R s-16QAM= R blog(16)=20000 bpsR s-QPSK = R b/log(4)=10000 bps

Nng lng bit trung bnh cho 16-QAM v QPSK c xc nh nh sau:Eavr-16QAM= P/R s= 1,6.10-6/20000=0,8.10-10

Eavr-QPSK = P/R s= 1,6.10-6/10000=1,6.10-10

T s tn hiu trn tp m cho hai h thng nh sau:2 Eavr-16QAM/N0=2x0,8.10-10/10-10= 1,62 Eavr-QPSK /N0=2x1,6,10-10/10-10= 3,2

Tra bng trong ph lc ta c:P b-16QAM=5,48.10-2

P b-QPSK = 7.10-4

b) Bng thng Nyquist ca hai h thng c xc nh nh sau:

B N-16QAM=(1+)R s-16QAM=1,2x20000bps=24000bpsB N-QPSK =(1+)R s-QPSK =1,2x10000bps=12000bps

Bi 17

h thng 16QAM t c xc sut li bit ging nh h thng QPSK ta cn tng cng sut cho hthng 16QAM ln bao nhiu ln.Gii

t c xc sut li bit ging nh QPSK ta cn t c t s tn hiu trn tp m nh sau:

2 Eavr-16QAM/N0= 2P/(R s-16QAMN0)= 3,2P=1,6xR s-16QAM N0=1,6x20000x10-10= 3,2x10-6

Nh vy ta cn pht cng sut cho 16QAM gp hai ln cng sut cho QPSK hai h thng c cng xc sut l

bit.

CHNG 4

Bi 1

Mt bn tin 3 bit c truyn trn h thng BPSK v t s tn hiu trn tp m thu c bng 7 dB.a) Tnh xc sut 2 bit mc li b) Bn tin c m ha sao cho t m tng ln 5 bit. Xc sut 2 bit mc li bng bao nhiu. Gi thit rng cnsut pht trong hai trng hp a) v b) l nh nhau.

Gii

17


18/52

E b/No=107/10=5 b0

2E 2x5 3,16 N

= = . Tra bng trong ph lc ta c P b=7,9.10-4. i vi ba bit mc li 2

ta c s t hp li hai bit trong 3 bit nh sau:

3 3! 32 2!(3 2)!

= = .

Xc sut li 2 bit nh sau:

( ) ( )2 3 2e b b3

p P 1 P2

= ( ) ( )24 4

3 7,9x10 1 7,9x10 = 76,4.10trong P b, P b2 v (1-P b)3-2 l xc sut li mt bit, xc sut li 2 bit v xc sut mt bit khng mc li. b) Tc bit sau m ha tng v bng: R bc=R b/r, trong R b l tc bit khng m ha v r=3/5 l t l m. Nng lng bit cho trng hp m ho bng:E bc=P/R bc=rP/R b=rE b. V th

bc b

0 0

2E 2rE 10x3/ 5 2, 45 N N

= = =

Tra bng trong ph lc ta c xc sut li bit khi m ha bng:P bc=7,14.10-3

Bi 2

Tm xc sut li ban tin cho h thng truyn dn trong bi 1 cho:a) Trng hp khng m ha (PM) b) Trng hp m ha (PMc)

Giia) PM=1-(1-P b)3=1-(1-7,9.10-4)3=2,37.10-3

b) V khong cch Hamming cc tiu dmin=3, nn m c th sa c mt bit. Bn tin s ch b li khi khi mcli 2,3,4,5 bit, v th:

( ) ( )5

k 5 k Mc bc bc

k 2

5P P 1 P

k

=

= Hay mt cch gn ng nu ta ch xt thnh phn th nht:

PMc ( ) ( )2 3 bc bc5

P 1 P2 ( ) ( )

2 33 357,14.10 1 7,14.10

2 =

=4,99.10-4

Ta nhn thy rng nh m ha xc sut li bn tin c ci thin

( 2,37.10-3) (4,99.10-4)=4,75 ln

Bi 3

Bng thng cho h thng c m ha trong bi 1 tng ln bao nhiu ln so vi h thng khng m ha.

Gii

18


19/52

Bng thng Nyquist cho h thng khng m ha nh sau: B N=(1+)R b, trong l h s dc b lc. Bngthng Nyquist cho h thng m ha nh sau: B N=(1+)R bcx(1/r), trong r l t l m. Vy s ln bng thng hthng m ha tng so vi h thng khng m ha nh sau: R bc/(R bxr)=5/3 ln

Bi 4Mt h thng iu ch BPSK c tc bit R b=4800bps. T s tn hiu trn tp m thu: E b/N0=6dB.

a) Tm xc sut li bit (P b) v li bn tin (PM) cho h thng khng m ha, trong bn tin di 11 bit

b) Tm xc sut li bit (P bc) v li bn tin (PMc) cho h thng m ha, cho m khi (15,11) sa li n

Gii

a)0,8 b

0

2E 2x10 3,55 N

= = P b=1,9.10-4

PM=1-(1-P b)11=1-(1-1,9,10-4)11=2,1.10-3

b)0,8 b

0

2E 11r 2x10 x 3,04 N 15

= = P bc=1,18.10-3

( )15 15 k k

MC bc bck 2

15P P (1 P )

2

=

= ( )23 3 15 215

1,18.10 (1 1,18.10 )2

=1,44.10-4

Tha s ci thin bng: (2,1.10-3)/(1,44.10-4)=14,58

Bi 5

Cho mt b to m khi tuyn tnh c ma trn to m sau:

0 1 1 1 0 0G 1 0 1 0 1 0

1 1 0 0 0 1

=

a) Tm cc t m

b) Tm Syndrome trong trng hp t m pht l "111" cn t m thu l "110"

Gii

19


20/52

c=mG=

0 0 0 0 0 0 0 0 00 0 1 1 1 0 0 0 10 1 0 1 0 1 0 1 0

0 1 1 1 0 00 1 1 0 1 1 0 1 1

1 0 1 0 1 01 0 0 0 1 1 1 0 0

1 1 0 0 0 11 0 1 1 0 1 1 0 11 1 0 1 1 0 1 1 01 1 1 0 0 0 1 1 1

=

b) [ ]Tk n PIH = = MT

3I P ; 3

1 0 0I 0 1 0

0 0 1

=

;PT=

0 1 11 0 11 1 0

H=

1 0 0 0 1 10 1 0 1 0 10 0 1 1 1 0

HT=

1 0 00 1 00 0 10 1 11 0 11 1 0

s=yHT=[0 0 0 1 1 0]

1 0 00 1 00 0 10 1 11 0 11 1 0

=[ ]1 1 0

Bi 6

Mt b to m vng (7,4) c a thc to m g(x)= 1+x+x3 v bn tin u vo 101. Tm t m u ra.

Gii

a thc bn tin c dng sau: m(x)=1+x2. Nhn a thc bn tin vi x7-4=x3 ta c: x3+x5.Chia tch trn cho a thc to m nhn c phn db(x):

35 3

2

5 3 2

2

x x 1x xx

x x x

x

+ ++

+ +

Vy b(x)=x2; c(x) =b(x) + xn-k m(x)= x2 +x3+x5c=[001101

20


21/52

Bi 7

Mt b to m vng c a thc to m g(x)=1+x2+x3.

a) Thit k s b to m

b)Kim tra hot ng cu n vi bn tinm=[1010].

Gii

a) S b m ha nh sau

D1 D2 D3

SW1

SW2

Vo

Ra1010

b) Ni dung thanh ghi dch khi bn tin vo (10110)

Dch Bit vo Ni dung thanh ghi (D1D2D3)

1234

1011

000 (trng thi u)101111011010

Bi 8

Mt b to m xon vi chui to m sau:g1 = ( g0,1 , g1,1 , g2,1) = (1,0,1)

g2 = ( g0,2 , g1,2 , g2,2) = (1,1,1)a) Thit k s b) Tnh ton chui u ra theo bng khi cho chui u vo m=[101011], trong bit ngoi cng bn tri l bitvo b to m u tin

Giia) S b to m nh sau:

21


22/52

D1 D2Vo

Ra

b) Tnh ton chui m u ra theo bng

Chui bn tin m 1 0 1 0 1 11/ B sung hai bit ui: m' 0 0 1 0 1 0 1 1 0 02/ tr mt bit: m'.x 0 0 1 0 1 0 1 1 0 03/ tr hai bit: m'.x2 0 0 1 0 1 0 1 1 0 0 c1 = m'+m'x2 1 0 0 0 0 1 1 1 c2 = m'+m'x+m'x2 1 1 0 1 0 0 0 1 c 11 01 00 01 00 10 1011

Bi 9(Tip) Tm chui m u ra theo phng php chui to m.

GiiTrin khai phng trnh (4.40) cho ta chui bit u ra.

c=

= = M

,j p,j -p0

g m j 1,2; p=0,1,2; =0,1, .., 7; pl ll

l l

g1 = ( g0,1 , g1,1 , g2,1) = (1,0,1)g2 = ( g0,2 , g1,2 , g2,2) = (1,1,1)m=[m0, m1, m2, m3,m4, m5]= [101011],

i vi nhnh trn j=1 ta c:

c0,1 = g0,1 m0 = 1 1 = 1c1,1 = g0,1 m1 + g1,1 m0 = 10 + 0 1 = 0c2,1 = g0,1 m2 + g1,1 m1 + g2,1 m0= 11 + 00 + 11= 0c3,1 = g0,1 m3 + g1,1 m2 + g2,1 m1 = 10 + 01 + 10 = 0c4,1 = g0,1 m4 + g1,1 m3 + g2,1 m2 = 11 + 00 + 11 = 0c5,1 = g0,1 m5 + g1,1 m4+g2,1m3 = 11 + 01 +10 = 1c6,1 = g0,1 m6+ g1,1 m5 + g2,1 m4 = 10 +01+11 = 1

c7,1 = g0,1 m7+ g1,1 m6 + g2,1 m5 = 10 +00+11 = 1

Vy chui bit u ra ca nhnh trn l :

c1 = (10000111)

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23/52

Tng t i vi nhnh di j=2 ta c:,c0,2 = g0,2 m0 = 1 1 = 1

c1,2 = g0,2 m1 + g1,2 m0 = 10 + 1 1 = 1

c2,2 = g0,2 m2 + g1,2 m1 + g2,2 m0= 11 + 10 + 11= 0

c3,2 = g0,2 m3 + g1,2 m2 + g2,2 m1 = 10 + 11 + 10 = 1

c4,2 = g0,2 m4 + g0,2 m3 + g2,2 m2 = 11 + 10 + 11 = 0

c5,2 = g0,2 m5 + g1,2 m4+g2,2m3 = 11 + 11+ 10 = 0

c6,2 = g0,2 m6+ g1,2 m5 + g2,3 m4 = 10 +11+11 = 0c7,2 = g0,2 m7+ g1,2 m6 + g2,3 m5 = 10 +10+11 = 1

Vy chui bit u ra ca nhnh di s l:

c2 = (11010001)

Chui bit u ra ca b lp m l ghp chung ca hai chuic1, c2 nh sau:

c = (11 01 00 01 00 10 1011 )c 16 bit.

Bi 10

(Tip) Tm chui m u ra theo phng php a thc to m.

Bng cch s dng a thc to m cc phng trnh (4.42), (4.43) v (4.44), ta c th vit: [101011],m(x) = 1 + x2 + x4+x5 g1(x) = 1 + x2

g2(x) = 1+x + x2 c1(x) = (1 + x2 + x4+x5) ( 1 + x2) = 1+ x2 + x2+x4+x4+x6 + x5 +x7

=1+ x5 + x6 + x7

hay: c1 = (c0,1, c1,1, c2,1, c3,1, c4,1, c5,1, c6,1, c7,1) = ( 1 0 0 0 01 1 1)

c2(x) = (1 + x2 + x4+x5)( 1 + x + x2 ) = 1 + x + x2+ x2+x3 + x4 + x4 +x5+ x6+x5+x6+ x7

=1+x+x3+x7

hay: c2 = (c0,2, c1,2, c2,2, c3,2, c4,2, c5,2, c7,2) = ( 1 1 0 1 0 0 01 )Cui cng ta c chui u ra l ghp xen ca hai chui trn nh sau:

c = ( 11 01 00 01 0010 1011 )

Bi 11(Tip) Tm chui m u ra theo biu li

23


24/52

S0=0000

11

11

00

01

10

10

01

00

11

11

00

01

10

10

01

00

11

11

00

01

10

10

01

00

11

11

00

01

10

10

01

00

11

11

00

01

10

10

01

00

11

11

00

01

10

10

01

00

11

11

00

01

10

10

01

S1=10

S2=01

S3=11

m 1 0 1 0 1 1 0

c 11 01 00 01 00 10 10

Cc bit vo

Cc k hiu ra

00

11

11

00

01

10

10

01

0

11

T biu li ta thy ng dn chui k hiu ra bt u t trng thi s=(00) v kt thc trng thi ny. d nh vy v hai bit ui bng "00".

Bi 12(Tip). Khi chui k hiu thu c bng:v=[11 01 00 01 00 10 11 11]. Tm khong cch Hamming gia

chui k hiu thu v k hiu pht.

Gii

Ta biu din khong cch Hamming ca chui k hiu thu cho tng nhnh trn biu li trn hnh sau.

S0=002

0

0

2

1

1

1

1

1

1

1

1

0

2

2

0

0

2

2

0

0

0

0

0

1

1

1

1

0

2

2

0

0

2

2

0

1

1

1

1

1

1

1

1

2

0

0

2

2

0

0

2

1

1

1

1

S1=10

S2=01

S3=11

v 11 01 00 01 00 10 11Cc k hiu thu

2

0

0

2

1

1

1

1

11

T hnh trn ta thy khong cch Hamming bng 1.Bi 13

(Tip). Biu th cc ng dn sng st sau ln th nht hi nhp cc cp ng dn.

Gii

24


25/52

S0=00

0

0

20

2

2

0

0

0

0

0

S1=10

S2=01

S3=11

Bi 14

(Tip). Biu th cc ng dn sng sau ln th hai hi nhp cc cp ng dn.S0=00

0

0

20

0

0

0

0

0

S1=10

S2=01

S3=11

1

1

1

0

0

Bi 15Mt b to m xon vi cc a thc to m sau:

g1(x) = 1+x+x2

g2 (x)= 1+x2

a) Thit k s b) V biu trng thic) V biu li

Giia)

25


26/52

D1 D2Vo

Ra

b)

S0=00

S1=10 S2=01

S3=11

00

11 11

00

10

01 01

10 Bi 16(Tip)a) V biu li

b) Tm chui k hiu ra theo biu li khi chui bit vo l: m=[101011].

Giia)

26


27/52

S0=0000

11

11

00

10

01

01

10

00

11

11

00

10

01

01

10

00

11

11

00

10

01

01

10

S1=10

S2=01

S3=11

b)

S0=0000

11

11

00

10

01

01

10

00

11

11

00

10

01

01

10

00

11

11

00

10

01

01

10

00

11

11

00

10

01

01

10

00

11

11

00

10

01

01

10

00

11

11

00

10

01

01

10

00

11

11

00

10

01

01

10

S1=10

S2=01

S3=11

m 1 0 1 0 1 1 0

c 11 10 00 10 00 01 01

Cc bit vo

00

11

11

00

10

01

01

10

0

11

Bi 17

Cho a thc ta m sau:g(x)=1+x2

a) Thit k b m ha xon h thng SC b) V biu li

c) Tm chui k hiu ra khi chui bit vo l m=[100].

Gii

a)

27


28/52

D1 D1

dk

ck

Vora

b)

a=0 0 00

11

01

10

00

11

01

10

00

11

01

10

00

11

01

10

00

11

01

10

00

11

01

10

b=1 0

c=0 1

d=1 1

c)

a=0 000

11

01

10

00

11

01

10

00

11

01

10

00

11

01

10

00

11

01

10

00

11

01

10

b=1 0

c=0 1

d=1 1 11 00 01

d 1 0 0

C

Chui k hiu ra l: C=[110001]

28


29/52

Bi 18Cho a thc to m sau:

g(x) += + +

2

211,

1 x

x x

a) Thit k b to m xon hi quy RSC b) V biu li

Giia)

dk

dk

ck

D D

0k S

1k S

b)

S0=0000

11

11

00

10

01

01

10

00

11

11

00

10

01

01

10

00

11

11

00

10

01

01

10

S1=10

S2=01

S3=11

Bi 19

(Tip). Chui bit s liu vo b m ha RSC ld=[100]a) Tm chui k hiu u ra ca b m ha da trn biu li. b) Chui k hiu lng cc a ln iu ch.

Giia)

29


30/52

a=0 000

11

01

10

00

11

01

10

00

11

01

10

00

11

01

10

00

11

01

10

00

11

01

10

b=1 0

c=0 1

d=1 1

11 00 01

d 1 0 0

CC={1,1,0,0,0,1} b) Chui lng cc c xc nh theo cng thc:

ak =1-2dk bk =1-2ck

C={-1,-1,+1,+1,+1,-1}

Bi 20

(Tip) Vi gi thit2 1 = v chui k hiu thu c ti u vo b gii m MAP ti cc thi imk=[0,1,2] nh sau: R=[(-0,5;-0,2), (1,5;1,2),(0,8;-0,4)]. Gi thit xc sut pht dk =0 v dk =1 l nh nhau. Tnhton s o nhnh choa) k=0 b) k=1c) k=2

a) T phng trnh (4.71) ,ta c:

( ) = + , , , ,21( ) exp ( )k i k i k k i k k im x a y b m

. V xc sut pht dk =0 v dk =1 bng nhau, nn =, 1/2k i .

Gi thit2=1. R 1=(-0,5;-0,2)

( )= = + 1,0 1,01(00) (10) exp ( 0,5 1) ( 0,2 1)2

=0,25

( )= = + 1,1 1,11(00) (10) exp ( 0,5 1) ( 0,2 1)2

= 1

( )= = + 1,0 1,01(01) (11) exp ( 0,5 1) ( 0,2 1)2

=0,37

( )= = + 1,1 1,11(01) (11) exp ( 0,5 1) ( 0,2 1)2

=0,67

R 2=(1,5;1,2)

30


31/52

( )= = + 2,0 2,01(00) (10) exp (1,5 1) (1,2 1)2

=7,4

( )= = + 2,1 2,11(00) (10) exp (1,5 1) (1,2 1)2

=0,03

( )= = + 2,0 2,01(01) (11) exp (1,5 1) (1,2 1)2

= 0,67

( )= = + 2,1 2,11(01) (11) exp (1,5 1) (1,2 1)2

= 0.37

R 3=(0,8;-0,4)

( )= = + 3,0 3,01(00) (10) exp (0,8 1) ( 0,4 1)2

=0,75

( )= = + 3,1 3,11(00) (10) exp (0,8 1) ( 0,4 1)2

=0,34

( )= = + 3,0 3,01(01) (11) exp (0,8 1) ( 0,4 1)2

= 1,66

( )= = + 3,1 3,11(01) (11) exp (0,8 1) ( 0,4 1)2

= 0,15

Bi 21

(Tip). Tnh ton s o trng thi thun (Forward State Metric).

Gii


=

=1

1 j , j0

( ) (b ( )) (b ( ))k k k j j

m m m

Gi thit b gii m bt u t m=00, k=1

= = 1 1(00) 1 ( ) 0 00v m khi m k=2

m=00: = + =2 1 1,0 1 1,0 1,0(00) (00) (00) (01) (01) (00) =0,25

m=10: = + =2 1 1,1 1 1,1 1,1(10) (00) (00) (01) (01) (00)

=1m=01: = + =2 1 1,0 1 1,0(01) (10) (10) (11) (11) 0 m=11: = + =2 1 1,1 1 1,1(11) (10) (10) (11) (11) 0

k=3m=00: = + =3 2 2,0 2 2,0(00) (00) (00) (01) (01) 1,85m=10: = +3 2 2,1 2 2,1(10) (00) (00) (01) (01) =0,07m=01: = +3 2 2,0 2 2,0(01) (10) (10) (11) (11) =7,4m=11: = +3 2 2,1 2 2,1(11) (10) (10) (11) (11) =0,3

k=4m=00: = + =4 3 3,0 3 3,0(00) (00) (00) (01) (01) 13,67

31


32/52

m=10: = +4 3 3,1 3 3,1(10) (00) (00) (01) (01) =1,74m=01: = +4 3 3,0 3 3,0(01) (10) (10) (11) (11) =0,55m=11: = +4 3 3,1 3 3,1(11) (10) (10) (11) (11) =0,07

Bi 22

(Tip). Tnh ton s o trng thi ngc (Backward State Metric).

Gii


k (m) +=

=1

, 10

( ) ( ( , ))k j k j

m f j m

k=4

m=00 4(00)=1m=10 4(10)=0m=01 4(01)= 0m=11 4(11)= 0

k=3m=00 = +3 3,0 4 3,1 4(00) (00) (00) (10) (10)

= 0,75x1+0,34x0=0,75m=10 = +3 3,0 4 3,1 4(10) (01) (01) (10) (11)

= 0,75x0+0,34x0=0m=01 = +3 3,0 4 3,1 4(01) (00) (00) (10) (10)

= 1,66x1+0,15x0=1,66m=11 = +3 3,0 4 3,1 4(11) (01) (01) (11) (11)

= 1,66x0+0,15x0=0 k=2

m=00 = +2 2,0 3 2,1 3(00) (00) (00) (10) (10) = 7,4x0,75+0,3x0=5,55

m=10 = +2 2,0 3 2,1 3(10) (01) (01) (10) (11) = 7,4x1,66+0,67x0=12,5

m=01 = +2 2,0 3 2,1 3(01) (00) (00) (10) (10) = 0,67x0,75+0,37x0=0,5

m=11 = +2 2,0 3 2,1 3(11) (01) (01) (11) (11) = 0,67x1,66+0,37x0=1,11

k=1m=00 = +1 1,0 2 1,1 2(00) (00) (00) (10) (10)

= 0,25x5,55+12,28x1=13,67

m=10 = +1 1,0 2 1,1 2(10) (01) (01) (10) (11) = 0,25x0,5+1x1,11=1,24

m=01 = +1 1,0 2 1,1 2(01) (00) (00) (10) (10)

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= 0,37x5,55+0,67x12,28=10,28m=11 = +1 1,0 2 1,1 2(11) (01) (01) (11) (11)

= 0,5x0,37+0,67x1,11=0,93

Bi 23(Tip).a) in cc s o tm c trong cc bi trc ln biu li b) Tnh L( k d ) v tm c tnh nhn c sau gii m.

Giia)

a=0 0

b=1 0

c=0 1

d=1 1

k=1 k=2 k=3 k=40,25

0,25

1

1

0,370,67

1,660,15

0,75

1,660,34

0,15

1 =

0 =

0 =

0 =

0,25 =

1 =

0 =

0 =

1,85 =

0,07 =

7,4 =

0,3 =

13,67 =

1,74 =

0,55 =

0,07 =

1 =

0 =

0 =

0 =

0,75 =

0 =

0 =

1,66 =

5,55 =

12,28 =

0,5 =

1,11 =

13,67 =

1,24 =

10,28 =

0,93 =

0,37

0,67

7,4

0,3

7,4

0,3

0,670,37

0,67

0,37

0,75

0,34

b) T phng trnh (4.67) ta c:

+

+

=

,0 1,0

,1 1,1

( ) ( ) ( (0, ))( ) l n

( ) ( ) ( (1, ))

k k k m

k

k k k m

m m f mL d

m m f m

k=1 = = = 1 11 0,25 5,55 ( ) ln 2,18 11 1 12,28 x x

L d d x x

k=2+ = = = 2 2

0,25 7,2 0,75 1 7,4 1,66 ( ) ln 00

x x x x L d d

k=3+ = = = 2 3

1,85 0,75 1 7,4 1,66 1 ( ) ln 00

x x x x L d d

CHNG 5

Bi 1

Cho s b ngu nhin ha t ng b c a thc to m g(x)=1+x4+x5 vi iu kin ban u thanhghi dch l 10101.

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a) V s b ngu nhin ha b) Chng minh rng lung s ton khng sau ngu nhin ha s c bin i thnh lung s c s s "1" v ss "0" gn bng nhau.

Giia) S b ngu nhin ho t ng b

D E F G H

A B

C

0 11 0

A

C&D

B

1 010

0 11 0

BH

G

10

1 0

b) Lung bit u ra b ngu nhin ha cho 24 bit lung s ton khng u vo c xc nh theo bng y.

Th tdch bit

Vo Ra Thanh ghi dchA B C D E F G H

/K banu

- 1 - 1 0 1 0 1

1 0 1 1 1 1 0 1 02 0 1 1 1 1 1 0 13 0 1 1 1 1 1 1 04 0 1 1 1 1 1 1 15 0 0 0 0 1 1 1 16 0 0 0 0 0 1 1 17 0 0 0 0 0 0 1 18 0 0 0 0 0 0 0 19 0 1 1 1 0 0 0 010 0 0 0 0 1 0 0 011 0 0 0 0 0 1 0 012 0 0 0 0 0 0 1 013 0 1 1 1 0 0 0 114 0 1 1 1 1 0 0 015 0 0 0 0 1 1 0 016 0 0 0 0 0 1 1 017 0 1 1 1 0 0 1 118 0 0 0 0 1 0 0 119 0 1 1 1 0 1 0 020 0 0 0 0 1 0 1 021 0 1 1 1 0 1 0 122 0 I I I I 0 1 023 0 1 1 1 1 1 0 124 0 1 1 1 1 1 1 0

Chui u ra C=[111100001000110010101111] c 13 con s 1 v 11 con s khng.

Bi 2

2. Cho s b gii ngu nhin t ng b c a thc to m g(x)=1+x4+x5 vi iu kin ban u thanh ghi dchl 10101.a) V s b gii ngu nhin

b) Chng minh rng lung s ton khng c ngu nhin ha sau khi c gii ngu nhin s tr li lung ton khng.

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35/52

Gii

a) S b gii ngu nhin t ng b

D E F G H

AB

C

0 11 0

C

A

B

1 010

0 11 0

BH

G

10

1 0

1 0

b) 24 bit trong lung s u ra b gii ngu nhin c xc inhj theo bng di y. khng.

Th tdch bit

Vo Ra Thanh ghi dchC B A D E F G H

/K banu

- 1 - 1 0 1 0 1

1 1 1 0 1 1 0 1 02 1 1 0 1 1 1 0 13 1 1 0 1 1 1 1 04 1 1 0 1 1 1 1 15 0 0 0 0 1 1 1 16 0 0 0 0 0 1 1 17 0 0 0 0 0 0 1 18 0 0 0 0 0 0 0 19 1 1 0 1 0 0 0 010 0 0 0 0 1 0 0 011 0 0 0 0 0 1 0 012 0 0 0 0 0 0 1 013 1 1 0 1 0 0 0 114 1 1 0 1 1 0 0 015 0 0 0 0 1 1 0 016 0 0 0 0 0 1 1 017 1 1 0 1 0 0 1 118 0 0 0 0 1 0 0 119 1 1 0 1 0 1 0 020 0 0 0 0 1 0 1 021 1 1 0 1 0 1 0 122 I I 0 I I 0 1 023 1 1 0 1 1 1 0 1

24 1 1 0 1 1 1 1 0Lung bit u ra b gii ngu nhin: A=[000000000000000000000000]

Bi 3Thit k mch logic cho b m ha vi sai trn hnh 5.8.

Gii

Ta lp bng logic sau.

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(t-T)00

00

010

90

110

180

100

270

AB

000

00

0

00

01

11

10

090

0180

0270

01 11 10090 0180

0270

01 11 10 00090 0180

027000

11 10 00 010180

0270

0900010 00 01 11

0270

00 090

0180

1 2u ( t T)u ( t T

(t)1 2u ( t ) u

Ct u ca bng logic trn th hin quan h gia cp bit u vo b m ha vi sai v pha ca tn hic iu ch

Dng u ca bng th hin quan h ca cp bit thng tin (AB) cn m ha vi sai pha(t) (t) (t T) =

Bng phi di 4x4 th hin quan h gia cp bit u ra b m ha vi sai v pha ca tn hiu c iu ch

Nhn xt cc ct ca bng th hin quan h gia cp bit u ra ca b m ha vi pha ca tn hiu ich ta thy:

Ct th nht:1 1 2 2u (t) ABu (t T);u (t) ABu (t T)= = Ct th hai: 1 2 2 1u (t) ABu (t T);u (t) ABu (t T)= = Ct th ba: 1 1 2 2u (t) ABu (t T);u (t) ABu (t T)= = Ct th t: 1 2 2 1u (t) ABu (t T);u (t) ABu (t T)= = Vy mch logic ca b m ha vi sai nh sau:

1 1 2 1 2u (t ) ABu (t T) ABu ( t T) ABu (t T) ABu (t T)= + + + 2 2 1 2 1u (t) ABu (t T) ABu (t T) ABu (t T) ABu (t T)= + + +

Bi 4Thit k mch logic cho b gii m vi sai QPSK trn hnh 5.8.

Gii

Ta lp bng logic sau.

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38/52

Ta c : sign sin(4 ) sign sin si gncos sign sin( / 4)signcos( / 4) = + + S l thuc signsin(4 ) vo c th hin trn h ta c cc di y:

++ + + +

-+ + - +

++ + - +

-- + - -

+- - - -

-- - + -

+

+ - + -

-+ - + +

Nu ta chuyn du vo mc logic nh sau : "+""0", "-""1", ta c th biu din quan h du vi sai pha dng xung nh phn nh trn hnh v dc y.

1

0 0 /2/4 3/23/4 5/4

Nu chn cc thi im i du t dng sang m chnh pha ta c sai pha nh sau:k

4 4

= + .

Bi 6

(Tip) Thit k mch logic cho vng kha pha.

Gii

Ta s dng cc bin i sau y:

( )2signsin ' sign sin ' cos '4 2

+ = +

( )2signcos ' sign sin ' cos '4 2 + =

trong ' (t); = (t) l gc pha iu ch.T phn tch trn ta c th thit k mch logic nh hnh di y.

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40/52

Lung nh phn nhnh I01

Lung nh phn nhnh Q

01

/ 2

TLO1

B iu ch BPSK 1

B iu ch BPSK 1(nhnh I)Tham kho 1

B iu ch BPSK 2(nhnh Q)

Tham kho 2

0001

11 10

Bi 9

Cho mt h thng truyn dn iu ch BPSK.a) Thit k b gii iu ch BPSK s dng xuyn Diod b) Phn tch dng sng trong qu trnh gii iu ch.

Gii

a)Tn hiu thu trung

tn iu ch BPSK

Lung nh phn

0

1

0(t) 0 =

0

(t) 180 =

RLO1

Tn hiu dao ngni thu trung tn

0

"1"

180 0"0"

0

Thamkho

b)

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Dng sng khi bit '"0" tng ng vi goc pha iu ch 0(t) 0 = c truynTham kho

Tn hiu

u ra nh phn =tham kho x tn hiu

0(t) 0

=

Sau b lc u ra c gi tr dng

Dng sng khi bit '"1" tng ng vi goc pha iu ch 0(t) 180 = c truyn

Tham kho

Tn hiu

u ra nh phn =tham kho x tn hiu

0(t) 180 =

Sau b lc u ra c gi tr m

Bi 10

Cho mt h thng truyn dn iu ch QPSK.a) Thit k b gii iu ch BPSK s dng xuyn Diod b) Phn tch dng sng trong qu trnh gii iu ch.

Gii

a)

Lung nh phn nhnh I01

Lung nh phn nhnh Q

01

RLO1

/ 2

Tn hiu thutrung tn iu

ch QPSK

b) Ta xt cho trng hp gc pha tham kho=0 (knh I)

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Tham kho=0

Tn hiu thuvi(t)= /4

Tn hiu thu

vi (t)=-/4

Tn hiu nhiphn

u ra nh phn sau b lc dng 0

Tn hiu thuvi(t)=3/4

Tn hiu thuvi(t)=5/4

Tham kho=0

u ra nh phn sau b lc m 1

Bi 11

Cho my thu i tn:a) Thit k b bin i h tn c loi tr tn s nh gng vi gi thit tn s dao ng ni f RLO2 nh hn tn ssng mang f c. b) Gii thch hot ng b bin i h tn loi tr tn s nh.

Gii

a)

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imagef cf RLO2f

IFf IFf

B trn

B trn

0

90

RLO2

0

90RLO2cos( t)

RLO2cos( t)

Mch ghp3dB

Mch chia cngsut 3dB

ccos( t (t)) +imagecos( t)

0ccos( t (t) 90 ) + +

0imagecos( t 90 ) +

A+A

B+B

C+Cccos( t (t)) +

imagecos( t)

Tn hiu thu

b)Thnh phn tn hiu Thnh phn nh gng

u ra b trn th nhtc RLO2cos(( )t (t)) + RLO imagecos(( )t)

0c RLO2A cos(( )t (t) 90 )= + +

0RLO imageA' cos(( )t 90 )= +090

u ra b trn th hai0

c RLO2B cos(( )t (t) 90 )= + +

RLO imagecos(( )t)

0

RLO imageB' cos(( )t 90 )=

Cng 0c RLO2C A B 2cos(( )t (t) 90 )= + = + + C=A+B=0

Bi 12Cho mt h thng truyn dn 34,368 Mbps. Cc bit khai thc c ghp ti b ghp lung v tuyn,

trong c mi khung c 35 bit s liu v mt bit khai thc.a) Tm tc u ra b ghp lung v tuyn v tc lung khai thc b) Thit k b ghp lung v tuyn.

Gii

a) Tc u ra b ghp khung: R=34,368 Mbps x 36/3535,35Mbps; Tc lung khai thc bng: R kt=R-R b=35,35 -343690,982Mbps b)

B nh c tc vit v c

khc nhau

Lung bit34,368 Mbps

ng h vit34,368 MHz

AND

AND

ong h c 35,350 MHz

vi c mt l trong 36 xung

Xung iu khin cccng AND, 0,982 MHz

Khungghp

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Bai 13(Tip). Thit k b to xung c c c l.

B so pha

Trit mt bittrong s 36 bit: 36VCXO

B lcthp tn

34,368 MHz

35,35 MHz c l

35,35 MHz

VCXO: B dao ng thch anh iu khin bng in p

Bi 14

a) Thit k b phn knh v tuyn cho lung s trong bi 12. b) Thit k mch to ng h u 34,368 MHz

Gii

a)

B nh c tc vit v c

khc nhau

Lung bit35,35 Mb ps

ng h vit35,35 MHz c c l

AND

AND

ong h c u 34,368 MHz

Xung iu khin cccng AND, 0,982 MHz

Lung s liu

Lung khai thc

b)

B so pha

VCXO

B lcthp tn

34,368 MHzc lm u

34,348 MHz u

ng h 35,35MHz c l

Bi 15.Thit k b phn nhnh siu cao tn cho mt h thng pht thu vi ba s gm ba my pht v ba my thu s dng

ba cp tn s thu pht :. . . .i i 1 1 3 2 3 3(f f ) (f f ), (f f ), (f f )= , trong

.if l tn s pht cnif l tn s thu.

Gii

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46/52

Bi 2 Mt h thng v tuyn s c cng sut pht 3 W, tn s pht 2 GHz, anten pht v anten thu c ng

knh 1 m v hiu sut anten 0,55.a)Tnh h s khuych i anten b) Tnh EIRP theo dBmc) Tnh cng sut thu theo dBm nu c ly thng tin 10 km v ch c suy hao khng gian t do.

Gii

a)

292

8Df x1x2.1010lgG 10lg 10lg 0,55c 3.10

= = =23,82dBi

b) EIRP= 10lg3000 [dBm]+ 23,82dBi=34,8dBm+23,82=56,62dBmc) Suy hao khng gian t do:

[ ] 10lg s s L dB L= = 92,5+20lgf[GHz]+20lgd[km]=92,5+10lg2+10lg10=105,5dBPRX=EIRP-Ls=56,62dBm-105,5dB-48,88dBm

Bi 3

Mt h thng v tinh qung b c EIRP=57dBW, tn s 12,5GHz, ch c tn hao khng gian t do, tc tn hiu s bng 5.107 bps. My thu nc nh c nhit tp m T=600K v i hi t s tn hiu trn tp mE b/N0=10 dB. Tm bn knh cho anten thu ti thiu p ng yu cu trn.

Gii

Cng sut thu ti thiu c tnh nh sau:

10/10 b RX

0 b

E P 10 10 N kTR

= = = PRX=10kTR b=10x1,38.10-23x600x5.107

=4,14.10-12W-114dBW

Do khong cch t qu o a tnh n mt t l 36.000km nn suy hao khng gian t do:[ ] 10lg s s L dB L= = 92,5+20lgf[GHz]+20lgd[km]=92,5+20lg12,5+20lg36.000

=200dB

H s khuych i anten thu ti thiu c tnh nh sau:

G2=PRX-EIRP+Ls=-114-57+200=29dBi

D=8

2,99

c 3.1010 / 0,55 38.f .15,5.10

= =0,23m

Bi 4

Mt b khuych i c tr khng vo v ra l 50 m, h s khuych i l 60 dB, bng thng l 10kHz.Khi mt in tr 50 m u vo u vo, tp m trung bnh qun phng ti u ra l 100v, tm nhit tpm hiu dng cu b khuych i.

Gii

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Cng sut tp m u ra b khuych i: Nout= ANi+Na

Do u vo phi khng nn: Ni=kTif= 1,38.10-23x290x10.103=4.10-17W Nout=(100.10-6)2/50=2.10-10W Na=2.10-10-4.10-17x106=1,6.10-10WTp m b khuych i quy i vo u vo:

Nai=1,6.10-10/106=1,6.10-16=kTaf Ta=1,6.10-16/(1,38.10-23x10.103)=1,16.103K=1160K

Bi 5Mt b khuych i c h s tp m 4dB, bng thng 500 KHz v tr khng vo 50 m. Tnh in p

tn hiu u vo cn thit c SNR out u ra bng 1, khi u vo b khuych i c ni n in tr 50m ti nhit 290K.

Gii

Cng sut tp m u vo: Ni=kTif=1,38.10-23x290x500.103=2.10-15W.T s tn hiu trn tp m u vo: SNR in=NFxSNR out=100,4x1= 2,5.Cng sut tn hiu u vo tnh nh sau: Pi/Ni =2,5Pi=2,5x2.10-15=5.10-15W

Ta c: Pi=(Ui)2/R U= 15iPR 5.10 x50= =5.10-7V=0,5V

Bi 6

Mt h thng thng tin v tinh c cc thng s sau: tn s truyn dn 3GHz, iu ch BPSK, xc su bit li 10-3, tc bit 100bps, d tr ng truyn 3dB, EIRP=100W, khuych i anten thu 10dB, khong cch pht thu 40.000km.a) Tnh mt ph cng sut tp m cho php cc i quy i vo u vo my thu theo W/Hz

b) Tnh nhit tp m cho php cc i ca my thu, nu nhit tp m anten l 290 K c) Tnh h s tp m cho php cc i ca my thu theo dB.

Giia)

Tra bng trong ph lc 1 ta c u= b0 req

E2 N

=3,1(E b/N0)req=0,5x3,12=4,8

(E b/N0)Rx=Mx(E b/N0)req= 100,3x4,8=9,6.Suy hao khng gian t do:

[ ] 10lg s s L dB L= = 92,5+20lgf[GHz]+20lgd[km]=92,5+20lg3+20lg40.000=194dBPRX=EIRP-Ls+G2=10lg100-194+10=-164dBW

(E b/N0)Rx= RX b 0

PR N =9,6 N0=PRX/(9,6xR b)= 10

-16,4/(9,6x100)

= 10-18/(9,6.100,4)=4.10-20W/Hz b) N0=kTsTs=N0/k= 4.10-20/1,38.10-23= 2,9.103=2900K TRX=Ts-290=2900-290=2610K c) TRX=(NF-1)290 NF=TRX/290 +1=99,5dB

Bi 7

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B tin khuych i my thu c h s tp m 13 dB, khuych i 60 dB v bng thng 2MHz. Nhit tp m anten 490K v cng sut u vo l 10-12W.a) Tm nhit tp m b tin khuych i theo Kelvin b) Tm nhit h thng theo Kelvinc) Tm SNR out theo dB.

Gii

a) TRX=(NF-1)290=(101,3-1)290=5496K b) Ts=490+5496=5886 K c) SNR in= P/Ni=Pi/(kTif)=10-12/(1,38.10-23x490x2.106)=0,74.102=7416,7dB

NF[dB]=SNR in[dB]-SNR out[dB]SNR out=16,7dB-13dB=3,7dB

Bi 8

Gi thit my thu c cc thng s sau: khuych i=50dB, h s tp m=10dB, bng thng=500MHz,cng sut vo = 50.10-12W, nhit tp m ngun TA=10K, tn hao phi=0dB. Bn c yu cu mc thmmt b tin khuych i vi khuych i 20 dB v bng thng 500MHz. Tm h s tp m cn thit t ci thin t s tn hiu trn tp m ton h thng l 10dB.

GiiK hiu Ni, Na l cng sut tp m ngun v cng sut tp m my thu, ta c cng sut tp m h thng thucha mc thm b tin khuych i quy i u vo my thu : N1=Ni+Na = k.TA.f+ k.(NF1-1). 290.f

= k.f.(TA+(NF1-1).290)=1,38.10-23x500.106(10+(101-1).290)=1,8.10-11W

T s tn hiu trn tp m ban u l: SNR 1=50.10-12/1,8.10-11=2,84,47dB

T s tn hiu trn tp m yu cu sau khi mc thm b tin khuych i:SNR 2=10+4,47=14,47dB28K hiu Np l cng sut tp m ca b tin khuych i, ta c cng sut tp m h thng thu sau khi mthm b tin khuych i quy i vo u vo my thu: N2=Ni+N p+10-2 Na= k.f.(TA+10-2(NF1-1).290)+ N p

=1,38.10-23x500.106(10+10-2(10-1).290)+N p=25.10-14W+N pSNR 2=28=Pi/N2=50.10-12/(25.10-14+N p) N p=50.10-12/28 -25.10-14=1,54.10-12K hiu h s tp m ca b tin khuych i l NF p, ta c: N p=k(NF p-1).290.f=1,38.10-23(NF p-1)x290x500.106=1,54.10-12 NF p=1,54/(1,38x2,9x0,5)+1=1,772,5dB

Bi 9

Mt my thu gm ba tng: tng vo l b tin khuych i c h s khuych i 20 dB v h s tp 6dB. Tng th hai l cp ni vi tn hao 3 dB. Tng ngoi cng l b khuych i c h s khuych i 60 dBv h s tp m 16 dB.a) Tm h s tp m tng ca my thu b) Lp li a) khi loi b b tin khuych i.

Gii

a) K hiu NF1, NF2, NF2 l h s tp m ca tng th nht, th hai v th ba tng ng, k hiu A1, A2 l h skhuych i cu tng vo v tng ra, k hiu L l suy hao ca cp ni, t phng trnh (7.13) v (7.10) ta c:

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32tol 1

1 1

L(NF 1) NF 1 NF NFA A

= + + = 311 1

L(NF 1)L 1 NFA A

+ +

=0,3 0,3 1.6

0,32 2

10 1 10 (10 1)1010 10

+ + = 2 22 1 2(40 1)210 10 + + =2,794,46dB

b)

tol 2 3 NF NF L(NF 1)= + =100,3+100,3(101,6-1)=2+2(40-1)=8019dB

Bi 10

Tm nhit tp m h thng TS cho php cc i m bo xc sut li bit 2.10-4 i vi s liuR b=10kbps. Cc thng s ng truyn nh sau: tn s pht 12GHz, EIRP=10dBW, khuych i anten thu 0 dB,kiu iu ch BPSK nht qun, cc tn hao khc bng khng, khong cch pht thu l 100km.

Gii

Tra cu bng trong ph lc 1, ta c:

b

0

2Eu N

= =3,4E b/N0=5,78

Tn hao trong khng gian t do:Ls=92,5+20lgf[GHz]+20lgd[km]=92,5+20lg12+20lg100=143dBCng sut thu: PRX=EIRP-Ls+G2=10-143=-133dBWE b/N0=PRX/(R b N0)=5,78 N0=PRX/(5,78R b)=10-13,3/(5,78x10.103)

=10-17/(5,75x2)=8,6.10-19W/Hz N0=k.TsTs=N0/k=8,6.10-19/1,38.10-23=6,232.104K

Bi 11

Cn thit k mt my thu nhiu tng c Ttol=300K. Gi thit nhit tp m v khuych i cu cctng 2,3,4 l: T2=600K, T3=T4=2000K, A2=13 dB v A3=A4=20dB.Tm khuych i A1 ca tng u trong iu kin:a) T1=200K, 230K, 265K, 290K, 300K . b) Dng th ph thuc A1 vo T1

Giia)

Ttol=T1+T2/A1+T3/(A1.A2)+T4/(A1A2A3)Ttol-T1 = (1/A1)[T2+T3/A2+T4/ (A2A3)]A1=[T2+T3/A2+T4/ (A2A3)]/(Ttol-T1)= [600+2000/101,3+2000/ (101,3x102)]/(300-T1)= 701/(300-T1)

T1,K 200 230 265 290 300A1 7 10 20 70 A1[dB] 8,5dB 10dB 13dB 18,5dB

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b)8

10

12

14

16

18

20

1A ,dB

200 220 240 260 280 300 1T ,K

Bi 12

Mt my thu c khuych i 80dB, nhit tp m 3000k c ni n anten c nhit tp m 600Ka) Tm cng sut tp m ngun trong bng 40MHz b) Tm Tm cng sut tp m my thu quy i vo u vo my thuc) Tm cng sut tp m u ra my thu trong bng 40MHz.

Giia) Cng sut tp m ngun: Ni=kTAf=1,38.10-23x 500x40.106= 2,79.10-13W b) Cng sut tp m my thu quy i vo u vo:

Nri=kTr f=1,38.10-23x3000x40.106=1,66.10-12Wc) Cng sut tp m u ra my thu:

Nout= A(Ni + Nri)= 108(2,79.10-13+1,66.10-12)=19,39.10-5W

Bi 13

Mt b khuych i 10 dB, h s tp m 3dB c ni trc tip n anten thu (khng c tn hao cni gia chng). Sau khuych i ny l mt cp ni c tn hao 10 dB. Gi s cng sut u vo l 10pW,

nhit anten l 290K, bng thng l 0,25GHz. Tm:a) SNR u vo, u ra b khuych i b) u ra cp c tn hao.Gii

a) K hiu t s tn hiu trn tp m u vo v u ra b khuych i l SNR i v SNR out, ta c:SNR i=Pi/(kTAf)= 10-9/(1,38.10-23x290x0,25.109)=100030dBSNR out=SNR i-NF=30-3=27dB

b) K hiu SNR L , HFL lt s tn hiu trn tp m u ra cp tn hao v h s tp m cp tn hao, ta c:SNR L=SNR out-NFL=SNR out-L=27-10dB=17dB

Bi 14

50


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Mt my thu c h s tp m 13 dB c ni n anten qua cp 300 m di 25m c tn hao 10dB trn100m..a) Tm h s tp m tng ca cp ni v my thu b) Gi s mt b tin khuych i 20 dB vi h s tp m 3dB c ni gia cp v my thu, tm h s tp tng ca cp, b tin khuch i v my thuc) Tn h s tp m tng nu b tin khuych i c u vo gia anten v cp ni.

Gii

a) K hiu NF1=L, NF2 l h s tp m ca cp ni v my thu, ta c:

Ntol=L+L(NF2-1)=LNF2Ntol[dB]= L+NF2 -1=25x10/100+ 13-1=14,5dB b) K hiu NFr v A l h s tp m b tin khuych i v khuych i ca my thu, ta c:

Ntol= L+L(NFr -1)+L(NF2-1)/A=100,25+100,25(100,3-1)+100,25(101,3-1)/102 =3,95,9dB

c) Ntol= NFr + (L-1)/A+ L(NF2-1)/A=100,3+ (102,5-1)/102 +100,25(101,3-1)/102

=2,343,7dB

Bi 15

Mt h thng thng tin v tinh s dng my pht cng sut 20W, tn s 8 GHZ, anten parabol ngknh 1m. Khang cch n trm mt t l 10.000 km. H thng thu mt t s dng anten ng knh 2,5m vc nhit tp m h thng 100K. Gi thit rng cc anten c hiu sut=0,55. Coi rng cc tn hao khc Lo bng 2dB.a) Tnh tc s liu cho php cc i nu iu ch BPSK c s dng v xc sut li bit bng 2.10-4

b) Lp li a) vi gi thit tn s pht xung t v tinh l 2GHz.Gii

a) G1=

2921

1 8D f x1x8.1010lg G 10lg 10lg 0,55c 3.10

= = =35,86dBi

G2=

2922

2 8D f x2,5x8.1010lg G 10lg 10lg 0,55c 3.10

= = =46,42dBi

Tn hao khng gian t do:Ls=92,5+20lgf[GHz]+20lgd[km]=92,5+20lg8+20lg10.000=190,56dBCng sut thu ti trm mt t bng:

PRX=PTx+G1+G2-Ls-Lo=10lg20+35,86+46,42-190,56-2=-97,28dBW

N0=kTs=1,38.10-23x100=1,38.10-21W

Tra bng trong ph lc 1 ta c: b0

2E N

=3,4E b/N0=5,78

PRX/(R b N0)=5,78R b=PRX/(5,78N0)=10-9,7/(5,78x1,38.10-21)=0,025.1012 bps

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= 25 Gbps

b)

G1=

2921

1 8D f x1x2.1010lg G 10lg 10lg 0,55c

3.10

= =

=23,82dBi

2922

2 8D f x2,5x2.1010lgG 10lg 10lg 0,55c 3.10

= = =31,78dBi

Ls=92,5+20lg2+20lg10.000=175,5

PRX=PTx+G1+G2-Ls-Lo=10lg20+23,82+31,78-175,5-2=-109dBW

R b=PRX/(5,78N0)= 10-10,09

/(5,78x1,38.10-21

)=0,1. 1011

=10Gbps. Lu : Cc tc bit ni trn ch tha mn iu kin nng lng ch khng tha mn iu kin bng thng vbng tn Nyquist B N =10GHz trong khi tn s mang l 2GHz .

Vo tuyen so

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