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HOC VIN CNG NGH BU CHINH VIN THNG

KHOA: Khoa Vin thng

NGN HNG CU HI THI T LUN

Tn hc phn: Truyn dn s (Digital Transmission)

M hc phn: TEL 1420

Ngnh o to: in t truyn thng

Trnh o to: H i hc Chnh quy

1. Ngn hang cu hoi thi

Cu hi loi 2 im

Cu hi 2.1: Trnh by cc m hnh knh truyn trong h thng truyn dn s?

Cu hi 2.2: V s khi v trnh by nguyn tc m ha v gii m ca phng php iu ch xung m vi sai thch ng ADPCM?

Cu hi 2.3: V s khi v trnh by nguyn tc m ha v gii m ca phng php iu ch xung m vi sai DPCM?

Cu hi 2.4: V s khi v trnh by nguyn tc m ha v gii m ca phng php m ha d on tuyn tnh LPC?

Cu hi 2.5: Trnh by nguyn tc hot ng v ng dng ca 2 phng php sa li FEC v ARQ?

Cu hi 2.6: Mt ngun tin ri rc trong mt h thng thng tin s pht ra cc k hiu vi tc 5000 k hiu/giy. Cc k hiu cng vi xc sut xut hin ca chng c cho nh bng sau:

iX X1 X2 X3 X4 X5 X6

iXp 0,1 0,21 0,13 0,33 0,15 0,08

a. Xc nh entropy H(X) [bit] v tc thng tin (information rate) ca ngun?

b. Hy thc hin m ho ngun trn theo phng php t m c chiu di c nh. Phng php ny c hiu qa khng? Ti sao?

c. Nu cc k hiu ny c m ho sang cc bit nh phn dng phng php m ho Huffman, hy thc hin qu trnh m ho v cho bit hiu sut b m?

Mu 2

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Cu hi 2.7: Cho ngun khng nh A 5; 3; 1;0;1;3;5 vi xc sut tng ng

0,05;0,1;0,1;0,15;0,05;0,25;0,3

a. Tm Entropy ca ngun?

b. Nu ngun c m ho sang cc bit nh phn dng phng php m ho Fano, hy thc hin qu trnh m ho v cho bit hiu sut b m?

c. Gi s ngun c lng t theo nguyn tc:

4)5()3(

0)1()0()1(

4)3(5(

QQ

QQQ

QQ

Tm Entropy ca ngun lng t?

Cu hoi 2.8: Cho chui nh phn sau:

00010010000001100001000000010000001010000100000011010000000110

Hy thc hin m ha ngun Lempel-Ziv?

Cu hi 2.9: Nu ngha ca hm tc mo R(D). V th nh gi hm tc mo

cho knh i xng M mc khi MPD

V 1

1log)1(loglog)( 222

M

DDDDMDR

Vi gi tr ca M l 2, 4, 8 v 16. MP xc sut li bt.

Cu hi 2.10: Mt tn hiu tng t c dng:

x(t) 15cos(1500t / 4) 25 sin(2500t / 5)

a. Xc nh khong thi gian cho php ln nht gia cc xung ly mu lin tip nhau m bo qu trnh khi phc li tn hiu l tt nht?.

b. Nu 8 bt c s dng m ha cho gi tr mi mt xung ly mu, xc nh tc u ra ca lung d liu ny?

c. Nu lung d liu ny c pht qua knh nhiu c SNR = 25dB. Xc nh bng thng cn thit m bo xc sut li truyn dn nh nht?

Cu hi 2.11: Nu ngha ca m ng truyn? Chuyn chui bit nh phn sau thnh m B8ZS, HDB3?

11000010000011100000000110

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Cu hi 2.12: Cho chui d liu nh phn b = [1, 0, 0, 1, 1] vi tc KbpsRb 1 . Hy v

dng sng biu din chui tn hiu cc m ng dy sau:

NRZ n cc; NRZ lng cc; RZ n cc; RZ lng cc; NRZ Manchester?

Cu hi 2.13: V s khi v trnh by nguyn l hot ng ca b ghp knh TDM?

Cu hi 2.14: Trnh by qu trnh ghp lung 140 Mbit/s vo khung STM-1?

Cu hi 2.15: V v m t cu trc khung tn hiu STM-1. Tnh tc truyn tn hiu ca khung?

Cu hi 2.16: V v gii thch by s iu ch v gii iu ch BPSK?

Cu hi 2.17: V v gii thch by s iu ch v gii iu ch QPSK?

Cu hi 2.18: V v gii thch by s iu ch v gii iu ch M-ASK?

Cu hi 2.19: V v gii thch by s iu ch v gii iu ch 16QAM?

Cu hi 2.20: V v gii thch by s iu ch v gii iu ch 64QAM

Cu hi 2.21: So snh cc thng s hiu nng ca cc s BPSK, QPSK, 16QAM, 64QAM?

Cu hi 2.22: Mt h thng iu ch 2-ASK c tc bt 1000 bps, xc sut li bt = . tng tc bt 3000bps, thit b 2-ASK c thay th bng thit b 8-ASK vi cng 1 cng sut pht.Tnh xc sut li bt ?

Cu hi 2.23: Cho h thng BPSK truyn qua knh AWGN vi mt ph cng sut

, , T l chu k ca bt v A l bin tn hiu. Xc nh gi tr ca A xc sut li bt t c 10-6, nu tc d liu l:

a.10kbps

b. 100kbps

c. 1Mbps

Cu hi 2.24: H thng di ng c di knh ng ln t 810 - 826 MHz v di knh ng

xung l 940 956 MHz, gi s rng c 90% rng bng dnh cho knh

lu lng, h thng h tr 1150 cuc gi cng mt lc dng k thut FDMA,

s iu ch c hiu sut ph l 1.68bps/Hz. Knh truyn i hi dng m

sa li FEC . Tm gii hn trn v tc ca b m thoai?

Cu hi 2.25: B m ha c s lng bit phn b theo vai tr quan trng ng gp vo cht

lng tn hiu. M ha tin hnh trn on 20 ms (260 bit li ra b m ha).

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50 bit u (loi 1) l quan trng nht c bo v CRC 10 bit v FEC 1/2.

132 bit tip theo (loi 2) bo v CRC 5 bit v 78 bit cui khng c bo v

li. Tnh tc d liu knh tng cng t c?

Cu hi loi 3 im

Cu hi 3.1:

Trnh by cc phn t c bn v cc tham s hiu nng ca h thng truyn dn s?

Cu hi 3.2: So snh v phn tch u nhc im ca mt s cc phng php m ha dng sng: DM, PCM, DPCM?

Cu hi 3.3: Trnh by nguyn tc hot ng ca cc k thut ARQ dng v ch (Stop and Wait ARQ), ARQ quay ngc N t m (Go back N ARQ), ARQ la chn vic lp li (Selective repeat ARQ)?

Cu hi 3.4: Tnh hiu sut ca phng php pht li theo c ch ARQ dng v ch cho tuyn thng tin v tinh. Gi thit khong cch t v tinh ti mt t l

36000 km, vn tc truyn sng trong khng kh l sm /10.3 8 , tc thng tin l 56 Kbps v khung c kch thc 4000 bits. Tng t tnh hiu sut ca ARQ, nhng s dng kt ni trong mng LAN vi khong cch gia hai

trm l 100 m, vn tc truyn sng trn cp ng l sm /10.2 8 , tc truyn thng tin l 10 Mbps v khung c kch thc 500 bits?

Cu hi 3.5: Phn loi m ng truyn v phn tch cc yu t cn xem xt khi chn m ng truyn?

Cu hi 3.6: V s khi, nu chc nng cc khi trong b ghp SDH tiu chun?

Cu hi 3.7: Trnh by qu trnh ghp 63 lung tn hiu 2048Kb/s thnh khung STM-1? Da vo cu trc khung STM-1, tnh tc bt STM-1?

Cu hi 3.8: V s khi v m t nguyn tc hot ng ca b ghp knh PCM-30. Hy xc nh tc u ra ca b ghp ny?

Cu hi 3.9: V v trnh by cu trc khung b ghp 34/140?

Cho bit gi tr ca 4 bt u ca cc phn khung 2 (PK2), phn khung 3 (PK3), phn khung 4 (PK4), phn khung 5 (PK5) v phn khung 6 (PK6) u bng 1010. Hy xc nh tng s bit tin ca mi lung nhnh trong khung?

Cu hi 3.10: Trnh by cu to ca con tr AU-4. Nu cch m ha a ch cc Byte ti trng trong khung STM-1. Cho bit gi tr con tr khi khng chn l 557, xc nh ti thi im chn dng v sau khi chn dng: 10 bt gi tr con tr tng ng l bao nhiu?

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Cu hi 3.11: Trnh by cu to ca con tr TU-3. Nu cch m ha a ch cc Byte ti trng trong khung VC-4. Cho bit gi tr con tr TU-3 th hai khi khng chn l 449, xc nh 10 bt gi tr con tr trong khung chn dng v sau khi chn dng?

Cu hi 3.12: Nu ngha ca vic iu ch tn hiu? V gin iu ch 4- ASK (Gin gm: Mc tn hiu s, sng mang, tn hiu sau iu ch) ca dy bt u vo l 00011011010011?

Cu hi 3.13: Nu ngha ca vic iu ch tn hiu? V gin tn hiu iu ch PSK (Gin gm: Mc tn hiu s, sng mang, tn hiu sau iu ch) ca dy bt u vo l 100110110101110?

Cu hi 3.14: Trnh by nguyn tc hot ng v so snh u nhc im ca hai phng php ng b k hiu vng kn v vng h ti pha thu?

Cu hi 3.15: Trnh by cu trc v nguyn tc hot ng ca mch vng kha pha PLL.Vai tr ca PLL trong ng b h thng truyn dn s?

Cu hi 3.16: Trnh by cu trc v nguyn tc hot ng ca mch vng kha pha s DPLL. So snh vi PLL?

Cu hi 3.17: Trnh by s khi ca mt b m ha Turbo. ng dng ca m ha Turbo?

Cu hi 3.18: Cho lung bt nh phn 0,1,0,0,0,1,1,0nb . Hy biu din tn hiu tng ng vi cc m ng truyn sau: (a) Unipolar NRZ; (b) Unipolar RZ; (c) Polar NRZ; (d) Polar RZ; (e) AMI-NRZ; (f) AMI-RZ; (g) Manchester; Gi s A l trng thi khi to. nh x cc dng nguyn tc sau:

0 -A

1 +A

Cu hi 3.19: Cho 2 k thut iu ch s nh hnh v. Ln lt vi mi k thut xc nh khong cch ti thiu gia 2 k hiu (symbol) dmin, nng lng trung bnh trn mt k hiu Eav v nng lng trung bnh trn 1 bt truyn Eb. K thut no hiu qu hn gii thch ti sao? Phn tch cc tham s so snh gia cc k thut iu ch?

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A -A x x x x

-3A 3A

11 10 00 01

x

x x

x10 00

0111

A

Ax

x x

x10 00

0111

A

A

a. 4-ASK

b. 4-QAM

Cu hi 3.20: Mt h thng truyn dn QAM c tc 2400 symbols/sec (baud). Nhiu

cng c gi s l trng v phn b Gauss. Xc nh b 0E / N t c

xc sut li 10-5 vi tc bit tng ng l:

a. 2400 bit/sec

b. 4800 bit/sec

c. 9600 bit/sec

d. 19200 bit/sec

e. V th biu din quan h gia b 0E / N v s bit trn mt symbol,

nu nhn xt?

Cu hi 3.21: Cho chm sao tn hiu 4-PSK v 8-PSK nh hnh v. Xc nh bn knh 1r v

2r ca vng trn bit rng khong cch gia 2 im trn chm sao l d.T

kt qu hy xc nh cn tng cng sut pht ca 8-PSK ln bao nhiu ln so vi 4-PSK 2 h thng c cng xc sut li bt?

Cu hi 3.22: Cho m tch chp (2, 1, 2) vi cc a thc sinh:

g(1)

= (1, 1, 1); g(2)

= (1, 0, 1)

a. V s m ha?

b. V s li m v gin trng thi?

c. Gi s chui d liu vo 1010. Hy xc nh d liu m ha?

d. Gi s d liu thu c l Y. Hy thc hin vic gii m dng gii thut Viterbi trong cc trng hp sau: Y = 1110001011; Y = 1110101011; Y = 1110111001?

Cu hi 3.23: Cho m tch chp (3, 1, 2) vi cc a thc sinh:

g(1)

= (1,1,1); g(2)

= (1,0,1); g(3)

= (1,1,0)

a. V s m ha?

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b. V s li m v gin trng thi?

c. Gi s chui d liu vo 00110. Hy xc nh d liu m ha?

Cu hi 3.24: Cho m Hamming (7, 4) vi:

1 1 2 3c m m m

2 2 3 4c m m m

3 1 2 4c m m m

a. V s khi b m ha v gii thch?

b. Tm t m vi chui bit tin tc vo M = [0 1 1 1]?

c. Gi s li trn knh truyn v my thu c t m Y = [0 1 1 1 0 1

0]. Xy dng s khi gii m Syndrome v gii thch. Chng

minh my thu sa c li n?

Cu hi 3.25: Mt h thng thng tin s c s khi c m t nh hnh v.

pha pht, ngi ta s dng m tch chp (3, 1, 2) c xc nh bi ba a thc sinh:

g0(D) = 1 + D + D2

g1(D) = 1 + D2

g2(D) = 1 + D

V pha thu, phng php gii m c dng l gii thut Viterbi.

a. V s thc hin b m ho?

b. Xy dng s li m v gin trng thi cho m tch chp trn?

c. Trong trng hp h thng b nh hng ca nhiu lm chui d liu thu c l: Y = (110101011101110), hy xc nh chui bit tin tc thu c M?

Chui bit tin tc thu c M

T m thu c Y

T m X Chui bit

tin tc M

Channel

Noise

Viterbi

Decoder

Convolutional

Encoder

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2. xut cac phng an t hp cu hoi thi thanh cac thi: thi nn tng hp theo phng n (2+2+3+3)

3. Hng dn cn thit khac:

- Sinh vin khng c s dng ti liu ngoi tr cc bng Gaussian, Q-fuction km theo thi

Ngn hng cu hi thi ny c thng qua b mn v nhm cn b ging dy hc phn.

H Ni, ngy 18 thng 11 nm 2013

Trng Khoa Trng B mn Giang vin chu tr bin son

TS. Nguyn Tin Ban TS. L Nht Thng

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