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NGN HNG THIMn: L THUY
T THNG TIN
S tn ch : 4SDNG CHO NGNH IN T- VIN THNG V CNG NGH THNG TIN
HI HC TXA
CHNG I: NHNG VN CHUNG V KHI NIM CBN
1/ Chn cu ng sau :a Tin lun c biu din di dng sb Tm nh, bn nhc, bc th . . . khng phi l cc tinc Tin l mt nh x lin tc n ngi nhnd Tin l dng vt cht biu din hoc th hin thng tin.
2/ Chn pht biu ng trong nhng cu sau :a Thng tin l nhng tnh cht xc nh ca vt cht m con ngi nhn c t th gii vt cht
bn ngoib Thng tin khng th xut hin di dng hnh nhc Thng tin tn ti mt cch ch quan, ph thuc vo h th cm.d Thng tin khng th xut hin di dng m thanh
3/ Mn l thuyt thng tin bao gm vic nghin cu:a Vai tr ca thng tin trong k thutb Cc qu trnh truyn tin v L thuyt m ha.c L thuyt ton xc sut ng dng trong truyn tin.d Cch chng nhiu phi tuyn trong v tuyn in
4/ Chn cu ng nht v ngun tina Ngun tin l ni sn ra tinb Ngun tin l tp hp cc tin c xc sut v k hiu nh nhauc Ngun tin lin tc sinh ra tp tin ri rc.
d Ngun tin ri rc sinh ra tp tin lin tc.
5/ Chn cu ng nht vng truyn tina L mi trng Vt l, trong tn hiu truyn i t my pht sang my thub L mi trng Vt l m bo an ton thng tinc L mi trng Vt l trong tn hiu truyn i t my pht sang my thu khng lm mt
thng tin ca tn hiu.d ng truyn tin chnh l knh truyn tin.
6/ bin i mt tn hiu lin tc theo bin v theo thi gian thnh tn hiu s, chng ta cnthc hin qu trnh no sau y:
a Ri rc ha theo trc thi gian v lng t ha theo trc bin b Gii m d liuc M ha d liu.d Lng t ha theo trc thi gian v ri rc ha theo trc bin
HC VIN CNG NGH BU CHNH VIN THNGKm10ng Nguyn Tri, Hng-H Ty
Tel: (04).5541221; Fax: (04).5540587Website: http://www.e-ptit.edu.vn; E-mail: dhtx@e-ptit.edu.vn
http://www.e-ptit.edu.vn/mailto:dhtx@e-ptit.edu.vnmailto:dhtx@e-ptit.edu.vnhttp://www.e-ptit.edu.vn/7/31/2019 NHDT LTTT
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CHNG II: TN HIU V NHIU
1/ Chn cu ng v tn hiu:a Tn hiu l mt nh x lin tc n ngi nhnb Tm nh, bn nhc, bc th . . . khng phi l cc tinc Tin lun c biu din di dng sd Tn hiu l qu trnh ngu nhin .
2/ Chn pht biu ng nht vc trng thng k :a c trng cho cc qu trnh ngu nhin chnh l cc quy lut thng k v cc c trng thng
kb K vng, phng sai, hm t tng quan, hm tng quan l cc quy lut thng kc Cc hm phn b v mt phn b l nhng c trng thng kd Tn hiu v nhiu khng phi l qu trnh ngu nhin theo quan im thng k
3/ Chn cu ng nht v hm t tng quan :a Hm t tng quan l quy lut thng k th hin ca qu trnh ngu nhin.
b Hm t tng quan x 1 2R (t , t ) lun c tnh bng biu thc sau[ ]{ }21 2( , ) ( ) ( )x xR t t M X t m t=
c Hm t tng quan x 1 2R (t , t ) c trng cho s ph thuc thng k gia hai gi trhai thiim thuc cng mt
d Hm t tng quan x 1 2R (t , t ) lun bng phng sai ( )xD t vi mi t
4/ Vic biu din mt tn hiu gii hp thnh tng ca hai tn hiu iu bin bin thin chm s lmcho vic phn tch mch v tuyn in di tc ng ca n trnn phc tp
a ngb Sai
5/ Ngi ta gi tn hiu gii rng nu b rng ph ca n tho mn bt ng thc sau:
0
1
. Cc tn hiu iu tn, iu xung, iu xung ct, manip tn s, manip pha, l cc tn hiu
gii rng.a Saib ng
6/ Chn cu ng v cng thc xc nh mt ph cng sut
a
22
TT T1
E x (t)dt S ( ) d2
= =
l cng thc xc nh mt ph cng sut ca cc qutrnh ngu nhin.
b
2
T
T
S ( )
G ( )T
= l cng thc xc nh mt ph cng sut ca cc qu trnh ngu nhin.
c { }
2
T
xT
S ( )
G( ) M G ( ) M lim T
= = l cng thc xc nh mt ph cng sut ca cc qutrnh ngu nhin.
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d
2
T
x TT T
S ( )
G ( ) lim G ( ) limT
= =l cng thc xc nh mt ph cng sut ca cc qu
trnh ngu nhin.
7/ Chn cu ng v cng thc quan h gia mt ph cng sut v hm t tng quan
a
2T
x TT T
S ( )
G ( ) lim G ( ) limT
= =
b{ }
2
T
xT
S ( )
G( ) M G ( ) M limT
= =
c
22
TT T1
E x (t)dt S ( ) d2
= =
d
jG( ) R( )e d
=
8/ Trong trng hp h thng tuyn tnh thng c suy gim th nhng thi im t >> t 0 = 0 (thiim t tc ng vo), qu trnh ngu nhin u ra sc coi l dng. Khi hm t tng quanv mt ph cng sut ca qu trnh ngu nhin u ra s lin h vi nhau theo biu thc sau :
jra ra
1R ( ) G ( )e d
2
=
a Saib ng
9/ ( )*aS t l hm lin hp phc ca ( )aS t : ( ) ( ) ( )
= +aS t x t jx t l tn hiu gii tch.
ng bao ca tn hiu gii tch c th biu din bng cng thc sau: ( ) ( ) ( )= *a aA t S t .S t
a Saib ng
10/ Mt mch v tuyn in tuyn tnh c tham s khng i v c tnh truyn t dng ch nht(hnh di) chu tc ng ca tp m trng dng. Tm hm t tng quan ca tp m ra theo cng
thc
2
0ra
0
GR ( ) K( ) cos d
2
=
ta c kt qu no ?
GV()
2N0
a.
0 1 0 2
| K() |
b.
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a
2ra ra 0
sin2R ( ) cos2
=
b
22
ra ra ra v1 1
R (0) G ( )d P K( ) G ( )d2 2
= = = =
c
jra ra
1R ( ) G ( )e d2
=
d 0=raR
11/ Cho qu trnh ngu nhin dng c biu thc sau: ( ) ( )= + 0X t A cos 2 f t
Trong A = const, 0f = const, l i lng ngu nhin c phn bu trong khong ( ) , .
Tnh k vng { }( )M X t theo cng thc { }( ) ( ) ( )M X t X t w d
= ta c gi tr no di y
a { }( ) 0M x t =
b { }( ) 2M x t =
c { }( ) 1M x t =
d { }( ) 1M x t =
12/ Cho qu trnh ngu nhin dng c biu thc sau: ( ) ( )= + 0X t A cos 2 f t Trong A =
const, 0f = const, l i lng ngu nhin c phn bu trong khong ( ) , . Tnh hm ttng quan 1 2( , )R t t theo biu thc { }1 2( , ) ( ). ( )R t t M X t X t = + ta c gi tr no di y:
a 21 2 0( , ) cos 2R t t A f =
b 1 2( , ) 0R t t =
c 21 2 01
( , ) cos 22
R t t A f =
d 21 2 0( , ) cos 2R t t A f =
13/ Tn hiu in bo ngu nhin X(t) nhn cc gi tr + a; - a vi xc sut nh nhau v bng 1/2. Cn
xc sut trong khong c N bc nhy l: ( ) ( ) = >N
P N, e 0N!
(theo phn b Poisson). T cc gi thit trn tnh c hm t tng quan 2 2( )xR a e
= . Khi
mt ph cng sut ( )xG ca X(t) c tnh theo cng thc0
( ) 2 ( )cosx xG R d
= ta c gi
tr no sau y:
a2
2 2
4( )
4xa
G
=+
b2
2 2
4( )
4x
aG
+=
+
c ( ) 0xG =
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d2
2 2
4( )
4xa
G
=
14/ Mt qu trnh ngu nhin dng c hm t tng quan: 2 0( ) cosxR e
=
Khi mt ph cng sut ca cc qu trnh ngu nhin trn l
a2
2 20 0
2 2
( ) ( ) ( )xG
= + + + +
b 22 2
0 0
2 2( )
( ) ( )xG
= + + +
c 22 2
0 0
1 1( )
( ) ( )xG
= + + +
d 22 2
0 0
1 1( )
( ) ( )xG
= + + + +
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CHNG 3: CSL THUYT THNG TIN THNG K
1/ Khi nim lng tin c nh ngha da trn:a Nng lng ca tn hiu mang tinb bt nh ca tinc ngha ca tin
d Nng lng ca tn hiu mang tin v ngha ca tin
2/ Chn pht biu ng nht v Entropy ca ngun tin, H(X):a L i lng c trng cho bt nh trung bnh ca ngun tinb c tnh theo cng thc ( ) ( ) log ( )
x X
H X P X P x
=
c t cc tiu khi ngun l ng xc sutd t cc i khi ngun l ng xc sut
3/ Chn pht biu sai v bt nha bt nh ca php chn t l nghch vi xc sut chn mt phn t
b bt nh gn lin vi bn cht ngu nhin ca php chnc bt nh ca mt phn t c gi tr 1 bit khi xc sut chn phn t l 1d bt nh cn c gi l lng thng tin ring ca bin c tin
4/ Entropy ca ngun ri rc nh phn ( ) ( )H(A) p log p 1 p log 1 p=
Khi p=1/2 th H(A) t max Chn cu ng v ax( )mH A
a ax( )mH A =3/2 bt;
b ax( )mH A =1/2 bt;
c ax( )mH A =1 bt ;
d ax( )mH A =0
5/ Trong mt trn thi u bng Quc t, i tuyn Vit Nam thng i tuyn Brazin, thng tin nyc bt nh l
a bng 0 ;b V cng lnc nh hn 0;d ln hn 0
6/ Hc sinh A c thnh tch 12 nm lin t danh hiu hc sinh gii, hc sinh B lc hc km Thi ttnghip ph thng trung hc, hc sinh A trt, hc sinh B th khoa Thng tin v hc sinh B thkhoa, hc sinh A trt c bt nh l:
a bng 0 ;b V cng lnc nh hn 0;d ln hn 0
7/ Chn ngu nhin mt trong cc st 0 n 7 c xc sut nhnhau Khi xc sut ca scchn ngu nhin l:
a 7b 1/8
c 8d -7
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8/Mt thit b v tuyn in gm 16 khi c tin cy nhnhau v c mc ni tip Gi sc mtkhi no b hng, khi xc sut ca mt khi hng l:
a 1/16b 16c -16d -1/16
9/B t lkh52 qun (khng kfng teo), A rt ra mt qun bi bt k Xc sut v qun bi m A rt l:
a Bng 1/52b Nh hn 5c Ln hn 5 nh hn 6d Bng 1/52
10/ Chn cu sai trong cc cu sau :a bt nh s trthnh thng tin khi n b th tiub bt nh chnh l thng tinc Lng thng tin = bt nh tin nghim + bt nh hu nghim
d Lng thng tin = bt nh tin nghim - bt nh hu nghim
11/ Chn cu ng sau :a Lng thng tin = thng tin tin nghim - thng tin hu nghimb Thng tin hu nghim chnh l thng tin ringc Lng thng tin = thng tin hu nghim - thng tin tin nghimd Lng thng tin = thng tin tin nghim + thng tin hu nghim
12/ Chn cu sai trong cc cu sau :a Thng tin tin nghim (k hiu I(
kx )) c xc nh theo cng thc sau:I(
kx ) = log P(
k);
b Thng tin tin nghim cn gi l lng thng tin ring;c Thng tin tin nghim (k hiu I(k
x )) c xc nh theo cng thc sau:I(k
x ) =- log P(k
x );
d Thng tin tin nghim cn gi l lng thng tin ring c xc nh theo cng thc sau :
I(k) =
1log
( )k
P x;
13/ ( , )k l
I x y l lng thng tin cho vk
x dol
y mang li c tnh bng cng thc no sau y :
a1 1
log log( ) ( / )k k lP x P x y
b ( ) ( / )j j kI y I y x c
1log log ( / )
( ) k lkP x y
P x
d( / )
log( )
l k
l
P y x
P y
14/ ( / ) log ( / )k l k l I x y p x y= l thng tin hu nghim v k ( / ) 1k lp x y = khi vic truyn tin khng
b nhiu. Chn cu sai trong nhng cu v ( / )k lI x y di y:
a ( / )k lI x y =0 khi knh khng c nhiu
b ( / )k lI x y l lng tin b mt i do nhiuc ( / )k lI x y l lng tin c iu kin
d ( / )k lI x y =1/2 khi knh khng c nhiu
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15/ ( / )k lI x y l thng tin hu nghim v kx ( / ) 1k lp x y = khi vic truyn tin khng b nhiu Chn cu
sai trong nhng cu v ( / )k lI x y di y:
a ( / )k lI x y l lng tin b tn hao do nhiu
b ( / )k lI x y = 0 khi knh khng c nhiu
c ( / )k l
I x y >1/2 khi knh khng c nhiu
d ( / )k lI x y l lng thng tin v kx khi bit ly
16/ ( / )k lI x y l lng thng tin ring ca kx khi bit ly v ( / )k lI x y = 0 khi khng c nhiu Cu
ny ng hay sai ?a ngb Sai
17/ Chn cu sai trong nhng cu sau:a Lng tin cn li ca k sau khi nhn c ly k hiu l ( / )k lI x y
b ( , )k lI x y l lng tin ring ca kv ly
c Lng tin ( )kI x l lng tin ban u ca kd Lng tin ( )kI x l lng tin ban u ca kx , Lng tin cn li ca kx sau khi nhn c
ly k hiu l ( / )k lI x y
18/ Cho tini
x c xc sut l ( ) 0,5i
P x = , lng tin ring ( )i
I x ca tin ny bng cc i lng no
di y :a 4 btb 1 btc 1/4 btd 2 bt
19/ Cho tinic xc sut l ( ) 1/ 4
iP x = , lng tin ring ( )
iI x ca tin ny bng cc i lng no
di y :a 2 btb 4 btc 3 btd 1/2 bte 1/4 bt
20/ Cho tini
x c xc sut l ( ) 1/ 8i
P x = , lng tin ring ( )i
I x ca tin ny bng cc i lng no
di y :a 5 btb 3 btc 4 btd 1/4 bt
21/ Cho tini
x c xc sut l ( ) 1/16i
P x = , lng tin ring ( )i
I x ca tin ny bng cc i lng no
di y:a 1/4 btb 2 bt
c 3 btd 4 bt
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22/ Cho tini
x c xc sut l ( ) 1/ 27i
P x = , lng tin ring ( )i
I x ca tin ny bng cc i lng no
di y:a 2log7 btb Log1/9 bt;c Log27 bt;d Log1/27 bt;
23/ Cho tin i c xc sut l ( ) 1/ 9iP x = , lng tin ring ( )iI x ca tin ny bng cc i lng nodi y :
a Log9 btb Log1/3 btc 2log3 btd Log1/9 bt
24/ Cho tini
x c xc sut l ( ) 1/ 25i
P x = , lng tin ring ( )i
I x ca tin ny bng cc i lng no
di y :a Log2/5 bt
b 2log5 btc Log1/25 btd - Log25 bt
25/ Tm cu sai trong nhng cu di ya bt ngca tin
ix trong ngun tin
NX c tnh bng entropy ca lp tin
itrong ngun
tinN
X
b bt nh ca tin v lng tin v ngha tri ngc nhau nhng v gi tr li bng nhauc bt nh ca tin v lng tin c ngha nh nhau nhng gi tr khc nhaud Lng tin trung bnh c hiu l lng tin trung bnh trong mt tin bt k ca ngun tin
cho
26/Lng thng tin ring ( btnh) ca mt bin ngu nhin kx l ( )kI Chn biu thc sai trong cc biu thc di y
a ( ) ( )k kI ln px x= n vo l bit;b ( ) ( )k kI lgp x= n vo l hart;c ( ) ( )k kI ln p x= n vo l nat;d ( ) ( )k 2 kI log p x= n vo l bt
27/ Lng thng tin ring ( bt nh) ca mt bin ngu nhin kl ( )kI c tnh bng biu thcno di y :
a ( ) ( )k kI k ln px x= ;b ( ) ( )k kI ln px x= n vo l bit;c ( ) ( )k 2 kI log px x= n vo l nat;d ( ) ( )k kI lg px x= n vo l hart;
28/ Lng thng tin ring ( btnh) ca mt bin ngu nhin k l ( )kI x c tnh nh sau( ) ( )k kI k ln px x= , trong k l h s t l Tm cu sai v cch chn k trong cc cu di y :a Chn k = 1 ta c ( ) ( )k kI ln p x=
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b Chn k =-1 ta c ( ) ( )k kI ln p x= ;
c Chn1
kln10
= ta c ( ) ( )k kI lg px x= ;
d Chn1
kln 2
= ta c ( ) ( )k 2 kI log px x=
29/ Entropy ca ngun tin ri rc A l trung bnh thng k ca lng thng tin ring ca cc tin thucA K hiu:
( )1H A ; ( ) ( )1 iH A M I a= vi ( ) ( ) ( )
1 2 s
1 2 s
a a ... aA
p a p a ... p a
=
( )i0 p a 1 ;
( )s
ii 1
p a 1=
= ; ( )1H A c tnh bng biu thc no di y:
a ( ) ( ) ( )s
1 i i
i 1
H A p a log p a
=
= (bt) ;
b ( ) ( ) ( )s 1
1 i ii 1
H A p a log p a
=
= (bt) ;
c ( ) ( ) ( )s
1 i ii 0
H A p a log p a=
= (bt) ;
d ( ) ( ) ( )s
1 i ii 1
H A p a log p a=
= (bt) ;
30/ Entropy ca ngun tin ri rc A l trung bnh thng k ca lng thng tin ring ca cc tin thucA
K hiu: ( )1H A ; ( ) ( )1 iH A M I a= vi ( ) ( ) ( )
1 2 s
1 2 s
a a ... aA
p a p a ... p a
=
( )i0 p a 1 ; ( )s
ii 1
p a 1=
= ( )1H A c tnh bng biu thc no :
a ( ) ( ) ( )s
1 i ii 1
H A p a log p a=
= (bt) ;
b ( ) ( ) ( )s
1 i ii 0
H A p a log p a=
= (bt)
c ( ) ( )( )
s
1 ii 1 i
1H A p a log
p a== (bt)
d ( ) ( ) ( )s 1
1 i ii 1
H A p a logp a+
=
= (bt) ;
31/ Entropy ca ngun tin ri rc A l trung bnh thng k ca lng thng tin ring ca cc tin thucA
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K hiu: ( )1H A ; ( ) ( )1 iH A M I a= vi ( ) ( ) ( )
0 1 s 1
0 1 s 1
a a ... aA
p a p a ... p a
=
( )i0 p a 1 ; ( )s 1
ii 0
p a 1
=
= ( )1H A c tnh nh sau:
a ( ) ( ) ( )
s
1 i ii 0H A p a log p a== (bt)
b ( ) ( ) ( )s 1
1 i ii 1
H A p a log p a
=
= (bt)
c ( ) ( ) ( )s
1 i ii 1
H A p a log p a=
= (bt) ;
d ( ) ( ) ( )s 1
1 i ii 0
H A p a logp a
=
= (bt);
32/ A v B l hai trng bin c bt k, Entropy ca 2 trng bin cng thi C=AB l H(AB)Trong cc tnh cht ca H(AB) di y, tnh cht no sai:
a ( ) ( ) ( )H AB H A H B A= + ;b ( ) ( ) ( )H AB H B H A B= + ;
c ( ) ( ) ( )s
1 i ii 1
H A p a log p a=
= (bt) ;
d ( ) ( ) ( )s 1
1 i ii 0
H A p a logp a
=
= (bt);
33/ Entropy c iu kin v 1 trng tin A khi r trng tin B l H(A/B)Trong cc tnh cht ca H(A/B) di y, tnh cht no sai
a ( ) ( )H A B H B/ A ;b ( )0 H A B ;c ( ) ( )H A H A B d ( ) ( )H A B H A ;
34/ Entropy c iu kin v 1 trng tin B khi r trng tin A l ( )H B/ A , Tnh cht no ca( )H B/ A di y l nga ( )0 H B/ A ;b ( )0 H B/ A ;c ( ) ( )H A B H B/ A ;d ( ) ( )H B/ A H A ;
35/ Entropy ca trng bin cng thi H(AB) c tnh bng cng thc no sau y
a H(A) + H(A/B);b H(A) + H(B)c H(A) + H(B) - H(A/B) - H(B/A);d H(B) + H(A/B);
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36/ Lng thng tin cho trung bnh (k hiu l I(A,B) c cc tnh cht no sau y
a ( ) ( )I A,B H A= khi knh c nhiu;b ( ) ( )0 I A,B H A ;c ( ) ( )H A I A,B 0 ;
d ( ) ( )I A,B H A= khi knh c nhiu;
37/ Lng thng tin cho trung bnh (k hiu l I(A,B) Trong cc tnh cht di y, tnh cht nosai:
a ( ) ( )I A,B H A ;b ( ) ( )I A,B H A= khi knh c nhiu;c ( )0 I A,B ;d ( ) ( )I A,B H A= khi knh khng c nhiu;
38/ Lng thng tin cho trung bnh (k hiu l I(A,B) c cc tnh cht no sau ya ( )I A,B 1= khi knh bt;b ( )0 I A,B v ( )I A,B 0= khi knh bt;c ( )0 I A,B d ( ) ( )I A,B H A= khi knh c nhiu;
39/ Lng thng tin cho trung bnh (k hiu l I(A,B) ), trong cc tnh cht di y ca I(A,B), tnhcht no sai
a ( ) ( )I A,B H A= khi knh khng c nhiu;b ( )I A,B 0= khi knh bt;c ( ) ( )H A I A,B ;d ( )I A,B 1 khi knh bt;
40/ Lng thng tin cho trung bnh (k hiu l I(A,B)) , tm biu thc sai trong cc biu thc diy
a I(A,B) = H(A) - H(A/B);
b I(A,B) = ( ) ( )( )
s t
i ji ji 1 j 1 i
p a bp a b log p a= =
c I(A,B) = H(A) - H(B/A);d I(A,B) = H(B) - H(B/A) ;
41/ Mnh no sau y saia H(A/B) H(A) ;b H(A,B) H(A) + H(B)c I(A,B) = H(A) + H(B) + H(AB);d I(A,B) = H(A) + H(B) - H(AB);
42/ Chn ngu nhin mt trong cc s t 0 n 7 c xc sut nhnhau btnh ca scchn ngu nhin l
a 1/8 bt;
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b -3 bt;c 3 bt;d 8 bt;
43/ Mt thit b v tuyn in gm 16 khi c tin cy nh nhau v c mc ni tip Gi s cmt khi no b hng, bt nh ca khi hng l:
a 1/16 bt;
b 16 bt;c 1/4 bt;d 4 bt;
44/ B t lkh52 qun (khng k fng teo), A rt ra mt qun bi bt k bt nh v qun bim A rt l:
a Nh hn 5 bt;b Ln hn 5 nh hn 6 bt;c Bng 6 btd Ln hn 6 bt ;
45/ Mt hp c 8 ng tin kim loi , trong c 02 ng tin 500 ng; 02 ng tin 1000 ng, 2ng tin 2000 v 2 ng tin 5000 ng Chn ngu nhin 1 trong 8 ng tin Khi xc sut cang tin c chn ngu nhin l:
a 8 bt ;b -1/2 bt;c 1/8 bt;d 1/4 bt ;
46/ Mt hp c 8 ng tin kim loi , trong c 02 ng tin 500; 02 ng tin 1000, 2 ng tin2000 v 2 ng tin 5000 Chn ngu nhin 1 trong 8 ng tin Khi bt nh ca ng tinc chn ngu nhin l:
a 1/4 bt;b 8 bt ;c -1/2 bt;d 2 bt;
47/ Cho ngun tin X = {x1, x2, x3} vi cc xc sut ln lt l {1/2, 1/4, 1/4}, Entropy ca ngun tinH(X) c tnh l:
a 1 1 1log2 + log4 + log82 4 4
b 1 1 1log2 + log4 + log42 4 2
c 1 1 1log2 + log4 + log42 4 4
;
d 1 1 1log2 + log4 + log42 2 4
48/ Cho ngun tin X = {x1, x2, x3, x4, x5, x6, x7, x8, x9} vi cc xc sut ln lt l {1/4, 1/8, 1/8,1/8, 1/16, 1/16, 1/8, 1/16, 1/16}. Trong cc kt qu tnh Entropy di y, kt qu no sai:
a 1 1 1log4 + 4 log8 + 4 log164 8 8
b 1 3log4 + log2 + log22 2
c 1 1 1log4 + 4 log8 + 4 log164 8 16
d 1 3log4 + log2 + log24 2
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49/Entropy ( )1H A ca ngun ri rc
( ) ( ) ( )1 2 s
1 2 s
a a ... aA
p a p a ... p a
=
vi ( )i0 p a 1 ; ( )
s
ii 1
p a 1=
= Trong cc tnh cht ca H(A) di y tnh cht no l sai
a Khi ( )kp a 1= , ( )ip a 0= vi i k th ( ) ( )1 1 minH A H A 1= = b Ngun tin ri rc A c s du ng xc sut cho entropy cc i ( )1 maxH A logs=
c Khi ( )kp a 1= , ( )ip a 0= vi i k th ( ) ( )1 1 minH A H A 0= =
d Entropy ca ngun ri rc A l mt i lng gii ni ( )10 H a logs
50/ Cho ngun ri rc A
( ) ( ) ( )1 2 s
1 2 s
a a ... aA
p a p a ... p a
=
vi ( )i0 p a 1 ; ( )
s
ii 1
p a 1=
=
Gi entropy ca ngun A l ( )1H A , trong cc biu thc tnh ( )1 maxH A logs di y, biuthc no sai:
a ( )1 maxH A logs = ( )s
ii 1 i
1p a log logs
p(a )=+
b ( )1 maxH A logs = ( )s
ii 1 i
1p a log logs
p(a )=
;
c ( )1 maxH A logs = ( )s s
i i
i 1 i 1i
1p a log p(a )logs
p(a )= = ;
d ( )1 maxH A logs = ( )s
ii 1 i
1p a log logs
p(a )= ;
51/ Cho ngun ri rc A
( ) ( ) ( )1 2 s
1 2 s
a a ... aA
p a p a ... p a
=
vi ( )i0 p a 1 ; ( )
s
ii 1
p a 1=
= Nu ngun A c s du ng xc sut , khi biu thc no di y l sai
a ( )
s s
ii 1 i 1
1p a 0s= =+ =
bi
1p(a )
s=
1 i s ;
c ( )s s
ii 1 i 1
1p a 0
s= = = ;
d 1 1H (A) logs 0 H (A) logs
52/ Kh nng thng qua ca knh ri rc C l gi tr cc i ca lng thng tin cho trung bnh
truyn qua knh trong mt n v thi gian ly theo mi kh nng c th c ca ngun tin A
( ) ( )' ' kA A
C max I A,B v max I A,B= = (bps);
KA
C' .C v m I(A,B)i C = ax=
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K biu th s du m knh truyn c (c truyn qua knh) trong mt n v thi gianI(A,B) l lng thng tin truyn qua knh trong mt n v thi gian C c gi l kh nngthng qua ca knh i vi mi du C c cc tnh cht no di y :
a C 0, C = 0 khi A v B c lp ; C Klogs , ' kC v logs= khi knh khng nhiub C = 0 khi v ch khi A v B c nhiu
c
'
kC v logs= khi knh c nhiud ' kC v logs= khi cc knh c lp
53/ I(A,B) l lng thng tin trung bnh c truyn qua knh ri rc c tnh cht : I(A,B) H(A)V mt snh ngha :
( ) ( )' ' kA A
C max I A,B v max I A,B= = ,
KA
C' .C v m I(A,B)i C = ax=
K biu th s du m knh truyn c trong mt n v thi gian. T cc tnh cht v nhngha trn cho bit cc biu thc di y, biu thc no sai
a K I(A,B) K H(A);
b K I(A,B) K H(A);
c K maxI(A,B) K maxH(A)
d max( K I(A,B)) max( K H(A));
54/ Cho ngun ri rc ch c hai du:1 2
1 2
a aA
p(a ) p(a )
=
Ngun ri rc nh phn l ngun A trn tho mn iu kin sau
1
2
a "0" v ) p
a "1" v ) 1 p1
2
i xc sut p(a
i xc sut p(a
=
= Khi ngun ri rc nh phn A c th vit biu thc no
a1 2
p 1 pA
a a
=
b1
2
a pA
a 1 p
=
c 1 0Ap 1 p
=
d1 2a aA
p 1 p
=
55/ 1 2a a
Ap 1 p
=
l ngun ri rc nh phn Tho mn iu kin
1
2
a "0" v ) p
a "1" v ) 1 p1
2
i xc sut p(a
i xc sut p(a
=
= Khi Entropy ( )1H A c tnh bng cng thc no sau y
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a ( ) ( )p log p 1 p log 1 p
b ( ) ( )p log p 1 p log 1 p ;
c ( ) ( )p 1 p log 1 p p log ;
d ( ) ( )p log p 1 p log 1 p
56/ ( )lH A / b l lng thng tin tn hao trung bnh ca mi tin u pht khi u thu thu cjb ( ) ( ) ( )
s
l i l i li 1
H A / b p a / b log p a b=
= ; iH(B / a ) l lng thng tin ring trung bnh cha trong mi tin u thu khi u pht pht i mt tin
ia c tnh theo cng thct
i i ij 1
H(B/ a ) / a ) log p / a )j jp(b (b=
=
Trong trng hp knh bt (b nhiu tuyt i) ta c biu thc no di y l sai :
a ( ) ( )jH A b H A H(B)= + b ( ) ( )iH B a H B= ;c ( ) ( )H B A H B=
d ( ) ( )jH A b H A= = ( )H A B
57/ ( )lH A / b l lng thng tin tn hao trung bnh ca mi tin u pht khi u thu thu c
jb
( ) ( ) ( )s
l i l i li 1
H A / b p a / b logp a b=
= ; iH(B / a ) l lng thng tin ring trung bnh chatrong mi tin u thu khi u pht pht i mt tin c tnh theo cng thc:
t
i i ij 1
H(B/ a ) / a ) logp / a )j jp(b (b=
=
Trong trng hp knh khng nhiu biu thc no di y l ng :
a ( )kH A b H(A / A) 0= = ;b ( )kH A b H(B/ A) 0= = c ( )kH A b H(B/ B) 0= = ;d ( )kH A b H(A / B) 0= = ;
58/ ( )lH A / b l lng thng tin tn hao trung bnh ca mi tin u pht khi u thu thu c
bj . ( ) ( ) ( )s
l i l i li 1
H A / b p a / b logp a b=
= ; iH(B / a ) l lng thng tin ring trung bnh chatrong mi tin u thu khi u pht pht i mt tin aic tnh theo cng thc sau:
t
i i ij 1
H(B/ a ) / a )log p / a )j jp(b (b=
=
Trong trng hp b nhiu tuyt i, A v B l c lp nhau suy ra :
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i j ip(a / b ) p(a )= ; j i jp(b / a ) p(b )= i j i jp(a b ) p(a )p(b ) = , khi ta c biu thc no sau
y l ng:
a ( ) ( ) ( )s
j i ii 1
H A / b p a log p a=
= v ( ) ( ) ( )t
i j jj 1
H B/ a p b logp b=
=
b ( ) ( ) ( )
t
i j jj 1H B/ a p b log p a== c ( ) ( ) ( )
s
j i ii 1
H A / b p a log p b=
= v ( ) ( ) ( )t
i j jj 1
H B/a p b log p a=
=
d ( ) ( ) ( )s
j i ii 1
H A / b p a log p b=
=
59/ Entropy c iu kin v 1 trng tin A khi r trng tin B l ( )H A B , c xc nh theocng thc sau:
( ) ( ) ( )s t
i j i ji 1 j 1
H A B p a b log p a b= =
= Trong trng hp b nhiu tuyt i, A v B l c lp nhau, suy ra :
i j ip(a / b ) p(a )= ; j i jp(b / a ) p(b )= i j i jp(a b ) p(a )p(b ) = , khi biu thc no sau y l
ng:
a ( ) ( ) ( )s t
i j i ji 1 j 1
H A B p a b log p a b= =
=
b ( ) ( ) ( ) ( )t s
j i ij 1 i 1H A B p b p a log p a= ==
c ( ) ( ) ( )s t
i j i ji 1 j 1
H A B p a / b log p a b= =
=
d ( ) ( ) ( ) ( )t s
j i ij 1 i 1
H A B p b p a log p a= =
=
60/ Entropy c iu kin v 1 trng tin B khi r trng tin A l ( )H B A , c xc nh theo
cng thc sau: ( ) ( ) ( )s t
j i j ii 1 j 1
H B/ A p b a log p b a= =
= Trong trng hp b nhiu tuyt i, A v B l c lp nhau, suy ra :
i j ip(a / b ) p(a )= ; j i jp(b / a ) p(b )= i j i jp(a b ) p(a )p(b ) = Khi biu thc no sau y l ng
a ( ) ( ) ( ) ( )s t
i j ji 1 j 1
H B/ A p a p b logp b= =
= ;
b( ) ( ) ( ) ( )
t s
j i ij 1 i 1
H B/ A p b p a logp a= =
=
;
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c ( ) ( ) ( )s t
i j i ji 1 j 1
H B/ A p a b log p a b= =
= ;
d ( ) ( ) ( )s t
i j i ji 1 j 1
H B/ A p a / b log p a b= =
=
61/ Entropy c iu kin v 1 trng tin A khi r trng tin B l ( )H A B , c xc nh theo
cng thc sau: ( ) ( ) ( )s t
i j i ji 1 j 1
H A B p a b log p a b= =
= Trong trng hp b nhiu tuyt i, A v B l c lp nhau, suy ra :
i j ip(a / b ) p(a )= ; j i jp(b / a ) p(b )= i j i jp(a b ) p(a )p(b ) = , khi biu thc no sau y l
ng:
a ( ) ( ) ( ) ( )t s
j i ij 1 i 1
H A B p b p a logp a H(A)= =
= =
b( ) ( ) ( )
s t
i j i ji 1 j 1
H A B p a b log p a b H(A)= =
= = ;
c ( ) ( ) ( )s t
i j i ji 1 j 1
H A B p a / b log p a b H(A)= =
= =
d ( ) ( ) ( ) ( )t s
j i ij 1 i 1
H A B p b p a log p a H(A)= =
= =
62/ Entropy c iu kin v 1 trng tin B khi r trng tin A l ( )H B A , c xc nh theocng thc sau:
( ) ( ) ( )s t
j i j ii 1 j 1
H B/ A p b a log p b a= =
= Trong trng hp b nhiu tuyt i, A v B l c lp nhau, suy ra :
i j ip(a / b ) p(a )= ; j i jp(b / a ) p(b )= i j i jp(a b ) p(a )p(b ) = , khi biu thc no sau y l
ng:
a ( ) ( ) ( )s t
i j i ji 1 j 1
H B/ A p a / b log p a b H(B)= =
= =
b ( ) ( ) ( ) ( )t s
j i ij 1 i 1
H B/ A p b p a log p a H(B)= =
= =
c ( ) ( ) ( ) ( )s t
i j ji 1 j 1
H B/ A p a p b log p b H(B)= =
= =
d ( ) ( ) ( )s t
i j i ji 1 j 1
H B/ A p a b logp a b H(B)= =
= =
63/ Entropy c iu kin v 1 trng tin A khi r trng tin B l ( )H A B , c xc nh theocng thc sau:
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( ) ( ) ( )s t
i j i ji 1 j 1
H A B p a b log p a b= =
= vi i ji jj
p(a b )p(a / b )
p(b )=
T cng thc ny c th khai trin ( )H A B thnh cng thc no sau y:
a ( ) ( )s t
i j i ji 1 j 1
p a b logp b a= =
b ( ) ( )s t
i j i ji 1 j 1
p a / b logp a ,b= =
c ( ) ( )t s
j i j i jj 1 i 1
p(b ) p a / b logp a b= =
;
d ( ) ( )s t
i j i ji 1 j 1
p a / b logp a b= =
;
64/ Lng thng tin cho trung bnh (k hiu l I(A,B)) :
i jI(A,B) M I(a ,b )
= vi ( )( )
( )i j
i ji
p a bI a ,b log
p a= Xc sut c thng tin i j
I(a ,b )l
i jp(a b ) Do c th vit I(A,B)) bng cng thc no sau y:
a ( )( )( )
s ti j
i ji 1 j 1
i
p a bp a b log
p a= = ;
b ( )( )
( )
s ti j
i ji 1 j 1 i
p a bp a b log
p a= = ;
c ( )( )
( )
s ti j
i ji 1 j 1 i
p a bp a / b log
p a= = ;
d ( )( )( )
s ti j
i ji 1 j 1 i j
p a bp a b log
p a b= =
65/ Lng thng tin cho trung bnh (k hiu l I(A,B)) c vit thnh :
( ) ( )( )
( )
s ti j
i ji 1 j 1 i
p a bI A,B p a b log
p a= == Khi c th khai trin I(A,B) thnh
I(A,B)= ( ) ( ) ( )s t
i j i j ii 1 j 1
p a b logp a b logp a= =
a Saib ng
66/ Lng thng tin cho trung bnh (k hiu l I(A,B)) c vit thnh :
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( ) ( )( )
( )
s ti j
i ji 1 j 1 i
p a bI A,B p a b log
p a= == Khi c th khai trin I(A,B) thnh
I(A,B)= ( ) ( ) ( ) ( )s t s t
i j i j i j ii 1 j 1 i 1 j 1
p a b logp a b p a b logp a= = = =
a ngb Sai
67/ Lng thng tin cho trung bnh (k hiu l I(A,B)) c vit thnh :
( ) ( )( )
( )
s ti j
i ji 1 j 1 i
p a bI A,B p a b log
p a= == Khi c th khai trin I(A,B) thnh
I(A,B)= ( ) ( ) ( )s t
i j i j ii 1 j 1
p a b logp a b p b= =
a Sai
b ng
68/ Lng thng tin cho trung bnh (k hiu l I(A,B)) c vit thnh :
( ) ( )( )
( )
s ti j
i ji 1 j 1 i
p a bI A,B p a b log
p a= == Trong cc biu thc khai trin I(A,B) di y, biu thc
no sai
a I(A,B) = ( ) ( ) ( )s t
i j i i ji 1 j 1
p a b logp a logp a b= =
;
b I(A,B) = ( ) ( ) ( )s t
i j i j ii 1 j 1
p a b logp a b logp b= =
+ ;
c I(A,B) = ( ) ( ) ( ) ( )s t s t
i j i j i j ii 1 j 1 i 1 j 1
p a b logp a b p a b logp a= = = =
;
d I(A,B) = ( ) ( ) ( )s t
i j i j ii 1 j 1
p a b logp a b logp a= =
;
69/ Lng thng tin cho trung bnh (k hiu l I(A,B)) c vit thnh :
( ) ( ) ( )( )
s t i ji j
i 1 j 1 i
p a bI A,B p a b logp a= =
=
Entropy c iu kin v 1 trng tin A khi r trng tin B l ( )H A B , c xc nh theo cngthc sau:
( ) ( ) ( )s t
i j i ji 1 j 1
H A B p a b logp a b= =
= Khi trong cc kt qu tnh I(A,B), kt qu no sai
a I(A,B) = H(A) - H(A/B) ;
b I(A,B) = H(A)+ H(A/B) ;c I(A,B) = H(A) + H(B) - H(AB);d I(A,B) = H(B) - H(B/A);
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70/ Xt 2 trng s kin A v B sau :( ) ( )
ji
i j
baA i 1,s ;B j 1, t
p a p b
= = = =
Khi , trng s kin ng thi C A.B= Nu A v B l c lp th C c th vit thnh biu thcno di y:
a ( ) ( )i j
i j
a / b
C p a / p b
=
;
b( ) ( )
i j
i j
a bC
p a p b
+ =
+
cji
ji
baC
p(b )p(a )
=
;
d ( ) ( )
i j
i j
a b
C p a p b
= ;
71/ Xt 2 trng s kin A v B sau :( ) ( )
ji
i j
baA i 1,s ;B j 1, t
p a p b
= = = =
Trng s kin ng thi C A.B= c entropy H(C) c tnh bng cng thc no di y:
a ( ) ( ) ( )s t
i j i ji 1 j 1
H C p a b log p a b= =
=
b ( ) ( ) ( )s t
i j i ji 0 j 0
H C p a b log p a b= =
=
c ( ) ( ) ( )s 1 t 1
i j i ji 1 j 1
H C p a b log p a b
= =
= ;
d ( ) ( ) ( )s t
i j i ji 1 j 1
H C p a b log p a b= =
= ;
72/ Chn cu sai :
a Xc sut xut hin cng ln, lng tin thu c cng lnb Mt tin x c xc sut xut hin l p(x), nu p(x) cng nh th lng tin khi nhn c tin nycng s cng ln
c Nu p(x) cng ln th 1/p(x) cng nhd Mt tin x c xc sut xut hin l p(x), nu p(x) cng ln th lng tin khi nhn c tin ny
cng s cng nh
73/ Chn cu sai sau :a Xc sut xut p(x) cng ln th lng tin khi nhn c tin ny cng s cng lnb Xc sut xut hin ca mt tin t l nghch vi bt ngkhi nhn c mt tinc Xc sut xut p(x) cng ln th lng tin khi nhn c tin ny cng s cng nh
d Lng tin ca mt tin t l thun vi s kh nng ca mt tin v t l nghch vi xc sutxut hin ca tin
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74/ Lng tin c iu kin hu nghim v Kx ( thng tin ring v Kx sau khi c y ) c nh nghal ( / ) log ( / )
k l k l I x y p x y=
Chn cu sai trong cc cu sau :a Xc sut ( / ) 1
k lP x y = ch xy ra khi knh truyn khng c nhiu
b Khi ( / ) 1k l
P x y th ( / ) 1k lI x y v ngc li
c Xc sut ( / ) 1/ 2k lP x y = ( / ) 1k lI x y = bt ,d Khi ( / ) 1k l
P x y th ( / ) 0k l
I x y v ngc li
75/ Lng thng tin v Kx khi r tin y l( / )
( , ) log( )k l
k l
k
p x yI x y
p x=
Chn cu sai sau :
a ( / ) 1k l
p x y = tc l l lng tin ca Kx c truyn nguyn vn
b Nu ( / ) 0k l
p x y = suy ra1
( , ) log( )k l
k
I x yp x
=
c Nu ( / ) 1k l
p x y = suy ra 1( , ) log( )k l
k
I x yp x
=
d Nu ( / ) 1k l
p x y = , c ngha l khi y nhn c tin th chc chn Kx pht
76/ Lng thng tin hu nghim v Kx ( thng tin ring v Kx sau khi c y ) c vit l :1
( / ) log( / )k l
k l
I x yp x y
= . Lng thng tin ring v Kx l1
( ) log( )k
k
I xp x
=
Lng thng tin cho v Kx do y mang li l :1 1
( , ) log log
( ) ( / )k l
k k l
I x y
p x p x y
=
Tm cu sai sau :a Lng thng tin ring bng tng lng thng tin cho v lng thng tin hu nghimb Lng thng tin ring c th mc Tng lng thng tin cho v lng thng tin hu nghim bng lng thng tin ringd Lng thng tin ring lun dng
77/ Cho 2ngun tin A v B c cc xc sut tng ng l :
1 2 3 4a a a aA1/ 2 1/ 4 1/8 1/8
=
;
1 2 3 4b b b bB0,5 0,25 0,125 0,125
=
Entropy ca ngun A ( k hiu l H(A)), entropy ca ngun B ( k hiu l H(B)) c quan h theo cch thc no di y
a H(A)=H(B);b H(A) > H(B);c H(B)>H(A);d H(A)=2H(B)
78/ Cho ngun tin A c cc xc sut tng ng l :
1 2 3 4a a a aA1/ 2 1/ 4 1/8 1/8
=
; Khi Entropy ca ngun A ( k hiu l H(A)) bng cc i lngno di ya H(A) = 1,85 bt;b H(A) = 1,75 bt ;
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c H(A) = 1,7 bt;d H(A) = 1,65 bt ;
79/ Cho ngun tin A c cc xc sut tng ng l :
1 2 3 4 5a a a a aA0.45 0.2 0.15 0.1 0.1
=
;
Khi Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no di ya H(A) = 2,75 bt;b H(A) = 2,06 bt;c H(A) = 2,7 bt;d H(A) = 2,85 bt;
80/ Cho ngun tin A c cc xc sut tng ng l :
1 2 3 4 5a a a a aA0.4 0.25 0.15 0.1 0.1
=
;
Khi Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no di ya H(A) = 2,85 bt;b H(A) = 2,7 bt;c H(A) = 2,75 bt;d H(A) = 2,1 bt;
81/ Cho ngun tin A c cc xc sut tng ng l :
1 2 3 4 5a a a a aA0.4 0.25 0.15 0.15 0.05
=
;
Khi Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no di y
a H(A) = 2,85 bt;b H(A) = 2,7 bt;c H(A) = 2,06 bt;d H(A) = 2,07 bt;
82/ Cho ngun tin A c cc xc sut tng ng l :
1 2 3 4 5a a a a aA0.4 0.25 0.2 0.1 0.05
=
;
Khi Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no di ya H(A) = 2,06 bt;
b H(A) = 2,85 bt;c H(A) = 2,04 bt;d H(A) = 2,07 bt;
83/ Cho ngun tin A c cc xc sut tng ng l :
1 2 3 4 5a a a a aA0.4 0.3 0.15 0.1 0.05
=
;
Khi Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no di ya H(A) = 2,04 bt;b H(A) = 2,07 bt;c H(A) = 2,06 bt;d H(A) = 2,01 bt;
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84/ Gi s ngun tin A c cc xc sut tng ng l :
1 2 3 4 5a a a a aA0.4 0.3 0.2 0.05 0.05
=
;
Khi Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no di ya H(A) = 1,95 bt;b H(A) = 2,07 bt;
c H(A) = 2,01 bt;d H(A) = 2,04 bt;
85/ Gi s ngun tin A c cc xc sut tng ng l :
1 2 3 4 5a a a a aA0.35 0.35 0.2 0.05 0.05
=
;
Khi Entropy ca ngun A (k hiu l H(A)) gn bng cc i lng no di ya H(A) = 2,04 bt;b H(A) = 2,01 bt;c H(A) = 2,07 bt;d H(A) = 1,96 bt;
86/ Gi s ngun tin A c cc xc sut tng ng l :
1 2 3 4 5a a a a aA0.35 0.3 0.25 0.05 0.05
=
;
Khi Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no di ya H(A) = 2,01 bt;b H(A) = 1,9 bt;c H(A) = 1,98 bt;
d H(A) = 2,04 bt;
87/ Gi s ngun tin A c cc xc sut tng ng l :
1 2 3 4 5a a a a aA0.35 0.3 0.2 0.1 0.05
=
;
Khi Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no di ya H(A) = 2,06 bt;b H(A) = 1,98 bt;c H(A) = 2,01 bt;d H(A) = 1,9 bt;
88/ Gi s ngun tin A c cc xc sut tng ng l :
1 2 3 4 5a a a a aA0.3 0.3 0.2 0.1 0.1
=
;
Khi Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no di ya H(A) = 1,9 bt;b H(A) = 2,06 bt;c H(A) = 2,17 bt;d H(A) = 1,98 bt;
89/ Gi s ngun tin A v B c cc xc sut tng ng l :
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1 2 3 4 5 6 7 8a a a a a a a aA1/ 4 1/ 8 1/16 1/16 1/ 4 1/ 8 1/16 1/16
=
;
1 2 3 4 5 6 7 8b b b b b b b bB0, 25 0,125 0, 0625 0, 0625 0, 25 0,125 0, 0625 0, 0625
=
Entropy ca ngun A ( k hiu l H(A)), entropy ca ngun B ( k hiu l H(B)) c quan h theo cch thc no di y
a H(A) = 2H(B)b H(A)=H(B) ;c H(B)>H(A);d H(A) > H(B);
90/ Gi s ngun tin A c cc xc sut tng ng l :
1 2 3 4 5 6 7 8a a a a a a a aA1/ 4 1/8 1/16 1/16 1/ 4 1/8 1/16 1/16
=
;
Khi Entropy ca ngun A ( k hiu l H(A)) bng cc i lng no di y
a H(A) = 2,7 bt;b H(A) = 2,75 bt;c H(A) = 2,85 bt;d H(A) = 2,80 bt;
91/ Gi s ngun tin B c cc xc sut tng ng l :
1 2 3 4 5 6 7 8b b b b b b b bB0,25 0,125 0,0625 0,0625 0,25 0,125 0,0625 0,0625
=
;Khi Entropy ca ngun B ( k hiu l H(B)) bng cc i lng no di y
a H(B) = 2,80 bt;
b H(B) = 2,7 bt;c H(B) = 2,75 bt;d H(B) = 2,85 bt;
92/ Cho mt knh nh phn nh hnh bn:
p(b1/a1) = pda1 b1
a2 b2p(b2/a2) = pd
p(b1/a2)= ps
Trong :Phn b xc sut ca tin u ra p( 1b )c tnh theo cng thc sau :
2
1 11
( ) ( ) ( / )i i
i
p b p a p b a=
=
T cng thc ny, c th khai trin p( 1b ) thnh cng thc no di y:
a ( ) ( ) ( ) ( ) ( )1 1 1 1 1 1 2p b p a .p b a p a .p b a= + ;
b ( ) ( ) ( ) ( ) ( )1 2 1 1 2 1 2p b p a .p b a p a .p b a= + ;c ( ) ( ) ( ) ( ) ( )1 1 1 1 2 1 2p b p a .p b a p a .p b a= + ;
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95/ Cho knh i xng nh phn nh hnh bn:
p(b1/a1) = pda1 b1
a2 b2p(b2/a2) = pd
p(b1/a2)= ps
Bit
( )1p a p= ; ( )2 1p a p= ( ) ( )1 2 2 1 1s dp b a p b a p p= = =
p(b1/a
1) = p(b
2/a
2) = p
d
( ) ( ) ( )s s s sH B A p log p 1 p log 1 p = +
T truyn tn hiu cho knh 1kv T= v ( ) ( ) ( )'
A A
1 1C max I A,B max H B H B AT T= =
Khi '
'max
C
Cc tnh theo biu thc no di y
a ( ) ( )'
s s s s'max
C1 p log p 1 p log 1 p
C= +
b ( ) ( )'
s s s s'max
C1 p log p 1 p log 1 p
C= +
c ( ) ( )'
s s s s'max
C 1 p log p 1 p log 1 pC
= + +
d ( ) ( )'
d d d d'max
C1 p log p 1 p log 1 p
C= + +
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CHNG 4: CSL THUYT M
1/ Chn cu ng v m haa M ha l php bin i mi dng tn hiu thnh tn hiu lin tcb M ha l nh x che du thng tin
c M ha l mt nh x 1- 1 t tp cc tin ri rc ia ln tp cc t min
i ;in
i i: af d M ha l php bin i dng s nh phn thnh dng tn hiu ban u
2/ Chn cu ng v m :a M l mt tp cc t m c lp nn mt cch ngu nhinb M l qu trnh phc hi tin tcc M l nh x che du thng tind M (hay b m) l sn phm ca php m ha
3/ Chn cu ng v m :a M l qu trnh phc hi tin tc
b M l mt tp cc t m c lp nn theo mt lut nhc M (hay b m) l sn phm ca php bin i tn hiu ri rcd M (hay b m) l sn phm ca php bin i tn hiu lin tc
4/ di t m in l s cc du m cn thit dng m ha cho tin ia Chn cu ng v di tm
a di t m in l mt s phcb Nu in const= vi mi i th mi t m u c cng di B m tng ng c gi l b
m u
c di t m in cng ln th php m ha cng ti ud di t m in l mt s nguyn c th m hoc dng
5/ di t m in l s cc du m cn thit dng m ha cho tin ia Chn cu sai v di t m
a Nu in const= vi mi i , khi b m tng ng c gi l b m ub di t m in l mt s nguyn lun ln hn hoc bng 1c Nu in const= i c ngha l tt c cc t m u c cng di
d di t minl mt s nguyn c th m
6/ di t m in l s cc du m cn thit dng m ha cho tin ia Chn cu ng v di t m
a Nu i jn n th b m tng ng c gi l b m khng u
b di t m in cng ln th php m ha cng ti uc Nu in = 1 th b m tng ng c gi l b m khng ud Nu in const= i , c ngha l php m ha l ti u
7/ S cc du m khc nhau (v gi tr) c s dng trong b m c gi l cs m Ta k hiugi tr ny l m Chn cu sai vcc du m m:
a Nu m = 2 th b m tng ng c gi l m nh phn
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b Nu m = 3 th b m tng ng c gi l m tam phnc Nu m = 0 th b m tng ng c gi l m ud Nu m = p th b m tng ng c gi l m p phn
8/ Gi s c t m 7i 0 1 1 0 1 0 1 = Chn cu ng nht v t m7i
a T m 7i c di bng 4
b T m 7i trong b m nh phn c m=2 ( tc l c 2 du m l 0 v 1) v c di l 7
c T m 7i c 7 du m
d T m 7i c 3 du m
9/ Gi s c t m 7i 0 1 1 0 1 0 1 = Chn cu ng v t m7i
a T m 7i c 7 du m
b T m 7
i trong b m nh phn c m=3, tc l c 3 du m l 0
c T m 7i c di bng 7
d T m 7i c di bng 3
10/ Chn cu ng v di t m
a di trung bnh ca t m n l k vng ca i lng ngu nhin in c xc nh nh
sau : ( )s
i ii 1
n p a n=
=
b di trung bnh ca t m n l k vng ca i lng ngu nhin in c xc nh nh
sau : ( )s
i ii 1
n p a n=
= c di t m in l s cc du m cn thit dng gii m cho tin ia d di t m in l S cc du m khc nhau (v gi tr) c s dng trong gii m
11/ Ni dung ca nh l m ho th nht ca Shannon (i vi m nh phn) c pht biu nh sau:Lun lun c thxy dngc mt php m ho cc tin ri rc c hiu qu m di trung bnhca tm c thnh tu , nhng khng nh hn entropie xc nh bi cc c tnh thng k ca
ngun . Chn cu ng v di t m :a S t m nh nhtb S t m khng ic Chiu di trung bnh ca cc t m nh hn hoc bng entropy ca ngund Chiu di trung bnh cc t m nh nht trong tt c cc cch m ha
12/ Ni dung ca nh l m ho th nht ca Shannon (i vi m nh phn) c pht biu nh sau:Lun lun c th xy dng c mt php m ho cc tin ri rc c hiu qu m di trung bnhca t m c th nh tu , nhng khng nh hn entropie xc nh bi cc c tnh thng k cangun. Chn cu sai v di t m :
a Chiu di trung bnh cc t m tho mn h thc ( ) ( )
s
i i 1i 1n p a n H A==
b Chiu di trung bnh cc t m nh nht trong tt c cc cch m hac Chiu di trung bnh ca cc t m ln hn hoc bng entropy ca ngun
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d Chiu di trung bnh cc t m tho mn h thc ( ) ( )s
i i 1i 0
n p a n H A=
= =
13/ Khong cch gia hai t m bt k ni vnj l s cc du m khc nhau tnh theo cng mt v
tr gia hai t m ny, k hiu ( )n ni jd , Gi s 7i 0 1 1 0 1 0 1 = ; 7j 1 0 0 1 1 1 0 = Khong cch gia 2 t m ni v
nj l ( )7 7i jd , bng cc i lng no di y
a 1;b 6 ;c 12;d 14;
14/ Khong cch gia hai t m bt k ni vnj l s cc du m khc nhau tnh theo cng mt v
tr gia hai t m ny, k hiu
( )n n
i j
d , . Tm biu thc sai v khong cch m d trong cc biu
thc sau:
a ( )n ni jn d , ;b ( )n ni jd , 0 c ( ) ( )n n n ni j j id , d , ;d ( ) ( )n n n ni j j id , d , =
15/ Khong cch gia hai t m bt k ni vnj l s cc du m khc nhau tnh theo cng mt v
tr gia hai t m ny, k hiu ( )n ni jd , Tm biu thc sai v khong cch m d trong cc biuthc sau:
a ( ) ( ) ( )n n n n n ni j j k i k d , d , d , + ;b ( )n ni jd , 0 ;c
( )n n
i jd , 0
d ( )n ni jn d ,
16/ Khong cch gia hai t m bt k ni vnj l s cc du m khc nhau tnh theo cng mt v
tr gia hai t m ny, k hiu ( )n ni jd , . Chn cu ng sau :a ( )n ni j0 d , 1 ;
b ( ) ( ) ( )n n n n n ni j j k i k d , d , d , + =
c ( )n ni j1 d , ;
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d ( )n ni jd , 0 ; ( )n ni jd , 0 = khi n ni j ;
17/ Trng s ca mt t m ( )niW l s cc du m khc khng trong t m V d:7i 0 1 1 0 1 0 1 = th ( )7iW 4 = . Chn cu ng v cc tnh cht ca trng s ( )niW :a ( ) ( )n n n ni j i jd , w + b ( )niW 0 c ( ) ( )n n n ni j i jd , W + ;d ( ) ( )n n n ni j i jd , W = + v ( )ni0 W 1 ;
18/ Trng s ca mt t m
( )n
iW l s cc du m khc khng trong t m Gi s
7i 0 1 1 0 1 0 1 = , th trng s ( )7iW bng s no di ya 4 ;b 3;c 5 ;d 7 ;
19/ Coi mi t m ni l mt vctn chiu trong mt khng gian tuyn tnh n chiu nV , khi phpcng c thc hin gia hai t m tng t nh php cng gia hai vcttng ng c thc hin
trn trng nh phn GF(2) Php cng theo modulo 2 ny c m t nh sau:
Cho ( )7i 0 1 1 0 1 0 1 0, 1, 1, 0, 1, 0, 1 =
( )7j 1 0 0 1 1 1 0 1, 0, 0, 1, 1, 1, 0 =
Khi 7 7 7k i j = + bng gi tr no di ya (0 1, 1, 1, 0, 1, 1);b (1 1, 1, 1, 0, 1, 1);c (1 1, 1, 1, 1, 1, 1);d (1 1, 1, 1, 0, 1, 0);
20/ Coi mi t m ni l mt vctn chiu trong mt khng gian tuyn tnh n chiu nV , khi phpcng c thc hin gia hai t m tng t nh php cng gia hai vcttng ng c thc hintrn trng nh phn GF(2) Php cng theo modulo 2 ny c m t nh sau:
Cho ( )8i 1 1 1 0 1 0 1 0 1, 1, 1, 0, 1, 0, 1, 0 =
( )8j 1 0 0 1 1 1 0 1 1, 0, 0, 1, 1, 1, 0, 1 =
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b n =1,51;
c n =1,35;
d n =1,81;
27/ Sau khi thc hin m ha ngun ri rc A ( c entropy l H(A)=1,4 ) c th tnh c di
trung bnh n Vi mi cch m ha khc nhau s tnh c n khc nhau Trong cc kt qu m ha,
gi tr n no di y c gi l ti u nht:
a n =1,51;
b n =1,45;
c n =1,55;
d n =1,75;
28/ Sau khi thc hin m ha ngun ri rc A (c entropy l H(A)=1,9) c th tnh c di trung
bnh n Vi mi cch m ha khc nhau s tnh c n khc nhau Trong cc kt qu m ha, gi tr
n no di y c gi l ti u nht:
a n =1,975;
b n =1,91;
c n =1,905;
d n =1,95;
29/ Sau khi thc hin m ha ngun ri rc A ( c entropy l H(A)=1,905 ) c th tnh c di
trung bnh n Vi mi cch m ha khc nhau s tnh c n khc nhau Trong cc kt qu m ha,
gi tr n no di y c gi l ti u nht:a n =1,906;
b n =1,907;
c n =1,91;
d n =1,95;
30/ Sau khi thc hin m ha ngun ri rc A (c entropy l H(A)=2,01 bt), c th tnh c di
trung bnh n Vi mi cch m ha khc nhau s tnh c n khc nhau Trong cc kt qu m ha,
gi tr n no di y c gi l ti u:a n =2,07;
b n =2,11;
c n =2,09;
d n =202;
31/ Cho m Cyclic C(n, k) = C(7,4) c a thc sinh lg(x) tha g(x) c bc 4 ;b g(x) l mt c ca x 7 4+ + 1
c g(x) c bc 5 ;d T g(x) c th xc nh c ma trn sinh h thng G v ma trn kim tra H cho b m
32/ Cho m Cyclic C(n, k) = C(7,4), c a thc sinh g(x) l
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a 4( ) 1g x x x= + + ;
b 2 4( ) 1g x x x x= + + +
c 2 3( ) 1g x x x= + + ;
d 2 3 4( ) 1g x x x x= + + +
33/ Cho m Cyclic C(n, k) = C(7,4), c a thc sinh g(x) l
a 2 3( ) 1g x x x x= + + + ;b 3( ) 1g x x x= + + ;
c 2 4( ) 1g x x x= + +
d 3 4( ) 1g x x x= + +
34/ Cho m Cyclic C(n, k) = C(7,3) , tm cu sai vg(x)
a g(x) khng l c ca7x + 1 ;
b T g(x) c th xc nh c ma trn sinh h thng G v ma trn kim tra H cho b mc g(x) c bc 4 ;
d g(x) l mt c ca 7x + 1 ;
35/ Cho m Cyclic C(n, k) = C(7,3), c a thc sinh g(x) l
a4( ) 1g x x x= + +
b2 4( ) 1g x x x x= + + + ;
c2 3 5( ) 1g x x x x= + + +
d2 3( ) 1g x x x= + + ;
36/ Cho m Cyclic C(n, k) = C(7,3), c a thc sinh g(x) la
5( ) 1g x x x= + +
b2 3 5( ) 1g x x x x= + + +
c4( ) 1g x x x= + + ;
d2 4( ) 1g x x x x= + + + ;
37/ Pht biu sau ng hay sai : Cho m Cyclic C(n, k) = C(7,3) ,g(x) l a thc bc 4a Saib ng
38/ Cho m tuyn tnh (7,4) S nhcn thit nhb m la 7;b 28;c 4;d 112;
39/ Cho m Xyclic (7,4) S cc nhcn thit nhb m la 7;b 112c 28;
d 8
40/ Cho m tuyn tnh (7,3) S nhcn thit nhb m la 7;
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b 21;c 3;d 56;
41/ Cho m Xyclic (7,3) S cc nhcn thit nhb m la 3b 56
c 21;d 7;
42/ Cho m tuyn tnh (n,k) S nhcn thit nhb m l
akn.2
b kn;c n+k;d n;
43/ Cho m Xyclic (n,k) S cc nhcn thit nhb m l
a kn;b n+k;
ckn.2
d n7;
44/ Chn cu ng sau:a Cc dng tuyn tnh ca k bin c lp 1 2 k, , ,x x x l cc biu thc c dng:
( )k
1 k i ii 1
, ,f x x a x=
= Trong : i Fa , F l mt trng
b Cc dng tuyn tnh ca k bin c lp 1 2 k, , ,x x x l cc biu thc c dng:
( )k
1 k i ii 1
, , )f x x (a + x=
= Trong : i Fa , F l mt trng
c Cc dng tuyn tnh ca k bin c lp 1 2 k, , ,x x x l cc biu thc c dng:
( )k
1 k i ii 1
, ,f x x a - x=
= Trong : i Fa , F l mt trng
d Cc dng tuyn tnh ca k bin c lp 1 2 k, , ,x x x l cc biu thc c dng:
( )k
1 k ii 1
, , xif x x a=
= Trong : i Fa , F l mt trng
45/ Chn cu sai v m tuyn tnh :
a k2 cc vtkhc nhau l tt c cc t hp tuyn tnh c th c ca k vcthng ny
G.H 0= Trong : r n k= b Trong i s tuyn tnh ta bit rng vi mi G s tn ti ma trn r nH tha mn:
c Trong i s tuyn tnh ta bit rng vi mi G s tn ti ma trn r nH tha mn:
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d m t m tuyn tnh, c th s dng ma trn sinh k,nG , ma trn ny cha k vcthng
c lp tuyn tnh to nn khng gian m ( )n,kV TG.H 0= Trong : r n k=
46/ m t m tuyn tnh, c th s dng ma trn sinh k,nG Trong i s tuyn tnh ta bit rng
vi mik,n
G s tn ti ma trnr n
H
tha mn: TG.H 0= Chn cu sai sau :
a k,nG l ma trn k hng, n ct
b TH c gi l ma trn k hng, n ct ;c r nH l ma trn r hng, n ct ;
d TH c gi l ma trn chuyn v ca H
47/ Khi xy dng mt m tuyn tnh ( )0n,k,d ngi ta phi tm c cc m c tha nh nhngli c kh nng khng ch sai ln. Ngi ta thng xy dng m ny da trn cc bi ton ti u Tmcu sai trong cc cu di y:
a Vi k v 0d xc nh, ta phi tm c m c di vi t m l ln nhtTng ng vi biton ny ta c gii hn n = k
b Vi n v k xc nh, ta phi tm c m c khong cch 0d l ln nht Tng ng vi bi
ton ny ta c gii hn Plotkin sau:k 1
0i
i 0
dn
2
=
c Vi k v 0d xc nh, ta phi tm c m c di vi t m l nh nhtTng ng vi bi
ton ny ta c gii hn Griesmer sau:k 1
0 k
n.2d
2 1
d Vi n v s sai khi sa t xc nh, ta phi tm c m c s du thng tin k l ln nht (hay sdu tha l nh nht)Tng ng vi bi ton ny ta c gii hn Hamming
sau:t
n k in
i 0
2 C
=
48/ Chn nh ngha sai v m xyclic trong cc nh ngha sau
a M xyclic (n, k) l Ideal ( )I g X= ca vnh a thc [ ] n2Z x X 1+ b M xyclic (n, k) l mt b m m a thc sinh c bc r = n+k
c M xyclic l mt b m tuyn tnhd M xyclic l mt b m , m nu ( )a X l mt t m th dch vng ca ( )a X cng l mt t
m thuc b m ny
49/ Chn cu ng ca nh l v kh nng sa saia M u nh phn c tha (D > 0) vi khong cch Hamming 0d 4= c kh nng sa c
t sai tho mn iu kin: 0d 1
t2
+
b M u nh phn c tha (D > 0) vi khong cch Hamming 0d 1> c kh nng pht hint sai tho mn iu kin 0t d 1
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c M u nh phn c tha (D > 0) vi khong cch Hamming 0d 3 c kh nng sa c
t sai tho mn iu kin: 0d 1
t2
v c kh nng pht hin t sai tho mn iu kin
0t d 1 d M u nh phn c tha (D > 0) vi khong cch Hamming 0d 1 c kh nng sa c
t sai tho mn iu kin: 0d 1t2
50/ Tnh l m ho th 1 ca Shannon i vi m nh phn ta c : ( ) ( )s
i i 1i 1
n p a n H A=
= T biu thc trn tm cu ng nht trong cc biu thc sau :
a ( ) ( )s
i ii 1
n p a log p a=
= ;
b ( ) ( ) ( )
s s
i i i ii 1 i 1
p a n p a logp a= == ;
c ( ) ( ) ( )s s
i i i ii 1 i 1
p a n p a logp a= =
;
d ( ) ( ) ( )i i i ip a n p a logp a=
51/ Theo nh l m ho th 1 ca Shannon i vi m nh phn ta c:
( ) ( ) ( ) ( )s s
i i 1 i ii 1 i 1
n p a n H A p a log p a= =
= =
T biu thc ny suy ra di t m in v xc sut ( )ip a lin h vi nhau: ( )i i1n log
p a (*)
T (*) tm cu ng sau v nguyn tc lp m tit kim:a Cc t m c di ln sc dng m ha cho cc tin c xc sut lnb Cc t m c di n t l thun vi xc sut Pc Cc tin c xc sut xut hin ln c m ha bng cc t m c di nh v ngc li cc
tin c xc sut xut hin nhc m ha bng cc t m c di lnd Cc t m c di nh sc dng m ha cho cc tin c xc sut nh
52/ Cho ngun tin{ }1 2 3 4 5
, , , ,X x x x x x= vi cc xc sut ln lt l {1/2, 1/4, 1/8, 1/16, 1/16}
Bit 1 c m ha thnh 0, 2x c m ha thnh 10, 3x c m ha thnh 110, 4x c m ha
thnh 1110, 5x c m ha thnh 1111 B m ti u cho ngun trn c chiu di trung bnh tnh
theo cng thc : ( )5
i ii 1
n p x n=
= l :a 1,88b 1,90c 1,875d 1,925
53/ Yu cu ca php m ha: nhng t m c di nh hn khng trng vi phn u ca t m c di ln hn Cc tin c xc sut xut hin ln hn c m ha bng cc t m c di nh vngc li.
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Cho ngun tin { }1 2 3 4 5, , , ,X x x x x x= vi cc xc sut ln lt l {1/2, 1/4, 1/8, 1/16, 1/16}
Bit 1x c m ha thnh 0, 2x c m ha thnh 10 Chn cu ng di y m ha cho 3x
a 101b 011c 110d 100
54/ Yu cu ca php m ha: nhng t m c di nh hn khng trng vi phn u ca t m c di ln hn Cc tin c xc sut xut hin ln hn c m ha bng cc t m c di nh vngc li.Cho ngun tin { }1 2 3 4 5, , , ,X x x x x x= vi cc xc sut ln lt l {1/2, 1/4, 1/8, 1/16, 1/16}
Bit 1x c m ha thnh 0, 2 c m ha thnh 11, 3x c m ha thnh 100
Chn cu ng di y m ha cho 4x
a 000b 110c 001d 1010
55/ Yu cu ca php m ha: nhng t m c di nh hn khng trng vi phn u ca t m c di ln hn Cc tin c xc sut xut hin ln hn c m ha bng cc t m c di nh vngc li Cho ngun tin { }1 2 3 4 5, , , ,X x x x x x= vi cc xc sut ln lt l {1/2, 1/4, 1/8, 1/16, 1/16}
Bit 1 c m ha thnh 0, 2x c m ha thnh 10, 3x c m ha thnh 110, 4x c m ha
thnh 1110. Chn cu ng di y m ha cho 5x
a 1010b 1111c 101d 110
56/ Cho m Cyclic C(7, 4) c a thc sinh l 3( ) 1g x x x= + + v a thc sinh G sau :3
2 4
2 3 5
3 4 6
x x
x x x
x x x
x x x
+ +
+ + + + + +
1
G =
Ma trn no sau y l mt ma trn sinh G ng vi m Cyclic C(7,4) trn
a
1101000
0110100
0011010
0001101
b
1010100
0101100
0010110
0001011
c
1000100
01000000011001
0001011
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d
1010100
0101010
0010101
0001101
57/ Cho m Cyclic C(7, 4) c a thc sinh l 2 3( ) 1g x x x= + + v a thc sinh G sau2 3
3 4
2 4 5
3 5 6
x x
x x x
x x x
x x x
+ +
+ + + + + +
1
G =
Ma trn no sau y l mt ma trn sinh G ng vi m Cyclic C(7,4) trn
a
1010100
0101010
0010101
0001101
b
1011000
0101100
0010110
0001011
c
1010100
0101100
00101100001011
d
1000100
0100000
0011001
0001011
58/ Cho m Cyclic C(7, 4) c a thc sinh l : 3( ) 1g x x x= + + v ma trn sinh G. T ma trn sinh Gtnh c ma trn kim tra H bn:
2 3 4
3 4 5
2 4 5 6
1 x x xH x x x x
x x x x
+ + +
= + + + + + +
Chuyn ma trn kim tra H sang dng khng gian tuyn tnh. Ma trn no sau y l mt ma trn kimtra H ( dng khng gian tuyn tnh)
a
1011100
0101110
0010111
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Tnh di m trung bnh theo biu thc( )
5
i ii 1
n n p a=
=
ta c gi tr no :
a 4;b 3;c 2,1;d 1,2;
67/ Gi s sau khi thc hin m ha, cc tini
a vi xc sut tng ng ( )i
P a c m ha thnh cc
m nh phn c di mi
n tng ng nh bng sau:
Tnh di m trung bnh theo biu thc ( )5
i ii 1
n n p a=
= ta c gi tr no:a 1,8;b 4;c 2,4;d 3;
68/ Gi s sau khi thc hin m ha, cc tini
a vi xc sut tng ng ( )i
P a c m ha thnh cc
m nh phn c di mi
n
tng ng nh bng sau :
Tnh di m trung bnh theo biu thc ( )5
i ii 1
n n p a=
= ta c gi tr no :a 1,8;b 2,1;c 4;d 2,75;
69/ Gi s sau khi thc hin m ha, cc tini
a vi xc sut tng ng ( )i
P a c m ha thnh cc
m nh phn c di mi
n
tng ng nh bng sau :
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Tnh di m trung bnh theo biu thc
( )
5
i ii 1
n n p a=
=
ta c gi tr no :
a 2,1 du m;b 1,8 du m ;c 2,75 du m;d 2,45 du m;
70/ Gi s sau khi thc hin m ha, cc tini
a vi xc sut tng ng ( )i
P a c m ha thnh cc
m nh phn c di mi
n tng ng nh bng sau :
Tnh di m trung bnh theo biu thc ( )5
i ii 1
n n p a=
= ta c gi tr no :a 2,5;b 2,45;c 1,8;d 2,75;
71/ Mt dy tin { }1 2, ,..., nX x x x= vi ; 1ix X i n = Lng tin I(X) cha trong dy tin X s l:
1 2
1 1 1( ) log log .... log
( ) ( ) ( )n
I XP x P x P x
= + + + Gi s cho ngun { }1 2 3 4 5, , , ,X x x x x x= vi cc xc
sut ln lt l {1/2, 1/4, 1/8, 1/16, 1/16}. Lng tin I(X) cha trong dy tin X={ }1 2 1 1 3 4 1 1 5x x x x x x x x x l:
a 17 bt ;b 15 bt;c 18 bt;
d 16 bt;
72/ Khi xy dng mt m tuyn tnh ( )0n,k,d ngi ta phi tm c cc m c tha nhnhng li c kh nng khng ch sai ln Ngi ta thng xy dng m ny da trn cc bi ton tiu Tm cu sai trong cc cu di y:
a Vi k v 0d xc nh, ta phi tm c m c di vi t m l ln nht. Tng ng vi biton ny ta c gii hn n = k
b Vi k v 0d xc nh, ta phi tm c m c di vi t m l nh nht. Tng ng vi biton ny ta c gii hn Griesmer sau:
k 1
0ii 0
dn 2
=
c Vi n v s sai khi sa t xc nh, ta phi tm c m c s du thng tin k l ln nht (hay sdu tha l nh nht) Tng ng
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d 1 2 3 4 51111 1110 110 10 0
a a a a a
76/ Thc hin m ha Shannon - Fano ngun ri rc A sau :
1 2 3 4 5
( ) 1/ 4 1/2 1/ 8 1/16 1/16i
i
a a a a a aA
p a
= =
sc kt qu no sau y:
a 1 2 3 4 510 0 110 1110 1111
a a a a a
b 1 2 3 4 50 10 111 1110 1111
a a a a a
c 1 2 3 4 50 01 11 111 1111
a a a a a
;
d 1 2 3 4 50 10 110 1110 1111
a a a a a
;
77/ Thc hin m ha Shannon - Fano ngun ri rc A sau :
1 2 3 4 5
( ) 1/ 8 1/4 1/ 2 1/16 1/16i
i
a a a a a aA
p a
= =
sc kt qu no sau y:
a 1 2 3 4 50 10 110 1110 1111
a a a a a
;
b 1 2 3 4 5110 10 0 1110 1111
a a a a a
c1 2 3 4 5
0 11 110 1110 1111
a a a a a ;
d 1 2 3 4 50 01 11 111 1111
a a a a a
;
78/ Thc hin m ha Shannon - Fano ngun ri rc A sau :
1 2 3 4 5
( ) 1/16 1/4 1/ 8 1/ 2 1/16i
i
a a a a a aA
p a
= =
sc kt qu no sau y:
a 1 2 3 4 5
0 10 110 1110 1111
a a a a a
;
b 1 2 3 4 51110 10 110 0 1111
a a a a a
c 1 2 3 4 50 01 11 111 1111
a a a a a
;
d 1 2 3 4 50 11 110 1110 1111
a a a a a
;
79/ Thc hin m ha Shannon - Fano ngun ri rc A sau :
1 2 3 4 5
( ) 1/ 2 1/4 1/16 1/ 8 1/16i
i
a a a a a aA
p a
= = sc kt qu no sau y:
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a 1 2 3 4 50 10 1110 110 1111
a a a a a
;
b 1 2 3 4 50 01 11 111 1111
a a a a a
;
c 1 2 3 4 5
0 11 110 1110 1111
a a a a a
;
d 1 2 3 4 50 10 111 1110 1111
a a a a a
80/ Thc hin m ha Shannon - Fano ngun ri rc A sau :
1 2 3 4 5
( ) 1/ 2 1/8 1/ 4 1/16 1/16i
i
a a a a a aA
p a
= =
sc kt qu no sau y:
a 1 2 3 4 50 10 110 1110 1111
a a a a a
;
b 1 2 3 4 50 11 01 111 1111a a a a a
;
c 1 2 3 4 50 10 111 1110 1111
a a a a a
d 1 2 3 4 50 11 110 1110 1111
a a a a a
81/ Thc hin m ha Shannon - Fano ngun ri rc A sau :
1 2 3 4 5
( ) 1/ 2 1/16 1/ 8 1/ 4 1/16i
i
a a a a a a
A p a
= = sc kt qu no sau y:
a 1 2 3 4 50 01 11 111 1111
a a a a a
;
b 1 2 3 4 50 10 111 1110 1111
a a a a a
c 1 2 3 4 50 1110 110 10 1111
a a a a a
;
d
1 2 3 4 5
0 10 110 1110 1111
a a a a a
;
82/ Thc hin m ha Shannon - Fano ngun ri rc A sau :
1 2 3 4 5
( ) 1/ 2 1/4 1/16 1/16 1/ 8i
i
a a a a a aA
p a
= =
sc kt qu no sau y:
a 1 2 3 4 50 10 1111 1110 110
a a a a a
b 1 2 3 4 50 10 110 1110 1111
a a a a a
;
c 1 2 3 4 50 01 11 111 1111
a a a a a
;
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d 1 2 3 4 50 11 110 1110 1111
a a a a a
;
83/ Thc hin m ha Shannon - Fano ngun ri rc A sau :
1 2 3 4 5
( ) 1/16 1/4 1/ 8 1/16 1/ 2i
i
a a a a a aA
p a
= =
sc kt qu no sau y:
a 1 2 3 4 50 10 110 1110 1111
a a a a a
;
b 1 2 3 4 50 11 110 1110 1111
a a a a a
;
c 1 2 3 4 50 01 11 111 1111
a a a a a
;
d 1 2 3 4 51111 10 110 1110 0
a a a a a
84/ Gi s sau khi thc hin m ha Shannon - Fano ngun ri rc A :
1 4 5 2 3
( ) 1/ 2 1/4 1/ 8 1/16 1/16i
i
a a a a a aA
p a
= =
Ta c kt qu m ho sau : 1 4 5 2 30 10 110 1110 1111
a a a a a
di t m trung bnh n v entropy H(A)
c tnh theo biu thc
( )5
i ii 1
n p a n=
= v ( ) ( ) ( )5
i ii 1
H A p a logp a=
= s c kt qu no sau y :
a n =H(A)=2,875;
b n =H(A)=1,875;
c n =1,875 v H(A)=1,95;
d n =1,95 v H(A)=1,875
85/ Thc hin m ha Shannon - Fano ngun ri rc A sau :
1 2 3 4 5
( ) 0.5 0.25 0.125 0.0625 0.0625i
i
a a a a a aA
p a
= =
sc kt qu no sau y:
a 1 2 3 4 50 01 11 111 1111a a a a a
;
b 1 2 3 4 50 10 110 1110 1111
a a a a a
c 1 2 3 4 50 10 111 1110 1111
a a a a a
d 1 2 3 4 50 11 110 1110 1111
a a a a a
;
86/ Thc hin m ha Shannon - Fano ngun ri rc A sau :
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1 2 3 4 5
( ) 0.25 0.5 0.125 0.0625 0.0625i
i
a a a a a aA
p a
= =
sc kt qu no sau y:
a 1 2 3 4 50 11 110 1110 1111
a a a a a
;
b 1 2 3 4 5
10 0 110 1110 1111
a a a a a
;
c 1 2 3 4 50 10 110 1110 1111
a a a a a
d 1 2 3 4 50 10 111 1110 1111
a a a a a
87/ Thc hin m ha Shannon - Fano ngun ri rc A sau :
1 2 3 4 5
( ) 0.125 0.25 0.5 0.0625 0.0625i
i
a a a a a aA
p a
= =
sc kt qu no sau y:
a 1 2 3 4 50 10 110 1110 1111
a a a a a
b 1 2 3 4 5110 10 0 1110 1111
a a a a a
c 1 2 3 4 50 01 11 111 1111
a a a a a
;
d 1 2 3 4 50 10 111 1110 1111
a a a a a
88/ Thc hin m ha Shannon - Fano ngun ri rc A sau :
1 2 3 4 5
( ) 0.0625 0.25 0.125 0.5 0.0625i
i
a a a a a aA
p a
= =
sc kt qu no sau y:
a 1 2 3 4 51110 10 110 0 1111
a a a a a
b 1 2 3 4 50 01 11 111 1111
a a a a a
;
c1 2 3 4 5
0 11 110 1110 1111
a a a a a
;
d 1 2 3 4 50 10 110 1110 1111
a a a a a
89/ Thc hin m ha Shannon - Fano ngun ri rc A sau :
1 2 3 4 5
( ) 0.0625 0.25 0.125 0.0625 0i
i
a a a a a aA
p a
= =
sc kt qu no sau y:
a 1 2 3 4 5
0 11 110 1110 1111
a a a a a
;
b 1 2 3 4 50 10 110 1110 1111
a a a a a
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c 1 2 3 4 51111 10 110 1110 0
a a a a a
d 1 2 3 4 50 01 11 111 1111
a a a a a
;
90/ Thc hin m ha Shannon - Fano ngun ri rc A sau :
1 2 3 4 5
( ) 0.5 0.125 0.25 0.0625 0.0625i
i
a a a a a aA
p a
= =
sc kt qu no sau y:
a 1 2 3 4 50 01 11 111 1111
a a a a a
;
b 1 2 3 4 50 10 111 1110 1111
a a a a a
c 1 2 3 4 50 11 110 1110 1111
a a a a a
;
d 1 2 3 4 50 110 10 1110 1111a a a a a
91/ Thc hin m ha Shannon - Fano ngun ri rc A sau :
1 2 3 4 5
( ) 0.5 0.0625 0.125 0.25 0.0625i
i
a a a a a aA
p a
= =
sc kt qu no sau y:
a 1 2 3 4 50 11 110 1110 1111
a a a a a
;
b1 2 3 4 5
1 00 010 0111 0101
a a a a a
c 1 2 3 4 50 10 111 1110 1111
a a a a a
d 1 2 3 4 50 1110 110 10 1111
a a a a a
92/ Thc hin m ha Shannon - Fano ngun ri rc A sau :
1 2 3 4 5
( ) 0.5 0.0625 0.125 0.0625 0.25
i
i
a a a a a aA
p a
= =
sc kt qu no sau y:
a 1 2 3 4 50 10 110 1110 1111
a a a a a
b 1 2 3 4 50 10 111 1110 1111
a a a a a
c 1 2 3 4 50 01 11 111 1111
a a a a a
;
d 1 2 3 4 50 1111 110 1110 10
a a a a a
;
93/ Thc hin m ha Shannon - Fano ngun ri rc A sau :
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1 2 3 4 5
( ) 0.5 0.25 0.0625 0.125 0.0625i
i
a a a a a aA
p a
= =
sc kt qu no sau y:
a 1 2 3 4 50 01 11 111 1111
a a a a a
;
b 1 2 3 4 5
0 10 110 1110 1111
a a a a a
c 1 2 3 4 50 10 1110 110 1111
a a a a a
d 1 2 3 4 50 11 110 1110 1111
a a a a a
;
94/ Gi s sau khi thc hin m ha Shannon - Fano ngun ri rc A :
1 4 5 2 3
( ) 0.5 0.25 0.125 0.0625 0.0625i
i
a a a a a aA
p a
= =
Ta c kt qu m ho sau : 1 4 5 2 30 10 110 1111 1110
a a a a a
di t m trung bnh n v entropy H(A)
c tnh theo biu thc ( )5
i ii 1
n p a n=
= v ( ) ( ) ( )5
i ii 1
H A p a log p a=
= s c kt qu no sauy :
a n =H(A)=1,875;
b n =H(A)=2,875;
c n =1,95 v H(A)=1,875;
d n =1,875 v H(A)=1,95;
95/ Gi s sau khi thc hin m ha Shannon - Fano ngun ri rc A :
1 2 3 4 5 6 7 8
( ) 0.25 0.125 0.0625 0.0625 0.25 0.125 0.0625 0.0625i
i
a a a a a a a a aA
p a
= =
Ta c kt qu m ho sau : 1 5 2 6 3 4 7 800 01 100 101 1100 1101 1110 1111
a a a a a a a a
di t m trung bnh n v entropy H(A) c tnh theo biu thc ( )8
i i
i 1
n p a n
=
= v
( ) ( ) ( )8
i ii 1
H A p a log p a=
= s c kt qu no sau y :
a n =1,875 v H(A)=1,95;
b n =1,95 v H(A)=1,875 ;
c n =H(A)=1,875;
d n =H(A)=2,75;
96/ Gi s sau khi thc hin m ha Shannon - Fano ngun ri rc A :
1 2 3 4 5 6 7 8
( ) 1/ 4 1/ 8 1/16 1/16 1/ 4 1/ 8 1/16 1/16i
i
a a a a a a a a aA
p a
= =
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Ta c kt qu m ho sau : 1 5 2 6 3 4 7 800 01 100 101 1100 1101 1110 1111
a a a a a a a a
di t m trung bnh n v entropy H(A) c tnh theo biu thc ( )8
i ii 1
n p a n=
= v
( ) ( ) ( )8
i ii 1
H A p a log p a=
=
s c kt qu no sau y :
a n =1,95 v H(A)=1,875 ;
b n =H(A)=2,75;
c n =H(A)=1,875;
d n =1,875 v H(A)=1,95;
97/ Gi s sau khi thc hin m ha ngun ri rc A :
1 4 5 2 3
( ) 0.5 0.25 0.125 0.0625 0.0625i
i
a a a a a aA
p a
= =
Ta c kt qu m ho sau : 1 4 5 2 3
0 10 110 1111 11100
a a a a a
di t m trung bnh n v entropy H(A)
c tnh theo biu thc: ( )5
i ii 1
n p a n=
= v ( ) ( ) ( )5
i ii 1
H A p a logp a=
= s c kt qu no sauy :
a n =H(A)=2,875;
b n =H(A)=1,875;
cn
=1,875 v H(A)=1,95;d n =1,9375 v H(A)=1,875;
98/ Gi s sau khi thc hin m ha ngun ri rc A :
1 4 5 2 3
( ) 1/ 2 1/4 1/ 8 1/16 1/16i
i
a a a a a aA
p a
= =
Ta c kt qu m ho sau : 1 4 5 2 30 10 110 1111 11100
a a a a a
di t m trung bnh n v entropy H(A)
c tnh theo biu thc
( )
5
i ii 1
n p a n=
=
v
( ) ( ) ( )
5
i ii 1
H A p a log p a=
=
s c kt qu no sau
y :
a n =1,875 v H(A)=1,95;
b n =H(A)=1,875;
c n =1,9375 v H(A)=1,875;
d n =H(A)=2,875;
99/ Gi s sau khi thc hin m ha ngun ri rc A :
1 4 5 2 3( ) 1/ 2 1/4 1/ 8 1/16 1/16
i
i
a a a a a aA
p a
= =
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Ta c kt qu m ho sau : 1 4 5 2 300 10 110 1111 1110
a a a a a
di t m trung bnh n v entropy H(A)
c tnh theo biu thc ( )5
i ii 1
n p a n=
= v ( ) ( ) ( )5
i ii 1
H A p a log p a=
= s c kt qu no sauy :
a n =2,5 v H(A)=1,85;b n =1,9375 v H(A)=1,87;
c n =2,375 v H(A)=1,875 ;
d n =2 v H(A)=1,9;
103/ Gi s sau khi thc hin m ha ngun ri rc A :
1 4 5 2 3
( ) 0.5 0.25 0.125 0.0625 0.0625i
i
a a a a a aA
p a
= =
Ta c kt qu m ho sau :1 4 5 2 3
00 10 1100 1111 1110
a a a a a
di t m trung bnhn
v entropy H(A)
c tnh theo biu thc ( )5
i ii 1
n p a n=
= v ( ) ( ) ( )5
i ii 1
H A p a log p a=
= s c kt qu no sauy :
a n =2,5 v H(A)=1,875;
b n =2,375 v H(A)=1,975
c n =2,55 v H(A)=1,775;
d n =2 v H(A)=1,675;
104/ Cho m Cyclic C(7,4) c a thc sinh l 3( ) 1g x x x= + + tng ng a thc thng tin
( ) 3a x = x + x S dng thut ton 4 bc thit lp t m h thng ta c kt qu no di y :a 1100101b 0101010;c 1000110;d 0110111;
105/ Cho m Cyclic C(7,4) c a thc sinh l 3( ) 1g x x x= + + tng ng a thc thng tin
( )2
a x = 1+ x S dng thut ton 4 bc thit lp t m h thng ta c kt qu no di y:a 0101010;b 1000110;c 0011010;d 0110111;
106/ Cho m Cyclic C(7,4) c a thc sinh l 3( ) 1g x x x= + + tng ng a thc thng tin
( ) 2a x = x + x S dng thut ton 4 bc thit lp t m h thng, sc kt qu no di y :a 0101010;b 1110010;c 1000110;d 0110111;
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107/ Cho m Cyclic C(7,4) c a thc sinh l 3( ) 1g x x x= + + tng ng a thc thng tin
( ) 3a x = 1+ x S dng thut ton 4 bc thit lp t m h thng, sc kt qu no di y :a 0101010;b 1000110;c 1110010;d 0111001;
108/ Cho m Cyclic C(7,4) c a thc sinh l 3( ) 1g x x x= + + tng ng a thc thng tin
( ) 2 3a x = x + x S dng thut ton 4 bc thit lp t m h thng, sc kt qu no diy:
a 0111001;b 1000110;c 1110010;d 0100011;
109/ Cho m Cyclic C(7,4) c a thc sinh l 3( ) 1g x x x= + + tng ng a thc thng tin
( ) 2 3a x = 1+ x + x S dng thut ton 4 bc thit lp t m h thng, sc kt qu no diy :
a 0100011;b 0111001;c 1110010;d 1001011;
110/ Cho m Cyclic C(7,4) c a thc sinh l 3( ) 1g x x x= + + tng ng a thc thng tin
( )a x = 1+ x S dng thut ton 4 bc thit lp t m h thng, sc kt qu no di y :
a 1001011;b 0100011;c 1011100;d 1110010;
111/ Cho m Cyclic C(7,4) c a thc sinh l 3( ) 1g x x x= + + tng ng a thc thng tin
( ) 2a x = 1+ x + x S dng thut ton 4 bc thit lp t m h thng, sc kt qu no diy :
a 1001011;b 0101110;
c 1011100;d 1110010;
112/ Cho m Cyclic C(7,4) c a thc sinh l 3( ) 1g x x x= + + tng ng a thc thng tin
( ) 3x+2a x = x + x S dng thut ton 4 bc thit lp t m h thng, sc kt qu no diy :
a 1011100;b 0101110;c 1001011;d 0010111;
113/ Cho m Cyclic C(7,4) c a thc sinh l 3( ) 1g x x x= + + tng ng a thc thng tin
( ) 3xa x = S dng thut ton 4 bc thit lp t m h thng, sc kt qu no di y :
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a 0101110;b 1001011;c 1010001d 1011100;
114/ Cho m Cyclic C(7,4) c a thc sinh l 3( ) 1g x x x= + + tng ng a thc thng tin
( )a x = x
S dng thut ton 4 bc thit lp t m h thng, sc kt qu no di y :a 0010111;b 0110100;c 0101110;d 1011100;
115/ Cho m Cyclic C(7,4) c a thc sinh l 3( ) 1g x x x= + + tng ng a thc thng tin
( ) 2a x = x S dng thut ton 4 bc thit lp t m h thng, sc kt qu no di y :a 1110010;b 0101110;
c 0110100;d 0010111;
116/ Cho m Cyclic C(7,4) c a thc sinh l 2 3( ) 1g x x x= + + tng ng a thc thng tin
( ) 3a x = x + x S dng thut ton 4 bc thit lp t m h thng ta c kt qu no di y:a 0101110;b 0110111;c 1000101;d 0101010;
117/ Cho m Cyclic C(7,4) c a thc sinh l 2 3( ) 1g x x x= + + tng ng a thc thng tin
( ) 2a x = 1+ x S dng thut ton 4 bc thit lp t m h thng ta c kt qu no di y:a 0011010;b 1000110;c 0101010;d 0111010;
118/ Cho m Cyclic C(7,4) c a thc sinh l 2 3( ) 1g x x x= + + tng ng a thc thng tin
( ) 2a x = x + x S dng thut ton 4 bc thit lp t m h thng, sc kt qu no di y :
a 1110010;b 0110111;c 1000110;d 0010110;
119/ Cho m Cyclic C(7,4) c a thc sinh l 2 3( ) 1g x x x= + + tng ng a thc thng tin
( ) 3a x = 1+ x S dng thut ton 4 bc thit lp t m h thng, sc kt qu no di y :a 1101001;b 0101010;c 0111001;
d 1000110;
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120/ Cho m Cyclic C(7,4) c a thc sinh l 2 3( ) 1g x x x= + + tng ng a thc thng tin
( ) 2 3a x = x + x S dng thut ton 4 bc thit lp t m h thng, sc kt qu no diy:
a 1010011;b 0111001;c 1000110;
d 0100011;
121/ Cho m Cyclic C(7,4) c a thc sinh l 2 3( ) 1g x x x= + + tng ng a thc thng tin
( ) 2 3a x = 1+ x + x S dng thut ton 4 bc thit lp t m h thng, sc kt qu no diy :
a 0001011;b 0100011;c 0111001;d 1110010;
122/ Cho m Cyclic C(7,4) c a thc sinh l 2 3( ) 1g x x x= + + tng ng a thc thng tin( )a x = 1+ x S dng thut ton 4 bc thit lp t m h thng, sc kt qu no di y :a 1110010;b 1001011;c 0101100;d 0100011;
123/ Cho m Cyclic C(7,4) c a thc sinh l 2 3( ) 1g x x x= + + tng ng a thc thng tin
( ) 2a x = 1+ x + x S dng thut ton 4 bc thit lp t m h thng, sc kt qu no di
y :a 1001011;b 1001110;c 1110010;d 1011100;
124/ Cho m Cyclic C(7,4) c a thc sinh l 2 3( ) 1g x x x= + + tng ng a thc thng tin
( ) 3x+2a x = x + x S dng thut ton 4 bc thit lp t m h thng, sc kt qu no diy :
a 1001011;
b 0100111;c 1011100;d 0101110;
125/ Cho m Cyclic C(7,4) c a thc sinh l 2 3( ) 1g x x x= + + tng ng a thc thng tin
( ) 3xa x = S dng thut ton 4 bc thit lp t m h thng, sc kt qu no di y :a 1001011;b 0111011c 1011100;d 0101110;
126/ Cho m Cyclic C(7,4) c a thc sinh l 2 3( ) 1g x x x= + + tng ng a thc thng tin
( )a x = x S dng thut ton 4 bc thit lp t m h thng, sc kt qu no di y :
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a 1110100;b 1011100;c 0010111;d 0101110;
127/ Cho m Cyclic C(7,4) c a thc sinh l 2 3( ) 1g x x x= + + tng ng a thc thng tin
( )2
a x = x S dng thut ton 4 bc thit lp t m h thng, sc kt qu no di y :a 0101110;
b 0110100;c 1110110;d 0010111;
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CHNG 5 : L THUYT THU TI U
1/ Thu tn hiu khi c nhiu l mt bi ton thng k:a ngb Sai
2/ Nhim v ca my thu l phi chn li gii, do my thu cn c gi l s gii Yu cu cas gii l phi tm li gii ng (pht i ta phi tm c i ) Trong thc t c rt nhiu sgii Trong tt c cc s gii c th c th ti mt s bo m xc sut nhn ln phi ng l lnnht (xc sut gii sai l b nht) S ny c gi l s gii ti u. Khi ngi ta gi Mythuxy dng theo s gii ti u c gi l my thu ti u Kt lun ny ng hay sai ?
a Saib ng
3/ ( )( )
( )
i/i
pu
p i l
l
l
>
i = 1,mVi l quy tc gii ti u vit di dng hm hp l. Chn cu
ng trong cc cu sau:
a Nu mi tn hiu gi i u ng xc sut th ( )/i u 0 il l = Vi b Nu mi tn hiu gi i u ng xc sut th ( )/i u 1 il l > Vi c Nu mi tn hiu gi i u ng xc sut th ( )/i u 1 il l < Vi d Nu mi tn hiu gi i u ng xc sut th ( )/i u 1 il l = Vi
4/ nh ngha b lc phi hp tuyn tnh thng : i vi mt tn hiu xc nh, mt mch tuyn
tnh thngm bo t s rara
N
S = cc i mt thi im quan st no y sc gi l
mch lc phi hp tuyn tnh thng ca tn hiu Trong ra l t sgia cng sut trungbnh ca nhiu u ra b lc y v cng sutnh ca tn hiunh ngha ny ng hay sai ?
a Saib ng
5/ Qu trnh x l tn hiu trong my thu ti u c gi l x l ti u tn hiu X l nhn li giic xc sut sai b nht Chn cu ng nht v x l ti u cc tn hiu:
a Da vo cc tiu chun ti u, bng cng c thng k ton hc ngi ta xc nh c quy tc
gii ti u, do ngy nay l thuyt truyn tin cho php bng ton hc tng hp c mythu ti u.b Vic tng hp cc my thu (xy dng s gii) ch cn c vo cc tiu chun cht lng
mang tnh cht chc nng m khng mang tnh cht thng k.c nh hng ca nhiu ln cht lng ca my thu ti u chc tnh theo t s tn/tpd Vic tng hp my thu ti u l da vo trc gic, kinh nghim, th nghim
6/ Gi s i l tn hiu gi i, c xc sut ( )ip - c gi l xc sut tin nghim my thu tanhn c ( )u t , t ( )u t qua s gii ta s c li gii l no Nh vy l c gi i vimt xc sut ( )p / ul - c gi l xc sut hu nghim. Do xc sut gii sai s l:
( ) ( )p sai / u, 1 p / ul l = Xt hai s gii:- T ( )u t cho ta 1 - gi l s (1)
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- T ( )u t cho ta 2 - gi l s (2)Nu ( ) ( )1 2p sai / u, p sai / u, < ta rt ra kt lun s (2) ti u hn (1) Khng nh ny nghay sai ?
a Saib ng
7/ Tn hiu tng qut c dng: ( ) ( ) ( )( )i 0i 0C t C t cos t t= + + Chn cu sai trong cc cu sau:
a Vic x l ti u tn hiu khng ph thuc ng bao ( )0iC t v tn s tc thi
( )( )
id t
tdt
= +
b Khi x l ti u tn hiu ta cn bit ng bao ( )0iC t v tn s tc thi ( )( )
id t
tdt
= +
c Nu vic thu ( )iC t cn bit 0 (iu chnh h thng thu) th c gi l thu kt hp Nu
vic thu ( )iC t khng cn bit 0 (iu chnh h thng thu) th c gi l thu khng kthp
d Thc t khi thay i s lm cho 0 thay i ch bin thin t nhng cng lm cho 0 thay i rt mnh Khi ta phi chuyn sang thu khng kt hp.
8/ Cho u vo mch tuyn tnh thng mt dao ng c dng: ( ) ( ) ( )iy t C t n t= + Trong
( )iC t l th hin ca tn hiu pht i (cn c gi l tn hiu ti) ( )n t l nhiu cng, trng, chun
Bi ton tng hp mch l tm biu thc gii tch ca hm truyn phc ( )iK ca mch tuyn tnh
thng sao cho mt thi im quan st (dao ng nhn c) no ra t max, p dng
cng thc bin i ngc Fourier tnh c ( )2
ra iv0
1S 2 d
Nf f
Trong ( )ivS l mt
ph (bin) phc ca th hin tn hiu u vo mch tuyn tnh
Theo nh l Parseval, ta c: iramax0
E
N = (*)
trong ( )2
i ivE S 2 df f
= l nng lng ca tn hiu ti
Chn cu sai trong nhng nhn xt sau :
a T (*) chng t t s rara
S
N =
ch ph thuc vo nng lng ca tn hiu v bi ton pht
hin dng ca tn hiu l khng quan trngb T (*) chng t : T s gia cng sut nh ca tn hiu v cng sut trung bnh ca nhiu
u ra b lc y ch ph thuc vo nng lng
c T (*) chng t t s rara
S
N =
ch ph thuc vo nng lng ca tn hiu m hon ton
khng ph thuc vo dng ca n
d T (*) chng t t s rara
SN =
hon ton ph thuc vo dng ca n
ca tn hiu m hon ton khng ph thuc vo dng ca n.
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9/ Cho knh nh phn, i xng, khng nhc nhiu cng, trng, chun theo m hnh sau:
1
2 2
1 ( )1 1p /
( )2 2p /
( )1 2p / ( )2 1
p b /
( )11
p2
=
( )21
p2
=
Xc sut sai ton phn sp ( xc sut sai khng iu kin) l
( ) ( ) ( ) ( )= + s 1 2 1 2 1 2p p .p / p .p /
Gi s tnh c ( )
= =
0
P T
4N 2
2 1
0
P T1p / exp d
2 2G2
khi sp s bng biu thc no di y:
a
= +
s0
P Tp 1
2G
b
= +
s0
P T1p
2 2G
c
= +
s0
P T1 1p
2 2 2G
d
=
s0
P Tp 1
2G
10/ Ti u vo b lc tuyn tnh tc ng tn hiu: x(t) = s(t) + n(t)Trong n(t) l tp m trng, chun, dng Cn s(t) l xung th tn c lp vi n(t) v c dng:
( )( ) =
>
A t TA.e t T
s t0 t T
Hm truyn = * j T0K (j ) kS (j )e
Hm truyn ca b lc ( = * j T
0K (j ) kS (j )e ) sao cho t s tn trn tp u ra ca b lc t cci s l biu thc no di y
a = + 0kA
K (j )A j
b =+ 0kA
K (j )A j
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