nhĐt ly thuyet thong tin
TRANSCRIPT
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Chn cu ng sau : d Tin l dng vt cht biu din hoc thhin thng tinChn pht biu ng trong nhng cu sau : a Thng tin l nhng tnhcht xc nh ca vt cht m con ngi nhn c t th gii vtcht bn ngoi
Mn l thuyt thng tin bao gm vic nghin cu: b Cc qu trnhtruyn tin v L thuyt m haChn cu ng nht v ngun tin a Ngun tin l ni sn ra tinChn cu ng nht v ng truyn tin a L mi trng Vt l,trong tn hiu truyn i t my pht sang my thu bin i mt tn hiu lin tc theo bin v theo thi gian thnh tn hius, chng ta cnthc hin qu trnh no sau y: a Ri rc ha theo trc thi gianv lng t ha theo trc bin Chn cu ng v tn hiu: d Tn hiu l qu trnh ngu nhin .Chn pht biu ng nht v c trng thng k : a c trng cho
cc qu trnh ngu nhin chnh l cc quy lut thng k v cc ctrng thng kChn cu ng nht v hm t tng quan : c Hm t tngquan 1 2( , )x R t t c trng cho s ph thuc thng k gia hai gi tr hai thi im thuc cng mtVic biu din mt tn hiu gii hp thnh tng ca hai tn hiu iu binbin thin chm s lmcho vic phn tch mch v tuyn in di tc ng ca n tr nn phc tp
b SaiNgi ta gi tn hiu gii rng nu b rng ph ca n tho mn bt ng
thc sau: 10
. Cc tn hiu iu tn, iu xung, iu xung ct, manip tn
s, manip pha, l cc tn hiu gii rng. b ngChn cu ng v cng thc xc nh mt ph cng sut c
2
( )
( ) ( ) limT
xT
S
G M G M T
l cng thc xc nh mt ph cng
sut ca cc qutrnh ngu nhin.Chn cu ng v cng thc quan h gia mt ph cng sut v hm t
tng quan d ( ) ( )j
G R e d
Trong trng hp h thng tuyn tnh th ng c suy gim th nhng thiim t >> t0= 0 (thi im t tc ng vo), qu trnh ngu nhin u ras c coi l dng. Khi hm t tng quan v mt ph cng sut caqu trnh ngu nhin u ra s lin h vi nhau theo biu thc sau
1( ) ( )
2
j
ra ra R G e d
b ng
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*( )aS t l hm lin hp phc ca ( ) : ( ) ( ) ( )a aS t S t x t j x 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 b ng
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 thc2
0
0
( ) ( ) cos2
ra
G R K d
ta c kt qu no ?
Hnh chng 2 cu 10 a 20
sin2( ) cos/ 2
ra raR
Cho qu trnh ngu nhin dng c biu thc sau: 0( ) cos(2 ) X t A f t Trong A = const, f0= const, l i lng ngu nhin c phn b u 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
Cho qu trnh ngu nhin dng c biu thc sau: 0( ) cos(2 ) X t A f t Trong A = const, f0= const, l i lng ngu nhin c phn b u trongkhong ( , ) . Tnh hm t tng 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: c
2
1 2 0
1( , ) cos 2
2 R t t A f
Tn hiu in bo ngu nhin X(t) nhn cc gi tr + a; - a vi xc sut nhnhau v bng 1/2. Cn xc sut trong khong c N bc nhy l:
( )( , )
!
N
P N eN
>0 (theo phn b Poisson). T cc gi thit trn tnh c
hm t tng quan 2 2( )x R 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( )
4x
aG
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
d 2 2 20 0
2 2( )
( ) ( )xG
Khi nim lng tin c nh ngha da trn: b bt nh ca tinChn pht biu ng nht v Entropy ca ngun tin, H(X): a Li lng c trng cho bt nh trung bnh ca ngun tin
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Chn pht biu sai v bt nh c bt nh ca mt phn t cgi tr 1 bit khi xc sut chn phn t l 1Entropy ca ngun ri rc nh phn H(A) =plog p(1p)log(1p).Khi p=1/2th H(A) t max Chn cu ng v H(A)max c H(A)max= 1 btTrong mt trn thi u bng Quc t, i tuyn Vit Nam thng i tuyn
Brazin, thng tin ny c bt nh l : b V cng lnHc sinh A c thnh tch 12 nm lin t danh hiu hc sinh gii, hc sinh Blc hc km Thi ttnghip ph thng trung hc, hc sinh A trt, hc sinh B th khoa Thngtin v hc sinh B th khoa, hc sinh A trt c bt nh l: b
V cng lnChn ngu nhin mt trong cc s t 0 n 7 c xc sut nh nhau Khi xc sut ca s cchn ngu nhin l: b 1/8Mt thit b v tuyn in gm 16 khi c tin cy nh nhau v c mcni tip Gi s c mt khi no b hng, khi xc sut ca mt khi
hng l: a 1/16B t l kh 52 qun (khng k fng teo), A rt ra mt qun bi bt k Xcsut v qun bi m A rt l: a Bng 1/52Chn cu sai trong cc cu sau : c Lng thng tin = bt nh tinnghim + bt nh hu nghimChn cu ng sau : a Lng thng tin = thng tin tin nghim -thng tin hu nghimChn cu sai trong cc cu sau : a Thng tin tin nghim (k hiuI(Xk)) c xc nh theo cng thc sau:I(Xk) = log P(Xk)I(Xk,Yl) l lng thng tin cho vXkdo Ylmang li c tnh bng cng thc
no sau y a
1 1
log log( ) ( / )k k l P x P x y
I(Xk/Yl)= - log ( / )k l P x y l thng tin hu nghim v kx ( / )k lp x y = 1 khi victruyn tin khngb nhiu. Chn cu sai trong nhng cu v I(Xk/Yl) di y: d I(Xk/Yl) =1/2 khi knh khng c nhiuI(Xk/Yl) l thng tin hu nghim v kx ( / )k l p x y = 1 khi vic truyn tin khng bnhiu Chn cusai trong nhng cu v I(Xk/Yl) di y: c I(Xk/Yl) >1/2 khi knhkhng c nhiu
I(Xk/Yl) l lng thng tin ring ca kx khi bit ly v I(Xk/Yl) = 0 khi khngc nhiu Cuny ng hay sai ? a ngChn cu sai trong nhng cu sau: b I(Xk,Yl) l lng tin ring ca kx
v ly
Cho tin ( )ix c xc sut l ( )iP x 0,5 lng tin ring I ( )ix ca tin ny bng cci lng no
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di y : b 1 btCho tin ( )ix c xc sut l ( )iP x 1/ 4 lng tin ring I ( )ix ca tin ny bng cci lng nodi y : a 2 btCho tin ( )ix c xc sut l ( )iP x 1/ 8 lng tin ring I ( )ix ca tin ny bng cc
i lng nodi y : b 3 btCho tin ( )ix c xc sut l ( )iP x 1/16 lng tin ring I ( )ix ca tin ny bng cci lng nodi y: d 4 btCho tin ( )ix c xc sut l ( )iP x 1/ 27 lng tin ring I ( )ix ca tin ny bngcc i lng nodi y: c Log27 bt;Cho tin ( )ix c xc sut l ( )iP x 1/ 9 lng tin ring I ( )ix ca tin ny bng cci lng no di y : c 2log3 bt
Cho tin ( )ix c xc sut l ( )iP x 1/ 25 lng tin ring I ( )ix ca tin ny bngcc i lng no di y : b 2log5 btTm cu sai trong nhng cu di y c bt nh ca tin vlng tin c ngha nh nhau nhng gi tr khc nhauLng thng tin ring ( bt nh) ca mt bin ngu nhin kx l ( )kI x . Chn
biu thc sai trong cc biu thc di y a ( ) ln ( )k k I x p x n vo l bitLng thng tin ring ( bt nh) ca mt bin ngu nhin kx l ( )kI x c
tnh bng biu thc no di y : a ( ) ln ( )k k I x k p x
Lng thng tin ring ( bt nh) ca mt bin ngu nhin kx l ( )kI x ctnh nh sau ( ) ln ( )k k I x k p x , trong k l h s t l. Tm cu sai v cchchn k trong cc cu di y : a Chn k = 1 ta c ( ) ln ( )k k I x p xEntropy ca ngun tin ri rc A l trung bnh thng k ca lng thng tin
ring ca cc tin thuc A. K hiu: 1( );H A 1( ) ( )iH A M I a
vi1
1( )
aA
p a
2
2( )
a
p a
....
.... ( )s
s
a
p a
0p(ai) 1 ;
1
( ) 1s
i
i
p a
; 1( )H A c tnh bng
biu thc no di y d 1 1( ) ( ) log ( )
s
i ii
H A p a p a (bt)
Entropy ca ngun tin ri rc A l trung bnh thng k ca lng thng tin
ring ca cc tin thuc A. K hiu: 1( );H A 1( ) ( )iH A M I a
vi1
1( )
aA
p a
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2
2( )
a
p a
....
.... ( )s
s
a
p a
0p(ai) 1 ;
1
( ) 1s
i
i
p a
; 1( )H A c tnh bng
biu thc no : c 11
1( ) ( ) log
( )
s
i
i i
H A p ap a
(bt)
Entropy ca ngun tin ri rc A l trung bnh thng k ca lng thng tin
ring ca cc tin thuc A. K hiu: 1( );H A 1( ) ( )iH A M I a
vi0
0( )
aA
p a
1
1( )
a
p a
....
....1
1( )
s
s
a
p a
0p(ai) 1 ;
1
1
( ) 1s
i
i
p a
; 1( )H A c tnh nh sau: d
1
1
0
( ) ( ) log ( )s
i i
i
H A p a p a
(bt)A v B l hai trng bin c bt k, Entropy ca 2 trng bin c ng thiC=AB l H(AB),
trong cc tnh cht ca H(AB) di y, tnh cht no sai: c1
1
( ) ( ) log ( )s
i i
i
H A p a p a
(bt)Entropy c iu kin v 1 trng tin A khi r trng tin B l H(A/B). Trongcc tnh cht ca H(A/B) di y, tnh cht no sai a H(A / B)H(B / A)Entropy c iu kin v 1 trng tin B khi r trng tin A l H(B/A),Tnhcht no ca H(B/A)di y l ng b 0H(B/A)Entropy ca trng bin c ng thi H(AB) c tnh bng cng thc nosau y d H(B) + H(A/B)Lng thng tin cho trung bnh (k hiu l I(A,B) c cc tnh cht no sauy b 0I(A,B)H(A)Lng thng tin cho trung bnh (k hiu l I(A,B) Trong cc tnh cht diy, tnh cht nosai: b I(A,B)=H(A) khi knh c nhiu;Lng thng tin cho trung bnh (k hiu l I(A,B) c cc tnh cht no sauy b 0I(A,B) v I(A,B)=0 khi knh b tLng thng tin cho trung bnh (k hiu l I(A,B) ), trong cc tnh cht diy ca I(A,B), tnh cht no sai d I(A,B)1 khi knh b t;Lng thng tin cho trung bnh (k hiu l I(A,B)) , tm biu thc sai trongcc biu thc di y c I(A,B) = H(A) - H(B/A)Mnh no sau y sai c I(A,B) = H(A) + H(B) + H(AB)Chn ngu nhin mt trong cc s t 0 n 7 c xc sut nh nhau btnh ca s c chn ngu nhin l c 3 btMt thit b v tuyn in gm 16 khi c tin cy nh nhau v c mcni tip Gi s c mt khi no b hng, bt nh ca khi hng l:
d 4 btB t l kh 52 qun (khng k fng teo), A rt ra mt qun bi bt k bt nh v qun bi
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m A rt l: b Ln hn 5 nh hn 6 btMt hp c 8 ng tin kim loi , trong c 02 ng tin 500 ng; 02 ngtin 1000 ng, 2ng tin 2000 v 2 ng tin 5000 ng Chn ngu nhin 1 trong 8 ngtin Khi xc sut ca ng tin c chn ngu nhin l: d
1/4 btMt hp c 8 ng tin kim loi , trong c 02 ng tin 500; 02 ng tin1000, 2 ng tin2000 v 2 ng tin 5000 Chn ngu nhin 1 trong 8 ng tin Khi bt nh ca ng tin c chn ngu nhin l: d 2 btCho ngun tin X = {x1, x2, x3} vi cc xc sut ln lt l {1/2, 1/4, 1/4},
Entropy ca ngun tin H(X) c tnh l c1 1 1
log 2 log 4 log 42 4 4
Cho ngun tin X = {x1, x2, x3, x4, x5, x6, x7, x8, x9} vi cc xc sut lnlt 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
Entropy 1( )H A ca ngun ri rc1
1( )
aA
p a
2
2( )
a
p a
....
.... ( )s
s
a
p a
0p(ai)
1 ;1
( ) 1s
i
i
p a
Trong cc tnh cht ca H(A) di y tnh cht no l saia Khi ( ) 1, ( ) 0k i p a p a ,vi Vi kth 1 1 min( ) ( ) 1H A H A
Cho ngun ri rc A1
1( )
aA
p a
2
2( )
a
p a
....
.... ( )s
s
a
p a
0p(ai) 1 ;
1
( ) 1s
i
i
p a
Gi entropy ca ngun A l 1( )H A , trong cc biu thc tnh 1 max( )H A
logs di y, biu thc no sai: a 1 max1
1( ) log ( ) log log
( )
s
i
i i
H A s p a sp a
Cho ngun ri rc A1
1( )
aA
p a
2
2( )
a
p a
....
.... ( )s
s
a
p a
0p(ai) 1 ;
1
( ) 1s
i
i
p a
Nu ngun A c s du ng xc sut , khi biu thc no di
y l sai a1 1
1 ( ) 0s s
i
i i
p as
Kh nng thng qua ca knh ri rc C l gi tr cc i ca lng thng tincho trung bnh truyn qua knh trong mt n v thi gian ly theo mi kh
nng c th c ca ngun tin A ' ' ' 'max ( , ) max ( , )( ); .k kA A
C I A B v I A B bps C v C
vi
max ( , ); kA
C I A B v biu th s du m knh truyn c (c truyn qua
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knh) trong mt n v thi gian I(A,B) l lng thng tin truyn qua knhtrong mt n v thi gian C c gi l kh nng thng qua ca knh i vimi du C c cc tnh cht no di y : a C 0, C = 0 khi Av B c lp ; C vklogs , = vklogs khi knh khng nhiuI(A,B) l lng thng tin trung bnh c truyn qua knh ri rc c tnh cht
: I(A,B) H(A)V mt s nh ngha : ' ' ' 'max ( , ) max ( , )( ); .k k
A AC I A B v I A B bps C v C
vi
max ( , ); kA
C I A B v biu th s du m knh truyn c trong mt n v thigian. T cc tnh cht v nh ngha trn cho bit cc biu thc di y,
biu thc no sai a( )
( , )H AK K
v I A B v
Cho ngun ri rc ch c hai du:1
1( )
aA
p a
2
2( )
a
p a
Ngun ri rc nh phn l
ngun A trn tho mn iu kin sau1
2
"0"
"1"
a
a
vi xc sut
1
2
( )
( ) 1
p a p
p a p
Khi
ngun ri rc nh phn A c th vit biu thc no d1
aA
p
2
1
a
p
1a
Ap
2
1
a
p
l ngun ri rc nh phn Tho mn iu kin
1
2
"0"
"1"
a
a
vi xc
sut1
2
( )
( ) 1
p a p
p a p
Khi Entropy 1( )H A c tnh bng cng thc no sau y
d p log p(1p)log(1p)1( / ) H A b l lng thng tin tn hao trung bnh ca mi tin u pht khi u
thu thu c jb 1 1 11
( / ) ( / ) log ( / ); ( / )s
i i i
i
H A b p a b p a b H B a
l lng thng tinring trung bnh cha trong mi tin u thu khi u pht pht i mt tin
ia c tnh theo cng thc1
( / ) ( / ) log ( / )t
i j i j i
j
H B a p b a p b a
Trong trng hpknh b t (b nhiu tuyt i) ta c biu thc no di y l sai : a
( / ) ( ) ( )jH A b H A H B 1( / ) H A b l lng thng tin tn hao trung bnh ca mi tin u pht khi u
thu thu c jb 1 1 11
( / ) ( / ) log ( / ); ( / )s
i i i
i
H A b p a b p a b H B a
l lng thng tinring trung bnh cha trong mi tin u thu khi u pht pht i mt tin
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c tnh theo cng thc:1
( / ) ( / ) log ( / )t
i j i j i
j
H B a p b a p b a
Trong trng hpknh khng nhiu biu thc no di y l ng : d
( / ) ( / ) 0kH A b H A B
1( / ) H A b l lng thng tin tn hao trung bnh ca mi tin u pht khi uthu thu c jb 1 1 1
1
( / ) ( / ) log ( / ); ( / )s
i i i
i
H A b p a b p a b H B a
l lng thng tinring trung bnh cha trong mi tin u thu khi u pht pht i mt tin
ia c tnh theo cng thc1
( / ) ( / ) log ( / )t
i j i j i
j
H B a p b a p b a
Trong trng hpb nhiu tuyt i, A v B l c lp nhau suy ra :
( / ) ( ); ( / ) ( ) ( ) ( ) ( )i j i j i j i j i jp a b p a p b a p b p a b p a p b khi ta c biu thc no
sau y l ng: a1
( / ) ( ) log ( )s
j i i
i
H A b p a p a
v ( / )i H B a
1
( ) log ( )s
j j
i
p b p b
Entropy c iu kin v 1 trng tin A khi r trng tin B l H(A/B), cxc nh theo
cng thc sau:1 1
( / ) ( ) log ( / )s t
i j i j
i j
H A B p a b p a b
Trong trng hp b nhiutuyt i, A v B l c lp nhau, suy ra :
( / ) ( ); ( / ) ( ) ( ) ( ) ( )i j i j i j i j i jp a b p a p b a p b p a b p a p b khi biu thc no sau y
l ng: d 1 1( / ) ( ) ( ) log ( )
t s
j i ij iH A B p b p a p a
Entropy c iu kin v 1 trng tin B khi r trng tin A l H(B/A), cxc nh theo
cng thc sau:1 1
( / ) ( ) log ( / )s t
j i j i
i j
H B A p b a p b a
Trong trng hp b nhiutuyt i, A v B l c lp nhau, suy ra :
( / ) ( ); ( / ) ( ) ( ) ( ) ( )i j i j i j i j i jp a b p a p b a p b p a b p a p b khi biu thc no sau y
l ng: a1 1
( / ) ( ) ( ) log ( )s t
i j j
i j
H B A p a p b p b
Entropy c iu kin v 1 trng tin A khi r trng tin B l H(A/B), cxc nh theo
cng thc sau:1 1
( / ) ( ) log ( / )s t
i j i j
i j
H A B p a b p a b
Trong trng hp b nhiutuyt i, A v B l c lp nhau, suy ra :
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( / ) ( ); ( / ) ( ) ( ) ( ) ( )i j i j i j i j i jp a b p a p b a p b p a b p a p b khi biu thc no sau y
l ng: a1 1
( / ) ( ) ( ) log ( ) ( )t s
j i i
j i
H A B p b p a p a H A
Entropy c iu kin v 1 trng tin B khi r trng tin A l H(B/A), c
xc nh theocng thc sau:
1 1
( / ) ( ) log ( / )s t
j i j i
i j
H B A p b a p b a
Trong trng hp b nhiutuyt i, A v B l c lp nhau, suy ra :
( / ) ( ); ( / ) ( ) ( ) ( ) ( )i j i j i j i j i jp a b p a p b a p b p a b p a p b khi biu thc no sau y
l ng: c1 1
( / ) ( ) ( ) log ( ) ( )s t
i j j
i j
H B A p a p b p b H B
Entropy c iu kin v 1 trng tin A khi r trng tin B l H(A B), cxc nh theo
cng thc sau: 1 1( / ) ( ) log ( / )
s t
i j i ji j
H A B p a b p a b vi
( )
( / ) ( )
i j
i j
j
p ab
a b p b T cngthc ny c th khai trin ( / ) H A B thnh cng thc no sau y: c
1 1
( ) ( / ) log ( / )t s
j i j i j
j i
p b p a b p a b
Lng thng tin cho trung bnh (k hiu l I(A,B)) ( , ) ( , )i jI A B M I a b
vi
( / )( , ) log
( )
i j
i j
i
p a b I a b
p a Xc sut c thng tin ( , )i j I a b l ( )i j p ab Do c th
vit I(A,B)) bng cng thc no sau y: b 1 1
( / )
( ) log ( )
s ti j
i ji j i
p a b
p ab p a Lng thng tin cho trung bnh (k hiu l I(A,B)) c vit thnh :
1 1
( / )( , ) ( ) log
( )
s ti j
i j
i j i
p a b I A B p a b
p a Khi c th khai trin I(A,B) thnh
1 1
( , ) ( ) log ( / ) ( )s t
i j i j i
i j
I A B p a b p a b p a
b ngLng thng tin cho trung bnh (k hiu l I(A,B)) c vit thnh :
1 1
( / )( , ) ( ) log
( )
s ti j
i j
i j i
p a b I A B p a b
p a Khi c th khai trin I(A,B) thnh
1 1 1 1
( , ) ( ) log ( / ) ( ) log ( ) s t s t
i j i j i j i
i j i j
I A B p a b p a b p a b p a
a ng
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Lng thng tin cho trung bnh (k hiu l I(A,B)) c vit thnh :
1 1
( / )( , ) ( ) log
( )
s ti j
i j
i j i
p a b I A B p a b
p a Khi c th khai trin I(A,B) thnh
1 1
( , ) ( ) log ( / ) ( )s t
i j i j j
i j
I A B p a b p a b p b
a SaiLng thng tin cho trung bnh (k hiu l I(A,B)) c vit thnh :
1 1
( / )( , ) ( ) log
( )
s ti j
i j
i j i
p a b I A B p a b
p a Trong cc biu thc khai trin I(A,B) di y,
biu thc
no sai b1 1
( , ) ( ) log ( / ) ( )s t
i j i j j
i j
I A B p a b p a b p b
Lng thng tin cho trung bnh (k hiu l I(A,B)) c vit thnh :
1 1
( / )( , ) ( ) log
( )
s ti j
i j
i j i
p a b I A B p a b
p a 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:
1 1
( / ) ( ) log ( / )s t
i j i j
i j
H A B p a b p a b
Khi trong cc kt qu tnh I(A,B), kt quno sai b I(A,B) = H(A)+ H(A/B)
Xt 2 trng s kin A v B sau : 1, ; 1,( ) ( )
ji
i j
ba A i s B j 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 thc no di y: d( ) ( )
i j
i j
a bC
p a p b
Xt 2 trng s kin A v B sau : 1, ; 1,( ) ( )
ji
i j
ba A i s B j 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: d1 1
( ) ( ) log ( )s t
i j i j
i j
H C p a b p a b
Chn cu sai : a Xc sut xut hin cng ln, lng tin thuc cng ln cng s cng nhChn cu sai sau : a Xc sut xut p(x) cng ln th lng tin khi
nhn c tin ny cng s cng lnLng tin c iu kin hu nghim v Kx ( thng tin ring v Kx sau khi c ly )c nh nghal ( / ) log ( / )k l k l I x y p x y Chn cu sai trong cc cu sau b Khi
( / )k lp x y 1 th ( / )k l I x y 1 v ngc li
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Lng thng tin v Kx khi r tin ly l( / )
( / ) log( )
k lk l
k
p x yI x y
p x Chn cu sai
sau : b Nu ( / )k l p x y = 0 suy ra1
( / ) log( )
k l
k
I x yp x
Lng thng tin hu nghim v Kx ( thng tin ring v Kx sau khi c ly ) cvit l :
1( / ) log
( )k l
k
I x yp x
Lng thng tin ring v Kx l1
( ) log( )
k
k
I xp x
Lng thng tin cho v Kx do ly mang li l :1 1
( / ) log log( ) ( / )
k l
k k l
I x y p x p x y
Tm cu sai sau : b Lng thng tin ring c th m
Cho 2 ngun tin A v B c cc xc sut tng ng l :1
1/ 2
aA
2
1/ 4
a 3
1/ 8
a
4 1;
1/ 8 0,5
a bB
2
0,25
b 3
0,125
b 4
0,125
b
Entropy ca ngun A ( k hiu l H(A)), entropy
ca ngun B ( k hiu l H(B)) c quan h theo cc h thc no di y a H(A)=H(B)
Cho ngun tin A c cc xc sut tng ng l :1
1/ 2
aA
2
1/ 4
a 3
1/ 8
a 4
1/ 8
a Khi
Entropy ca ngun A ( k hiu l H(A)) bng cc i lng no di ybH(A) = 1,75 bt
Cho ngun tin A c cc xc sut tng ng l :1
0,45
aA
2
0,2
a 3
0,15
a 4
0,1
a 5
0,1
a
Khi
Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no di yb H(A) = 2,06 bt
Cho ngun tin A c cc xc sut tng ng l :1
0,4
aA
2
0,25
a 3
0,15
a 4
0,1
a 5
0,1
a
Khi
Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no di yd H(A) = 2,1 bt
Cho ngun tin A c cc xc sut tng ng l :1
0,4
aA
2
0,25
a 3
0,15
a 4
0,15
a 5
0,05
a
Khi
Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no diy d H(A) = 2,07 bt
Cho ngun tin A c cc xc sut tng ng l :1
0,4
aA
2
0,25
a 3
0,2
a 4
0,1
a 5
0,05
a
Khi
Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no di yc H(A) = 2,04 bt
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Cho ngun tin A c cc xc sut tng ng l :1
0,4
aA
2
0,3
a 3
0,15
a 4
0,1
a 5
0,05
a
Khi
Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no di yd H(A) = 2,01 bt
Gi s ngun tin A c cc xc sut tng ng l : 10,4aA
2
0,3a 3
0,2a 4
0,05a 5
0,05a
Khi
Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no diy a H(A) = 1,95 bt
Gi s ngun tin A c cc xc sut tng ng l :1
0,35
aA
2
0,35
a 3
0,2
a 4
0,05
a 5
0,05
a
Khi Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng nodi y d H(A) = 1,96 bt
Gi s ngun tin A c cc xc sut tng ng l :1
0,35
aA
2
0,3
a 3
0,25
a 4
0,05
a 5
0,05
a
Khi Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng nodi y c H(A) = 1,98 bt
Gi s ngun tin A c cc xc sut tng ng l :1
0,35
aA
2
0,3
a3
0,2
a4
0,1
a 5
0,05
a
Khi
Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no diy a H(A) = 2,06 bt
Gi s ngun tin A c cc xc sut tng ng l :1
0,3
aA
2
0,3
a 3
0,2
a 4
0,1
a 5
0,1
a
Khi
Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no di y
c H(A) = 2,17 bt
Gi s ngun tin A v B c cc xc sut tng ng l :1
1/ 4
aA
2
1/ 8
a 3
1/16
a 4
1/16
a
5
1/ 4
a 6
1/ 8
a 7
1/16
a 8
1/16
a
1
0,25
bB
2
0,125
b3
0,0625
b4
0,0625
b5
0,25
b6
0,125
b7
0,0625
b 8
0,0625
b
Entropy
ca ngun A ( k hiu l H(A)), entropy ca ngun B ( k hiu l H(B)) cquan h theo cc h thc no di y d H(A) > H(B)
Gi s ngun tin A c cc xc sut tng ng l :1
1/ 4
aA
2
1/ 8
a 3
1/16
a 4
1/16
a 5
1/ 4
a 6
1/ 8
a
7
1/16
a 8
1/16
a
khi Entropy ca ngun A ( k hiu l H(A)) bng cc i lng
no di y b H(A) = 2,75 bt
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Gi s ngun tin B c cc xc sut tng ng l :1
0,25
bB
2
0,125
b 3
0,0625
b 4
0,0625
b
5
0,25
b 6
0,125
b 7
0,0625
b 8
0,0625
b
Khi Entropy ca ngun B ( k hiu l H(B)) bng
cc i lng no di y c H(B) = 2,75 btCho mt knh nh phn nh hnh bn: hnh chng 3 cu 92 Trong :Phn b xc sut ca tin u ra 1( )p b c tnh theo cng thc sau :
2
1 1
1
( ) ( ) ( / )i ii
p b p a p b a
T cng thc ny, c th khai trin 1( )p b thnh cng thcno di y: c 1 1 1 1 2 1 2 p(b ) p(a ).p(b / a ) p(a ).p(b / a ) Cho mt knh nh phn nh hnh bn: hnh chng 3 cu 93 Trong :Phn b xc sut ca tin u ra 2( )p b c tnh theo cng thc sau :
2
2 2
1
( ) ( ) ( / )i ii
p b p a p b a
T cng thc ny, c th khai trin 1( )p b thnh cng
thc no di y: b 2 1 2 1 2 2 2p(b ) p(a ).p(b / a ) p(a ).p(b / a ) Cho knh i xng nh phn nh hnh bn: hnh chng 3 cu 94 Bit
1 2( ) ; ( ) 1p a p p a p 1 2 2 1( / ) ( / ) 1s dp b a p b a p p 1 1 2 2( / ) ( / ) dp b a p b a p 2 2
1 1
( / ) ( ) ( / ) log ( / )i j i j ii j
H B A p a p b a p b a
Khai trin H(B/A) theo pd v ps s ckt qu no di y b ( / ) log (1 ) log(1 ) s s s s H B A p p p p Cho knh i xng nh phn nh hnh bn: hnh chng 3 cu 95 Bit
1 2( ) ; ( ) 1p a p p a p 1 2 2 1( / ) ( / ) 1s dp b a p b a p p 1 1 2 2( / ) ( / ) dp b a p b a p
( / ) log (1 ) log(1 ) s s s s H B A p p p p T truyn tn hiu cho knhvK= 1/T
v '1 1
max ( , ) max ( ) ( / )A A
C I A B H B H B AT T
Khi '
'
max
C
Cc tnh theo biu
thc no di y c'
'
max
1 log (1 ) log(1 ) s s s sC
p p p pC
Chn cu ng v m ha c M ha l mt nh x 1- 1 t tp cctin ri rc ia ln tp cc t m ; :i i
n n
i i if a Chn cu ng v m : d M (hay b m) l sn phm ca php mha
Chn cu ng v m : b M l mt tp cc t m c lp nntheo mt lut nh di t m ni l s cc du m cn thit dng m ha cho tin ai . Chncu ng v di t m b Nu ni = constvi mi i th mi tm u c cng di B m tng ng c gi l b m uChn cu sai v di t m d di t m ni l mt snguyn c th m
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di t m ni l s cc du m cn thit dng m ha cho tin ai . Chncu ng v di t m a Nu ninjth b m tng ng cgi l b m khng uS cc du m khc nhau (v gi tr) c s dng trong b m c gi lc s m Ta k hiu
gi tr ny l m . Chn cu sai v cc du m m: c Nu m = 0 thb m tng ng c gi l m uGi s c t m 7i =0 1 1 0 1 0 1Chn cu ng nht v t m
7
i b
T m 7i trong b m nh phn c m=2 ( tc l c 2 du m l0 v 1) v c di l 7Gi s c t m 7i =0 1 1 0 1 0 1 . Chn cu ng v t m
7
i c T
m 7i c di bng 7
Chn cu ng v di t m a di trung bnh ca t m n
l k vng ca i lng ngu nhin nic xc nh nh1
( )s
i i
i
n p a n
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 hocc tin ri rc c hiu qu m di trung bnh ca t m c th nh tu ,nhng khng nh hn entropie xc nh bi cc c tnh thng k cangun . Chn cu ng v di t m : d Chiu di trung bnhcc t m nh nht trong tt c cc cch m haNi 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 hocc tin ri rc c hiu qu m di trung bnh ca t m c th nh tu ,nhng khng nh hn entropie xc nh bi cc c tnh thng k ca
ngun. Chn cu sai v di t m : d Chiu di trung bnh
cc t m tho mn h thc 11
( ) ( )s
i i
i
n p a n H A
Khong cch gia hai t m bt k ni v
n
j l s cc du m khc nhau tnh
theo cng mt v
tr gia hai t m ny, k hiu d( ni ,n
j ) Gi sn
i =0 1 1 0 1 0 1;n
j =1 0 0
1 1 1 0 . Khong cch gia 2 t m ni vn
j l d((7
i ,7
j ) bng cc i lng
no di y b 6Khong cch gia hai t m bt k ni v
n
j l s cc du m khc nhau tnh
theo cng mt vtr gia hai t m ny, k hiu d( ni ,
n
j ) . Tm biu thc sai v khong cch
m d trong cc biu thc sau: c d( ni ,n
j ) d(n
i ,n
j )
Khong cch gia hai t m bt k ni vn
j l s cc du m khc nhau tnh
theo cng mt v
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tr gia hai t m ny, k hiu d( ni ,n
j ) Tm biu thc sai v khong cch
m d trong cc biu
thc sau: c d( ni ,n
j ) 0
Khong cch gia hai t m bt k ni vn
j l s cc du m khc nhau tnh
theo cng mt vtr gia hai t m ny, k hiu d( ni ,
n
j ) . Chn cu ng sau : d d(n
i ,n
j ) 0 ; d(n
i ,n
j ) = 0 khin
i n
j
Trng s ca mt t m W( ni ) l s cc du m khc khng trong t m V
d: ni = 0 1 1 0 1 0 1 th W(n
i ) = 4. Chn cu ng v cc tnh cht ca
trng s W( ni ) d d(n
i ,n
j ) = W(n
i +n
j ) v 0 W(n
i ) 1
Trng s ca mt t m W( ni ) l s cc du m khc khng trong t m .
Gi s ni = 0 1 1 0 1 0 1 th trng s W(n
i ) bng s no di y. a
4Coi mi t m ni l mt vct n chiu trong mt khng gian tuyn tnh nchiu Vn, 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 modulo2 ny c m t nh sau: hnh cu 19 chng 4 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 7
k i j bng gi tr no di y b (1 1, 1, 1, 0, 1, 1)Coi mi t m ni l mt vct n chiu trong mt khng gian tuyn tnh nchiu Vn, 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 modulo2 ny c m t nh sau: hnh cu 20 chng 4 Cho 8i =1 1 1 0 1 0 1 0
(1, 1, 1, 0, 1, 0, 1, 0) 7j = 1 0 0 1 1 1 0 1 (1, 0, 0, 1, 1, 1, 0, 1). Khi 8 8 8
k i j bng gi tr no di y c (0 1, 1, 1, 0, 1, 1, 1 )Chn cu ng v m tuyn tnh b M tuyn tnh di n l mm t m ca n c cc du m l cc dng tuyn tnhChn cu ng v m h thng tuyn tnh : b M h thng tuyntnh (n,k) l m tuyn tnh di n trong ta c th ch ra c v
tr ca k du thng tin trong t mChn cu ng v m tuyn tnh ngu nhin : b M tuyn tnh ngunhin l m tuyn tnh c cc du m c chn ngu nhin t ccdng tuyn tnh c th c m t m tuyn tnh, c th s dng ma trn sinh Gk,n Trong i s tuyntnh ta bit rng
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vi mi Gk,n s tn ti ma trn Hrxn tha mn: G.HT= 0 Chn cu ng sau :a H c gi l ma trn kim tra ca m tuyn tnh (n,k) v
H cha r vc t hng trc giao vi cc vc t hng ca GGi s sau khi thc hin m ha, tin aic m ha thnh 1001, khi di t m nica tin
ny l : a 4Sau khi thc hin m ha ngun ri rc A ( c entropy l H(A)=1,5 ) c thtnh 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:
b n =1,51Sau khi thc hin m ha ngun ri rc A ( c entropy l H(A)=1,4 ) c thtnh 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:
b n =1,45Sau 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:
c n =1,905Sau khi thc hin m ha ngun ri rc A ( c entropy l H(A)= 1,905) c thtnh 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 =1,906Sau khi thc hin m ha ngun ri rc A ( c entropy l H(A)= 2,01 bt) cth 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 gil ti u: d n =202Cho m Cyclic C(n, k) = C(7,4) c a thc sinh l g(x) th d T g(x) cth xc nh c ma trn sinh h thng G v ma trn kim tra Hcho b mCho m Cyclic C(n, k) = C(7,4), c a thc sinh g(x) l c g(x) = 1 +x2 +x3
Cho m Cyclic C(n, k) = C(7,4), c a thc sinh g(x) l b g(x) = 1 +x+x3
Cho m Cyclic C(n, k) = C(7,3) , tm cu sai v g(x) a g(x)khng l c cax7 + 1Cho m Cyclic C(n, k) = C(7,3), c a thc sinh g(x) l b g(x) = 1 +x+x2 +x4
Cho m Cyclic C(n, k) = C(7,3), c a thc sinh g(x) l d g(x) = 1 +x+x2 +x4
Pht biu sau ng hay sai : Cho m Cyclic C(n, k) = C(7,3) , g(x) l a thcbc 4 b ngCho m tuyn tnh (7,4) S nh cn thit nh b m l b 28
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Cho m Xyclic (7,4) S cc nh cn thit nh b m l a 7Cho m tuyn tnh (7,3) S nh cn thit nh b m l b 21Cho m Xyclic (7,3) S cc nh cn thit nh b m l d 7Cho m tuyn tnh (n,k) S nh cn thit nh b m l b knCho m Xyclic (n,k) S cc nh cn thit nh b m l d n7
Chn cu ng sau: b Cc dng tuyn tnh ca k bin c lpx1,x2,,xkl cc biu thc c dng: 1
1
( ,..., ) ( )k
k i i
i
f x x a x
Trong :
ia F , F l mt trngChn cu sai v m tuyn tnh : c Trong i s tuyn tnh ta bitrng vi mi G s tn ti ma trn Hrn tha mn m t m tuyn tnh, c th s dng ma trn sinh Gk,nTrong i s tuyntnh ta bit rng vi mi Gk,ns tn ti ma trn Hrn tha mn: G.HT = 0Chncu sai sau : b HTc gi l ma trn k hng, n ctKhi xy dng mt m tuyn tnh (n,k,d0)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 datrn cc bi ton ti u tm cu sai trong cc cu di y: d Vi n vs sai khi sa t xc nh, ta phi tm c m c s du thng tin kl ln nht (hay s du tha l nh nht) Tng ng vi bi ton
ny ta c gii hn Hamming sau:0
2t
n k i
n
i
C
Chn nh ngha sai v m xyclic trong cc nh ngha sau b M xyclic(n, k) l mt b m m a thc sinh c bc r = n+kChn cu ng ca nh l v kh nng sa sai c M u nh phn c tha (D > 0) vi khong cch Hamming d0 3c kh nng sa
c t sai tho mn iu kin: 0 12
dt v c kh nng pht hin t
sai tho mn iu kintd01T nh l m ho th 1 ca Shannon i vi m nh phn ta c :
1
1
( ) ( )s
i i
i
n p a n H A
T biu thc trn tm cu ng nht trong cc biu thc
sau : c1 1
( ) ( ) log ( )s s
i i i i
i i
p a n p a p a
Theo nh l m ho th 1 ca Shannon i vi m nh phn ta c:
1
1
( ) ( )
s
i i
i
n p a n H A
= 1 ( ) log ( )s
i i
i
p a p a
T biu thc ny suy ra di t m ni
v xc sut p(ai) lin h vi nhau:1
log (*)( )
i
i
np a
T (*) tm cu ng sau v
nguyn tc lp m tit kim: c Cc tin c xc sut xut hin lnc m ha bng cc t m c di nh v ngc li cc tin cxc sut xut hin nh c m ha bng cc t m c di ln
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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 x1c m ha thnh 0, x2c m ha thnh 10, x3cm ha thnh 110, x4c m ha thnh 1110, x5c m ha thnh 1111B m ti u cho ngun trn c chiu di trung bnh tnh theo cng thc :
5
1( )i i
in p x n
l : c 1,875Yu cu ca php m ha: nhng t m c di nh hn khng trng viphn u ca t m c di ln hn Cc tin c xc sut xut hin ln hnc m ha bng cc t m c di nh v ngc 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} Bitx1c m ha thnh 0, x2c m ha thnh 10 . Chn cu ng di y m ha cho x3 c 110Yu cu ca php m ha: nhng t m c di nh hn khng trng viphn u ca t m c di ln hn Cc tin c xc sut xut hin ln hnc m ha bng cc t m c di nh v ngc 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} Bitx1c m ha thnh 0, x2c m ha thnh 11, x3 c m ha thnh 100Chn cu ng di y m ha cho x4 d 1010Yu cu ca php m ha: nhng t m c di nh hn khng trng viphn u ca t m c di ln hn Cc tin c xc sut xut hin ln hnc m ha bng cc t m c di nh v ngc 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} Bitx1c m ha thnh 0, x2c m ha thnh 10, x3 c m ha thnh110, x4 c m ha thnh 1110. Chn cu ng di y m ha cho x5
b 1111
Cho m Cyclic C(7, 4) c a thc sinh l g(x) =1 +x+x3v a thc sinh G
sau :
3
2 4
2 3 5
3 4 6
1 x x
x x xG
x x x
x x x
Ma trn no sau y l mt ma trn sinh G ng vi m
Cyclic C(7,4) trn a
1101000
0110100
0011010
0001101
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Cho m Cyclic C(7, 4) c a thc sinh l g(x) = 1 +x2 +x3v a thc sinh G
sau
2 3
3 4
2 4 5
3 5 6
1 x x
x x xG
x x x
x x x
Ma trn no sau y l mt ma trn sinh G ng vi m
Cyclic C(7,4) trn b
1011000
0101100
0010110
0001011
Cho m Cyclic C(7, 4) c a thc sinh l : g(x) = 1 +x+x3v ma trn sinh
G. T ma trn sinh G tnh c ma trn kim tra H bn:
2 3 4
3 4 5
2 4 5 6
1 x x x
H x x x x
x x x x
Chuyn ma trn kim tra H sang dng khng gian tuyn tnh. Ma trn nosau y l mt ma trn kim tra H ( dng khng gian tuyn tnh) a1011100
0101110
0010111
Cho m Cyclic C(7, 4) c a thc sinh l : (x) = 1 +x2 +x3 v ma trn sinh GT ma trn sinh G
tnh c ma trn kim tra H bn :
2 4
2 3 5
2 3 4 6
1 x x x
H 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
kim tra H ( dng khng gian tuyn tnh) a
1110100
0111010
0011101
Tin ri rc ai A; RA: T m fi (x)tng ng vi ai Thut ton xy dng t mxyclic gmtheo 4 bc: M,N,P,Q c sp xp ngu nhin M: M t tin ai trong tp tin
cn m ha (gm 2
k
tin) bng mt a thc ai (X)vi deg ai (X) k
1. N: Chiaai (X).xn-k cho a thc sinh g(X) tm phn d ri (X). P: Nng bc ai (X) bngcch nhn n vi xn-k. . Q : Xy dng t m xyclic: ( ) ( ). ( )n ki i if X a X x r X
.Chn cc sp xp th t ng thut ton xy dng t m xyclic: a
M-P-N-Q
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8/14/2019 NHT Ly Thuyet Thong Tin
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Gi s sau khi thc hin m ha ngun ri rc A. Ta c kt qu m ho sau :A1 A4 A5 A2 A3
00 10 110 1111 1110
Gii m cho dy bt nhn c c dng
11100011010111100 s ckt qu no sau y: b A3-A1-A5-A4-A2-A1Gi s sau khi thc hin m ha ngun ri rc A . Ta c kt qu m ho sau :
A1 A4 A5 A2 A3
00 10 110 1111 1110
Gii m cho dy bt nhn c c dng
11110011010111100 s c. kt qu no sau y: d A2-A1-A5-A4-A2-A1Gi s sau khi thc hin m ha ngun ri rc A . Ta c kt qu m ho sau :
A1 A4 A5 A2 A3
00 10 110 1111 1110
Gii m cho dy bt nhn c c dng
1111001101011110010 s c kt qu no sau y: c A2-A1-
A5-A4-A2-A1-A4Gi s sau khi thc hin m ha ngun ri rc A . Ta c kt qu m ho sau :A1 A4 A5 A2 A3
00 10 110 1111 1110
Gii m cho dy bt nhn c c dng
11110011010111100110 s c kt qu no sau y: b A2-A1-A5-A4-A2-A1-A5Gi s sau khi thc hin m ha ngun ri rc A . Ta c kt qu m ho sau :
A1 A4 A5 A2 A3
00 10 110 1111 1110
Gii m cho dy bt nhn c c dng
1111001101011110011000 s c kt qu no sau y: b A2-A1-
A5-A4-A2-A1-A5-A1Gi s sau khi thc hin m ha, cc tin aivi xc sut tng ng P(ai )cm ha thnh ccm nh phn c di m nitng ng nh bng sau: hnh cu 66 chng 4
Tnh di m trung bnh theo biu thc5
1
( )i ii
n n p a
ta c gi tr no :c 2,1
Gi s sau khi thc hin m ha, cc tin aivi xc sut tng ng P(ai )cm ha thnh ccm nh phn c di m nitng ng nh bng sau: hnh cu 67 chng 4
Tnh di m trung bnh theo biu thc
5
1( )i i
in n p a
ta c gi tr no :
a 1,8Gi s sau khi thc hin m ha, cc tin aivi xc sut tng ng P(ai )cm ha thnh cc
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m nh phn c di m nitng ng nh bng sau: hnh cu 68 chng 4
Tnh di m trung bnh theo biu thc5
1
( )i ii
n n p a
ta c gi tr no :d 2,75
Gi s sau khi thc hin m ha, cc tin aivi xc sut tng ng P(ai )c
m ha thnh ccm nh phn c di m nitng ng nh bng sau: hnh cu 69 chng 4
Tnh di m trung bnh theo biu thc5
1
( )i ii
n n p a
ta c gi tr no :d 2,45 du m
Gi s sau khi thc hin m ha, cc tin aivi xc sut tng ng P(ai )cm ha thnh ccm nh phn c di m nitng ng nh bng sau: hnh cu 70 chng 4
Tnh di m trung bnh theo biu thc5
1
( )i ii
n n p a
ta c gi tr no :
a 2,5Mt dy tin 1 2, , ... n X x x x vi ; 1i x X i n Lng tin I(X) cha trong dy tin
X s l:1 2
1 1 1( ) log log ... log
( ) ( ) ( )nI X
p 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 1 2 1 1 3 4 1 1 5X x x x x x x x x x l: c 18 btKhi xy dng mt m tuyn tnh (n,k,d0 ) ngi ta phi tm c cc m c tha nh nhng li c kh nng khng ch sai ln Ngi ta thng xydng m ny da trn cc bi ton ti u Tm cu sai trong cc cu di y:
a Vi k v d0 xc nh, ta phi tm c m c di vi tm l ln nht. Tng ng vi bi ton ny ta c gii hn n = kChn cu ng ca nh l v kh nng pht hin sai sau c M unh phn c tha (D > 0) vi khong cch Hamming d0>1c khnng pht hin t sai tho mn iu kintd0-1kh nng sa c t
sai tho mn iu kin: 01
2
dt
Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1
1( ) 1/ 2
a aA
p a
21/ 4
a
3
1/ 8
a
4
1/16
a
5
1/16
a s c kt qu no sau y: d
1a
0
2
10
a
3
110
a
4
1110
a
5
1111
a
Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1
1( ) 1/16
a aA
p a
21/16
a
3
1/ 8
a 41/ 4
a
5
1/ 2
a
s c kt qu no sau y: d1a
1111
21110
a 3
110
a 4
10
a
5
0
a
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8/14/2019 NHT Ly Thuyet Thong Tin
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Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1
1( ) 1/ 4
a aA
p a
21/ 2
a
3
1/ 8
a 41/16
a
5
1/16
a
s c kt qu no sau y: a1a
10
20
a 3
110
a 4
1110
a
5
1111
a
Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1
1( ) 1/ 8
a aA
p a
21/ 4
a
3
1/ 2
a 41/16
a
5
1/16
a
s c kt qu no sau y: b1a
110
210
a 3
0
a 4
1110
a
5
1111
a
Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1
1( ) 1/16
a aA
p a
21/ 4
a
3
1/ 8
a 41/ 2
a
5
1/16
a
s c kt qu no sau y: b1a
1110
210
a 3
110
a 4
0
a
5
1111
a
Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1
1( ) 1/ 2
a aA
p a
21/ 4
a
3
1/16
a 41/ 8
a
5
1/16
a
s c kt qu no sau y: a1a
0
210
a 3
1110
a 4
110
a
5
1111
a
Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1
1( ) 1/ 2
a aA
p a
21/ 8
a
3
1/ 4
a 4
1/16
a
5
1/16
a
s c kt qu no sau y: b1a
0
2
11
a 3
01
a 4
111
a
5
1111
a
Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1
1( ) 1/ 2
a aA
p a
21/16
a
3
1/ 8
a 41/ 4
a
5
1/16
a
s c kt qu no sau y: c1a
0
21110
a 3
110
a 4
10
a
5
1111
a
Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1
1( ) 1/ 2
a aA
p a
21/ 4
a
3
1/16
a 4
1/16
a
5
1/ 8
a
s c kt qu no sau y: a1a
0
2
10
a 3
1111
a 4
1110
a
5
110
a
Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1
1( ) 1/16
a aA
p a
21/ 4
a
3
1/ 8
a 41/16
a
5
1/ 2
a
s c kt qu no sau y: d1a
1111
210
a 3
110
a 4
1110
a
5
0
a
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8/14/2019 NHT Ly Thuyet Thong Tin
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Gi s sau khi thc hin m ha Shannon - Fano ngun ri rc A :
1 1
1( ) 1/ 2
a aA
p a
2
1/ 4
a 3
1/ 8
a 4
1/16
a 5
1/16
a
Ta c kt qu m ho sau :1a
0
210
a 3
110
a 4
1110
a
5
1111
a di t m trung bnh
n
v entropy H(A) c tnh theo biu thc5
1
( )i ii
n p a n
v5
1
( ) ( ) log ( )i ii
H A p a p a
s c kt qu no sau y : b n=H(A)=1,875
Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1
1( ) 0,5
a aA
p a
2
0,25
a
3
0,125
a 4
0,625
a 5
0,625
a
s c kt qu no sau y: b1a
0
210
a 3
110
a 4
1110
a
5
1111
a
Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1
1( ) 0,25
a a
A p a
2
0,5
a
3
0,125
a 4
0,625
a 5
0,625
a
s c kt qu no sau y: b1a
10
20
a 3
110
a 4
1110
a
5
1111
a
Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1
1( ) 0,125
a aA
p a
2
0,25
a 3
0,5
a 4
0,625
a 5
0,625
a
s c kt qu no sau y: b1a
110
210
a 3
0
a 4
1110
a
5
1111
a
Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1
1( ) 0,625
a aA
p a
2
0,25
a3
0,125
a4
0,5
a 5
0,625
a
s c kt qu no sau y: a1a
1110
210
a 3
110
a 4
0
a
5
1111
a
Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1
1( ) 0,625
a a
A p a
2
0,25
a3
0,125
a4
0,625
a 5
0
a
s c kt qu no sau y: c1a
1111
210
a 3
110
a 4
1110
a
5
0
a
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8/14/2019 NHT Ly Thuyet Thong Tin
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Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1
1( ) 0,5
a aA
p a
2
0,125
a
3
0,25
a 4
0,625
a 5
0,625
a
s c kt qu no sau y: d1a
0
2110
a 3
10
a 4
1110
a
5
1111
a
Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1
1( ) 0,5
a aA
p a
2
0,625
a
3
0,125
a 4
0,25
a 5
0,625
a
s c kt qu no sau y: c1a
0
21110
a 3
110
a 4
10
a
5
1111
a
Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1
1( ) 0,5
a aA
p a
2
0,625
a
3
0,125
a 4
0,625
a 5
0,25
a
s c kt qu no sau y: d1a
0
21111
a 3110
a 4
1110
a
5
10
a
Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1
1( ) 0,5
a aA
p a
2
0,25
a
3
0,625
a4
0,125
a 5
0,625
a
s c kt qu no sau y: a1a
0
210
a 3
1110
a 4
110
a
5
1111
a
Gi s sau khi thc hin m ha Shannon - Fano ngun ri rc A :
1 1
1( ) 0,5
a aA
p a
2
0,25
a3
0,125
a4
0,625
a 5
0,625
a
Ta c kt qu m ho sau :1a
0
210
a 3
110
a
4
1111
a
5
1110
a
di t m trung bnh n v entropy H(A) c tnh theo biu
thc
5
1
( )i ii
n p a n
v5
1
( ) ( ) log ( )i ii
H A p a p a
c kt qu no sau y : an =H(A)=1,875
Gi s sau khi thc hin m ha Shannon - Fano ngun ri rc A :
1 1
1( ) 0,25
a aA
p a
2
0,125
a 3
0,625
a 4
0,625
a 5
0,25
a 6
0,125
a 7
0,625
a 8
0,625
a
Ta c kt qu m ho
sau :1a
00
5
01
a
2
100
a
6
101
a
3
1100
a
4
1101
a
7
1110
a
8
1111
a di t m trung bnh
n v
entropy H(A) c tnh theo biu thc8
1
( )i ii
n p a n
v8
1
( ) ( ) log ( )i ii
H A p a p a
c kt qu no sau y : a n =H(A)=2,75Gi s sau khi thc hin m ha Shannon - Fano ngun ri rc A :
1 1
1( ) 1/ 4
a aA
p a
2
1/ 8
a 3
1/16
a 4
1/16
a 5
1/ 4
a 6
1/ 8
a 7
1/16
a 8
1/16
a
Ta c kt qu m ho sau : 1a
00
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5
01
a 2
100
a 6
101
a 3
1100
a 4
1101
a 7
1110
a
8
1111
a
di t m trung bnh n v entropy
H(A) c tnh theo biu thc8
1
( )i ii
n p a n
v8
1
( ) ( ) log ( )i ii
H A p a p a
c kt
qu no sau y :b n =H(A)=2,75
Gi s sau khi thc hin m ha ngun ri rc A :1 1
1( ) 0,5
a aA
p a
4
0,25
a 5
0,125
a
2
0,625
a 3
0,625
a
Ta c kt qu m ho sau :1a
0
410
a 5
110
a 2
1111
a
3
11100
a
di t
m trung bnh n v entropy H(A) c tnh theo biu thc
5
1
( )i ii
n p a n
v5
1
( ) ( ) log ( )i ii
H A p a p a
c kt qu no sau y : d n =1,9375 vH(A)=1,875
Gi s sau khi thc hin m ha ngun ri rc A :1 1
1( ) 1/ 2
a aA
p a
4
1/ 4
a 5
1/ 8
a 2
1/16
a
3
1/16
a
Ta c kt qu m ho sau :1a
0
410
a 5
110
a 2
1111
a
3
11100
a
di t m trung
bnh n v entropy H(A) c tnh theo biu thc5
1
( )i ii
n p a n
v5
1
( ) ( ) log ( )i ii
H A p a p a
c kt qu no sau y : c n =1,9375 vH(A)=1,875
Gi s sau khi thc hin m ha ngun ri rc A :1 1
1( ) 1/ 2
a aA
p a
4
1/ 4
a 5
1/ 8
a 2
1/16
a
3
1/16
a
Ta c kt qu m ho sau :1a
00
410
a 5
110
a 2
1111
a
3
1100
a
di t m trung
bnh n v entropy H(A) c tnh theo biu thc5
1
( )i ii
n p a n
v5
1
( ) ( ) log ( )i ii
H A p a p a
c kt qu no sau y : b
n=2,375 v
H(A)=1,875
Gi s sau khi thc hin m ha ngun ri rc A :1 1
1( ) 1/ 2
a aA
p a
4
1/ 4
a 5
1/ 8
a 2
1/16
a
3
1/16
a
Ta c kt qu m ho sau :1a
0
410
a 5
1100
a 2
1111
a
3
1100
a
di t m trung
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bnh n v entropy H(A) c tnh theo biu thc5
1
( )i ii
n p a n
v5
1
( ) ( ) log ( )i ii
H A p a p a
c kt qu no sau y : c n =2 v H(A)=1,875
Gi s sau khi thc hin m ha ngun ri rc A : 1 11( ) 0,5
a aAp a
4
0,25a 5
0,125a
2
0,625
a 3
0,625
a
Ta c kt qu m ho sau :1a
0
410
a 5
1100
a 2
1111
a
3
1100
a
di t
m trung bnh n v entropy H(A) c tnh theo biu thc
5
1
( )i ii
n p a n
v5
1
( ) ( ) log ( )i ii
H A p a p a
c kt qu no sau y : d n =2 v H(A)=1,875
Gi s sau khi thc hin m ha ngun ri rc A :1 1
1( ) 0,5
a a
A p a
4
0,25
a5
0,125
a
2
0,625
a 3
0,625
a
Ta c kt qu m ho sau :1a
00
410
a 5
110
a 2
1111
a
3
1110
a
di t m
trung bnh n v entropy H(A) c tnh theo biu thc
5
1
( )i ii
n p a n
v5
1
( ) ( ) log ( )i ii
H A p a p a
c kt qu no sau y : c n =2,375 vH(A)=1,875
Gi s sau khi thc hin m ha ngun ri rc A : 1 11( ) 0,5
a aA
p a 4
0,25
a5
0,125
a
2
0,625
a 3
0,625
a
Ta c kt qu m ho sau :1a
00
410
a 5
1100
a 2
1111
a
3
1110
a
di t
m trung bnh n v entropy H(A) c tnh theo biu thc
5
1
( )i ii
n p a n
v5
1
( ) ( ) log ( )i ii
H A p a p a
c kt qu no sau y : a n =2,5 vH(A)=1,875
Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3
tng ng a thcthng tin a(x) = x + x3 S dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y : a 1100101Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thcthng tin a(x) = 1 + x2 S dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y: c 0011010
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8/14/2019 NHT Ly Thuyet Thong Tin
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Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thcthng tin a(x) = x + x2 S dng thut ton 4 bc thit lp t m h thngta c kt qu no di y: c 1000110Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thcthng tin a(x) = 1 + x3 S dng thut ton 4 bc thit lp t m h
thng ta c kt qu no di y : d 0111001Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thcthng tin a(x) = x2 + x3 S dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y : d 0100011Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thcthng tin a(x) = 1 + x2 + x3 S dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y : d 1001011Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thcthng tin a(x) = 1 + xS dng thut ton 4 bc thit lp t m h thngta c kt qu no di y: c 1011100Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thc
thng tin a(x) = 1 + x
+ x2
S dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y: b 0101110Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thcthng tin a(x) = x + x2 + x3 S dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y : d 0010111Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thcthng tin a(x) = x3 S dng thut ton 4 bc thit lp t m h thng tac kt qu no di y : c 1010001Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thcthng tin a(x) = x S dng thut ton 4 bc thit lp t m h thng tac kt qu no di y : b 0110100
Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3
tng ng a thcthng tin a(x) = x2 S dng thut ton 4 bc thit lp t m h thng tac kt qu no di y: a 1110010Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = x + x3 s dng thut ton 4 bc thit lp t m h thngta c kt qu no di y : c 1000101Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = 1 + x2 s dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y : d 0111010Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = x + x2 s dng thut ton 4 bc thit lp t m h thng
ta c kt qu no di y : d 0010110Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = x + x3 s dng thut ton 4 bc thit lp t m h thngta c kt qu no di y : a 1101001Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = x2 + x3 s dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y : a 1010011
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Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = 1 + x2 + x3 s dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y : a 0001011Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = 1 + x s dng thut ton 4 bc thit lp t m h thng
ta c kt qu no di y : c 0101100 Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = 1 + x + x2 s dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y : b 1001110Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = x + x2 + x3 s dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y : b 0100111Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = x3 s dng thut ton 4 bc thit lp t m h thng tac kt qu no di y : b 0111011Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thc
thng tin a(x) = x s dng thut ton 4 bc thit lp t m h thng tac kt qu no di y : a 1110100Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = x2 s dng thut ton 4 bc thit lp t m h thng tac kt qu no di y : c 1110110Thu tn hiu khi c nhiu l mt bi ton thng k: a ngNhim v ca my thu l phi chn li gii, do my thu cn c gi l s gii Yu cu ca s gii l phi tm li gii ng (pht i ta phi tmc i ) Trong thc t c rt nhiu s gii Trong tt c cc s gii cth c th ti mt s bo m xc sut nhn ln phi ng l ln nht(xc sut gii sai l b nht) S ny c gi l s gii ti u. Khi
ngi ta gi My thu xy dng theo s gii ti u c gi l my thu tiu Kt lun ny ng hay sai ? b ng
/
( )( )
( )
il i
l
pu
p
vi
1,
1
i m
i
l quy tc gii ti u vit di dng hm hp l.
Chn cung trong cc cu sau: b Nu mi tn hiu gi i u ng xcsut th / ( ) 1l i u vi i l nh ngha b lc phi hp tuyn tnh th ng : i vi mt tn hiu xc
nh, mt mch tuyn tnh th ng m bo t s rara
N
S
cc i mt
thi im quan st no y s c gi l mch lc phi hp tuyn tnh thng ca tn hiu Trong ra l t s gia cng sut trung bnh ca nhiu u ra b lc y v cng sut nh ca tn hiu nh ngha ny ng haysai ? a SaiQu trnh x l tn hiu trong my thu ti u c gi l x l ti u tn hiuX l nhn li gii c xc sut sai b nht Chn cu ng nht v x l tiu cc tn hiu: a Da vo cc tiu chun ti u, bng cng c
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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 hpc my thu ti u
Gi s i l tn hiu gi i, c xc sut p( i ))- c gi l xc sut tinnghim my thu ta nhn c u(t), t u(t)qua s gii ta s c li gii 1
no Nh vy l c gi i vi mt xc sut p( /l u )- c gi l xcsut hu nghim. Do xc sut gii sai s l:
p(sai/u, l ) =1p( /l u )Xt hai s gii: - T u(t)cho ta 1 - gi l s (1) - Tu(t)cho ta 2 - gi l s (2). Nu p(sai/u, 1 ) < p(sai/u, 2 ) ta rtra kt lun s (2) ti u hn (1) Khng nh ny ng hay sai ? a
SaiTn hiu tng qut c dng: 0 0( ) ( ) cos( ( ) )i iC t C t t t Chn cu sai trong cccu sau: a Vic x l ti u tn hiu khng ph thuc ng
bao 0 ( )iC t v tn s tc thi( )
( )id t
t dt
Cho u vo mch tuyn tnh th ng 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 tmbiu thc gii tch ca hm truyn phc Ki()ca mch tuyn tnh th ngsao cho mt thi im quan st (dao ng nhn c) no ra tmax, p dng cng thc bin i ngc Fourier tnh c
2
0
1(2 )ra ivS f df
N
Trong ( )ivS l mt ph (bin) phc ca th hin
tn hiu u vo mch tuyn tnh Theo nh l Parseval, ta c: max0
(*)iraEN
trong 2(2 )i iv E S f df
l nng lng ca tn hiu ti d T (*) chng
t t s rara
S
N
hon ton ph thuc vo dng ca n ca tn hiu m
hon ton khng ph thuc vo dng ca n.Cho knh nh phn, i xng, khng nh c nhiu cng, trng, chun theom hnh sau: hnh cu 9 chng 5 Xc sut sai ton phn sp ( xc sut sai
khng iu kin) l 1 2 1 2 1 2( ). ( / ) ( ). ( / )s p p p p p Gi s tnh c
04 2
2 1
0
1( / ) exp
2 22
P T
NP T
p dG
khi sp s bng biu thc no
di y: d0
12
s
P Tp
G
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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:( ).
( )0
A t T A es t
t T
t T
Hm truyn *0 ( ) ( )
j T K j kS j e Hm truyn
ca b lc (
*
0 ( ) ( )
j T
K j kS j e
) sao cho t s tn trn tp u ra ca b lct cc i s l biu thc no di y b 0 ( )
kAK j
A j