Giai Bai Tap Ly Thuyet Thong Tin_PTIT

NG QUANG HUY - D09VTH2 GVHD: NGUYN TH LAN ANH

Chng III: M THNG K TI U

Bi tp: Lp mt b m cho ngun tin u c s thng k nh sau.

Ui U1 U2 U3 U4 U5 U6 U7 U8 U9P(u) 0.34 0.2 0.19 0.1 0.07 0.04 0.03 0.02 0.01Bng m nh phn (m=2)theo phng php shannon. Tnh di trung bnh ca t m ntb v tnh kinh t ca t m: p=H(U)/Ntb.

GiiS m ha theo phng php m ha shannon:

di trung bnh t m: NTB=

Entropy ca tp tin: H(u)=

Ch s kinh t ca b m: p= =0,8279

111/13/2011

Ui P(ui) Pi ni Dng m nh phn ca pi T mU1 0.34 0 2 0 00U2 0.2 0.34 3 0.01010111 010U3 0.19 0.54 3 0.10001010 100U4 0.1 0.73 4 0.10111010 1011U5 0.07 0.83 4 0.11010100 1101U6 0.04 0.9 5 0.11100110 11100U7 0.03 0.94 6 0.11110000 111100U8 0.02 0.97 6 0.11111000 111110U9 0.01 0.99 7 0.11111101 1111110


__________________________________________________________________

Chng IV: M CHNG NHIU: M KHI

Bi tp: Ma trn sinh ca b m tuyn tnh (6,3) trn trng GF(2) l:

G=

a. Biu din G di dng chun Gch.

b. Lit k cc t m c c t G v Gch.

c. C bao nhiu t m c trng s hamming l 1,2,3,4,5,6,7.

Gii

a. Biu din G di dng chun Gch.

Gch=

b. Lit k cc t m ca G v Gch.

Vector mang tin a v= a.G w(v) v=a.Gch w(v)

000 000000 0 000000 0001 110110 4 001111 4010 110001 3 010110 3011 000111 3 011001 3100 011010 3 100011 3101 101100 3 101100 3110 101011 4 110101 4111 011101 4 111010 4

211/13/2011


c. C hai b m trn u c

1 t m c trng s l 0

0 t m c trng s l 1

0 t m c trng s l 2

4 t m c trng s l 3

3 t m c trng s l 4

0 t m c trng s l 5

0 t m c trng s l 6

__________________________________________________________________

Chng V: M VNG

Bi tp: Cho m vng (n,k)=(7,4)c ma trn sinh l g(x)=x3+x+1. Lit k tt c cc t m ca b m v cho bit b m c kh nng sa sai bao nhiu bit?

Giia. Cho d= (0001) ta tnh c:

V1=0001000+d s

V1=0001000+011=0001011

Quay vng t m v1 ta s c thm 6 t m na:V2=1000101

V3=1100010

V4=0110001

V5=1011000

311/13/2011


V6=0101100

V7=0010110

Cho d=(0011) ta tnh c:V=0011000+d sV=0011000+101=00111101Quay vng t m v ta s c thm 6 t m na:V9 =1001110

V10 =0100111

V11=1010011

V13 =1110100

V14=0111010

Cho d=(1111000) ta tinh c:

V15=1111000+d s

V15=1111000+111=1111111

Cho d=(0000) ta tnh c:

V0=0000000 + d s

b. Ta c th tnh d dng trng s Hamming ca b m l trng s nh nht ca t m khc khng : H(V)=3 nn b m c kh nng sa c tt cc mu sai 1bt.

BI TP L THUYT THNG TIN

Chng 2: Lng Tin

Trong mt tr s s vui ngi ta s 10 ch s t 0 n 9.

411/13/2011


Xc sut trng ca mi nhm l nh nhau.

a. Tnh lng tin ring ca tin: s trng gii l s 9

b. Tnh lng tin tng h gia tin: s trng gii l s 9 so vi tin: s trng gii l s chia ht cho 3.

c. Trong 10 tin trn gi U={u1, u2, u3, u4, u5, u6} vi ui l tin s i trung gii (i=0,1,.....,6).

Tm lng tin trung bnh ca tp tin U.

Gii

a. Gi p(9) l xc sut s 9 trng gii ta c:

p(9) = = 0.1

Vy I(9) =

b. Gi p(0-3-6-9) l xc sut trng gii chia ht cho 3.

Ta c p(0-3-6-9) =

Mc khc I(9/0-3-6-9) = =

=

Vy: I(9;0-3-6-9) = I(9)-I(9/0-3-6-9)

=

= 1.322

c. Lng tin trung bnh ca tp tin U:

I(U) = = - = = 2.325

511/13/2011


Chng 3: M Thng K Ti u

Cho ngun tin U c s thng k nh sau:

ui u1 u2 u3 u4 u5-6 u7-9 u10-18 u19p(ui) 0.338 0.32 0.13 0.1 0.018 0.01 0.005 0.01

Dng m Huffman kt hp m u m ha ngun tin trn vi c s m m=3. Tnh ntb v tnh kinh t ca t m:

Gii

S m ha theo phng php m ha Huffman:

ui p(ui) ni T mu1 0.338 1 0u2 0.32 1 1u3 0.13 2 20u4 0.1 2 21u5 0.018 4 2210u6 0.018 4 2211u7 0.01 4 2200u8 0.01 4 2201u9 0.01 4 2202u10 0.005 5 22200u11 0.005 5 22201u12 0.005 5 22202u13 0.005 5 22210u14 0.005 5 22211u15 0.005 5 22212u16 0.005 5 22220u17 0.005 5 22221u18 0.005 5 22222u19 0.001 4 2212

Phng php m ha nh sau:

Bc 1: sp xp cc tin c xc sut theo th t gim dn. Bc 2: ghp 3 tin c xc sut nh nht, tr thnh tin ph mi Bc 3: lp li ging bc 1( sp xp cc tin theo xc sut gim dn Bc 4: ghp 3 tin c xc sut nh nht tr thnh tin ph mi (cch trnh

by ny lp i lp li cho n khi cn 1 tin cui cng).

611/13/2011


di trung bnh t m: ntb =

Entropy ca tp tin: H(U) = -

Ch s kinh t ca b m: p = = = 0.9913

Chng 4: M Khi M Chng Nhiu

Ma trn sinh ca b m tuyn tnh (7,3) trn trng GF(2) l:

G =

d. Biu din G di dng chun Gch.

e. Xc nh ma trn th Hch.

f. Lit k cc t m c c t G v Gch.

g. Xc nh khong cch Hamming ca b m ny.

Gii

d.

Gch =

e. Ta c Hch = [-PT.In-k]

1 0 1 1 0 0 0

711/13/2011


0 1 1 0 1 0 0

1 1 1 0 0 1 0

1 1 0 0 0 0 1

f. Lit k cc t m ca G v Gch.


000 0000000 0 0000000 0001 1011100 4 0011110 4010 0101110 4 0100111 4011 1110010 4 0111001 4100 0010111 4 1001011 4101 1001011 4 1010101 4110 0111001 4 1101100 4111 1100101 4 1110010 4

g. C hai b m trn u c D=4 l trng s Hamming nh nht ca cc t m khc khng ca chng.

Chng 5: M Vng

a thc sinh ca b m vng hamming l: g(x)=1+x3+x4

a. Tm a thc th h(x) ca b m ny.

b. Thit k mch m ha thc hin qua h(x).

c. Xc nh cc bit th v t m nhn c tng ng vi chui bit mang tin d=(10000001011).

Gii

811/13/2011

Hch =


a. B m hamming c (n,k) = (2n-k-1,k) nn:

n = 2n-k-1 = 24-1 = 15.

T cng thc: h(x) = (xn+1)/g(x)

Ta tnh c: h(x) =

b. Khi d = (10000001011) hay d(x) = th:

v(x) = d(x).xn-k + d s ca

= + s d ca

= ( )

Trong x3+1 tng ng vi chui bit 1001 l cc bit th v t m nhn c tng ng vi v(x) = ( ) l v = (100000101110010).

BI TP MN L THUYT THNG TIN.

CHNG 1: LNG TIN

BI: Xc nh entropy khi sc sc. Gi s sc sc c ch to sao cho xc xut sut hin ca bt k mt no cng t l vi s chm trn mt sc sc.

GII:

Gi P(xi) l xc sut ca cc mt sc sc. Vi i=(1,2,3,4,5,6)

Entropy khi sc sc:

H(x) = 6

1 P(xi). I(xi)

= 21 1.log ( )21 21

- 22 2log ( )21 21

- 23 3log ( )21 21

-2

4 4log ( )21 21

- 25 5log ( )21 21

- 26 6log ( )21 21

= 0.209 + 0.323 + 0.401 + 0.455 + 0.492 + 0.516

911/13/2011


= 2.396 (bit)

CHNG 2: M THNG K TI U

BI: Lp mt b m cho ngun tin U c s thng k nh sau:

Ui U1 U2 U3 U4 U5 U6 U7 U8 U9P(ui) .34 .2 .19 .1 .07 .04 .03 .02 .01

Bng m nh phn (m = 2) theo phng php Shannon. Tnh di trung bnh ca t m v tnh kinh t ca b m.

GII:

S m ha theo phng php Shannon:

ui P(ui) Pi ni Dng nh phn ca Pi T mU1 .34 0 2 0 00U2 .2 .34 3 0.010 010U3 .19 .54 3 0.100 100U4 .1 .73 4 0.1011 1011U5 .07 .83 4 0.1101 1101U6 .04 .9 5 0.11100 11100U7 .03 .94 6 0.111100 111100U8 .02 .97 6 0.111110 111110U9 .01 .99 7 0.1111110 1111110

di trung bnh ca t m:

9

1( ) 3.1tb i i

in n p u

=

= =Entropy ca tp tin:

9

21

( ) ( ) log ( ) 2.5644i ii

H U p u p u=

= =Tnh kinh t ca b m:

1011/13/2011


( ) 2.5644 0.82793.1tb

H Un

= = =


Chng 2: Lng Tin

Xem cc con bi ca b bi 52 l to thnh 1 ngun tin ri rc. Tnh entropy ca mt l bi rt ngu nhin. Gi s b qua nc ca con bi by gi U={ Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Qeen, King}. Tnh entropy ca mt l bi rt ngu nhin trong trng hp ny, trong trng hp U=(bi c hnh, bi khng hnh).

Gii

d. H(U)= vi p(ui)= (i=1, 2, 3,.......,52)

=- =-52 =5.700

e. H(U)= vi p(ui)= (i=1, 2, 3,.......,13)

=- =-13 =3.700

f. Gi uh l tin bi c hnh th p(uh)=

Gi uoh l tin con bi khng hnh th p(uoh)=

Vy H(U)= p(uh)I(uh)+p(uoh)I(uoh)

=- p(uh) )-p( )

1111/13/2011


=- - = 0.779

Chng 5: M Vons

a thc sinh ca b am vng hamming l:

g(x)=1+x3+x4

d. Tm a thc th h(x) ca b am ny.

e. Thit k mch am ha thc hin qua h(x).

f. Xc nh cc bit th v t am nhn c tng ng vi chui bit mang tin d=(10000001011).

Gii

c. B am hamming c (n,k)=(2n-k-1,k) nn:

n=2n-k-1=24-1=15.

T cng thc: h(x)=(xn+1)/g(x)

Ta tnh c: h(x)=

d. Khi d=(10000001011) hay d(x)= th:

v(x)=d(x).xn-k+ d s ca

1211/13/2011


= + s d ca

=( )

Trong x3+1 tng ng vi chui bit 1001 l cc bit th v t am nhn c tng ng vi v(x)= ( ) l v=(100000101110010).


Chng 2: Lng Tin.

=

Chng 3: M Thng K Ti u.


Ui U1 U2 U3 U4 U5 U6 U7 U8 U9P(u) 0.34 0.2 0.19 0.1 0.07 0.04 0.03 0.02 0.01Bng m nh phn (m=2)theo phng php shannon. Tnh di trung bnh ca t m ntbv tnh kinh t ca t m:p=H(U)/NtbGiiS m ha theo phng php m ha shannon:

Ui P(ui) Pi ni Dng m nh phn ca pi T mU1 0.34 0 2 0 00U2 0.2 0.34 3 0.01010111 010U3 0.19 0.54 3 0.10001010 100U4 0.1 0.73 4 0.10111010 1011U5 0.07 0.83 4 0.11010100 1101U6 0.04 0.9 5 0.11100110 11100U7 0.03 0.94 6 0.11110000 111100U8 0.02 0.97 6 0.11111000 111110U9 0.01 0.99 7 0.11111101 1111110 di trung bnh t m: NTB=

1311/13/2011



Ch s kinh t ca b m: p= =0,8279

Chng 4: M Chng Nhiu : M Khi

Chng 5: M Vng.

Bi tp: cho m vng (n,k)=(7,4) c ma trn sinh l g(x)=x3+x+1. Lit k tt c cc t m ca b m v cho bit b m c kh nng sa sai bao nhiu bt.

Gii

A.cho d= (0001) ta tnh c:

V1=0001000+d s

V1=0001000+011=0001011

Quay vng t mv1 ta s cthm 6 t m na:V2=1000101

V3=1100010

V4=0110001

V5=1011000

V6=0101100

V7=0010110

Cho d=(0011)ta tnh c:V=0011000+d sV=0011000+101=00111101Quay vng t m v ta s cthm 6 t m na:

1411/13/2011


V9 =1001110

V10 =0100111

V11=1010011

V13 =1110100

V14=0111010


V15=1111000+d s

V15=1111000+111=1111111


V0=0000000+d s

B. ta c th tnh r rng trng s hamming ca b m l trng s nh nht ca t m khc khng : H(V)=3 nn b m c kh nng sa c tt cc mu sai 1bt.

__________________________________________________________________





Ui P(ui) Pi ni Dng m nh phn ca pi T mU1 0.34 0 2 0 00U2 0.2 0.34 3 0.01010111 010U3 0.19 0.54 3 0.10001010 100

1511/13/2011


U4 0.1 0.73 4 0.10111010 1011U5 0.07 0.83 4 0.11010100 1101U6 0.04 0.9 5 0.11100110 11100U7 0.03 0.94 6 0.11110000 111100U8 0.02 0.97 6 0.11111000 111110U9 0.01 0.99 7 0.11111101 1111110 di trung bnh t m: NTB=


Ch s kinh t ca b m: p= =0,8279__________________________________________________________________


Bi tp: Ma trn sinh ca b m tuyn tnh (6,3) trn trng GF(2) l:

G=

h. Biu din G di dng chun Gch.

i. Lit k cc t m c c t G v Gch.

j. C bao nhiu t m c trng s hamming l 1,2,3,4,5,6,7.

Gii

h. Biu din G di dng chun Gch.

Gch=

i. Lit k cc t m ca G v Gch.

1611/13/2011



000 000000 0 000000 0001 110110 4 001111 4010 110001 3 010110 3011 000111 3 011001 3100 011010 3 100011 3101 101100 3 101100 3110 101011 4 110101 4111 011101 4 111010 4

j. C hai b m trn u c

1 t m c trng s l 0

0 t m c trng s l 1

0 t m c trng s l 2

4 t m c trng s l 3

3 t m c trng s l 4

0 t m c trng s l 5

0 t m c trng s l 6

__________________________________________________________________

Chng V: M VNG


CHNG 1: LNG TIN.

1. Trong mt tr chi x s vui ngi ta x 10 ch s t 0 n 9.Xc xut trng ca mi s l nh nhau.a. Tnh lng tin ring ca tin : s trng gii l s 9.b. Tnh lng tin tng h gia tin : s trng gii l 9 so

1711/13/2011

(9 / 0 3 6 9)log(0 3 6 9)

pp

=

1/10log log44 /10

= =

10log10 log4 log 1.3224

= = =

6

0( ) ( ) ( )i i

iI U pu I u

=

= 60

1 1log10 10

1 .7.log1010

i==

=


vi tin :s trng gii l s chia ht cho 3.c. Trong 10 tin trn gi U=(u1,u2,u3,u4,u4,u5,u6) vi ui l tin

s i trng gii (i=0,1,,6).Tm lng tin trung bnh ca tp tin U.

Bi lma. Gi p(9) l xc sut s 9 trng gii ta c

p(9) = 1/10= 0.1

vy I(9)= -log p(9)= log 10= 3.322

b. Gi p(0-3-6-9) l xc sut s trng gii chia ht cho 3, ta c:

p(0-3-6-9) = 4/10= 0.4

Mt khc : I (9/0-3-6-9) = - log p(9/0-3-6-9)

Vy I(9; 0-3-6-9) = I(9) I(9/0-3-6-9)

c. Lng tin trung bnh ca tp tin U :

CHNG 2: M THNG K TI

1811/13/2011


Trong knh truyn tin nh phn c nhiu ngun X = {x0,x1} c p(x0) = 0.4 v p(x1) = 0.6. Xc sut truyn tin sai nhm p(y0/x1) = p(y1/x0) = 0.1, xc sut truyn tin ng p(y0/x0) = p(y1/x1) = 0.9, vi Y = {y0,y1}

a/ Tnh H(Y)

b/ Tnh H(Y/X)

c/ Tnh I(X;Y)

Bi lm:

a/ H(Y) = p(y0)I(y0) + p(y1)I(y1)

p dng cng thc ),()(

=

iiyxpxp

Ta c: P(y0) = p(y0,x0) + p(y0,x1)

= p(y0) p(y0/x0) + p(x1) p(y0/x1)

= 0,4.0,9 + 0,6.0,1

= 0,42

P(y1) = p(y1,x0) + p(y1,x1)

= p(x0) p(y1/x0) + p(x1) p(y1/x1)

= 0,4.0,1 + 0,6.0,9

= 0,58

H(Y) = -0,42log0,42 0,58log0,58

=0,8915

b/ =XY

xyIyxPXYH ),(),()/(

= p(x0,y0) I(y0/x0) + p(x0,y1) I(y1/x0) + p(x1,y0) I(y0/x1) + p(x1,y1) I(y1/x1)

M: p(x0,y0) = p(x0) p(y0/x0) = 0,4.0,9 = 0,36

p(x0,y1) = p(x0) p(y1/x0) = 0,4.0,1 = 0,04

1911/13/2011


p(x1,y0) = p(x1) p(y0/x1) = 0,6.0,1 = 0,06

p(x1,y1) = p(x1) p(y1/x1) = 0,6.0,9 = 0,54

I(y0/x0) = -log p(y0/x0) = -log0,9 = log10/9

I(y1/x0) = -log p(y1/x0) = -log0,1 = log10



Nn: H(Y/X) = 0,36log10/9 + 0,04log10 + 0,06log10 + 0,54log10/9

= 0.469

c/ =XY

yxIyxpYXI );(),();(

= p(x0,y0) I(x0;y0) + p(x0,y1) I(x0;y1) + p(x1,y0) I(x1;y0) + p(x1,y1)I(x1;y1)

M: I(x0;y0) = I(y0;x0) = I(y0) - I(y0/x0) = log p(y0/x0)/p(y0) = log0,9/0,42

I(x0;y1) = I(y1;x0) = I(y1) - I(y1/x0) = log p(y1/x0)/ p(y1) = log0,1/0,58



Vy I(X;Y) = 0,36log0,9/0,42 + 0.04log0,1/0,58 + 0,06log0,1/0,42 + 0,54log0,9/0,58

= 0,512

BI TP L THUYT THNG TIN__________________________________________________________________


2011/13/2011







Ch s kinh t ca b m: p= =0,8279__________________________________________________________________


__________________________________________________________________

Chng V: M VNG


Chng II

2111/13/2011


Xem cc con bi ca b bi 52 l to thnh 1 ngun tin ri rc, Tnh Entropy ca 1 l bi rt ngu nhin. Gi s b qua nc ca con bi by gi U={ Ace, 2,3,4,5,6,7,8,9,10,Jack, Queen, King}. Tnh Entropy ca 1 l bi rt ngu nhin trong trng hp ny, trong trng hp U={ bi c hnh, bi ko hnh}

Gii:

H(U)=i=152puiIui vi p(ui)=1/52 ( i= 1,2,3,.52)

=-i=152p(ui)log pui =-52.1/52 log (1/52)= log52 = 5700

b) H(U)=i=113puiIui vi p(ui)=1/13 ( i= 1,2,3,.13)

=-i=113p(ui)Iui =-13.1/13 log (1/13)= log13 = 3700

c) Gi uh l tin bi c hnh th p(uh)=3/13

Gi uoh l tin bi c hnh th p(uoh)=10/13

Vy H(U)= p(uh)I (uh)+ p(uoh)I (uoh)

=- p(uh)log p(uh)- )- p(uoh)log (uoh)

= 313log 313- 1013log1013

=0,779

Chng III

Lp 1 b m cho ngun tin U c s thng k nh sau:

Ui u1 u2 u3 u4 u5 u6 u7 u8 u9

P(ui) .34 .2 .19 .1 .07 .04 .03 .02 .01

Bng m nh phn the phng php shannon. Tnh di trung bnh ca t m ntb v tnh kinh t ca t m p=H(U)/ ntb

Gii:

S m ha theo phng php m ha shannon:

Ui P(ui) Pi ni Dng nh phn ca Pi T M

2211/13/2011


U1

U2

U3

U4

U5

U6

U7

U8

U9

.34

.2

.19

.1

.07

.04

.03

.02

.01

0

.34

.54

.73

.83

.9

.94

.97

.99

2

3

3

4

4

5

6

6

7

0

0,01010111

0,10001010

0,10111010

0,11010100

0,11100110

0,11110000

0,11111000

0,11111101

00

010

100

1011

1101

11100

111100

111110

1111110

di trung bnh t m : ntb= i=19ni pui=3.1

Entropy ca tp tin : H(U)=-i=19puilogui=2,5664

Ch s kinh t ca b m: = H(U)/ ntb=2.5664/3.1=0,8279Chng V

Chng minh rng trong 1 b m vng khi sai E(x) c th d c th mu sai Ei(x) chuyn dch vng I bit so vi E(x) cng c th d c

Gii:

Nu Ei(x) l 1 mu sai ko pht hin c th Ei(x) l 1 t m. T y theo n ca m vng th E(x) cng l 1 t m.Khi v m vng l m tuyn tnh nn 1 t m v(x) bt k sau khi chu tc ng sai ca E(x) s bin thnh 1 t m v(x) v khng pht hin sai c. iu ny mu thun vi gi thit l E(x) l mu sai pht hin c.


2311/13/2011


Bi 8 : chng minh rng

Bi gii :

= H(X) + H(Y/X) (pcm)

Bi 9 : Gi s c mt ngun tin ri rc xut hin p(u) khi u U. Vi mi tin u chn 1 t m c

di l(u) tha :

Chng minh rng di trung bnh

Tha

Bi gii:

2411/13/2011


Theo bi vi mi u ta c:

V p(u)>0 nn bt ng thc trn tng ng vi:

Cng tt c cc bt ng thc ca tng tin u theo v:

Hay

Chng III :M THNG K TI U

BI 1: cho b m c c s m m = 4 nh sau:

U={3,23,11,123,10}

a. V cy m v hnh kt cu ca b m U

b. V mt ta m

c. Lp 1 b m h thng 6 t m c cc t hp s ng. V hnh kt cu ca b m va lp

Bi gii:

a. Cy m :

----------------------------------------------------------------------------------------------------Mc 0

2 1

---------------------------------------------------------------------------------------------------Mc 1

3 1 2 0

--------------------------------------------------------------------------------------------------Mc 2

3

2511/13/2011


---------------------------------------------------------------------------------------------------Mc 3

hnh kt cu :

3

b. Mt ta m:

ui ni pi

3 1 3

23 2 14

11 2 5

123 3 57

10 2 1

c. T hp s ng : 23,123,10

T hp cui : 3,11

B m h thng c 6 t m :u1 =233, u2=1233, u3=101011, u4=2311, u5=12311, u6=102311

Bi 5 : lp 1 b m cho ngun tin U c s thng k nh sau:

ui u1 u2 u3 u4 u5 u6 u7 u8 u9

P(ui) .34 .2 .19 .1 .07 .04 .03 .02 .01

Bng m nh phn m=2, theo phng php Shannon,tnh di trung bnh ca t m ntb v tnh theo kinh t ca t m :

2611/13/2011

0

01

0


Bi gii :


ui P(ui) Pi ni Dng nh phn ca Pi T M

u1 .34 0 2 0 00

u2 .2 .34 3 0.01010111 010

u3 .19 .54 3 0.10001010 100

u4 .1 .73 4 0.1011010 1011

u5 .07 .83 4 0.11010100 11 01

u6 .04 .9 5 0.11100110 11100

u7 .03 .94 6 0.11110000 111100

u8 .02 .97 6 0.11111000 111110

u9 .01 .99 1 0.11111101 1111110

di trung bnh t m :

Etropy ca tp tin:

Ch s kinh t ca b m :

Bi 7 : cho ngun tin U c s thng k nh sau:

ui u1 u2 u3 u14

P(ui) .5 .25 .315 .31 . 0157 .0156

Dung m Huffman kt hp m u m ha ngun tin trn vi c s m m=2

Bi gii:

P(ui) S m ha T m

.5 0

2711/13/2011


.25 10

.315 11000

.31 11001

.31 11010

.31 11011

. 0157 111000

. 0157 111001

. 0157 111010

. 0157 111011

. 0157 111100

. 0157 111101

. 0157 111110

.0156 111111

Chng IV : M CHNG NHIU: M KHI

Bi 3 : ma trn Sinh ca b m tuyn tnh (7,3) trn ng GF (2) l :

a.Biu din G di dng chun Gch

b.Xc nh ma trn th Hch

c. Lit k cc t m c c t G v Gch

d. Xc nh khong cch Hamming ca b m ny

2811/13/2011


bi gii:

a.Ly ct 7 lm ct 1,ct 1 lm ct 3 ta c ma trn Sinh Gch:

b.Ta c :Hch [-PTIn-k] =

c.Lit k cc t m ca G v Gch:

Vector mang tin a

V=a.G w(v) v=a.Gch W(v)

000 0000000 0 0000000 0

001 1011100 4 0011110 4

010 0101110 4 0100111 4

011 1110010 4 0111001 4

100 0010111 4 1001011 4

101 1001011 4 1010101 4

110 0111001 4 1101100 4

2911/13/2011


111 1100101 4 1110010 4

a. C 2 b m trn u c D=4 l trng s Hamming nh nht ca cc t m khc khng ca chng

Chng 5 : M VNG

Bi 1: chng minh rng a thc sinh g(x) ca 1 b m vng l c s ca xn+1

Bi gii:

Gi s

cng l 1 a thc m.Khi :

V

T y :

Hay

Tc l (pcm)

Bi 5 : chng minh rng trong 1 b m vng khi mu sai E(x) c th d c th mu sai E(i)(x) chuyn dch vng i bit so vi E(x) cng c th d c

Bi gii:

Ta chng minh bng phn chng.

Nu E(i)(x) l 1 mu sai khng pht hin c th E(i)(x) l 1 t m. T y theo nh ngha ca m vng th E(x) cng l 1 t m. Khi v m vng l m tuyn tnh nn 1 t m v(x) bt k sau khi chu tc ng sai ca E(x) s bin thnh 1 t m v(x) v khng pht hin sai c. iu nay mu thun vi gi thit l E(x) l mu sai pht hin c.

BI TP L THUYT THNG TIN30

11/13/2011


Bi tp: Xem cc con bi ca b bi 52 l to thnh mt ngun tin ri rc. Tnh entropy ca mt l bi rt ngu nhin. Gi s b qua nc ca con bi by gi U={Ace,2,3,4,5,6,7,8,9,10, Jack, Queen, King}. Tnh entropy ca mt l bi rt ngu nhin trong trng hp ny, trong trng hp U={bi c hnh, bi khng c hnh}

Gii:

Cu a:

H(U)= vi = (i= 1,2,,52)

=

=-52

= 52

=5.7

Cu b:

H(U)= vi = (i= 1,2,,13)

=

=-13

3111/13/2011


= 13

=3.7

Cu c:

Gi =

Gi l tin bi khng c hnh =

Vy: H(U) = +

=

=

= 0.779

CHNG III: M THNG K TI U

Bi tp: Lp mt b m cho ngun tin U c s thng k:

0.34 0.2 0.19 0.1 0.07 0.04 0.03 0.02 0.01

3211/13/2011


Bng m nh phn (m=2)theo phng php Fano. Tnh di trung bnh ca t m ntbv tnh kinh t ca t m: =H(U)

Gii:


T m0.34 0 0 00 20.2 0 1 01 20.19 1 0 0 100 30.1 1 0 1 101 30.07 1 1 0 110 30.04 1 1 1 0 1110 40.03 1 1 1 1 0 11110 50.02 1 1 1 1 1 0 111110 60.01 1 1 1 1 1 1 111111 6

di trung bnh t m: =

Entropy ca tp tin: H(U)= =2.5664

Ch s kinh t ca b m: = = 0.9685

3311/13/2011


CHNG IV: M CHNG NHIU M KHI

Bi tp: Mt khng gian vector c to bi mi t hp tuyn tnh ca tp hp cc vector:

= 1101000

= 0110100

= 0100011

= 1110010

= 1010001

a. Cc vector b c c lp tuyn tnh khng?

b. Tm s chiu v c h ca khng gian vector ny

Gii:

Cu a: Cc vector ph thuc tuyn tnh v:

= 0

3411/13/2011


Cu b:

Gi l h s

Gi s tn ti sao cho:

= 0

1101000)+ (0110100)+ 0100011) + (1110010) = (0000000)

= 0

= 0

= 0

= 0

= 0

= 0

= 0

= = = = 0

Vy c lp tuyn tnh, c th biu din tuyn tnh theo nn khng gian vector c 4 chiu v mt c h ca n l

3511/13/2011


< >.

CHNG V: M VNG

Bi tp:Xt b m vng (7,3) c a thc sinh g(x)= 1+ + + . Cho v =

(0011101) l mt t m ca b m (7,3). Tm tt c cc t m bng cch

dng ng thc sau:

[v(i)T]=S[v(i-1)]T, i= 1,n-1

Vi v(k) l t m th k v:

S =

Gii:

3611/13/2011


Trc ht t : v(1)= (0011101)

[v(2)]T= [v(1)]T

=

=

Vy v(2)= (1001110). Tng t ta xc nh c:

v(3)= (0100111)

3711/13/2011


v(4)= (1010011)

v(5)= (1101001)

v(7)= (1110100)

v(8)= (0111010)

Cui cng b m c t m tm thng:

v(0)= (0000000)

3811/13/2011


3911/13/2011



Chng II: LNG TIN







Ch s kinh t ca b m: p= =0,8279__________________________________________________________________


__________________________________________________________________

4011/13/2011


Chng V: M VNG


Chng 2 lng tin

Chng 3:M thng k ti u:


Ui U1 U2 U3 U4 U5 U6 U7 U8 U9P(u) 0.34 0.2 0.19 0.1 0.07 0.04 0.03 0.02 0.01Bng m nh phn (m=2)theo phng php shannon. Tnh di trung bnh ca t m ntbv tnh kinh t ca t m:p=H(U)/Ntb


Ui P(ui) Pi ni Dng m nh phn ca pi T mU1 0.34 0 2 0 00U2 0.2 0.34 3 0.01010111 010U3 0.19 0.54 3 0.10001010 100U4 0.1 0.73 4 0.10111010 1011U5 0.07 0.83 4 0.11010100 1101U6 0.04 0.9 5 0.11100110 11100U7 0.03 0.94 6 0.11110000 111100U8 0.02 0.97 6 0.11111000 111110U9 0.01 0.99 7 0.11111101 1111110 di trung bnh t m:NTB=

Entropy ca tp tin:H(u)=

Ch s kinh t ca b m:p= =0,8279

4111/13/2011


Chng 4:m chng nhiu:m khi

Bi tp ma trn sinh ca b m tuyn tnh (6,3)c dng chu1n l

a.tm ma trn th hch ca b m.b.lit k tt c cc t m ca b m trn v xc nh khong cch hamming ca b m.c.xc nh hm cu trc trng s m A(z) v xc nh hm s cu tr trng s m ca b m trc giao sinh ra t ma trn HCH lB (z) bng hai phng php:trc tip v gian tip qua A(z).d.lp bng sp chunca b m ny vi ch rng bng c cha mu sai 2 bt v tnh syndrome ca tt c cc mu sai c th sa sai c.e,thit lp mch m ha,mch tinhsyndrome v mch sa sai ca b m.

GII

a Hch=(-pT,IN-K)=

HTch=

b lit k cc t m:

Vector mang tin a V= aGch W(v)000 000 000 0001 001 010 2010 010 111 4011 011 101 4100 100 100 2101 101 110 4110 110 011 4111 111 001 4

Vy khong cch hamming ca b m D=2 l trng s hamming nh nht ca cc t m khc khng.

C: Da vo bng lit k ta c:A(z)=1+2z2=5z4Xc nh trc tip b(z):Ta lit k cc tu8 m ca5 b m sinh ra bi hch

Vector mang tin a V= aGch W(v)000 000 000 0001 010 001 2010 011 010 3011 001 011 3

4211/13/2011


100 110 100 3101 100 101 3110 101 110 4111 111 111 6

Vy B(z)=1+z2+4z3+z4+z6

Chng 5:m vng:

Bi tp:cho m vng (n,k)=(7,4)c ma trn sinh l g(x)=x3+x+1. Lit k tt c cc t m ca b m v cho bit b m c kh nng sa sai bao nhiu bt.

GIIA.cho d= (0001) tatnh c:V1=0001000+d sV1=0001000+011=0001011Quay vng t mv1 ta s cthm 6 t m na:V2=1000101

V3=1100010

V4=0110001

V5=1011000

V6=0101100

V7=0010110

Cho d=(0011)ta tnh c:V=0011000+d sV=0011000+101=00111101Quay vng t m v ta s cthm 6 t m na:V9 =1001110

V10 =0100111

V11=1010011

V13 =1110100

V14=0111010


V15=1111000+d s

V15=1111000+111=1111111


4311/13/2011


V0=0000000+d s

b,ta c th tnh r rng trng s hamming ca b m l trng s nh nht ca t m khc khng : H(V)=3 nn b m c kh nng sa c tt cc mu sai 1bt



Chng 5: M Vons

a thc sinh ca b am vng hamming l:

g(x)=1+x3+x4

g. Tm a thc th h(x) ca b am ny.

h. Thit k mch am ha thc hin qua h(x).

i. Xc nh cc bit th v t am nhn c tng ng vi chui bit mang tin d=(10000001011).

Gii

e. B am hamming c (n,k)=(2n-k-1,k) nn:

4411/13/2011


n=2n-k-1=24-1=15.

T cng thc: h(x)=(xn+1)/g(x)

Ta tnh c: h(x)=

f. Khi d=(10000001011) hay d(x)= th:

v(x)=d(x).xn-k+ d s ca

= + s d ca

=( )

Trong x3+1 tng ng vi chui bit 1001 l cc bit th v t am nhn c tng ng vi v(x)= ( ) l v=(100000101110010).

BI TP MN L THUYT THNG TIN

CHNG 1: LNG TIN


Chng 3: M Thng K Ti u

Lp mt b m cho ngun tin U c s thng k nh sau:

ui u1 u2 u3 u4 u5 u6 u7 u8 u9P(ui) 0.34 0.2 0.19 0.1 0.07 0.04 0.03 0.02 0.01Bng m nh phn (m=2) theo phng php shannon. Tnh di trung bnh ca t m ntbv tnh kinh t ca t m: p=H(U)/ntb

Gii

4511/13/2011


S m ha theo phng php m ha shannon:ui p(ui) Pi ni Dng m nh phn ca Pi T mu1 0.34 0 2 0 00u2 0.2 0.34 3 0.01010111 010u3 0.19 0.54 3 0.10001010 100u4 0.1 0.73 4 0.10111010 1011u5 0.07 0.83 4 0.11010100 1101u6 0.04 0.9 5 0.11100110 11100u7 0.03 0.94 6 0.11110000 111100u8 0.02 0.97 6 0.11111000 111110u9 0.01 0.99 7 0.11111101 1111110 Chn ni tha iu kin 2-ni I(a) = - log(0.125)/log(2)

4611/13/2011


= 3

I(a;b) = I (a) I(a/b)

= -log p(u3) + log p( u3/ u3,u4,u5,u6)

= 3 + log 2 0.125/0.125*4

>> I= 3+log(0.125/4/0.125)/log(2)

= 1

I(a;c) = I (a) I(a/c)

= -log p(u3) + log p( u3/ u3,u4)

= 3 + log 2 0.125/0.125*2

>> I= 3+log(0.125/2/0.125)/log(2)

= 2

Chng 2 : M ha ngun tin

Bi 5 / tr


ui U1 U2 U3 U4 U5 U6 U7 U8 U9

P(ui) 0.34 0.2 0.19 0.1 0.07 0.04 0.03 0.02 0.01

Bng m nh phn (m=2) theo phng php Fano . Tnh di trung bnh ca t m ntb v tnh kinh t ca t m p=H(u)/ ntb

Gii

4711/13/2011


di trung bnh ca t m ntb l:

ntb = ni pi

= (0.34*2) + (0.2*2) + (0.19*3) + (0.1*3) + (0.07*3) +(0.04*4) +(0.03*5) +(0.02*6) + (0.01*6)

= 2.65

H(u) = - pi log2 pi

= - [ (0.34*log2 0.34)+(0.2*log2 0.2)+(0.19*log2 0.19)+(0.1*log2 0.1)+(0.07*log2 0.07)+(0.04*log2 0.04)+(0.03*log2 0.03)+(0.02*log2 0.03)+(0.01*log2 0.01) ]

H(u) = 2.5664

Ch s kinh t ca b m : p = H(u) / ntb = 2.5664 / 2.65 = 0.9684

Chng 3 : M ha knh truyn ( m khi Hamming)

Bi 2 / tr

Cho ma trn sinh ca b m tuyn tnh C (6,3) trn trng Galois

G = 0 1 1 0 1 0

4811/13/2011

Ui P(ui) Ln 1 Ln 2 Ln 3 Ln 4 Ln 5 Ln 6 ni T m

U1 0.34 0 0 2 00

U2 0.2 0 1 2 01

U3 0.19 1 0 0 3 100

U4 0.1 1 0 1 3 101

U5 0.07 1 1 0 3 110

U6 0.04 1 1 1 0 4 1110

U7 0.03 1 1 1 1 0 5 11110

U8 0.02 1 1 1 1 1 0 6 111110

U9 0.01 1 1 1 1 1 1 6 111111


1 1 0 0 0 1

1 1 0 1 1 0

Tnh Gch

C bao nhiu t m c trng s Hamming 0,1,2,3,4,5,6,7

Gii

Ly ct 3 lm ct 1, ct 6 lm ct 2 v ct 4 lm ct 3 ta c ma trn sinh Gch nh sau :

Gch = 1 0 0 0 1 1

0 1 0 1 1 0

0 0 1 1 1 1

Lit k cc t m c c t ma trn sinh G v ma trn sinh chun tc Gch

Vector mang tin a

V=aG W(v) V = a Gch W(v)

000 000000 0 000000 0

001 110110 4 001111 4

010 110001 3 010110 3

011 000111 3 011001 3

100 011010 3 100011 3

101 101100 3 101100 3

110 101011 4 110101 4

111 011101 4 111010 4

Vy b m c

1 t m c trng s l 0

0 t m c trng s l 1

0 t m c trng s l 2

4 t m c trng s l 3

4911/13/2011


3 t m c trng s l 4

0 t m c trng s l 5

0 t m c trng s l 6

0 t m c trng s l 7

Chng 3 : M vng

Bi 2 / tr

Cho b m vng (7,4) c a thc sinh l :

G(x) = x3+x2+1

a. Tm a thc th h(x) ca b m ny

b. Gi s nhn c t hp u(x) = x6+x5+x2+1 . Xc nh a thc syndrome ca u(x)

Gii

a.Ta c cng thc h(x) = xn+1 / g(x) v ta tnh c

h(x) = x7+1 / x3+x2+1 = x4+ x3+x2+1

vy h(x) = x4+ x3+x2+1

b.Gi s nhn c a thc u(x) = x6+x5+x2+1 , khi a thc syndrome ca n :

Su(x) = s d ca ( u(x) / g(x) ) = s d ca (x6+x5+x2+1 / x3+x2+1) = 0

Vy a thc syndrome ca u(x) l su(x) = 0

5011/13/2011



Chng 1 : Tin v lng tin

Bi 2.9 / tr 36

Cho tp tin U={ ui } vi ui=xiyizi ( i = 1,2,3,4,5)

i 1 2 3 4 5 6Ui 010 011 100 101 110 111P(ui) 0.25 0.25 0.125 0.125 0.125 0.125

Gi tin a l tin ui= u3 = 100, gi tin b l tin xi=1 ,gi tin c l tin yi=0 . Tnh I(a) , I(a;b) , I(a;c)

Gii

I(a) = - log 2 0.125

>> I(a) = - log(0.125)/log(2)

= 3

I(a;b) = I (a) I(a/b)

5111/13/2011


= -log p(u3) + log p( u3/ u3,u4,u5,u6)

= 3 + log 2 0.125/0.125*4

>> I= 3+log(0.125/4/0.125)/log(2)

= 1

I(a;c) = I (a) I(a/c)

= -log p(u3) + log p( u3/ u3,u4)

= 3 + log 2 0.125/0.125*2

>> I= 3+log(0.125/2/0.125)/log(2)

= 2

Chng 2 : M ha ngun tin

Bi 5 / tr


ui U1 U2 U3 U4 U5 U6 U7 U8 U9

P(ui) 0.34 0.2 0.19 0.1 0.07 0.04 0.03 0.02 0.01

Bng m nh phn (m=2) theo phng php Fano . Tnh di trung bnh ca t m ntb v tnh kinh t ca t m p=H(u)/ ntb

Gii

5211/13/2011


di trung bnh ca t m ntb l:

ntb = ni pi

= (0.34*2) + (0.2*2) + (0.19*3) + (0.1*3) + (0.07*3) +(0.04*4) +(0.03*5) +(0.02*6) + (0.01*6)

= 2.65

H(u) = - pi log2 pi

= - [ (0.34*log2 0.34)+(0.2*log2 0.2)+(0.19*log2 0.19)+(0.1*log2 0.1)+(0.07*log2 0.07)+(0.04*log2 0.04)+(0.03*log2 0.03)+(0.02*log2 0.03)+(0.01*log2 0.01) ]

H(u) = 2.5664

Ch s kinh t ca b m : p = H(u) / ntb = 2.5664 / 2.65 = 0.9684

Chng 3 : M ha knh truyn ( m khi Hamming)

Bi 2 / tr

Cho ma trn sinh ca b m tuyn tnh C (6,3) trn trng Galois

G = 0 1 1 0 1 0

5311/13/2011

Ui P(ui) Ln 1 Ln 2 Ln 3 Ln 4 Ln 5 Ln 6 ni T m

U1 0.34 0 0 2 00

U2 0.2 0 1 2 01

U3 0.19 1 0 0 3 100

U4 0.1 1 0 1 3 101

U5 0.07 1 1 0 3 110

U6 0.04 1 1 1 0 4 1110

U7 0.03 1 1 1 1 0 5 11110

U8 0.02 1 1 1 1 1 0 6 111110

U9 0.01 1 1 1 1 1 1 6 111111


1 1 0 0 0 1

1 1 0 1 1 0

Tnh Gch

C bao nhiu t m c trng s Hamming 0,1,2,3,4,5,6,7

Gii

Ly ct 3 lm ct 1, ct 6 lm ct 2 v ct 4 lm ct 3 ta c ma trn sinh Gch nh sau :

Gch = 1 0 0 0 1 1

0 1 0 1 1 0

0 0 1 1 1 1

Lit k cc t m c c t ma trn sinh G v ma trn sinh chun tc Gch

Vector mang tin a

V=aG W(v) V = a Gch W(v)

000 000000 0 000000 0

001 110110 4 001111 4

010 110001 3 010110 3

011 000111 3 011001 3

100 011010 3 100011 3

101 101100 3 101100 3

110 101011 4 110101 4

111 011101 4 111010 4

Vy b m c

1 t m c trng s l 0

0 t m c trng s l 1

0 t m c trng s l 2

4 t m c trng s l 3

5411/13/2011


3 t m c trng s l 4

0 t m c trng s l 5

0 t m c trng s l 6

0 t m c trng s l 7

Chng 3 : M vng

Bi 2 / tr

Cho b m vng (7,4) c a thc sinh l :

G(x) = x3+x2+1

c. Tm a thc th h(x) ca b m ny

d. Gi s nhn c t hp u(x) = x6+x5+x2+1 . Xc nh a thc syndrome ca u(x)

Gii

a.Ta c cng thc h(x) = xn+1 / g(x) v ta tnh c

h(x) = x7+1 / x3+x2+1 = x4+ x3+x2+1

vy h(x) = x4+ x3+x2+1

b.Gi s nhn c a thc u(x) = x6+x5+x2+1 , khi a thc syndrome ca n :

Su(x) = s d ca ( u(x) / g(x) ) = s d ca (x6+x5+x2+1 / x3+x2+1) = 0

Vy a thc syndrome ca u(x) l su(x) = 0

5511/13/2011



Chng II: LNG TIN

Bi tp: Gi s c mt ngun tin ri rc vi xc sut xut hin p khi u thuc U, vi mi tin u, chn mt t m c d di 1(u) tha:

Log(1/p(u))


__________________________________________________________________







Ch s kinh t ca b m: p= =0,8279__________________________________________________________________


__________________________________________________________________

Chng V: M VNG

5711/13/2011


Chng II : LNG TIN

Bi 8 : Chng minh rng

Bi gii :

= H(X) + H(Y/X) (pcm)

CHNG III: M THNG K TI U

Bi tp: Lp mt b m cho ngun tin U c s thng k:

0.34 0.2 0.19 0.1 0.07 0.04 0.03 0.02 0.01

Bng m nh phn (m=2)theo phng php Fano. Tnh di trung bnh ca t m ntbv tnh kinh t ca t m: =H(U)

Gii:

5811/13/2011



T m0.34 0 0 00 20.2 0 1 01 20.19 1 0 0 100 30.1 1 0 1 101 30.07 1 1 0 110 30.04 1 1 1 0 1110 40.03 1 1 1 1 0 11110 50.02 1 1 1 1 1 0 111110 60.01 1 1 1 1 1 1 111111 6

di trung bnh t m: =

Entropy ca tp tin:

H(U)= =2.5664

Ch s kinh t ca b m: = = 0.9685

CHNG IV : M CHNG NHIU: M KHI

CHNG V: M VNG59

11/13/2011


Bi tp:Xt b m vng (7,3) c a thc sinh g(x)= 1+ + + . Cho v =

(0011101) l mt t m ca b m (7,3). Tm tt c cc t m bng cch

dng ng thc sau:

[v(i)T]=S[v(i-1)]T, i= 1,n-1

Vi v(k) l t m th k v:

S =

Gii:

Trc ht t : v(1)= (0011101)

[v(2)]T= [v(1)]T

6011/13/2011


=

=

Vy v(2)= (1001110). Tng t ta xc nh c:

v(3)= (0100111)

v(4)= (1010011)

v(5)= (1101001)

v(7)= (1110100)

v(8)= (0111010)

Cui cng b m c t m tm thng:

v(0)= (0000000)


CHNG 1: LNG TIN

CHNG 3: M KHI

6111/13/2011


Chng II : LNG TIN


Bi gii :

= H(X) + H(Y/X) (pcm)



ui u1 u2 u3 u4 u5 u6 u7 u8 u9

P(ui) .34 .2 .19 .1 .07 .04 .03 .02 .01


Bi gii :

6211/13/2011




u1 .34 0 2 0 00

u2 .2 .34 3 0.01010111 010

u3 .19 .54 3 0.10001010 100

u4 .1 .73 4 0.1011010 1011

u5 .07 .83 4 0.11010100 11 01

u6 .04 .9 5 0.11100110 11100

u7 .03 .94 6 0.11110000 111100

u8 .02 .97 6 0.11111000 111110

u9 .01 .99 1 0.11111101 1111110

di trung bnh t m :

Etropy ca tp tin:








6311/13/2011


bi gii:




Vector mang tin a


000 0000000 0 0000000 0

001 1011100 4 0011110 4

010 0101110 4 0100111 4

011 1110010 4 0111001 4

100 0010111 4 1001011 4

101 1001011 4 1010101 4

110 0111001 4 1101100 4

111 1100101 4 1110010 4

b. C 2 b m trn u c D=4 l trng s Hamming nh nht ca cc t m khc khng ca chng

6411/13/2011


Chng 5 : M VNG


Bi gii:



Chng II: LNG TIN

Bi 6: Trong knh truyn tin nh phn c nhiu ngun X =

{x0,x1} c p(x0) = 0.4 v p(x1) = 0.6. Xc sut truyn tin sai nhm p(y0/x1) = p(y1/x0) = 0.1, xc sut truyn tin ng p(y0/x0) = p(y1/x1) = 0.9, vi Y = {y0,y1}

a/ Tnh H(Y)

b/ Tnh H(Y/X)

c/ Tnh I(X;Y)

Bi lm:

a/ H(Y) = p(y0)I(y0) + p(y1)I(y1)

p dng cng thc ),()(

=

iiyxpxp

Ta c: P(y0) = p(y0,x0) + p(y0,x1)

= p(y0) p(y0/x0) + p(x1) p(y0/x1)

= 0,4.0,9 + 0,6.0,1

6511/13/2011


= 0,42

P(y1) = p(y1,x0) + p(y1,x1)

= p(x0) p(y1/x0) + p(x1) p(y1/x1)

= 0,4.0,1 + 0,6.0,9

= 0,58

H(Y) = -0,42log0,42 0,58log0,58

=0,8915

b/ =XY

xyIyxPXYH ),(),()/(

= p(x0,y0) I(y0/x0) + p(x0,y1) I(y1/x0) + p(x1,y0) I(y0/x1) + p(x1,y1) I(y1/x1)

M: p(x0,y0) = p(x0) p(y0/x0) = 0,4.0,9 = 0,36

p(x0,y1) = p(x0) p(y1/x0) = 0,4.0,1 = 0,04

p(x1,y0) = p(x1) p(y0/x1) = 0,6.0,1 = 0,06

p(x1,y1) = p(x1) p(y1/x1) = 0,6.0,9 = 0,54





Nn: H(Y/X) = 0,36log10/9 + 0,04log10 + 0,06log10 + 0,54log10/9

= 0.469

c/ =XY

yxIyxpYXI );(),();(

= p(x0,y0) I(x0;y0) + p(x0,y1) I(x0;y1) + p(x1,y0) I(x1;y0) + p(x1,y1)I(x1;y1)

M: I(x0;y0) = I(y0;x0) = I(y0) - I(y0/x0) = log p(y0/x0)/p(y0) = log0,9/0,42


6611/13/2011




Vy I(X;Y) = 0,36log0,9/0,42 + 0.04log0,1/0,58 + 0,06log0,1/0,42 + 0,54log0,9/0,58

= 0,512


Bi 5: Lp mt b m cho ngun tin U c s thng k nh sau:

ui u1 u2 u3 u4 u5 u6 u7 u8 u9

P(ui) .34 .2 .19 .1 .07 .04 .03 .02 .01

Bng m nh phn(m = 2)theo phng php Shannon. Tnh di trung binh2cua3 t m ntbv tnh kinh t ca t m.

Bi lm:

S m ha theo phng php m ha Shannon:

ui P(ui) Pi ni Dng nh phn ca pi T m

u1 .34 0 2 0 00

u2 .2 .34 3 0,01010111 010

u3 .19 .54 3 0,10001010 100

u4 .1 .73 4 0,10111010 1011

6711/13/2011


u5 .07 .83 4 0,11010100 1101

u6 .04 .9 5 0,11100110 11100

u7 .03 .94 6 0,11110000 111100

u8 .02 .97 6 0,11111000 111110

u9 .01 .99 7 0,11111101 1111110

di trung bnh t m: 1,3)(9

1

== =

ii

itb upnn

Entropy ca tp tin: 5664,2)(log)()( 29

1

== =

ii

i upupUH

Ch s kinh t ca b m: 8279,01,35644,2)(

===

tbnUH


Bi 2: Ma trn ca b m tuyn tnh (6,3) trn trng GF (2) c cho bi:

=

1 1 0 1 1 01 1 0 0 0 10 1 1 0 1 0

G

a/ Hy biu din G di dng chunGch

6811/13/2011


b/ Lit k cc t m c c t 2 ma trn sinh G v Gch

c/ Co1bao nhiu t m c trng s hamming l 0,1,2,3,4,5,6?

Bi lm:

a/ Ly ct 3 lm ct 1, ct 6 lm ct 2 v ct 4 lm ct 3 ta c ma trn sinh Gch

=

0 0 1 1 1 10 1 0 1 1 01 0 0 0 1 1

c hG

b/ Lit k cc t m c c t ma trn sinh G v ma trn sinh Gch

Vector mang tin a V = aG W(v) V = aGch W(v)

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 1 1 1 0 1 1 0 4 0 0 1 1 1 1 4

0 1 0 1 1 0 0 0 1 3 0 1 0 1 1 0 3

0 1 1 0 0 0 1 1 1 3 0 1 1 0 0 1 3

1 0 0 0 1 1 0 1 0 3 0 1 1 0 0 1 3

1 0 1 1 0 1 1 0 0 3 1 0 1 1 0 0 3

1 1 0 1 0 1 0 1 1 4 1 1 0 1 0 1 4

1 1 1 0 1 1 1 0 1 4 1 1 1 0 1 0 4

c/ C 2 b m u c:

1 t m c trng s l 0

0 t m c trng s l 1

6911/13/2011


0 t m c trng s l 2

4 t m c trng s l 3

3 t m c trng s l 4

0 t m c trng s l 5

0 t m c trng s l 6

7011/13/2011


Chng V: M VNG

Bi 2: Chng minh rng trong 1 b m vng khi mu sai E(x) c th d c th mu sai E(i)(x) chuyn dch vng I bit so vi E(x) cng c th d c.

Bi lm:

Chng minh bng phn chng.

Nu E(i)(x)l mt mu sai khng pht hin c th E(i)(x) l mt t m. T y theo nh ngha ca m vng th E(x) cng l mt t m. Khi v m vng l m tuyn tnh nn mt t m v(x) bt k sau khi chu tc ng sai ca E(x) s bin thnh mt t m v(x) v khng pht hin sai c. iu ny mu thun vi gi thit l E(x) l mu sai pht hin c.


CHNG 1: LNG TIN

CHNG 3: M KHI

=

7111/13/2011


Chng II : LNG TIN

BI 2: Vi ngun tin ri rc gm 2 k t U = {a1,a2} v p(a1) = p

Chng minh rng H(U) = H(p) = p log (1/p) + log 1/(1-p) ln nht khi p=

Bi gii:

Xt hm s : f(x) = log = - log

C f( ) = -log e (ln ) = -log e (ln )

f ( ) =

trong khong (0,1] c f( .

Vy vi a,b

Du ng thc xy ra khi v ch khi a = b

Cho a = p v b = 1-p ta c :

=

Vy H(U) =


7211/13/2011


Bi gii :

= H(X) + H(Y/X) (pcm)


di l(u) tha :


Tha

Bi gii:


7311/13/2011




Hay



U={3,23,11,123,10}

d. V cy m v hnh kt cu ca b m U

e. V mt ta m

f. Lp 1 b m h thng 6 t m c cc t hp s ng. V hnh kt cu ca b m va lp

Bi gii:

d. Cy m :

----------------------------------------------------------------------------------------------------Mc 0

2 1

---------------------------------------------------------------------------------------------------Mc 1

3 1 2 0

--------------------------------------------------------------------------------------------------Mc 2

3

---------------------------------------------------------------------------------------------------Mc 3

7411/13/2011


hnh kt cu :

3

e. Mt ta m:

ui ni pi

3 1 3

23 2 14

11 2 5

123 3 57

10 2 1

f. T hp s ng : 23,123,10

T hp cui : 3,11



ui u1 u2 u3 u4 u5 u6 u7 u8 u9

P(ui) .34 .2 .19 .1 .07 .04 .03 .02 .01


Bi gii :

7511/13/2011

0

01

0




u1 .34 0 2 0 00

u2 .2 .34 3 0.01010111 010

u3 .19 .54 3 0.10001010 100

u4 .1 .73 4 0.1011010 1011

u5 .07 .83 4 0.11010100 11 01

u6 .04 .9 5 0.11100110 11100

u7 .03 .94 6 0.11110000 111100

u8 .02 .97 6 0.11111000 111110

u9 .01 .99 1 0.11111101 1111110

di trung bnh t m :

Etropy ca tp tin:



ui u1 u2 u3 u14

P(ui) .5 .25 .315 .31 . 0157 .0156


Bi gii:

P(ui) S m ha T m

.5 0

.25 10

7611/13/2011


.315 11000

.31 11001

.31 11010

.31 11011

. 0157 111000

. 0157 111001

. 0157 111010

. 0157 111011

. 0157 111100

. 0157 111101

. 0157 111110

.0156 111111







bi gii:

7711/13/2011





Vector mang tin a


000 0000000 0 0000000 0

001 1011100 4 0011110 4

010 0101110 4 0100111 4

011 1110010 4 0111001 4

100 0010111 4 1001011 4

101 1001011 4 1010101 4

110 0111001 4 1101100 4

111 1100101 4 1110010 4

7811/13/2011


c. C 2 b m trn u c D=4 l trng s Hamming nh nht ca cc t m khc khng ca chng

Chng 5 : M VNG


Bi gii:

Gi s


V

T y :

Hay

Tc l (pcm)


Bi gii:



BI TP L THUYT THNG TINChng III

7911/13/2011



Ui u1 u2 u3 u4 u5 u6 u7 u8 u9

P(ui) .34 .2 .19 .1 .07 .04 .03 .02 .01

Bng m nh phn the phng php shannon. Tnh di trung bnh ca t m ntb v tnh kinh t ca t m p=H(U)/ntb

Gii:


Ui P(ui)

Pi ni Dng nh phn ca Pi

T M

U1

U2

U3

U4

U5

U6

U7

U8

U9

.34

.2

.19

.1

.07

.04

.03

.02

.01

0

.34

.54

.73

.83

.9

.94

.97

.99

2

3

3

4

4

5

6

6

7

0

0,01010111

0,10001010

0,10111010

0,11010100

0,11100110

0,11110000

0,11111000

0,11111101

00

010

100

1011

1101

11100

111100

111110

1111110

di trung bnh t m :ntb= =3.1

Entropy ca tp tin : H(U)= =2,5664

Ch s kinh t ca b m: = H(U)/ntb=2.5664/3.1=0,8279Chng IV

8011/13/2011


Chng V

Chng minh rng trong 1 b m vng khi sai E(x) c th d c th mu sai Ei(x) chuyn dch vng I bit so vi E(x) cng c th d c

Gii:

Nu Ei(x) l 1 mu sai ko pht hin c th Ei(x) l 1 t m. T y theo n ca m vng th E(x) cng l 1 t m.Khi v m vng l m tuyn tnh nn 1 t m v(x) bt k sau khi chu tc ng sai ca E(x) s bin thnh 1 t m v(x) v khng pht hin sai c. iu ny mu thun vi gi thit l E(x) l mu sai pht hin c.

Chng II : LNG TIN

BI 2:

Vi ngun tin ri rc gm 2 k t U = {a1,a2} v p(a1) = p


Bi gii:


C f( ) = -log e (ln ) = -log e (ln )

f ( ) =


Vy vi a,b

8111/13/2011




=

Vy H(U) =


BI 1:

cho b m c c s m m = 4 nh sau:

U={3,23,11,123,10}

g. V cy m v hnh kt cu ca b m U

h. V mt ta m

i. Lp 1 b m h thng 6 t m c cc t hp s ng. V hnh kt cu ca b m va lp

Bi gii:

g. Cy m :

----------------------------------------------------------------------------------------------------Mc 0

2 1

---------------------------------------------------------------------------------------------------Mc 1

3 1 2 0

--------------------------------------------------------------------------------------------------Mc 2

8211/13/2011


3

---------------------------------------------------------------------------------------------------Mc 3

hnh kt cu :

3

h. Mt ta m:

ui ni pi

3 1 3

23 2 14

11 2 5

123 3 57

10 2 1

i. T hp s ng : 23,123,10

T hp cui : 3,11



Chng V : M VNG


Bi gii:

Gi s


8311/13/2011

0

01

0


V

T y :

Hay

Tc l (pcm)



Chng II: LNG TIN





Ui P(ui) Pi ni Dng m nh phn ca pi T mU1 0.34 0 2 0 00U2 0.2 0.34 3 0.01010111 010U3 0.19 0.54 3 0.10001010 100U4 0.1 0.73 4 0.10111010 1011U5 0.07 0.83 4 0.11010100 1101U6 0.04 0.9 5 0.11100110 11100

8411/13/2011


U7 0.03 0.94 6 0.11110000 111100U8 0.02 0.97 6 0.11111000 111110U9 0.01 0.99 7 0.11111101 1111110 di trung bnh t m: NTB=


Ch s kinh t ca b m: p= =0,8279__________________________________________________________________


__________________________________________________________________

Chng V: M VNG

BI TP L THUYT THNG TINChng II

Chng III


Ui u1 u2 u3 u4 u5 u6 u7 u8 u9

P(ui) .34 .2 .19 .1 .07 .04 .03 .02 .01

Bng m nh phn the phng php shannon. Tnh di trung bnh ca t m ntb v tnh kinh t ca t m p=H(U)/ntb

Gii:


Ui P(u Pi ni Dng nh phn T M

8511/13/2011


i) ca Pi

U1

U2

U3

U4

U5

U6

U7

U8

U9

.34

.2

.19

.1

.07

.04

.03

.02

.01

0

.34

.54

.73

.83

.9

.94

.97

.99

2

3

3

4

4

5

6

6

7

0

0,01010111

0,10001010

0,10111010

0,11010100

0,11100110

0,11110000

0,11111000

0,11111101

00

010

100

1011

1101

11100

111100

111110

1111110

di trung bnh t m :ntb= =3.1

Entropy ca tp tin : H(U)= =2,5664

Ch s kinh t ca b m: = H(U)/ntb=2.5664/3.1=0,8279

BI TP L THUYT THNG TIN-----------------o0o-----------------

CHNG I:

H THNG THNG TIN

8611/13/2011


CHNG II:

LNG TIN

BI 6: Trong knh truyn tin nh phn c nhiu ngun X={x0,x1} c p(x0)=0.4 v p(x1)=0.6. Xc sut truyn tin sai nhm p(y0, x1)=p(y1, x0)=0.1, Xc sut truyn tin ng p(y0, x0)= p(y1, x1)=0.9 vi Y= (y0, y1).

a.Tnh H(Y).

b.Tnh H(Y/X).

c.Tnh I(X;Y).

Gii.a. H(Y) = p(y0)I(y0) + p(y1)I(y1)

p dng cng thc :

Ta c p(y0)= p(y0, x0)+ p(y0, x1) = p( x0) p(y0 /x0)+ p( x1) p(y0 /x1)=0.4*0.9+0.6*0.1=0.42

p(y1)= p(y1, x0)+ p(y1, x1)= p( x0) p(y1 /x0)+ p( x1) p(y1 /x1)=0.4*0.1+0.6*0.9=0.58

H(Y)=-0.42log0.42-0.58log0.58=0.8915

8711/13/2011


b. H(Y/X) =

= p(x0, y0) I(y0 / x0) + p(x0, y1) I(y1 / x0)+ p(x1, y0) I(y0 / x1)+ p(x1, y1) I(y1 /x1)

M

p(x0, y0)= p(x0 ) p(y0 /x0)=0.4*0.9=0.36

p(x0, y1)= p(x0 ) p(y1 / x0)=0.4*0.1=0.04

p(x1, y0)= p(x1 ) p(y0 /x1)=0.6*0.1=0.06

p(x1, y1)= p(x1 ) p(y1 / x1)=0.6*0.9=0.54

I(y0 /x0)= - log p(y0 /x0)= - log 0.9 = log (10/9)

I(y1 /x0)= - log p(y1 /x0)= - log 0.1 = log 10

I(y0 /x1)= - log p(y0 /x1)= - log 0.1 = log 10

I(y01/x1)= - log p(y1 /x1)= - log 0.9 = log (10/9)

Nn H(Y/X) = 0.36log (10/9) + 0.04*log10 + 0.06*log10 + 0.54*log(10/9) = 0.469

c. I(X/Y) =

= p(x0, y0) I(x0 ;y0) + p(x0, y1) I(x0 ;y1) + p(x1, y0) I(x1 ;y0) + p(x1, y1) I(x1;y1)

M

I(x0 ;y0) = I(y0 ;x0)= I(y0)- I(y0 ;x0) = log (p(y0 ;x0)/ p(y0))=log(0.9/0.42)



I(x1;y1) = I(y1 ;x1)= I(y1)- I(y1 ;x1) = log (p(y1 ;x1)/ p(y1))=log(0.9/0.58)

8811/13/2011


Vy I(X;Y) = 0.36*log(0.9/0.42) + 0.04*log(0.1/0.58) + 0.06*log(0.1/042) + 0.54*log(0.9/0.58) = 0.512

CHNG III:

M THNG K TI U

Cu 2: Vi b m gc {00,01,100,1010,10110} phn thnh 3 t hp s ng 00,01,100 v hai t hp cui 1010 v 10110 thnh lp m h thng c tnh prefix. Hy xc nh G(nj) vi nj = 1,2,3,,9.

Gii:Theo bi:

n1=2, n2=2, n3=3, i=3

1 =4, 2 =4, k=2.

p dng cng thc:

G(nj)= g(nj- 1) + g(nj- 2)+..+ g(nj- 1)+ g(nj- k)

Ta c: G(nj)= g(nj- 4) + g(nj- 5) (1)

Mc khc: g(nj)= g(nj- n1) + g(nj- n2) + g(nj- n3)

Hay g(nj)= g(nj- 2) + g(nj- 2) + g(nj- 3)

8911/13/2011


g(nj)= 2*g(nj- 2) + g(nj- 3) (2)

Vy

g(1)=2g(-1)+g(-2)=0

g(2)=2g(0)+g(-1)=2

g(3)=2g(1)+g(0)=1

g(4)=2g(2)+g(1)=4

g(5)=2g(3)+g(2)=4

T y theo (1):

G(1)= g(-3) + g(-4)=0

G(2)= g(-2) + g(-3)=0

G(3)= g(-1) + g(-2)=0

G(4)= g(0) + g(-1)=1

G(5)= g(1) + g(0)=1

G(6)= g(2) + g(1)=2

G(7)= g(3) + g(2)=3

G(8)= g(4) + g(3)=5

G(9)= g(5) + g(4)=8

9011/13/2011


CHNG IV:

M CHNG NHIU (M KHI)

Cu 1: Mt khng gian vect c to bi mi t hp tuyn tnh ca tp hp cc vect sau:

b0=1101000

b1=0110100

b2=0100011

b3=1110010

b4=1010001

a. Cc vct b trn c c lp tuyn tnh hay khng?

b. Tm s chiu v c h ca khng gian vect ny.

Gii:

1. a. Cc vect b0, b1, b2, b3, b4 ph thuc tuyn tnh v d thy:

9111/13/2011


0 b0 +0b1 + b2+b3+b4=0

b.By gi gi s tn ti 0 ,1, 2, 3 sao cho:

0b0 + 1b1+ 2b2+ 3b3=0

0(1101000)+ 1(0110100)+ 2(0100011)+ 3(1110010)=(0000000)

0+ 3=0

0+ 1 + 2+ 3=0

0+ 3=0

0=0

1=0

2+ 3=0

2=0

0= 1= 2= 3=0

Vy b0, b1, b2 ,b3 c lp tuyn tnh, b4 c th biu din tuyn tnh theo b0, b1, b2, b3 nn khng gian vect c 4 chiu di v mt c h ca n l < b0, b1,b2, b3>.

9211/13/2011


9311/13/2011


CHNG V:

M VNG

Cu 8: Cho b m vng (n,k)=(7,4) c ma trn sinh l g(x)=x3+x+1. Lit k tt c cc t m ca b m v cho bit b m c kh nng sa c cc mu sai bao nhiu bit.

Gii:

8. a. Cho d=(0001) ta tnh c:

v1=0001000 + d s (0001000/1011)

v1=0001000 + 011=0001011

Quay vng t m v1 ta s c them 6 t m na:

v2=1000101

v3=1100010

v4=0110001

v5=1011000

v6=0101100

9411/13/2011


v7=0010110


v8=0011000 + d s(0011000/1011)

v8=0011000 + 101=0011101

Quay vng t m v8 ta s c thm 6 t m na;

v9=1001110

v10=0100111

v11=1010011

v12=1101001

v13=1110100

v14=0111010

Cho d=(11110) ta tnh c.

V15=1111000 + d s(1111000/1011)

V15=1111000 + 111=1111111


V0=0000000 + d s(1111000/1011)

V0=0000000 + 000=0000000

b.Ta c th tnh c d dng trng s Hamming ca b m l trng s nh nht ca cc t m khc khng: H(V) = 3 nn b m c kh nng sa c tt c cc mu sai 1 bit.

9511/13/2011


BI TP96

11/13/2011


L THUYT THNG TIN

CHNG 2: LNG TIN

1. Cho tp tin U = {ui} vi ui = xi yi zi

9711/13/2011


i 1 2 3 4 5 6 7 8

ui 000 001 010 011 100 101 110 111

p(ui) 0.25 0.25 0.125 0.0625 0.0625 0.0625 0.0625 0.0625

Gi tin ui = u6 = 101 l tin a

xi = 1 l tin b

xi = 1 , xi = 0 l tin c

xi = 1 , yi = 0 , zi = 1 l tin d

Tnh I(a) , I(a,b) , I(a,c) , I(a,d).

Bi lm:

I(a) = I(u6) = - log p(u6)

= - log 1/16 = log 1/16

I(a,b) = I(u6 ; xi=1)

= I(u6) - I(u6/ xi=1)

= log

= log

= log

= log

= log

9811/13/2011


= log

= log

= log 4

I(a,c) = I(u6 ; xi=1,yi=0)

= I(u6) - I(u6/ xi=1,yi=0)

= log

= log

= log

= log

= log

= log 8

I(a,d) = I(u6) - I(u6) = 0

9911/13/2011


CHNG 3: M THNG K TI U

1. Lp mt b m cho ngun tin U c thng k nh sau:

ui u1 u2 u3 u4 u5 u6 u7 u8 u9

p(ui) 0.34 0.2 0.19 0.1 0.07 0.04 0.03 0.02 0.01

Bng m nh phn (m=2) theo phng php Shannon. Tnh di trung bnh ca t m ntb v tnh kinh t ca t m:

= H(U)/ ntb

Bi lm:

S m ha theo phng php m ha Shannon:

ui p(ui) pi ni Dng nh phn ca pi T m

u1 0.34 0 2 0 00

u2 0.2 0.34 3 0,01010111 010

u3 0.19 0.54 3 0,10001010 100

u4 0.1 0.73 4 0,10111010 1011

u5 0.07 0.83 4 0,11010100 1101

u6 0.04 0.9 5 0,11100110 11100

u7 0.03 0.94 6 0,11110000 111100

u8 0.02 0.97 6 0,11111000 111110

u9 0.01 0.99 7 0,11111101 1111110

10011/13/2011


di trung bnh ca t m: ntb =


Ch s kinh t ca b m: =

2. Cho ngun tin U c s thng k nh sau:

ui u1 u2 u3 u4+ u6 u7+ u13 U14

p(ui) 0.5 0.25 0.315 0.31 0.0157 0.0156

Dng m Huffman kt hp m u m ha ngun tin trn vi c s m m = 2.

Bi lm:

S m ha theo phng php m ha Fano:

ui p(ui) T m ni

u1 0.34 0

0

0 2

u2 0.2 1 2

u3 0.19 1

1

1

1

1

1

1

0

0

0 3

u4 0.1 1 3

u5 0.07 1

1

1

1

1

0 3

u6 0.04 1

1

1

1

0 4

u7 0.03 1

1

1

0 5

u8 0.02 1

1

0 6

u9 0.01 1 6

10111/13/2011


di trung bnh ca t m: ntb =


Ch s kinh t ca b m: =

CHNG 5: M VNG

1. Cho b m vng (7,4) c a thc sinh l:

g(x) = x3+ x2+ 1

a. Tm a thc th H(x) ca b m ny.

b. Gi s nhn c t hp u(x) = x6+ x5+ x2+ 1. Xc nh a thc syndrome ca u(x).

Bi lm:

10211/13/2011


a. T cng thc:

h(x) =

ta tnh c h(x) = = x4+ x3 +x2+ 1

b. Gi s nhn c a thc u(x) = x6+ x5+ x2+ 1, khi a thc syndrome ca n :

Su(x) = s d ca = s d ca = 0

Vy a thc syndrome ca u(x) l Su(x) = 0.


Chng 3:M thng k ti u:


Ui U1 U2 U3 U4 U5 U6 U7 U8 U9P(u) 0.34 0.2 0.19 0.1 0.07 0.04 0.03 0.02 0.01Bng m nh phn (m=2)theo phng php shannon. Tnh di trung bnh ca t m ntbv tnh kinh t ca t m:p=H(U)/Ntb


Ui P(ui) Pi ni Dng m nh phn ca pi T mU1 0.34 0 2 0 00U2 0.2 0.34 3 0.01010111 010U3 0.19 0.54 3 0.10001010 100U4 0.1 0.73 4 0.10111010 1011U5 0.07 0.83 4 0.11010100 1101U6 0.04 0.9 5 0.11100110 11100

10311/13/2011


U7 0.03 0.94 6 0.11110000 111100U8 0.02 0.97 6 0.11111000 111110U9 0.01 0.99 7 0.11111101 1111110 di trung bnh t m:NTB=

Entropy ca tp tin:H(u)=

Ch s kinh t ca b m:p= =0,8279

Chng 5:m vng:Bi tp:cho m vng (n,k)=(7,4)c ma trn sinh l g(x)=x3+x+1. Lit k tt c cc t m ca b m v cho bit b m c kh nng sa sai bao nhiu bt.

GIIA.cho d= (0001) tatnh c:

V1=0001000+d s

V1=0001000+011=0001011

Quay vng t mv1 ta s cthm 6 t m na:V2=1000101

V3=1100010

V4=0110001

V5=1011000

V6=0101100

V7=0010110

Cho d=(0011)ta tnh c:V=0011000+d sV=0011000+101=00111101Quay vng t m v ta s cthm 6 t m na:V9 =1001110

V10 =0100111

V11=1010011

V13 =1110100

V14=0111010

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V15=1111000+d s

V15=1111000+111=1111111


V0=0000000+d s

b,ta c th tnh r rng trng s hamming ca b m l trng s nh nht ca t m khc khng : H(V)=3 nn b m c kh nng sa c tt cc mu sai 1bt

Trong cc bi tp logarit c tnh theo c s 2.

Bi 1-Chng LNG TIN

1. Trong mt tr chi x s vui ngi ta x 10 ch s t 0 n 9.

Xc xut trng ca mi s l nh nhau.a. Tnh lng tin ring ca tin : s trng gii l s 9.b. Tnh lng tin tng h gia tin : s trng gii l 9

so vi tin :s trng gii l s chia ht cho 3.c. Trong 10 tin trn gi U=(u1,u2,u3,u4,u4,u5,u6) vi ui l

tin s i trng gii (i=0,1,,6).Tm lng tin trung bnh ca tp tin U.

Bi lma. Gi p(9) l xc sut s 9 trng gii ta c

p(9) = 1/10= 0.1

vy I(9)= -log p(9)= log 10= 3.322

b. Gi p(0-3-6-9) l xc sut s trng gii chia ht cho 3, ta c:

p(0-3-6-9) = 4/10= 0.4

Mt khc : I (9/0-3-6-9) = - log p(9/0-3-6-9)105

11/13/2011

(9 / 0 3 6 9)log(0 3 6 9)

pp

=

1/10log log44 /10

= =

10log10 log4 log 1.3224

= = =

6

0( ) ( ) ( )i i

iI U pu I u

=

= 60

1 1log10 10

1 .7.log1010

i==

=


Vy I(9; 0-3-6-9) = I(9) I(9/0-3-6-9)

c. Lng tin trung bnh ca tp tin U :

Bi 2 Chng M CHNG NHIU

2. Ma trn sinh ca b gii m tuyn tnh (6,3) trn ng GF (2) c cho bi :

10611/13/2011

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

G

=

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

G

=


M

a. Hy biu din G di dng chun Gch.b. Lit k cc t m c c t 2 ma trn sinh G v Gch .c. C bao nhiu t m c trng s Hamming l

0,1,2,3,4,5,6,7Bi lm

a. Ly ct 3 lm ct 1,ct 6 lm ct 2 v ct 4 lm ct 3 ta c ma trn sinh Gch :

b. Lit k cc t m c c t ma trn sinh G v ma trn sinh Gch :

Vector mang tin a

V=aG w(v) v= a.Gch w(v)

000

001

010

000000

110110

110001

0

4

3

000000

001111

010110

0

4

3

10711/13/2011


011

100

101

110

111

000111

011010

101100

101011

011101

3

3

3

4

4

011001

100011

101100

110101

111010

3

3

3

4

4

c. C 2 b m u c :

1 t m c trng s l 0

0 t m c trng s l 1

0 t m c trng s l 2

4 t m c trng s l 3

3 t m c trng s l 4

0 t m c trng s l 5

0 t m c trng s l 6.

10811/13/2011


Bi 9 Chng M THNG K TI U

3. Cho ngun tin c s thng k nh sau :109

11/13/2011


Dng m Fano m ha ngun tin trn vi c s m m = 3. Tnh ntb v tnh kinh t t m :

P = H(U) / ntb

Bi lm

S m ha bng b m Fano :

11011/13/2011

19

0( ) 1.618tb i i

in n p u

=

= =

19

0

( ) ( ) log ( ) 1.597i ii

H U p u p u=

= =


di trung bnh t m :

Entropy ca tp tin :

11111/13/2011

( ) 1.597 0.9871.618tb

H Un

= = =



11211/13/2011


Bi 1 Chng M VNG

4. Chng minh rng a thc sinh g(x) ca mt m vng l c s ca xn +1

Gi s v(x) = v1xn-1 + v2xn-2 + +vn-1x + vn l mt a thc m khi v(x) = vnxn-1 + v1xn-2 + +vn-2x + vn-1 cng l mt a thc m . Khi :

v(x) = v1xn-1 + v2xn-2 + +vn-1x + vn M g(x)

v v(x) = vnxn-1 + v1xn-2 + +vn-2x + vn-1 M g(x)

T y : xv(x) v(x) = vnxn vn M g(x)

Hay : xn 1 M g(x)

y l iu cn chng minh.

11311/13/2011


Chng II : LNG TIN

BI 2: Vi ngun tin ri rc gm 2 k t U = {a1,a2} v p(a1) = p


Bi gii:


C f( ) = -log e (ln ) = -log e (ln )

f ( ) =


Vy vi a,b



=

Vy H(U) =


11411/13/2011


Bi gii :

= H(X) + H(Y/X) (pcm)


di l(u) tha :


Tha

Bi gii:



11511/13/2011



Hay



U={3,23,11,123,10}

j. V cy m v hnh kt cu ca b m U

k. V mt ta m

l. Lp 1 b m h thng 6 t m c cc t hp s ng. V hnh kt cu ca b m va lp

Bi gii:

j. Cy m :

----------------------------------------------------------------------------------------------------Mc 0

2 1

---------------------------------------------------------------------------------------------------Mc 1

3 1 2 0

--------------------------------------------------------------------------------------------------Mc 2

3

---------------------------------------------------------------------------------------------------Mc 3

hnh kt cu :

3

11611/13/2011

0

01

0


k. Mt ta m:

ui ni pi

3 1 3

23 2 14

11 2 5

123 3 57

10 2 1

l. T hp s ng : 23,123,10

T hp cui : 3,11



ui u1 u2 u3 u4 u5 u6 u7 u8 u9

P(ui) .34 .2 .19 .1 .07 .04 .03 .02 .01


Bi gii :



u1 .34 0 2 0 00

u2 .2 .34 3 0.01010111 010

11711/13/2011


u3 .19 .54 3 0.10001010 100

u4 .1 .73 4 0.1011010 1011

u5 .07 .83 4 0.11010100 11 01

u6 .04 .9 5 0.11100110 11100

u7 .03 .94 6 0.11110000 111100

u8 .02 .97 6 0.11111000 111110

u9 .01 .99 1 0.11111101 1111110

di trung bnh t m :

Etropy ca tp tin:



ui u1 u2 u3 u14

P(ui) .5 .25 .315 .31 . 0157 .0156


Bi gii:

P(ui) S m ha T m

.5 0

.25 10

.315 11000

.31 11001

.31 11010

11811/13/2011


.31 11011

. 0157 111000

. 0157 111001

. 0157 111010

. 0157 111011

. 0157 111100

. 0157 111101

. 0157 111110

.0156 111111







bi gii:

11911/13/2011





Vector mang tin a


000 0000000 0 0000000 0

001 1011100 4 0011110 4

010 0101110 4 0100111 4

011 1110010 4 0111001 4

100 0010111 4 1001011 4

101 1001011 4 1010101 4

110 0111001 4 1101100 4

111 1100101 4 1110010 4

12011/13/2011


d. C 2 b m trn u c D=4 l trng s Hamming nh nht ca cc t m khc khng ca chng

Chng 5 : M VNG


Bi gii:

Gi s


V

T y :

Hay

Tc l (pcm)


Bi gii:




12111/13/2011


Chng 5: M Vng

Ga s g(x) l a thc sinh ca mt b m vng c di t m n.Chng minh rng nu n l s nguyn nh nht sao cho xn + 1 chia ht cho g(x) th trng s Hamming ca b m khng nh hn 3.

GII

Ta chng minh bi ton bng phng php chng minh phn chng.Gi s b m nh vy c trng s Hamming nh hn 3,tc l bng 1 hoc 2.

a. Trng hp trng s Hamming bng 1 th tn ti mt t m v(x) sao cho w(v)=1.Khi trong t m v ch c duy nht mt k t bng 1,gi s l vi. Vy:

Vi(x) = vixn-i : g(x)

Khi g(x) khng th l c s ca xn + 1

b. Trng hp trng s Hamming bng 2 th tn ti mt t m v(x) sao cho w(v) = 2. Khi trong t m v c 2 k t bng 1, gi s l vi v vj. C th gi s i > j v d nhin n > i > j.Vy:

vi(x)= vixn-i + vjxn-j = xn-i + xn-j : g(x)

Hay: xn-i (xi-j + 1) : g(x)

T y: xi-j + 1 : g(x) trong khi i-j < n mu thun vi gi thit ban u n l s nh nht sao cho xn + 1 : g(x).

Vy trng s Hamming ca b m phi lun ln hn hoc bng 3.

12211/13/2011

Giai Bai Tap Ly Thuyet Thong Tin_PTIT

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