Bi 1: Cho X v Y l hai trng tin, vi v v xc sut xut hin ng thi ca cc tin cho trong bng sau: x=1x=2x=3

p(xi,y=1)1/41/121/6

p(xi,y=2)1/121/41/6

Tnh p(xi/yj) H(X,Y), H(X), H(Y), H(Y/X) v I(X;Y).

Bi 2: Cho X v Y l hai trng tin, vi v v xc sut xut hin ng thi ca cc tin cho trong bng sau: x=1x=2x=3

p(xi,y=1)0,10,20,2

p(xi,y=2)0,150,150.2

Tnh p(xi/yj) H(X,Y), H(X), H(Y), H(Y/X) v I(X;Y).: Cho xc sut xut hin cc tin p(xi) v xc sut c iu kin p(yj/xi) nh bng 1 v bng 2.Tnh p(xi,yj), H(X), H(Y), H(X,Y), H(Y/X), I(X;Y)Bng 1xix1x2x3

p(xi)0,20,30,5

Bng 2p(yj/xi)x1x2x3

y10,50,60,4

y20,50,40,6

Bi 3: Cho xc sut xut hin cc tin p(xi) v xc sut c iu kin p(yj/xi) nh bng 1 v bng 2.Tnh p(xi,yj), H(X), H(Y), H(X,Y), H(Y/X), I(X;Y)Bng 1xix1x2x3x4

p(xi)0,20,30,10,4

Bng 2p(yj/xi)x1x2x3 x4

y10,50,60,4 0,3

y20,50,40,6 0,7

Bi 4: Ma trn knh ca knh nh phn i xng c xa nh sau:

Cho xc sut xut hin tin p(x1) = 2/5.a- V m hnh knhb- Tnh H(X), H(Y), lng thng tin cho trung bnh truyn qua knh I(X;Y)V gin ca knh, vi ngun tin vo l X, ngun tin ra l (Y)Bi 5: Ma trn knh ca knh nh phn i xng c xa nh sau:

Cho xc sut xut hin tin p(x1) = 2/5.a- V m hnh knhb- Tnh H(X), H(Y), lng thng tin cho trung bnh truyn qua knh I(X;Y)V Gin ca knh, vi ngun tin vo l X, ngun tin ra l (Y)Bi 6: Cho X v Y l hai trng tin pha pht v pha thu 1 knh nh phn i xng khng nh, vi cc tham s cho trn hnh v.

a- Tnh p(xi,yj), H(X), H(Y), H(X,Y), H(Y/X) v I(X;Y).b- Tnh kh nng thng qua ca knh trn bit tc pht ca ngun vk=1200 baud.Bi 7: Cho X v Y l hai trng tin pha pht v pha thu 1 knh nh phn i xng khng nh, khng xa vi cc tham s cho trn hnh v.

a- Tnh p(xi,yj), H(X), H(Y), H(X,Y), H(Y/X) v I(X;Y).b- Tnh kh nng thng qua ca knh trn bit tcs pht ca ngun vk=2400 baud

Bi 8: Tn hiu s c iu ch cc mc vi xc sut tng ng l vi tc bit vo l 20 kbps.a- Tnh entropie ca cc im tn hiu sau iu ch b- Tnh rng bng thng ti thiu cn thit truyn tn hiu vi xc sut li c th b qua t s S/N=10dB.Tnh t s S/N ti thiu truyn tn hiu trn vi xc sut li c th b qua khi bng thng cho php l 10kHz.

Bi 9: : Mt ngun tin pht ra cc du dng m BCD vi cc gi tr xc sut nh cho trong Bng sau, vi tc 2400 baud.a) Xc nh tc tin (information rate) v tc d liu (data rate) ca ngun.b) p dng m ho bng cy Huffman cho ngun ny. Gii m chui bt k c di 16 bt.c) Tc d liu ca tn hiu sau khi m ho l bao nhiu. M ho t c h s nn l bao nhiu? Hy tnh hiu sut m ho cho c tn hiu BCD v tn hiu m ho ngun.d) Tnh bng thng ti thiu truyn lung d liu trn sau khi m Huffman qua knh Gauss t s SNR = 20 dB d liu thu c c xc sut li nh.Bng phn b xc sut cc tin ca ngun

ABCDEFG

0.200.250.030.150.300.050.02

T m BCD000001010011100101110

Bi 10: Mt ngun tin pht ra cc du dng m BCD vi cc gi tr xc sut nh cho trong Bng sau, vi tc 2400 baud.a) Xc nh tc tin (information rate) v tc d liu (data rate) ca ngun.b) p dng m ho Huffman cho ngun ny. Gii m chui bt k c di 16 bt.c) Tc d liu ca tn hiu sau khi m ho l bao nhiu. M ho t c h s nn l bao nhiu? Hy tnh hiu sut m ho cho c tn hiu BCD v tn hiu m ho ngun.d) Tnh bng thng ti thiu truyn lung d liu trn sau khi m Huffman qua knh Gauss t s SNR = 10 dB d liu thu c c xc sut li nh.

Bng phn b xc sut cc tin ca ngun

ABCDEFG

0.200.250.030.150.300.050.02

T m BCD000001010011100101110

Bi 11: Mt ngun tin pht ra cc du dng m BCD vi cc gi tr xc sut nh cho trong Bng di y, vi tc 4800 baud.a) Xc nh tc tin (information rate) v tc d liu (data rate) ca ngun.b) p dng m ho bng cy Huffman cho ngun ny. Gii m chui bt k c di 16 bt.c) Tc d liu ca tn hiu sau khi m ho l bao nhiu. M ho t c h s nn l bao nhiu? Hy tnh hiu sut m ho cho c tn hiu BCD v tn hiu m ho ngun.d) Tnh bng thng ti thiu truyn lung d liu trn sau khi m Huffman qua knh Gauss t s SNR = 10 dB d liu thu c c xc sut li nh.

Bng phn b xc sut cc tin ca ngun

ABCDEFG

0.250.100.100.150.250.050.10

T m BCD000001010011100101110

Bi 12 : Mt ngun tin pht ra cc du dng m BCD vi cc gi tr xc sut nh cho trong Bng di y, vi tc 4800 baud.a- Xc nh tc tin (information rate) v tc d liu (data rate) ca ngun.b-p dng m ho Huffman cho ngun ny. Gii m chui bt k c di 15 bt.c-Tc d liu ca tn hiu sau khi m ho l bao nhiu. M ho t c h s nn l bao nhiu? Hy tnh hiu sut m ho cho c tn hiu BCD v tn hiu m ho ngun.d- Tnh bng thng ti thiu truyn lung d liu trn sau khi m Huffman qua knh Gauss t s SNR = 20 dB d liu thu c c xc sut li nh.

Bng phn b xc sut cc tin ca ngun

ABCDEFG

0.250.100.100.150.250.050.10

T m BCD000001010011100101110

Bi 13: Mt ngun tin pht ra cc du dng m BCD vi cc gi tr xc sut nh cho trong Bng di y, vi tc 4800 baud.a- Xc nh tc tin (information rate) v tc d liu (data rate) ca ngun.b-p dng m ho bng cy Huffman cho ngun ny. Gii m chui bt k c di 16 bt.c-Tc d liu ca tn hiu sau khi m ho l bao nhiu. M ho t c h s nn l bao nhiu? Hy tnh hiu sut m ho cho c tn hiu BCD v tn hiu m ho ngun.d- Tnh bng thng ti thiu truyn lung d liu trn sau khi m Huffman qua knh Gauss t s SNR = 20 dB d liu thu c c xc sut li nh.

Bng phn b xc sut cc tin ca ngun

x1x2x3x4x5x6x7

1/47/321/83/321/163/161/16

T m BCD000001010011100101110

Bi 14: Mt ngun tin pht ra cc du dng m BCD vi cc gi tr xc sut p(xi) nh cho trong Bng di y, vi tc 3600 baud.a- Xc nh tc tin (information rate) v tc d liu (data rate) ca ngun.b-p dng m ho bng cy Huffman cho ngun ny. Gii m chui bt k c di 15 bt.c-Tc d liu ca tn hiu sau khi m ho l bao nhiu. M ho t c h s nn l bao nhiu? Hy tnh hiu sut m ho cho c tn hiu BCD v tn hiu m ho ngun.d- Tnh bng thng ti thiu truyn lung d liu trn sau khi m Huffman qua knh Gauss t s SNR = 20 dB d liu thu c c xc sut li nh.

Bng phn b xc sut cc tin ca ngun

x1x2x3x4x5x6x7

6/215/424/213/212/211/215/42

T m BCD000001010011100101110

Bi 15: Cho quy tc thnh lp m Hamming (7,4) nh sau:

a- Thnh lp ma trn kim tra (H) v ma trn sinh (G). b- Vit tt c cc t m ca b m, v hy bnh lun kh nng pht hin li v sa li ca m ny.c- Hy s dng m trn thc hin gii m bng syndrome vi mt t m khi b sai (s li sai nm trong kh nng sa li ca m).Gi s tc bt vo khi m ha l 16kbps tng ng vi gi tr SNR cho trc, tnh xc sut khng pht hin c ca m trn bit xc sut li Pe=0,025 v tc truyn tin hiu dng khi s dng m ny.Bi 16:Cho mt m C c xc nh bi ma trn kim tra:

a-Xc nh ma trn sinh G tng ng. b-Hy lit k tt c cc t m. Xc nh c ly Hamming ti thiu ca m. Hy bnh lun v kh nng sa li, kh nng pht hin li ca m.

c-Gi s l t m c gi i qua knh nh phn i xng, thu c vc-t , vi l vc-t li. Hy chng t rng syndrome ch ph thuc vo . T chn mt t m bt k, gy li v tr tu (trong kh nng sa li ca m). Chng t rng gii m bng syndrome c th sa c li .d-Hy tnh xc sut khng pht hin c sai v tc truyn tin hiu dng khi s dng m ny cho tc bt vo m ha l 16kbps v ti Pe= 0,025.Bi 17:Cho mt m C c xc nh bi ma trn sinh:

a-Xc nh ma trn sinh H tng ng. b-Hy lit k tt c cc t m. Xc nh c ly Hamming ti thiu ca m. Hy bnh lun v kh nng sa li, kh nng pht hin li ca m.

c-Gi s l t m c gi i qua knh nh phn i xng, thu c vc-t , vi l vc-t li. Hy chng t rng syndrome ch ph thuc vo . T chn mt t m bt k, gy li v tr tu (trong kh nng sa li ca m). Chng t rng gii m bng syndrome c th sa c li .d-Hy tnh xc sut khng pht hin c sai v tc truyn tin hiu dng khi s dng m ny cho knh Gauss c kh nng thng qua l 16kbps v ti Pe= 0,025.

trang 2 | 8

x1

x2

y2

y1

p(y1/x1)=2/3

p(y2/x2)=2/3

p(y1/x2)=1/3

p(y2/x1)=1/3

p(x1)=5/8

p(x2)=3/8

x1

x2

y2

y1

p(y1/x1)=4/5

p(y2/x2)=4/5

p(y1/x2)=1/5

p(y2/x1)=1/5

p(x1)=3/5

p(x2)=2/5