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N?i dung. Tn hi?u v d? li?uTruy?n d?n d? li?uC?u trc knh truy?n Tu?n t?Song songC?u trc truy?n B?t d?ng b??ng b?Cc phuong ki?m tra v pht hi?n l?iC?u hnh Giao ti?p V.24/EIA-232-FNn thng tin Phn h?p knh (Multiplexing)ADSL. Thu?t ng?. Thnh ph?n trong m hnh truy?n d? li?u
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1. CHUONG 3 CC K? THU?T CO B?N TRONG TRUY?N S? LI?U anhph@cse.hcmut.edu.vn
2. N?i dung Tn hi?u v d? li?u
Truy?n d?n d? li?u
C?u trc knh truy?n
Tu?n t?
Song song
C?u trc truy?n
B?t d?ng b?
?ng b?
Cc phuong ki?m tra v pht hi?n l?i
C?u hnh
Giao ti?p V.24/EIA-232-F
Nn thng tin
Phn h?p knh (Multiplexing)
ADSL
3. Thu?t ng? Thnh ph?n trong m hnh truy?n d? li?u (du?i gc d? v?t l)
Thi?t b?
Thi?t b? pht (Transmitter)
Thi?t b? thu (Receiver)
Mi tru?ng truy?n (Medium)
K?t n?i
K?t n?i tr?c ti?p (Direct link)
Khng c?n cc thi?t b? trung gian
K?t n?i di?m-di?m (Point-to-point)
K?t n?i tr?c ti?p
Ch? c 2 thi?t b? dng chung k?t n?i
K?t n?i nhi?u di?m (Multi-point)
= 2 thi?t b? dng chung k?t n?i
4. Ch? d? truy?n Simplex mode
Khng dng r?ng ri v khng th? g?i ngu?c l?i l?i ho?c tn hi?u di?u khi?n cho bn pht
Television, teletext, radio
Half-duplex mode
B? dm
Full-duplex mode
i?n tho?i
5. Truy?n d?n d? li?u D? li?u
Th?c th? mang thng tin
Analog
Cc gi tr? lin t?c trong m?t vi th?i kho?ng
e.g. m thanh, video
Digital
Cc gi tr? r?i r?c
e.g. van b?n, s? nguyn
Tn hi?u
Bi?u di?n di?n ho?c di?n t? c?a d? li?u
Analog
Bi?n lin t?c
Mi tru?ng lin t?c (wire, fiber optic, space)
Bang thng ti?ng ni 100Hz t?i 7kHz
Bang thng di?n tho?i 300Hz t?i 3400Hz
Digital
Dng 2 thnh ph?n m?t chi?u
Truy?n d?n
Trao d?i d? li?u thng qua vi?c x? l v lan truy?n tn hi?u
6. Tn hi?u mi?n th?i gian Tn hi?u lin t?c
Thay d?i lin t?c theo th?i gian
Tn hi?u r?i r?c
Thay d?i t?ng m?c theo th?i gian
Tn hi?u chu k?
M?u l?p l?i theo th?i gian
Tn hi?u khng tu?n hon
M?u khng l?p l?i theo th?i gian
7. Tn hi?u analog Ba d?c di?m chnh c?a tn hi?u analog bao g?m
Bin d? (Amplitute)
T?n s? (Frequency)
Pha (Phase)
Bin d? c?a tn hi?u analog
o d? m?nh c?a tn hi?u, don v?: decibel (dB) hay volts.
Bin d? cng l?n, tn hi?u cng c cu?ng d? m?nh.
Tn hi?u ti?ng ni - t? hello.
Ti?ng ni (speech) l m?t tn hi?u r?t ph?c t?p.
Ti?ng ni ch?a hng ngn t? h?p khc nhau c?a nhi?u tn hi?u.
8. T?n s? c?a tn hi?u analog T?c d? thay d?i c?a tn hi?u trong m?t giy, don v? Hz hay s? chu k? trong m?t giy (cycles per second)
Tn hi?u c t?n s? 30Hz ~ thay d?i 30 l?n trong m?t giy.
M?t chu k? l s? di chuy?n sng c?a tn hi?u t? di?m ngu?n b?t d?u cho d?n khi quay tr? v? l?i di?m ngu?n d.
9. Pha c?a tn hi?u analog T?c d? thay d?i quan h? c?a tn hi?u d?i v?i th?i gian, du?c m t? theo d? (degree)
S? d?ch pha x?y ra khi chu k? c?a tn hi?u chua k?t thc, v m?t chu k? m?i c?a tn hi?u b?t d?u tru?c khi chu k? tru?c d chua hon t?t
Tai ngu?i khng c?m nh?n du?c s? d?ch pha
Tn hi?u mang d? li?u b? ?nh hu?ng b?i s? d?ch pha
V d? cc m?i n?i khng hon h?o s? gy ra d?ch pha
10. Tn hi?u mi?n t?n s?
11. Thnh ph?n c?a ti?ng ni T?m t?n s? c kh? nang nghe 20Hz 20kHz
Ti?ng ni 100Hz 7kHz
D? dng chuy?n sang d?ng tn hi?u di?n t? d? truy?n d?n
Cc t?n s? v?i m lu?ng khc nhau du?c chuy?n thnh t?n s? di?n t? v?i di?n p khc nhau
T?m t?n s? gi?i h?n cho knh tho?i
300 3400Hz
12. Tn hi?u s? (digital) Tn hi?u s? bao g?m ch? hai tr?ng thi, du?c di?n t? v?i hai tr?ng thi ON hay OFF ho?c l 0 hay 1
Tn hi?u s? yu c?u kh? nang bang thng l?n hon tn hi?u analog.
13. Tn hi?u s? (digital) Cc v?n d? khi s? d?ng knh tho?i (voice channel) trong vi?c truy?n tn hi?u s?
M?t tn hi?u s? l m?t t? h?p c?a cc tn hi?u khc. ?c bi?t, tn hi?u s? c th? du?c bi?u di?n nhu sau
Signal = f + f3 + f5 +f7 +f9 +f11 +f13 ....f?
Do d m?t tn hi?u s? g?m 1 t?n s? co b?n (f), c?ng thm t?n s? 3f (hi t?n b?c 3), c?ng thm t?n s? 5f (hi t?n b?c 5),
N?u bin d? c?a t?n s? f, f3, f5, l a, a3, a5, th a = 3a3 = 5a5
? g?i tn hi?u s? qua knh truy?n tho?i, bang thng c?a knh truy?n ph?i cho php t?n s? co b?n f, t?n s? 3f v t?n s? 5f di qua m khng ?nh hu?ng nhi?u d?n cc t?n s? ny
y l yu c?u t?i thi?u d? bn nh?n nh?n dng du?c tn hi?u s?
14. Tn hi?u s? (digital) Truy?n 1 tn hi?u s? nh? phn t?c d? 2400bps trn knh tho?i c bang thng 3.1kHz
T?n s? co b?n: 1200Hz (thng thu?ng b?ng t?c d? bit)
Ch? c t?n s? co b?n di qua m khng b? thay d?i
15. D? li?u v tn hi?u Thu?ng dng tn hi?u s? cho d? li?u s? v tn hi?u analog cho d? li?u analog
C th? dng tn hi?u analog d? mang d? li?u s?
Modem
C th? dng tn hi?u s? d? mang d? li?u analog
Compact Disc audio
16. Truy?n d?n Truy?n d?n analog
Khng quan tm d?n n?i dung d? li?u du?c truy?n (s? ho?c tuong t?)
Suy gi?m khi truy?n xa
Dng b? khu?ch d?i (amplifier) d? truy?n d? li?u di xa
Khu?ch d?i c? tn hi?u l?n nhi?u
17. Analog data/Analog Signal
G?i bnh thu?ng ho?c m ha vo ph?n ph? khc
Analog data/Digital Signal
M ha dng b? codec d? t?o ra chu?i bit s?
Digital Data/Analog Signal
u?c m ha dng modem d? t?o ra t/h tuong t?
Digital Data/Digital Signal
Bi?u di?n tr?c ti?p d? li?u ho?c m ha d? t?o ra t/h s? c d?c tnh mong mu?n
Analog Signal/Analog Transmission
Lan truy?n thng qua cc b? khu?ch d?i, x? l t/h nhu nhau b?t k? d? li?u l s? ho?c tuong t?
Analog Signal/Digital Transmission
Gi? s? t/h bi?u di?n d? li?u s?, lan truy?n qua cc b? repeater
Digital Signal/Analog Transmission
Khng dng
Digital Signal/Digital Data
T/h l chu?i nh? phn lan truy?n qua cc b? repeater D? li?u, tn hi?u v truy?n d?n
18. Truy?n d?n s? Uu di?m
Cng ngh? s?
Cng ngh? LSI/VLSI lm gi?m gi thnh
Ton v?n d? li?u
Nhi?u v suy gi?m tn hi?u khng b? tch luy b?i cc repeater
Truy?n kho?ng cch xa hon trn cc du?ng truy?n km ch?t lu?ng
Hi?u qu? knh truy?n
TDM > FDM
B?o m?t
Cc k? thu?t m ha d? b?o m?t d? li?u d? p d?ng
Tch h?p
D? li?u s? v analog du?c x? l tuong t? nhau
19. Digital ? Digital Tn hi?u s?
Xung di?n p r?i r?c, khng lin t?c
M?i xung l m?t ph?n t? tn hi?u
D? li?u nh? phn du?c m ha thnh cc ph?n t? tn hi?u
20. Thu?t ng? Unipolar
T?t c? cc ph?n t? tn hi?u c cng d?u
Polar
M?t tr?ng thi logic du?c bi?u di?n b?ng m?c di?n p duong, tr?ng thi logic khc du?c bi?u di?n b?ng m?c di?n p m
T?c d? d? li?u (data rate)
T?c d? truy?n d?n d? li?u theo bps (bit per second)
? r?ng (chi?u di 1 bit)
Th?i gian (thi?t b? pht) dng d? truy?n 1 bit
T?c d? di?u ch?
T?c d? m?c tn hi?u thay d?i
on v? l baud = s? ph?n t? tn hi?u trong 1 giy
Mark v Space
Tuong ?ng v?i 1 v 0 nh? phn
21. Di?n gi?i tn hi?u C?n bi?t
?nh th?i c?a cc bit (khi no chng b?t d?u v k?t thc)
M?c tn hi?u
Y?u t? ?nh hu?ng d?n vi?c di?n gi?i tn hi?u
T? s? SNR
T?c d? d? li?u
Bang thng
22. Polar Encoding
23. Nonreturn to zero (NRZ) Nonreturn to Zero-Level (NRZ-L)
2 m?c di?n p khc nhau cho bit 1 v bit 0
i?n p khng thay d?i (khng c transition) khi khng c s? thay d?i tn hi?u
i?n p thay d?i (c transition) khi c s? thay d?i tn hi?u (t? 0?1 ho?c t? 1?0)
Nonreturn to Zero Inverted (NRZI)
NRZI cho cc bit 1
D? li?u du?c m ha can c? vo vi?c c hay khng s? thay d?i tn hi?u ? d?u th?i kho?ng bit.
Bit 1: du?c m ha b?ng s? thay d?i di?n p (c transition)
Bit 0: du?c m ha b?ng s? khng thay d?i di?n p (khng c transition)
24. Nonreturn to Zero (NRZ) M ha sai phn
D? li?u du?c bi?u di?n b?ng vi?c thay d?i tn hi?u (thay v b?ng m?c tn hi?u)
Nh?n bi?t s? thay d?i d? dng hon so v?i nh?n bi?t m?c
Trong cc h? th?ng truy?n d?n ph?c t?p, c?m gic c?c tnh d? dng b? m?t
Uu v nhu?c di?m c?a m ha NRZ
Uu
D? dng n?m b?t
Bang thng dng hi?u qu?
Nhu?c
C thnh ph?n m?t chi?u
Thi?u kh? nang d?ng b?
Dng trong vi?c ghi bang t?
t dng trong vi?c truy?n tn hi?u
25. Multilevel Binary Dng nhi?u hon 2 m?c
Bipolar-AMI (Alternate Mark Inversion)
Bit-0 du?c bi?u di?n b?ng khng c tn hi?u
Bit-1 du?c bi?u di?n b?ng xung duong hay xung m
Cc xung 1 thay d?i c?c tnh xen k?
Khng m?t d?ng b? khi d? li?u l m?t dy 1 di (dy 0 v?n b? v?n d? d?ng b?)
Khng c thnh ph?n m?t chi?u
Bang thng th?p
Pht hi?n l?i d? dng
Pseudoternary
1 du?c bi?u di?n b?ng khng c tn hi?u
0 du?c bi?u di?n b?ng xung duong m xen k? nhau
Khng c uu di?m v nhu?c di?m so v?i bipolar-AMI
26. Trade Off
Khng hi?u qu? b?ng NRZ
M?i ph?n t? t/h ch? bi?u di?n 1 bit
H? th?ng 3 m?c c th? bi?u di?n log23 = 1.58 bit
B? thu ph?i c kh? nang phn bi?t 3 m?c (+A, -A, 0)
C?n thm kho?ng 3dB cng su?t d? d?t du?c cng xc su?t bit l?i
Multilevel Binary
27. Biphase Manchester
Thay d?i ? gi?a th?i kho?ng bit
Thay d?i du?c dng nhu tn hi?u d?ng b? d? li?u
L?H bi?u di?n 1
H?L bi?u di?n 0
Dng trong IEEE 802.3
28. Biphase Differential Manchester
Thay d?i gi?a th?i kho?ng bit ch? dng cho d?ng b?
Thay d?i d?u th?i kho?ng bi?u di?n 0
Khng c thay d?i ? d?u th?i kho?ng bi?u di?n 1
Dng trong IEEE 802.5
29. Biphase Uu v nhu?c di?m
Nhu?c di?m
T?i thi?u c 1 thay d?i trong th?i kho?ng 1 bit v c th? c 2
T?c d? di?u ch? t?i da b?ng 2 l?n NRZ
C?n bang thng r?ng hon
Uu di?m
?ng b? d?a vo s? thay d?i ? gi?a th?i kho?ng bit (self clocking)
Khng c thnh ph?n m?t chi?u
Pht hi?n l?i
Khi thi?u s? thay d?i mong d?i
30. Biphase
31. Polar Encoding
32. Bi t?p
33. Bi t?p
34. Bi t?p
35. Bi t?p
36. Bi t?p
37. Bi t?p
38. Bi t?p
39. Bi t?p
40. Scrambling Dng k? thu?t scrambling d? thay th? cc chu?i t?o ra h?ng s? di?n p
Chu?i thay th?
Ph?i t?o ra d? s? thay d?i tn hi?u, dng cho vi?c d?ng b? ha
Ph?i du?c nh?n di?n b?i b? thu v thay th? tr? l?i chu?i ban d?u
Cng d? di nhu chu?i ban d?u
Khng c thnh ph?n m?t chi?u
Khng t?o ra chu?i di cc tn hi?u m?c 0
Khng gi?m t?c d? d? li?u
C kh? nang pht hi?n l?i
41. B8ZS B8ZS (Bipolar With 8 Zeros Substitution)
D?a trn bipolar-AMI
N?u c 8 s? 0 lin ti?p v xung di?n p cu?i cng tru?c d l duong, m thnh 000+0+
N?u c 8 s? 0 lin ti?p v xung di?n p cu?i cng tru?c d l m, m thnh 000+0+
Gy ra 2 vi ph?m m AMI
C th? l?m l?n v?i tc d?ng gy ra b?i nhi?u
B? thu pht hi?n v di?n gi?i chng thnh 8 s? 0 lin ti?p
42. B8ZS
43. HDB3 HDB3 (High Density Bipolar 3 Zeros)
D?a trn bipolar-AMI
Chu?i 4 s? 0 lin ti?p du?c thay th? theo quy lu?t nhu sau
44. HDB3
45. Bi t?p
46. Bi t?p
47. Bi t?p
48. Bi t?p
49. So snh cc phuong php m ha Ph? tn hi?u
Vi?c thi?u thnh ph?n t?n s? cao lm gi?m yu c?u v? bang thng
T?p trung cng su?t ? gi?a bang thng
?ng b?
?ng b? b? thu v b? pht
Tn hi?u d?ng b? ngo?i vi
Co ch? d?ng b? d?a trn tn hi?u
Kh? nang pht hi?n l?i
C th? du?c tch h?p trong co ch? m ha
Nhi?u v kh? nang mi?n nhi?m
Vi m t?t hon cc m khc
? ph?c t?p v chi ph
T?c d? tn hi?u cao hon (v do d t?c d? d? li?u cao hon) d?n t?i chi ph cao
Vi m di h?i t?c d? tn hi?u cao hon t?c d? d? li?u
50. ?ng d?ng
Dng d? truy?n d? li?u s? trn m?ng di?n tho?i cng c?ng
300Hz ? 3400Hz
Thi?t b?
MODEM (MOdulator-DEMulator)
K? thu?t
i?u bin: Amplitude-Shift Keying (ASK)
i?u t?n: Frequency-Shift Keying (FSK)
i?u pha: Phase-Shift Keying (PSK) Digital ? Analog
51. i?u bin (ASK) Dng 2 bin d? khc nhau c?a sng mang d? bi?u di?n 0 v 1 (thng thu?ng m?t bin d? b?ng 0)
S? d?ng m?t t?n s? sng mang duy nh?t
Phuong php ny ch? ph h?p trong truy?n s? li?u t?c d? th?p (~1200bps trn knh truy?n tho?i)
T?n s? c?a tn hi?u sng mang du?c dng ph? thu?c vo chu?n giao ti?p dang du?c s? d?ng
K? thu?t du?c dng trong cp quang
52. i?u bin (ASK)
53. i?u bin (ASK)
54. i?u t?n (FSK) Binary FSK (BFSK) S? d?ng hai t?n s? sng mang: t?n s? cao tuong ?ng m?c 1, t?n s? th?p tuong ?ng m?c 0.
t l?i hon so v?i ASK
u?c s? d?ng truy?n d? li?u t?c d? 1200bps hay th?p hon trn m?ng di?n tho?i
C th? dng t?n s? cao (3-30MHz) d? truy?n trn sng radio ho?c cp d?ng tr?c
55. i?u t?n (FSK) Binary FSK (BFSK)
56. i?u t?n (FSK) Multiple (FSK) Dng nhi?u hon 2 t?n s?
Bang thng du?c dng hi?u qu? hon
Kh? nang l?i nhi?u hon
M?i ph?n t? tn hi?u bi?u di?n nhi?u hon 1 bit d? li?u
57. i?u pha (PSK) S? d?ng m?t t?n s? sng mang v thay d?i pha c?a sng mang ny
PSK vi phn (differential PSK) thay d?i pha tuong d?i so v?i sng tru?c d (thay v so v?i sng tham chi?u c? d?nh)
Cho php m ha nhi?u bit trn m?i thay d?i tn hi?u sng mang (Phase Amplitude Modulation)
Phuong php ny thu?ng du?c dng trong truy?n d? li?u ? t?c d? 2400bps (2 bits per phase change - CCITT V.26) ho?c 4800bps (3 bits encoding per phase change - CCITT V.27) ho?c 9600bps (4 bits encoding per phase/amplitude change)
T?ng qut cho m ha NRZ-L
58. i?u pha (PSK)
59. i?u pha (PSK) Quadrature PSK (QPSK)
M-ary PSK
H? th?ng 64 v 256 tr?ng thi
C?i thi?n t?c d? d? li?u v?i bang thng khng d?i
Tang kh? nang ti?m ?n l?i
60. Hi?u su?t Bang thng
Bang thng ASK v PSK lin quan tr?c ti?p v?i t?c d? bit
BT = (1+r)R
Bang thng FSK c quan h? v?i t?c d? d? li?u d?i v?i cc t?n s? th?p, c quan h? v?i d? d?ch chuy?n c?a cc t?n s? di?u ch? d?i v?i t?n s? cao
BT = 2?F + (1+r)R
Tn hi?u nhi?u m?c
BT = (1+r)R/m = (1+r)R/log2M
Trong tru?ng h?p c l?i, t?c d? l?i c?a PSK v QPSK cao hon kho?ng 3dB so v?i ASK v FSK
61. Quadrature Amplitude Modulation (QAM) QAM du?c dng trong ADSL v m?t s? h? th?ng wireless
K?t h?p gi?a ASK v PSK
M? r?ng logic c?a QPSK
G?i d?ng th?i 2 tn hi?u khc nhau cng t?n s? mang
Dng 2 b?n sao c?a sng mang, m?t ci du?c d?ch di 90
M?i sng mang l ASK d du?c di?u ch?
2 tn hi?u d?c l?p trn cng mi tru?ng
Gi?i di?u ch? v k?t h?p cho d? li?u nh? phn ban d?u
62. Digital ? Analog
63. Analog ? Digital ?ng d?ng
Dng d? truy?n d? li?u tuong t? trn m?ng truy?n d? li?u s?
T?n d?ng cc uu di?m c?a truy?n d?n s? (thi?t b? r?, dng repeater, TDM, )
S? ha
D? li?u s? c th? truy?n dng NRZ-L hay cc lo?i m khc
Thi?t b?
CODEC (COder-DECoder)
K? thu?t
i?u ch? xung m: Pulse Code Modulation (PCM)
i?u ch? Delta: Delta Modulation (DM)
64. i?u ch? xung m (PCM) L thuy?t l?y m?u
N?u tn hi?u f(t) du?c l?y m?u d?u v?i t?c d? l?y m?u cao hon t?i thi?u 2 l?n t?n s? tn hi?u cao nh?t, th cc m?u thu du?c ch?a d? thng tin c?a tn hi?u ban d?u. T/h f(t) c th? du?c ti t?o, dng b? l?c thng th?p
Cng th?c Nyquist: N >= 2f
N: t?c d? l?y m?u
f: t?n s? c?a tn hi?u du?c l?y m?u
D? li?u ti?ng ni
Gi?i h?n t?n s? <4000Hz
T?c d? l?y m?u c?n thi?t 8000 m?u/giy
65. i?u ch? xung m (PCM) PAM (Pulse Amplitude Modulation)
Cc xung du?c l?y m?u ? t?n s? R=2B
Lu?ng t? ha cc xung PAM
Xc d?nh gi tr? c?a di?m du?c l?y m?u, roi vo kho?ng no th l?y gi tr? kho?ng d
Ty thu?c vo cc m?c lu?ng t? 2n (n l s? bit c?n thi?t d? s? ha 1 xung)
M ha d? li?u
Th?c hi?n cc thao tc m ha thng tin tru?c khi truy?n di
66. i?u ch? xung m (PCM)
67. i?u ch? xung m
68. Non-linear coding M?c lu?ng t? khng d?u
Gi?m mo tn hi?u
Companding (compressing-expanding)
69. i?u ch? Delta (DM) Tn hi?u tuong t? du?c x?p x? b?i hm b?c thang (staircase)
Hnh vi nh? phn
i ln hay xu?ng 1 m?c (?) t?i m?i th?i kho?ng l?y m?u
Hi?u su?t
? ti t?o ti?ng ni t?t
PCM - 128 m?c (7 bit)
Bang thng tho?i 4khz
C?n 8000 x 7 = 56kbps d?i v?i PCM
K? thu?t nn d? li?u c th? c?i thi?n thm
V d?: k? thu?t m xen khung (interframe coding) cho video
70. i?u ch? Delta (DM)
71. i?u ch? Delta (DM)
72. Analog ? Analog ?ng d?ng
Dng d? di?u ch? d? li?u tuong t?: thay d?i t?n s? truy?n (t?n s? cao hon truy?n d?n t?t hon)
Dng cho FDM
K? thu?t
i?u ch? bin: Amplitude Modulation (AM)
i?u ch? gc (Angle Modulation)
i?u ch? t?n s?: Frequency Modulation (FM)
i?u ch? pha: Phase Modulation (PM)
73. i?u ch? bin (AM) Bin d? c?a sng mang du?c thay d?i b?i bin d? c?a tn hi?u du?c truy?n di
s(t) = [1+nax(t)]cos(2?fct)
T?o ra t/h 2 bn (DSBTC), trong d ch? c?n c m?t bn
na<1 t/h bao l b?n sao c?a t/h ban d?u
na >1 t/h bao c?t tr?c th?i gian (thng tin b? m?t)
Pt = Pc(1+ na2/2)
Pt v Pc cng su?t t/h du?c truy?n di v t/h sng mang
na ch? s? di?u ch?, t? s? bin d? t/h du?c truy?n v sng mang
SSB v DSBSC
Uu di?m
D? hi?n th?c (di?u ch? v gi?i di?u ch?)
D? bi?n d?i tn hi?u sang cc gi?i bang t?n khc nhau
Khuy?t di?m
D? b? ?nh hu?ng c?a nhi?u
Khng s? d?ng hi?u qu? nang lu?ng
74. i?u ch? bin (AM)
75. i?u ch? t?n s? (FM)
76. i?u ch? gc s(t) = Accos[2?fct + ?(t)]
Phuong php di?u t?n s? (FM)
Tn hi?u du?c truy?n di thay d?i thnh ph?n t?n s? c?a sng mang t? l? v?i bin d? v t?n s? c?a tn hi?u truy?n di
?(t) = nfm(t)
Uu di?m
Kh b? ?nh hu?ng c?a nhi?u
S? d?ng hi?u qu? nang lu?ng
Khuy?t di?m
Tn hi?u du?c di?u ch? yu c?u bang thng r?ng hon nhi?u tn hi?u truy?n di ban d?u (d? li?u)
Hi?n th?c m?ch di?u ch? v gi?i di?u ch? ph?c t?p hon so v?i phuong php di?u bin
77. i?u ch? gc
78. i?u ch? gc Phuong php di?u ch? pha (PM)
?(t) = npm(t)
Tn hi?u truy?n di khng ?nh hu?ng d?n thnh ph?n bin d? v t?n s? m ch? lm thay d?i pha c?a sng mang
Ph? t?n s? c?a tn hi?u du?c di?u ch? theo phuong php di?u pha tuong t? nhu phuong php di?u t?n ? phuong php di?u pha cung c cc d?c di?m tuong t? phuong php di?u t?n
Tuy nhin, c hai l do phuong php di?u pha du?c d? ch?p nh?n hon
?i v?i bn nh?n: t?n s? c?a tn hi?u nh?n du?c l c? d?nh, ch? c pha thay d?i nn ch? c?n thi?t k? b? l?c t?n s? ch? cho m?t t?n s? duy nh?t thay v nhi?u t?n s? nhu trong phuong php di?u t?n ? gi?m chi ph thi?t k? v hi?n th?c m?ch
Trong tru?ng h?p tn hi?u di?u ch? ch? nh?n m?t s? gi tr? (nhu tn hi?u s?), m?ch di?u ch? v gi?i di?u ch? hi?n th?c theo phuong php di?u pha du?c don gi?n r?t nhi?u
79. Hi?u su?t Bang thng
AM
BT = 2B
FM&PM
BT = 2(?+1)B
FM&PM c?n bang thng l?n hon so v?i AM
80. C?u trc knh truy?n M d? li?u Baudot (Emile Baudot)
5 bit (32 m)
dng 2 m 5 bit (letter & figure) d? m h?t cc k t?, ch? s? v d?u
ASCII (American Standard Code for Information Interchange)
7 bit (128 m), bao g?m cc k t? ch? thu?ng v hoa, cc k t? ch? s?, cc k t? d?u ch?m cu v cc k t? d?c bi?t.
Ph? bi?n nh?t hi?n nay du?c s? d?ng trong giao ti?p d? li?u tu?n t?.
EBCDIC (Extended Binary Coded Decimal Interchange Code)
8 bit
u?c dng trong cc h? th?ng my tnh IBM
Unicode
16 ho?c 32 bit
H?a h?n du?c s? d?ng r?ng ri trong tuong lai
81. M Baudot
82. M ASCII
83. C?u trc knh truy?n Song song (Parallel)
M?i bit dng m?t du?ng truy?n ring. N?u c 8 bits du?c truy?n d?ng th?i s? yu c?u 8 du?ng truy?n d?c l?p
? truy?n d? li?u trn m?t du?ng truy?n song song, m?t knh truy?n ring du?c dng d? thng bo cho bn nh?n bi?t khi no d? li?u c s?n (clock signal)
C?n thm m?t knh truy?n khc d? bn nh?n bo cho bn g?i bi?t l d s?n sng d? nh?n d? li?u k? ti?p
84. C?u trc knh truy?n Tu?n t? (Serial)
T?t c? cc bit d?u du?c truy?n trn cng m?t du?ng truy?n, bit ny ti?p theo sau bit kia
Khng c?n cc du?ng truy?n ring cho tn hi?u d?ng b? v tn hi?u b?t tay (cc tn hi?u ny du?c m ha vo d? li?u truy?n di)
2 cch truy?n
B?t d?ng b?: m?i k t? du?c d?ng b? b?i start v stop bit
?ng b?: m?i kh?i k t? du?c d?ng b? dng c?
85. Truy?n b?t d?ng b? D? li?u du?c truy?n theo k t? (5 ? 8 bits)
Ch? c?n gi? d?ng b? trong m?t k t?
Ti d?ng b? cho m?i k t? m?i
Hnh vi
?i v?i dng d? li?u d?u, kho?ng cch gi?a cc k t? l d?ng nh?t (chi?u di c?a ph?n t? stop)
? tr?ng thi r?nh, b? thu pht hi?n s? chuy?n 1 ? 0
L?y m?u 7 kho?ng k? ti?p (chi?u di k t?)
?i vi?c chuy?n 1 ? 0 cho k t? k? ti?p
Hi?u su?t
on gi?n
R?
Ph t?n 2 ho?c 3 bit cho m?t k t? (~20%)
Thch h?p cho d? li?u v?i kho?ng tr?ng gi?a cc k t? l?n (d? li?u nh?p t? bn phm)
86. Truy?n b?t d?ng b?
87. Truy?n b?t d?ng b? ?ng b? bit
Chuy?n d?i 1 byte thng tin thnh/t? chu?i bit
PISO SIPO
Clock thu?ng m?t d?ng b?
B? thu thu?ng dng clock g?p N l?n clock c?a b? pht
88. Truy?n b?t d?ng b?
89. Truy?n b?t d?ng b? ?ng b? k t? (character synchronization): dng start v stop bit
90. Truy?n b?t d?ng b? khung ?ng b? khung (frame synchronization): dng cc k t? di?u khi?n (STX, ETX, DLE)
91. Truy?n b?t d?ng b? - t?c d? xung clock
92. Truy?n b?t d?ng b? - t?c d? xung clock
93. Truy?n b?t d?ng b? - t?c d? xung clock
94. Truy?n d?ng b? Truy?n khng c?n start/stop
Ph?i c tn hi?u d?ng b?
?ng b? bit (bit synchronization): s? d?ng cc phuong php sau
Clock encoding and extraction (Timestamp)
Tch h?p thng tin d?ng b? (clock) vo trong d? li?u truy?n
?u nh?n s? tch thng tin d?ng b? d?a vo d? li?u nh?n du?c
RZ, Manchester (NRZ signaling), differential Manchester
Digital Phase-Lock-Loop
Dng m?t du?ng tn hi?u d?ng b? ring bi?t
S? d?ng m?t ngu?n clock ?n d?nh du?c gi? d?ng b? v?i d? li?u d?n t?i noi nh?n
M ha thng tin ph?i d?m b?o c s? thay d?i bit trong m?t kho?ng th?i gian d? d? ngu?n clock du?c ti d?ng b?
C?n s? d?ng cc phuong php m ha nh? phn (AMI, HDB3, B8ZS)
Thch h?p khi truy?n m?t kho?ng cch ng?n
Tn hi?u d?ng b? d? b? suy gi?m trn du?ng truy?n
95. M ha v tch d? li?u d?ng b?
96. Digital Phase Lock Loop
97. Truy?n d?ng b? ?ng b? khung (frame synchronization): s? d?ng cc phuong php sau
Character-oriented synchronous transmission
Dng cc k t? di?u khi?n : SYN, STX, ETX, DLE.
Bit-orienter synchronous transmission
Dng cc m?u bit di?u khi?n (flag byte or flag pattern)
? bit stuffing problem
Hi?u qu? (ph t?n th?p) hon so v?i truy?n b?t d?ng b?
98. L?i
99. i?u khi?n l?i Mi tru?ng truy?n d?n b? nhi?u (di?n, t?, ) ? d? li?u nh?n c l?i
2 cch kh?c ph?c khi pht hi?n c l?i
Forward error control: thng tin s?a sai du?c thm vo cc k t? ho?c cc frame truy?n di, d? bn nh?n c th? pht hi?n khi no c l?i v l?i n?m ? du d? s?a (c kh? nang s?a l?i)
Feedback (backward) error control: thng tin s?a sai du?c thm vo cc k t? ho?c cc frame truy?n di ch? d? d? pht hi?n khi no c l?i (khng c kh? nang s?a l?i). Co ch? yu c?u truy?n l?i k t?/frame sai du?c dng trong tru?ng h?p ny
Co ch? pht hi?n l?i
BER: xc su?t 1 bit don b? l?i trong m?t th?i kho?ng nh?t d?nh
Phn lo?i l?i
Single-bit error nhi?u tr?ng
Burst error: chu?i cc bit lin ti?p b? l?i nhi?u xung, suy gi?m (khi truy?n v tuy?n)
100. Qu trnh pht hi?n sai
101. Parity Bit parity
Parity ch?n: (N + P) ph?i l m?t s? ch?n
Parity l?: (N + P) ph?i l m?t s? l?
N: t?ng s? bit 1 c trong d? li?u c?n ki?m tra l?i
P: gi tr? c?a bit parity, l 0 hay 1 sao cho t?ng s? bit 1 (N+P) lun l m?t s? ch?n (l?) ty theo phuong php parity ch?n hay l? tuong ?ng
102. Parity ?c di?m
Ch? d du?c l?i sai m?t s? l? bit, khng d du?c l?i sai m?t s? ch?n bit
Khng s?a du?c l?i
Hi?u su?t truy?n thng tin km, do s? bit thm vo d? d tm l?i chi?m t? l? l?n so v?i d? li?u truy?n di.
103. Block Sum Check Block Sum Check (BSC): s? d?ng parity hng v c?t
Khng s?a du?c sai, ch? s?a du?c sai khi s? bit sai trong d? li?u l m?t
D tm du?c t?t c? cc l?i sai m?t s? l? bit v h?u h?t cc l?i sai m?t s? ch?n bit.
Khng d du?c l?i sai m?t s? ch?n bit x?y ra d?ng th?i trn c? hng v c?t.
104. Block Sum Check Bi?n th?
Dng t?ng b 1 (1s-complement sum) thay cho t?ng modulo 2 (2-modulo sum)
Cc k t? trong block du?c truy?n du?c coi nhu cc s? nh? phn khng d?u
T?t hon phuong php modulo 2
105. Cyclic Redundancy Check Nguyn l
k-bit message
Bn pht t?o ra chu?i n bit FCS (Frame Check Sequence) sao cho frame g?i di (k+n bit) chia h?t cho 1 s? xc d?nh tru?c
Bn thu chia frame nh?n du?c cho cng 1 s? v n?u khng c ph?n du th c kh? nang khng c l?i
S? h?c modulo 2
Exlusive-or
106. Cyclic Redundancy Check Xc d?nh FCS
T: frame du?c truy?n (k+n bit)
M: message d? li?u c?n truy?n (k bit d?u c?a T)
F: FCS (n bit sau c?a T)
P: s? chia du?c xc d?nh tru?c (n+1 bit)
Ki?m tra
107. Cyclic Redundancy Check Cch khc d? xc d?nh FCS: da th?c
M = 111101 ? M(x) = X5 + X4 + X3 + X2 + 1
P = 1101 ? P = X3 + X2 + 1
? FCS c 3 bits (n = 3)
D? li?u d?ch tri n bits: 2nM(x) = X8 + X7 + X6 + X5 + X3
108. Cyclic Redundancy Check
109. Cyclic Redundancy Check P
Di hon 1 bit so v?i FCS mong mu?n
u?c ch?n ty thu?c vo lo?i l?i mong mu?n pht hi?n
Yu c?u t?i thi?u: msb v lsb ph?i l 1
Bi?u di?n l?i
Error=ngh?ch d?o bit (i.e. exclusive-or c?a bit d v?i 1)
Tr = T + E
T: frame du?c truy?n
Tr: frame nh?n du?c
E: error pattern v?i 1 t?i nh?ng v? tr l?i x?y ra
Error khng b? pht hi?n iff Tr chia h?t cho P (i.e. iff E chia h?t cho P)
Cc l?i du?c pht hi?n
T?t c? cc l?i bit don
T?t c? cc l?i kp n?u P c t nh?t 3 ton h?ng
M?t s? l? l?i b?t k? n?u P ch?a 1 th?a s? (X+1)
B?t k? l?i chm no m chi?u di c?a chm nh? hon chi?u di FCS
H?u h?t cc l?i chm l?n hon
110. Cyclic Redundancy Check 4 P du?c s? d?ng r?ng ri
CRC-12 = X12 + X11 + X3 + X2 + X + 1
12-bit FCS
6-bit characters
CRC-16 = X16 + X15 + X2 + 1
16-bit FCS
8-bit characters
US
CRC-CCITT = X16 + X12 + X5 +1
Europe
CRC-32 = X32 + X26 + X23 + X22 + X16 + X12 + X11 + X10 + X8 + X7 + X5 + X4 + X2 + X + 1
32-bit FCS
Point-point synchronous transmission, DoD apps
111. Cyclic Redundancy Check
112. Cyclic Redundancy Check
113. S?a l?i S?a cc l?i du?c pht hi?n thng thu?ng yu c?u truy?n l?i kh?i d? li?u
Khng thch h?p cho cc ?ng d?ng trao d?i d? li?u khng dy
BER cao
Truy?n l?i nhi?u
Th?i gian tr? truy?n l?n hon nhi?u so v?i th?i gian truy?n d? li?u (vd truy?n v? tinh)
Kh?i d? li?u du?c truy?n l?i b? l?i v nhi?u kh?i d? li?u khc ti?p theo
C?n thi?t ph?i s?a l?i d?a vo cc d? li?u nh?n du?c
114. Qu trnh s?a sai
115. Qu trnh s?a sai M?i kh?i d? li?u k bit du?c nh x? vo kh?i n bit (n>k)
T? m Codeword
Forward error correction (FEC) encoder
Codeword du?c truy?n di
Chu?i bit nh?n du?c tuong t? nhu chu?i du?c truy?n di, nhung c ch?a l?i
Codeword du?c g?i t?i b? gi?i m FEC
N?u khng c l?i, trch xu?t kh?i d? li?u ban d?u
M?t vi m?u l?i c th? du?c pht hi?n v s?a l?i
M?t vi m?u l?i c th? du?c pht hi?n nhung khng s?a du?c
M?t vi m?u l?i c th? khng du?c pht hi?n (t x?y ra)
FEC trch xu?t kh?i d? li?u sai
116. C?u hnh du?ng truy?n C?u hnh
S?p x?p v?t l cc tr?m trn mi tru?ng
C?u hnh truy?n th?ng
117. Giao ti?p Giao ti?p
Thi?t b? x? l d? li?u (DTE) thu?ng khng c cc phuong ti?n pht d? li?u
C?n m?t thi?t b? giao ti?p (DCE) v d?: modem, NIC,
DCE pht cc bit d? li?u trn mi tru?ng truy?n d?n
DCE trao d?i d? li?u v thng tin di?u khi?n v?i DTE
u?c th?c hi?n thng qua m?ch trao d?i
C?n m?t chu?n giao ti?p r rng
?c tnh
Co kh
K?t n?i
i?n
i?n p, d?nh th?i, m ha,
Ch?c nang
D? li?u, di?u khi?n, d?nh th?i, d?t,
Th? t?c
Chu?i cc s? ki?n
118. Chu?n V.24/EIA232F ITU-T v.24
Ch? d?c t? ch?c nang v th? t?c
Tham kh?o cc chu?n khc cho cc d?c tnh co kh v d?c tnh di?n
EIA-232-F (USA)
RS-232
?c tnh co kh: ISO 2110
?c tnh di?n: v.28
Ch?c nang: v.24
Th? t?c: v.24
119. K?t n?i V.24/EIA232 (DTE)
121. Local/Remote loopback testing
122. Nghi th?c V d? modem ring b?t d?ng b?
Khi modem du?c b?t ln v s?n sng, n (DCE) b?t tn hi?u DCE ready
Khi DTE s?n sng g?i d? li?u, n b?t tn hi?u Request To Send
C?m ch? d? nh?n d? li?u (n?u trong ch? d? truy?n half-duplex)
Modem dp l?i s?n sng b?ng tn hi?u Clear To Send
DTE g?i d? li?u
Khi d? li?u d?n, modem g?n vo DTE s? b?t tn hi?u Line Signal Detector v g?i d? li?u cho DTE
123. Ho?t d?ng quay s? (1)
124. Ho?t d?ng quay s? (2)
125. Ho?t d?ng quay s? (3)
126. Chu?n giao ti?p EIA RS232C
127. Trao d?i thng tin gi?a DCE v DTE Trao d?i thng tin gi?a DTE v DCE
Truy?n d? li?u (DTE?DCE)
B?t DTR v RTS
?i DSR
?i CTS
Truy?n d? li?u
Nh?n d? li?u (DCE?DTE)
B?t DTR
?i DSR
Nh?n d? li?u
128. C?u hnh dy d?n k?t n?i DTE ? DTE
129. Nn d? li?u Run-length encoding (packed decimal)
Dng cho message g?m cc ch? s?
M BCD thay v ASCII
C th? dng 4 bit th?p d?i v?i cc thng tin d?y d?
130. Differential encoding (relative encoding)
Encoding used if differences between values is much smaller than the values themselves
Original 1509 1506 1508 1510 1511 1509 1513
Encoded 1509 -3 +2 +2 +1 -2 +4
Send only the difference in magnitude
Character suppresion
Encoding used if 3 or more of same character found
Original AAAABBCCCCCDEEEEFF
Encoded A4B2C5D1EEEE4F2
Nn d? li?u
131. Nn d? li?u Huffman encoding (Statistical Methods)
?c di?m
y l m th?ng k (phuong php nn m t?i uu)
M ha d?a trn xc su?t s? d?ng c?a cc k t?
Nh?ng k t? du?c dng nhi?u nh?t s? c t? m ng?n nh?t
Khng c tnh prefix
Gi?i thu?t
S?p x?p cc ngu?n tin c xc su?t gi?m d?n
M?t c?p bit 0-1 du?c gn cho 2 ngu?n tin c xc su?t nh? nh?t
2 ngu?n tin ny du?c k?t h?p, t?o thnh ngu?n tin m?i c xc su?t b?ng t?ng xc su?t c?a 2 ngu?n tin thnh ph?n
S?p x?p l?i cc ngu?n tin theo th? t? gi?m d?n c?a xc su?t
Qu trnh trn du?c l?p l?i d?n khi 2 ngu?n tin cu?i cng du?c k?t h?p
T? m cho m?i ngu?n tin du?c vi?t theo th? t? t? g?c d?n ng?n
Chi?u di t? m trung bnh Lavg = ?li x pi
li : chi?u di ngu?n tin Xi
pi : xc su?t xu?t ngu?n tin Xi
132. Huffman code
133. Nn d? li?u Shannon-Fano encoding (Statistical Methods)
?c di?m
M t?i uu
Khng c tnh prefix
Gi?i thu?t
S?p x?p cc ngu?n tin theo th? t? gi?m d?n v? xc su?t
Chia cc ngu?n tin thnh hai ph?n c xc su?t x?p x? nhau v gn 0 cho ph?n trn, gn 1 cho ph?n du?i
L?p l?i bu?c trn cho m?i ph?n cho d?n khi ch? cn m?t ngu?n tin
Ghi ra cc t? m
134. Shannon Fano Cc ngu?n tin v xc su?t xu?t hi?n c?a cc ngu?n tin tuong ?ng
X1 (30%), X2 (20%), X3 (10%), X4 (10%), X5 (20%), X6 (5%), X7 (3%), X8 (2%)
Lavg = 2.0,3+2.0,2+3.0,2+3.0,1+3.0,1+4.0,05+5.0,03+5.0,02 = 2,65 bits
135. Phn h?p knh (Multiplexing)
136. Frequency Division Multiplexing (FDM) Phuong php ny ch? hi?n th?c du?c khi bang thng mi tru?ng truy?n l?n hon bang thng m tn hi?u du?c truy?n yu c?u
Nhi?u tn hi?u c th? du?c truy?n d?ng th?i n?u m?i tn hi?u du?c di?u ch? trn m?t t?n s? sng mang
Cc t?n s? sng mang khc nhau sao cho bang thng c?a cc tn hi?u du?c di?u ch? khng trng l?p nhau (guard bands)
V d? broadcast radio
Knh truy?n du?c c?p pht ngay c? khi khng c d? li?u (c?p pht tinh)
137. FDM
138. FDM
139. FDM
140. FDM
141. FDM
142. Wavelength Division Multiplexing Nhi?u chm nh sng v?i t?n s? khc nhau
Truy?n trong cp quang
M?t d?ng c?a FDM
M?i mu nh sng (chi?u di sng khc nhau) du?c truy?n trn knh d? li?u ring bi?t
1997 t?i Bell Labs
100 chm nh snh
M?i chm t?c d? 10 Gbps
1 terabit per second (Tbps)
H? th?ng thuong m?i hi?n t?i c 160 knh, m?i knh 10 Gbps
Phng th nghi?m (Alcatel) c th? c 256 knh v?i t?c d? 39.8 Gbps m?i knh
10.1 Tbps
Trn 100km
143. Ho?t d?ng WDM Cng ki?n trc t?ng qut nhu cc FDM khc
Ngu?n sng t?o ra cc chm laser v?i t?n s? khc nhau
Nhi?u chm sng k?t h?p v?i nhau d? lan truy?n trn cng m?t cp quang
B? khu?ch d?i quang h?c
Khu?ch d?i t?t c? chi?u di sng khc nhau
Thng thu?ng kho?ng cch ~10km
Phn knh t?i dch d?n
Thng thu?ng t?m chi?u di sng 1550nm
200MHz per channel
Hi?n t?i ln d?n 50GHz
144. Dense Wavelength Division Multiplexing DWDM
Chua c d?nh nghia chnh th?c (chua chu?n ha)
Cc knh st nhau hon WDM
200GHz
145. Time Division Multiplexing (TDM) Synchronous TDM
Phuong php ny ch? hi?n th?c du?c khi t?c d? d? li?u (bang thng,) mi tru?ng truy?n l?n hon t?c d? d? li?u m tn hi?u du?c truy?n yu c?u
Nhi?u tn hi?u (c? analog v digital) c th? du?c truy?n d?ng th?i trn cng m?t du?ng truy?n b?ng cch dan xen cc ph?n c?a m?i tn hi?u theo th?i gian (time slot)
Time slot du?c gn tru?c v tinh (time slot du?c c?p pht ngay c? khi khng c d? li?u d? truy?n)
Time slot c th? du?c gn khng d?ng d?u gi?a cc ngu?n d? li?u
146. TDM
147. TDM
148. TDM
149. TDM i?u khi?n lin k?t Khng c?n header v tailer
Khng c?n cc nghi th?c di?u khi?n lin k?t d? li?u (cho ton b? du?ng truy?n phn/h?p)
i?u khi?n dng
T?c d? d? li?u c?a du?ng truy?n phn/h?p c? d?nh
N?u c m?t knh khng th? nh?n d? li?u, cc knh khc v?n ti?p t?c
Ngu?n pht tuong ?ng ph?i ngung
? b? knh tr?ng (empty slot)
i?u khi?n l?i
L?i du?c pht hi?n v x? l b?i t?ng knh ring bi?t
150. TDM i?u khi?n lin k?t
151. TDM Framing Khng c c? (flag) ho?c cc k t? SYNC d? dng khung cc b TDM
Ph?i c co ch? d?ng b?
Co ch? dng khung s?
M?t bit di?u khi?n du?c thm vo m?i b TDM
Cc bit di?u khi?n ny t?o thnh m?t knh khc knh di?u khi?n
Dng m?u bit d?nh d?ng trn knh di?u khi?n
V d? m?u 01010101, khc v?i knh d? li?u
So snh m?u bit d?n trn t?ng knh v?i m?u bit m?u bit d?ng b?
152. TDM pulse stuffing V?n d?: d?ng b? cc ngu?n d? li?u khc nhau
Tn hi?u clock trn cc ngu?n d? li?u khc nhau b? tri (drift)
T?c d? d? li?u c?a cc ngu?n d? li?u khc nhau khng quan h? theo m?t t? l? don gi?n
Gi?i php Pulse Stuffing
T?c d? d? li?u d?u ra (khng tnh cc bit khung) cao hon t?ng cc t?c d? d?u vo
Chn thm cc bit/xung khng c nghia vo m?i tn hi?u d?u vo cho d?n khi n b?ng v?i clock c?c b?
Cc bit/xung du?c thm vo t?i nh?ng v? tr c? d?nh (bi?t tru?c) trong khung v n s? b? lo?i b? khi d?n b? phn knh
153. TDM ngu?n tuong t? v ngu?n s?
154. TDM h? th?ng truy?n mang Phn c?p TDM
USA/Canada/Japan dng m?t h? th?ng
ITU-T (chu Au) dng m?t h? th?ng khc (nhung tuong t?)
H? th?ng M? xy d?ng d?a trn d?nh d?ng DS-1
24 knh du?c phn h?p
M?i khung c 8 bit/knh v 1 bit khung
? 193 bit/khung
?i v?i truy?n tho?i, m?i knh ch?a m?t t? c?a d? li?u du?c s? ha (PCM, 8000 m?u/giy)
T?c d? d? li?u 8000 x 193 = 1.544Mbps
5 trong s? 6 khung c cc m?u PCM 8 bit
Khung th? 6 ch?a m?t t? PCM 7 bit v m?t bit tn hi?u
Cc bit tn hi?u t?o thnh m?t dng (stream) cho m?i knh d? di?u khi?n v ch?a thng tin tm du?ng
?nh d?ng tuong t? cho d? li?u s?
23 knh d? li?u (7 bit/khung v 1 bit ch? th? cho d? li?u ho?c di?u khi?n h? th?ng)
Knh th? 24 dng d? d?ng b?
DS-1 c th? dng h?n h?p d? li?u s? v tuong t?
Dng c? 24 knh
Khng c k t? d?ng b?
155. TDM T1 vs. E1
156. TDM T1 D?ch v? s? m?c 0 (DS0) = 64 kbps
T1 = 24 knh tho?i = d?ch v? s? m?c 1 (DS1)
TDM
ng khung don gi?n: thm 101010 (1 bit/khung)
B?t k? chu?i khc ? ti d?ng b?
157. TDM du?ng truy?n E1 E1
Dng ? chu u, tuong t? nhu T1 (dng ? M?)
C 32 bytes trong m?t khung di 125s = 2048 Mbps
30 knh du?c dng cho d? li?u
1 knh dng d? d?ng b?
1 knh dng d? bo hi?u (di?u khi?n)
159. TDM phn c?p
160. TDM b?t d?ng b? TDM th?ng k/thng minh
Trong TDM d?ng b?, nhi?u slot c th? b? b? tr?ng
TDM b?t d?ng b? c?p pht time slot d?ng ty theo nhu c?u
B? h?p knh qut cc du?ng nh?p v t?p h?p d? li?u cho d?n khi d?y khung
T?c d? d? li?u ra th?p hon t?c d? cc du?ng nh?p g?p l?i
C th? gy v?n d? trong th?i gian cao di?m
?m cc du?ng nh?p
Gi? kch thu?c b? d?m t?i thi?u d? gi?m th?i gian tr?
161. TDM b?t d?ng b? mc t?c d? d? li?u t?i da c?a du?ng truy?n trung k?
mi t?c d? d? li?u t?i da c?a ngu?n th? i
pi xc xu?t d? li?u c?a ngu?n th? i
mc c th? nh? hon t?ng cc mi
?pimi < mc
Nguyn t?c 80%
Bao nhiu terminal t?c d? 9600bps c th? dng chung du?ng truy?n 56Kbps khi dng k? thu?t TDM ho?c STDM (pi l 75%)
162. Kch thu?c b? d?m v th?i gian tr?
163. TDM b?t d?ng b? - d?nh d?ng khung
164. Asymmetrical Digital Subscriber Line ADSL
xDSL
High data rate DSL
Single line DSL
Very high data rate DSL
Lin k?t gi?a thu bao v m?ng
u?ng thu bao
Hi?n t?i dng cp twisted pair
C th? c bang thng l?n hon
1 MHz ho?c l?n hon
B?t d?i x?ng
T?c d? dng d? li?u xu?ng (downstream) l?n hon t?c d? dng d? li?u ln (upstream)
FDM
25kHz th?p nh?t cho tho?i
Plain old telephone service (POTS)
Dng k? thu?t lo?i b? echo (echo cancellation) ho?c FDM d? cho 2 bang t?n
Dng FDM trong cc bang t?n
Ph?m vi 5.5km
165. C?u hnh knh truy?n ADSL
166. Discrete Multitone DMT
Nhi?u tn hi?u sng mang ? cc t?n s? khc nhau
Vi bit trn m?i knh
Knh ph? 4kHz
G?i t/h test v dng knh ph? v?i t? s? SNR t?t hon
256 knh ph? downstream m?i knh 4kHz (60kbps)
15.36MHz
Impairments bring this down to 1.5Mbps to 9Mbps
167. DMT Transmitter
168. ?c thm W. Stallings, Data and Computer Communications (7th edition), Prentice Hall 2004, chapters 3, 5, 6, 8
Web pages from ITU-T on V. specification
Cc website v? ADSL v SONET