3.6 Intersymbol interference. 1 Your site here

Size: px

Start display at page:

Download "3.6 Intersymbol interference. 1 Your site here"

Shannon Curtis
6 years ago
Views:

1 3.6 Intersymbol intererence 1

2 3.6 Intersymbol intererence what is intersymbol intererence and what cause ISI 1. The absolute bandwidth o rectangular multilevel pulses is ininite. The channels bandwidth is limited. 2. The channels property is not plat. Pulses are iltered improperly as they through channel, they will spread in time. 2

3 Intersymbol intererence Input waveorm w in (t) individual impulse response received waveorm w out (t)(pulse response sum) 0 t 0 Ts t 0 t 码间串扰 0 t Sample points (transmitter clock) 0 t 0 Fig 3-23 Ts sample point (receiver clock) Ts sample point (receiver clock) t 3

4 problem Intersymbol intererence How can we restrict the bandwidth and still not introduce ISI? Nyquist discovered three dierent methods or pulse shaping that could be used to eliminate ISI. 4

5 Intersymbol intererence The digital signaling system : w in (t) send ilter channel(ilter) w c (t) receiver ilter w out (t) h T ( t ) h C ( t ) h R ( t ) H T ( ) H C ( ) H R ( ) recovered rounded pulse (to sample and decode circuit) The equivalent impulse response is h e (t)=h(t)*h T (t)*h C (t)*h R (t) The equivalent system transer unction : H e ( )=H( ) H T ( t ) H C ( ) H R ( ) 5

6 Nyquist s First Method (zero ISI) I the equivalent system impulse response satisies the condition h e (kt s +τ)= It will eliminating ISI C, k=0 0, k 0 Where : C is a nonzero constant, K is an integer, T s is the symbol (sample ) clocking period, τis the oset in the receiver sampling clock times, compared with the clock times o the input symbols. 6

7 Nyquist s First Method (zero ISI) I we choose a (sinx)/x unction or h e (t), the impulse response satisies Nyquist s irst criterion or zero ISI. sinπ t h e ( t) = π s t s where s =1/ T s sinπ st he ( t) = π t s H ( ) = e 1 s s -3T s -2T s -T s T s 2T s 3T s - s /2 s /2 7

8 Nyquist s First Method (zero ISI) I the transmit and receive ilters are designed so that the overall transer unction is H e ( ) = 1 s There will be no ISI, urthermore, the absolute bandwidth o this transer unction is B = / 2 s s Diiculties: sin π s t h e ( t ) = π t -3T s -2T s -T s T s 2T s 3T s - s /2 s /2 H e ( ) is diicult to approximate because o the steep skirts in the ilter transer unction The synchronization o the clock in the decoding sampling circuit has to be 8almost perect. s H e ( ) = 1 s s

9 Nyquist s First Method (zero ISI) Because o these diiculties, we are orced to consider other pulse shapes The idea is to ind pulse shapes that go through zero at adjacent sampling points, and yet have an envelope that decays much aster than 1/x, so that clock jitter in the sampling times does not cause appreciable ISI Solution: Raised cosine-rollo Nyquist ilter 9

Raised cosine-rollo Nyquist ilter The raised cosine-rollo Nyquist ilter has the transer unction 1, 1 = 1 + 2 0, π ( cos 10 ) 1 H e ( ),

10 Raised cosine-rollo Nyquist ilter The raised cosine-rollo Nyquist ilter has the transer unction 1, 1 = , π ( cos 10 ) 1 H e ( ), 1 2 Δ Where B is the absolute bandwidth Δ = B 0, 1 = 0 Δ 0 is 6dB bandwidth o the ilter The rollo actor is deined: r = Δ 0, < < > 1 B < B

11 Raised cosine-rollo Nyquist ilter The corresponding impulse response is: h e ( t) = F 1 sin 2π 0t cos 2π Δt [ H ( )] = 2 e 0 2π 0 t 1 (4 Δ t) 2 Frequency and time response or dierent rollo actor As the absolute bandwidth is increased (r=0.5 or 1): 1. The iltering requirements are relaxed. 2. The clock timing requirements are also relaxed. 11

12 Raised cosine-rollo Nyquist ilter T s = 1/(2 0 ) D = 1/ Ts = = D / 2 The 6-dB bandwidth o the raised cosine-rollo ilter 0 is designed to be hal the symbol (baud) rate. The baud rate that the raised cosine-rollo system can support without ISI r Δ = = Δ B / B = 1+ r D 2B = 1 + r 12

13 Raised cosine-rollo Nyquist ilter Example 3-1 Assume that a binary digital signal, with Polar NRZ signaling, is pass through a communication system with a raised cosine-rollo iltering charcteristic. Let the rollo actor be the bit rate o the digital signal is 64 kbit/s. Determine the absolute bandwidth o the iltered digital signal. 13

14 Raised cosine-rollo Nyquist ilter The raised cosine-rollo ilter is only one o a more general class o ilters that satisy Nyquist s irst criterion The general class o ilters that satisy Nyquist s irst criterion---- Nyquist ilter. 14

15 Nnyquist ilter A ilter is said to be a Nyquist ilter i the eective transer unction is H e ) = 0, 2 + Y ( ) 0 ( 0 15, < 2 elsewhere where Y( ) is a real unction that is even symmetric about =0 Y( - )=Y( ), < 2 0 And Y( ) is odd symmetric about = 0 Y( ) = -Y( + 0 ), < 0 Then there will be no ISI at the system output i the symbol rate is D = = s 2 0

16 Nnyquist ilter H e ) = 0, 2 + Y ( ) 0 ( 0, < 2 elsewhere 16

17 Nyquist second and third methods or control o ISI Nyquist s second method (ISI control) Allows some ISI to be introduced in a controlled way, so that it can be canceled out the receiver and the data can be recovered without error i no noise is present. Nyquist s third method (ISI control) The eect o ISI is eliminated by choosing h e (t): the area under h e (t) pulse within the desired symbol interval, Ts, is not zero, but the areas under h e (t) in adjacent symbol intervals are zero. 17

18 3.7 Dierential pulse code modulation 18

19 Dierential pulse code modulation The reason o we use DPCM There is a lot o redundancy in the signal samples. The bandwidth and the dynamic range o a PCM system are wasted Solution To transmit the dierence in adjacent sample values. That is, to use Dierential pulse code modulation (DPCM) Method To use prediction ilter 19 x n k = i= 1 a i x n i

20 Dierential pulse code modulation Prediction ilter may be realized by using a tapped delay line to orm a transversal ilter Y(nT s ) Delay T s Delay T s Delay T s The output samples are a 1 a 2. a l. a k z( nt s ) = K l = 1 a l y( nt s lt s ) z(nt s ) In simpliied notation: z n = K l= 1 a l y n l 20

21 Dierential pulse code modulation The irst coniguration using prediction rom samples o input signal 21

22 Dierential pulse code modulation The second coniguration using prediction rom quantized dierential signal 22

23 DPCM -- Eects o noise DPCM, like PCM, ollows the 6-dB rule S n db = 6.02n +α Unlike companded PCM, the αor DPCM varies over a wide range, depending on the Properties o the input analog signal. or DPCM speech: -3<α<15 The DPCM perormance may be compared with that or PCM For the same SNR, DPCM could require 3 or 4 ewer bits per sample than companded PCM. 23

24 DPCM standard A 32-Kbits s DPCM CCITT standard: To use 4-bit quantization at an 8- Ksample s rate or encoding 3.2-KHz bandwidth VF signals. A 64-Kbits s DPCM CCITT standard: To use 4-bit quantization and 16- Ksample s or encoding audio signals that have a 7-KHz bandwidth. 24

25 3.8 Delta modulation 25

26 Delta modulation DM Delta Modulation. It is a special case o DPCM. +v c Cn=1 Characteristics: There are only two quantizing levels Only one bit is transmitted per sample. Cn=0 -v c 26

27 DM system waveorms 27

28 Granular noise & slope overload noise Slope overload noise Granular noise wish δ wish δ Slope overload noise will decrease as δ increase. Granular noise will decrease as δ decrease. 28

29 Granular noise & slope overload noise Granular noise & slope overload noise 29

30 Granular noise & slope overload noise Example 3-5: Design o a DM system. problem: ind the step size δ required to prevent slope overload noise or the case when the input signal is a sine wave. w( t) = Asinω t a δ t 30

31 SNR or the DM system the granular noise power in the analog output signal band: N = < n 2 2 B δ > = p B n ( ) d = 3 s From eq.(3-84),with equality: B N = 4π 2 3 A 2 2 s 2 a B The signal power is (or a sine-wave test signal) 2 2 A S =< w ( t) >= 2 31

32 SNR or the DM system The resulting average signal-toquantizing noise ratio: S N out 3 = 8π 2 3 s 2 a B a B s --- the DM sampling requency ---the requency o the sinusoidal input ---the bandwidth o the receiving system Attention: This Eq. is valid only or sinusoidal-type signal 32

33 Adaptive Delta modulation and continuously variable slope Delta modulation Adaptive Delta modulation ADM : the step size vary as a unction o time as the input waveorm changes. When signal δ When signal δ 33

34 Adaptive Delta modulation and continuously variable slope Delta modulation Method 1 The step size may be adapted by examining the DM pulses at the transmitter output. When the DM pulses consists o a string o pulses with the same polarity, the step size is increased until the DM pulses begin to alternate in polarity, then the step size is decreased, and so on. 34

35 Adaptive Delta modulation and continuously variable slope Delta modulation step-size Algorithm: Data Sequence x x 0 1 x X: don t care Number o Successive Binary 1 s or 0 s Step-size Algorithm 2 4 (d) δ δ δ δ 35

36 continuously variable slope delta modulation (CVSD) Method 2 CVSD is another variation o ADM An integrator (instead o accumulator) is used, so that z(t) is made continuously variable Product The Motorola MC34115 The Motorola MC

37 Summary Question Which is better, PCM or DM? The answer depends on the criterion used or comparison and the type o message. To have a relatively simple, low-cost system, DM may be the best To have a high output SNR, PCM probably the best To interace existing equipment, compatibility, PCM has the advantage. 37

38 3.9 Time-Division Multiplexing (TDM) 38

39 Time-Division Multiplexing (TDM) Why we must use TDM? Aims: to make use o the channel bandwidth to achieve high spectral eiciency What is the TDM? TDM (Time-division multiplexing) is the time interleaving o samples rom several sources so that the inormation orm these sources can be transmitted serially over a single communication channel. 39

40 Time-Division Multiplexing (TDM) three analog sources are multiplexed over a PCM system. 40

41 Time-Division Multiplexing (TDM) The pulse width o the TDM PAM: T s 3 = 1 3 s The pulse width o the TDM PCM: T s 3 n = 1 3n s TDM PAM Ts 41 Ts 3

42 Frame synchronization. Aims o the rame sync. : To make the received multiplexed data can be sorted and directed to the appropriate output channel at the TDM receiver. The rame sync. Signal can be provided to the receiver demultiplexer by: Sending a rame sync signal over a separate channel Deriving the rame sync rom the TDM signal itsel 42

43 Frame synchronization. Frame synchronization word: A segmented bits data stream which obeys some rules. Usually, it should be unique in the data stream, or at least, the appear probability is very small. s1 s2 sk Ch. 1 data Ch. 2 data Ch. N data s1 s2 sk 43

44 Time-Division Multiplexing (TDM) Example 3.6 Design a time-division multiplexer that will accommodate 11 sources, assume that the sources have the ollowing speciications: Source 1. analog, 2-kHz bandwidth Source 2. analog, 4-kHz bandwidth Source 3. analog, 2-kHz bandwidth Sources digital, synchronous at 7200 bits/s. 44

45 Time-Division Multiplexing (TDM) Example 45

46 Time-Division Multiplexing (TDM) The preceding example illustrates the main advantage o TDM: It can easily accommodate both analog and digital sources. Unortunately, when analog signals are converted to digital signals without redundancy reduction, they consume a great deal o digital system capacity. 46

47 TDM hierarchy Two categories: TDM used in digital computer system The output rate has been standardized to 1.2, 2.4, 3.6, 4.8, 7.2, 9.6, 14.4, 19.2, 28.8 kb/s. and to 10 and 100 to 1000Mb/s, 10Gb/s. TDM used by common carrier North American digital TDM hierarchy Europe digital TDM hierarchy (CCITT TDM) 47

48 TDM hierarchy North American digital TDM hierarchy: ( T1 TDM system ) 48

49 TDM hierarchy 24-VF analog telephone signals are converted to a DS-1 (1.544 Mbit/s) data stream The sampling rate used on each o the 24-VF analog signals is 8 khz Each analog sample is encoded into an 8-bit PCM word There are 8*24=192 bits o data, plus one bit is added or rame synchronization, yielding a total o 193 bits per ram. 49

50 TDM hierarchy Europe digital TDM hierarchy: (CCITT TDM standard) 50

EEE482F: Problem Set 1

EEE482F: Problem Set 1 1. A digital source emits 1.0 and 0.0V levels with a probability of 0.2 each, and +3.0 and +4.0V levels with a probability of 0.3 each. Evaluate the average information of the source.