Summary: ISI. No ISI condition in time. II Nyquist theorem. Ideal low pass filter. Raised cosine filters. TX filters

Transcription

1 UORIAL ON DIGIAL MODULAIONS Part 7: Intersymbol interference [last modified: ] Roberto Garello, Politecnico di orino Free download at: (personal use only)

2 Part 7: Intersymbol interference Summary: ISI No ISI condition in time II Nyquist theorem Ideal low pass filter Raised cosine filters X filters R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 2

3 ime domain constraint { ( ),..., ( ),..., ( ) i m } M = s t s t s t Until now: all the constellation signals have finite time domain 0 t < R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 3

4 Consequence here is no intersymbol interference: every symbol can be independently analyzed. s( t) = ( s[0]( t) s[]( t)... s[ i]( t)... r( t) = ( r[0]( t) r[]( t)... r[ i]( t)... R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 4

5 Example 0 t 0 t < s(t) 2 3 t R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 5

6 ISI What happens if we remove this time domain constraint? s( t) = ( s[0]( t) s[]( t)... s[ i]( t)... r( t) = ( r[0]( t) r[]( t)... r[ i]( t)... he signal received in a given interval also depends on other transmitted signals. R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 6

7 Example t < 2 s(t) 2 3 R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 7

8 ISI In this example, it is not possible to separate the symbol intervals. Does this happen for any signal without the time constraint? R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 8

9 Analytical procedure Let us consider a mono-dimensional constellation with basis signal p(t). We have: [ ] ( ) s( t) = α n p t n n Let us transmit s(t) over an ideal channel: r( t) = s( t) By computing the projection of r(t) on p(t-n) we should obtain α[n]. R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 9

10 ( ) α [ ] ( ) ( ) r( t) p t n dt = n p t n p t n dt = [ ] α [ ] ( ) ( ) = α n + m p t m p t n dt m n n ISI We obtain α[n] + a second term that depends on all the other symbols: ISI R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 0

11 With time domain contraint If p(t) has time domain 0 t < All the integrals are zero p t m p t n dt = m n ( ) ( ) 0 here is NO ISI r( t) p t n dt n ( ) = α [ ] R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ]

12 Without time domain contraint here is NO ISI if and only if p ( t m ) p ( t n ) dt = 0 m n All the shifted versions of the basis signal must be orthogonal. R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 2

13 Example p(t) 0 t < p(t-) p(t-2) p(t+) t t t t A pulse p(t) with time-domain constraint is automatically orthogonal to all its shifted versions R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 3

14 Example 2 p(t) 0 t < 2 p(t-) p(t+) t t t he pulse p(t) is not orthogonal to these two shifted versions R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 4

15 NO ISI condition Does exist any pulse p(t) without the time domain constraint 0 t < that satisfy the No ISI condition? ( ) ( ) 0 p t m p t n dt = m n R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 5

16 he function x Let us introduce ( ) ( ) x( τ ) = p t p t n dt Clearly, the No ISI condition is equivalent to: x(0) = (p(t) has unitary energy) x( τ ) = 0 τ = n n 0 R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 6

17 he NO ISI condition in time If p(t) satisfies the No ISI condition, then x(t) has this time behavior: x(t) t R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 7

18 he time domain contraint Clearly, an infinite number of functions x(t) have this properties. he key question is if some of them do not have the time domain constraint If p(t) has the time domain constraint, then its Fourier transform P(f) has infinite frequency response Since x( τ ) = p t p t n dt then ( ) ( ) X f P f P f * ( ) = ( ) ( ) Also X(f) has infinite frequency response R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 8

19 Looking for X(f) Do exist any X(f) with finite frequency response that satisfies the No ISI condition? R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 9

20 II Nyquist theorem X(f) satisfies the No ISI condition in time if and only i i X f = Given X(f), the sum of all its replicas centered around multiples of / must be a constant: R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 20

21 Proof ( ) = δ ( ) x( t) δ t n 0 n i X ( f ) δ f = n i i X f = R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 2

22 Nyquist criterion, case Case : f max < 2 X ( f ) 2 f max f max 2 2 In this case it is impossible to satisfy the Nyquist criterion (holes at frequencies n/2) n n X f = R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 22

23 Nyquist criterion, case 2 Case 2: f max = 2 X ( f ) 2 f max f max 2 2 R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 23

24 Nyquist criterion, case 2 Case 2: f max = 2 A solution exists: the ideal low pass filter X ( f ) 2 2 It satisfies the frequency-domain Nyquist criterion 2 2 n n X f = R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 24

25 Ideal low pass filter he ideal low pass filter x( t) = sin( πt / ) ( πt / ) Note that it clearly satisfies the time-domain Nyquist criterion x( i ) = if i = 0 x( i ) = 0 if i 0 t/ R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 25

26 Ideal low pass filter he ideal low pass filter X ( f ) 2 2 his is the waveform satisfying Nyquist with the smaller bandwidth occupation R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 26

27 Ideal low pass filter he ideal low pass filter represents the best possible solution. Unfortunately it is not possible to implement it. It has an infinite impulse response. Even if we truncate it, it is not possible to obtain a good approximation. R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 27

28 Nyquist criterion, case 3 Case 3: f max > 2 X ( f ) f max 2 f max 2 here are a lot of solutions. 2 R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 28

29 Raised cosine filters Raised cosine filters Example (very important for applications) x( t) = sin( πt / ) cos( απt / ) 2 ( πt / ) (2 αt / ) roll-off coefficient α 0 α Note that. hey clearly satisfy the time-domain Nyquist criterion x( i ) = if i = 0 x( i ) = 0 if i 0 2. For α=0 we obtain the ideal low pass filter R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 29

30 Raised cosine filters.0 alfa=0.2 alfa= x( t) = sin( πt / ) cos( απt / ) 2 ( πt / ) (2 αt / ) t/ R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 30

31 Raised cosine filters X(f)/.0 alfa=0.2 alfa=0.8 Frequency response f ( α) X ( f ) = for f 2 π ( α) ( + α) X ( f ) = sin f for f 2 α ( + α) X ( f ) = 0 for f 2 R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 3

32 Raised cosine filter Raised cosine filters x( t) = sin( πt / ) cos( απt / ) 2 ( πt / ) (2 αt / ) hey can be approximated very well if α is not too low. he impulse response is infinite and must be truncated secondary lobes. Can be very low (FIR filters). (Example, FIR digital filter, windowing design, 6 intervals = 64 taps with sampling frequency 4R) R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 32

33 X filter Remember that we have studied the properties of the function x(t), where ( ) ( ) x( τ ) = p t p t n dt X f P f P f * ( ) = ( ) ( ) Given X(f) equal to the ideal low pass filter or the raised cosine, what is p(t)? R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 33

34 X filter If we look for an even function p(t)=p(-t) P f P f * ( ) = ( ) [ P f ] 2 X ( f ) = ( ) P( f ) = X ( f ) R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 34

35 Ideal low pass X filter For the ideal low pass filter x( t) = sin( πt / ) ( πt / ) We obtain p( t) = sin( πt / ) πt ( πt / ) If X(f) is the ideal low pass filter than P(f) is the ideal low pass filter, too. R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 35

36 Ideal low pass X filter ransmission filter p(t): ideal low pass filter Minimal bandwidth occupancy equal to 2 P( f ) X ( f ) 2 2 he best possible solution in terms of bandwidth ideal benchmark. R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 36

37 RRC X filter For the raised cosine filter x( t) = sin( πt / ) cos( απt / ) 2 ( πt / ) (2 αt / ) We obtain t t t sin( π ( α)) + 4α cos( π ( + α)) t t π ( (4 α ) ) Root Raised Cosine p( t) = (RRC) 2 R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 37

38 RRC X filter ransmission filter p(t): root raised cosine filter Bandwidth occupancy equal to ( ) 2 +α We will approximate P(f) as a trapezoid, too: P( f ) X ( f ) ( +α ) 2 ( +α ) 2 ( +α ) R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 38

39 RRC X filter Can be well approximated by a FIR filter, if α is not too low (typically, α 0.2) R. Garello. utorial on digital modulations - Part 7 intersymbol interference [ ] 39