R. Garello. Tutorial on digital modulations - Part 9b m-pam [ ] 2

Transcription

1 TUTORIAL ON DIGITAL MODULATIONS Part 9: m-pam [ ] 26] Roerto Garello, Politecnico di Torino Free download at: (personal use only) 1

2 m-pam constellations: characteristics 1. Base-and modulations 2. One-dimensional signal space 3. m signals, symmetrical with respect to the origin 4. Information associated to the impulse amplitude PAM=Pulse Amplitude Modulation R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 2

3 m-pam constellation: : constellation SIGNAL SET M = { s ( ) ( )} m i t = α i p t i= 1 Basis signal 1 (t)=p(t) (d=1) VECTOR SET M = { s ( ) } m i = α i i= 1 R Typically (symmetric points): M = { s = ( ( m 1) α), s = ( ( m 3) α),..., s = ( + ( m 3) α), s = ( + ( m 1) α)} R 1 2 m 1 m k = log ( m) 2 T = kt R = R k R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 3

4 m-pam constellation: : constellation Example: 4-PAM constellation M = { s = (-3α), s = (-α), s = ( + α ), s = ( +3α)} R s = ( α ) s2 = ( α ) 0 s3 = ( + α ) 1 3 s ( α ) 4 = + 3 1( t) R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 4

5 m-pam constellation: : constellation Example: 8-PAM constellation M = { s = (-7α), s = (-5α), s = (-3α), s = (-α), s = ( + α ), s = ( +3α ), s = ( +5α ), s = ( +7α)} R s = ( α ) s2 = ( 5α ) s3 = ( 3α ) s4 = ( α ) s5 = ( + α ) s6 = ( + 3α ) s7 = ( + 5α ) s8 = ( + 7α ) ( t ) R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 5

6 m-pam constellation: : inary laelling e : Hk M It is always possile to uild a Gray laeling 4-PAM: 01/ s 11/ s / s / s4 ( 3α ) ( α ) 0 ( +α ) ( ) +3α ( 1 t ) 110 / s1 111/ s 101/ s PAM: 100 / s4 000 / s 001/ s / s7 010 / s8 ( 7α ) ( 5α ) ( 3α ) ( α ) ( + α ) ( + 3α ) ( + 5α ) ( +7α ) 0 1( t) R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 6

7 m-pam constellation: : transmitted waveform u T Example: 4-PAM 1 p( t) = PT ( t) T T 2 T 3 T 4 T 5 T 6 T 7 T 8 T s T ( t ) 3 α T α T α T T 2T 3T 4T 3 α T R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 7

8 m-pam constellation: : transmitted waveform Example: 4-PAM p( t) = RRC α = R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 8

9 m-pam constellation: andwidth and spectral efficiency Case 1: p(t) = ideal low pass filter Total andwidth (ideal case) R R / k B id = = 2 2 R Spectral efficiency (ideal case) R η id = = 2 k ps / Hz B id R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 9

10 m-pam constellation: andwidth and spectral efficiency Case 2: p(t) = RRC filter roll off α Total andwidth R R / k B = (1 + α) = (1 + α) 2 2 R(1 + α ) Spectral efficiency η R B = = 2k (1 + α) ps / Hz R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 10

11 Exercize Given a aseand channel with andwidth B up to 4000 Hz, compute the maximum it rate R we can transmit over it with a 256-PAM constellation in the two cases: Ideal low pass filter RRC filter with α=0.25 R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 11

12 m-pam constellation: : modulator p ( t ) e ( ) u = ( v [ n ]) + ( a [ n ]) + T T + s ( t ) = a [ n ] p ( t nt ) n = Equal to 2-PAM, ut we have m possile levels: a[ n] { ( m 1) α, ( m 3) α,..., + ( m 3) α, + ( m 1) α } R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 12

13 m-pam constellation: : demodulator q ( t ) r ( t ) y ( t ) ρ 1[ n] ML CRITERION s R[ n ] v R [ n ] e ( ) Symol synchronization R t 0 + = 1 / T n T Equal to 2-PAM, ut we have m possile levels: a[ n] { ( m 1) α, ( m 3) α,..., + ( m 3) α, + ( m 1) α } R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 13

14 m-pam constellation: : eye diagram 4-PAM, p(t) = RRC with α =0.5 R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 14

15 m-pam constellation: : eye diagram 8-PAM, p(t) = RRC with α =0.5 R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 15

16 m-pam constellation: : error proaility Comparison: 2-PAM vs. 4-PAM 1 E 2 PAM: P ( e) = erfc 2 N E 4 PAM: P ( e) erfc 8 5 N 0 The 2-PAM constellation has etter performance The constellation gain is in the order of 10 log(5/2) = 4 db R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 16

17 m-pam constellation: : error proaility E-3 1E-4 1E-5 Comparison: 2-PAM vs. 4-PAM 2-PAM 4-PAM BER 1E-6 1E-7 1E-8 1E-9 1E-10 1E-11 1E E/N0 [db] 17 R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 17

18 m-pam constellation: : error proaility The BER performance are (asymptotically) dominated y the normalized minimum distance 2 1 w 1 d E min min P ( e) erfc 2 k 4 E N 0 At the parity of E/N0 value, the constellation with larger performs etter since it achieves a low it error proaility (the erfc is a decreasing function of its argument). d E 2 min R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 18

19 m-pam constellation: : error proaility By increasing the numer of signals of an m-pam, the parameter decreases, and also the BER performance decreases. d E 2 min R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 19

20 It is easy to understand this y looking at this picture: d E 2 min When m increases decreases At the parity of d min, signals with larger energy are added E increases R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 20

21 m-pam constellation: : error proaility It is also possile to verify this y computing the asymptotic approximation for a generic 2 k -PAM constellation: P ( e) m 1 3k E erfc 2 mk m 1 N 0 R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 21

22 m-pam constellation: : error proaility BER E-3 1E-4 1E-5 1E-6 1E-7 1E-8 1E-9 1E-10 1E-11 2-PAM 4-PAM 8-PAM 16-PAM 32-PAM 64-PAM 128-PAM 256-PAM The performance decrease for increasing m 1E E/N0 [db] R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 22

23 m-pam constellation: performance/spectral efficiency trade-off Given a aseand channel with andwidth B and an m-pam constellation, y increasing the numer of signals m=2 k we increase the spectral efficiency η = R / B = 2 k ps / Hz then we can transmit a higher it rate R. id Unfortunately, the performance decrease: fixed a BER value, the signal-to-noise ratio E /N 0 necessary to achieve it increases with m. R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 23

24 Example Suppose B=4kHz. With a (ideal) 2-PAM we transmit R = 8 kps With a (ideal) 256-PAM we transmit R = 64 kps However, fixed a target BER (e.g. BER=1e-10), a 256-PAM requires a larger (34 db of difference!). ratio E /N 0 As an example, for a telephone copper line, the attenuation increases with the square of line length. As an example, at the parity of transmitted power, the line distance covered y a 256-PAM is much lower than the distance covered y the 2-PAM (y a factor of 50!) R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 24

25 Typical applications Baseand modulations cale transmission They were used for V90 telephone modems (using the first 4kHz andwidth, a 256-PAM was used) They are used for Ethernet connections. As an example, 10G-BaseT standard over 4 pairs has an availale andwidth up to f max =500 MHz and uses a 16-PAM modulation. R. Garello. Tutorial on digital modulations - Part 9 m-pam [ ] 25