Bài giảng Nhập môn Kỹ thuật truyền thông: Bài 10 - PGS. Tạ Hải Tùng

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Bài giảng "Nhập môn Kỹ thuật truyền thông: Bài 10 - Không gian tín hiệu PSK (Phase Shift Keying)" trình bày các nội dung chính sau đây: Đặc điểm của 2-PSK; 2-PSK dạng sóng truyền; Xác xuất lỗi 2-PSK;... Mời các bạn cùng tham khảo!

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Nội dung Text: Bài giảng Nhập môn Kỹ thuật truyền thông: Bài 10 - PGS. Tạ Hải Tùng

Nhập môn Kỹ thuật Truyền thông Phần 2: Các kỹ thuật điều chế số (Digital Modulations) Bài 10: Không gian tín hiệu PSK (Phase Shift Keying) PGS. Tạ Hải Tùng 1
2-PSK: characteristics 1. Bandpass modulation 2. One-dimensional signal space and antipodal binary constellation (equal to 2-PAM) 3. TX filter p(t)cos(2f0t) 4. Information associated to the carrier phase = Phase Shift Keying 2
2-PSK: constellation SIGNAL SET M  {s1 (t )   p(t ) cos(2 f 0t ) , s2 (t )   p(t ) cos(2 f 0t ) } Information associated to the impulse amplitude BUT we can also write SIGNAL SET M  {s1 (t )   p(t ) cos(2 f 0t ) , s2 (t )   p(t ) cos(2 f 0t   ) } Information associated to the carrier phase 3
2-PSK: constellation Versor b1(t)=p(t)cos(2πf0t) (d=1) VECTOR SET M  {s1  ( ) , s2  ( )}  R s 2     0 s1     b1 (t ) 4
2-PSK: binary labeling (example) e : H1  M e(1)  s1 e(0)  s 2 0 / s2 1/ s1    0    b1 (t ) 5
2-PSK: transmitted waveform m  2  k 1 R  Rb T  Tb  Transmitted waveform s(t )   a[n]b (t  nT ) n  1 where a[n] { ,  } b1 (t )  p(t ) cos(2 f 0t ) 6
2-PSK: transmitted waveform 1 f 0  2 Rb example for p (t )  PT (t ) T  T 1 1 0 1 0 0 1 1 .2 1 .0 0 .8 0 .6 0 .4 0 .2 0 .0 -0 .2 -0 .4 -0 .6 -0 .8 -1 .0 -1 .2 0 1 2 3 4 5 6 7 8 t/T 7
2-PSK: bandwidth and spectral efficiency Case 1: p(t) = ideal low pass filter  f0 f0 R R Total bandwidth Bid  R  Rb (ideal case) Spectral efficiency Rb id   1 bps / Hz (ideal case) Bid 8
2-PSK: bandwidth and spectral efficiency Case 2: p(t) = RRC filter with roll off   f0 f0 R(1) R(1) Total bandwidth B  R (1   )  Rb (1   ) Spectral efficiency Rb 1   bps / Hz B (1   ) 9
Exercize Given a bandpass channel with bandwidth B = 4000 Hz, centred around f0=2 GHz, compute the maximum bit rate Rb we can transmit over it with a 2- PSK constellation in the two cases: • Ideal low pass filter • RRC filter with =0.25 10
2-PSK: modulator The transmitted waveform is given by s (t )   a[n]b1 (t  nT ) n Where b1 (t )  p(t ) cos(2 f 0t ) Then we must generate s (t )   a[n] p (t  nT ) cos(2 f 0 (t  nT )) n We choose f0 multiple of R=1/T It follows cos(2 f 0 (t  nT ))  cos(2 f 0t  2 f 0 nT )  cos(2 f 0t ) Then we can generate   s (t )    a[n] p (t  nT )  cos(2 f 0t )  n  11
2-PSK: modulator c o s 2 f 0t  p (t ) e( )   sT (t )    a[ n ] p (t  nT )  cos  2 f 0t   n  u T  ( u T [ n ]) ( a [ n ])  n a[ n ] p (t  n T ) 12
2-PSK: demodulator Given the received signal  (t ) the received symbol is obtained by projecting it on the versor b1 (t )  p (t ) cos(2 f 0t )   [0]    (t )b (t )dt    (t ) p(t ) cos(2 f t )dt 1 0   This projection could be computed by using a matched filter q (t )  b1 (T  t )  p (T  t ) cos(2 f 0 (T  t )) 13
2-PSK: demodulator As an alternative, we can work as follows: 1. Given the received signal  (t ) multiply it by cos(2 f 0t ) 2. Use a filter matched to p(t): q '(t )  p (T  t ) (t)  '(t) q '(t) y(t) [n] t 0  nT cos  2 f 0 t  R  1/T 14
2-PSK: demodulator (t)  '(t) q '(t) y(t) [n] t 0  nT cos  2 f 0 t  R  1/ T By sampling the matched filter output waveform we obtain   y (t )    '( )q '(t   )d    ( ) cos(2 f  ) p(T  t   )d   0  y (t  T )    ( ) cos(2 f  ) p( )d  [0]  0 15
2-PSK: demodulator r(t) p(t) y(t) 1[n] ML sR[n] uR[ CRITERION e( ) Carrier t 0  nT synchronization cos  2 f 0 t  Symbol Synchronization R  1/T 16
2-PSK: interpretation We generate a baseband signal v(t )   a[n]p (t  nT ) n Multiplication by cosine shifts the spectrum around f0 s (t )  v(t ) cos(2 f 0t ) 17
2-PSK: interpretation At the receiver side, multiplication by cosine generates 2 1  cos(2  2 f 0  t )  s (t ) cos(2 f 0t )  v(t ) cos(2 f 0t ) cos(2 f 0t )  v(t ) cos (2 f 0t )  v(t )    2  This signal enters the matched filter q(t)=p(T-t). It is a low pass filter: the high frequency component around 2f0 is eliminated Only the baseband component v(t )   a[n]p (t  nT ) survives n The matched filter output is then equal to a[n] when sampled at t0  nT 18
2-PSK: analytic signal The 2-PSK transmitted waveform   s (t )    a[n] p (t  nT )  cos(2 f 0t )  n  can be written as   s (t )  Re  s(t )   Re   a[n] p (t  nT ) e j 2 f0t    n  Where  s (t ) is called the analytic signal associated with s(t) 19
2-PSK: Eye diagram 2-PSK constellation with RRC filter ( =0.5) 20