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! | 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 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
