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- 1. Pulse Shaping Sy(w)=|P(w)|^2Sx(w) Last class: – Sx(w) is improved by the different line codes. – p(t) is assumed to be square How about improving p(t) and P(w) – Reduce the bandwidth – Reduce interferences to other bands – Remove Inter-symbol-interference (ISI) EE 541/451 Fall 2006
- 2. ISI Example sequence sent 1 0 1 sequence received 1 1(!) 1 Signal received Threshold 0 t -3T -2T -T 0 T 2T 3T 4T 5T Sequence of three pulses (1, 0, 1) sent at a rate 1/TEE 541/451 Fall 2006
- 3. Baseband binary data transmission system. ISI arises when the channel is dispersive Frequency limited -> time unlimited -> ISI Time limited -> bandwidth unlimited -> bandpass channel -> time unlimited -> ISI p(t) EE 541/451 Fall 2006
- 4. ISI First term : contribution of the i-th transmitted bit. Second term : ISI – residual effect of all other transmitted bits. We wish to design transmit and receiver filters to minimize the ISI. When the signal-to-noise ratio is high, as is the case in a telephone system, the operation of the system is largely limited by ISI rather than noise. EE 541/451 Fall 2006
- 5. ISI Nyquist three criteria – Pulse amplitudes can be detected correctly despite pulse spreading or overlapping, if there is no ISI at the decision- making instants x 1: At sampling points, no ISI x 2: At threshold, no ISI x 3: Areas within symbol period is zero, then no ISI – At least 14 points in the finals x 4 point for questions x 10 point like the homework EE 541/451 Fall 2006
- 6. 1st Nyquist Criterion: Time domainp(t): impulse response of a transmission system (infinite length) p(t) 1 shaping function 0 no ISI ! t 1 =T 2 fN t0 2t0 Equally spaced zeros, -1 1 interval =T 2 fn EE 541/451 Fall 2006
- 7. 1st Nyquist Criterion: Time domainSuppose 1/T is the sample rateThe necessary and sufficient condition for p(t) to satisfy 1, ( n = 0 ) p( nT ) = 0, ( n ≠ 0 )Is that its Fourier transform P(f) satisfy ∞ ∑ P( f + m T ) = T m = −∞ EE 541/451 Fall 2006
- 8. 1st Nyquist Criterion: Frequency domain ∞ ∑ P( f + m T ) = T m = −∞ f 0 fa = 2 f N 4 fN (limited bandwidth) EE 541/451 Fall 2006
- 9. Proof ∞Fourier Transform p( t ) = ∫ P ( f ) exp( j 2πft ) df −∞ ∞At t=T p( nT ) = ∫ P( f ) exp( j 2πfnT ) df −∞ ∞ ( 2 m +1) p( nT ) = ∑ ∫( P( f ) exp( j 2πfnT ) df 2T 2 m −1) 2T m = −∞ ∞ ∑∫ P( f + m T ) exp( j 2πfnT ) df 1 2T = −1 2T m = −∞ ∞ ∑ P( f + m T ) exp( j 2πfnT ) df 1 2T =∫ −1 2T m = −∞ ∞ =∫ 1 2T B( f ) exp( j 2πfnT ) df B( f ) = ∑ P( f + m T ) −1 2T m = −∞ EE 541/451 Fall 2006
- 10. Proof ∞ ∞B( f ) = ∑ P( f + m T ) B( f ) = ∑ b exp( j 2πnfT ) n m = −∞ n = −∞ B ( f ) exp( − j 2πnfT ) 1 2T bn = T ∫ −1 2Tbn = Tp ( − nT ) T ( n = 0) bn = 0 ( n ≠ 0) ∞ B( f ) = T ∑ P( f + m T ) = T m = −∞ EE 541/451 Fall 2006
- 11. Sample rate vs. bandwidth W is the bandwidth of P(f) When 1/T > 2W, no function to satisfy Nyquist condition. P(f) EE 541/451 Fall 2006
- 12. Sample rate vs. bandwidth When 1/T = 2W, rectangular function satisfy Nyquist condition sin πt T πt T , ( f < W ) p( t ) = = sinc P( f ) = , πt T 0, ( otherwise ) 1 0.8 0.6 0.4 Spectra 0.2 0 -0.2 -0.4 0 1 2 3 4 5 6 S bcarrier N m k u u ber EE 541/451 Fall 2006
- 13. Sample rate vs. bandwidth When 1/T < 2W, numbers of choices to satisfy Nyquist condition A typical one is the raised cosine function EE 541/451 Fall 2006
- 14. Cosine rolloff/Raised cosine filter Slightly notation different from the book. But it is the same sin(π T ) cos(rπ T ) t t prc 0 (t ) = ⋅ π Tt 1 − (2 r T ) 2 t r : rolloff factor 0 ≤ r ≤1 1 f ≤ (1 − r ) 21TPrc 0 ( j 2πf ) = 1 2 [1 + cos( π 2r ( πTf + r − 1)) ] if 1 2T (1 − r ) ≤ f ≤ 1 2T (1 + r ) 0 f ≥ 1 2T (1 + r ) EE 541/451 Fall 2006
- 15. Raised cosine shaping Tradeoff: higher r, higher bandwidth, but smoother in time. P(ω) π W r=0 r = 0.25 r = 0.50 r = 0.75 r = 1.00 p(t) W 2w ω π π − + W W 0 0 t EE 541/451 Fall 2006
- 16. Cosine rolloff filter: Bandwidth efficiency Vestigial spectrum Example 7.1 data rate 1/ T 2 bit/s β rc = = = bandwidth (1 + r ) / 2T 1 + r Hz bit/s 2 bit/s 1 ≤ < 2 Hz (1 + r ) Hz 2nd Nyquist (r=1) r=0 EE 541/451 Fall 2006
- 17. 2nd Nyquist Criterion Values at the pulse edge are distortionless p(t) =0.5, when t= -T/2 or T/2; p(t)=0, when t=(2k-1)T/2, k≠0,1 -1/T ≤ f ≤ 1/T ∞ Pr ( f ) = Re[ ∑( −1) n P ( f + n / T )] = T cos( fT / 2) n =−∞ ∞ PI ( f ) = Im[ ∑( −1) n P ( f + n / T )] = 0 n =−∞ EE 541/451 Fall 2006
- 18. ExampleEE 541/451 Fall 2006
- 19. 3rd Nyquist Criterion Within each symbol period, the integration of signal (area) is proportional to the integration of the transmit signal (area) ( wt ) / 2 π sin( wT / 2) , w ≤ T P ( w) = 0, π w > T π /T 1 ( wt / 2) p (t ) = ∫/ T sin( wT / 2) e dw jwt 2π −π 2 n +1T 1, n=0 A = ∫2 n−1 p(t )dt = 2 2 T 0, n≠0 EE 541/451 Fall 2006
- 20. Cosine rolloff filter: Eye pattern2nd Nyquist 1st Nyquist: 1st Nyquist: 2nd Nyquist: 2nd Nyquist:1st Nyquist 1st Nyquist: 1st Nyquist: 2nd Nyquist: 2nd Nyquist: EE 541/451 Fall 2006
- 21. Example Duobinary Pulse – p(nTb)=1, n=0,1 – p(nTb)=1, otherwise Interpretation of received signal – 2: 11 – -2: 00 – 0: 01 or 10 depends on the previous transmission EE 541/451 Fall 2006
- 22. Duobinary signaling Duobinary signaling (class I partial response) EE 541/451 Fall 2006
- 23. Duobinary signal and Nyguist Criteria Nyguist second criteria: but twice the bandwidth EE 541/451 Fall 2006
- 24. Differential Coding The response of a pulse is spread over more than one signaling interval. The response is partial in any signaling interval. Detection : – Major drawback : error propagation. To avoid error propagation, need deferential coding (precoding). EE 541/451 Fall 2006
- 25. Modified duobinary signaling Modified duobinary signaling – In duobinary signaling, H(f) is nonzero at the origin. – We can correct this deficiency by using the class IV partial response. EE 541/451 Fall 2006
- 26. Modified duobinary signaling Spectrum EE 541/451 Fall 2006
- 27. Modified duobinary signaling Time Sequency: interpretation of receiving 2, 0, and -2? EE 541/451 Fall 2006
- 28. Pulse Generation Generalized form ofcorrelative-levelcoding(partial response signaling)Figure 7.18 EE 541/451 Fall 2006

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