Principles of Digital Data Transmission Chapter 7 Lectured by Dr. Yun Q. Shi Dept. of Electrical & Computer Engr. New Jersey Institute of Technology [email protected] Text used for the course: <Modern Digital and Analog Communication Systems>, 4th Edition, Lathi and Ding, Oxford Introduction • A significant portion of communication in 1990s was in analog. • It has been replaced rapidly by digital communication. • Now most of the communications become digital, with analog communication playing a minor role. • This chapter addresses the framework (several major aspects) of digital data transmission • All we learnt in this course so far is utilized here. • Hence, this chapter is naturally a conclusion of this course. Dr. Shi Lathi & Ding-Digital Commun 2 1 Digital Communication Systems: Source • Digital Communication Systems: Figure 7.1 Source encoder, Baseband modulation (Line coding), Digital carrier modulation, Multiplexer, Channel, Regenerative repeater, Receiver (detection). • Source: The input – A sequence of digits – We discuss mainly: The binary case (two symbols) – Later in this chapter: The M-ary case (M symbols) more general case Dr. Shi Lathi & Ding-Digital Commun 3 Figure 7.1 Fundamental building blocks of digital communication systems Figure 7.1 Fundamental building blocks of digital communication systems Dr. Shi Lathi & Ding-Digital Commun 4 2 Line Coder (Transmission Coder) • The output of a multiplexer is coded into electrical pulses or waveforms for the purpose of transmission over the channel. • This process is called a line coding or transmission coding • Many possible ways of assigning waveforms (pulses) to the digital data. 1 • On-off: Dr. Shi ↔ a 0 ↔ pulse p (t) no pulse Lathi & Ding-Digital Commun 5 Line Coder … • Polar: 1 ↔ ↔ 0 a a pulse pulse p (t) − p (t) • Bipolar (pseudoternary or alternate mark inversion (AMI)) 1 ↔ 0 ↔ a pulse p (t) or - p (t) depending on whether the previous 1 is encoded by -p (t) or p (t) no pulse In short, pulses representing consecutive 1’s alternate in sign. • All the above three could use half-width pulses. It is possible to select other width. • Figure 7.1 (parts a, b and c). Dr. Shi Lathi & Ding-Digital Commun 6 3 Line Codes Dr. Shi Lathi & Ding-Digital Commun 7 Line Coder … • Full width pulses are often used in some applications. – i.e., the pulse amplitude is held to a constant value throughout the pulse interval (it does not have a chance to go to zero before the next pulse begins). – These schemes are called non return-to-zero (NRZ) in contrast to return-to-zero (RZ) – Figure 7.2 shows • An On-off NRZ signal • A Polar NRZ signal in d and e parts of the figure Dr. Shi Lathi & Ding-Digital Commun 8 4 Multiplexer • Usually, the capacity of a practical channel >> the data rate of individual sources. • To utilize this capacity effectively, we combine several sources through a digital multiplexer using the process of interleaving. • One way: A channel is time-shared by several messages simultaneously. Dr. Shi Lathi & Ding-Digital Commun 9 Regenerative Repeater • Used at regularly spaced intervals along a digital transmission line to: 1) detect the incoming digital signal and 2) regenerate new clean pulses for further transmission along the line. • This process periodically eliminates, and thereby combats the accumulation of noise and signal distortion along the transmission path. • The periodic timing information (the clock signal at R b Hz) is required to sample the incoming signal at a repeater. • R b (rate): pulses/sec Dr. Shi Lathi & Ding-Digital Commun 10 5 Regenerative Repeater… • The clock signal can be extracted from the received signal. – e.g., the polar signal when rectified ⇒ a clock signal at Rb Hz • The on-off signal = A periodic signal at Rb + a polar signal (Figure 7.) • When the periodic signal is applied to a resonant circuit tuned to Rb Hz, the output, a sinusoid of Rb Hz, can be used for timing. Dr. Shi Lathi & Ding-Digital Commun 11 On-off Signal Decomposition Dr. Shi Lathi & Ding-Digital Commun 12 6 Regenerative Repeater… rectified • The bipolar signal an on-off signal ⇒ the clock signal can also be extracted. • The timing signal (the output of the resonant circuit) is sensitive to the incoming pattern, sometimes. – e.g. in on-off, or 0 bipolar ↔ no pulse • If there are too many zeros in a sequence, will have problem. – no signal at the input of the resonant circuit for a while – sinusoids output of the resonant circuit starts decaying • Polar scheme has no such a problem. Dr. Shi Lathi & Ding-Digital Commun 13 Transparent Line Code • A line code in which the bit pattern does not affect the accuracy of the timing information is said to be a transparent line code. – The polar scheme is transparent. – The on-off and bipolar are not transparent (nontransparent). Dr. Shi Lathi & Ding-Digital Commun 14 7 Desired Properties of Line Codes 1. Transmission bandwidth: as small as possible 2. Power efficiency: for a specific bandwidth and detection error rate, transmitted power should be as small as possible. 3. Error detection and correction capability: as strong as possible 4. Favorable power spectral density: desirable to have zero PSD at (dc), called DC null, because of ac coupling requires this. If at dc, ω = 0 , PSD ≠ 0 , then dc wanders in the pulse stream, which is not desired. 5. Adequate timing content: should be possible to extract the clock signal (timing information). 6. Transparency Dr. Shi Lathi & Ding-Digital Commun 15 PSD of Various Line Codes • Consider a general PAM signal, as shown in Figure 7.4 (b): Tb = 1 t = kTb Rb Rb – the k th pulse in the pulse train y (t ) = ak p (t ) • p (t): the basic pulse • P (ω) [P(f)]: Fourier spectrum of p (t) • a k: arbitrary and random – The on-off, polar, bipolar line codes are all special cases of the general pulse train y(t), a k = 0, +1, -1 Dr. Shi Lathi & Ding-Digital Commun 16 8 A Random PAM signal All 3 line codes can be represented this way New approach to determine PSD of y(t) Dr. Shi Lathi & Ding-Digital Commun 17 PSD of Various Line Codes … • How to determine the PSD? – Figure 7.4 – An attractive general approach Rx (τ ), S x (ω ) need to be studied once Then for different p(t), ⇒ different PSD – x(t): an impulse train – xˆ (t ) : a rectangular pulse train ( ε ⋅ hk = a k ) when ε → 0, xˆ (t ) → x(t ) (Figure 7.5) Dr. Shi Lathi & Ding-Digital Commun 18 9 Dr. Shi Lathi & Ding-Digital Commun 19 PSD of Various Line Codes … R0 = ak2 ⇒ time average of a k2 R1 = ak .ak +1 ,L L, Rn = ak .ak +n 1 RX (τ ) = Tb S X (ω ) = 1 Tb ∞ ∑ R δ (τ − nT ) n = −∞ n b ∞ ∑ Rn e− jnωTb = n = −∞ 1 Tb ∞ [ ] R + 2 R cos( n ω T ) ∑ 0 n b n =1 S y (ω ) = P (ω ) S X (ω ) 2 Different line codes Dr. Shi ⇒ different P(ω ) ⇒ different S y (ω ) Lathi & Ding-Digital Commun 20 10 Polar Signaling 1 ↔ p(t) 0 ↔ -p(t) } ⋅ ak = ±1, equally R0 = ak2 = 1 R1 = ak .ak +1 = 0 Rn = 0, n ≥ 1 likely, ak2 = 1 Q ak .ak +1 = 1 or − 1 equally likely Similar reasoning S y (ω ) = P (ω ) .S X (ω ) 2 1 = P(ω ) . Tb 2 Dr. Shi Lathi & Ding-Digital Commun 21 Polar Signaling… • To be specific, assume p(t) is a rectangular pulse of width T b (a half-width rectangular 2 pulse) t p (t ) = re c t T b 2 T ω Tb P ( ω ) = b s in c 2 4 T ω Tb S y ( ω ) = b s in c 2 4 4 Dr. Shi Lathi & Ding-Digital Commun 1st zero − cros sin g : ω⋅Tb 4 = π ⇒ω = or ⇒ f = 4π Tb 2 = 2Rb Tb 22 11 PSD of polar signal (half-width rectangle) Dr. Shi Lathi & Ding-Digital Commun 23 Polar Signaling … • Comment 1: • • • • The essential bandwidth of the signal (main lobe) = 2 Rb For a full-width pulse, ⇒ Rb “Polar Signaling is not bandwidth efficient.” Comment 2: “No error-detection or error-correction capability.” Comment 3: “Non-zero PSD at dc( ω = 0)” ⇒ This will rule out the use of ac coupling in transmission. Comment 4: “Most efficient scheme from the power requirement viewpoint” for a given power, the detection-error probability for a polar scheme is the smallest possible. Comment 5: “Transparent” Dr. Shi Lathi & Ding-Digital Commun 24 12 Achieving DC Null in PSD by Pulse Shaping Q P (ω ) = ∫ ∴ P (0) = ∫ ∞ −∞ ∞ −∞ p ( t ) e − j ω t dt p ( t )dt ⇒ If the area under p(t) = 0, P(0) = 0 . Dr. Shi Lathi & Ding-Digital Commun 25 Manchester (split-phase, twinned-binary) Signal Dr. Shi Lathi & Ding-Digital Commun 26 13 On-off Signaling • 1 ↔ p(t) 0 ↔ no pulse R 0 R n S X = 1 2 1 = 4 = (ω ) = ∞ 1 1 + 2Tb 4Tb 1 1 + 4Tb 4Tb Dr. Shi n ≥ 1 for all ∞ ∑ e ∑ e − jn ω T b n = −∞ n ≠0 − jn ω T b n = −∞ Lathi & Ding-Digital Commun 27 On-off Signaling S X (ω ) = 1 2π + 4Tb 4 T b2 ∞ ∑ n = −∞ δ ω − 2π n Tb P (ω ) 2π ∞ 2π n S y (ω ) = 1 + δ ω − ∑ 4Tb Tb n = −∞ Tb For a half-width rectangular pulse, 2 S y (ω ) = Tb 2π ω Tb sinc 2 1 + T 16 4 b ∞ ∑ n = −∞ δ ω − 2π n Tb Note: 1. For some derivation, refer to the next four pages. 2. PSD is shown in Figure 7.8. 3. PSD consists of both a discrete part and a continuous part. Dr. Shi Lathi & Ding-Digital Commun 28 14 Some Derivation (Lathi’s book, pp. 58-59, Example 2.12) • A unit impulse train: FS g (t) = ∞ ∑ n = −∞ D ne jn ω b t δ T (t ) = CombT (t ) = g (t ) 0 = = 0 1 Tb ∞ ∑ e n = −∞ 1 2 + Tb Tb jn ω b t ∞ ∑ n = −∞ cos( n ω b t ) 1 ∞ ∑ δ ( f − nf 0 ) T 0 n = −∞ 2π ∞ 2π G ( ω ) = FT { g ( t )} = ) δ (ω − n ∑ T 0 n = −∞ T0 G ( f ) = FT { g ( t )} = ≅ ω 0 COMB Dr. Shi ω0 (ω ) Lathi & Ding-Digital Commun 29 Example 2.12 Dr. Shi Lathi & Ding-Digital Commun 30 15 Some Derivation… • Fact 1 (Lathi’s book, p.83, Eq. (3.20a, 3.20b): 1↔ δ(f ) e j 2πf bt ↔ δ ( f − f b ) • Fact 2: [e jω bt ] ↔ δ (ω − ωb ) Also, (t ) ↔ 1 δ (t − nT b ) ↔ δ Dr. Shi e − jn ω T b Lathi & Ding-Digital Commun 31 Some Derivation … • Fact 3: ( ) ∞ ∑ δ t − nT b n1=4−∞ 42443 FS = ∞ jnω t 1 2π b ω = ∑ e b T n = −∞ T b 1b4 44442444443 b FT 644744 8 ∞ − jnω t b ∑e n = −∞ Dr. Shi b FT = ω ( ∞ ∑ δ ω − nω b b n = −∞ Lathi & Ding-Digital Commun ) 32 16 PSD of An On-off Signal Dr. Shi Lathi & Ding-Digital Commun 33 On-off Signaling • This is reasonable since, as shown in Fig. 7.2, an on-off signal can be expressed as the sum of a polar and a periodic component • Comment 1: For a given transmitted power, it is less immune to noise interference. – Q Noise immunity ∝ difference of amplitudes representing binary 0 and 1. – If a pulse of amplitude 1 or –1 has energy E, then a pulse of amplitude 2 has energy (2)2E = 4E. Dr. Shi Lathi & Ding-Digital Commun 34 17 On-off Signaling … • For polar:Q 1 Tb ∴ • For on-off: digits are transmitted per second polar signal power 1 2E = power = 4 E 2 T b Tb 1 E = E = Tb Tb Power is now as large as twice of above. • Comment 2: Not transparent Dr. Shi Lathi & Ding-Digital Commun 35 Bipolar Signaling (Pseudoternary or Alternate Mark Inverted (AMI)) A. 0 ↔ no pulse 1 ↔ p(t) or –p(t) depending on whether the previous 1 was transmitted by –p(t) or p(t) B. [p(t), 0 , -p(t)]: In reality, it is ternary signaling. C. Merit: a dc null in PSD demerit: not transparent Dr. Shi Lathi & Ding-Digital Commun 36 18 Bipolar Signaling… 1 N N 1 ( ± 1) 2 + (0) = N →∞ N 2 2 2 k 1 N 3N 1 R (1) = lim ( − 1) + (0) = − N →∞ N 4 4 4 111, 101, 110, 011, 010, 001, 000 ak ak + n R (2) D. R ( 0 ) = Nlim →∞ R ( 2 ) = lim N →∞ R n = lim N →∞ 1 N 1 N ∑a 2 k = lim 1 N N 5N (1) + ( − 1) + (0) = 0 N 8 8 8 ∑a k a k + n = 0, n > 1 k Dr. Shi Lathi & Ding-Digital Commun 37 Bipolar Signaling… E. S y (ω ) = P(ω ) S X (ω ) 2 P (ω ) ∞ R n e − jnω Tb = ∑ Tb n = −∞ 2 P (ω ) = Tb 2 P (ω ) = Tb P (ω ) = 2 ⋅ Tb Dr. Shi 2 ∞ R + 2 R n cos nω Tb ∑ 0 n =1 2 (7.10b) (7.10c) 1 1 + 2 ⋅ ( − ) cos ω T b 4 2 P (ω ) (1 − co s ω Tb ) = Tb 2 Lathi & Ding-Digital Commun ω Tb sin 2 2 38 19 Bipolar Sampling … ⇒ As ω=0 (dc) regardless of P(ω) A dc null (desirable for ac coupling) S y (ω ) = 0 sin 2 ( ω Tb 2 )= 0 at ω = ⇒ b a n d w id th = R b H z , 2π 1 ,∴ f = = Rb Tb Tb reg a rd less For a half-width rectangular pulse, Dr. Shi ⇒ Zero: of P (ω ) t Tb , P (t ) = rect 2 T b 2 ω Tb P (ω ) = s in c 2 4 T b Lathi & Ding-Digital Commun S y (ω ) = 39 Tb ωT ωT sinc 2 b sin 2 b 4 4 2 ɷ1Tb=4π 4π ω1 = Tb f1 = 2 Rb ɷ2Tb=2π 2π ω2 = Tb f 2 = Rb ⇒ 1 Rb = Tb essential ⋅ bandwidth Half that of polar or on-off signaling. Twice that of theoretical minimum bandwidth (channel b/w). Table 3.1: Dr. Shi ω t ωτ rect ↔ τ sin c = τ sin c 2 τ 2 τ Lathi & Ding-Digital Commun 40 20 Dr. Shi Lathi & Ding-Digital Commun 41 Merits of Bipolar Signaling 1.Spectrum has a dc dull. 2.Bandwidth is not excessive. 3.It has single-error-detection capability. Since, a single detection error ⇒ a violation of the alternation pulse rule. Dr. Shi Lathi & Ding-Digital Commun 42 21 Demerits of Bipolar Signaling 1. Requires twice as much power as a polar signaling. Distinction between A, -A, 0 vs. Distinction between A/2, -A/2 2. Not transparent Dr. Shi Lathi & Ding-Digital Commun 43 Pulse Shaping S y (ω ) = P (ω ) S X (ω ) 2 = P (ω ) Tb 2 ( R0 + 2 ∞ ∑R n cos nωTb ) [Slide 19] n =1 • Different signaling (line coding, SX (ω) ) ⇒ diff.S y (ω ) • Different pulse shaping (P(ω)) ⇒ diff. S y (ω ) An additional factor. • Intersymbol Interference (ISI) (refer to the next slide) – Time-limited pulses (i.e., truncation in time domain) not band-limited in frequency domain – Not time-limited, causes problem band-limited signal (i.e., truncation in frequency domain) Dr. Shi Lathi & Ding-Digital Commun 44 22 Intersymbol Interference (ISI) in Detection B. Sklar’s Digital Communications 1988 Dr. Shi Lathi & Ding-Digital Commun Back to slide 44 45 Phase Shaping… Intersymbol Interference (ISI) • Whether we begin with time-limited pulses or band-limited pulses, it appears that ISI cannot be avoided. An inherent problem in the finite bandwidth transmission. • Fortunately, there is a way to get away: be able to detect pulse amplitudes correctly. Dr. Shi Lathi & Ding-Digital Commun 46 23 Nyquist Criterion for Zero ISI • The first method proposed by Nyquist. 1, t = 0 1 p (t ) = 0, t = ± nTb Tb = Rb (7.22) T b : separation between successive transmitted pulses • Example 6.1 (p.256) p(t ) = sinc(2π Bt ), B= 1 1 = Rb 2Tb 2 Since 2πBTb=π, Tb is the 1st zero crossing. • Minimum bandwidth pulse that satisfies Nyquist Criterion Fig. 7.10 (b and c) Dr. Shi Lathi & Ding-Digital Commun 47 Minimum Bandwidth Pulse (Nyquist Criterion) Dr. Shi Lathi & Ding-Digital Commun 48 24 Nyquist Criterion for Zero ISI… 1 p (t ) = sinc (π Rb t ) = 0 P (ω ) = ω 1 rect Rb 2π R b Table 3.1: t=0 t = ± nTb 1 Tb = Rb bandwidth = Rb 2 FT ω Sinc (Wt ) ↔ rect π 2W W Dr. Shi Lathi & Ding-Digital Commun 49 Problems of This Method • Impractical 1. − ∞ < t < ∞ 2. It decays too slowly at a rate 1/t. ⇒ Any small timing problem (deviation) ⇒ ISI • Solution: Find a pulse satisfying Nyquist criterion (7.22), but decays faster than 1/t. • Nyquist: Such a pulse requires a bandwidth k⋅ Dr. Shi Rb , 2 1≤ k ≤ 2 Lathi & Ding-Digital Commun 50 25 p (t ) ↔ P (ω ) • Let – The bandwidth of P(ω) is in (Rb /2, Rb) – p(t) satisfies Nyquist criterion Equation (7.22) • Sampling p(t) every Tb seconds by multiplying p(t) by an impulse train δ T (t ). b ⇒ b ⇒ p (t )δ p (t) = T b (t) = δ (t) FT 1 T b ∞ ∑ P (ω − n ω b P (ω − n ω n = −∞ Equation G (ω ) = ) = 1 ( 6 .4 ) 1 T s ∞ ∑ n = −∞ s ) Dr. Shi Lathi & Ding-Digital Commun 51 Dr. Shi Lathi & Ding-Digital Commun 52 26 Derivation • |P(ω)| is odd symmetric in y1-o-y2 system. Dr. Shi Lathi & Ding-Digital Commun 53 Derivation… ω b • The bandwidth of P(ω) is 2 + ω x – ω x : the excess bandwidth – γ roll − offfactor = = excess bandwidth theoretica l min imum bandwidth ωx ω = 2⋅ x ωb ωb 0 ≤ r ≤1 2 R rR R • The bandwidth of P(ω) is BT = 2b + 2b = ( 1 + γ ) 2b • P(ω), thus derived, is called a Vestigial Spectrum. Dr. Shi Lathi & Ding-Digital Commun 54 27 Realizability • A physically realizable system, h(t), must be causal, i.e., h(t) = 0 , for t < 0. (The n.c. & s.c.) • In the frequency domain, the n.c. & s.c. is known as Paley-Wiener criterion ∞ ln H (ω ) −∞ 1+ω 2 ∫ dω < ∞ • Note that, for a physically realizable system, H(ω) may be zero at some discrete frequencies. But, it cannot be zero over any finite band • So, ideal filters are clearly unrealizable. Dr. Shi Lathi & Ding-Digital Commun 55 • From this point of view, the vestigial spectrum P(ω) is unrealizable. • However, since the vestigial roll-off characteristic is gradual, it can be more closely approximated by a practical filter. • One family of spectra that satisfies the Nyquist criterion is ωb 1 / 2 P (ω ) = 0 1 Dr. Shi ) π (ω − 2 1 − sin 2ω x Lathi & Ding-Digital Commun ω− ω > ω < ωb 2 ωb 2 ωb 2 < ωx +ωx −ωx 56 28 Realizability … ωx = 0 ω ωx = b 4 ωx = ωb 2 (r = 0) ( r = 0 .5 ) ( r = 1) Shown in the figure on next slide • Comment 1: Increasing ω x (or r) improves p(t), i.e., more gradual cutoff reduces the oscillatory nature of p(t), p(t) decays more rapidly. Dr. Shi Lathi & Ding-Digital Commun 57 Pulses satisfying Nyquist Criterion Dr. Shi Lathi & Ding-Digital Commun 58 29 Realizability… • Comment 2: As γ = 1, i.e., ω x = ωb 2 ω ω 1 rect P (ω ) = 1 + cos 2 2 Rb 4πRb ω ω rect = cos 2 4 Rb 4πRb • This characteristic is known as: The raised-cosine characteristic The full-cosine roll-off characteristic Dr. Shi Lathi & Ding-Digital Commun 59 Realizability… p (t ) = Rb A. B. C. D. co s(π R b t ) S in c (π R b t ) 1 − 4 R 2bt 2 Bandwidth is Rb (γ =1) p(0) = Rb p(t) = 0 at all the signaling instants & at points midway between all the signaling instants p(t) decays rapidly as 1/t3 ⇒ relatively insensitive to derivations of Rb , sampling rate, timing jitter and so on. Dr. Shi Lathi & Ding-Digital Commun 60 30 Realizability… E. Closely realizable F. Can also be used as a duobinary pulse. G. H c (ω ) : channel transfer factor Pi ( k ) : transmitted pulse Pi (ω ) ⋅ H c (ω ) = P (ω ) Received pulse at the detector input should be P(ω) [p(t)]. Dr. Shi Lathi & Ding-Digital Commun 61 Signaling with Controlled ISI: Partial Response Signals • The second method proposed by Nyquist to overcome ISI. n = 0,1 1 • Duobinary pulse: p(nTb ) = 0 for all other n Dr. Shi Lathi & Ding-Digital Commun 62 31 Signaling with Controlled ISI… • Use polar signaling 1 ↔ p (t ) 0 ↔ − p (t ) • The received signal is sampled at t = nTb p(t) = 0 for all n except n= 0,1. • Clearly, such a pulse causes zero ISI with all the pulses except the succeeding pulses. Dr. Shi Lathi & Ding-Digital Commun 63 Signaling with Controlled ISI… • Consider two such successive pulses located at 0 and Tb. – If both pulses are positive, the sample value at t = Tb ⇒ 2. – If both pulses are negative, ⇒ -2 – If pulses are of opposite polarity, ⇒ 0 • Decision Rule: If the sample value at t = Tb is • Positive ⇒ 1, 1 • Negative ⇒ 0, 0 • Zero ⇒ 0, 1 or 1, 0 Dr. Shi Lathi & Ding-Digital Commun 64 32 Signaling with Controlled ISI… • Table 7.1 Transmit ed Seq. Sample of x(Tb) Detected seq. Dr. Shi 1 1 0 1 1 0 0 0 1 0 1 1 1 1 2 0 0 2 0 -2 -2 0 0 0 2 2 1 1 0 1 1 0 0 0 1 0 1 1 1 Lathi & Ding-Digital Commun 65 Signaling with Controlled ISI: Duobinary Pulses • If the pulse bandwidth is restricted to be it can be shown that the p(t) must be: p (t) = Rb 2 sin( π R b t ) π R b t (1 − R b t ) ω 2 P (ω ) = cos Rb 2 Rb rect ω 2π R b − e j ω 2 Rb ωb • Note: The raised-cosine pulse with ω = 2 , ( r = 1), satisfies the condition (7.3b) to be signaling with controlled ISI – refer to Figure 7.13 (slides 57). x Dr. Shi Lathi & Ding-Digital Commun 66 33 Signaling with Controlled ISI Duobinary Pulses… Dr. Shi Lathi & Ding-Digital Commun 67 Use of Differential Coding • For the controlled ISI method, 0-valued sample ⇒ 0 to 1 or 1 to 0 transition. • An error may be propagated. • Differential coding helps. • In differential coding, “1” is transmitted by a pulse identical to that used for the previous bit. “0” is transmitted by a pulse negative to that used for the previous bit. • Useful in systems that have no sense of absolute polarity. Fig 7.17 Dr. Shi Lathi & Ding-Digital Commun 68 34 Differential Code Dr. Shi Lathi & Ding-Digital Commun 69 Scrambling • Purpose: A scrambler tends to make the data more random by – Removing long strings of 1’s or 0’s – Removing periodic data strings. – Used for preventing unauthorized access to the data. • Example. (Structure) Scrambler: S ⊕ D 3T ⊕ D 5T = T D nT : ⊕: T delayed by n units Modulo 2 Sum 1⊕1 = 0 0⊕0=0 Dr. Shi → Incomplete version 1⊕ 0 = 1 0 ⊕1 = 1 Lathi & Ding-Digital Commun 70 35 Dr. Shi Lathi & Ding-Digital Commun 71 Scrambling… • Descrambler: Incomplete version R = T ⊕ (D 3 ⊕ D 5 )T = [1 ⊕ ( D 3 ⊕ D 5 )] T = (1 ⊕ F ) T ∆ F = D Dr. Shi 3 ⊕ D5 Lathi & Ding-Digital Commun 72 36 Example 7.2 • The data stream 101010100000111 is fed to the scrambler in Fig. 7.19a. Find the scrambler output T, assuming the initial content of the registers to be zero. – From Fig. 7.19a we observer that initially T = S, and the sequence S enters the register and is returned as through the feedback path. This ( D 3 ⊕ D 5 ) S = FS new sequence FS again enters the register and is returned as F 2 S , and so on. Hence, (7.41) T = S ⊕ FS ⊕ F 2 S ⊕ F 3 S ⊕ ... = (1 ⊕ F ⊕ F 2 ⊕ F 3 ⊕ ...) S Complete version Dr. Shi Lathi & Ding-Digital Commun 73 Example 7.2 … • Recognizing that We have ( )( F = D3 ⊕ D5 ) F 2 = D 3 ⊕ D 5 D 3 ⊕ D 5 = D 6 ⊕ D10 ⊕ D 8 ⊕ D8 • Because modulo-2 addition of any sequence with itself is zero, D 8 ⊕ D 8 = 0 and F 2 = D 6 ⊕ D10 • Similarly, ( )( ) F 3 = D 6 ⊕ D10 D 3 ⊕ D 5 = D 9 ⊕ D11 ⊕ D13 ⊕ D15 Dr. Shi Lathi & Ding-Digital Commun 74 37 Example 7.2 … …… and so on …… Hence, T = (1⊕ D3 ⊕ D5 ⊕ D6 ⊕ D9 ⊕ D10 ⊕ D11 ⊕ D12 ⊕ D13 ⊕ D15 ⊕...)S Because DnS is simply the sequence S delayed by n bits, various terms in the preceding equation correspond to the following sequences: Dr. Shi Lathi & Ding-Digital Commun 75 Example 7.2 … S = 1010101000 00111 D 3 S = 0001010101 00000111 D 5 S = 0000010101 0100000111 D 6 S = 0000001010 1010000011 1 D 9 S = 0000000001 0101010000 0111 D 10 S = 0000000000 1010101000 00111 D 11 S = 0000000000 0101010100 000111 D 12 S = 0000000000 0010101010 0000111 D 13 S = 0000000000 0001010101 00000111 D 15 S = 0000000000 000001010101000 00111 T = 1011100011 01001 Dr. Shi Lathi & Ding-Digital Commun 76 38 Example 7.2 … • Note that the input sequence contains the periodic sequence 10101010…, as well as a long string of 0’s. • The scrambler output effectively removes the periodic component as well as the long strings of 0’s. • The input sequence has 15 digits. The scrambler output up to the 15th digit only is shown, because all the output digits beyond 15 depend on the input digits beyond 15, which are not given. • We can verify that the descrambler output is indeed S when this sequence T is applied at its input. • Homework: Learn to be able to determine where no terms needs to be considered. Dr. Shi Lathi & Ding-Digital Commun 77 Regenerative Repeater • Three functions: 1. Reshaping incoming pulses using an equalizer. 2. Extracting timing information Required to sample incoming pulses at optimum instants. 3. Making decision based on the pulse samples Dr. Shi Lathi & Ding-Digital Commun 78 39 Regenerative Repeater… Dr. Shi Lathi & Ding-Digital Commun 79 Preamplifier and Equalizer • A pulse train is attenuated and distorted by transmission medium, say, dispersion caused by an attenuation of high-frequency components. • Restoration of high frequency components ⇒ increase of channel noise • Fortunately, digital signals are more robust. – Considerable pulse dispersion can be tolerated • Main concern: Pulse dispersion ⇒ ISI ⇒ increase error probability in detection Dr. Shi Lathi & Ding-Digital Commun 80 40 Zero-Forcing Equalizer • Detection decision is based solely on sample values. ⇒ No need to eliminate ISI for all t. All that is needed is to eliminate or minimize ISI at their respective sampling instants only. • This could be done by using the transversal-filter equalizer, which forces the equalizer output pulse to have zero values at the sampling (decisionmaking ) instant. (refer to the diagram in next slide) Dr. Shi Lathi & Ding-Digital Commun 81 Dr. Shi Lathi & Ding-Digital Commun 82 41 Zero-Forcing Equalizer … c0 =1 ck = 0 ∀k ≠ 0 • Let tap setting ⇒ P0 (t ) = Pr (t − NTb ) P0 (t ) = Pr (t ) If ignore delay. • Fig 7.22 (b) indicates a problem : a1, a-1, a2, a-2, …are not negligible due to dispersion. • Now, want to force a1= a-1= a2= a-2 =…= 0 • Consider ck’s assume other values (other tap setting). Dr. Shi Lathi & Ding-Digital Commun 83 Zero-Forcing Equalizer … • Consider ck’s assume other values (other tap setting). N p0 ( t ) = ∑c n pr ( t − nTb ) n =− N N p0 ( kTb ) = ∑c n pr [(k − n )Tb ], k = 0, ± 1, ± 2 ,... n =− N • For simplicity of notation: p0 ( k ) = ∞ ∑c n =−∞ n pr ( k − n) • Nyquist Criterion: Dr. Shi p0 ( k ) = 0 ∀k ≠ 0 p0 ( k ) = 1 for k = 0 Lathi & Ding-Digital Commun 84 42 Zero-Forcing Equalizer … • ⇒ A set of infinitely many simultaneous equations: 2N+1: cn’s Impossible to solve this set of equations. • If, however, we specify the values of p0(k) only at 2N+1 points: k =0 1 p0 ( k ) = 0 k = ±1,±2,...,± N then a unique solution exists. Dr. Shi Lathi & Ding-Digital Commun 85 Zero-Forcing Equalizer … • Meaning: Zero ISI at the sampling instants of N preceding and N succeeding pulses. • Since pulse amplitude decays rapidly, ISI beyond the Nth pulse is not significant for N>2 in general. Dr. Shi Lathi & Ding-Digital Commun 86 43 Zero-Forcing Equalizer … • Page 7.37 Think about when n = 2, 1, 0, -1, -2; an example in next slide Dr. Shi Lathi & Ding-Digital Commun 87 Zero-Forcing Equalizer … • Page 7.37 Dr. Shi Lathi & Ding-Digital Commun 88 44 Eye Diagram • A convenient way to study ISI on an oscilloscope • Figure Dr. Shi Lathi & Ding-Digital Commun 89 Timing Extraction • • • The received signal needs to be sampled at precise instants. Timing is necessary. Three ways for synchronization: 1. 2. 3. Dr. Shi Derivation from a primary or a secondary standard (a master timing source exists, both transmitter and receiver follow the master). Transmitting a separate synchronizing signal (pilot clock) Self synchronization (timing information is extracted from the received signal itself). Lathi & Ding-Digital Commun 90 45 Timing Extraction … • Way 1: Suitable for large volumes of data high speed comm. Systems. High cost. • Way 2: Part of channel capacity is used to transmit timing information. Suitable for a large available capacity. • Way 3: Very efficient. • Examples: – On-off signaling (decomposition: Fig. 7.2) – Bipolar signaling rectification on-off signaling Dr. Shi Lathi & Ding-Digital Commun 91 Timing Jitter • Small random deviation of the incoming pulses from their ideal locations. • Always present, even in the most sophisticated communication systems. • Jitter reduction is necessary about every 200 miles in a long digital link to keep the maximum jitter within reasonable limits. • Figure 7.23 Dr. Shi Lathi & Ding-Digital Commun 92 46 Figure 7.23 Dr. Shi Lathi & Ding-Digital Commun 93 Detection –Error Probability • Received signal = desired + AWGN Example: Polar signaling and transmission (Fig. 7.24, the next slide) Ap : Peak signal value • Polar: Instead of ± Ap , received signal = ± Ap + n • Because of symmetry, the detection threshold = 0. – If sample value > 0, ⇒ “1” – If sample value < 0, ⇒ “0” – Error in detection may take place Dr. Shi Lathi & Ding-Digital Commun 94 47 Figure 7.24 Dr. Shi Lathi & Ding-Digital Commun 95 Error Probability for Polar Signal • With respect to Figure 7.24 P (∈ / 0 ) = probability that n > Ap P (∈ / 1) = probability that n < -Ap p (n) = n: AWGN ⇒ P(∈ 0) = = Dr. Shi 1 2π 1 ∞ Ap ∫σ n e 1 σ n 2π − 1 2 ( x) 2 ∫ ∞ Ap 2π σ n e 1 n − 2 σn e 1 n − 2 σ n 2 2 dn x= n σn Ap dx = Q σn Lathi & Ding-Digital Commun 96 48 Error Probability for Polar Signal… 1 Q( y ) = 2π where • Similarly, ∫ ∞ y e − x2 2 dx Ap P(∈ 1) = Q σn − Ap to − ∞ Symmetric • Summary: P (∈) = P (∈,0) + P (∈,1) Here, assume P(0)= P(1)=1/2. = P (∈ 0) P(0) + P (∈ 1) P(1) = A 1 [P(∈ 0) + P(∈ 1)] = Q p 2 σn • Q(x): Complementary error function: erfc (x) 1 Q ( x) ≅ x 2π Dr. Shi x2 0 .7 − 1 − 2 e 2 , x > 2 x Lathi & Ding-Digital Commun 97 Error Probability for Polar Signal… • If Ap σn = k , P(∈ 0) = P(∈ 1) = P(∈) = Q(k ) k Q(k)=P(ε) 1 0.1587 e.g. 3.16 ×10−5 ⇒ one of wrongly Dr. Shi 2 3 4 5 6 0.0227 0.00135 3.16 ×10−5 2.87 ×10 −7 9 . 9 × 10 − 9 3.16 × 105 pulses will be possibly detected Lathi & Ding-Digital Commun 98 49 Error Probability for On-off Signals • In this case, we have to distinguish between Ap and 0. 1 Ap • Threshold: 2 • Error Probabilities: A 1 P(∈ 0) = prob. of P(∈ 1) = prob. of P(∈) = n> p Ap = Q σ 2 n Ap −1 n< Ap = Q 2 2σ n A 1 [P(∈ 0) + P(∈ 1)] = Q p 2 2σ n 2 Also assume 0 and 1 equally probable P(∈) > P(ε ) (in the case of polar signaling) Dr. Shi Lathi & Ding-Digital Commun 99 Dr. Shi Lathi & Ding-Digital Commun 100 50 Error Probability for Bipolar Signals 1 ↔ Ap or − Ap 0 ↔ 0v ⇒ A A p p If the detected sample value: ∈ − 2 , 2 ↔ 0 Ap P(∈ 0) = prob. n > 2 Ap A + prob. n < − p = prob. n > 2 2 o.w. ↔ 1 Ap A = 2Q p = 2 prob. n > 2 2σ n Dr. Shi Lathi & Ding-Digital Commun 101 Error Prob. for Bipolar Signals… Ap when P (∈ 1) = prob . n < − positive pulse used 2 Ap when negative or prob . n > pulse used 2 Ap = Q 2σ n 1 P (∈) = P (∈ 0 ) P ( 0 ) + P (∈ 1) P (1) = [P (∈ 0 ) + P (∈ 1) ] 2 Ap > P (ε ) = 1 . 5Q for on − off signaling 2σ n Dr. Shi Lathi & Ding-Digital Commun 102 51 Detection Error Probability… Ap Q σn A on-off is in middle, Q p 2σ n A bipolar is the worst, 1.5Q p 2σ n • Summary, polar is the best, Assumption: “0” and “1” Equally likely • Another factor: P(∈) decreases exponentially with the signal power. Dr. Shi Lathi & Ding-Digital Commun 103 Comparison among three line codes • To obtain: P(∈) = 0.286 × 10−6 Ap • We need: σn Ap σn Ap σn =5 for polar case = 10 Q Q(5) = 0.286 × 10−6 Ap Q P (∈) = Q 2 σ n for bipolar case Q P (∈) = 1.5Q Ap 2σ n −6 = 0.286 × 10 for on-off case = 10 . 16 c Dr. Shi Lathi & Ding-Digital Commun Ap 2σ n = 5.08 104 52 Comparison among three line codes… Ap • For the same error rate, required SNR σ n , Polar < On-off < Bipolar • For the same A SNR σ p n (from previous example) , the caused error rate Polar < On-off < Bipolar (from next example) • Polar is most efficient in terms of SNR vs. error rate • Between, on-off and bipolar: Only 16% improvement of on-off over bipolar • Performance of bipolar case ≈ performance of on-off case in terms of error rate. Dr. Shi Lathi & Ding-Digital Commun 105 Example a) Polar binary pulses are received with peak amplitude Ap = 1 mV. The channel noise rms amplitude is 192.3 µV. Threshold detection is used, and 1 and 0 are equally likely. b) Find the error probability for (i) the polar case, and the on-off case (ii) the bipolar case if pulses of the same shape as in part (a) are used, but their amplitudes are adjusted so that the transmitted power is the same as in part (a) Dr. Shi Lathi & Ding-Digital Commun 106 53 Example … a) For the polar case Ap σn = 10 −3 = 5 .2 192.3(10 −6 ) From Table 8.2 (4th ed), we find P (∈) = Q ( 5 . 2 ) = 0 . 9964 × 10 − 7 b) Because half the bits are transmitted by nopulse, there are, on the average, only half as many pulses in the on-off case (compared to the polar). Now, doubling the pulse energy is accomplished by multiplying the pulse by 2 (in order to keep the power dissipation same) Dr. Shi Lathi & Ding-Digital Commun 107 Example… Thus, for on-off Ap is 2 times the Ap in the polar case. •Therefore, from Equation (7.53) A P(∈) = Q p = Q (3.68) = 1.66 × 10 − 4 2σ n •As seen earlier, for a given power, the Ap for both the on-off and the bipolar cases are identical. Hence, from equation (7.54) Ap P (∈) = 1.5Q 2σ n Dr. Shi = 1.749 × 10 −4 Lathi & Ding-Digital Commun 108 54 M-ary Communication • Digital communications use only a finite number of symbols • Information transmitted by each symbol increases with M. IM = log2 M bits IM : information transmitted by an M-ary symbol • Transmitted power increases as M2 , i.e., to increase the rate of communication by a factor of log2 M , the power required increases as M2. (see an example below) Dr. Shi Lathi & Ding-Digital Commun 109 M-ary Communication… • Most of the terrestrial digital telephone network: Binary • The subscriber loop portion of the integrated services digital network (ISDN) uses the quarternary code 2BIQ shown in Figure. 7.28 Dr. Shi Lathi & Ding-Digital Commun 110 55 Dr. Shi Lathi & Ding-Digital Commun 111 Pulse Shaping in Multi-amplitude Case • Nyquist Criterion can be used for M-ary case Controlled ISI • Figure 7.28: One possible M-ary Scheme • Another Scheme: Use M orthogonal pulses: ϕ1 (t ), ϕ 2 (t ),..., ϕ M (t ) • Definition: Tb c i = j ϕ i ( t )ϕ j ( t ) = 0 0 i ≠ j ∫ Dr. Shi Lathi & Ding-Digital Commun 112 56 Pulse Shaping in Multi-amplitude Case • The figure in next slide: One example in M orthogonal pulses: 2π ⋅ k ⋅ t , 0 < t < Tb k = 1, 2,..., M sin Tb ϕk (t ) = 0, otherwise • In the set, pulse frequency: k 1 1 2 M : , ,..., Tb Tb Tb Tb M times that of the binary scheme (see an example below) Dr. Shi Lathi & Ding-Digital Commun 113 Dr. Shi Lathi & Ding-Digital Commun 114 57 Pulse Shaping in Multi-amplitude Case • In general, it can be shown that the bandwidth of an orthogonal M-ary scheme is M times that of the binary scheme – In an M-ary orthogonal scheme, the rate of communication is increased by a factor of log 2 M at the cost of an increase in transmission bandwidth by a factor of M. Dr. Shi Lathi & Ding-Digital Commun 115 Digital Carrier Systems • So far: Baseband digital systems – Signals are transmitted directly without any shift in frequency. – Suitable for transmission over wires, cables, optical fibers. • Baseband signals cannot be transmitted over a radio link or satellites. – Since it needs impractically large size for antennas – Modulation (Shifting signal spectrum to higher frequencies is needed) Dr. Shi Lathi & Ding-Digital Commun 116 58 Digital Carrier Systems… • A spectrum shift to higher frequencies is also required when transmitting several messages simultaneously by sharing the large bandwidth of the transmission medium, • FDM: Frequency-division Multiplexing. (FDMA: Frequency-division Multiplexing Access) Dr. Shi Lathi & Ding-Digital Commun 117 Several Types of Modulation • Amplitude-Shift Keying (ASK), also known as onoff keying (OOK) m(t ) cos(ω c t ) : m(t ) : on-off baseband signal (modulating signal) cos(ω c t ) Dr. Shi : carrier Lathi & Ding-Digital Commun 118 59 Modulation Types… • Phase-Shift Keying (PSK) m(t ) : polar Signal 1 ↔ p(t) cos(ωct),0 ↔ − p(t) cos(ωct) = p(t) cos(π −ωct) = p(t) cos(ωct −π ) Dr. Shi Lathi & Ding-Digital Commun 119 Modulation Types… • Frequency-Shift Keying (FSK) – Frequency is modulation by the base band signal. 1 ↔ ω c1 ,0 ↔ ω c0 – Information bit resides in the carrier frequency. Dr. Shi Lathi & Ding-Digital Commun 120 60 PSD of ASK, PSK & FSK Dr. Shi Lathi & Ding-Digital Commun 121 FSK • FSK signal may be viewed as a sum of two interleaved ASK signals, one with a modulating frequency ω c , the other, ω c 0 1 – Spectrum of FSK = Sum of the two ASK – Bandwidth of FSK is higher than that of ASK or PSK Dr. Shi Lathi & Ding-Digital Commun 122 61 Demodulation A. Review of Analog Demodulation • Baseband signal and bandpass signal The end of the 2nd part of the slides of Ch. 2 and Ch. 3 • Double-sideband Suppressed Carrier (DSB-SC) Modulation m(t) ↔ M(ω) message signal 1 cos(ωct) ↔ [δ (ω +ωc ) +δ (ω −ωc )] carrier 2 1 m(t)cos(ωct) ↔ [ M(ω +ωc ) + M(ω −ωc )] modulated signal 2 Dr. Shi Lathi & Ding-Digital Commun 123 Figure 4.1 (a), (b), (c) DSB-SC Modulation 124 62 DSB-SC Demodulation • Demodulation: Synchronization Detection (Coherent Detection) • Figure 4.1 (e) and (d) Using a carrier of exactly the same frequency & pulse Dr. Shi 125 DSB-SC Demodulation … [m(t ) cos(ωct )]cos(ωct ) = e(t ) = 1 [m(t ) + m(t ) cos(2ωct )] 2 1 1 M (ω ) + [M (ω + 2ωc ) + M (ω − 2ωc )] 2 4 1 Aftera LPF ⇒ M (ω ) 2 E (ω ) = 1 m(t ) 2 Dr. Shi 126 63 Figure 4.1 (d), (e) DSB-SC demodulation 127 AM Signal • Motivation: – Requirements of a carrier in frequency and phase synchronism with carrier at the transmitter is too sophisticated and quite costly – Sending the carrier ⇒ a. eases the situation b. requires more power in transmission c. suitable for broadcasting Dr. Shi 128 64 Amplitude Modulation (AM) ϕ AM (t ) = A cos(ω c t ) + m(t ) cos(ω c t ) = [A + m(t )]cos(ω c t ) ϕ AM (t ) ⇔ 1 [M (ω + ω c ) + M (ω − ω c )] 2 + π ⋅ A[δ (ω + ω c ) + δ (ω − ω c )] Dr. Shi 129 • AM Signal and its Envelope 130 65 • Demodulation of AM signals – Rectifier Detection: Figure 4.11 Output: – Envelope Detector: 1 π m(t ) Figure 4.12 Output: A + m(t ) Dr. Shi 131 132 66 133 B. Back to Digital Communication System: Demodulation • Demodulation of Digital-Modulated Signals is similar to Demodulation of Analog-Modulated Signals • Demodulation of ASK Synchronous (Coherent) Detection Non-coherent Detection: Say, envelope detection Dr. Shi Lathi & Ding-Digital Commun 134 67 Demodulation of PSK • Cannot be demodulated non-coherently (envelope detection). – Since, envelope is the same for 0 and 1 • Can be demodulated coherently. • PSK may be demodulated non-coherently if use: Differential coherent PSK (DPSK) Dr. Shi Lathi & Ding-Digital Commun 135 Differential Encoding • Differential coding: – “1” encoded by the same pulse used to encode the previous data bit no transition – “0” encoded by the negative of pulse used to encode the previous data bit transition • Figure 7.30 (a) • Modulated signal consists of pulses ± A cos(ω ct ) with a possible sign ambiguity Dr. Shi Lathi & Ding-Digital Commun 136 68 Differential Encoding • Figure Dr. Shi Lathi & Ding-Digital Commun 137 Differential Encoding… • Tb : one-bit interval • If the received pulse is identical to the previous pulse 1, A2 Y (t ) = A2 cos 2 (ω c t ) = [1 + cos(2ω ct )] 2 z (t ) = 2 A 2 • If 0 is transmitted Y (t ) = − A2 cos 2 (ω c t ) = − z (t ) = − Dr. Shi A2 [1 + cos(2ω ct )] 2 A2 2 Lathi & Ding-Digital Commun 138 69 Demodulation of FSK • FSK can be viewed as two interleaved ASK signals with carrier frequency ω c and ω c , respectively. • Hence, FSK can be detected by non-coherent detection (envelope) or coherent detection technique 0 Dr. Shi Lathi & Ding-Digital Commun 1 139 Demodulation of FSK Dr. Shi Lathi & Ding-Digital Commun 140 70 Demodulation of FSK • In (a), non-coherent detection (envelope), ωc – H0(ω) 0 Turned to – H1(ω) respectively ωc 1 • Comparison: above or bottom Whose output is large 0 or 1 Dr. Shi Lathi & Ding-Digital Commun 141 71
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