Chapter 6: Baseband Data Transmission A First Course in Digital Communications Ha H. Nguyen and E. Shwedyk February 2009 A First Course in Digital Communications 1/28
Chapter 6: Baseband Data Transmission Introduction Bits are mapped into two voltage levels for direct transmission without any frequency translation. Various baseband signaling techniques (line codes) were developed to satisfy typical criteria: Signal interference and noise immunity 1 Signal spectrum 2 Signal synchronization capability 3 Error detection capability 4 Cost and complexity of transmitter and receiver 5 implementations Four baseband signaling schemes to be considered: nonreturn-to-zero-level (NRZ-L), return-to-zero (RZ), bi-phase-level or Manchester, and delay modulation or Miller. A First Course in Digital Communications 2/28
Chapter 6: Baseband Data Transmission Baseband Signaling Schemes I Binary Dat a 1 0 1 1 0 0 1 0 1 Clock T b V (a) NRZ Code Time 0 V (b) NRZ - L 0 − V A First Course in Digital Communications 3/28
Chapter 6: Baseband Data Transmission Baseband Signaling Schemes II V (c) RZ Code 0 V (d) RZ - L 0 − V V (e) Bi - Phase 0 A First Course in Digital Communications 4/28
Chapter 6: Baseband Data Transmission Baseband Signaling Schemes III V (f) Bi - Phase - L 0 − V V (g) Miller Code 0 V (h) Miller - L 0 − V A First Course in Digital Communications 5/28
Chapter 6: Baseband Data Transmission Miller Code Has at least one transition every two bit interval and there is never more than two transitions every two bit interval. Bit “1” is encoded by a transition in the middle of the bit interval. Depending on the previous bit this transition may be either upward or downward. Bit “0” is encoded by a transition at the beginning of the bit interval if the previous bit is “0”. If the previous bit is “1”, then there is no transition. V (f) Bi - Phase - L 0 − V V (g) Miller Code 0 V A First Course in Digital Communications 6/28
Chapter 6: Baseband Data Transmission NRZ-L Code s 1 t s 2 t ( ) ( ) ☎ ☎ ☎ ☎ ✆ ✝ ✞ ✟ ✟ ✠ ✝ φ 1 t ( ) V T T ✄ 1 ✄ b b T 0 0 b ✡ − V T 0 b � ✂ ☛ ☞ ✌ ✁ ⇐ Choose 0 ✑ Choose 1 T T s 1 t s 2 t ( ) ( ) φ 1 t ( ) − E E 0 NRZ - L NRZ - L ✍ ✏ ✎ �� � P [ error ] NRZ-L = Q 2 E NRZ-L /N 0 . A First Course in Digital Communications 7/28
Chapter 6: Baseband Data Transmission RZ-L Code φ 1 t φ 2 t ( ) ( ) s 1 t s 2 t ( ) ( ) ✖ ✖ ✖ ✖ ✚ ✙ ✛ ✗ ✗ ✘ ✙ T 1 b V T ✤ T T ✤ 2 b b b T T ✕ T ✕ 2 0 0 b b b 0 0 − T − T 1 1 ✜ ✢ ✣ b b − V ✒ ✔ − V ✓ φ 2 t ( ) E s 2 t ( ) RZ - L �� � Choose 0 P [ error ] RZ-L = Q E RZ-L /N 0 . T Choose 1 T s 1 t ( ) 0 φ 1 t ( ) E RZ - L ✥ ✧ ✦ A First Course in Digital Communications 8/28
Chapter 6: Baseband Data Transmission Bi-Phase-Level (Bi φ -L) Code s 1 t s 2 t φ 1 t ( ) ( ) ( ) ✬ ✬ ✬ ✬ ✭ ✮ ✯ ✰ ✰ ✱ ✮ V V T 1 b T T ✫ ✫ ✲ b b 0 0 T 0 b − V − V − T 1 b ★ ✪ ✳ ✴ ✵ ✩ ⇐ Choose 0 ✹ Choose 1 T T s 1 t s 2 t ( ) ( ) φ 1 t ( ) − E E 0 Bi φ Bi φ - L - L ✶ ✸ ✷ �� � P [ error ] Bi φ -L = Q 2 E Bi φ -L /N 0 . A First Course in Digital Communications 9/28
Chapter 6: Baseband Data Transmission Miller-Level (M-L) I s 1 t ✻ ✻ ✻ ✻ s 2 t ( ) ( ) ✿ ✾ ❀ ✼ ✼ ✽ ✾ V V T ✺ ✺ b T 0 0 b − V s 3 t ✻ ✻ s 4 t ✻ ✻ ( ) ( ) ✿ ✾ ❀ ✼ ✼ ✽ ✾ V T ✺ ✺ b 0 0 T b − V ❁ ❃ − V ❂ φ 2 t φ 1 t ( ) ( ) T T 1 1 b b T ❄ ❄ b 0 0 T b ❅ ❆ ❇ − T 1 b A First Course in Digital Communications 10/28
Chapter 6: Baseband Data Transmission Miller-Level (M-L) II ❉ ❊❊ ❋ ● ❍ ❈ ■ φ 2 t ( ) s 2 t ( ) ❲ ❫ ❱ ❪ s 3 t s 1 t ❯ ❭ ( ) ( ) ❚ ❬ ❙ ❩ ❙ ❩ ❘ ❨ φ 1 t ( ) ◗ ❳ 0 s 4 t ( ) ❏ ❑ ▲▲▼ ◆ ❖ P ❴ ❵ ❛ A First Course in Digital Communications 11/28
Chapter 6: Baseband Data Transmission Miller-Level (M-L) III φ 2 t r ( ) 2 s 2 t r ( ) ˆ 2 0 E . 5 M - L s 1 t s 3 t ( ) ( ) φ 1 t 0 ( ) 45 0 r E 1 M - L s 4 t ( ) r ˆ 1 �� 2 � �� P [ error ] M-L = 1 − 1 − Q E M-L /N 0 A First Course in Digital Communications 12/28
Chapter 6: Baseband Data Transmission Performance Comparison −1 10 −2 10 M−L RZ−L −3 10 P [error] NRZ−L and Bi φ −L −4 10 −5 10 −6 10 0 2 4 6 8 10 12 14 E b / N 0 (dB) E NRZ-L = E RZ-L = E Bi φ -L = E M-L = V 2 T b ≡ E b (joules/bit) . �� � 2 E b P [ error ] NRZ-L = P [ error ] Bi φ -L = Q , N 0 �� �� � � E b E b P [ error ] RZ-L = Q , P [ error ] M-L ≈ 2 Q . N 0 N 0 A First Course in Digital Communications 13/28
Chapter 6: Baseband Data Transmission Optimum Sequence Demodulation for Miller Signaling Sequence demodulation exploits memory in Miller modulation. Example: The four Miller signals have unit energy and the projections of the received signals on to φ 1 ( t ) and φ 2 ( t ) are � � � � r (1) = − 0 . 2 , r (1) r (2) = +0 . 2 , r (2) = − 0 . 4 , = − 0 . 8 , 1 2 1 2 � � � � r (3) = − 0 . 61 , r (3) r (4) = − 1 . 1 , r (4) = +0 . 5 , = +0 . 1 . 1 2 1 2 Transmitted signal Distance squared 0 → T b T b → 2 T b 2 T b → 3 T b 3 T b → 4 T b s 1 ( t ) 1.6 1.28 2.8421 4.42 s 2 ( t ) 2.0 3.28 0.6221 2.02 s 3 ( t ) 0.8 2.08 0.4021 0.02 s 4 ( t ) 0.4 0.08 2.6221 2.42 { s 4 ( t ) , s 4 ( t ) , s 3 ( t ) , s 3 ( t ) } is not a valid transmitted sequence! A First Course in Digital Communications 14/28
Chapter 6: Baseband Data Transmission Assume that a total of n bits are transmitted. Each n -bit pattern results in a transmitted signal over 0 ≤ t ≤ nT b . Denote the entire transmitted signal over the time interval [0 , nT b ] as S i ( t ) , i = 1 , 2 , . . . , M = 2 n . Write S i ( t ) = � n j =1 S ij ( t ) where S ij ( t ) is one of the four possible signals used in Miller code in the bit interval [( j − 1) T b , jT b ] and zero elsewhere. Received signal over [0 , nT b ] is r ( t ) = � n j =1 r j ( t ) where r j ( t ) = r ( t ) in the interval [( j − 1) T b , jT b ] and zero elsewhere. Distance (squared) from S i ( t ) to r ( t ) : � nT b � jT b n � d 2 [ r ( t ) − S i ( t )] 2 d t = [ r j ( t ) − S ij ( t )] 2 d t = i 0 ( j − 1) T b j =1 n �� � 2 � � 2 r ( j ) − S ( j ) � r ( j ) − S ( j ) � = + . 1 i 1 2 i 2 j =1 Direct evaluation of the M = 2 n distances is impossible! A First Course in Digital Communications 15/28
Chapter 6: Baseband Data Transmission State Diagram State of a system: Information from the past we need at the present time, which together with the present input allows us to determine the system’s output for any future input. ❦ ❤ ♦ ♣ ❤ ❧ ♠ ♥ ❜ ❝ s 2 t s 2 t 1 ( ) 1 ( ) q t ❤ ❤ r ♣ s ✐ ♥ ♠ ♥ ❧ s 4 t 1 ( ) s 1 t 0 ( ) ❜ ❡ ❜ ❢ s 1 t 0 ( ) ❣ ❤ ✐ ❤ ❥ s 3 t 0 ( ) s 3 t s 4 t 0 ( ) 1 ( ) ❜ ❞ A First Course in Digital Communications 16/28
Chapter 6: Baseband Data Transmission Trellis Diagram ❷ ⑤ ❹ ④ ⑤ ⑩ ❸ ⑨ ❺ ❻ ❽ ④ ⑤ ⑤ ➂ ④ ➃ ➄ ❾❿ ⑩ ➀ ➁ ⑦ ➀ ⑩ ❿ ❷ ⑤ ❹ ④ ⑤ ⑩ ❸ ⑨ ❺ ❼ ③ ④ ⑤ ⑥ ⑦ ⑧ ⑨ ⑦ ⑩ ❶ ⑦ s 1 t ( ) s 3 t ( ) ⋅ ⋅ ⋅ ✉ ✈ ✇ ✈① ② s 2 t ( ) s 4 t ( ) T T T T 0 2 3 4 b b b b r ( 1 ) r ( 1 ) r ( 2 ) r ( 2 ) r r r ( 4 ) r ( 4 ) ( 3 ) ( 3 ) , , , , 1 2 1 2 1 2 1 2 − − + − − + − + 0 . 2 , 0 . 4 0 . 2 , 0 . 8 . 61 , 0 . 5 1 . 1 , 0 . 1 A First Course in Digital Communications 17/28
Chapter 6: Baseband Data Transmission Viterbi Algorithm Step 1: Start from the initial state ( s 1 ( t ) in our case). Step 2: In each bit interval, calculate the branch metric , which is the distance squared between the received signal in that interval with the signal corresponding to each possible branch. Add this branch metric to the previous metrics to get the partial path metric for each partial path up to that bit interval. Step 3: If there are two partial paths entering the same state, discard the one that has a larger partial path metric and call the remaining path the survivor . Step 4: Extend only the survivor paths to the next interval. Repeat Steps 2 to 4 till the end of the sequence. A First Course in Digital Communications 18/28
Chapter 6: Baseband Data Transmission Example 6.2 I s 1 t ( ) ➊ ➋ ➌ s 3 t ( ) ➎ ➏ ➏ ➐ ➑ ➐ ➒ ➅ ➆ ➇ ➆ ➈ ➉ ➓ ➔ → ➣ ➇ ➆ ➈ ➍ ➋ ➊ s 2 t ( ) s 4 t ( ) T 0 b � � � � r (1) = − 0 . 2 , r (1) r (2) = +0 . 2 , r (2) = − 0 . 4 , = − 0 . 8 , 1 2 1 2 � � � � r (3) = − 0 . 61 , r (3) r (4) = − 1 . 1 , r (4) = +0 . 5 , = +0 . 1 . 1 2 1 2 A First Course in Digital Communications 19/28
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