A First Course in Digital Communications Ha H. Nguyen and E. Shwedyk - - PowerPoint PPT Presentation

SLIDE 1

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

SLIDE 2

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:

1

Signal interference and noise immunity

2

Signal spectrum

3

Signal synchronization capability

4

Error detection capability

5

Cost and complexity of transmitter and receiver 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

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Chapter 6: Baseband Data Transmission

Baseband Signaling Schemes I

1 1 1 1 1 a Binary Dat V V V −

b

T Time Clock Code NRZ (a) L

• NRZ

(b)

A First Course in Digital Communications 3/28

SLIDE 4

Chapter 6: Baseband Data Transmission

Baseband Signaling Schemes II

V V V − V Code RZ (c) L

• RZ

(d) Phase

• Bi

(e)

A First Course in Digital Communications 4/28

SLIDE 5

Chapter 6: Baseband Data Transmission

Baseband Signaling Schemes III

V V − V V V − L

• Phase
• Bi

(f) Code Miller (g) L

• Miller

(h)

A First Course in Digital Communications 5/28

SLIDE 6

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 V − V V L

• Phase
• Bi

(f) Code Miller (g)

A First Course in Digital Communications 6/28

SLIDE 7

Chapter 6: Baseband Data Transmission

NRZ-L Code

• ✁

✂ ✄

) (

1 t

s

b

T V −

☎ ✆ ✝ ✞ ✟ ☎ ✄

) (

2 t

s V

b

T

☎ ✟ ✠ ✝ ☎ ✡

) (

1 t

φ

b

T 1

b

T

☛ ☞ ✌

) (

1 t

φ ) (

1 t

s ) (

2 t

s

✍ ✎ ✏

L

• NRZ

E

L

• NRZ

E −

T T

1 Choose Choose

✑

⇐

P[error]NRZ-L = Q

• 2ENRZ-L/N0
• .

A First Course in Digital Communications 7/28

SLIDE 8

Chapter 6: Baseband Data Transmission

RZ-L Code

✒ ✓ ✔

V −

✕

V

b

T ) (

2 t

s

✖ ✗ ✘ ✙ ✖

) (

1 t

s

✕

V −

b

T

✖ ✚ ✙ ✛ ✗ ✖

2

b

T

✜ ✢ ✣

) (

1 t

φ

✤

b

T

b

T 1 −

✤

b

T 1

b

T

b

T 1 − ) (

2 t

φ 2

b

T

) (

1 t

φ ) (

1 t

s ) (

2 t

s ) (

2 t

φ

L

• RZ

E

L

• RZ

E

T

1 Choose

T

Choose

✥ ✦ ✧

P[error]RZ-L = Q

• ERZ-L/N0
• .

A First Course in Digital Communications 8/28

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Chapter 6: Baseband Data Transmission

Bi-Phase-Level (Biφ-L) Code

★ ✩ ✪

) (

1 t

s

✫

V −

b

T

✬ ✭ ✮ ✯ ✰ ✬

V V −

✫

V

b

T ) (

2 t

s

✬ ✰ ✱ ✮ ✬ ✲

b

T 1

b

T

b

T 1 − ) (

1 t

φ

✳ ✴ ✵

) (

1 t

φ ) (

1 t

s ) (

2 t

s

✶ ✷ ✸

T T

1 Choose Choose

✹

⇐

L

• Biφ

E

L

• Biφ

E −

P[error]Biφ-L = Q

• 2EBiφ-L/N0
• .

A First Course in Digital Communications 9/28

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Chapter 6: Baseband Data Transmission

Miller-Level (M-L) I

V −

✺

V

b

T ) (

2 t

s

✻ ✼ ✽ ✾ ✻ ✺

) (

1 t

s

b

T

✻ ✿ ✾ ❀ ✼ ✻

V V −

✺

) (

4 t

s

b

T

✺

) (

3 t

s V −

b

T

✻ ✼ ✽ ✾ ✻ ✻ ✿ ✾ ❀ ✼ ✻

V

❁ ❂ ❃ ❄

) (

1 t

φ

b

T 1

b

T

b

T 1 −

❄

) (

2 t

φ

b

T

b

T 1

❅ ❆ ❇

A First Course in Digital Communications 10/28

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Chapter 6: Baseband Data Transmission

Miller-Level (M-L) II

) (

1 t

φ ) (

1 t

s ) (

2 t

s ) (

2 t

φ ) (

3 t

s ) (

4 t

s

❈ ❉ ❊❊ ❋ ● ❍ ■ ❏ ❑ ▲▲▼ ◆ ❖ P ◗ ❘ ❙ ❙ ❚ ❯ ❱ ❲ ❳ ❨ ❩ ❩ ❬ ❭ ❪ ❫ ❴ ❵ ❛

A First Course in Digital Communications 11/28

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Chapter 6: Baseband Data Transmission

Miller-Level (M-L) III

) (

1 t

φ ) (

1 t

s ) (

2 t

s ) (

2 t

φ ) (

3 t

s ) (

4 t

s

L

• M

E

1

r

2

r

1

ˆ r

2

ˆ r

L

• M

5 . 0 E 45

P[error]M-L = 1 −

• 1 − Q
• EM-L/N0

2

A First Course in Digital Communications 12/28

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Chapter 6: Baseband Data Transmission

Performance Comparison

2 4 6 8 10 12 14 10

−6

10

−5

10

−4

10

−3

10

−2

10

−1

Eb/N0 (dB) P[error] NRZ−L and Biφ−L RZ−L M−L

ENRZ-L = ERZ-L = EBiφ-L = EM-L = V 2Tb ≡ Eb (joules/bit). P [error]NRZ-L = P [error]Biφ-L = Q

• 2Eb

N0

• ,

P [error]RZ-L = Q

• Eb

N0

• , P [error]M-L ≈ 2Q
• Eb

N0

• .

A First Course in Digital Communications 13/28

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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 φ1(t) and φ2(t) are

• r(1)

1

= −0.2, r(1)

2

= −0.4

• ,
• r(2)

1

= +0.2, r(2)

2

= −0.8

• ,
• r(3)

1

= −0.61, r(3)

2

= +0.5

• ,
• r(4)

1

= −1.1, r(4)

2

= +0.1

• .

Transmitted signal Distance squared 0 → Tb Tb → 2Tb 2Tb → 3Tb 3Tb → 4Tb s1(t) 1.6 1.28 2.8421 4.42 s2(t) 2.0 3.28 0.6221 2.02 s3(t) 0.8 2.08 0.4021 0.02 s4(t) 0.4 0.08 2.6221 2.42 {s4(t), s4(t), s3(t), s3(t)} is not a valid transmitted sequence!

A First Course in Digital Communications 14/28

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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 ≤ nTb. Denote the entire transmitted signal over the time interval [0, nTb] as Si(t), i = 1, 2, . . . , M = 2n. Write Si(t) = n

j=1 Sij(t) where Sij(t) is one of the four

possible signals used in Miller code in the bit interval [(j − 1)Tb, jTb] and zero elsewhere. Received signal over [0, nTb] is r(t) = n

j=1 rj(t) where

rj(t) = r(t) in the interval [(j − 1)Tb, jTb] and zero elsewhere. Distance (squared) from Si(t) to r(t): d2

i

= nTb [r(t) − Si(t)]2dt =

n

• j=1

jTb

(j−1)Tb

[rj(t) − Sij(t)]2dt =

n

• j=1
• r(j)

1

− S(j)

i1

2 +

• r(j)

2

− S(j)

i2

2 . Direct evaluation of the M = 2n distances is impossible!

A First Course in Digital Communications 15/28

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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.

❜ ❝ ❜ ❞ ❜ ❡ ❜ ❢

) ( 1

2 t

s ) ( 1

2 t

s ) ( 1

4 t

s ) (

1 t

s ) (

1 t

s ) (

3 t

s ) (

3 t

s ) ( 1

4 t

s

❣ ❤ ✐ ❤ ❥ ❦ ❧ ♠ ♥ ❤ ♦ ♣ ❤ q ♥ ❤ ♠ ♥ ❤ r ♣ s ❧ ✐ t

A First Course in Digital Communications 16/28

SLIDE 17

Chapter 6: Baseband Data Transmission

Trellis Diagram

✉ ✈ ✇ ✈① ② ③ ④ ⑤ ⑥ ⑦ ⑧ ⑨ ⑦ ⑩ ❶ ⑦

⋅ ⋅ ⋅

❷ ⑩ ❸ ⑨ ⑤ ❹ ④ ⑤ ❺ ❻ ❷ ⑩ ❸ ⑨ ⑤ ❹ ④ ⑤ ❺ ❼

4 . , 2 . ,

) 1 ( 2 ) 1 ( 1

− − r r

b

T

b

T 2

b

T 3

b

T 4 8 . , 2 . ,

) 2 ( 2 ) 2 ( 1

− + r r 5 . , 61 . ,

) 3 ( 2 ) 3 ( 1

+ − r r 1 . , 1 . 1 ,

) 4 ( 2 ) 4 ( 1

+ − r r

❽ ❾❿ ⑩ ➀ ➁ ④ ⑤ ⑤ ⑦ ➂ ➀ ④ ➃ ⑩ ❿ ➄

) (

1 t

s ) (

2 t

s ) (

3 t

s ) (

4 t

s

A First Course in Digital Communications 17/28

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Chapter 6: Baseband Data Transmission

Viterbi Algorithm

Step 1: Start from the initial state (s1(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

SLIDE 19

Chapter 6: Baseband Data Transmission

Example 6.2 I

➅ ➆ ➇ ➆ ➈ ➉

b

T ) (

1 t

s ) (

2 t

s ) (

3 t

s ) (

4 t

s

➊ ➋ ➌ ➍ ➋ ➊ ➎ ➏ ➏ ➐ ➑ ➐ ➒ ➓ ➇ ➆ ➈ ➔ → ➣

• r(1)

1

= −0.2, r(1)

2

= −0.4

• ,
• r(2)

1

= +0.2, r(2)

2

= −0.8

• ,
• r(3)

1

= −0.61, r(3)

2

= +0.5

• ,
• r(4)

1

= −1.1, r(4)

2

= +0.1

• .

A First Course in Digital Communications 19/28

SLIDE 20

Chapter 6: Baseband Data Transmission

Example 6.2 II

↔ ↕➙ ↕➛ ➜

b

T

b

T 2 ) (

1 t

s ) (

2 t

s ) (

3 t

s ) (

4 t

s

➝ ➞ ➟ ➠ ➡ ➞ ➟ ➠ ➝ ➞ ➟ ➠ ➟ ➞ ➠ ➠ ➟ ➞ ➠ ➝ ➞ ➟

• r(1)

1

= −0.2, r(1)

2

= −0.4

• ,
• r(2)

1

= +0.2, r(2)

2

= −0.8

• ,
• r(3)

1

= −0.61, r(3)

2

= +0.5

• ,
• r(4)

1

= −1.1, r(4)

2

= +0.1

• .

A First Course in Digital Communications 20/28

SLIDE 21

Chapter 6: Baseband Data Transmission

Example 6.2 III

➢ ➤ ➥ ➤ ➦ ➧

b

T

b

T 2 ) (

1 t

s ) (

2 t

s ) (

3 t

s ) (

4 t

s

➨ ➩ ➫ ➭ ➯ ➩ ➫ ➭ ➫ ➩ ➭ ➭ ➲ ➳ ➳ ➵ ➸ ➵ ➺ ➻ ➥ ➤ ➦ ➼ ➽ ➾

• r(1)

1

= −0.2, r(1)

2

= −0.4

• ,
• r(2)

1

= +0.2, r(2)

2

= −0.8

• ,
• r(3)

1

= −0.61, r(3)

2

= +0.5

• ,
• r(4)

1

= −1.1, r(4)

2

= +0.1

• .

A First Course in Digital Communications 21/28

SLIDE 22

Chapter 6: Baseband Data Transmission

Example 6.2 IV

➚ ➪➶ ➪ ➹ ➘

b

T

b

T 2

b

T 3 ) (

1 t

s ) (

2 t

s ) (

3 t

s ) (

4 t

s

➴ ➷ ➬ ➮ ➮ ➱ ➮ ➷ ✃ ❐ ➮ ➱ ➱ ➷ ❒ ❮ ➮ ➱ ❰ ➷ ➬ ❮ ➮ ➱ Ï Ð Ð Ñ Ò Ñ Ó Ô ➶ ➪ ➹ Õ Ö ×

• r(1)

1

= −0.2, r(1)

2

= −0.4

• ,
• r(2)

1

= +0.2, r(2)

2

= −0.8

• ,
• r(3)

1

= −0.61, r(3)

2

= +0.5

• ,
• r(4)

1

= −1.1, r(4)

2

= +0.1

• .

A First Course in Digital Communications 22/28

SLIDE 23

Chapter 6: Baseband Data Transmission

Example 6.2 V

Ø Ù Ú Ù Û Ü

b

T

b

T 2

b

T 3

b

T 4 ) (

1 t

s ) (

2 t

s ) (

3 t

s ) (

4 t

s

Ý Þ ß à á â â Þ ã á á â ã Þ ä å á â æ Þ ß á á â ç è è é ê é ë ì Ú ÙÛ í î ï

• r(1)

1

= −0.2, r(1)

2

= −0.4

• ,
• r(2)

1

= +0.2, r(2)

2

= −0.8

• ,
• r(3)

1

= −0.61, r(3)

2

= +0.5

• ,
• r(4)

1

= −1.1, r(4)

2

= +0.1

• .

A First Course in Digital Communications 23/28

SLIDE 24

Chapter 6: Baseband Data Transmission

Symbol-by-Symbol vs. Sequence Demodulation

2 4 6 8 10 12 14 10

−6

10

−5

10

−4

10

−3

10

−2

10

−1

10 Eb/N0 (dB) P[error] Symbol−by−symbol demodulation (analytical result) Sequence demodulation (simulation result)

2 dB gain at the error probability of 10−2 and 0.5 dB at 10−6.

A First Course in Digital Communications 24/28

SLIDE 25

Chapter 6: Baseband Data Transmission

Spectrum I

SNRZ-L(f) E = 1 Tb (1 − 2P)2δ(f) + 4P(1 − P)sin2(πfTb) (πfTb)2 . SBiφ(f) E = 1 Tb (1 − 2P)2

∞

• n=−∞

n=0

2 nπ 2 δ

• f − n

Tb

• +

4P(1 − P)sin4(πfTb/2) (πfTb/2)2 . SM-L(f) E = 1 2θ2(17 + 8 cos 8θ)(23 − 2 cos θ − 22 cos 2θ − 12 cos 3θ +5 cos 4θ + 12 cos 5θ + 2 cos 6θ − 8 cos 7θ + 2 cos 8θ), where θ = πfTb and P2 = P1 = 0.5.

A First Course in Digital Communications 25/28

SLIDE 26

Chapter 6: Baseband Data Transmission

Spectrum II

0.5 1 1.5 2 0.5 1 1.5 2 2.5 Normalized frequency (fTb) Normalized PSD (S s(f)/E) NRZ−L Biφ−L Miller−L

A First Course in Digital Communications 26/28

SLIDE 27

Chapter 6: Baseband Data Transmission

Differential NRZ-L Modulation

In differential modulation, the signal transmitted in one bit interval is relative to the one transmitted in the previous interval. If the present bit is a “1”, then transmit a level that is

• pposite to that of the previous interval.

If the present bit is a “0”, then stay at the same level.

ð ñ ò ò ó ô ó õö ñ ÷ ø ù õ ú û ü ó ô ý þÿ

• ✁

✂ û ü ✄ ø ÷ ö û ô

k

b

1

z−

ð ó ø ÷ ☎

1 k k k

d b d − = ⊕

1 k

d − Transmitted signal ( )

T

s t

ù ✆ ú ø ✄ ✝ ñ ✞ ó

• ✟

þ

If bk = 1 then dk = dk−1, implying a level change and if bk = 0 then dk = dk−1, which means no level change.

A First Course in Digital Communications 27/28

SLIDE 28

Chapter 6: Baseband Data Transmission

Demodulation of Differential NRZ-L

Exclusive-OR Differential Decoder

( )

d

b

T

t

• ∫

b

t kT = ( ) ( ) ( )

T

t s t t = + r w

k k

V = ± + r w

1

D D

≥ ≤

1

z− ˆ

k

d ˆ

k

b

1

ˆ

k

d − Delay

b

T 1

b

T t

First determine dk and call this estimate ˆ dk. To recover bk, note that dk ⊕ dk−1 = bk ⊕ dk−1 ⊕ dk−1 = bk The receiver uses ˆ dk−1 instead of dk−1. If ˆ dk is in error, there will be two errors in the sequence {ˆ bk}.

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