2/6/01

Receivers, Antennas, and Signals

Modulation and Coding Professor David H. Staelin

Lec14.10-1

2/6/01

Multi-Phase-Shift Keying, “MPSK”

F1

ing Step

S2(t)T

“EyeDiagram”

S1(t)

E/No(dB)

B/R

Small Loss

M = 2 4

8 3216

QPSK is bandwidth efficient, but has little E/No penalty.

0

Bandwidth/bit = B/R

[Hz] bits/sec

1.0 0.5

BPSK: Binary Phase-Shift Keying

E/No(dB) (E is J/bit)

M = 16 (4 bits/baud)

10-2

10-4

10-6

0

BER (Bit Error Rate)

Pe for Ideal

Receiver

Lec14.10-2

QPSK: Quadrature Phase-Shift Key

10 20 30

M = 2 phases M = 4 (2 bits) M = 8 (3 bits)

10 20

2/6/01

Error Probabilities for Binary Signaling

3 dB

1.0 0.5

10-1

10-2

10-3

10-4

10-5

10-6

10-7

2 4 6 8 10 12 14 (Eb/No) (dB)

Pe

P e =

Pro

babi

lity

of B

it E

rror

, L.W. Couch II, (4th Edition), Page 351

Matched filter reception of unipolar baseband, coherent OOK or coherent FSK

Matched filter reception of bipolar baseband, BPSK or QPSK

Q Eb /No

Q 2(Eb /No )

F2

bE = average energy per limit

Lec14.10-3

–1 0

Source: Digital and Analog Communication Systems

2/6/01

Phasor Diagrams S(t) = Re{Sejωt}

Equal-NoiseContour QPSK

Re{S}

Decision Region for H1

Phasor

Im{S} Im{S}

16-PSK

Re{S} Decision Region for H1

M = 16, Amplitudeand Phase

Im{S}

Re{S}

Rectangular

H1

Re{S}

Hexagonal

Superior toRectangular

M = 19 Im{S}

H1

F3Lec14.10-4

2/6/01

Baud Boundary

Want symbols to be orthogonalwithin and between windows

Overlapping Window Functions

Si(f)

0 fi

f

Narrowband with Lower Sidelobes

Lower Sidelobes

Window FunctionTypical

Baud

Si(f)

0 fi

f

Broader Channel Bandwidth

F4

Intersymbol Interference MPSK Examples:

Boxcar Envelope

Typical Baud

Si(f)

0 fi

f Sidelobes Interfere with Adjacent Channels

Spectrum for Channel i

Lec14.10-5

2/6/01

Performance Degradation Due to Interchannel Interfence

10 Signal degradation (dB) same as required signal boost for constant Pe

1

0.1 2

∆f/R for given E/No

N = 1 Interfering Channel

“Tamed FM” (window overlap) Boxcar (QPSK)

e

o becomes negligible)

BPS

N = ∞

N = ∞

∆f/R10-8

10-2

BER

N = 1

“Tamed FM”

BER o)

F5

Channel spacing (Hz)

Lec14.10-6

0 0.5 1 Closer channel spacing requires more signal power to maintain PRecover by boosting signal power (works until N

Interfering Channels

0.4 0.5 0.6 0.7

(for a given E/N

2/6/01

Error Reduction via Channel Coding

We want Pe → ⇒ o → ∞ using prior methods

Theorem: Pe →

∼21 dB e.g. 3-kHz phone at 9600 bps requires S/N ≥ 10 dB ~

Shannon’s Channel Capacity Theorem

C = B LOG2 /N) bits/sec

[Hz] oB

Average Signal Power

H1Lec14.10-7

0 (banking, etc.) E/N

0 if channel capacity “C” not exceeded, in bits/sec

Examples: S/N = 10 yields C = 3B (3 bits/Hz), S/N = 127 yields C = 7B (7 bits/Hz)

(1 + S

Noise Power = N

(Shannon showed “can,” not “how”)

2/6/01

Channel Codes

Channel codes are our principal approach to letting R → C with acceptable Pe

e

Solomon Golomb: has gotten to be quite a rarity; to combat the terror of serious error, use bits of appropriate parity.

H2

Definitions:

Lec14.10-8

1. “Channel codes” reduce P

2. “Source codes” reduce redundancy

3. “Cryptographic” codes conceal

A message with content and clarity

Coding Delays Message and Increases Bandwidth

Can show: Pe ≤ 2–Tk(C,R), T = time delay in coding process

e.g. use M = 2RT possible messages in T sec.(RT = #bits in T sec; “block coding”)

use M = 2RT frequencies spaced at ∼1/T Hz

then B = 2RT/T (can → ∞!)

Lec14.10-9 H32/6/01

2/6/01

Minimum S/No for Pe → 0

Therefore

S/No ≥ e → 0 as B → ∞

H4

( ) ( ) bits/secRNS 0ePo →∞ ≥= ∆ = B→∞

Pe{Bit Error}

0.1 0.69

1 E/No →

(J/bit) (W/Hz)

∞ = log2 M1 10-1

10-2

10-3

10-4

10-5

20 = log2 M (M ≅ 106)

M ≅ 2

Typically 2 ≤ M ≤ 64

frequencies

Lec14.10-10

0.69 R for P

2 ln C lim C : show Can

10 M = number of

2/6/01

Error Detection K + R Code

Blocks: message check bits

K bits R bits

Simple parity check – xxx…x P where P ∋ Σ 1’s = even or odd

(2 standards)

(half are illegal here).

K bits R = 1 bit

H5Lec14.10-11

i.e. = A single bit error transforms its block to “illegal” message set

2/6/01

Error Correction Code

1 m2…mK ….m

Any of these K + R bits can be erroneous

Receive: m1 m2……………mCorrect: m1 m2……………m

Sum (modulo 2) = 0’s if no error → 0 0 0

Consider locations of “1”s in K + R slots of Sum

we need K + R ≥ K + LOG2 its/block

Original

Number of Slots for

1-bit error

“No Error” Message

ˆ ˆ ˆ

H6Lec14.10-12

Message = m Checks = mK + 1 K + R

K + R

K + R

If we wish to detect and correct 0 or 1 bit error in the block of K + R bits, (K + R + 1) b

Information

2/6/01

Single-Bit Error Correction

K = R ≥ R/(R +K) 1 2 0.67 2 3 3 3 4 3 5 4 100 7 103 ~10 106 ~20

Not too efficient

R = Check bits needed ≤ 1 error

H7

we need K + R ≥ K + LOG2 its/block

Original

Number of Slots for

1-bit error

“No Error” Message

Lec14.10-13

0.6 0.5 0.4 0.4 0.07 0.01 0.002

to detect and fix in a block of K + R

If we wish to detect and correct 0 or 1 bit error in the block of K + R bits, (K + R + 1) b

Information

2/6/01

Two-Bit Error Correction

) 2

“No Error” Message

K = R ≥ R/(R +K) 5 7 103 ∼20 0.02 106 ∼40 0.004

R = Check bits needed ≤ 2 errors

We need K + R ≥ K + LOG2 1 + K + R +

1 Error

J1Lec14.10-14

(K + R)(K + R - 1

0.6 to detect and fix in a block of K + R bits

If we wish to correct two errors:

2 Errors

2/6/01

Implementation: Single-Error Correction ∆

(Note: C1 ⊕ C1 = m1 ⊕ m2 ⊕ m3 ⊕ C1 ≡ 0)∆

C2 = m1 ⊕ m2 ⊕ m4 ∆ C3 = m1 ⊕ m3 ⊕ m4

∆

J2

Block = m1 m2 m3 m4 C1 C2 C3

(K = 4, R = 3) (4 message bits, 3 check bits)

If no errors (Note: m1 ⊕ m2 ⊕ m3 = C1)

1 1 1 0 1 0 0 m1 0 1 1 0 1 0 1 0 m2 = 0 1 0 1 1 0 0 1 m3 0

m4 C1 C2 C3

∆

= H∆

Modulo-2 matrix

multiply

Q =∆

1 = m1 ⊕ m2 ⊕ m3 ⊕ 0 1 0 0 1 1 1 0

“Sum, modulo-2” Truth Table

∆

Lec14.10-15

HQ = 0 defines legal codewords Q

Let C

2/6/01

Implementation: Single-Error Correction

Can even rearrange transmitted word so: 0 0 0 1 1 1 1 C3 0 1 1 0 0 1 1 C2 1 0 1 0 1 0 1 m4 1 2 3 4 5 6 7 C1

m3 m2 m1

of error location “L”

J3

1, C2, and C3 3

1 1 ⇒ Error in m3 Note that Hi3 = 0

1 1

≡ 0 ≠ 0

Interpret to yield error-free Q from R

Lec14.10-16

HQ = 0 defines legal codewords Q

L =

= E = Binary representation

Only 1/8 of all 7-bit words are legal because C are each correct only half the time and (0.5) = 1/8

Say HR = 0

Suppose transmitted Q is legal and received R = Q + E then HR = HQ + HE

2/6/01

Pe Benefits of Channel Coding

⇒ less E/No

Pe → 6 × 10-4 (per bit; depends on modulation)

4 7

e

Benefits depend on Pe(E/No) relation

7 2p{2 errors in 7 bits @ 6 × 10-4} = p{no error}5 2 ≅ 8 × 10-6

(6 × 10-4)2 !

Compare new p{block error} 8 × 10-6 to 4 × 10-5 without coding

J4

Suppose Pe = 10-5, then P{error in 4-bit word} = 1 – ( -5)4 ≅ 4 × 10-5

P{no errors}(no-coding case)

Lec14.10-17

If we add 3 bits to block (4 + 3 = 7) for single-error correction, and send it in the same time (2.4 dB loss)

Alternatively, reduce power and maintain P

• p{error}

7 • 6/2

Coding reduced block errors by a factor of 5 with same transmitter power

1 – 10

2/6/01

Benefits of Soft Decisions Soft decisions can yield ∼2 dB SNR improvement for same Pe

incorrectly. Choose the one bit for which the decision was least clear.

e.g.

v A B C G H

i = 1

i = 2

“Soft Decision:” Say 8 Levels

J5Lec14.10-18

Example: Parity bit implies one of n bits was received

D E F

“Hard Decision:” 2 Alternatives

2/6/01

Constraint Length

employ overlappingblocks (sliding window)

One advantage: accommodates soft decisions Here each message bit impacts 3 output bits and thereforeimpacts decoder decisions impacting 3 or more reconstructedbits, so soft decisions help identify erroneous bits.

J6

2R bits/sec output

3-bit shift register Sum modulo 2

Example:

InR (bits/sec)

Lec14.10-19

Convolutional Codes

Convolutional codes

This is a “rate 1/2, constraint-length-3 convolutional coder”

2/6/01

Receivers, Antennas, and Signals

Modulation and Coding Professor David H. Staelin

Lec14.10-20

2/6/01

e.g. Fading from deep vigorous multipath

Σ xi cos ωt + yi sin ωt (sum of N phasors, one per path)

N

i = 1

Im{z} z2 (t)(filtered)

Deep Fadet00 Re{z}

i = 1z

K1

z

i and yi are independent g.r.v.z.m. Im{z} P{|z|}

0 0Re{z} |z|eσ2 1

σ

Lec14.10-21

Rayleigh Fading Channels

Consider multipath with output signal z(t) =

Rayleigh fading: x

2/6/01

Variance of Re{z}, Im{z} ≡ σ2

⟨|z|⟩ = σ π/2

⟨|z|2⟩ = σ2(2 – π/2)

|z| – ⟨|z|⟩ 2 ≅ 2σ/3 ≠ f(N)

P{|z| > zo} = e–(zo/σ)2/2

K2

i and yi are independent g.r.v.z.m.

Im{z} P{|z|}

0 0 z

Re{z} |z|eσ2

1

σ

Lec14.10-22

Rayleigh Fading Channels Rayleigh fading: x

2/6/01

Effect of Fading on Pe(Eb/No) Pe curve increases and flattens when there is fading

Pe

Eb/No(dB)

Pe New Pe{Eb/No} Relation After Fading

Eb(t) Fading history, increases Pe(t)

t

Deep fades produce bursts of errors (error clusters)

p{Eb/No}

K3Lec14.10-23

2/6/01

Remedies for Error Bursts

AAA…A AAA…A

(Tolerate adjacent errors better than random ones)

so then block error-correct the symbols A:

≥ 2 independent paths

K4

Error burst, hits only,one bit per block

Burst hits fewer bits per block

t

Lec14.10-24

2. General error-correcting codes

A…

4. Reed-Solomon codes

e.g. multivalue symbols A (say 4 bits each, 16 possibilities)

1. Diversity – Space – Frequency – Polarization

3. Same, plus interleaving:

2/6/01

Remedies for Error Bursts Fading flattens Pe(Eb/No) curve, so potential coding gain can exceed 10 dB sometimes

10 dB 10-8

10-2

Pe Increase in Eb required to accommodate coding

Reduction in Pe using coding

Eb/No (dB)

New Pe for coded fading channel

Note: Coding gain greater for flatter Pe(Eb/No)

Coding Gain

K5

Flatter Pe(Eb/No) for fading channel

Lec14.10-25