transmission planning mod 4

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Section 3 - Module 1 - Page 13FL 42104 AAAA WBZZA Edition 2 - July 2005

All rights reserved 2005, Alcatel

- RADIO NETWORK PLANNING

3.1 Appendix3FL 42104 AAAA WBZZA Edition 2 - July 2005

Appendix


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Network Planning - Appendix

All rights reserved 2005, Alcatel- RADIO NETWORK PLANNING

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Blank Page

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Objectives

To be able to understand the modulation concepts.

To be able in an example to calculate the unavailability objective

due to the equipment failures. To be able to understand the general concepts of the

M.21xx series and the differences between G.821/826and M.21xx recommendations.


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Table of Contents

Switch to notes view! Page

1 Refresh on modulation concepts 7Modulation Concepts 8

BB Transmission 10Bandwidth Formula 11Modulated Signal Spectrum 122-PSK 174-PSK 2016-QAM 2216-TCM 27Performances Versus Noise 30Exercise 31Main Modulation Types Characteristics 32Thermal Noise (C/N versus BER) 33Comparison of Different Mod. Schemes 37Roll-off calculation example 39Blank Page 40

2 Equipment unavailability 41Introduction 43Unavailability objective 44Unavailability of a non-protected section (1+0) 47Unavailability of a protected section (1+1) 50

3 M.21xx-series Recommendations 51End of Module 54


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Table of Contents [cont.]

Switch to notes view!



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1 Refresh on modulation concepts


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Modulation Concepts

Why modulation?

Modulation is necessary to occupy RF narrow bandwidth!

Without modulation (BB transmission) the occupied bandwidth is:

where: f b = bit rate

= roll-off factor

( )12

fBw b +=


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BB Transmission [cont.]

Ideal Transmission Channel

Att. = constant

Rx

Att.

f

f0

0

Tx

-


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BB Transmission

Real Transmission Channel

Att. = Kost.Att.

f0

Tx Att. =

fc

Rx

32fc

t

2fc

1

Att. = Kost.Att.

f0

Att. =

fc t

1T =

2 13

T TT T

2 13

1fb

T =

fb = Bit rate frequency

1=

1fb

2

2fc

2fc

2fc 2=

fbfc


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Bandwidth Formula

= 1.0

= 1.0

= 0.3

= 0.1

0 < < 1

R(f)

-fC

0.1

r(t)

C

-2fC

0.3

+fC +2fC

a

Antisymmetrical Freq. Responce

ac

Roll Off = =

R(f)

Ideal Freq. Responce

-T-2T-3T-4T 0 +T +2T +3T +4T

Bw = Bw = f b

Bw = (1+ )

fb

2

fb

2

-fc +fc


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Modulated Signal Spectrum

V

f

MOD

70 MHz

LO

IF

f0

Bw = 2fc

fc 70+fc

f0

7070-fc

B

2fc


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2-PSK [cont.]

2 PSK Modulator

2 PSK Demodulator

DIFF.

DEC.

100111

Data

L.O.

IF

IF signal

BTF

1 0

B A

DIFF.

ENC.

100111

Data

L.O.

IF

IF signal

PostConversion

Filter

2 PSKMixer

BTF


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2-PSK [cont.]

2-PSK Waveforms - Modulator

DATA IN

1 1 0 1 0 1 1 0 0

CARRIER

IF OUTPUT

+V

-V


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2-PSK [cont.]

2-PSK Waveforms - Demodulator

DATA OUT

1 1 0 1 0 1 1 0 0

CARRIER

IF INPUT

DEMODULATED SIGNAL

-V

+V


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2-PSK [cont.]

Absolute Coding Differential Coding

0 = B 0 = No change in the phase of the carrier

1 = A 1 = 180 change in the phase of the carrier

BA1 0

A A

1

B

0

A

1

B

1

B

0

A

1

Switch

A A B B A B B A

0 1 0 1 1 0 1

B A B B A B B A

1 1 0 1 1 0 1

RX

ON

TX B

0


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2-PSK

BTF Binary Transversal Filter (digital filter)

IN

H(f)

T5

IN

XA10

T5

XA5

T5

XA2

T5

XA5

A10

X

OUT

A

A/10

A/5T/5

A/2T/5

A/5T/5

A/10T/5

fN-fN-2fN

=1

0.4

0

fN(1+ ) 2f N

OUT

H(t)

1W

-1

2W-

12W

+1W

+


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4-PSK [cont.]

4-PSK Modulator1 0

DIFFER.

ENCODER

IF

PostConvertion

Filter

2 PSKMixer

BTF

L.O.

90

L.O.

90

BTF

0010111

2 PSKMixer

SP

L.O.

RFBranching

Filter

Bw=fb (1+ ) Bw= f s (1+ )

fs

0

1

2 PSK f s = fb

4 PSK f s =fb2

22

8 PSK =3

23

16 PSK =4

24

B (10) A (00)

C(11) D (01)

fsfb

fsfb


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4-PSK [cont.]

Differential Coding

B B

00

B

B

D

B

C

C

D

B

D

Switch

D

11 10 01 11 01 01

ON

= No change

01 = -90 changeTX C

A

A (00)

10 01 11 01 0100

RX B B B C BC10 = +90 change

11 = -180 change

001001110101.........

D (01)

B (10)

C (11)

- +


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4-PSK

4-PSK Demodulator

2 PSKMixer

BTF

L.O.

90

L.O.

90

BTF

2 PSK

Mixer

P

S

IF DIFFER.

DECODER

Y1

X1

Y1

X1DecisionCircuit

DecisionCircuit


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16-QAM [cont.]

16-QAM Modulator

11

10

01

00

0100 1110

Vy

Vx

Y1

X1

Y2

X2

Y

X

1 1 +3V

1 0 +1V

0 1 -1V

0 0 -3V

BTF

L.O.

90

L.O.

90

BTF

S

P

IFDIFFER.

ENCODER

X2 X2

2RX1 X1

Y2

Y1

X2

X1

Y2

Y1

FEC

X2

X1

Y2

Y1

2R

Y2

2R

Y2

Y1 Y1

2R

X2

X1


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16-QAM

16-QAM Demodulator

BTF

L.O.

90

L.O.

90

BTF

P

S

IF DIFFER.

DECODER

X2X2

DecisionCircuit

DecisionCircuit

X2

X1X1X1

Y2Y2Y2

Y1Y1Y1


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16-TCM [cont.]

16-TCM Modulator

BTF

L.O.

90

L.O.

90

BTF

S

P

IF

DIFFER.

CONVOL.

X2 X2

2RX1 X1

Y2

Y1

X2

X1

Y2

Y1

MAPPING

X2

X1

Y2

Y1

2R

Y2

2R

Y2

Y1 Y1

2R

X2

X1

+

ENCODER


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16-TCM [cont.]

16-TCM Demodulator

BTF

L.O.

90

L.O.

90

BTF

P

S

IF

DIFFER.DECODER

X2X2

DecisionCircuit

DecisionCircuit

X2

X1X1X1

Y2Y2Y2

Y1Y1Y1

VITERBIDECODER

+


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16-TCM [cont.]

TCM Principles - State Diagram (Example with 8-TCM)

SP

ab

S0S1

c

CONVOLUTIONAL ENCODER

S0 S1

0 0

b c0 / 0

S0 S1

0 1S0 S1

1 1

b c0 / 1

S0 S1

1 0

b c1 / 0

b c1 / 1

b c1 / 0

b c0 / 0

b c0 / 1

b c1 / 1


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16-TCM [cont.]

TCM Principles - Mapping (Example with 8-TCM)

1

0

7

65

4

3

2

a

0 1 2 3 4 5 6

0 0 0 0 1 1 1

b 0 0 1 1 0 0 1

0 1 0 1 0 1 0c

7

1

1

1


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16-TCM

TCM Principles - Trellis Diagram (Example with 8-TCM)

04

0

4

0

4

0b=0

T0 T1 T2 T3

3

7

b=1

b=0

1

5

2

6

5

1

b=1

37

2

6

62

04

1

5

37

0

0 1

1 0

1 1

S0 S1


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Performances Versus Noise [cont.]

2-PSK

C

A

= Carrier

N = Noise B

Threshold

1 1

CN

C+N

Errors depend of the distance between two points.

We have "ERROR" if N > C N > 1


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Performances Versus Noise [cont.]

4-PSK

2 PSK and 4 PSK have the same performance versus noise, but for this reason is never used2 PSK due to its double bandwidth

B A

C D

1

1

Two DifferentThreshold

2

2= 0.7

2

If the Noise (N) is:

you have error

N > 0.7

ModulationType

2 PSK

4 PSK

ErrorCondition

N > 1

N > 0.7

Bandwidth

BW

BW

2(-3dB)

SymbolFreq. (fs)

fb

fb

2

Noise Power (N) = Amplitde x Bandwidth


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Performances Versus Noise

DEMODULATOR

IF data

DETECTOR

ERROR

10-6

S

N= 13.5 dB

10-6

4 PSK

S

N= 18.6 dB

10-6

8PSK

SN = 20.5 dB10-6

16 QAM

SN = 26.5 dB10-6

64 QAM


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Main Modulation Types Characteristics

4 PSK

0

8 PSK

0

16 QAM

2.5

64 QAM

3.7

Modulation type

Position of Vectorial modulationstates (levels) at equal peakpower (Cmax)

Peak-to-Mean power ratio (dB)

R/2 R/3 R/4 R/6Nyquist Bandwidth (Bny)Symbol frequency (S)(R = Binary information capacity)

2 3 4 6Modulation efficiency (bit/sec/Hz)

(Theoretical)

S/N (dB)(Theoretical at BER = 10-6)

13.5 18.6 20.5 26.5


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Thermal Noise (C/N versus BER)

1 1 0 (normalized)2 PSK

v C/N (20log v/)Mod.

1 0.70 +3.1 dB4 PSK

1 0.38 +8.4 dB8 PSK

1 0. 19 +14.2 dB16 PSK

0.7 0.23 +9.7 dB16 QAM

0.6 0.10 +15.6 dB64 QAM

0.6 0.047 +22.1 dB256 QAM

16 QAM

Phase level

decisionthreshold

I

v

Q

v

Q8 PSK

I

= noise voltage

v = carrier peak voltage


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Comparison of Different Mod. Schemes [cont.]

Bit/s

(Hz)6

4

2

10 15 20 25 W (dB)

2 2

4

8

4

8

16

16

BER = 10-6QAM

FSK

64

32

16 QAM 16 PSK

PSK


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10-10

5 W(dB)

10-9

10 15 20 25

10-8

10-7

10-6

10-5

10-4

10-3

10-2

16QAM16PSK

2PSK4PSK

8PSK32PSK

64QAM


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Comparison of different modulation schemes

(Theoretical WandS/N values at 10-6 BER; calculated values may have slightly different assumptions)

a) Basic modulation scheme

(1) As an example, error

correction with redundancy (r)

of 6.7% is used for calculation

in this Table.

System Variants W

(dB)

S/N

(dB)

Nyquist

Bandwidth (bn)

FSK 2-state FSK with discriminator detection 13.4 13.4 B

3-state FSK (duo-binary) 15.9 15.9 B

4-state FSK 20.1 23.1 B/2

PSK 2-state PSK with coherent detection 10.5 10.5 B

4-state PSK with coherent detection 10.5 13.5 B/2



QAM 16-QAM with coherent detection 17.0 20.5 B/4

32-QAM with coherent detection 18.9 23.5 B/5





Basic modulation schemes with FEC

QAM 16-QAM with coherent detection 13.9 17.6 B/4*(1+r)

with 32-QAM with coherent detection 15.6 20.6 B/5*(1+r)

block 64-QAM with coherent detection 19.4 23.8 B/6*(1+r)

codes(1)

128-QAM with coherent detection 21.1 26.7 B/7*(1+r)




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Comparison of Different Mod. Schemes

B) Coded modulation scheme

System Variants W

(dB)

S/N

(dB)

Nyquist

Bandwidth (bn)(1)

BCM(2)

16 BCM - 8D (QAM. One step partition) 15.3 18.5 B/3.75

80 BCM - 8D (QAM. One step partition) 23.5 28.4 B/6




TCM(3)

16 TCM - 2D 12.1 14.3 B/3

32 TCM - 2D 13.9 17.6 B/4

64 TCM - 4D 18.3 21.9 B/5.5

128 TCM - 2D 19.0 23.6 B/6

128 TCM - 4D 20.0 24.9 B/6.5

512 TCM - 2D 23.8 29.8 B/8

512 TCM - 4D 24.8 31.1 B/8.5

MLCM(4)

32-MLCM - 2D (QAM) 14.1 18.3 B/4.5

64-MLCM - 2D (QAM) 18.1 21.7 B/5.5

128-MLCM - 2D (QAM) 19.6 24.5 B/6.5

(1) The bit rate B does not include code redundancy.(2) The block code length is half the number of the BCM signal dimensions.(3) The performances depend upon the implemented decoding algorithm.

In this example, an optimum number is used.(4) In this example, convolutional code is used for lower 2 levels and block codes are used for the third level to

give overall redundancies as those of 4D-TCM. Specially redundancies on the two convolutional codedlevels are 3/2, 8/7 and 24/23 on the block coded third level.


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Roll-off calculation example [cont.]

Example 1Available bandwidth = 40 MHzTransmitted stream = 34 Mbit/s

Modulation type = 2 PSK Roll-off = ?

BW = fb (1+K)40 = 34 (1+ K)a = 40/34-1 = 0.05

RELATIONSHIP BETWEEN fb and fs AS FUNCTION OF THE MODULATION TYPE

2 PSK fs = fb fb = 34 Mbit/s fs = 34 MHz4 PSK fs = fb/2 fb = 34 Mbit/s fs = 17 MHz

8 PSK fs = fb/3 fb = 34 Mbit/s fs = 11.3 MHz16 QAM fs = fb/4 fb = 34 Mbit/s fs = 8.5 MHz


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Roll-off calculation example

Example 2Available bandwidth = 20 MHzTransmitted stream = 140 Mbit/s

Modulation type = ?

BW = fb/nn = fb/BW = 140/20 = 7

27 = 128 128 QAM with K = 028 = 256 256 QAM with K = 1


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Blank Page



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2 Equipment unavailability


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Introduction [cont.]

Unavailability = Part of the time in which the link is out of order.

Where:

MTTR = Mean Time To Repair

MTBF = Mean Time Between Failures

MTBMTTR

MTTR

U+

=

Equipment unavailability


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Introduction

By supposing:

Failures statistically independent

MTTR


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Unavailability objective

EQUIPMENT UNAVAILABILITY OBJECTIVE

for HRDP (L = 2500 km) is supposed to be 1/3 of the total unavailability:

Ueq. < 0.1% = 0.001

The HRDP consists of 9 switching sections (section length = 280 km approx.)

For one-direction of the link only:

Ueq.s1 < 55.10-6

4eq.

eq.s 101.19

UU

=


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Unavailability of a non-protected section (1+0) [cont.]

Suppose that a radio section consists of:

1 Tx Terminal

1 Rx Terminal

5 Repeaters (egual each other)


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Unavailability of a non-protected section (1+0) [cont.]

1+0 radio section: 6 hops, 5 repeater stations

Mod. Tx

PSU

Z'

Rx Dem

PSU

Mod Tx Rx Dem

PSU

Z

L = 50 km L = 50 km


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Unavailability of a non-protected section (1+0)

UTx Term. = UTerm. Mod + UTx + UPSU

URep. = URx + URep. Dem + URep. Mod + UTx + UPSU

URx Term. = URx + UTerm. Dem + UPSU

Unavailability of the non-protected section (uni-directional) (points Z-Z):

US(1+0) = UTx Term + 5 URep. + URx Term


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Unavailability of a protected section (1+1) [cont.]

TS = Tx part of the switching system, the failure of which causes the totalunavailability of the section.

RS = Rx part of the switching system, the failure of which causes the totalunavailability of the section.

Lp = Part of the switching system, the failure of which doesnt allow the regularoperation of the switching system.

MTBFs = Global MTBF of the switching system series part.

MTBFp = Global MTBF of the switching system parallel part.

US

US

R'

TS

R

RS

Lp


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Unavailability of a protected section (1+1) [cont.]

1+1 radio section: 6 hops, 5 repeater stations

Mod. Tx

PSU

Z'

Rx Dem

PSU

Mod Tx Rx Dem

PSU

Z

L = 50 km L = 50 km

Mod. Tx

PSU

Z'

Rx Dem

PSU

Mod Tx Rx Dem

PSU

Z

R' R

LOGIC


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Unavailability of a protected section (1+1)

Global unavailability of the 1+1 protected section:

The section is unavailable due to:

failures of the 2 channels

failure of the series part of the switching system

failure of a channel and of the parallel part of the switchingsystem

( ) ( ) ( ) ( )0.5$UU$UUU 01sparser2 01s11s ++= +++


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3 M.21xx-series Recommendations


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General concepts [cont.]

Differences between Recommendations G.821/G.826 and the M.21xx series start with theirdifferent origins:

G-series Recommendations are from ITU-T Study Group 13 (General networkissues);

M-series are from Study Group 4 (Network Maintenance and TMN).

Main differences:

G.821/G.826 define long-term performance objectives to be met.

G.821/G.826 requirevery long test intervals (one month).

The M-series Recommendations are particularly useful when bringing-into-servicenew transmission equipment. They are intended to assure that the requirements ofthe G series are met in every case.

As a general rule, the requirements of the M-series are tougher than those of theG-series.

For practical reasons, the M.21xx-series Recommendations allow short testintervals.


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General concepts [cont.]

Media independent (ITU-T)

M.2100 for PDH paths sections and transmission systems

M.2110 how to apply M.2100 and M.2101 for BIS (Bring-Into-Service)

M.2120 how to apply M.2100 and M.2101 for maintenance

M.2101 for SDH paths and multiplex section

Radio specific (ITU-R)

F.1330 for parts of international PDH and SDH paths and sections.


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