32
26/11/2007 1 CDMA: An Introduction Βαγγέλης Δούρος EY0619 [email protected]

Cdma introduction pr

Embed Size (px)

Citation preview

Page 2: Cdma introduction pr

2

Motivation

Why do we need CDMA?Which are its basic principles?What do we earn by using it?

Page 3: Cdma introduction pr

3

Outline

Introduction CDMA Basics CDMA Deeper CDMA ProblemsCDMA BenefitsConclusions

Page 4: Cdma introduction pr

4

Why Multiple Access?

Goal: Multiple use of a shared medium.Multiplex channels in three dimensions:

Time (t)–

Frequency (f)–

Code (c)

Page 5: Cdma introduction pr

5

Frequency Division Multiple Access (FDMA)

Cocktail Party Analogy: People create teams and discuss. There is a distance among them. (-) Requires guard band between channels(-) Waste of bandwidth if traffic is distributed unevenlyExample: broadcast radio

Page 6: Cdma introduction pr

6

Time Division Multiple Access (TDMA)

Cocktail Party Analogy: People have access to the same room but each of them waits for his turn to speak.(-) Precise synchronization necessary

Page 7: Cdma introduction pr

7

Time and Frequency Division Multiple Access

(-) Precise coordination requiredExample: GSM

Page 8: Cdma introduction pr

8

Code Division Multiple Access (CDMA)

Cocktail Party Analogy: All people are in the same room together. They can all be talking the same time!Example: UMTS

Page 9: Cdma introduction pr

9

Spread Spectrum

PN: Pseudo-Noise code sequence spread/ despread the signal. Modulation:

FSK Frequency hopped (FH) multiple access)–

PSK Direct sequence (DS) multiple access)

D mD

Page 10: Cdma introduction pr

10

FHMA

Bandwidth divides in non overlapping bands.Signal shifts from band to band in predefined intervals.Receiver is synchronized with the transmitter(+) less interferenceUse: Bluetooth

Page 11: Cdma introduction pr

11

CDMA Basics (1)

Each station is assigned a unique m-bit code (chip sequence) To send bit 1, station sends chip sequence.To send bit 0, station sends the complement.Example: 1 MHz band with 100 stations. FDMA

Each station a 10kHz band–

Rate:10 kbps (Assume that you can send 1bit/Hz)CDMA

Each station uses the whole 1MHz band 106 cps.–

If <100 chips per bit Rate >10 kbps

Page 12: Cdma introduction pr

12

CDMA Basics (2)

Let or

Compare any pair of these sequences-vectors…Multiply any pair of these sequences-vectors…Two chips S,T are orthogonal IFF

0 0S T S T−

• = ⇔ • =

0S T• =

1

1 0m

i ii

S T S Tm =

• = =∑2 2

1 1 1

1 1 1 ( 1) 1m m m

i i ii i i

S S S S Sm m m= = =

• = = = ± =∑ ∑ ∑_ _

1 1

1 1 1( 1) ( ) 1m m

iii i

S S S S mm m m= =

• = = − = − = −∑ ∑

Page 13: Cdma introduction pr

13

CDMA Basics (3)

Let A, B, C, D transmit correspondingly bit 1,0,1,_.

Assume that:–

All stations are perfectly synchronous.–

All codes are pair wise orthogonal (aren’t they?).–

If two or more stations transmit simultaneously, the bipolar signals add up linearly.

Receiver “understands” How can the receiver “understand” what bit station C send?

_

( 1 1 3 3 1 1 1 1)S A B C= + + = − + − + + − − +

Page 14: Cdma introduction pr

14

CDMA Basics (4)

Let’s compute the normalized inner product S•C

Right! By accident??

Remember: All codes are pair wise orthogonal!

1 ( 1 +1 -3 +3 +1 -1 -1 +1) (-1 +1 -1 +1 +1 +1 -1 -1)=8

1 1= (1 1 3 3 1 1 1 1) 8 18 8

S C• = − ⋅

+ + + + − + − = =

( ) 0 0 1 1S C A B C C A C B C C C− −

• = + + • = • + • + • = + + =

Page 15: Cdma introduction pr

15

CDMA Basics (5)

Reverse way…Think that each chip sequence arrives separatelyReceiver separately computes each inner product

It keeps only the non-orthogonal pair, i.e. the right bit

0 (1)

0 (2)1 (3)

0 0 1 1

A C

B CC CS

• =

• =• == + + =

Page 16: Cdma introduction pr

16

CDMA Deeper (1)

More advanced analysis…Sender:

Receiver:

( ) cos(2 ( ))( ) {0, }( ) { 1, 1}( ) cos(2 ( )) ( )

d c

c

s t A f t tt

c ts t A f t t c t

π θθ π

π θ

= +

∈ +∈ − += +

( ) ( ) cos(2 ( )) ( ) ( )cos(2 ( )) ( )

c

c d

s t c t A f t t c t c tA f t t s t

π θπ θ= + =

= + =

Page 17: Cdma introduction pr

17

CDMA Deeper (2)

With orthogonal codes, we can safely decode the coding signals.Noise?R’=R+N, N: m-digit noise vector and N=(±a ±a…±a)Decode …No problem if chipping codes are balanced (same ±)

R' S=(R+N) S=S S +(orthogonal codes) S +N S= 0 ?1+ +• • • • •

R' S=(R+N) S=S S +(orthogonal codes) S +N S==1+0+(a a a a)(+1+1-1-1)=1+0+0=1• • • • •

Page 18: Cdma introduction pr

18

CDMA Deeper (3)

How many codes can we construct with m chips?m (why?)If m=2k, Walsh-Hadamard codes can be constructed recursively!The set of codes of length 1 is For each code we have two codesCode Tree:

0 { }C = < + >

kc C< >∈_

k+1{ > <c c } Cc c< > ∈

0

1

2

{ }{ , }{ , - - , - - , - - }

CCC

= < + >

= < + + > < + − >= < + + + + > < + + > < + + > < + + >

Page 19: Cdma introduction pr

19

CDMA Deeper (4)

Correlation: Determines similarity between two sets of data. –

Possible values1 sequences are similar0 no relationship between them. -1 one is the mirror of the other

Cross correlation: Compare two sequences from different sourcesAuto correlation: Compare a sequence with itself after a time-intervalWalsh Codes: No cross correlation – Low auto correlationPN sequences: Low cross correlation – Low auto Correlation

Page 20: Cdma introduction pr

20

CDMA Deeper (5)

We cannot have more than m orthogonal codes.Let m + k stations and m chips… Idea: Use PN Sequences.

?: the sum of the k random variables that are either 1 or -1.But PN Sequences = low cross correlation. ? should be 0. Experimental evaluation: For k=m=128, decoding is correct more than 80%.

( random codes) S+(m-1 orhogonal codes?

) =1+ +0R S S S k S• = • + • •

Page 21: Cdma introduction pr

21

CDMA Problems (1)

All stations are received with the same power level…

In reality… users may be received with very different powers!Near-far ProblemSolutions:

Empirical rule: Each MS transmits with the reverse power that it receives from the BS

Power Control!Open LoopFast Closed Loop

Page 22: Cdma introduction pr

22

CDMA Problems (2)

Bad Properties of Walsh CodesPerfect Synchronization of all users required.Impossible……is nothing! Use a long enough known chip sequence.

But…In a multipath channel, delayed copies may be received, which are not orthogonal any longer. Self-Interference.

Page 23: Cdma introduction pr

23

CDMA Problems (3)

So far…(-)tight synchronization(-)self-interference (-)Near-far problem(-)Higher complexity of sender/ receiver…How did Qualcomm convince people to use this stuff??

Page 24: Cdma introduction pr

24

CDMA Benefits (1)

Unlike FDMA and TDMA, CDMA does not rely on orthogonal frequency and time slots!

Page 25: Cdma introduction pr

25

CDMA Benefits (2)

In TDMA and FDMA systems–

Nothing to send time/frequency slot is wasted–

Dynamic allocation is very difficult

In CDMA systems–

Nothing to send less interference–

Transmit ~half times doubles the capacity

Page 26: Cdma introduction pr

26

CDMA Benefits (3)

FDMA-TDMA use sectors to decrease the reuse distanceCDMA use sectors to increase capacity (triple it)!

Page 27: Cdma introduction pr

27

CDMA Benefits (4)

Why handoff?Types

“Hard”–

“Soft”

Page 28: Cdma introduction pr

28

CDMA Benefits (5)

Break-Before-MakeEach MS communicates with only one BS each time(+)Reduced dropped calls

Hard Handoff

Page 29: Cdma introduction pr

29

CDMA Benefits (6)

Each MS communicates with more than one BS each timeUse Signal Strength to decide where to connect.Make-Before-Break(+++) no dropped calls

Soft Handoff

Page 30: Cdma introduction pr

30

CDMA Benefits (7)

Capacity…TDMA-FDMA: bandwidth limitedCDMA: interference limitedCDMA’s capacity is bigger.How?Long Story…

Page 31: Cdma introduction pr

31

Conclusions

Back to the start…Why do we need CDMA?

Introduction

Which are its basic principles?–

CDMA Basics –

CDMA Deeper

What do we earn by using it?–

CDMA Problems–

CDMA Benefits

Page 32: Cdma introduction pr

32

Ευχαριστώ!