Different “Flavors” of OFDM

Different “Flavors” of OFDM

There are different “flavors” of OFDM according what we put in the Prefix:

data P data P data P time

Prefix

Three main choices:

• CP-OFDM with Cyclic Prefix (CP)• ZP-OFDM with Zero Prefix (ZP)• TDS-OFDM (Time Domain Synchronous) with Pseudo-Random Prefix

Prefix Prefix

CP-OFDM with Cyclic Prefix

• The most used: IEEE802.11, 802.16, Digital Video Broadcasting in Europe and many others• Advantages: Simple to implement CP good for synchronization (since it repeats)

• Disadvantages:CP discarded (waste of transmitted power) possible nulls at subcarriers in fading channels

data CP

Reason for Null Carrier in CP

Let’s follow one subcarrier:

Steady stateCP

[ ]h n

Transient

2jk nN

kX e 22 jk n

NkH k X e

N

• With CP, at the receiver we discard the transient and just look at steady state;• if the frequency response at the subcarriers frequency is zero (deep fading), then we completely loose that data of that subcarrier.

channel

ZP-OFDM with Zero Prefix

• Used in some standards (“WiMedia UWB” Personal Area Network for multimedia, short range, file transfer)• Advantages: in principle, there is never a null, if properly implemented no power loss in ZPsuitable for Blind Equalization (see later)

• Disadvantages:“proper implementation” cannot use FFT and is very inefficient keeps turning on and off: not good for components.

data ZP

Reference: B. Muquet, Z. Wang, G.B. Giannakis, M. deCourville, P. Duhamel,” Cyclic Prefix or Zero Padding for Wireless Multicarrier Transmission?”, IEEE Transactions on Communications, Vol 50, no 12, December 2002

Reason for Never a Null Carrier in ZP

Let’s follow one subcarrier corresponding to deep fading:

Steady state

ZP[ ]h n

Transients

2jk nN

kX e

• No Inter Block Interference (IBI) due to the ZP• With ZP, you do not discard anything;• if the frequency response at the subcarriers frequency is zero (deep fading), then we still have a transient response, no matter what (most likely it will have low energy, but never zero)

channel

Time Domain Synchronous TDS-OFDM with Pseudo-random Prefix (PP)

• In Chinese Digital TV standard (DTMB)• Advantages: Excellent Synchronization Excellent channel estimation

• Disadvantages:Slightly higher complexity (but worth it)Applicable to long OFDM frames (such as Digital Broadcasting)

data PP

Reference: M. Liu, M. Crussiere, J.F. EHeard, “A Novel Data Aided Channel Estimation wit Reduced Complexity for TDS OFDM Systems,” to appear.

OFDM-ZP and Channel Equalization

Channel Equalization in general (not OFDM yet).1. Trained:

[ ]s n

[ ]w n

[ ]y n[̂ ]s n[ ]h n [ ]g n

Channel

Equalizer

time

[ ]s n

dataTraining data

Training data

estimator

ReceiverIt is based on training data, known at the receiver.

2. Blind Equalization (general):

No training data (something like “no hands!”)

[ ]s n

[ ]w n

[ ]y n

[̂ ]s n[ ]h n [ ]g n

Channel

Equalizer

estimator

Receiver

How do we do Blind Equalization in general?We need to exploit features of the signal. Mainly two approaches:

• Constant Modulus (for BPSK and QPSK signals):

[ ]s n [ ]y n[̂ ]s n[ ]h n [ ]g n

Channel

Equalizer

estimator

[ ] 1 for all s n nIf QPSK or BPSK:

Determine which minimizes[ ]g n22[̂ ] 1

n

s n

Problem: non quadratic minimization and likely it converges to local minima

[ ]w n

Better Approach to general Blind Equalization:• Subspace method: the received signal is in a subspace determined by the channel.;• One approach: Fractionally Spaced Equalizers:

[ ]y n

[ ]w n

[ ]d m2 ( )H z

[ ]d m

sF

( )d t Transmitter,Channel,Receiver

symbol rate 2 sF

[ ]y nM-QAM

DAC

Sample at twice the symbol rate

Same as:

At the receiver, separate the two data streams (even and odd samples):

[ ]d m

sF

Transmitter,Channel,Receiver

2 sF

[ ]y n

M-QAM

DAC

2

2

0[ ]y m

1[ ]y m1z

See a discrete time model

[ ]y n

2 1 20 1( )H z H z z H z

Take the Polyphase decomposition of the channel and ignore the noise (for simplicity):

[ ]d m2

20 ( )H z 2

2

0[ ]y m

1[ ]y m21( )H z 1z 1z

Apply Noble Identitites

20 ( )H z 22 0 ( )H z=

2 20 ( )z H z 22 =

10 ( )z H z

1 20 ( )z H z 22 =

“zero”0 ( )H z22 1z = =

“zero”

[ ]d m 0 ( )H z 0[ ]y m

1[ ]y m11( )z H z

[ ]y n[ ]d m2

2

2

0[ ]y m

1[ ]y m1z

( )H z

DAC+Transmitter+Channel+Receiver+ADC

They are the same!!!

Apply z-Transforms: 0 0

1 1

( ) ( ) ( )( ) ( ) ( )

Y z H z D zzY z H z D z

Multiply both: 1 0 1 0

0 1 0 1

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

H z Y z H z H z D zH z zY z H z H z D z

Right Hand Sides are the same. Then :

1 0 0 1( ) ( ) ( ) ( ) 0H z Y z H z zY z

Back in time domain:

1 0 0 1[ ] [ ] [ ] [ 1 ] 0h y m h y m

This relates the channel parameters to the received data without knowledge of the transmitted message.

Example. Take a first order case:

1( ) [0] [1]H z h h z Polyphase decomposition:

0 1( ) [0], ( ) [1]H z h H z h

Then:

1[1] [2 ] [0] [2 1] 0h y m h y m

In vector form:

[1][2 ] [2 1] 0

[0]h

y m y mh

Compute Channel parameters from received signal:

* **

[0] [1][2 ][2 ] [2 1]

[1] [0][2 1]yy yy

yyyy yy

r ry mR E y m y m

r ry m

Then the channel impulse response is proportional to the eigenvector corresponding to the smallest eigenvalue (zero if no noise) of yyR

[2 ]y m

[2 1]y m

This means that the received signal ‘’lives” in a subspace.The channel parameters “live” in the orthogonal subspace.

[1][0]hh

[2 ][2 1]y my m

noise

Received signal

Channel parameters

Mod and Demod with ZP OFDM

[ ]h n[ , ]x n i [ , ]y n iN L

i-1

N L N L

Take one OFDM Symbol (with index i ) :

[ , ] [ ]* [ , ] [ , ], 0,..., 1y n i h n x n i w n i n N L

[ , ], 0,..., 1x n i n N Transmittedsignal

Channel

Received data

[ ], 0,..., 1h n n L

i i+1 i-1 i i+1

Define the 2N points FFT, by zero padding

2 [ ] [ ] , 0,..., 2 1

[ ] [ ] , 0,..., 2 1

[ ] [ ] , 0,..., 2 1

NX k FFT x n k N

H k FFT h n k N

Y k FFT y n k N

Due to the zero padding, convolution and circular convolution are the same:

2[ ] [ ] [ ] [ ], 0,...2 1NY k H k X k W k k N

Recall the transmitted data (drop the block index “i” for convenience:

[ ] [ ] , 0,..., 1X k FFT x n k N

Fact (easy to show):

2 [2 ] [ ], 0,..., 1NX m X m m N

Demodulation:*

2 2

[2 ] [2 ] [2 ]ˆ [ ] , 0,..., 1| [2 ] | [2 ]W

H m Y m Y mX m m NH m H m

N-IFFT +ZP P/S

[ ]x n

[ ]X m TX

2N-FFT S/P[2 ]Y m RXChoose

even indices

ZP OFDM: one approach to Mod. and Demod.

Blind Equalization with ZP OFDM

2 1

220

10 [ ] [ ] [ ]2

N

NNDFT x n v n V k X

N

data

N N

See the zero padded [ ]x n

Define:0 if 0 1

[ ]1 if 2 1

n Nv n

N n N

1

[ ]x n

[ ]v n

Then: for all [ ] [ ] 0x n v n 0,..., 2 1n N

Recall that DFT of the product is the circular convolution of the DFT’s:

n

n

22 12 /

( 1) 1 if 1,..., 2 1[ ] [ ] 1

if 0

kN jk n

N jk N

n N

k NV k DFT v n e e

N k

where:

Notice that for k even, non zero.Then:

[ ] 0V k

2 1

220

1

2 220

0 2 1 [ ]

[0] [2 1] 2 1 2 [2 ] 0

N

NN

N

N NN

V m X

V X m V m X

2 1m even odd

This relates even and odd frequency components:

1

2 220

1[2 1] 2( ) 1 [2 ]N

N NNX m V m X

N

0,..., 1m N

Since (neglect the noise and put back block index “i”):

2[ , ] [ ] [ , ], 0,...2 1NY k i H k X k i k N

This implies that, for each data block i for m=0,…,N-1

1

20

[2 1, ] 1 [2 , ]2( ) 1[2 1] [2 ]

N

N

Y m i Y iV mH m N H m

In matrix form, for the i-th received data block :

1 1[ ] [ ]o o e eH y i VH y i

In matrix form, for the i-th received data block :

1 1[ ] [ ]o o e eH y i VH y i Where we define:

1 , [3],..., [2 1],..., [2 1] , oH diag H H H m H N N N 0 , [2],..., [2 ],..., [2 2] , eH diag H H H m H N N N

[ ] [1, ] [2 1, ] [2 1, ] , 1Toy i Y i Y m i Y N i N

[ ] [0, ] [2 , ] [2 2, ] , 1Tey i Y i Y m i Y N i N

2

1[ , ] 2( ) 1 , , 0,..., 1N

V m V m m NN

a) the NxN diagonal matrices of even and odd 2N DFT components of the channel:

b) The Nx1 vectors of even and odd 2N DFT components of each received block:

c) The NxN matrix of this term defined earlier:

1 1[ ] [ ], 1,...,o o e eH y i VH y i i M N

This expression relates the received data blocks with the channel frequency response.Now see how to actually compute the channel frequency response.First collect a M received data blocks:

“Pack” all the se vectors in a matrix:

1 1[1] [ ] [1] [ ]o o o e e eH y y M VH y y M

1 1o o e eH Y VH Y

N M N M

1 1o o e eH Y VH Y

1 * 1 *T To o e e e eH Y Y VH Y Y

11 * * 1T To o e e e eH Y Y Y Y VH

1* *T To e e e e oY Y Y Y H H V

Multiply both sides on the right by :*TeY

Multiply both sides on the right by : 1*Te eY Y

Start with:

and you get:

This relates the channel freq. response H with the received signal Y.

Summarize it so far:1. Take M>N ofdm received frames :

[ , ]y n i

0,..., 1n P

1i i M

2. For each frame, take the 2N point FFT by zero padding:

[ , ]y n i

[ , ]Y k i

[ , ]Y k i

0,..., 2 1k N

1i i M

3. Separate even and odd subcarrier indices and “pack” them in two NxM matrices:

[1]ey

[1]oy

[ ]ey i

[ ]oy i

1i i M

[[

] eY] oY

Now we want to compute the channel from the expression

1* *T To e e e e oY Y Y Y H H V

1* *T T

o e e eR Y Y Y Y

Define:

Since are diagonal matrices, here is how this expression looks like:

,e oH H

[0] 0 0[ , ] 0 [2 ] 0

0 0 [2 2]

[1] 0 0 = 0 [2 1] 0 [ , ]

0 0 [2 1]

HR m H

H N

HH m V m

H N

Equate the m-th row on both sides (any one):

[ , ] [2 ] [2 1] [ , ]R m H H m V m , 0,..., 1m N

[ ,0] [ , ] [ , 1][0] [2 ] [2 2] [2 1][ ,0] [ , ] [ , 1]V m V m V m NH H H N H mR m R m R m N

Just a scaling constant!

Demodulation:For the i-th block. Take any arbitrary

2 2[2 , ] [2 , ] [ , ][ , ][2 ] [2 1] [ , ]

N NY i Y i R mX iH H m V m

Given just one known symbol you determine .

[2 1]H m

Time Domain Synchronous TDS-OFDM with Pseudo-random Prefix (PP)

• The PP facilitates synchronization and channel estimation

DFT Data Block PP

Pseudo Noise

Pre- amble

Post- amble

• The PP has its own Cyclic Prefix, both at the beginning (Pre-amble) and the end (Post-amble);• The Pseudo Noise (PN) changes for every frame.

Application in Chinese Digital Terrestrial Television Broadcasting (DTTB). In this standard the PN is an m-sequence of length N=255 BPSK symbols.

DFT Data Block PP 3780

255

420

83 82

Post- amble: repeat first 82 PN samples

Pre- amble: repeat last 83 PN samples

A B CC A

LL n

[ ]p n

M

In general (make the pre- and post- amble the same lengths for simplicity):

A B CC A *L

= [ ]y nL n

0

[ ]h n

Guard Interval Channel

[ ]p n

M

A B C = [ ]y nn

0

[ ]h n[ ]p n

M

M

Fact:[ ] [ ], for 0,..., 1[ ] [ ], for ,..., 1y n y n n My n y n M n M M L

Due to the repetitions, linear convolutions and circular convolutions of the Guard Interval are the same:

L

A B CC A * = [ ]r nL n

0 0

*[( ) ]Mp n[ ]y n

M

A B C = [ ]r nn

0 0M

M

Fact: [ ] [ ], for ,..., 1r n r n n M L M

Now see the guard interval at the receiver and correlate with shifted PN:

[( ) ]* [ ]Mp n L h n

A

[( ) ] [ ]Mp n L h n

DATA

C

L

B

Define:*[( ) ]Mp n

L2M L

Then:*[ ] [( ) ] [ ] [( ) ]M Mr n p n L h n p n

But:1

* *

0

1*

0

1* 2

0

[( ) ] [( 2 ) ] [( ) [( ) ]

change to obtain [( ) ] [( ) ]

[( ) [( ) ] [ ( )]

M

M M M M

M

M M

M

M M p

p n L p L n p L p n

L p p n L

p p L n M n M L

Therefore: 2[ ] [ ( )]pr n M h n M L

2[ ] [ ( )], for 1pr n M h n M L M L n M and:

*[( ) ]Mp n

[ ]y n

Received data

[ ]r n

DFT of DATA DFT of DATA GI GI

0n

[ ]y n

20[ ] [ ], for 0 1pr n M L n M h n n L

Algorithm for Channel Estimation in TDS-OFDM: