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Different “Flavors” of OFDM. There are different “flavors” of OFDM according what we put in the Prefix:. Prefix. Prefix. Prefix. data. P. data. P. data. P. time. Three main choices: CP-OFDM with Cyclic Prefix (CP) ZP-OFDM with Zero Prefix (ZP) - PowerPoint PPT Presentation
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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: