Chapter 5 Simulations and Comparisons
5.2 Channel Estimation for 802.11n-like Systems
5.2.2 Channel Estimation for All-pilot Preamble MIMO OFDM
MIMO OFDM
When we further investigate channel estimation in all-pilot preamble MIMO OFDM system, the parameter must be chosen carefully. As mentioned in
Chapter 4, equals to the number of most significant taps in time domain. To illustrate this property, we consider another channel model other than Model 1 and 2.
This channel has a larger delay spread that can demonstrate the influence of . The K0
K0
K0
delay taps of this channel is at 0, 4, 5, and 11 samples. The power attenuation in dB is 0, 4, 5, and 8, respectively.
0 10 20 30 40 50 60 70
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Delayed Tap Number
Magnitude
Real impulse response of the channel Estimated channel impulse response
(a)
0 10 20 30 40 50 60 70
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Delayed Tap Number
Magnitude
Real impulse response of the channel Estimated channel impulse response
(b)
0 10 20 30 40 50 60 70 0
0.2 0.4 0.6 0.8 1 1.2 1.4
Delayed Tap Number
Magnitude
Real impulse response of the channel Estimated channel impulse response
(c)
Figure 5.4 Estimated channel responses for three different (a) =15 (b)
=7 (c) =4
K0 K0
K0 K0
From the observation of previous simulations, is a critical factor of performance. To verify the influence of , a static two-ray channel is used for
simulations. In order to evaluate performances of various , the multipath power of each tap is set to a large value to emphasize their difference. The delay taps of this channel is at 0 and 14 samples. The power attenuation in dB is 0 and 3. In Figure 5.5, MSE versus plot is illustrated under noiseless condition with the static two-ray channel. One can see that MSE has the lowest value when equals to 15. The
K0
K0
K0
K0
K0
MSE is large when is smaller than 15 because a significant tap is lost. The MSE increases with when is greater than 15.
K0
K0 K0
0 10 20 30 40 50 60 70
-300 -250 -200 -150 -100 -50 0 50
K0
MSE in dB
Figure 5.5 Averaged MSE versus of an all-pilot preamble system, NT=2, assumed a static two-ray model
K0
0 5 10 15 20 25 30 10-5
10-4 10-3 10-2 10-1 100
SNR in dB
BER in logarithm scale
K0=15, Tx=2 K0=12, Tx=2 K0=10, Tx=2 K0=18, Tx=2 K0=20, Tx=2 K0=22, Tx=2 K0=24, Tx=2
Figure 5.6 BER comparison due to various values, all-pilot-preamble, with NT=2, NR =1, assumed a static two-ray model
K0
The BER comparison is shown in Figure 5.6. The BER versus SNR curves show the same trend as the MSE curve. The curve =15 outperforms all other curves. From this result, one can conclude that the best choice of is length of the channel.
K0
K0
In the following simulations, Rayleigh fading channels are assumed. The power delay profiles are set to Model 1 and Model 2. In a single packet, there are 10 OFDM symbols. From Figure 5.4 to 5.6, we can obtain two conclusions. First, should be
large enough to include all the delay taps in time domain. In 5.4 (b) and 5.4 (c), is smaller than the maximum delay spread (8 in this case). Therefore, there are more noticeable mismatches between estimated taps and the actual taps, than in the case of
K0
K0
enough . Another conclusion is that should not be too large. A more than
enough will include noise and CCI on those null taps. For example, all four major taps are well caught by the extra channel estimator as shown in Figure 5.4 (a), but additional interference can also be observed on those 15 taps ( =15). The mentioned two conditions cause the performance degradation in the channel estimation.
K0 K0
K0
K0
0 5 10 15 20 25 30
10-6 10-5 10-4 10-3 10-2 10-1 100
SNR in dB
BER in logarithm scale
K0=64, Tx=2 K0=48, Tx=2 K0=36, Tx=2 K0=24, Tx=2 K0=12, Tx=2 K0=8, Tx=2 K0=4, Tx=2
(a)
0 5 10 15 20 25 30 10-7
10-6 10-5 10-4 10-3 10-2 10-1 100
SNR in dB
BER in logarithm scale
K0=64, Tx=2 K0=48, Tx=2 K0=36, Tx=2 K0=24, Tx=2 K0=12, Tx=2 K0=8, Tx=2 K0=4, Tx=2
(b)
Figure 5.7 BER comparison due to various values, all-pilot-preamble, with NT=2, NR =1(a) Model 1 (b) Model 2
K0
0 5 10 15 20 25 30
10-6 10-5 10-4 10-3 10-2 10-1 100
SNR in dB
BER in logarithm scale
K0=64, Tx=3 K0=48, Tx=3 K0=36, Tx=3 K0=24, Tx=3 K0=12, Tx=3 K0=8, Tx=3 K0=4, Tx=3
(a)
0 5 10 15 20 25 30 10-6
10-5 10-4 10-3 10-2 10-1 100
SNR in dB
BER in logarithm scale
K0=64, Tx=3 K0=48, Tx=3 K0=36, Tx=3 K0=24, Tx=3 K0=12, Tx=3 K0=8, Tx=3 K0=4, Tx=3
(b)
Figure 5.8 BER comparison due to various values, all-pilot-preamble, with NT=3, NR =1 (a) Model 1 (b) Model 2
K0
0 5 10 15 20 25 30
10-5 10-4 10-3 10-2 10-1 100
SNR in dB
BER in logarithm scale
K0=64, Tx=4 K0=48, Tx=4 K0=36, Tx=4 K0=24, Tx=4 K0=12, Tx=4 K0=8, Tx=4 K0=4, Tx=4
(a)
0 5 10 15 20 25 30 10-4
10-3 10-2 10-1 100
SNR in dB
BER in logarithm scale
K0=64, Tx=4 K0=48, Tx=4 K0=36, Tx=4 K0=24, Tx=4 K0=12, Tx=4 K0=8, Tx=4 K0=4, Tx=4
(b)
Figure 5.9 BER comparison due to various values, all-pilot-preamble, with NT=4, NR =1 (a) Model 1 (b) Model 2
K0
In Figure 5.7 (a) and (b), BER versus SNR curves with different are shown for NT=2. According to previous statements, inadequate ’s have worse
performances than that of exact . However, the lengths of Model 1 and Model 2 indoor wireless channels are not very long, and the multipath powers of last few paths are quite small (smaller than -13 dB). The small taps are not very significant compared with additive noise. For this reason, the performance degradation with small is not obviously in our simulations. When we discuss the simulation results, it is noted that the maximum delay spread of is 3 in Model 1, and 11 in Model 2. In Figure 5.7 (a), =4 provides best performance among several different . In
K0
K0
K0
K0
K0 K0
(b), =8 outperforms other ’s. As a result, decision of is crucial to the
performance of the channel estimation. Set to the maximum delay spread is a good choice. In Figure 5.8 and 5.9, one can see that the influence of CCI increases when the number of transmission antennas is larger. There are significant error floors in high SNR condition while the number is 4. In Figure 5.10 to 5.12, the cases of multiple receiver antennas are simulated. One can see that the receiver diversity provides better performance while multiple receiver antennas are used. For example, the error floor in Figure 5.9(a) is improved in Figure 5.12.
K0 K0 K0
K0
0 5 10 15 20 25 30
10-6 10-5 10-4 10-3 10-2 10-1 100
SNR in dB
BER in logarithm scale
K0=36, Tx=2, Rx=2 K0=24, Tx=2, Rx=2 K0=12, Tx=2, Rx=2 K0=8, Tx=2, Rx=2
Figure 5.10 BER comparison due to various values, all-pilot-preamble, NT=2, NR =2 (Model 1)
K0
0 2 4 6 8 10 12 14 10-6
10-5 10-4 10-3 10-2 10-1
SNR in dB
BER in logarithm scale
K0=36, Tx=3 K0=24, Tx=3 K0=12, Tx=3 K0=8, Tx=3
Figure 5.11 BER comparison due to various values, all-pilot-preamble, NT=3, NR =3 (Model 1)
K0
0 2 4 6 8 10 12
10-6 10-5 10-4 10-3 10-2 10-1
SNR in dB
BER in logarithm scale
K0=36, Tx=4 K0=24, Tx=4 K0=12, Tx=4 K0=8, Tx=4
Figure 5.12 BER comparison due to various values, all-pilot-preamble, NT=4, NR =4 (Model 1)
K0
0 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 -60
-50 -40 -30 -20 -10 0 10 20 30 40
K0
MSE in dB
MSE of Estimated Channels (Model 1), Tx=2
(a)
0 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64
-40 -30 -20 -10 0 10 20 30 40
K0
MSE in dB
MSE of Estimated Channels (Model 2), Tx=2
(b)
Figure 5.13 Averaged MSE versus of all-pilot preamble system, NT=2 (a) Model 1 (b) Model 2
K0
0 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 -40
-20 0 20 40 60 80 100 120
K0
MSE in dB
MSE of Estimated Channels (Model 1), Tx=3
(a)
0 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64
-20 0 20 40 60 80 100 120
K0
MSE in dB
MSE of Estimated Channels (Model 2), Tx=3
(b)
Figure 5.14 Averaged MSE versus of all-pilot preamble system, NT=3 (a) Model 1 (b) Model 2
K0
0 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 -40
-20 0 20 40 60 80 100 120
K0
MSE in dB
MSE of Estimated Channels (Model 1), Tx=4
(a)
0 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64
-20 0 20 40 60 80 100
K0
MSE in dB
MSE of Estimated Channels (Model 2), Tx=4
(b)
Figure 5.15 Averaged MSE versus of all-pilot preamble system, with NT=4 (a) Model 1 (b) Model 2
K0
The MSE versus plot is also shown in Figure 5.13 to Figure 5.15. In the figures, the channels are estimated with SNR set to 40dB. MSE in this plot is the mean of both estimated channels from two antennas. In this case, co-channel interference dominates the MSE performance. One can obtain that if is smaller than the max delay spread (4 in Model 1 and 12 in Model 2), the MSE is relative large due to insufficient . If is chosen to be close to the max delay spread, the
performance is relative good. As grows, the estimated channels include more taps with interference. For this reason, MSE gets larger in this region.
K0
K0
K0 K0
K0
In Figure 5.16 to Figure 5.19, converge curves of the proposed modification for the LS estimator in 4.2.3.2 are presented. In every OFDM packet, the channel length is estimated again. The curves illustrate the converge of corresponding MSE and values of . In Figure 5.16 to 5.19, one can see that the MSE converge to low value after a few iterations. For the case of four receiver antennas, the curve can’t converge in present simulations. The improvement will be one of the future works.
K0
0 5 10 15 20 25 30
Average MSE of estimated channels (dB)
Converage MSE curve of proposed method (Model 1)
0 5 10 15 20 25 30
Corresponding K0 curve of proposed method (Model 1)
Figure 5.16 MSE and curves versus iteration no. of proposed decision algorithm,
Average MSE of estimated channels (dB)
Converage MSE curve of proposed method (Model 2)
0 5 10 15 20 25 30
Corresponding K0 curve of proposed method (Model 2)
Figure 5.17 MSE and curves versus iteration no. of proposed decision
algorithm,
K0 K0
T 2
N = (Model 2)
0 5 10 15 20 25 30
Average MSE of estimated channels (dB)
Converage MSE curve of proposed method (Model 1)
0 5 10 15 20 25 30
Corresponding K0 curve of proposed method (Model 1)
Figure 5.18 MSE and curves versus iteration no. of proposed decision algorithm,
Average MSE of estimated channels (dB)
Converage MSE curve of proposed method (Model 2)
0 5 10 15 20 25 30
Corresponding K0 curve of proposed method (Model 2)
Figure 5.19 MSE and curves versus iteration no. of proposed decision algorithm,
K0 K0
T 3
N = (Model 2)