There are several problems need to be addressed in this technique and we hope we can solve them in the future.
(1.) The channel state may not be static, i.e. the channel model may be time-varying.
In this case, the SVM techniques proposed here may not be used directly due to the SMO algorithm and the QP problems.
(2.) When the size of M is large, the simulation will take large memories and time.
This is due to the SMO and DAG algorithm and it may be better to solve the QP problem once at all or to handle the training set in advance.
In practice, the size of M will be large. For instance, in [9], it set M = 256. Also, when we apply this technique to Flash-OFDM, this problem will appear again.
(3.) In this thesis, we do not indicate the optimal kernel and the choice of parameter given kernel. We believe that the problem is ongoing research for the moment.
(4.) To combine the SVM-based receiver with error correction code will appear some problems. One of them is the size of M , which is discussed in (2). For instant, the size of M may be 204 for RS(204, 188). Another is the soft input for the error correcting decoder. In our thesis, the output of the receiver is one of {1, 2, 3, . . . M}.
It could not be used to be the soft input of the decoder directly. In this case, it may rely on the notion of regression, which is out of scope here.
Appendix A
The Objective Function for SMO Algorithm
In the section, we derive the (3.9). By SMO algorithm, one can break a large QP problem into a series of smallest possible QP problems, and the smallest possible QP problem can be solved analytically, which avoids using a time-consuming numerical QP optimization as an inner loop. Also, the matrix computation is avoided, so the amount of memory required for SMO is reduced. Hence, one can solve a large QP problem efficiently.
In this algorithm, since the η = 0 is possible, one need some other steps to make the algorithm work. In those steps, one need to know the equation (3.9). Hence we derive the equation for completing the SMO algorithm here.
We begin these deductions with equation (3.3).
JD = 1
2K11(κ − sα2)2+1
2K22α22+ sK12(κ − sα2)α2+ y1(κ − sα2)v1
+y2α2v2− (κ − sα2) − α2+ residual terms
= 1
2(K11+ K22− 2K12)α22 + (−K11κs + sK12κ − y1sv1+ y2v2+ s − 1)α2 +1
2K11κ2+ y1κv1− κ + residual terms (3.1) where vi =
PN j=3
yjαj∗Kij = ui− b∗− y1α1∗K1i− y2α∗2K2i
Consider y1sv1 and y2v2 separably,
y1sv1 = y1s(u1− b∗− y1α∗1K11− y2α∗2K21)
= y1su1− y21sα∗1K11− s2α∗2K21− y1sb∗
= y2u1− (sκ − α∗2)K11− α∗2K21− y2b∗ y2v2 = y2(u2− b∗− y1α∗1K12− y2α∗2K22)
= y2u2− sα∗1K12− α∗2K22− y2b∗
= y2u2− (sκ − α∗2)K12− α∗2K22− y2b∗
where
α∗1 + sα∗2 = κ =⇒ sα∗1+ α∗2 = sκ (3.2) and
y1s = y21y2 = y2 (3.3)
Then consider
−y1sv1 + y2v2 = y2(u2 − u1) − α2∗(K11+ K22− 2K12) + sκK11− sκK12
= y2(u2 − u1) − α2∗η + sκK11− sκK12
Therefore,
−K11κs + sK12κ − y1sv1+ y2v2+ s − 1 = y2(u2− u1) − α∗2η + s − 1 (3.4)
= y2(u2− u1− y2+ y1) − α∗2η (3.5)
= y2(E2− E1) − α∗2η (3.6)
Finally, substitute (3.4) into (3.1)
JD = 1
2ηα22+ (y2(E2− E1) − α∗2η)α2+ Const.
where Cons.=residual terms+12K11κ2+ y1κv1− κ.
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簡 歷：
姓 名：劉人仰
居 住 地：台北市
研 究 方 向：Machine Learning
學 經 歷：
中山大學電機工程學系 (1999～2003)
交通大學電信工程碩士班 (2003～2005)
Graduate Course：
Information Theorem
Adaptive Signal processing
Detection and Estimation Theory Digital Signal Processing
Computer Network Algorithm
Matrix Computation Statistic
Real Analysis (I)