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Assume that a set of initial training samples D0 f x i ; z i ji 1; . . . ; Ng is given before the learning gets started. To obtain an initial eigenspace model W ; Uk ; Lk ; N , the conventional PCA is applied to D0 and the smallest dimensionx ality k of the eigenspace is determined such that the accumulation ratio is larger than q. Since a proper q is usually unknown and often depends on training data, the crossvalidation technique can be applied to determining q (Ozawa et al., 2008). However, let us assume that a proper q is given in advance. The initial training is shown in Pseudo-code 2.

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Then the electrical law governing a component describes how the variables of its enclosing cross-sections relate to each other Note that there are usually several ways of selecting the cross-sections and that an appropriate selection is crucial to the complexity of the network However, selecting appropriate cross-sections is easier than finding good node mergers in the approach discussed above Given a set of cross-sections we construct the join tree as follows: the separator sets form, not surprisingly, the node separators (cf Section 422) For each circuit part (containing one component) we create a (clique) node containing the union of the separator sets of the bounding cross-sections In addition, we create a node for each component, comprising the variables needed to describe its behavior, and connect it to the node corresponding to the circuit part the component is in.

Equation (3.8.8) can be further solved to nd the times t0:5 and t0:9 for v 0:5 and v 0:9, respectively, as t0:5 RC 0:377 0:693 RT CT RT CT t0:9 RC 1:02 2:3 RT CT RT CT 3:8:9 3:8:10

An interesting observation can be made about Figure 725, where a detailed view of the non-normalized histograms of the extrinsic informations for 0 and for 1 obtained by Monte Carlo simulations over the LOG MAP BCJR decoder of the turbo code of Example 74 is presented Bold lines describe the extrinsic estimates for 1 and dotted lines describe the extrinsic estimates for 0 The non-normalized histogram of the highest peak value corresponds to E b /N0 = 02 dB, and then lower peak value histograms correspond in decreasing order to E b /N0 = 03, 04, 05 and 06 dB The circled line highlights the non-normalized histograms for 0 and for 1 when E b /N0 = 06 dB, which is the rst case for which the two peak values have different abscissa values.

Now at this value of E b /N0 = 06 dB, the BER performance curve of Figure 721, which relates to the same turbo code, is in the waterfall region that starts at E b /N0 = 05 dB and ends at about 15 dB For values of E b /N0 lower than 05 dB, the non-normalized histograms for 0 and 1 have their peaks at the same value of the abscissa, as seen in Figure 725 Abscissa values of the peaks of the histograms in Figure 725 for low values of E b /N0 are coincident due to the quantization effects of the histogram representation This means anyway that parameters E0 and E1 are in this case practically equal.

The peak value of the histogram for 0 is at the abscissa value E0 that corresponds to the abscissa for the probability density function for 0 , p E ( / X = 1), which is of the form p E ( / X = 1) = 2 2 e (( + E0 x) )/2 E0 / 2 E0 The peak value of the histogram for 1 is at the abscissa value E1 that corresponds to the abscissa for the probability density function for 1 , p E ( / X = +1), 2 2 which is of the form p E ( / X = +1) = e (( E1 x) )/2 E1 / 2 E1 .

It will be observed by means of these simulations that when the values of E0 and E1 are similar, the code is in the BER performance region where the coded scheme is worse than uncoded transmission As the.

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