Uniform Channel Coefficients

The previous section derived an EM algorithm for the case of deterministic channel coefficients, using an impulse function for $f({\bf h})$. In this section we analyze a similar situation, where no information about ${\bf h}$ is available, but here we use a uniform distribution for $f({\bf h})$. Although impulse and uniform distributions for $f({\bf h})$ do not seem similar, they are similar in this application because the location of the impulse (mean of ${\bf h}$) is unknown in the previous section.

If $f({\bf h})$ is uniform over some finite interval, then computation of ${\bf g}$ and ${\bf G}$ could be performed numerically using $f({\bf h }\vert{\bf r},{\bf B})$ from (12). But if we let the domain of ${\bf h}$ be infinite, then we can compute ${\bf g}$ and ${\bf G}$ analytically. In this case $f({\bf h})$ is improper, since it would have to be zero for the area under it to be 1. Nevertheless, assuming that $f({\bf h})$ is uniform and of infinite extent is useful, since it is practically equivalent to letting $f({\bf h})$ be uniform over some range of several standard deviations for which the exponential term in (12) is significant. Outside of some finite region, the exponential term in (12) is practically zero.

So we propose that letting $f({\bf h}) = 1$ in (12) is reasonable, which reduces (12) to:

$$% latex2html id marker 826 f({\bf h}\vert{\bf r},{\bf B}) = K... ...}} \ e^{- \frac {\vert{\bf r}-{\bf Bh}\vert^2} {\sigma^2} } .$$ (24)

To put this into a more useful form, expand the norm-squared term from (24):

 

$$\vert{\bf r} - {\bf B}{\bf h} \vert^2 {\bf r}^{*T} {\bf r} - {\bf r}^{*T} {\bf B}{\bf h} - {\bf h}^{*T}{\bf B}^{*T} {\bf r} + {\bf h}^{*T} {\bf B}^{*T} {\bf B} {\bf h}$$ =

$${\bf r}^{*T} {\bf r} - {\bf r}^{*T} {\bf B}{\bf h} - {\bf h}^{*T}{\bf B}^{*T} {\bf r}$$ =

$$+ ({\bf h}-{\bf g})^{*T} {\bf B}^{*T} {\bf B} ({\bf h}-{\bf g})$$ (25)

$$+ \ {\bf h}^{*T} {\bf B}^{*T} {\bf B} {\bf g} + {\bf g}^{*T} {\bf B}^{*T} {\bf B} {\bf h} - {\bf g}^{*T} {\bf B}^{*T} {\bf B} {\bf g} ,$$

for any ${\bf g}$.

Now, if we let ${\bf g} = {\bf B}^+{\bf r}$, and if ${\bf B}$ has full rank so that ${\bf B}^+ = ({\bf B}^{*T} {\bf B})^{-1} {\bf B}^{*T}$, then (25) reduces to:

$${\vert{\bf r} - {\bf B}{\bf h} \vert^2} = {\bf r}^{*T} {\bf r... ...{\bf h}-{\bf g})^{*T} {\bf B}^{*T} {\bf B} ({\bf h}-{\bf g}) ,$$ (26)

and (24) may be written as:

$$f({\bf h}\vert{\bf r},{\bf B}) \doteq e^{- \frac {({\bf h}-{\... ...)^{*T} {\bf B}^{*T} {\bf B} ({\bf h}-{\bf g})} {\sigma^2} } .$$ (27)

This shows that, given ${\bf r}$ and ${\bf B}$, ${\bf h}$ has a joint Gaussian distribution with mean ${\bf g} = {\bf B}^+{\bf r}$ and covariance:

$${\bf G} = \left[ \frac {{\bf B}^{*T} {\bf B}} {{\sigma}^2} \right] ^ {-1} .$$ (28)

If ${\bf B}$ does not have full rank then some simplification of (25) into a form similar to (26) may be performed, but that is not pursued here. In the case of data values which do not contain 0's, e.g. (-1,1) data values, ${\bf B}$ will always have full rank if a_j = 0 for $j \le 0$, due to the 0's in the first L-1 rows which make the columns of ${\bf B}$ linearly independent.

Therefore, for the case of uniformly distributed channel coefficients, the EM algorithm is almost the same as for deterministic channel coefficients, except that the Viterbi algorithm incremental cost function from (18), which includes ${\bf G}$, is used instead of (23), and ${\bf G}$ is given by (28).

To initialize ${\bf B}$, we can initialize ${\bf g}$ and ${\bf G}$, then use the Viterbi algorithm to estimate ${\bf A}$, then set ${\bf B}$ to this estimate of ${\bf A}$. The initial estimate of ${\bf G}$depends on the set of transmitted sequence values and initial channel conditions. For example, in the case of (-1,1) transmitted data values, with a_j = 0 for $j \le 0$, ${\bf G} =$diag ${ \left[ \frac {\sigma^2} {N-i+1} \right]}, ~ i=1,\ldots,L$may be used, based on just the diagonal elements of ${\bf B}^{*T} {\bf B}$.