Kim H. Parker edited There_is_no_physical_reason__.tex  over 8 years ago

Commit id: f24c1977db73605e86fbbcd924493277fb56de70

deletions | additions      

       

There is no physical reason that this value of $\rho c$ is the best estimate. Since the forward and backward components of pressure change share their origins in the contraction and relaxation of the myocardium, it would be expected that they contain mutual information. In addition to this mutual information arising from their common mechanistic origin, we argue that the use of the 'wrong' value of $\rho c$ introduces extra mutual information through too high or too low a dependency upon $dP$ or $dU$. For example, if $\rho c$ is very small $dP_\pm$ will both depend almost totally on $dP$ and $I(dP_+,dP_-)$ will approach $H(dP)$ whereas if it is very large $dP_\pm$ will depend almost totally on $dU$ and $I(dP_+,dP_-)$ will approach $H(dU)$. For the 'right' value of $\rho c$, $I(dP_+,dP_-)$ should be closer to $I(dP,dU)$, the mutual information between the measured pressure and velocity due to their shared mechanically deterministic origin. Because both $P$ and $U$ contain measurement noise, we do not expect that $I(dP_+,dP_-)$ to be equal to $I(dP,dU)$, only approach it. By this argument, the value of $\rho c$ that minimises the mutual information still contains the mutual information between the forward and backward waves due to their origin mechanically but minimises the extra mutual information arising from the definition of $dP_\pm$, thereby giving us a way to estimate the local wave speed.  In order to utilise this procedure we must estimate the various probabilities. For convenience in the following discussion we will identify $X = dP_+$ and $Y = dP_-$ measured over an integral number of cardiac periods. $x$ is therefore the forward change of pressure and $y$ is the backward change in pressure calculated at any particular time. The probability density functions of $x$ and $y$ can be estimated from their histograms