Dynamic Assessment of Baroreflex Control of Heart Rate During Induction of Propofol Anesthesia Using a Point Process Method

Abstract

In this article, we present a point process method to assess dynamic baroreflex sensitivity (BRS) by estimating the baroreflex gain as focal component of a simplified closed-loop model of the cardiovascular system. Specifically, an inverse Gaussian probability distribution is used to model the heartbeat interval, whereas the instantaneous mean is identified by linear and bilinear bivariate regressions on both the previous R−R intervals (RR) and blood pressure (BP) beat-to-beat measures. The instantaneous baroreflex gain is estimated as the feedback branch of the loop with a point-process filter, while the $\hbox{RR}\to\hbox{BP}$ feedforward transfer function representing heart contractility and vasculature effects is simultaneously estimated by a recursive least-squares filter. These two closed-loop gains provide a direct assessment of baroreflex control of heart rate (HR). In addition, the dynamic coherence, cross bispectrum, and their power ratio can also be estimated. All statistical indices provide a valuable quantitative assessment of the interaction between heartbeat dynamics and hemodynamics. To illustrate the application, we have applied the proposed point process model to experimental recordings from 11 healthy subjects in order to monitor cardiovascular regulation under propofol anesthesia. We present quantitative results during transient periods, as well as statistical analyses on steady-state epochs before and after propofol administration. Our findings validate the ability of the algorithm to provide a reliable and fast-tracking assessment of BRS, and show a clear overall reduction in baroreflex gain from the baseline period to the start of propofol anesthesia, confirming that instantaneous evaluation of arterial baroreflex control of HR may yield important implications in clinical practice, particularly during anesthesia and in postoperative care.

Appendix

For clarity of proof of Eq. (11), we assume that inputs u(t) and x(t) are both stationary and have zero means (in our case, we only model the “de-meaned” RR and BP signals with {a _i } and {b _j }). The line of logic is similar to the one that was presented in the literature,38 which only considered a univariate input. We first decompose the output y(t) into three (two linear and one bilinear) terms

$$ y(t)=\sum_i a_i x(t-i) +\sum_j b_j u(t-j) +\sum_k\sum_l h_{kl} x(t-k) u(t-l). $$

From the assumption that ${\mathbb{E}}[x(t)]=0$ and ${\mathbb{E}}[u(t)]=0,$ it further follows that

$$ \begin{aligned} {\mathbb{E}}[y(t)]=&\sum_k\sum_l h_{kl} {\mathbb{E}}[x(t-k) u(t-l)] \\ =& \sum_k\sum_l h_{kl}C_{xu}(l-k)\\ =& {\frac{1}{(2\pi)^3}} \sum_k\sum_l \int \int {\mathcal{H}}(f_1,f_2)e^{j2 k \pi f_1}e^{j2 l \pi f_2} df_1df_2 \int {\mathcal{C}}_{xu}(f) e^{j2 (l-k) \pi f } df \\ =& {\frac{1}{2\pi}}\int {\mathcal{H}}(f,-f) {\mathcal{C}}_{xu} (f)df,\\ \end{aligned} $$

(A.1)

where the expectation operation is averaged on the argument over time.

Next, we compute the cross third-order cumulant statistic between x(t) and y(t) (viz. cross bicovariance):

$$ \begin{aligned} C_{xxy}(\tau_1,\tau_2)=&{\mathbb{E}}\{x(t+\tau_1)x(t+\tau_2)[y(t)- {\mathbb{E}}\{y(t)\}]\}\\ =& {\mathbb{E}}\{x(t+\tau_1)x(t+\tau_2)y(t)\} -{\mathbb{E}}\{y(t)\} {\mathbb{E}}\{x(t+\tau_1)x(t+\tau_2)\}\\ =& \sum_{j}b_j {\mathbb{E}}\{x(t+\tau_1)x(t+\tau_2) u(t- j) \} \\ &+\sum_k\sum_l h_{kl} {\mathbb{E}}\{x(t+\tau_1)x(t+\tau_2)x(t-k) u(t-l) \} -{\mathbb{E}}\{y(t)\} C_{xx}(\tau_1-\tau_2)\\ \end{aligned} $$

where we have used the fact that if x(t) is zero-mean Gaussian, then the third cumulant statistic C _xxx (τ₁, τ₂) is zero, such that $\sum_i a_i {\mathbb{E}}\{x(t+\tau_1)x(t+\tau_2) x(t- i) \}=0.$ Furthermore, if u(t) and x(t) are zero-mean jointly Gaussian distributed, and the odd-moment statistic $C_{xxu}(\tau_1,\tau_2)\approx 0$ (because of the symmetry of Gaussian distribution), then the following relationship holds38,51:

$$ {\mathbb{E}}[x_1x_2x_3 u_4]= {\mathbb{E}}[x_1x_2]{\mathbb{E}}[x_3u_4]+ {\mathbb{E}}[x_1x_3]{\mathbb{E}}[x_2u_4]+ {\mathbb{E}}[x_2x_3]{\mathbb{E}}[x_1u_4], $$

(A.2)

and

$$ C_{xxy}(\tau_1,\tau_2)= \sum_k\sum_l h_{kl} \left\{ {\mathbb{E}}\{x(t+\tau_1)x(t+\tau_2)x(t-k) u(t-l) \} - C_{xu}(l-k) C_{xx}(\tau_1-\tau_2) \right\}. $$

(A.3)

In light of Eqs. (A.2) and (A.3), we obtain

$$ \begin{aligned} C_{xxy}(\tau_1,\tau_2)=& 2 \sum_k\sum_l h(k-\tau_1, l-\tau_2) C_{xx}(k) C_{xu}(l)\\ =& {\frac{2}{(2\pi)^2}} \int\int {\mathcal{H}}(-f_1,-f_2) e^{j2\pi \tau_1 f_1} e^{j2\pi \tau_2 f_2} {\mathcal{Q}}_{\rm RR}(f_1){\mathcal{C}}_{xu}(f_2) df_1 df_2\\ \end{aligned} $$

Finally, we compute the two-dimensional Fourier transform of C _xxy (τ₁, τ₂) to obtain the cross bispectrum ${\mathcal{C}}_{xxy} (f_1,f_2):$

$$ \begin{aligned} {\mathcal{C}}_{xxy} (f_1,f_2) =& \int C_{xxy}(\tau_1,\tau_2) e^{-j2\pi \tau_1 f_1} e^{-j2\pi \tau_2 f_2} d\tau_1 d\tau_2\\ =& 2{\mathcal{H}}(-f_1,-f_2) {\mathcal{Q}}_{\rm RR}(f_1) {\mathcal{C}}_{xu}(f_2),\\ \end{aligned} $$

(A.4)

which then completes the proof of Eq. (11).