小儿左心辅助心电触发方式下辅助延迟时间的探讨

The Delay Time of Pediatric LVAD of ECG-Triggering Mode

心电是触发左心辅助同步控制的一个重要的信号源 ,而辅助延迟时间是辅助循环中的一个重要环节。适当的延迟既可以使左心辅助装置 (L VAD)有充分的收缩时间 ,又可以避免 L VAD与心脏在收缩期的冲突 ;如果延迟时间不适当、尤其是在左心房引流不充分时 ,L VAD与心脏的冲突在短时间内就可造成急性左心衰、肺水肿。本文旨在寻找一个经验公式以有效地指导 L VAD延迟时间。目的 :以 R- Ao代表心电 QRS波峰点与主动脉瓣关闭结束的间隔时间 ,以 RR代表 RR间期 ,找出 RR与 R- Ao的相关方程 :R- Ao=A× (RR) n+B。通过加权平均法计算RR的平均值 ,再求出 R- Ao预期值 ,对 R- Ao做安全较正后 ,以此做为辅助的延迟时间。方法 :选择 5～ 14岁行升主动脉导管的小儿患者 ,经筛选 ,测量 45 7例心导管记录图中心电部分的 RR间期以及与此心电对应的主动脉压力波形 ,测量主动脉下降波切迹与 QRS波起始点间的间隔 ,以 R- Ao为因变量 ,以 (RR) n为自变量 ,绘散点图 ,找出相关模型 ,并求出相关方程。结论 :当心率小于 12 0次 / min时 ,相关系数为 0 .733,回归系数为 - 0 .182 ,P值为 2 .2 94E-78;截距为 1.0 70 ,P值为 4.83E- 10 9,相关方程为 :R- Ao=(- 0 .182 ) RR1 /3 +1.0 70。

Electrocardiac signal is one of the most important signals which is used to trigger ventricular assist device (VAD), and the delay time of VAD assistance is very important to get a satisfied result. Proper delay will give VAD relatively enough time to assist, avoiding left heart failure caused by the collision of the heart and VAD during systolic phase. This becomes much more important when the left atrium drainage is insufficient. The aim of our study is to set up an equation to calculate the delay time by RR interval. We try to set up an equation about RR and R-Ao like: R-Ao=A×(RR) n+B(A and B are constant). RR represents the RR interval and R-Ao represents the duration of the period between the peak point of QRS and the point of aortic valve closing; First, calculate RR according to weighting average method, and then, calculate the anticipant R-Ao according to the before-mentioned equation. After adjustment, R-Ao will be used as assistance delay time. R-R interval was measured in 457 selected pediatric patients who were undergiong left heart catheterization and who did not have arrhythmias. From the ECG recording during catheterization, R-R interval was measured while R-Ao was obtained from aortic pressure wave chart; Plot graphs with R-Ao as dependent variable and (RR) n as independent variable; find out correlating model and calculate the arguments A and B of R-Ao=A×(RR) n+B. The results showed that the relation between (RR) 1/3 and R-Ao is the most significant, the relation coefficient is 0.733, the regress coefficient is -0.182 (P< 0.001) and the interception is 1.070. This means that R-Ao=(-0.182)(RR)1/3 + 1.070. The likelyhood degrees of different sections differ markedly. When heart rate is less than 120 beats per min. the relation argument is about 0.733 while 0.45 when heart rate is more than 120 beats per min, Therefore, we can use the equation R-Ao=(-0.182)(RR) 1/3+1.070 to calculate R-Ao when heart rate is less than 120 beats per min.$$$$