来自前肺静脉壁隔离-A-Q-RATE POWER试验的最佳高功率除设置的见解
Piotr Gardziejczyk1, Roman Piotrowski2, Martyna Skrzyńska-Kowalczyk3
1Division of Clinical Electrophysiology, Department of Internal Medicine and Cardiology, Medical University of Warsaw, Lindleya 4, 02-005, Warsaw, Poland. piotrgardziejczyk@gmail.com.
概括
对于肺静脉隔离 (PVI) 的非常高功率短时间射频 (RF) 切除,与高功率短时间 (HPSD) 设置相比,需要更多的触摸应用,特别是在右肺静脉中. 混合方法可以减少对额外射频应用的需求,而不会增加程序时间.
科学领域:
- 心脏病学 心脏病学
- 电子生理学 电子生理学
- 医疗器械 医疗器械
背景情况:
- 非常高功率短持续 (vHPSD) 射频 (RF) 损伤比高功率短持续 (HPSD) 或常规设置的损伤浅.
- vHPSD射频应用可能无法在前肺静脉 (PV) -左心房 (LA) 壁交叉点的较厚区域实现转移性.
- 心房动 (AF) 除疗效取决于完整的肺静脉隔离 (PVI).
研究的目的:
- 为了比较使用vHPSD对PVI的急性疗效与使用AI引导的HPSD射频切除在前PV方面.
- 评估是否需要额外的射频应用来实现AF患者的完整PVI.
- 评估两种废除策略的手术和光学时间.
主要方法:
- 潜在的,双中心的,随机的A-Q-RATE权力试验.
- 在前PV方面分配给vHPSD或HPSD废除的患者;后PVs用vHPSD废除.
- 主要结果:需要在前方光伏方面进行额外的射频应用,以实现完整的PVI.
主要成果:
- 七十名患者随机分组:35名为vHPSD,35名为HPSD.
- vHPSD组需要显著更多的触摸射频应用 (46%对19%,p < 0.001),特别是在正确的光伏 (57%对20%,p = 0.001).
- 程序的持续时间,LA停留时间和光镜时间在各组之间是相似的.
结论:
- 使用HPSD用于前方光伏方面和vHPSD用于后方方面的混合策略与较低的额外射频应用需求有关.
- 这种混合方法可以实现完全的PVI,而不会延长手术或光镜的持续时间.
- 与HPSD相比,仅在前PV方面进行vHPSD废除对于实现完全PVI的效果较低.
相关概念视频
01:16Maximum Power Transfer
Numerous practical applications within engineering disciplines, such as telecommunications, necessitate optimizing power delivery to a connected load. This pursuit, however, entails inherent internal losses, which can either equal or exceed the power supplied to the load. The Thevenin equivalent circuit is helpful in finding the maximum power a linear circuit can deliver to a load. It is assumed in this context that the load resistance can be adjusted.
By substituting the entire circuit with...
By substituting the entire circuit with...
01:20The Maximum Power Transfer Theorem
Consider a linear AC Thevenin equivalent circuit connected to a load impedance.
The load connected draws the current, and the circuit delivers the power to the load. The alternating current flowing through the load is determined using the rectangular form of voltages, currents, network impedance, and load impedance. The average power delivered to the load is obtained from the product of the square of current and load resistance.
The load connected draws the current, and the circuit delivers the power to the load. The alternating current flowing through the load is determined using the rectangular form of voltages, currents, network impedance, and load impedance. The average power delivered to the load is obtained from the product of the square of current and load resistance.