FX2重新枚举会发生什么

你好，
我正试图了解当FX2有电源但USB被拔出后再插入的时候会发生什么。我们的固件使用VANDYAX示例作为骨架代码。我原本以为，如果用户拔出USB电缆并将它插回（电源是单独提供的，而不是通过USB），那么TDyIn（）将被再次调用。然而，事实并非如此。我知道，当USB被拔出时，FX2将进入休眠模式，当USB电缆***入到主机时，它会“唤醒”。
我的问题是，考虑到我们的FX2和FPGA是如何集成的，我需要发送一个信号从微控制器来重置FPGA当这个重新枚举事件发生，我不知道什么样的程序将被调用一次，当USB电缆插入。我发现，当“重新枚举一个新设备”时，调用TDyIn（）的地方，但显然在连接到主机的USB电缆中插入和插入不是那个事件吗？
欢迎任何智慧！
大肆宣传



     以下为原文
   Hi,
    I'm trying to understand exactly what happens when the FX2 has power but the USB is unplugged and then re-plugged back in.   Our firmware uses the vend_ax example as skeleton code.   I had thought originally that if the user unplugs the USB cable and plugs it back in (power is supplied separately; not through USB) that TD_Init() would be called again.  However, that appears to not be the case.  I understand that when the USB is unplugged the FX2 will go into sleep mode and that it "wakes up" when the USB cable to the host is plugged back in.  
    My issue is, given how our FX2 and FPGA are integrated, I need to send a signal from the microcontroller to reset the FPGA when this re-enumeration event happens and I'm not sure what routine will be called once when the USB cable is inserted. I found somewhere that TD_Init() is called when "re-enumerating a new device" but apparently unplugging and plugging in USB cable connecting to the host is not that event?
    Any wisdom welcome!
    -Hyped