问题1:请问AD单元手册12页的滤波时间2ns是指哪个滤波器呢？这个2ns是如何计算出来的呢？
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对于Dummy的问题，Errata中的描述如下：
从描述的内容看，问题出现在初始化阶段，如果系统可以接受初始化时的IDDA，那么不进行Dummy操作也可以
在例程Example_2803xAdcSoc中，有如下配置：
EALLOW;
AdcRegs.ADCCTL1.bit.INTPULSEPOS = 1; //ADCINT1 trips after AdcResults latch
AdcRegs.INTSEL1N2.bit.INT1E = 1; //Enabled ADCINT1
AdcRegs.INTSEL1N2.bit.INT1CONT = 0; //Disable ADCINT1 Continuous mode
AdcRegs.INTSEL1N2.bit.INT1SEL = 2; //setup EOC2 to trigger ADCINT1 to fire
AdcRegs.ADCSOC0CTL.bit.CHSEL = 4; //set SOC0 channel select to ADCINA4(dummy sample for rev0 errata workaround)
AdcRegs.ADCSOC1CTL.bit.CHSEL = 4; //set SOC1 channel select to ADCINA4
AdcRegs.ADCSOC2CTL.bit.CHSEL = 2; //set SOC2 channel select to ADCINA2
AdcRegs.ADCSOC0CTL.bit.TRIGSEL = 5; //set SOC0 start trigger on EPWM1A, due to round-robin SOC0 converts first then SOC1, then SOC2
AdcRegs.ADCSOC1CTL.bit.TRIGSEL = 5; //set SOC1 start trigger on EPWM1A, due to round-robin SOC0 converts first then SOC1, then SOC2
AdcRegs.ADCSOC2CTL.bit.TRIGSEL = 5; //set SOC2 start trigger on EPWM1A, due to round-robin SOC0 converts first then SOC1, then SOC2
AdcRegs.ADCSOC0CTL.bit.ACQPS = 6; //set SOC0 S/H Window to 7 ADC Clock Cycles, (6 ACQPS plus 1)
AdcRegs.ADCSOC1CTL.bit.ACQPS = 6; //set SOC1 S/H Window to 7 ADC Clock Cycles, (6 ACQPS plus 1)
AdcRegs.ADCSOC2CTL.bit.ACQPS = 6; //set SOC2 S/H Window to 7 ADC Clock Cycles, (6 ACQPS plus 1)
EDIS;
问题2：如上所示，每次触发3个通道，第一个通道的数据是没有用的，这样会占用一些采样时间，如果没有Dummy，是不是也是可以？
