PWM输出有异常情况发生

请具体描述你的问题，比如如何配置PWM，产生问题的图是怎么样的？

整个程序的思想是AD 实时采样，用PWM实时输出AD的采样值，把PWM当DA用。

PWM的配置如下：

EPwm1Regs.TBPRD =CC_TBPRD; 
EPwm1Regs.TBCTL.all = 0;   
EPwm1Regs.TBCTL.bit.CLKDIV = TB_DIV2; 
EPwm1Regs.TBCTL.bit.HSPCLKDIV = 0; 
EPwm1Regs.TBCTL.bit.CTRMODE = TB_COUNT_UP; 

EPwm1Regs.CMPCTL.bit.SHDWAMODE=CC_SHADOW ;
EPwm1Regs.CMPCTL.bit.LOADAMODE=CC_CTR_PRD;

EPwm1Regs.AQCTLA.bit.CAU=AQ_SET;
EPwm1Regs.AQCTLA.bit.PRD=AQ_CLEAR;

EPwm1Regs.CMPA.half.CMPA=0;

下面是AD中断的程序，意思是在每个AD采样中断中修改一次PWM的CMPA的值。CMPA与AD采样值有一个对应的关系，使得PWM的占空比与AD采样值成正比关系。AD采样率与PWM同频。AD配置以及中断配置这里没有写，可以正常工作。

interrupt void adc1_isr(void)
{
AD_result_A0= Adc1Result.ADCRESULT0;

Value_CMPA1=CC_TBPRD-AD_result_A0*Value_num;

EPwm1Regs.CMPA.half.CMPA=Value_CMPA1;

Adc1Regs.ADCINTFLGCLR.bit.ADCINT1 = 1;
PieCtrlRegs.PIEACK.all = PIEACK_GROUP1;

}

（为了清晰描述这个问题，不考虑AD过采样）现在AD用10K的采样速度采集5K的的输入信号，也就是输入信号一个周期有两个采样值，对应的PWM占空比应该变化两次。但是出现了下图中的问题，在某个地方，CMPA的值与理论上采样值不对应的情况。现在已证实AD采样值是没有错误的，揣测是CMPA的值有的时候没有装载到当前寄存器上或者是什么原因。

5k的输入信号是什么波形，另外触发采样点在哪里？怎么设置的？

输入信号时5K的正弦信号。

用的是F28M35x开发板，下面是C28x整个的程序

#include "DSP28x_Project.h" // Device Headerfile and Examples Include File
#include <string.h>

#define C28_FREQ 150 //C28 CPU frequency in MHz
#define CC_TBPRD 7500

extern Uint16 RamfuncsLoadStart;
extern Uint16 RamfuncsLoadSize;
extern Uint16 RamfuncsRunStart;

float Value_num;

interrupt void adc1_isr(void);
//interrupt void Epwm1_INT_isr(void);

Uint16 AD_result_A0;
Uint16 Value_CMPA1;

void main(void)
{
InitSysCtrl(); // Init C28 core
EALLOW;
GpioG1CtrlRegs.GPADIR.bit.GPIO4 = 1; // GPIO4 as output
GpioCtrlRegs.GPAMUX1.bit.GPIO0 = 1; // GPIO0 = ePWM1A
EDIS;

memcpy(&RamfuncsRunStart, &RamfuncsLoadStart, (size_t)&RamfuncsLoadSize);

InitFlash();

Value_num=CC_TBPRD/4095.0;

DINT;
InitPieCtrl();
IER=0x0000;
IFR=0x0000;
InitPieVectTable();
EALLOW; // This is needed to write to EALLOW protected registers
PieVectTable.ADCINT1 = &adc1_isr;
EDIS;

InitCpuTimers();
ConfigCpuTimer(&CpuTimer0, C28_FREQ, 100); //SOC
//Enable INT
PieCtrlRegs.PIEIER1.bit.INTx1 = 1;
IER |= M_INT1;
EINT; // Enable Global interrupt INTM
ERTM; // Enable Global realtime interrupt DBGM

//configure EPwm1
EPwm1Regs.TBPRD =CC_TBPRD; //
EPwm1Regs.TBCTL.all = 0; // clear all bits in TBCTL
EPwm1Regs.TBCTL.bit.CLKDIV = TB_DIV2; // CLKDIV = 2
EPwm1Regs.TBCTL.bit.HSPCLKDIV = 0; // HSPCLKDIV =1
EPwm1Regs.TBCTL.bit.CTRMODE = TB_COUNT_UP; // count mode = up-mode
//fpwm =fcpu / ( TBPRD * CLKDIV * HSPCLKDIV)

EPwm1Regs.CMPCTL.bit.SHDWAMODE=CC_SHADOW ;
EPwm1Regs.CMPCTL.bit.LOADAMODE=CC_CTR_PRD;

EPwm1Regs.AQCTLA.bit.CAU=AQ_SET;
EPwm1Regs.AQCTLA.bit.PRD=AQ_CLEAR;

EPwm1Regs.CMPA.half.CMPA=0;

//configure ADC 
InitAdc1();
EALLOW;
Adc1Regs.ADCCTL2.bit.ADCNONOVERLAP = 1; 
Adc1Regs.ADCCTL1.bit.INTPULSEPOS = 1; // ADCINT1 trips after AdcResults latch
Adc1Regs.INTSEL1N2.bit.INT1E = 1; // Enabled ADCINT1
Adc1Regs.INTSEL1N2.bit.INT1CONT = 0; // Disable ADCINT1 Continuous mode
Adc1Regs.INTSEL1N2.bit.INT1SEL = 0; // setup EOC0 to trigger ADCINT1 to fire

Adc1Regs.ADCSOC0CTL.bit.CHSEL = 0; // set SOC0 channel select to ADC1in_B0
Adc1Regs.ADCSOC0CTL.bit.TRIGSEL = 5; // Set SOC0 start trigger to trigger1 // ADC Trigger 1 of the ADC
Adc1Regs.ADCSOC0CTL.bit.ACQPS =6; // set SOC0 S/H Window to 7 ADC

AnalogSysctrlRegs.TRIG1SEL.all = 1; // Assigning TINT0 to ADC-TRIGGER 1 // Clock Cycles, (6 ACQPS + 1)
EDIS;

CpuTimer0Regs.TCR.bit.TSS = 0; // start T0
while(1)
{

}
}

interrupt void adc1_isr(void)
{
AD_result_A0= Adc1Result.ADCRESULT0;

Value_CMPA1=CC_TBPRD-AD_result_A0*Value_num;

EPwm1Regs.CMPA.half.CMPA=Value_CMPA1;

Adc1Regs.ADCINTFLGCLR.bit.ADCINT1 = 1;
PieCtrlRegs.PIEACK.all = PIEACK_GROUP1;

GpioG1DataRegs.GPADAT.bit.GPIO4 ^= 1;
}

放了一个GPIO04 的电平反转在AD中断中看AD采样的频率是否正常，可以看到频率是5K，脉宽有5ns到50ns的变化。  

最后的原因是 PWM分频后的TBCLK与 AD 的trigger Timer0 有频差