步进电机梯形精度问题

[zhaoyang6]

问题：该算法每运行一次时会多走一个脉冲。
解决办法： 修改bsp_stepper_T_speed.c中的void speed_decision()，将STOP状态拿出if(i==2)的条件判断，修改为如下：
`
if(i==2) // 2次，说明已经输出一个完整脉冲
{
i=0; // 清零定时器中断次数计数值
switch(srd.run_state)
{
/步进电机停止状态/
//在这里判断是否在停止状态，会在每一次运行后多一个脉冲
// case STOP:
// step_count = 0;
// rest = 0;

			// 		// 关闭通道
			// 		HAL_TIM_OC_Stop_IT(&TIM_TimeBaseStructure,MOTOR_PUL_CHANNEL_x);

			// 		status.running = FALSE;
			// 		break;
			/*步进电机加速状态*/
			case ACCEL:
					TIM_CCxChannelCmd(MOTOR_PUL_TIM, MOTOR_PUL_CHANNEL_x, TIM_CCx_ENABLE);
					StepperCounter(srd.dir);
					step_count++;
					srd.accel_count++;
					new_step_delay = srd.step_delay - (((2 * (long)srd.step_delay) 
													 + rest)/(4 * srd.accel_count + 1));
					rest = ((2 * (long)srd.step_delay)+rest)%(4 * srd.accel_count + 1);
					//检查是够应该开始减速
					if(step_count >= srd.decel_start) {
						srd.accel_count = srd.decel_val;
						srd.run_state = DECEL;
					}
					//检查是否到达期望的最大速度
					else if(new_step_delay <= srd.min_delay) {
						last_accel_delay = new_step_delay;
						new_step_delay = srd.min_delay;
						rest = 0;
						srd.run_state = RUN;
					}
					break;
			/*步进电机最大速度运行状态*/
			case RUN:

		      StepperCounter(srd.dir);//根据脉冲计数位置
					step_count++;
					new_step_delay = srd.min_delay;
					//检查是否需要开始减速
					if(step_count >= srd.decel_start) 
					{
						srd.accel_count = srd.decel_val;
						//以最后一次加速的延时作为开始减速的延时
						new_step_delay = last_accel_delay;
						srd.run_state = DECEL;
					}
					break;
			/*步进电机减速状态*/
			case DECEL:

		      StepperCounter(srd.dir);
					step_count++;
					srd.accel_count++;
					new_step_delay = srd.step_delay - (((2 * (long)srd.step_delay) 
													 + rest)/(4 * srd.accel_count + 1));
					rest = ((2 * (long)srd.step_delay)+rest)%(4 * srd.accel_count + 1);
					//检查是否为最后一步
					if(srd.accel_count >= 0)
					{
						srd.run_state = STOP;
					}
					break;
		}
  if(srd.run_state == STOP){
      step_count = 0;
      rest = 0;
      // 关闭通道
      HAL_TIM_OC_Stop_IT(&TIM_TimeBaseStructure,MOTOR_PUL_CHANNEL_x);
      status.running = FALSE;
  }

`