EQEP 的TI例程问题：

typedef struct {int theta_elec;         // Output: Motor Electrical angle (Q15)
                int theta_mech;         // Output: Motor Mechanical Angle (Q15)
                int DirectionQep;       // Output: Motor rotation direction (Q0)
                int QEP_cnt_idx;        // Variable: Encoder counter index (Q0)
                int theta_raw;          // Variable: Raw angle from Timer 2 (Q0)
                int mech_scaler;        // Parameter: 0.9999/total count, total count = 4000 (Q26)
                int pole_pairs;         // Parameter: Number of pole pairs (Q0)
                int cal_angle;          // Parameter: Raw angular offset between encoder and phase a (Q0)
                int index_sync_flag;    // Output: Index sync status (Q0)

                Uint32 SpeedScaler;     // Parameter :  Scaler converting 1/N cycles to a GLOBAL_Q speed (Q0) – independently with global Q
                _iq Speed_pr;           // Output :  speed in per-unit
                Uint32 BaseRpm;         // Parameter : Scaler converting GLOBAL_Q speed to rpm (Q0) speed – independently with global Q
                int32 SpeedRpm_pr;      // Output : speed in r.p.m. (Q0) – independently with global Q

                _iq  oldpos;            // Input: Electrical angle (pu)
                _iq Speed_fr;           // Output :  speed in per-unit
                int32 SpeedRpm_fr;      // Output : Speed in rpm  (Q0) – independently with global Q
                void (*init)();         // Pointer to the init funcion
                void (*calc)();         // Pointer to the calc funtion
                }  POSSPEED;

红色参数设置的作用是什么？初始化时  BaseRpm=6000 是不是选择电机的转速是6000转每分钟？

// 2. **min rpm ** = selected at 10 rpm based on CCPS prescaler options available (128 is greatest)
//
// 3. **SpeedRpm_pr**
//    SpeedRpm_pr = X/(t2-t1)                                                    – Equation 4
//    where X = QCAPCTL [UPPS]/4000 rev. (position relative to Index in 1 revolution)
// If  max/base speed = 6000 rpm: //               6000 = (32/4000)/[(t2-t1)/(200MHz/64)]
//           where 32 = QCAPCTL [UPPS] (Unit timeout – once every 32 edges)
//            32/4000 = position in 1 revolution (position as a fraction of 1 revolution)
//   t2-t1/(200MHz/64), t2-t1= # of QCAPCLK cycles, and
//      QCAPCLK cycle = 1/(200MHz/64)
//                    = QCPRDLAT
//     // So:       6000 rpm = [32(200MHz/64)*60s/min]/[4000(t2-t1)]
//              t2-t1 = [32(200MHz/64)*60s/min]/(4000*6000rpm)           – Equation 5
//                    = 250 CAPCLK cycles = maximum (t2-t1) = SpeedScaler
//
// Divide both sides by (t2-t1), and:
//            1 = 32/(t2-t1) = [32(200MHz/64)*60 s/min]/(4000*6000rpm)]/(t2-t1)
//       Because (t2-t1) must be < 250 for QPOSCNT increment:
//               250/(t2-t1) < 1 for CW rotation
//       And because (t2-t1) must be >-250 for QPOSCNT decrement:
//                250/(t2-t1)> -1 for CCW rotation
//     //       eed_pr = 250/(t2-t1)
//             or [32(200MHz/64)*60 s/min]/(4000*6000rpm)]/(t2-t1)  – Equation 6
//
// To convert speed_pr to RPM:
// Multiply Equation 6 by 6000rpm:
// SpeedRpm_fr  = 6000rpm * [32(200MHz/64)*60 s/min]/[4000*6000rpm*(t2-t1)]
//              = [32(200MHz/64)*60 s/min]/[4000*(t2-t1)]
//              or [(32/4000)rev * 60 s/min]/[(t2-t1)(QCPRDLAT)]- Final Equation

计算低速的公式推导，懂的大神能否详细说明一下。