作者 Sam Wang & River Liang    说明,本文对比8位MCU的位操作在系统升级到M0+内核的MCU后所带来的影响,可以作为客户对升级MCU时,对代码及RAM上资源的评估使用. 一)简单的I/O翻转对比.对比条件:MCU为MC9S08PT与KE02,开发平台CW10.3                   1.使用PT的代码如下:          if (!PORT_PTAD_PTAD0){             PORT_PTAD_PTAD0=1;         }else{             PORT_PTAD_PTAD0=0;}       PT的代码编译后占用9个Byte。                000002D4 000004   BRSET  0,PORT_PTAD,PORT_PTAD                000002D7 1000     BSET   0,PORT_PTAD                000002D9 202E     BRA    *+48       ;abs = 0x0309                000002DB 1100     BCLR   0,PORT_PTAD       2.KE是基于ARM的M0+内核，使用的代码如下                  if(GPIOA_PDOR & 0x1)                     { GPIOA_PCOR = 0x1;}                  else                     { GPIOA_PSOR = 0X1;}       编译后结果为                00000706:   ldr r3,[pc,#24]                00000708:   ldr r2,[r3,#0]                0000070a:   movs r3,#1                0000070c:   ands r3,r2                0000070e:   beq main+0x6c (0x718)       ; 0x00000718                00000710:   ldr r3,[pc,#12]                00000712:   movs r2,#1                00000714:   str r2,[r3,#8]                00000716:   b main+0x20 (0x6cc)       ; 0x000006cc                00000718:   ldr r3,[pc,#4]                0000071a:   movs r2,#1                0000071c:   str r2,[r3,#4]       这段M0+内核的代码编译后占用24个Byte       3.KE系列是Freescale在M0+的基础上加入了位操作引擎BME，用以优化ARM内核的位操作性能，使用BME功能的代码如下                     #define PTA0_SET   (void) (*((volatile unsigned char *)(0x4C000000+(0<<21)+0xF004))) //?==0                                                       //LAS1      第0位       GPIOA_PSOR地址的A0-A19                     #define PTA0_CLR   (void)(*((volatile unsigned char *)(0x4C000000+(0<<21)+0xFF008)))                                                      //LAS1      第0位       GPIOA_PCOR地址的A0-A19                     #define PTA0                *((volatile unsigned char *)(0x50000000+(0<<23)+(0<<19)+0xF000))                                                     //UBFX           第0位     1位           GPIOA_PDOR 地址的A0-A18                if (!(PTA0))                     {PTA0_SET; }                else                     {PTA0_CLR;}           KE的BME代码编译结果如下：                    165                if (!(PTA0)){                 00000998:   ldr r3,[pc,#24]                0000099a:   ldrb r3,[r3,#0]                0000099c:   uxtb r3,r3                0000099e:   cmp r3,#0                000009a0:   bne RTC_IRQHandler+0x18 (0x9a8); 0x000009a8                  166                PTA0_SET;                                             //Using BME                000009a2:   ldr r3,[pc,#20]                000009a4:   ldrb r3,[r3,#0]                000009a6:   b RTC_IRQHandler+0x1c (0x9ac); 0x000009ac                  168                PTA0_CLR;      //Using FASTER GPIO                000009a8:   ldr r3,[pc,#16]                000009aa:   ldrb r3,[r3,#0]       代码编译后占用20个Byte         4, CW里面有设置可以优化C编译器，具体路径在Project->Proteries->C/C++ Build->Setting->GCC C Complier->Optimization           优化后共用16个Byte                  165         if (!(PTA0)){                 0000091e:   ldr r3,[pc,#20]                00000920:   ldrb r3,[r3,#0]                00000922:   cmp r3,#0                00000924:   bne RTC_IRQHandler+0x12 (0x92a); 0x0000092a                  166                         PTA0_SET;                                             //Using BME                00000926:   ldr r3,[pc,#16]                00000928:   b RTC_IRQHandler+0x12 (0x92c); 0x0000092c                  168                         PTA0_CLR;      //Using FASTER GPIO                0000092a:   ldr r3,[pc,#16]                0000092c:   ldrb r3,[r3,#0]       5, 结果     如果单纯靠M0+内核访问寄存器，KE代码的占用空间与PT的比为24：9        如果使用KE的BME功能，代码与PT的比为16:9（使用了BME）        在判断Bit时, KE使用代码与PT的比为8:3        单单设置一个Bit时KE与PT代码占比为4:2 因此在M0+等ARM核上进行位操作，其效率比8位单片机低，使用了BME功能后，可以有效提高位操作的性能。 二)典型变量的位操作. 对比条件:MCU为MC9S08PT与KE02,开发平台CW10.3                测试代码:if (xx&1){              xx&=0xFE;       }else{             xx|=1;}       1,设置XX在0 page时,其与上面的I/O翻转结果一样,代码为9个BYTES    2,在KE中，编译结果如下,设置优化前,需要52个Bytes的代码量,26个执行周期.                                                     if (xx&1){                00000a52:   ldr r3,[pc,#64]                00000a54:   ldrb r3,[r3,#0]                00000a56:   uxtb r3,r3                00000a58:   mov r2,r3                00000a5a:   movs r3,#1                00000a5c:   ands r3,r2                00000a5e:   uxtb r3,r3                00000a60:   cmp r3,#0                00000a62:   beq main+0x5a (0xa76)       ; 0x00000a76                     200                         xx&=0xFe;                00000a64:   ldr r3,[pc,#44]                00000a66:   ldrb r3,[r3,#0]                00000a68:   uxtb r3,r3                00000a6a:   movs r2,#1                00000a6c:   bics r3,r2                00000a6e:   uxtb r2,r3                00000a70:   ldr r3,[pc,#32]                00000a72:   strb r2,[r3,#0]                     203         }}                00000a74:   b main+0x28 (0xa44)       ; 0x00000a44                     202                         xx|=1;                00000a76:   ldr r3,[pc,#28]                00000a78:   ldrb r3,[r3,#0]                00000a7a:   uxtb r3,r3                00000a7c:   movs r2,#1                00000a7e:   orrs r3,r2                00000a80:   uxtb r2,r3                00000a82:   ldr r3,[pc,#16]                00000a84:   strb r2,[r3,#0]                     203         }}       3, 设置优化后,需要22/20个Bytes的代码量,11/10个执行周期.                ldr r3,[pc,#40]                     199         if (xx&1){                0000095e:   movs r2,#1                     197         xx++;                00000960:   ldrb r1,[r3,#0]                00000962:   adds r1,#1                00000964:   uxtb r1,r1                00000966:   strb r1,[r3,#0]                     199         if (xx&1){                00000968:   ldrb r1,[r3,#0]                0000096a:   tst r1,r2                0000096c:   beq main+0x34 (0x974)       ; 0x00000974                     200                         xx&=0xFe;                0000096e:   ldrb r1,[r3,#0]                00000970:   bics r1,r2                00000972:   b main+0x34 (0x978)       ; 0x00000978                     202                         xx|=1;                00000974:   ldrb r1,[r3,#0]                00000976:   orrs r1,r2                00000978:   strb r1,[r3,#0]                0000097a:   b main+0x20 (0x960)       ; 0x00000960 如果采用以空间换时间的话,其参考代码如下.                 if (xx==0){                                 xx=1;                 }else{                                 xx=0;  }       4, 如考虑中断嵌套的话,还令需要4个Byte代码。       5, 结果 KE使用代码与PT的比为至少为20:9。      在判断Bit时, KE使用代码与PT的比为8:3.       6,使用BYTE替换Bit, 编译结果,设置优化前,需要22个BYTES.                     197         if (xx==0){                000009e8:   ldr r3,[pc,#44]                000009ea:   ldrb r3,[r3,#0]                000009ec:   cmp r3,#0                000009ee:   bne main+0x38 (0x9f8)       ; 0x000009f8                     198                         xx=1;                000009f0:   ldr r3,[pc,#36]                000009f2:   movs r2,#1                000009f4:   strb r2,[r3,#0]                000009f6:   b main+0x3e (0x9fe)       ; 0x000009fe                     200                         xx=0;                000009f8:   ldr r3,[pc,#28]                000009fa:   movs r2,#0                000009fc:   strb r2,[r3,#0]    7,设置优化后,需要16/14个BYTES的代码量.                     197         xx++;                0000095c:   ldr r3,[pc,#36]                     202                         xx=1;                0000095e:   movs r1,#1                     197         xx++;                00000960:   ldrb r0,[r3,#0]                00000962:   adds r0,#1                00000964:   uxtb r0,r0                00000966:   strb r0,[r3,#0]                     199         if (xx){                00000968:   ldrb r0,[r3,#0]                0000096a:   cmp r0,#0                0000096c:   beq main+0x32 (0x972)       ; 0x00000972                     200                         xx=0;                0000096e:   strb r2,[r3,#0]                00000970:   b main+0x20 (0x960)       ; 0x00000960                     202                         xx=1;                00000972:   strb r1,[r3,#0]                00000974:   b main+0x20 (0x960)       ; 0x00000960       8, 结果 ,在RAM的空间允许的情况下,KE使用代码与PT的比为至少为12:9. 三) 8 bit变量加1       1,在PT中对8 bit变量加1,只需要4个BYTES.    24:                 XX++; 00000014 450000   LDHX   #XX       00000017 7C       INC    ,X       2,M0+的8 bit变量加1,设置优化前,需要14个BYTES                     197         xx++;                00000a44:   ldr r3,[pc,#48]                00000a46:   ldrb r3,[r3,#0]                00000a48:   uxtb r3,r3                00000a4a:   adds r3,#1                00000a4c:   uxtb r2,r3                00000a4e:   ldr r3,[pc,#40]                00000a50:   strb r2,[r3,#0]       3,而如果使用优化设置,那么要12个BYTES                     197         xx++;                0000095c:   ldr r3,[pc,#36]                     202                         xx=1;                0000095e:   movs r1,#1                     197         xx++;                00000960:   ldrb r0,[r3,#0]                00000962:   adds r0,#1                00000964:   uxtb r0,r0                00000966:   strb r0,[r3,#0]       4, 结果 , 在8 bit变量加1时,KE使用代码与PT的比为至少为12:4，但这是32bitARM内核操作8bit变量都普遍存在效率变低的现象。 四) 16位+8位加法       1, 8 bit 编译结果,需要8个BYTES.                0000008 320000    LDHX   xx                0000000B AF01     AIX    #1                0000000D 960000   STHX   xx       2, M0+ 编译结果,设置优化前,需要10个BYTES.                00000a44:   ldr r3,[pc,#44]                00000a46:   ldr r3,[r3,#0]                00000a48:   adds r2,r3,#1                00000a4a:   ldr r3,[pc,#40]                00000a4c:   str r2,[r3,#0].       3, M0+ 编译结果,设置优化后,需要8个BYTES.                0000095c:   ldr r3,[pc,#20]                0000095c:   ldr r3,[pc,#20]                0000095e:   ldr r2,[r3,#0]                00000960:   adds r2,#1                00000962:   str r2,[r3,#0]       4,结果,M0+在16位加法时能够达到8bit单片机的效率，结果相同. 五)结论     因此用户在移植PT(或其它8 bit MCU)代码到KE02时,要选型时需要充分考虑客户原先代码具体运算情况,理论上存在使用KE后代码变大的情况.   但是使用KE等32bitM0+内核时可以在16bit或以上的乘、加运算时获得更好的效率，占用更小的代码空间和运算时间。   另外KE对GPIO的控制寄存器比PT多了一些功能,可以一次操作多个I/O,是不错的功能.

在KE系列MCU中提供了多种寄存器用于实现GPIO的控制：    -PDOR寄存器，用于写入或读取IO的输出状态    -PSOR寄存器，用于置位IO口    -PCOR寄存器，用于清零IO口    -PTOR寄存器，用于翻转IO口    -PDIR寄存器，用于读取IO口的输入状态 当我们想要将PTA0置1时，有多种方法可以选择：    1. 直接操作寄存器，PDOR或PSOR都可以实现：          GPIOA->PDOR |= 0x0001；          GPIOA->PSOR |= 0x0001；          直接操作寄存器的效率更高，但可读性较差。    2. 使用官方的库函数操作       GPIO_PinSet(GPIOA, GPIO_PTA0);       库函数的可读性很好，但显得有些啰嗦，字符较多。 通过KE的BME来实现GPIO的操作能够很好的解决上面的问题，只用将附件中的头文件gpio_bitdef.h包含到工程里，再调用里边的宏定义就可以了。 对PTA0置位和清零可以使用下面的语句： POUTA0 = -1; POUTA0 = 0; 读取PTA0的输入状态则可以使用： tmp = PINA0; 上面的语句是不是看上去简洁了很多呢。 实际上上面GPIO的读写指令，是通过BME的BFI（位域插入）和BFX（位域提取）指令来实现的。 -其中ADDR是存储空间内的地址，我们最终操作的还是GPIO的寄存器，因此在两个指令中分别取GPIOA的PDOR寄存器地址和PDIR寄存器地址。 -bit则表示需要插入或提取位域的起始位置，由于这里是PTA0，PTA0位于寄存器的最低位，因此这里填入了0。 -width则表示需要插入或提取位域的宽度，我们只对单个管脚进行操作，也就是单个位进行操作，宽度自然就是1了。 需要注意的是，BFI（位域插入）指令在插入时，是将对应位插入到目的地址。因此，如果直接为POUTxx赋值为1的话，有可能出现错误。 POUTA0 = 0x01;//正确 POUTA1 = 0x02;//正确 POUTA2 = 0x04;//正确 POUTA1 = 0x01;//错误 为了避免这种情况，我们可以在IO口需要置位时，直接将POUTxx赋值为-1，即0xFFFF FFFF，这样保证了每一位的值都为1。 #define BME_BFI(ADDR,bit,width)        (*(volatile uint32_t *)((((uint32_t)ADDR&0xFFFF))   \                                   | (5 <<28)  \                                   | ((bit)<<23) | ((width-1))<<19)) #define BME_BFX(ADDR,bit,width)        (*(volatile uint32_t *)(((uint32_t)ADDR&0xFFFF)    \                                   | (5 <<28)  \                                   | ((bit)<<23) | ((width-1))<<19))  #define POUTA0 BME_BFI(&GPIOA->PDOR,0,1) #define PINA0 BME_BFX(&GPIOA->PDIR,0,1) ‍‍‍‍‍‍‍‍‍‍

1 Abstract Stepper motor can be controlled by the electrical pulse signal with the open loop system, it use the electrical pulse signal realize the angular movement or linear movement.  The speed and position of the stepper motor is determined by the pulse frequent and the pulse number. Stepper motor can be used in the low speed area application, with higher work efficiency and low noise. KE02 is the 5V kinetis E series MCU, it is based on ARM Cortex M0+ core, KE series are designed to maintain high robustness for complex electrical noise environment and high reliability application. For these advantages, KE02 is fit the Stepper motor control application. This document is mainly about how use the KE02 realize the Stepper motor speed, step and direction control. It can use the UART in the PC to control the Stepper motor speed. The following picture is the control diagram.                                                                              Fig.1 2. Motor control parameter calculation      Just as Fig.1 shows, KE02 should control the EN, DIR, PWM signal to the motor driver, then realize the stepper motor control. EN is the motor driver enable signal, 0 is enable, 1 is disable; DIR is the stepper motor direction control, 0, clockwise, 1 anticlockwise; PWM is the pulse signal to control the step and speed for the stepper motor.       Stepper motor is 1.8’, it means a round have 360’/1.8’= 200 steps. But because the Motor driver have the divider, it is 32, so one stepper motor round should have 200*32 = 6400 steps.       KE02 system, it use the external 10Mhz crystal, and configure both core and bus frequent to 20Mhz,  it use FTM0 module as the motor pulse generate module, bus clock with 32 prescale used as the FTM0 clock source, choose up counter. If need to change the motor speed and control step, just control the FTM PWM frequent and PWM counter. For Stepper motor, one FTM period means one motor step. From the reference manual of KE02, we get that, the FTM period in up counting mode is: (MOD-CNTIN+1)*period of the FTM counter clock, if want to change the frequent of motor, just calculate the MOD of FTM is ok, then count the number of the FTM cycle, now assume CNTIN =0, then: Tftm= (32/20Mhz)*(MOD+1) From the Stepper Motor and it’s driver, we get that one step time is : Tmstep= 60/(V*6400) V is the speed of Motor, the unit is round/minute. Because Tftm=Tmstep, then we know: MOD= (60/(V*6400))*(20Mhz/32)-1                     (F1) In this document, we calculate the speed of 150 round/minute, 110 round/minute, 80 round/minute, 50 round/minute and 0.1 round/minute, according to (F1), we can get the MOD for each speed as the following: 150 round/minute   MOD=38 110 round/minute   MOD=52 80 round/minute     MOD=72 50 round/minute     MOD=116 0.1 round/minute    MOD=58592 If each speed need to do 10 Stepper motor round, then just control the speed counter number to: 10*6400=64000. 3. MCU pin assignment PTF0 : DIR PTF1 : EN PTA0 : PWM PTC6 : UART1_RX PTC7 : UART1_TX 4. code writing (1)FTM initial code void STEEPMOTOR_PWM_Init(uint16 MODdata) {                    SIM_SCGC |= SIM_SCGC_FTM0_MASK;                 FTM0_SC = 0;                 FTM0_C0SC = 0 ;                 FTM0_C0SC = FTM_CnSC_MSB_MASK |FTM_CnSC_ELSA_MASK ;                 FTM0_C0V = 0;                 FTM0_C0V = MODdata>>1;                 FTM0_MOD=MODdata;                 FTM0_SC |=FTM_SC_CLKS(1) | FTM_SC_PS(5) | FTM_SC_TOIE_MASK                 enable_irq(17); //enable interrupt } MODdata can choose the different Stepper motor speed, eg, 150 round/minute, MODdata is 38. (2) interrupt service function void FTM0_IRQHandler(void) {                                 FTM0_SC  &= ~FTM_SC_TOF_MASK;//clear the TOF flag. roundcount++;                 if(roundcount >= 64000) {FTM0_C0SC = 0x00; FTM0_SC &= ~(FTM_SC_TOIE_MASK);} } It can used for the step counter, and when reach the 10round, then stop the motor( stop the FTM output). (3)Speed choose with UART input void Motor_Speed_GPIO_CTRL_30round(void) {                 char motormode=0;                 uint32 COMPDATA=0;                                 printf("\n 1 for 150 round/minute\n\r");                 printf("\n 2 for 110 round/minute\n\r");                 printf("\n 3 for 80 round/minute\n\r");                    printf("\n 4 for 50 round/minute\n\r");                    printf("\n 5 for 0.1 round/minute\n\r");                 motormode = UART_getchar(PC_TERM_PORT);                                     switch(motormode)                                 {                                    case '1':                                                       STEEPMOTOR_PWM_Init(38);//150 round/minute                                                         break;                                   case '2':                                                       STEEPMOTOR_PWM_Init(52);//110 round/minute                                                         break;                                   case '3':                                                                       STEEPMOTOR_PWM_Init(72);//80 round/minute                                                          break;                                   case '4':                                                                       STEEPMOTOR_PWM_Init(116);//50 round/minute                                                         break;                                   case '5':                                                                       STEEPMOTOR_PWM_Init(58592);//0.1 round/minute                                                         break;                                                                                default: break;                                 }                                 while( roundcount < 64000 ) {} //10 round                                 Disable_PWM;                                 printf("\n %c 10round PWM is finished ", motormode);                                 roundcount=0; } 5 DEMO About the test code, please find it from the attachment.