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请问28335怎么生成一个复杂的pwm波?
它的高低电平情况是: 1111000 1111000 1111000 1111000 1111000 1111000 1111000 0001111 0001111 0001111 0001111 0001111 0001111 0001111 |
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12个回答
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加个变频的代码
最佳答案
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ePWM模块,看你需要哪种调制方式,都能输出啊
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这个变换这么多次 也能用pwm模块吗?不是只有zro prd cmpa cmpb可以用来动作吗? |
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客气了
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谢谢支持
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这是我中断部分里的,但是就是出不来波形,感觉原理上没出错,请版主帮忙看下 interrupt void epwm3_timer_isr(void) { EPwm3TimerIntCount++; switch(EPwm3Regs.CMPA.half.CMPA) { case 1435: EPwm3Regs.CMPA.half.CMPA=3348; break; case 3348: EPwm3Regs.CMPA.half.CMPA=4783; break; case 4783: EPwm3Regs.CMPA.half.CMPA=6696; break; case 6696: EPwm3Regs.CMPA.half.CMPA=8131; break; case 8131: EPwm3Regs.CMPA.half.CMPA=10044; break; case 10044: EPwm3Regs.CMPA.half.CMPA=11479; break; case 11479: EPwm3Regs.CMPA.half.CMPA=13392; break; case 13392: EPwm3Regs.CMPA.half.CMPA=14827; break; case 14827: EPwm3Regs.CMPA.half.CMPA=16740; break; case 16740: EPwm3Regs.CMPA.half.CMPA=18175; break; case 18175: EPwm3Regs.CMPA.half.CMPA=20088; break; case 20088: EPwm3Regs.CMPA.half.CMPA=21522; break; case 21522: EPwm3Regs.CMPA.half.CMPA=25349; break; case 25349: EPwm3Regs.CMPA.half.CMPA=26783; break; case 26783: EPwm3Regs.CMPA.half.CMPA=28697; break; case 28697: EPwm3Regs.CMPA.half.CMPA=30131; break; case 30131: EPwm3Regs.CMPA.half.CMPA=32045; break; case 32045: EPwm3Regs.CMPA.half.CMPA=33479; break; case 33479: EPwm3Regs.CMPA.half.CMPA=35392; break; case 35392: EPwm3Regs.CMPA.half.CMPA=36827; break; case 36827: EPwm3Regs.CMPA.half.CMPA=38740; break; case 38740: EPwm3Regs.CMPA.half.CMPA=40175; break; case 40175: EPwm3Regs.CMPA.half.CMPA=42088; break; case 42088: EPwm3Regs.CMPA.half.CMPA=43523; break; case 43523: EPwm3Regs.CMPA.half.CMPA=45436; break; case 45436: EPwm3Regs.CMPA.half.CMPA=1435; break; default: break; } EPwm3Regs.ETCLR.bit.INT = 1; PieCtrlRegs.PIEACK.all = PIEACK_GROUP3; } |
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EPwm3Regs.AQCTLA.bit.ZRO=AQ_SET; EPwm3Regs.AQCTLA.bit.CAU=AQ_TOGGLE; EPwm3Regs.ETSEL.bit.INTSEL = ET_CTRU_CMPA; EPwm3Regs.ETSEL.bit.INTEN = PWM3_INT_ENABLE; EPwm3Regs.ETPS.bit.INTPRD = ET_1ST; 这是我pwmsetup里的 是增计数模式 |
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客气了
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谢谢支持
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一起加油
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