單片機波形發(fā)生器報告(存儲版)
【正文】 波形發(fā)生器的概述波形發(fā)生器：顧名思義，一個能產(chǎn)生各種波形的儀器。P1口連接按鍵鍵盤，作為波形的選擇和其他擴展功能的輸入口。 本設計任務設計要求： 一、基本功能： 1 、 可產(chǎn)生多種波形，如正弦波、三角波、鋸齒波、方波； 2 、 各種波形可通過按鍵選擇； 二、擴展功能： 1 、 可調(diào)節(jié)信號的頻率、占空比等參數(shù)； 2 、 其他自行增加的功能；擴展部分：按鍵ss6分別是控制方波的占空比增、減；ss8分別是控制波形的頻率增、減。（2）學習了如何使用DAC083LM339芯片和用Proteus 7 Professional仿真。（3）鋸齒波：也使用查表法。 硬件設計 設計圖示：仿真的總電路圖仿真的正弦波效果圖：方波的效果圖：鋸齒波的效果圖：三角波的效果圖：焊接出來的成品：所用到的電子元件介紹：（1）51單片機的內(nèi)部結構基本組成部分：1 一個8位的CPU2 128B或256B單元內(nèi)數(shù)據(jù)存儲器（RAM）3 4KB或8KB片內(nèi)程序存儲器（ROM或EPROM）4 4個8位并行I/O接口P0~P3。P0能夠用于外部程序數(shù)據(jù)存儲器，它可以被定義為數(shù)據(jù)/地址的第八位。P2口當用于外部程序存儲器或16位地址外部數(shù)據(jù)存儲器進行存取時，P2口輸出地址的高八位。 ALE/PROG：當訪問外部存儲器時，地址鎖存允許的輸出電平用于鎖存地址的地位字節(jié)。如果微處理器在外部執(zhí)行狀態(tài)ALE禁止，置位無效。 XTAL2：來自反向振蕩器的輸出。XFER：傳送控制信號，低電平有效。 nA (Typ)? Low Input Offset Voltage: 177。sbit s6=P1^5。uchar code sjtab[256]=//三角波的數(shù)字量表{0x00,0x02,0x04,0x06,0x08,0x0a,0x0c,0x0e,0x10,0x12,0x14,0x16,0x18,0x1a,0x1c,0x1e,0x20,0x22,0x24,0x26,0x28,0x2a,0x2c,0x2e,0x30,0x32,0x34,0x36,0x38,0x3a,0x3c,0x3e,0x40,0x42,0x44,0x46,0x48,0x4a,0x4c,0x4e,0x50,0x52,0x54,0x56,0x58,0x5a,0x5c,0x5e,0x60,0x62,0x64,0x66,0x68,0x6a,0x6c,0x6e,0x70,0x72,0x74,0x76,0x78,0x7a,0x7c,0x7e,0x80,0x82,0x84,0x86,0x88,0x8a,0x8c,0x8e,0x90,0x92,0x94,0x96,0x98,0x9a,0x9c,0x9e,0xa0,0xa2,0xa4,0xa6,0xa8,0xaa,0xac,0xae,0xb0,0xb2,0xb4,0xb6,0xb8,0xba,0xbc,0xbe,0xc0,0xc2,0xc4,0xc6,0xc8,0xca,0xcc,0xce,0xd0,0xd2,0xd4,0xd6,0xd8,0xda,0xdc,0xde,0xe0,0xe2,0xe4,0xe6,0xe8,0xea,0xec,0xee,0xf0,0xf2,0xf4,0xf6,0xf8,0xfa,0xfc,0xfe,0xfe,0xfc,0xfa,0xf8,0xf6,0xf4,0xf2,0xf0,0xee,0xec,0xea,0xe8,0xe6,0xe4,0xe2,0xe0,0xde,0xdc,0xda,0xd8,0xd6,0xd4,0xd2,0xd0,0xce,0xcc,0xca,0xc8,0xc6,0xc4,0xc2,0xc0,0xbe,0xbc,0xba,0xb8,0xb6,0xb4,0xb2,0xb0,0xae,0xac,0xaa,0xa8,0xa6,0xa4,0xa2,0xa0,0x9e,0x9c,0x9a,0x98,0x96,0x94,0x92,0x90,0x8e,0x8c,0x8a,0x88,0x86,0x84,0x82,0x80,0x7e,0x7c,0x7a,0x78,0x76,0x74,0x72,0x70,0x6e,0x6c,0x6a,0x68,0x66,0x64,0x62,0x60,0x5e,0x5c,0x5a,0x58,0x56,0x54,0x52,0x50,0x4e,0x4c,0x4a,0x48,0x46,0x44,0x42,0x40,0x3e,0x3c,0x3a,0x38,0x36,0x34,0x32,0x30,0x2e,0x2c,0x2a,0x28,0x26,0x24,0x22,0x20,0x1e,0x1c,0x1a,0x18,0x16,0x14,0x12,0x10,0x0e,0x0c,0x0a,0x08,0x06,0x04,0x02,0x00}。}void square(){ P2=0xff。 n=n+p。 n=0。 n=0。//確實一下是否真的松手了 } else if(s6==0)//占空比減小 {
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