文件名称:ht6221
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1 解码的关键是如何识别0 和 1 从位的定义我们可以发现0 1 均以 0.56ms
的低电平开始不同的是高电平的宽度不同0 为 0.56ms, 1 为 1.68ms,所以
必须根据高电平的宽度区别0 和 1 如果从 0.56ms 低电平过后开始延时
0.56ms 以后 若读到的电平为低说明该位为0 反之则为1 为了可靠起
见 延时必须比0.56ms 长些 但又不能超过1.12ms,否则如果该位为0 读到的
已是下一位的高电平因此取1.12ms+0.56ms /2=0.84ms 最为可靠一般取0.84ms
左右均可
2 根据码的格式应该等待9ms 的起始码和4.5ms 的结果码完成后才能读码
3 从上述两点我们可得到解码程序的流程图-A decoder, the key is how to identify 0 and 1 from the definition of bits we can find 01 are 0.56ms
The difference is that the beginning of the low-level high the width of the different 0 for 0.56ms, 1 is 1.68ms, so
Must be based on the width of the high difference between 0 and 1, if, after starting from the low latency 0.56ms
0.56ms later read that if a lower level shows that the bit is 0 the other hand, compared with one in order to reliably from
See Delay must be longer than 0.56ms but no more than 1.12ms, otherwise if the bit is 0 read
Is next to high so he 1.12ms+0.56 ms/2 = 0.84ms the most reliable in general take 0.84ms
So can
2 According to the format code should wait until the beginning of 9ms code and 4.5ms after the completion of the result code before reading
3 from the above two points that we receive decoding process flow chart
的低电平开始不同的是高电平的宽度不同0 为 0.56ms, 1 为 1.68ms,所以
必须根据高电平的宽度区别0 和 1 如果从 0.56ms 低电平过后开始延时
0.56ms 以后 若读到的电平为低说明该位为0 反之则为1 为了可靠起
见 延时必须比0.56ms 长些 但又不能超过1.12ms,否则如果该位为0 读到的
已是下一位的高电平因此取1.12ms+0.56ms /2=0.84ms 最为可靠一般取0.84ms
左右均可
2 根据码的格式应该等待9ms 的起始码和4.5ms 的结果码完成后才能读码
3 从上述两点我们可得到解码程序的流程图-A decoder, the key is how to identify 0 and 1 from the definition of bits we can find 01 are 0.56ms
The difference is that the beginning of the low-level high the width of the different 0 for 0.56ms, 1 is 1.68ms, so
Must be based on the width of the high difference between 0 and 1, if, after starting from the low latency 0.56ms
0.56ms later read that if a lower level shows that the bit is 0 the other hand, compared with one in order to reliably from
See Delay must be longer than 0.56ms but no more than 1.12ms, otherwise if the bit is 0 read
Is next to high so he 1.12ms+0.56 ms/2 = 0.84ms the most reliable in general take 0.84ms
So can
2 According to the format code should wait until the beginning of 9ms code and 4.5ms after the completion of the result code before reading
3 from the above two points that we receive decoding process flow chart
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