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需要设计一个模块来实现ADC给定的12位数字输出的平方。
以上来自于百度翻译 以下为原文 It is desired to design a module to achieve the square of a given 12 bit digital output from the adc |
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由于ADC的输出不是一个12位总线,而是包含在两个寄存器中,所以您首先需要一些机制来将该值获取到组件中,通常是DMA的工作。
平方的几个选择,最简单的第一: EOC信号触发你的组件(包括一个中断和一些软件)来获得这个值并计算平方。这将只在几个kHz的低采样率下工作。 这里是HTTP://www. CyPress?应用=论坛&;ID = 2492 = &;摆脱76712you找到一些信息如何访问PSoC内部组件,“乘累加”可以在18(平方!!!!)时钟周期,但是是相当复杂的修改你的需要。 最后的建议是使用Verilog程序数据通路来做这项工作。这可以做任务(移位加)在14到20个时钟周期(猜测)。 获取DMA的示例、视频和附注 为组件创建视频 当你决定要采取“硬”的方式使用硬件(笑)看“通路” 在本页的右上角是一个关键词的搜索会让你一堆文件和大量的用户的帖子告诉你可能会遇到的麻烦。(再次微笑) 所以,你必须要做的第一件事就是读了很多! 鲍勃 以上来自于百度翻译 以下为原文 Since the output of an ADC is not a 12-bit bus but contained in two registers, you first need some mechanism to get that value into your component which is usually a job for DMA. Several choices for squaring, easiest first: The eoc-signal triggers your component (consisting of an interrupt and some software) to get that value and calculate the square. This will work only at lower sample-rates as some few kHz. Here http://www.cypress.com/?app=forum&id=2492&rID=76712 you find some informations how to access a PSoC-internal component, a "multiply and accumulator" which could do the squaring within 18(!!!) clock cycles but is quite more complex to modify to your needs. Last suggestion is to use VeriLog and program a datapath to do that job. This could do that task (shift and add) in 14 to 20 (guessed) clock cycles. Get the examples, videos and appnotes for DMA Get videos for component creation When you decide to take "The Hard Way" using hardware (smile) look at "DataPath" In the right-hand upper corner of this page is a keyword search which wil lead you to a bunch of documents and a large number of user-posts telling where you might run into troubles. (smile again) So, what you've got to do first is to read a lot! Bob |
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是什么原因想要在硬件中做到这一点?做你想做的事与方后来什么?PSoC3中的ADC不是那么快,为什么软件不够?
以上来自于百度翻译 以下为原文 What is the reason to want to do this in hardware? What do you want to do with the square later on? The ADC in the PSoC3 isn't _that_ fast, why is software not sufficient? |
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我猜你想把它发送到一个DAC。否则,你可以使用软件的采样时间之间执行乘法。
如果输入和输出并不是一个大问题之间的延迟,你能做的multpliation SW并把它发送到DAC。只是一个想法。 以上来自于百度翻译 以下为原文 My guess is that you want to send it out to a DAC. otherwise, you can use SW to perform the multiplication between the sampling time. If the delay between the input and output is not a big issue, you can do the multpliation in SW and send it out to the DAC. Just an idea. |
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所以每个人都不必继续猜测,你在尝试什么
计算,多项式,rms,速度你需要,甚至应用 看另一种方法是否更直接有效。 问候,Dana。 以上来自于百度翻译 以下为原文 So everyone does not have to keep guessing, what are you trying to compute, polynomial, RMS, speed you need, even application to see if another approach more direct and efficient. Regards, Dana. |
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是的,水晶球今年很便宜,但是性能很差(宽泛的微笑)
鲍勃 以上来自于百度翻译 以下为原文 Yeah, crystal balls are cheap this year, but poor performance (Broad Smile) Bob |
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几个月前,一些公司销售水晶标示为澳大利亚香港水晶重贴现奥地利水晶。-)
以上来自于百度翻译 以下为原文 A few months ago, some company sell crystal labeled as AUSTRALIAN CRYSTAL in Hongkong as heavyly discounted AUSTRIAN CRYSTAL. :-) |
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我的朋友有一件衬衫,上面写着“奥地利没有袋鼠”。好像是同一家商店。
鲍勃 以上来自于百度翻译 以下为原文 My friend has got a shirt with "There are no Kangaroos in Austria" on it. Seems to be the same shop. Bob |
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在奥地利凯恩霍夫动物园有袋鼠,不知道他们是否仍然有。
以上来自于百度翻译 以下为原文 Kernhof Zoo in Austria had some kangaroo. |
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