如何在斯巴达3ADSP中找到64位数据的平方根

好的，所以你的输入格式是64.0，你想要32.4格式输出正确吗？
我怀疑你在网上找不到你想要的东西，所以你需要找到
这样的例子并修改它们以满足您的确切需求。
例如，这是一个8.0输入，4.4输出样本设计，显示了一种方法： 
http://www.engr.usask.ca/classes/EE/431/Verilog%20Files/better_square_root.v
您需要将设计扩展到所需的宽度，然后展开逻辑以达到您的要求
延迟要求。
这是思考的另一个例子。
同样，每个输出需要大约1个时钟周期
位（最糟糕的情况）。
您可以展开循环 
http://academic.csuohio.edu/yuc/comp-f08/sqrt-virtex.v
这是另一个页面，您可以使用36.0输入平方根算法
去适应。
它纯粹是组合的。
您需要将其修改为64位，产生32.4输出。
所以，假设你找不到更好的起点......一个方法就是对待你的
输入为64.8输入格式，即用8个零填充它。
执行整数平方根后，
你应该得到32.4输出。
一旦你使核心算法工作，看看你如何管道它以改善它的时钟
达到延迟目标时的频率，即分散6个流水线阶段。
祝你好运，希望你能找到更多建议！
约翰普罗塞纳
在原帖中查看解决方案



以下为原文

OK, so your input format is 64.0 and you want 32.4 format ouput correct?
 
I suspect you you won't find exactly what you want on the web, so you'll need to find
so examples and modify them to fit your exact needs.  
 
For example, here is an 8.0 input, 4.4 output sample design that shows one approach:
    http://www.engr.usask.ca/classes/EE/431/Verilog%20Files/better_square_root.v
You'd need to extend the design to the width you want and unroll the logic to reach your
latency requirement.
 
Here's another example for thought.  Again, it will take about 1 clock cycle per output
bit (worst case).  You may be able to unroll the loop
  http://academic.csuohio.edu/yuc/comp-f08/sqrt-virtex.v
 
Here's another page that has a sample 36.0 input square-root algorithm that you may be able
to adapt.  it is purely combinatorial.  You'd need to modify it to be 64 bits producing a 32.4 output.
 
So, assuming you can't find any better starting points....  One aproach would be to treat your
input as a 64.8 input format, ie, pad it with 8 zeroes.  After you perform an integer square root,
you should get a 32.4 output.
 
Once you get the core algorithm working, look into how you can pipeline it to improve it's clock
frequency while meeting your latency goal, ie, scatter 6 pipeline stages in it.
 
Good luck, I hope you can find more suggestions!
 
John Providenza
 
View solution in original post

你能提供更多信息吗？
- 什么是数据格式 - 二进制还是实数？
- 语言偏好 -  Verilog或VHDL
- 需要时钟速度
- 允许的延迟
约翰普罗塞纳



以下为原文

Can you provide more info?
 
- what is the data format - binary or real?
- language preference - Verilog or VHDL
- clock speed required
- permissable latency
 
John Providenza

数据是二进制格式，我使用的是vhdl，我的时钟源是125 MHz Raj



以下为原文

data is in binary format ,i am using vhdl and my clock souce is of 125 MHzRaj

如果你真的需要帮助，你需要在你的要求中多做一些工作。
好的，你有二进制数。
你需要一个整数结果，即floor（sqrt（n））或
你需要一些小数位吗？
如果是这样，你想要多少分数位？
什么样的延迟？
你需要在一个时钟周期内得到答案吗？
多个周期？
您是否使用Google查找内核？
那里有很多信息。
我找到了一个Verilog核心，可以作为设计的弹簧板，取决于你的
需要。
我在过去构建了一个32位sqrt功能，具有6个时钟延迟和一个
尽管每个时钟输出1次。
那么你的需求是什么？
约翰普罗塞纳



以下为原文

You need to put a little more work into your requirements if you really want help.
 
OK, you have a binary number.  Do you need  an integer resul, ie, floor(sqrt(n)) or
do you need some fractional bits?  If so, how many fraction bits do you want?
 
What kind of latency?  Do you need the answer in one clock cycle or can it take
multiple cycles?
 
Have you used Google to look for cores? There's a lot of information available there.
I found a Verilog core that could be a spring board for a design, depending on your
needs.  I've built a 32 bit sqrt function in the past that had a 6 clock latency with a
thoughput of 1 operation per clock.
 
So what are all your needs?
 
John Providenza

海约翰
感谢您的回复
在这种情况下，我想要squarerootof 64位二进制值与4分数位。
在输出的情况下，我可以等待最多5个时钟周期（时钟频率为125 MHz）。为了找到64位数据的平方根，我经历了不同的网站和书籍，我设计了一个程序，但没有得到输出。
该逻辑的问题在于没有资源来处理64位数据。
我在xilinx网站上看到了一个示例代码，找到了32位的sqrt。
根据我的观点，逻辑不适合这种情况。
这就是为什么我在这个论坛上发布我的问题。
我想现在有了这个概念，希望你能帮助我。
拉吉
拉吉



以下为原文

hai John
thank you for reply
 
in this case i want squarerootof 64 bit binary value  with 4 fraction bits. in the case of output i can wait up to 5 clock cycles(clock frequency is 125 MHz).for finding squareroot of 64 bit data i went through different sites and books and  i designed one program but didnt get the output. the problem with that logic is that there is no resource to handle 64bit data .  I saw a example code in xilinx website to find sqrt of 32 bit no.  as per my concept that logic is not suitable for this case . that why i posted my problem in this forum. i think now got the concept and hope you can help me.
 
raj
Raj

好的，所以你的输入格式是64.0，你想要32.4格式输出正确吗？
我怀疑你在网上找不到你想要的东西，所以你需要找到
这样的例子并修改它们以满足您的确切需求。
例如，这是一个8.0输入，4.4输出样本设计，显示了一种方法： 
http://www.engr.usask.ca/classes/EE/431/Verilog%20Files/better_square_root.v
您需要将设计扩展到所需的宽度，然后展开逻辑以达到您的要求
延迟要求。
这是思考的另一个例子。
同样，每个输出需要大约1个时钟周期
位（最糟糕的情况）。
您可以展开循环 
http://academic.csuohio.edu/yuc/comp-f08/sqrt-virtex.v
这是另一个页面，您可以使用36.0输入平方根算法
去适应。
它纯粹是组合的。
您需要将其修改为64位，产生32.4输出。
所以，假设你找不到更好的起点......一个方法就是对待你的
输入为64.8输入格式，即用8个零填充它。
执行整数平方根后，
你应该得到32.4输出。
一旦你使核心算法工作，看看你如何管道它以改善它的时钟
达到延迟目标时的频率，即分散6个流水线阶段。
祝你好运，希望你能找到更多建议！
约翰普罗塞纳



以下为原文

OK, so your input format is 64.0 and you want 32.4 format ouput correct?
 
I suspect you you won't find exactly what you want on the web, so you'll need to find
so examples and modify them to fit your exact needs.  
 
For example, here is an 8.0 input, 4.4 output sample design that shows one approach:
    http://www.engr.usask.ca/classes/EE/431/Verilog%20Files/better_square_root.v
You'd need to extend the design to the width you want and unroll the logic to reach your
latency requirement.
 
Here's another example for thought.  Again, it will take about 1 clock cycle per output
bit (worst case).  You may be able to unroll the loop
  http://academic.csuohio.edu/yuc/comp-f08/sqrt-virtex.v
 
Here's another page that has a sample 36.0 input square-root algorithm that you may be able
to adapt.  it is purely combinatorial.  You'd need to modify it to be 64 bits producing a 32.4 output.
 
So, assuming you can't find any better starting points....  One aproach would be to treat your
input as a 64.8 input format, ie, pad it with 8 zeroes.  After you perform an integer square root,
you should get a 32.4 output.
 
Once you get the core algorithm working, look into how you can pipeline it to improve it's clock
frequency while meeting your latency goal, ie, scatter 6 pipeline stages in it.
 
Good luck, I hope you can find more suggestions!
 
John Providenza
 

海约翰
谢谢你
拉吉



以下为原文

hai john
 thank u
Raj