How to take the floating point adder output
I have a piece of code in VHDL here : Floating Point(IP Core) Adder.
I want to take the result_fp to the output signal test. The result_fp is not affecting the test signal.
Line 142 : test <= STD_LOGIC_VECTOR(b_fp);
I have tried with
test <= STD_LOGIC_VECTOR(result_fp);
as well as
test <= result_fp;
Not working.
How can I solve this.
clockIn input is fed with a square wave of 50 ns period.
clk_fp is fed with a square wave of 10 ns period(faster clock).
Timing diagram:
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Regards ,Consider the code below, its serves your requirement, make changes accordingly
data:
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Hi,
How to take the cron job log in HP UX
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| | | | |
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Subscribe to our free e-mail newsletter or RSS feed to get all updates. You can Email this page to a friend.
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January 15th, 2008 at 12:21 am
You might note that all files normally found in /etc/init.d, /etc/rc*.d, and others are all in /sbin in HP-UX 10.20 and up.
Note too that your formatting is off in the description of the /etc/rc.config.d/cron file. Also, if you run /usr/bin/cron instead of /sbin/init.d/cron start, the /etc/rc.config.d/cron file is not used.
Regards,
SatyaI guess it would :)
btw, how did this piece miss from the contents pasted by the OP ?
The doc has given all details of CRON and missed where to see the logfiles. :)
****Added****
I tried man crontab in windows, i dont see it.i overlook most of the things, i would have done the same ;)
Edited by: Bobcatalog on Oct 23, 2008 9:40 AM -
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How to get the entry point in the ActiveX/COM adaptor
below is the description of the Demo of using the ActiveX/COM to call new seq .
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Attachments:
QQ图片20140115200924.jpg 58 KB
QQ图片20140115201708.jpg 100 KBThis is a duplicate post of this:
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The BIG failure of the floating point computation
The Big failure of the floating point computation .... Mmmm
You are writing some code .. writing a simple function ... and using type Double for your computation.
You are then expecting to get a result from your function that is at least close to the exact value ...
Are you right ??
Let see this from an example.
In my example, I will approximate the value of pi. To do so, I will inscribe a circle into a polygon, starting with an hexagon, and compute the half perimeter of the polygon. this will give me an approximation for pi.
Then I will in a loop doubling the number of sides of that polygon. Therefore, each iteration will give me a better approximation.
I will perform this twice, using the same algorithm and the same equation ... with the only difference that I will write that equation in two different form
Since I don't want to throw at you equations and algorithm without explanation, here the idea:
(I wrote that with Word to make it easier to read)
====================
Simple enough ...
It is important to understand that the two forms of the equation are mathematically always equal for a given value of "t" ... Since it is in fact the exact same equation written in two different way.
Now let put these two equations in code
Public Class Form1
Private Sub Form1_Load(sender As Object, e As EventArgs) Handles MyBase.Load
RichTextBox1.Font = New Font("Consolas", 9)
RichTextBox2.Font = New Font("Consolas", 9)
TextBox1.Font = New Font("Consolas", 12)
TextBox1.TextAlign = HorizontalAlignment.Center
TextBox1.Text = "3.14159265358979323846264338327..."
Dim tt As Double
Dim Pi As Double
'===============================================================
'Using First Form of the equation
'===============================================================
'start with an hexagon
tt = 1 / Math.Sqrt(3)
Pi = 6 * tt
PrintPi(Pi, 0, RichTextBox1)
'Now we will double the number of sides of the polygon 25 times
For n = 1 To 25
tt = (Math.Sqrt((tt ^ 2) + 1) - 1) / tt
Pi = 6 * (2 ^ n) * tt
PrintPi(Pi, n, RichTextBox1)
Next
'===============================================================
'Using Second Form of the equation
'===============================================================
'start with an hexagon
tt = 1 / Math.Sqrt(3)
Pi = 6 * tt
PrintPi(Pi, 0, RichTextBox2)
'Now we will double the number of sides of the polygon 25 times
For n = 1 To 25
tt = tt / (Math.Sqrt((tt ^ 2) + 1) + 1)
Pi = 6 * (2 ^ n) * tt
PrintPi(Pi, n, RichTextBox2)
Next
End Sub
Private Sub PrintPi(t As Double, n As Integer, RTB As RichTextBox)
Dim S As String = t.ToString("#.00000000000000")
RTB.AppendText(S & " " & Format((6 * (2 ^ n)), "#,##0").PadLeft(13) & " sides polygon")
Dim i As Integer = 0
While S(i) = TextBox1.Text(i)
i += 1
End While
Dim CS = RTB.GetFirstCharIndexFromLine(RTB.Lines.Count - 1)
RTB.SelectionStart = CS
RTB.SelectionLength = i
RTB.SelectionColor = Color.Red
RTB.AppendText(vbCrLf)
End Sub
End Class
The results:
The text box contains the real value of PI.
The set of results on the left were obtain with the first form of the equation .. on the right with the second form of the equation
The red digits show the digits that are exact for pi.
On the right, where we used the second form of the equation, we see that the result converge nicely toward Pi as the number of sides of the polygon increases.
But on the left, with the first form of the equation, we see the after just a few iterations, the function stop converging and then start diverging from the expected value.
What is wrong ... did I made an error in the first form of the equation?
Well probably not since this first form of the equation is the one you will find in your math book.
So, what up here ??
The problem is this:
What is happening is that at each iteration when using the first form, I subtract 1 from the radical, This subtraction always gives a result smaller than 1. Since the type double has a fixed number of digits on the left of the decimal
point, at each iteration I am loosing precision caused by rounding.
And after only 25 iterations, I have accumulate such a big rounding error that even the digit on the left of the decimal point is wrong.
When using the second form of the equation, I am adding 1 to the radical, therefore the value grows and I get no lost of precision.
So, what should we learn from this ?
Well, ... we should at least remember that when using floating point to compute a formula, even a simple one, as I show here, we should always check the exactitude of the result. There are some functions that a floating point is unable to evaluate.I manually (yes, manually) did the computations with calc.exe. It has a higher accuracy. PI after 25 iterations is 3.1415926535897934934541990520762.
This means tt = 0.000000015604459512183037864437694609544 compared to 0.0000000138636291675699 computed by the code.
Armin
Manually ...
You did better than Archimedes.
He only got to the 96 sides polygon and gave up. Than he said that PI was 3.1427 -
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