Hi all.

I get a hex value from a device through serial port over modbus protocol. but I don't know how to convert it to equivalent fraction decimal.

for example:

I get : 41C41DC3

and it is : 24.51453

is any body familiar with algorithm?

also if there is a algorithm to convert binary to fraction decimal it would help.

or if there is any vb.net function to do that?

Thanks : Saeid_Yousefpour [at] yahoo.com

I get a hex value from a device through serial port over modbus protocol. but I don't know how to convert it to equivalent fraction decimal.

for example:

I get : 41C41DC3

and it is : 24.51453

is any body familiar with algorithm?

also if there is a algorithm to convert binary to fraction decimal it would help.

or if there is any vb.net function to do that?

Thanks : Saeid_Yousefpour [at] yahoo.com

It looks as if the value is encoded in IEEE 32-bit floating-point format.

The first bit is a sign bit - 1 = -, 0 = +

The next 8 bits are the exponent with a offset of 127 In your case

41C = s100 0001 1nnn and the exponent = 131 - 127 = 4.

The remainder is a normalised value between 1 and 2 - 01 and 10 in binary.

Since the first bit is always 1 it is always present and is not transmitted.

So the remaining bits are the binary fraction part - the first bit is worth 0.5, the 2nd bit 0.25, the 3rd 0.125 and so on.

Just looking at the first few bits, the value is (.)100 0100 0001 1100 = 0.5 + 0 x 0.25 + 0 x 0.125 + 0x 0.0625 + 1 x .03125 = 0.5 + .03125 = 0.53125

Add this to the 1 and the mantissa part is 1.53125. Multiply this by 2 to the power of the exponent = =2^4 = 16 and you get 1.53125 x 16 = 24.5 to 3 sig figs. Add in the rest of the bits and you'll get something approaching your value.

To recap,

The first bit is a sign bit s.

The next 8 bits represent 127+ n where n is the power of 2 to multiply by.

The mantissa is 1.bbb bbbb bbbb bbbb bbbb bbbb where each succeeding \b is weighted at 1/2 the value of the one before. The value represented is s 1. bbb bbbb bbbb bbbb bbbb bbbb x 2^n.

Hope this helps.

Cheers,

Bruce.

The first bit is a sign bit - 1 = -, 0 = +

The next 8 bits are the exponent with a offset of 127 In your case

41C = s100 0001 1nnn and the exponent = 131 - 127 = 4.

The remainder is a normalised value between 1 and 2 - 01 and 10 in binary.

Since the first bit is always 1 it is always present and is not transmitted.

So the remaining bits are the binary fraction part - the first bit is worth 0.5, the 2nd bit 0.25, the 3rd 0.125 and so on.

Just looking at the first few bits, the value is (.)100 0100 0001 1100 = 0.5 + 0 x 0.25 + 0 x 0.125 + 0x 0.0625 + 1 x .03125 = 0.5 + .03125 = 0.53125

Add this to the 1 and the mantissa part is 1.53125. Multiply this by 2 to the power of the exponent = =2^4 = 16 and you get 1.53125 x 16 = 24.5 to 3 sig figs. Add in the rest of the bits and you'll get something approaching your value.

To recap,

The first bit is a sign bit s.

The next 8 bits represent 127+ n where n is the power of 2 to multiply by.

The mantissa is 1.bbb bbbb bbbb bbbb bbbb bbbb where each succeeding \b is weighted at 1/2 the value of the one before. The value represented is s 1. bbb bbbb bbbb bbbb bbbb bbbb x 2^n.

Hope this helps.

Cheers,

Bruce.

Thank you Bruce. it works very good.

That is a single precision floating point number in "big-endian" (network order) byte order. I don't know if you can do this in VB DotNet, but in Python this is very easy.

Typing the following:

struct.unpack('!f', '\x41\xc4\x1d\xc3')

results in:

(24.514532089233398,)

As for binary to floating point, that is what the above is doing. The "hex" number is just a convenient way of representing a binary number in a printable format.

Typing the following:

struct.unpack('!f', '\x41\xc4\x1d\xc3')

results in:

(24.514532089233398,)

As for binary to floating point, that is what the above is doing. The "hex" number is just a convenient way of representing a binary number in a printable format.

I suppose the format is IEEE_754-1985

http://en.wikipedia.org/wiki/IEEE_754-1985#Single-precision_32-bit

It might be necessary to reverse the byte order depending on which type of CPU is used.

http://en.wikipedia.org/wiki/IEEE_754-1985#Single-precision_32-bit

It might be necessary to reverse the byte order depending on which type of CPU is used.

Hi, There are a lot of examples and information about floating point number's.

I have created a simple vbscript that you should beable to use;

'number to be converted

'HexNumber = 1103371715 'as a decimal

HexNumber = &h41C41DC3 'as hex

'work out exponent get bits 23-30

exponent = HexNumber

For n = 0 To 22

exponent = exponent / 2

Next

exponent = exponent - 127

exponent = 2 ^ Int(exponent)

'Debug.Write Int(exponent) & vbcrlf

'work out Mantissa bits 0-22

'check each bit inturn and store result

decimalpart = 1

mantissa = 1

For n = 22 To 0 Step -1

decimalpart = decimalpart / 2

testbit = 2 ^ n

If (Hexnumber And testbit) Then

'Debug.Write testbit & " " & n & vbCrLf

mantissa = mantissa + decimalpart

End If

Next

Result = exponent * mantissa

MsgBox Result

Kevin

I have created a simple vbscript that you should beable to use;

'number to be converted

'HexNumber = 1103371715 'as a decimal

HexNumber = &h41C41DC3 'as hex

'work out exponent get bits 23-30

exponent = HexNumber

For n = 0 To 22

exponent = exponent / 2

Next

exponent = exponent - 127

exponent = 2 ^ Int(exponent)

'Debug.Write Int(exponent) & vbcrlf

'work out Mantissa bits 0-22

'check each bit inturn and store result

decimalpart = 1

mantissa = 1

For n = 22 To 0 Step -1

decimalpart = decimalpart / 2

testbit = 2 ^ n

If (Hexnumber And testbit) Then

'Debug.Write testbit & " " & n & vbCrLf

mantissa = mantissa + decimalpart

End If

Next

Result = exponent * mantissa

MsgBox Result

Kevin

1-> reverse it like this C31DC4412

2-> convert it into binary its in ascii Hex

3-> then just assign the value to a float variable

2-> convert it into binary its in ascii Hex

3-> then just assign the value to a float variable

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