real in vhdl

[Altera_Forum]

hi all  

i want to ask a question and i don't know if it possible in vhdl or how to make it. if i have an eqution such as   x= m(a)+a*max(a)+b*sd_d(a) and the values of a and b is as  a=1  b=1.5 --------  a=1  b=1.75 --------  a=1.5  b=0 --------- and i want to substitue the values of a,b every time and check the x value with comparator how i make it in vhdl.any idea please

[Altera_Forum]

The values of a and b that you give can be represented using "fractional integer format". Assuming the values of a and b can be signed values, you can use a 4-bit integer in Q1.2 format. This format can represent the following values; 

  Binary Q1.2 Value ------ ---------- 01.11 1.75 01.10 1.50 01.01 1.25 01.00 1.00 00.11 0.75 00.10 0.50 00.01 0.25 00.00 0.00 11.11 -0.25 11.10 -0.50 11.01 -0.75 11.00 -1.00 10.11 -1.25 10.10 -1.50 10.01 -1.75 10.00 -2.00   You can use these numbers, along with the knowledge that a multiplication of two Q1.2 numbers yields a Q3.4 value (eg., -2.00 x -2.00 = +4.00 = 0100.0000b) and addition results in a Q3.2 number, to determine the bit-growth in your equation. The resulting value of x can either be used directly, or convergent rounded to whatever bit-width you need for your design.  If none of this makes any sense, read these notes ...  http://www.ovro.caltech.edu/~dwh/correlator/pdf/esc-104paper_hawkins.pdf  Cheers, Dave

[Altera_Forum]

you mean ichould convert this constant values first to binary and then to integer to multiply with integer signal

[Altera_Forum]

--- Quote Start ---  you mean i could convert this constant values first to binary and then to integer to multiply with integer signal  --- Quote End ---   Yes. There is no "rule" that says your constants must use floating-point. If fixed-point numbers represent the fractional part with the same precision as floating-point, then you will get the same answer. The math is the same.  Cheers, Dave