Hi! Experts, Could anyone please help me to implement e(t^2/l^2) in VHDL? where t is continuous time and l is a constant. Thank you very much in advance. Regards, Akilan.T

for synthesis or just a testbench? and do you mean e^(t^2/l^2)? You will probably have to implement it via a look up table.

--- Quote Start --- for synthesis or just a testbench? and do you mean e^(t^2/l^2)? You will probably have to implement it via a look up table. --- Quote End --- Tricky, Yes that is right e^(t^2/l^2). Then, I want to sysnthesis it and to place into a FPGA. Look up table means have to place all the values into a long array?

yes, but remember it has to map to physical hardware, so make sure you follow the coding guidlines for ROMS. You'll also have to make sure t isnt too many bits. t at 20bits per word will be about the limit for the larger devices.

If your application is not critical you can use a small LUT e.g. 256 points then interpolate (linearly) between two values as they occur by finding the difference between input (t) value and nearst LUT values. i.e. some t values will point directly, most others will point in between LUT points so a basic calculation is needed to adjust the result.

--- Quote Start --- If your application is not critical you can use a small LUT e.g. 256 points then interpolate (linearly) between two values as they occur by finding the difference between input (t) value and nearst LUT values. i.e. some t values will point directly, most others will point in between LUT points so a basic calculation is needed to adjust the result. --- Quote End --- Altera Guru, Infact, my design is critical. output need to be very precise. Is ther anyway for that? Now, I got some points then linearize them (y = mx + c) with a particular time interval, then write code like below. if t = 0 then e = 1; elsif t > 0 and t <= 1 then e = -0.5 * x + 1; ...... end. Could I implement in this way? Thank you. Akilan.T

Help: How can I implement e(t^2/l^2) in VHDL

12 Replies

Altera_Forum
Honored Contributor
15 years ago
for synthesis or just a testbench?
and do you mean e^(t^2/l^2)?

You will probably have to implement it via a look up table.
Altera_Forum
Honored Contributor
15 years ago
--- Quote Start ---
for synthesis or just a testbench?
and do you mean e^(t^2/l^2)?

You will probably have to implement it via a look up table.
--- Quote End ---

Tricky, Yes that is right e^(t^2/l^2). Then, I want to sysnthesis it and to place into a FPGA. Look up table means have to place all the values into a long array?
Altera_Forum
Honored Contributor
15 years ago
yes, but remember it has to map to physical hardware, so make sure you follow the coding guidlines for ROMS. You'll also have to make sure t isnt too many bits. t at 20bits per word will be about the limit for the larger devices.
Altera_Forum
Honored Contributor
15 years ago
If your application is not critical you can use a small LUT e.g. 256 points then interpolate (linearly) between two values as they occur by finding the difference between input (t) value and nearst LUT values. i.e. some t values will point directly, most others will point in between LUT points so a basic calculation is needed to adjust the result.
Altera_Forum
Honored Contributor
15 years ago
--- Quote Start ---
If your application is not critical you can use a small LUT e.g. 256 points then interpolate (linearly) between two values as they occur by finding the difference between input (t) value and nearst LUT values. i.e. some t values will point directly, most others will point in between LUT points so a basic calculation is needed to adjust the result.
--- Quote End ---

Altera Guru,

Infact, my design is critical. output need to be very precise. Is ther anyway for that?

Now, I got some points then linearize them (y = mx + c) with a particular time interval, then write code like below.

if t = 0 then
e = 1;
elsif t > 0 and t <= 1 then
e = -0.5 * x + 1;

......

end.

Could I implement in this way?

Thank you.

Akilan.T
Altera_Forum
Honored Contributor
15 years ago
Writing the piecewise linear characteristics as "hard coded" if .. then .. else structure requests all design resources from logic cells. Using a LUT would use the input value as index to a ROM and calculate the interpolated value based on the ROM content.
Altera_Forum
Honored Contributor
15 years ago
To implement LUT interpolation follow these lines:

youe input (T) is say represented by 16 bits unsigned value. Use its 8 MSBs as address to LUT. This picks up the nearest LUT point (lower) then what is left of your (T) value (remainder of 8 LSBs) indicate how much you should add taking into account your scaling. You need to find out (LUT2 - LUT1) where LUT1 is the value addressed at by 8 MSBs. Then translate the (T) excess into LUT extra thus your final value = LUT1 + 8 LSB value * (LUT2-LUT1) / delta (T) of two LUT points* scale factor. (remember delta y/delta x is gradient of any draph)
Altera_Forum
Honored Contributor
15 years ago
--- Quote Start ---
Could anyone please help me to implement e(t^2/l^2) in VHDL? where t is continuous time and l is a constant.
--- Quote End ---

What is the maximum value for 't' and what is the resolution of 't' , and the range for 'i'? What precision do you need for the result?
If you want to use a Look-Up Table you will need ((tmax / resolution(t)) * precision) memory bits. If that exceeds the number of available you will have to find a (mathematical) solution to compute the result.
Altera_Forum
Honored Contributor
15 years ago
--- Quote Start ---
To implement LUT interpolation:
--- Quote End ---

Here the content of LUT must be 1s and 0s right?, but I have the generated values for e^(x) in floating point values where I take x = 0:0.001:10. could you please show me some directions or any links to such sample code?

Thank you.
Altera_Forum
Honored Contributor
15 years ago
Your numbers are fixed point not floating point. Fixed point means applying a fixed scaling to integer numbers. Most popular are binary (2^n) scaling factors. Number ranges and resolution can be different for x and y of a function and need to be defined according to your accuracy requirements.

By the way, do you actually mean divergent exp(t^2/l^2) or gauss function exp(-t^2/l^2)?

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Help: How can I implement e(t^2/l^2) in VHDL

12 Replies

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