- **VHDL**
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- - **exponential function**
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exponential functionI need to find the value of e^(-x/y); where y is the constant and x keeps changing.
Does anyone know how to find this value on Xilinx FPGAs. Is there any core available for it? or any Algorithm would also do. Thanks in advance. :) |

Re: exponential functionOn Jan 16, 3:58*am, blue39 <manjeet.rat...@gmail.com> wrote:
> I need to find the value of e^(-x/y); where y is the constant and x keepschanging. > > Does anyone know how to find this value on Xilinx FPGAs. Is there any core available for it? or any Algorithm would also do. > > Thanks in advance. > :) Like all things, having an idea of what X is: complex/real, fixed/ floating point, range & resolution, bandwith and allowable latency to get the answer (and what is its kind/range/resolution?) are going to impact the choice of method to do the calculation. Also, is Y REALLY a constant, or is it a parameter that changes very infrequently, but you don't want to have to re-synthesize, place & route to change it? Without knowing the above, and since you only have one real-time input, I would hesitantly suggest a look-up table, and depending on range/resolution of the input and the result, linear interpolation between table entries. Depending on the target architecture (whether block ROMs are available), you may be able to populate the table with a VHDL initialization function for the constant array that is the table. Andy |

Re: exponential functionAndy wrote:
> On Jan 16, 3:58 am, blue39 <manjeet.rat...@gmail.com> wrote: >> I need to find the value of e^(-x/y); where y is the constant and x keeps changing. >> >> Does anyone know how to find this value on Xilinx FPGAs. Is there any core available for it? or any Algorithm would also do. >> >> Thanks in advance. >> :) > > Like all things, having an idea of what X is: complex/real, fixed/ > floating point, range & resolution, bandwith and allowable latency to > get the answer (and what is its kind/range/resolution?) are going to > impact the choice of method to do the calculation. > > Also, is Y REALLY a constant, or is it a parameter that changes very > infrequently, but you don't want to have to re-synthesize, place & > route to change it? > > Without knowing the above, and since you only have one real-time > input, I would hesitantly suggest a look-up table, and depending on > range/resolution of the input and the result, linear interpolation > between table entries. Depending on the target architecture (whether > block ROMs are available), you may be able to populate the table with > a VHDL initialization function for the constant array that is the > table. > > Andy Also think about re-writing this equation as 2^(-x/Z) where Z is another constant = Y * ln(2). Then the integer portion of x/Z (use a multiplier i.e. x * (1/Z)) indicates the binary point position, and you only need a look-up on the fractional portion of x/Z. This method can give good precision over a wide input range. -- Gabor |

Re: exponential functionHi guys,
In e^(x/y), y = 0.007568, x = fraction(0.nnnnnnn). The bit width of each x and y will be 16 bits each. Similarly, I want the output also of 16 bit wide. Y is not a constant in pure sense as it will be changing very "in"frequently. I am using Virtex-6 SX series FPGA to have more number of DSP Slices. One more thing, I want this operation to finish in at max 2 clock cycles of frequency ranging between 300 to 400 MHz. I guess we can forget about y = 0.007568 for a moment, as I'm thinking of getting the value of (x/y) prior to e^ operation. Now how can I find the value of e^(0.fraction) with latency of 2 or less and frequency of 300-400 MHz. -BLue |

Re: exponential functionHi guys,
In e^(x/y), y = 0.007568, x = fraction(0.nnnnnnn). The bit width of each x and y will be 16 bits each. Similarly, I want the output also of 16 bit wide. Y is not a constant in pure sense as it will be changing very "in"frequently. I am using Virtex-6 SX series FPGA to have more number of DSP Slices. One more thing, I want this operation to finish in at max 2 clock cycles of frequency ranging between 300 to 400 MHz. I guess we can forget about y = 0.007568 for a moment, as I'm thinking of getting the value of (x/y) prior to e^ operation. How can I find the value of e^(0.fraction) with latency of 2 or less and frequency of 300-400 MHz? Thanks in advance :) -BLue |

Re: exponential functionblue39 wrote:
> Hi guys, > > In e^(x/y), y = 0.007568, x = fraction(0.nnnnnnn). The bit width of each x and y will be 16 bits each. Similarly, I want the output also of 16 bit wide. Y is not a constant in pure sense as it will be changing very "in"frequently. > > I am using Virtex-6 SX series FPGA to have more number of DSP Slices. > > One more thing, I want this operation to finish in at max 2 clock cycles of frequency ranging between 300 to 400 MHz. > > I guess we can forget about y = 0.007568 for a moment, as I'm thinking of getting the value of (x/y) prior to e^ operation. > > How can I find the value of e^(0.fraction) with latency of 2 or less and frequency of 300-400 MHz? > > > Thanks in advance :) > -BLue That really depends on the hardware you're targetting. In a Xilinx FPGA, I'd say your only hope is to use a look-up table (block RAM) and then if you needed interpolation, you could use a multiplier - but getting the interpolation to run at 300+ MHz might be tough since after the block RAM you only have one more cycle to meet your 2 clock latency. Do you really need such a low latency or do you really mean that you need to start a new operation at least every two cycles. There's a big difference, and it would certainly be easy to run pipelined with even a new operation starting on every clock as long as you can live with a few cycles of latency (Xilinx DSP48 multipliers run fastest with 3 cycle latency, +1 for BRAM = 4 cycles). If you're targetting an ASIC, then there are probably a lot more options. -- Gabor |

Re: exponential functionOn 1/17/2013 9:27 AM, GaborSzakacs wrote:
> blue39 wrote: >> Hi guys, >> In e^(x/y), y = 0.007568, x = fraction(0.nnnnnnn). The bit width of >> each x and y will be 16 bits each. Similarly, I want the output also >> of 16 bit wide. Y is not a constant in pure sense as it will be >> changing very "in"frequently. >> I am using Virtex-6 SX series FPGA to have more number of DSP Slices. >> One more thing, I want this operation to finish in at max 2 clock >> cycles of frequency ranging between 300 to 400 MHz. >> I guess we can forget about y = 0.007568 for a moment, as I'm thinking >> of getting the value of (x/y) prior to e^ operation. >> How can I find the value of e^(0.fraction) with latency of 2 or less >> and frequency of 300-400 MHz? >> >> >> Thanks in advance :) >> -BLue > > That really depends on the hardware you're targetting. In a Xilinx > FPGA, I'd say your only hope is to use a look-up table (block RAM) > and then if you needed interpolation, you could use a multiplier - > but getting the interpolation to run at 300+ MHz might be tough since > after the block RAM you only have one more cycle to meet your 2 clock > latency. Do you really need such a low latency or do you really mean > that you need to start a new operation at least every two cycles. > There's a big difference, and it would certainly be easy to run > pipelined with even a new operation starting on every clock as long > as you can live with a few cycles of latency (Xilinx DSP48 multipliers > run fastest with 3 cycle latency, +1 for BRAM = 4 cycles). > > If you're targetting an ASIC, then there are probably a lot more > options. > > -- Gabor I thought your other response was more interesting. Potentially the exponent can be broken into smaller pieces using much smaller lookup tables, efficiently implemented in block RAM. If the value of the exponent is broken into two 8 bit pieces and a value returned from two separate 256 word, 16 bit lookup tables, these values can be multiplied to get an exact result. e^(a+b) = e^a * e^b. This should be faster and simpler than an interpolation. How fast can a multiply be done in this part? Rick |

Re: exponential functionOn 1/18/2013 10:09 PM, rickman wrote:
> On 1/17/2013 9:27 AM, GaborSzakacs wrote: >> blue39 wrote: >>> Hi guys, >>> In e^(x/y), y = 0.007568, x = fraction(0.nnnnnnn). The bit width of >>> each x and y will be 16 bits each. Similarly, I want the output also >>> of 16 bit wide. Y is not a constant in pure sense as it will be >>> changing very "in"frequently. >>> I am using Virtex-6 SX series FPGA to have more number of DSP Slices. >>> One more thing, I want this operation to finish in at max 2 clock >>> cycles of frequency ranging between 300 to 400 MHz. >>> I guess we can forget about y = 0.007568 for a moment, as I'm thinking >>> of getting the value of (x/y) prior to e^ operation. >>> How can I find the value of e^(0.fraction) with latency of 2 or less >>> and frequency of 300-400 MHz? >>> >>> >>> Thanks in advance :) >>> -BLue >> >> That really depends on the hardware you're targetting. In a Xilinx >> FPGA, I'd say your only hope is to use a look-up table (block RAM) >> and then if you needed interpolation, you could use a multiplier - >> but getting the interpolation to run at 300+ MHz might be tough since >> after the block RAM you only have one more cycle to meet your 2 clock >> latency. Do you really need such a low latency or do you really mean >> that you need to start a new operation at least every two cycles. >> There's a big difference, and it would certainly be easy to run >> pipelined with even a new operation starting on every clock as long >> as you can live with a few cycles of latency (Xilinx DSP48 multipliers >> run fastest with 3 cycle latency, +1 for BRAM = 4 cycles). >> >> If you're targetting an ASIC, then there are probably a lot more >> options. >> >> -- Gabor > > I thought your other response was more interesting. Potentially the > exponent can be broken into smaller pieces using much smaller lookup > tables, efficiently implemented in block RAM. If the value of the > exponent is broken into two 8 bit pieces and a value returned from two > separate 256 word, 16 bit lookup tables, these values can be multiplied > to get an exact result. e^(a+b) = e^a * e^b. This should be faster and > simpler than an interpolation. > > How fast can a multiply be done in this part? > > Rick That's a good point. And the lower bits "b" represent a small enough number that e^b can be approximated as 1+b so you don't need a second lookup table. The multipliers in the DSP blocks are quite fast, and I know they can do 500 MHz when you use full pipelining, but I'm not sure if they can even to 300 MHz without pipelining. You'd have to run it through the tools to see. -- Gabor |

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