Re: Tri to Sine diode shaper
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Posted by The Phantom on June 13, 2008, 1:57 am
Please log in for more thread options >>Can someone please direct me to a passive circuit using breakpoint
>>diodes, preferably no more than six, to shape a sine wave from a 5Vpp >>triangle? The available power source is single supply 6V. >> >>I did check the net, but had no luck with finding a basic approach. >> >>Mike Kendall >
>Patent:3737642 shows one way to do it. > >General question comes to mind... has anyone written a program that >would derive the "best" piecewise linear fit to a function, f(x), for >a given number of segments? > > ...Jim Thompson A nifty way to do it if the function you're trying to fit has an analytical derivative is this: Create a stepped approximation to the derivative of the function you're trying to fit; in this case find a stepped approximation to a cosine. Integrate the stepped approximation and you will get a linear segment approximation. The slopes of the linear segments are, of course, the values of the steps. The start and finish point of each linear segment is the same as the start and finish of each associated step. It may not be the exact minimum error fit, but it's really fast to compute and it's good enough for the sort of thing the OP wants. If you want better accuracy, generate more steps. See the graphic over on ABSE. | ||||||||||||||||||||||
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Posted by The Phantom on June 13, 2008, 12:11 pm
Please log in for more thread options >wrote:
> >>On Wed, 11 Jun 2008 17:54:52 -0700, Jim Thompson
>> >>>On Wed, 11 Jun 2008 09:33:12 +1000, Mike Kendall
>>> >>>>Can someone please direct me to a passive circuit using breakpoint >>>>diodes, preferably no more than six, to shape a sine wave from a 5Vpp >>>>triangle? The available power source is single supply 6V. >>>> >>>>I did check the net, but had no luck with finding a basic approach. >>>> >>>>Mike Kendall >>> >>>Patent:3737642 shows one way to do it. >>> >>>General question comes to mind... has anyone written a program that >>>would derive the "best" piecewise linear fit to a function, f(x), for >>>a given number of segments? >>> >>> ...Jim Thompson >>
>>A nifty way to do it if the function you're trying to fit has an analytical >>derivative is this: >> >>Create a stepped approximation to the derivative of the function you're >>trying to fit; in this case find a stepped approximation to a cosine. >> >>Integrate the stepped approximation and you will get a linear segment >>approximation. >> >>The slopes of the linear segments are, of course, the values of the steps. >>The start and finish point of each linear segment is the same as the start >>and finish of each associated step. >> >>It may not be the exact minimum error fit, but it's really fast to compute >>and it's good enough for the sort of thing the OP wants. If you want >>better accuracy, generate more steps. >> >>See the graphic over on ABSE. >
>I see an EXE and a bunch of RAR files over on ABSE. What do I do with >them ?:-) Those aren't for you. You want the one called "Linear segment approximation method". >
> ...Jim Thompson | ||||||||||||||||||||||
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Posted by The Phantom on June 13, 2008, 3:01 pm
Please log in for more thread options On Fri, 13 Jun 2008 09:34:19 -0700, Jim Thompson wrote
<SNIP>
>>
>>Those aren't for you. You want the one called "Linear segment approximation >>method". >> >>>
>>> ...Jim Thompson >
>OK. What process did you use to space the segments? (I'm not >conversant in Mathematica.) > > ...Jim Thompson You just plot a certain function of t and you will get the steps. For the first sine graph, use: (Ceiling(y+Cos(t))/y + Floor(y+Cos(t))/y)/2 where y is the number of steps you want. You will get a good approximation for any value of y, but if you make it an integer plus .66, you will get slightly lower distortion than otherwise. For the second sine graph, the one with a flat on top, use: (Round(y+Cos(t))/y where y is the number of steps you want. Make y an integer plus .16 for best results. For positive numbers: Ceiling(z) means the integer just greater than z (the integer part of z plus 1). Floor(z) means the integer just less than z (the integer part of z). Round(z) means the integer part of (z + .5). Make the obvious changes for negative numbers. | ||||||||||||||||||||||
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Posted by The Phantom on June 15, 2008, 7:24 pm
Please log in for more thread options On Sun, 15 Jun 2008 14:54:49 -0700, Jim Thompson
<SNIP>
I think there are good reasons to use the round-to-even method.
> >Mr. Phantom, > >Please comment on this discussion... > >http://mathforum.org/library/drmath/view/71202.html > >Which convention is generally followed with negative numbers? > > ...Jim Thompson See: http://en.wikipedia.org/wiki/Rounding Also, Knuth in Vol 2 of "The Art of Computer Programming" discusses rounding. | ||||||||||||||||||||||
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