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Francois Normandin

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Posts posted by Francois Normandin

  1. QUOTE(shoneill @ Feb 15 2008, 07:00 AM)

    The algorithm is also called "the travelling salesman" and it's anything other than trivial.

    I agree that the traveling salesman problem is anything but trivial. However, I don't think this problem is an equivalent. The traveling salesman's solution is to find the shortest route possible to visit all points on a map. It's been extensively studied in many engineering problems such as reducing time to solder thousands of components on a circuit board. But if I understand Daku's problem, I tend to confirm Yuri33's assessment that we can discard any points in the middle as they will not give a result anywhere close to the maximum distance between any two points on the periphery.

  2. QUOTE(Dirk J. @ Jan 30 2008, 04:42 AM)

    Isn't there a theoretical curve you could fit?

    I mean, it looks like a 'simple' second order step resonse (...)

    Good idea Dirk... it could be as simple as a RLC type circuit equivalent. More info on the system that produces this step response would be required.

  3. QUOTE(jbrohan @ Jan 29 2008, 03:06 PM)

    Can some mathematically inclined member point me in a better direction.

    If you have an 8th-degree polynomial fit, why not do a 2nd derivative on the algebraic solution? It all depends on how good is the fit compared to original signal, but algebraic solution wouldn't have bumps...

    y = ax^8 + bx^7 + cx^6 + dx^5 + ex^4 + fx^3 + gx^2 + hx + i

    dy/dx = 8ax^7 + 7bx^6 + 6cx^5 + 5dx^4 + 4ex^3 + 3fx^2 + 2gx + h

    d(dy)/dx^2 = 56ax^6 + 42bx^5 + 30cx^4 + 20dx^3 + 12ex^2 + 6fx + 2g

    Use an algorithm to find the zeroes of your second derivative (inflection points), then keep only those values whose first derivatives are positive. (i.e. where the slope stops increasing and starts decreading). After finding "x", the slope can be found by solving dy/dx for this value...

    :lightbulb: Hope this helps!

  4. QUOTE(Gabi1 @ Jan 25 2008, 04:24 PM)

    now next step: why and how? why strict, what does it mean for a class specifier, and how to find it?Thanks a bunch!

    To create a strict type reference of a waveform chart, right-click on waveform chart and create a reference. Then, in the block diagram, right-click on the reference you just created and create a constant. This newly created reference will be "strict" type.

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