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Cubic Spline Interpolation Tutorial

  1. Jul 31, 2007 #1


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    Attached below are two cubic spline tutorials:

    1. Explanation of the classic tri-diagonal cubic spline formulation. Included are 2 example problems.

    2. Extension to parametric cubic splines. Included are 2 example problems

    Attached Files:

    Last edited by a moderator: Mar 3, 2008
  2. jcsd
  3. Oct 19, 2007 #2

    After a very long search in google i found this tutorial.

    And it was the first one, which describes parametric cubic splines in a good way.

    Thank you very much :-)

    Now i can lay streets through my landscape :-)
  4. Oct 31, 2007 #3
    Sigh....Thank You!

    I was missing something very stupid and now I know what it is!
  5. Jan 15, 2008 #4
    Thank you for tutorials. This has helped me very mush.
  6. Mar 2, 2008 #5
    Cubic Spline Interpolation of a Circle

    Hey there -
    Thanks for the great tutorials - they really helped me! I'm trying to duplicate your results for cubic interpolation of a circle with 4 points and I got the same solution for the 2nd derivatives in the x and y directions. However, when I solve for the coefficients and plot the cubic polynomials I can't seem to get the same result as you - Heres the code that calculates the coefficients (in MATLAB)

    mx - x''
    my - y''
    Sx - x coefficients
    Sy - y coefficients
    s - arc length [0 0.25 0.5 0.75 1]

    for i = 1:n-1
    Sx(i,1) = (mx(i+1) - mx(i))/6*h;
    Sx(i,2) = mx(i)/2;
    Sx(i,3) = (x(i+1) - x(i))/h - h*(mx(i+1) + 2*mx(i))/6;
    Sx(i,4) = x(i);

    Sy(i,1) = (my(i+1) - my(i))/6*h;
    Sy(i,2) = my(i)/2;
    Sy(i,3) = (y(i+1) - y(i))/h - h*(my(i+1) + 2*my(i))/6;
    Sy(i,4) = y(i);


    %Plotting the function
    ds = 0.01;
    for i = 1:n-1
    for k = s(i):ds:s(i+1)
    genpnx(m) = Sx(i,1)*(k - s(i))^3 + Sx(i,2)*(k-s(i))^2 + Sx(i,3)*(k-s(i)) + Sx(i,4);
    genpny(m) = Sy(i,1)*(k - s(i))^3 + Sy(i,2)*(k-s(i))^2 + Sy(i,3)*(k-s(i)) + Sy(i,4);
    m = m+1;


    Do you have any idea what I might be doing wrong? (Ive attached an image of what the spline looks like for 4 points)


    Attached Files:

    Last edited: Mar 2, 2008
  7. Mar 3, 2008 #6


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    Though I'm not a MATLAB person, your expressions for calculating the coefficients look correct. Make sure you are using 0.25 for h. Also, for plotting, t should vary from 0 to 0.25 for each segment.

    Here are the coefficient values I got for the 1st segment:

    ax = 32
    bx = -24
    cx = 0
    dx = 1

    ay = -32
    by = 0
    cy = 6
    dy = 0

    Actually, this isn't the best way to fit a circle with cubic polynomials. It is better to fit a quarter circle with a single parametric cubic. The tutorial has been updated - check it out.
    Last edited: Mar 3, 2008
  8. Mar 3, 2008 #7

    Thanks for the coefficients - they helped me figure out what the problem was - a syntax problem - the correct expression for the coefficients are

    Sx(i,1) = (mx(i+1) - mx(i))/(6*h)

    while I had

    Sx(i,1) = (mx(i+1) - mx(i))/6*h

    Thanks a ton!
  9. Mar 3, 2008 #8


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    Glad to help. Fyi, the problem can be made even simpler by making h = 1, meaning s = [0 1 2 3 4]. Second deriveratives are x" = [-3 0 3 -3], y" = [0 -3 0 3 0] and plotting can be done for t varying from 0 to 1 for each segment.

    I encourage you to review the updated tutorial on circle fitting.
  10. Mar 12, 2008 #9


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    Attached is an update to the Cubic Spline Tutorial.

    To Staff,

    Would someone be willing to replace the 1st attachment in this thread with the below? Thanks.
    Last edited: Dec 17, 2008
  11. Nov 22, 2008 #10
    thanks a lot buddy.
  12. Dec 17, 2008 #11
    Hi hotvette,
    I have been looking for a good tutorial on Cubic Splines, nice work. I have found what I think might be an error in one of your examples, however. Example Problem #1 shows c2 = 0, but when I plug in the values:
    y3 = 1
    y2 = 0.125
    h2 = .5
    y3" = 0
    y2" = 4.5
    I come up with c2 = 1, not 0. Plugging 0.5 into the second polynomial with c2 = 1, I get y2 = 1, thus resulting in a discontinuity with the first polynomial. I'm too lazy to try to work out the root of the problem, but I figured I'd flag the discrepancy.
  13. Dec 17, 2008 #12


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    Agree that C2 = 1.0 (typo), but I think the evaluation of the 2nd poly is OK:

    [tex]y = a2(x - x_2)^3 + b2(x - x_2)^2 + c2(x - x_2) + d2[/tex]

    Update with typo fixed is attached.

    Attached Files:

  14. Dec 17, 2008 #13
    Yep, my mistake. I used ax^3 + bx^2 + cx + d rather than using (x-x0), etc.
    Thanks again for the excellent tutorial.
  15. Dec 29, 2008 #14
    For the cubic spline interpolation if we have
    A=(x2-x)/h; B=(x-x1/h); C=1/6*(A^3-A)*h^2; D=1/6*(B^3-B)*h^2;
    y=A y1 + B y2 + C y1'' + D y2''

    What is the interpolation error of above?

    For example for Hermitian cubic spline it is smaller than 5/384 ||f^{(4)}|| h^4
  16. Apr 8, 2009 #15
    Thanks very much my friend! The tutorial for the parametric C Splines was very helpful to me!
    You see, every book I ve read doesn t say how we can relate the derivatives of the actual curve to the derivatives of the parametric equations of x & y. Ok, its easy to proof with the chain rule but what about the values of these derivatives on bounds!! Simply, x'(t)/y'(t) must be constant on bounds, so give x'(t),y'(t) any value you want to keep this ratio constant! HAHAHAH, i feel stupid...
    Very helpful!
    Thanks again!
  17. Oct 16, 2009 #16
    Thanks! Just what i was looking for! :)
  18. Nov 15, 2009 #17
    Any reading suggestion to smooth out a curve using spline or similar approach.
  19. Nov 24, 2009 #18
    great tuto, very precise. Thanks a lot
  20. Jul 19, 2010 #19
    That was helpful
  21. Jul 21, 2010 #20
    I have trouble implementing the interpolation. not sure what is wrong. appreciate if someone can point out which part I'm wrong. Im pretty sure I have the coefficients correct but somehow im using the spline equations incorrectly (my guess).
    So here are my original x and y
    x y
    0.90 1.30
    1.30 1.50
    1.90 1.85
    2.10 2.10
    3.00 1.95
    3.80 0.40
    4.30 0.25

    Here are my coefficients
    a b c d
    Equation 1 -0.282 0.000 0.545 1.30
    Equation 2 1.046 -0.338 0.410 1.50
    Equation 3 -4.815 1.545 1.134 1.85
    Equation 4 -0.162 -1.344 1.174 2.10
    Equation 5 1.757 -1.780 -1.638 1.95
    Equation 6 -1.625 2.437 -1.112 0.40

    Now If I want to interpolate x=1, my formula would read
    y=-0.282*(1-0.9)^3 + 0.000*(1-0.9)^2 +0.545*(1-0.9) + 1.30

    I think the answer is wrong because I validated it against the utility at http://www.akiti.ca/CubicSpline.html.. Would appreciate if someone can tell me if my coefficients are correct and if i am using the formulas correctly.

  22. Jul 22, 2010 #21


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    Using natural boundary conditions (f" = 0 at either end) I derived the same result as you. The utility doesn't state what algorithm is being used nor what assumptions are being made on boundary conditions. The routines mentioned in the credits refer to Hermite interpolation. If that's what is being used by the utility, then it's a different algorithm than what I posted. From what I read, Hermite interpolation isn't C2 compatible at spline boundaries whereas classic cubic spline interpolation is. Bottom line is that there are many interpolation methods. There is no right one.

    The other possibility is that the utility is performing cubic spline interpolation but is making some assumption about the end boundary conditions. Just for kicks I was able to replicate the result of the utility by adjusting the slope at the first point to 0.811 and the slope of the last point to 2.226. I checked several points in between and they all matched.

    Black box utilities are great because they can be easy to use, but a caveat is that you may not know what they are doing.
    Last edited: Jul 23, 2010
  23. Jul 24, 2010 #22


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    I believe I've unraveled the mystery of the black box utility. It appears to use cubic splines with end boundary conditions called 'not-a-knot', which is continuous 3rd derivatives in the end pairs of splines. See page 5 in the following link:


    This means the cubic term (a in the tutorial) is the same in the end pairs of splines. If we equate the cubic terms of the 1st two splines, the resulting boundary condition after some rearranging is:

    -h2y1'' + (h1+h2)y2'' - h1y3'' = 0

    There is a similiar condition at the other end. If these are used as the end boundary conditions, the resulting splines produce the same results as the utility. Fyi, according to the link, MATLAB uses this boundary condition.

    An interesting way to confirm would be to obtain 4 interpolated points from the utility (2 end points are given so you'd just need 2 intermediate points), from which you could easily derive the coefficients of the cubic polynomial and compare to the ones calculated using the algorithm in the cubic spline tutorial (and not-a-knot boundary conditions). They should match.
    Last edited: Jul 25, 2010
  24. Aug 8, 2010 #23
    Hi all,

    I have a stupid question regarding calculating the derivative of cubic spline. I'm currently looking at the code here http://www.ee.ucl.ac.uk/~mflanaga/java/CubicSpline.java at method calcDeriv(), and I do not understand what are these 2 lines are for

    Any help would be appreciated.

  25. Oct 8, 2010 #24
    Hi All,
    I am trying to convert a file that has three arrays of 528 samples into three arrays of 301.
    The first is log spaced frequency points the second is the impedance values at each frequency and the last is phase values. I have the routines spline and splint. My problem is the the second to last sample [299] is off scale. The last sample [300] is correct. I know it is some kind of boundary problem, but it is beyond me to figure it out.

    fr,re,ph are arrays[1..528] of float

    For i:=1 to 301 do
    For i:=1 to 301 do
    For i:=0 to 300 do
  26. Oct 11, 2010 #25
    Thanks a lot
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