Handling linear interpolation of periodic functions

Question

I have a periodic functions and I have the values of this function at discrete points. So I have: f(t_i) for some t_i, i\el(0,n) Now between these discrete points I want to lineary interpolate the function such that: f(t) = f(t(i)) + (t - t(i)) ( f(t(i+1)) - f(t(i)) )/( t(i+1)-t(i) ) for t\el(t(i),t(i+1)).

The implementation (I am using Fortran notation): I have arrays t_n(0:n + 1) to represent t(i) at the given discrete points, f_n(0:n) to represent the function values at those points, and f_lin(n + 1) which is just a precalculation of the linear interpolation term --- f_lin(i) = ( f(t(i)) - f(t(i - 1) )/( t(i)-t(i - 1) ).

My implementation works fine as long as I am within the first period but after that I run into problems. The code goes as:

ALLOCATE ( t_n(0:n + 1), f_n(0:n + 1), f_lin(n + 1) )
t_n(0) = 0.D0
t = 0.D0
f_n(0) = f(t)
DO WHILE i = 1, n
    t_n(i) = t*dt
    ! some other code that calculates f_n(i) actually
    f_lin(i) = ( f_n(i) - f_n(i - 1) )/( t_n(i) - t_n(i - 1) )
END DO
t_n(n + 1) = period
f_lin(n + 1) = ( f_n(0) - f_n(n) )/( t_n(n + 1) - t_n(n) )
i = 0
j = 0
DO WHILE (t .LT. tmax)
    ! this tests whether I go to the higher interval where I have the discrete values
    IF ( t_n(i + 1) .LE. MOD(t, period) ) i = i + 1
    f_t = f_n(i) + ( t - t_n(i) )*f_lin(i + 1)
    j = j + 1
    t = t + dt
    ! some other code
END DO

This only works for the first period, to make it work in other periods as well, I think this would work:

DO WHILE (t .LT. tmax)
    ! this tests whether I go to the higher interval where I have the discrete values
    IF ( MOD(t, period) .GE. t_n(i + 1) ) THEN
        i = i + 1
    ELSE IF ( (MOD(t, period) .LT. t_n(i)) .AND. ( i .EQ. n ) ) THEN
        i = 0
    END IF
    f_t = f_n(i) + ( t - t_n(i) )*f_lin(i + 1)
    j = j + 1
    t = t + dt
    ! some other code
END DO

I have not tried it yet because it looks very complicated. I believe that there must be a simple way to handle this issue (I can probably solve the problem with using several IF statements but I think I am missing something really simple). The problem is that during the last interval between t(n) and t(0), modulus of t can be both larger than t(n) or smaller.

I was also thinking about nested loops the top would be going through periods but I do not like that solution much either (especially because tmax is not generally a multiple of period).

What would the easiest but also the fastest (the execution speed) solution to this be?

Answer 1

My implementation works fine as long as I am within the first period but after that I run into problems.

You should never have to use values outside the first period.

You know the period T, so when given a value, you store its equivalent in the first period (remainder of division by T). When asked for a value at a point X, you calculate the remainder of X/T and interpolate the value for that. Since the function is periodic, that will also be the value in X.