Error using interpolation line 64

How do i get around this error with says: Error using interp (line 64) Length of data sequence must be at least 9. You either need more data or a shorter filter (L). I can't really add more data, that was the whole point of my interpolation. Any help?

[the cyclist]

You haven't given us much to help you. Can you post a small, self-contained piece of code that will allow us to replicate the error for ourselves.

Answer 1

[Mark O'Sullivan]

If you have a look at the interp .m file the first paragraph states this:

function [odata,b] = interp(idata,r,l,cutoff)
%INTERP Resample data at a higher rate using lowpass interpolation.
%   Y = INTERP(X,R) resamples the sequence in vector X at R times
%   the original sample rate.  The resulting resampled vector Y is
%   R times longer, LENGTH(Y) = R*LENGTH(X).
%
%   A symmetric filter, B, allows the original data to pass through
%   unchanged and interpolates between so that the mean square error
%   between them and their ideal values is minimized.
%   Y = INTERP(X,R,L,CUTOFF) allows specification of arguments
%   L and CUTOFF which otherwise default to 4 and .5 respectively.
%   2*L is the number of original sample values used to perform the
%   interpolation.  Ideally L should be less than or equal to 10.
%   The length of B is 2*L*R+1. The signal is assumed to be band
%   limited with cutoff frequency 0 < CUTOFF <= 1.0. 
%   [Y,B] = INTERP(X,R,L,CUTOFF) returns the coefficients of the
%   interpolation filter B.

In other words, the filter needs more than 2*L numbers of original sample points in order to perform the interpolation. (by default is 4 so you need 4*2+1 = 9 samples)

To get around this, you can define the value of L when using the function.

i.e.

interpData = interp(originalData, 50, 2)
% This means you require at least 5 original samples

or

interpData = interp(originalData, 50, 1)
% This means you require at least 3 original samples

Hope this helps :)