MATLAB: How to find the interpolation error when I use INTERP in the Signal Processing Toolbox

errorinterpinterpolationprocessingsignalSignal Processing Toolbox

I was wondering if there was a way to calculate interpolation error when using INTERP in the Signal Processing Toolbox.

Best Answer

One way to find the interpolation error given the following:

t = 0:0.001:1; % Time vector
x = sin(2*pi*30*t) + sin(2*pi*60*t);
r = 4;
y = interp(x,r);

is to use:

e = x - y(1:r:end);

"e" is the difference between the original data, "x", and the interpolated data, "y", at the original discrete times.

Related Solutions

MATLAB: How to fix error about interp1

One way (without seeing your data, so this is a guess on my part):

y85 = find(y <= 85, 1, 'last');
yi = interp1(y(y85-1:y85+1),x(y85-1:y85+1),xi, 'linear','extrap');

That is how I would do it. This assumes there is only 1 where ‘y’ is approximately 85.

Note — This is obviously UNTESTED CODE. It should work, if there is one ‘y’ near 85.

EDIT — Added the 'extrap' option to make my approach more robust.

MATLAB: Does the INTERP1 function in MATLAB handle complex data with the “cubic” method

This enhancement has been incorporated in Release 2006a (R2006a). For previous product releases, read below for any possible workarounds:

The cubic interpolation algorithm is not intended for complex data and hence does not behave well under certain circumstances when the data is complex. As a workaround, use the spline interpolation instead.

Y_spline = interp1(X,Y,Xi,'spline');
plot(X,abs(Y),'b',Xi,abs(Y_spline),'r')

Alternatively, you can split Y into real and imaginary parts, and then interpolate.

Yr = real(Y);
Yi = imag(Y);
Y_cubic = interp1(X,Yr,Xi,'cubic')+i*interp1(X,Yi,Xi,'cubic');
plot(X,abs(Y),'b',Xi,abs(Y_cubic),'r')

Best Answer

Related Solutions

MATLAB: How to fix error about interp1

MATLAB: Does the INTERP1 function in MATLAB handle complex data with the “cubic” method

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