blue is how I tried to sinc interpolate. why would something like this happen?
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$\begingroup$ Maybe do it in Frequency Domain - dsp.stackexchange.com/questions/25131. $\endgroup$– MarkCommented Sep 1, 2021 at 8:06
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2 Answers
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Since Sinc based Interpolation requires you to know the data at any point. Hence it is not feasible.
You might do a Truncated Sinc Interpolation.
The artifacts you're seeing can be caused by a kernel which is too short or the parameters aren't good.
In order to create a good Sinc kernel you need to know things about the Band Width of the signal and the Sampling Rate.
Did you took those into account?
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$\begingroup$ I just zoomed to a part of the signal. the signal goes to zero and it is sampled often enough $\endgroup$ Commented Aug 9, 2015 at 11:06
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$\begingroup$ Could you post the MAT file of the data and the M file you are running? $\endgroup$– RoyiCommented Aug 9, 2015 at 11:18
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$\begingroup$ I tried extending the sinc I convolved my samples with 100 times, it gave acceptable results. $\endgroup$ Commented Aug 9, 2015 at 13:37
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1$\begingroup$ How boundaries are handled is also important aspect. $\endgroup$ Commented Feb 13, 2022 at 21:03
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t=linspace(-.5,.5,256);
x=exp(-pi*t.^2*16).*(sin(2*pi*40*t)+0.154*cos(2*pi*47*t)-1.454*cos(2*pi*27*t));
figure;plot(t,x)
tt=linspace(-.5,.5,256*8-7);
xorj=exp(-pi*tt.^2*16).*(sin(2*pi*40*tt)+0.154*cos(2*pi*47*tt)-1.454*cos(2*pi*27*tt));
figure;plot(tt,xorj)
xf=SincInt(x,8,1);
sh=0;
xf=[ xf(1+sh:end) zeros(1,sh)];
figure;plot(tt,xf)
hold on;plot(tt,xorj)
figure;plot(tf,xf-xorj)
function f = SincInt( f,k ,varargin)
% function f = SincInt( f,k ,varargin)
% varargin=1 to keep the beginning and the end the same
nargin=length(varargin);
N=length(f);
f=[zeros(1,N*(k-1)) f];
for i = 1: N
f(k*i-k+1:k*i)= [zeros(1,k-1) f(N*(k-1)+i)];
end
f=conv(f,sinc(-60:1/k:60),'same');
if(nargin==1 && varargin{1})
f=[f(k:length(f)) ];
end
end