# Unexpected Result When Using Sinc Interpolation blue is how I tried to sinc interpolate. why would something like this happen?

Since Sinc based Interpolation requires you to know the data at any point it can not be done.

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?

• I just zoomed to a part of the signal. the signal goes to zero and it is sampled often enough – grdgfgr Aug 9 '15 at 11:06
• Could you post the MAT file of the data and the M file you are running? – Royi Aug 9 '15 at 11:18
• pastebin.com/sLYWrH5q – grdgfgr Aug 9 '15 at 13:32
• I tried extending the sinc I convolved my samples with 100 times, it gave acceptable results. – grdgfgr Aug 9 '15 at 13:37
• Dra, i don't think the problem is from not choosing good parameters. grd just ain't doing it right. looks like the OP might be using $|\operatorname{sinc}(\cdot)|$ instead. – robert bristow-johnson Aug 10 '15 at 0:49
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