CODEWITHSUNDEEP
matlabmathsignal-processinginterpolationcoordinate-transformation
Feel-ipe
Feel-ipe

Reputation: 1

Using the sinc function for interpolation

For a university assignment, I have to interpolate some points that I had previously sampled from a wave. It works as long as I interpolate them with the interp1 function.

My teacher wants us to interpolate it using the sinc function, so I googled it and the mathworks documentation says I should do this:

rng default

t = 1:10;
x = randn(size(t))';
ts = linspace(-5,15,600);
[Ts,T] = ndgrid(ts,t);
y = sinc(Ts - T)*x;

plot(t,x,'o',ts,y)
xlabel Time, ylabel Signal
legend('Sampled','Interpolated','Location','SouthWest')
legend boxoff

It works, but since I have to interpolate points that are in a reduced domain I modified the code to try and interpolate points that are in the range from 0 to 1: 

rng shuffle

t = 0:0.1:1;
x = randn(size(t))';
ts = linspace(-1,2,600);
[Ts,T] = ndgrid(ts,t);
y = sinc(Ts - T)*x;

plot(t,x,'o',ts,y)
xlabel Time, ylabel Signal
legend('Sampled','Interpolated','Location','SouthWest')
legend boxoff

And it just doesn't work properly, I get this from the above:

Figure

Could you please tell me what am I doing wrong, or if sinc only works in "big" domains?

Upvotes: 0

Views: 4405

Answers (1)

m7913d
m7913d

Reputation: 11064

You should also scale your sinc function in the time direction to fit your data:

dt = 0.1;
y = sinc((Ts - T)/dt)*x;

Explanation

Because you want to interpolate between your data point, you should be sure that the interpolation function (f) of the other data points is zero at the current data point:

f(k*dt) = 0    for all integers k != 0

It is known that

sinc(k) = 0    for all integers k != 0

Therefore your interpolation function should be

f(t) = sinc(t/dt);

Upvotes: 5

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