Working of interpolation or decimation

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Hello,
Can someone explain how the interpolation or decimation can be used to fit the number of samples between the two signals if both the signals doesn't have the same number of samples.
Thanks.
  1 个评论
Gova ReDDy
Gova ReDDy 2014-1-9
编辑:Gova ReDDy 2014-1-9
Can I know what is interpolation and decimation.And where they can be implemented.

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Image Analyst
Image Analyst 2014-1-9
See this demo:
% Make signal #1
t1 = linspace(-2*pi, 2*pi, 200);
period = 1.5;
y1 = sin(2*pi*t1/period);
plot(t1, y1, 'bo-', 'LineWidth', 6, 'MarkerSize', 20);
grid on;
% Enlarge figure to full screen.
set(gcf, 'units','normalized','outerposition',[0 0 1 1]);
% Make signal #2
t2 = linspace(-2*pi, 2*pi, 130);
period = 1.5;
y2 = sin(2*pi*t2/period);
hold on;
plot(t2, y2, 'ys-', 'LineWidth', 4);
% Interpolate signal 2 up to have same number of samples as signal 1
y2Interp = interp1(t2, y2, t1);
plot(t1, y2Interp, 'rd-', 'LineWidth', 2);
legend('signal 1', 'signal 2', 'signal 2 interpolated');
% Now get signal half way between the signal #1
% and the interpolated signal #2
signal3 = 0.5 * y1 + 0.5 * y2Interp;
I trust you can do the case where you want to interpolate y1 down to the lesser number of samples that y2 has - it's straightforward.
  6 个评论
Image Analyst
Image Analyst 2014-1-16
You're not using interp correctly - look at the first two input arguments - they are the same and they should not be.
Gova ReDDy
Gova ReDDy 2014-1-16
编辑:Gova ReDDy 2014-1-16
Thanks IA ,
And if I am not wrong then the correct way should be the below and yes as you said,it is not giving the same values for both signals
Here the s1 samples< s2
s1_inter=interp1(1:length(s1), s1,1:length(s2));
%%It is adding Nan values from the length of s1 to the length of s2 making the samples in (or the length of) both same.%%
s2_inter=interp1(1:length(s2), s2,1:length(s1));
%%It is clipping off the values from length of s1 to the length of s2 making the samples in (or the length of) both same%%

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