Matched filtering syntax problem

Hello, i am trying to apply a Matchedfilter on a chirp signal on Matlab, and since im a beginner i couldnt solve the problem i had with this syntax. Can anyone help me plz?

t=0:0.001:2;
y= chirp(t,19,10,24);
x= step(y);
z= phased.MatchedFilter('filter',getMatchedFilter(y));
a= step(z,x);
plot(a);

Answer 1

[Honglei Chen]

You are mixing several things here. y in your script is a signal vector, not a waveform, so it does not work with the step() method.

You can do either

t=0:0.001:2;
y= chirp(t,19,10,24);
a= filter(fliplr(y),1,y);
plot(a);

or

t=0:0.001:2;
y= chirp(t,19,10,24);
myFilter = phased.MatchedFilter('Coefficients',flipud(y(:)));
a= step(myFilter,y(:));
plot(real(a));

Note that MatchedFilter handles complex signal and assumes each column as a channel. It also handles the state in case you want to stream the filtering operation.

HTH

[Hana]

Thank you Honglei. I tried this script (matched filtering+ cross correlation)

t=0:0.1:10;
y= chirp(t,19,10,40);

k=awgn(y,10);
a= filter(fliplr(y),1,y);

subplot(4,1,1);
plot(y);

subplot(4,1,2);
plot(a);

s= xcorr(y,y);

subplot(4,1,3);
plot(s);

but im not sure of what i got here, the matched filter signal doesnt look right.

[Honglei Chen]

These are just matched filter results. In this case, filter and xcorr does essentially the same thing. However, note that the horizontal axes are different. For xcorr, it outputs all lags, from -(N-1) to (N-1), where N is the length of the input signal. That's why you see the peak at the middle. On the other hand, filter only outputs N samples so you see the peak, at the end. The result of filter is basically the first half of the result using xcorr. You could argue that xcorr result is more intuitive but in real applications input and output are often the same length so filter makes more sense there. If you want to see the rest, just pass in another zero frame, you'll see the rest, e.g.,

a = filter(fliplr(y),1,y);
a1 = filter(fliplr(y),1,zeros(size(y)));

HTH