How to create a delayed version of a cosine signal?
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I have defined a cosine signal with 100 Hz frequency using the code:
st=1/1000; % sampling frequency Fs=1000;
t=[0:999]*st;
cos(2*pi*100*t);
I want to create a second cosine signal of the same frequency and length but with a time delay of 0.0050 seconds. I tried creating a separate vector and using that time vector to define the signal.
t1=[50:1049]*st;
cos(2*pi*100*t1)
But when I use the cross correlation function to verify the delay between the signals it shows that the delay is zero.
Please help me out here. Thank you
Réponses (2)
Andrew Reibold
le 24 Juil 2014
1 vote
Your second cosine function definitely IS shifting. Its just that you are shifting it over exactly 5 periods, so it lies on top of itself again.
I'm not sure which cross correlation function you are talking about, but perhaps it only only looks at delay between overlapping periods, in your case which may be 0. Does that make sense?
2 commentaires
Andrew Reibold
le 24 Juil 2014
PS. You can verify it is shifting visually with a simple plot :)
figure;
plot(t,cos(2*pi*100*t),t1,cos(2*pi*100*t));
Varun
le 25 Juil 2014
Azzi Abdelmalek
le 24 Juil 2014
delay=5*st
y1=cos(2*pi*100*(t-delay)) % delayed signal
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