# How to create a delayed version of a cosine signal?

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Varun on 24 Jul 2014
Commented: Varun on 25 Jul 2014
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.

Andrew Reibold on 24 Jul 2014
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?
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Varun on 25 Jul 2014
if s1 is the first signal and the s2 is the delayed signal then:
[c,lag]=xcorr(s1,s2);
[~,I] = max(abs(c));
T = lag(I)
plot(lag,c/max(c))
This is the piece of code im using to verify the delay. When I was using two different time vectors t and t1 the peak of the cross correlation plot which indicates the sample delay was zero. When I tried it using Azzis method the plot shows that the sample delay is 5 which should be correct. However the T value displayed by matlab obtained by T=lag(I) is still zero. When I plotted the two signals using the same time vector t I can see that at t=0, the first signal has a value of 1 while the second has a value of -1 indicating a 180 degree phase shift which should happen for a 0.05 second delay. This is also confirmed when I take the FFT of the two signals and plot the phase spectrum. So despite the T value being zero, Can I now accept that my second signal is in fact delayed by 0.05 seconds as required?

Azzi Abdelmalek on 24 Jul 2014
delay=5*st
y1=cos(2*pi*100*(t-delay)) % delayed signal