Auto-correlation function for a time signal












0












$begingroup$


For a time signal $x(t)= cos^2(2pi t)$, solve for the Auto-correlation Function.



Basically I have been following a guide on how to compute the auto-correlation function and am getting hung-up on this Trig equation. The steps I have taken so far are




  1. $cos^2(2pi t) = (1/2)[1+cos(4pi t)]$


  2. the integral I used to solve for the correlation function (1/T) in front of the integral
    My main question is if this is the correct process, how would I go about solving this integral when the time delay is present in the second cosine?










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  • $begingroup$
    I'm not sure what your question is?
    $endgroup$
    – Mattos
    Dec 4 '18 at 23:26


















0












$begingroup$


For a time signal $x(t)= cos^2(2pi t)$, solve for the Auto-correlation Function.



Basically I have been following a guide on how to compute the auto-correlation function and am getting hung-up on this Trig equation. The steps I have taken so far are




  1. $cos^2(2pi t) = (1/2)[1+cos(4pi t)]$


  2. the integral I used to solve for the correlation function (1/T) in front of the integral
    My main question is if this is the correct process, how would I go about solving this integral when the time delay is present in the second cosine?










share|cite|improve this question











$endgroup$












  • $begingroup$
    I'm not sure what your question is?
    $endgroup$
    – Mattos
    Dec 4 '18 at 23:26
















0












0








0





$begingroup$


For a time signal $x(t)= cos^2(2pi t)$, solve for the Auto-correlation Function.



Basically I have been following a guide on how to compute the auto-correlation function and am getting hung-up on this Trig equation. The steps I have taken so far are




  1. $cos^2(2pi t) = (1/2)[1+cos(4pi t)]$


  2. the integral I used to solve for the correlation function (1/T) in front of the integral
    My main question is if this is the correct process, how would I go about solving this integral when the time delay is present in the second cosine?










share|cite|improve this question











$endgroup$




For a time signal $x(t)= cos^2(2pi t)$, solve for the Auto-correlation Function.



Basically I have been following a guide on how to compute the auto-correlation function and am getting hung-up on this Trig equation. The steps I have taken so far are




  1. $cos^2(2pi t) = (1/2)[1+cos(4pi t)]$


  2. the integral I used to solve for the correlation function (1/T) in front of the integral
    My main question is if this is the correct process, how would I go about solving this integral when the time delay is present in the second cosine?







correlation data-analysis






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share|cite|improve this question













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share|cite|improve this question








edited Dec 4 '18 at 23:18









Carmeister

2,8442924




2,8442924










asked Dec 4 '18 at 23:04









JamesJames

1




1












  • $begingroup$
    I'm not sure what your question is?
    $endgroup$
    – Mattos
    Dec 4 '18 at 23:26




















  • $begingroup$
    I'm not sure what your question is?
    $endgroup$
    – Mattos
    Dec 4 '18 at 23:26


















$begingroup$
I'm not sure what your question is?
$endgroup$
– Mattos
Dec 4 '18 at 23:26






$begingroup$
I'm not sure what your question is?
$endgroup$
– Mattos
Dec 4 '18 at 23:26












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