Auto-correlation function for a time signal
$begingroup$
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
- $cos^2(2pi t) = (1/2)[1+cos(4pi t)]$
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
edited Dec 4 '18 at 23:18
Carmeister
2,7992923
asked Dec 4 '18 at 23:04
JamesJames
1
$endgroup$
add a comment |
$begingroup$
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
- $cos^2(2pi t) = (1/2)[1+cos(4pi t)]$
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
edited Dec 4 '18 at 23:18
Carmeister
2,7992923
asked Dec 4 '18 at 23:04
JamesJames
1
$endgroup$
$begingroup$
I'm not sure what your question is?
$endgroup$
– Mattos
Dec 4 '18 at 23:26
add a comment |
$begingroup$
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
- $cos^2(2pi t) = (1/2)[1+cos(4pi t)]$
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
edited Dec 4 '18 at 23:18
Carmeister
2,7992923
asked Dec 4 '18 at 23:04
JamesJames
1
$endgroup$
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
- $cos^2(2pi t) = (1/2)[1+cos(4pi t)]$
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
correlation data-analysis
edited Dec 4 '18 at 23:18
Carmeister
2,7992923
asked Dec 4 '18 at 23:04
JamesJames
1
edited Dec 4 '18 at 23:18
Carmeister
2,7992923
asked Dec 4 '18 at 23:04
JamesJames
1
edited Dec 4 '18 at 23:18
Carmeister
2,7992923
edited Dec 4 '18 at 23:18
Carmeister
2,7992923
edited Dec 4 '18 at 23:18
Carmeister
2,7992923
2,7992923
asked Dec 4 '18 at 23:04
JamesJames
1
asked Dec 4 '18 at 23:04
JamesJames
1
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
I'm not sure what your question is?
$endgroup$
– Mattos
Dec 4 '18 at 23:26