Expectation of terms in double summation

AI Thread Summary
The discussion centers on calculating the time average (expectation) of a double summation involving a cosine function. The equation presented involves terms where f(k,j) and θ(k,j) are functions of k and j. It is noted that the cosine terms average to zero unless j equals k, but in this case, j never equals k within the summation. Consequently, the overall average simplifies to zero. The participants confirm the correctness of this conclusion based on the provided insights.
singhofmpl
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Does anybody help me how to find the average (expectation) of terms involving double summation? Here is the equation which I'm trying solve.

[\tex]E\Big[2\sum_{k=0}^{N-2}\sum_{j=k+1}^{N-1}f(k,j)\cos[2\pi(j-k)t-\theta_{k,j}]\Big][\tex]
where f(k,j) and [\tex]\theta_{k,j}[\tex] are some function of k and j.
 
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singhofmpl said:
does anybody help me how to find the average (expectation) of terms involving double summation? Here is the equation which I'm trying solve.

e\big[2\sum_{k=0}^{n-2}\sum_{j=k+1}^{n-1}f(k,j)\cos[2\pi(j-k)t-\theta_{k,j}]\big]
where f(k,j) and \theta_{k,j} are some function of k and j.

[ latex ] expresion [ /latex ]
 
Are you wanting to take the time-average? Have you been given the equation for doing that -- it involves doing an integral with respect to time...
 
Redbelly98 said:
Are you wanting to take the time-average? Have you been given the equation for doing that -- it involves doing an integral with respect to time...

Yes I want to take the time average of the said expression. Its not part of any assignment. I came across this equation while deriving PAPR for the QAM OFDM signal. It does not involve any integral.
 
The cosine terms all have a time-average of 0, except when j=k. As long as the f's and θ's are time-independent, this simplifies things greatly.

When j=k, the time-average of cos[2π(j-k)t - θk,j]=cos(θk,j). However, it appears that j is never equal to k in this summation, so we are left with 0 for the average.
 
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Redbelly98 said:
The cosine terms all have a time-average of 0, except when j=k. As long as the f's and θ's are time-independent, this simplifies things greatly.

When j=k, the time-average of cos[2π(j-k)t - θk,j]=cos(θk,j). However, it appears that j is never equal to k in this summation, so we are left with 0 for the average.

Dear Sir
Thanks a lot for your prompt response. After your comment now I've got the confirmation of my answer. Thanks a lot once again.
 
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