Expectation of terms in double summation

[singhofmpl]

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.

 

[Xitami]

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 ]

 

[Redbelly98]

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...

 

[singhofmpl]

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.

 

[Redbelly98]

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 - θ_k,j]=cos(θ_k,j). However, it appears that j is never equal to k in this summation, so we are left with 0 for the average.

 

[singhofmpl]

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 - θ_k,j]=cos(θ_k,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.