Expectation of terms in double summation

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Discussion Overview

The discussion revolves around finding the average (expectation) of terms involving a double summation in a mathematical expression. The context includes theoretical exploration related to signal processing, specifically in deriving Peak-to-Average Power Ratio (PAPR) for QAM OFDM signals.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant presents an equation involving a double summation and seeks assistance in calculating its expectation.
  • Another participant questions whether the average being sought is a time-average and mentions that it typically involves an integral with respect to time.
  • A participant confirms they are looking for a time-average and clarifies that the equation is not part of an assignment but rather a derivation related to signal processing.
  • One participant notes that the cosine terms generally have a time-average of 0, except when j equals k, and suggests that since j is never equal to k in the summation, the average would be 0.
  • A later reply expresses gratitude for the clarification regarding the average, indicating that the response confirmed their understanding of the situation.

Areas of Agreement / Disagreement

Participants express some agreement on the nature of the time-average of the cosine terms, particularly that they average to 0 when j is not equal to k. However, there is no consensus on the broader implications or any additional methods for calculating the expectation.

Contextual Notes

The discussion does not resolve the mathematical steps involved in calculating the expectation or the specific conditions under which the average is taken. There are also limitations regarding the assumptions about the functions f(k,j) and θ(k,j) being time-independent.

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.

[itex]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][/itex]
where f(k,j) and [itex]\theta_{k,j}[/itex] 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.
 
Last edited:
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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