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Integral of SINC^2 f

Discussion in 'Mathematics and Physics' started by derick007, Feb 28, 2017.

  1. derick007

    derick007 Member

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    ∞​
    Ifsin^2 (f) / (f)^2 df =π
    -∞​

    ∞​
    and sin^2 (πf) / (πf)^2 df = 1
    -∞​

    Then does

    sin^2 (πfT) / (πfT)^2 df = 1/T
    -∞
     
    Last edited: Feb 28, 2017
  2. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Where do you see proof of that second statement:
    integral(sin(nf)^2/(nf)^2,f,-inf,inf)=1

    Try a couple different n's.
     
  3. derick007

    derick007 Member

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    Could we just clarify - there aren't any n's, only pi's (π).
    Apologies but the size and type of text makes the pi's look like n's.

    I found the second statement on line - I will try to find it again.
     
  4. dave

    Dave New Member

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  5. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Oh ok, ha ha, my reading mistake there :)
    Wow those pi's do look like n's a lot.

    No need to look it up again, it is apparent that the second formula is correct.

    For the third formula i am getting:
    -1/(2*pi^2*f*T^2)

    but this needs to be verified. Perhaps plug in a few T's and see if it always works out ok. It's usually easy to find a value that does not work if the formula is not right.
     
  6. derick007

    derick007 Member

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    My maths is very limited - I tried to integrate the sinc ^2 function by parts to no avail.

    I searched the internet for solutions to the integral and came across those above for sinc^2 (f) (= π) and sinc^2 (π f) (=f) and from these results wondered if sinc^2 (πfT) = 1/T ?

    I am trying to derive formula for the noise equivalent bandwidth of a filter with H(f) = h T sinc(πfT) and am fairly convinced = 1/2T, if the integral of sinc^2 (πfT) = 1/T.

    If this proves to be correct I will try to find the integral of sinc^2 (π(f-fc)T) ( I think it = 2/T, but would like proof). Again I will do this by searching the internet starting with BORWEIN.
     
  7. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Ok well the way i did it was i broke it down trigonometrically first then went on to do the integration on the two resulting terms. I did not double check the results though. After checking i see i must have done something wrong.

    Having checked the results, i am getting 1/T now too.

    Oh wait, i get 1/abs(T) actually not just 1/T.
     
    Last edited: Mar 5, 2017
  8. derick007

    derick007 Member

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    Hi

    It sounds as though you were able to solve this by using integration by parts ?

    As said before, I did some research on the internet and it appeared it could only be solved by using contour integrals or complex integrals whatever that means.

    I am not too bothered about the fact you get 1/abs(T) as I am dealing with time and frequencies.

    Thanks for your time, checking this out - much appreciated.

    I am not having much success with the second integration of sinc^2 (π(f-fc)T).

    I am thinking it must also be 1/T as from any integration point of view the only difference between it and the first integration is that this one is displaced in frequency by fc. The areas under the curves should be exactly the same ?
     
  9. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    I would tend to think that too.

    Contour integration would probably mean considering f to be a complex variable with real and imaginary parts:
    f=a+b*i

    You'd have to look into how to do that for this problem.

    If you try different values for T in the first problem you had, you quickly see that 1/abs(T) is probably right and of course for values that are only positive 1/T is probably right. So if you do the same for your new problem you should see the same. This works on a try by try basis because it's easier to integrate when T or Fc is a constant numerical value like 1,2,3,... etc. So it should work for every positive value tried for T or Fc after you replace it and then do the integration by your usual method. Granted this isnt a definite proof, but may still be useful for a given range of the variable.
     

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