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Hi I want to change the variable $(n - m)$ into $p$ in the sumation bellow but I don't know how to change the indices of the sumation and how to convert the 3 sigmas into 2 sigmas.$J_0$ is zero order bessel function of first kind. and $a$ and $b$ are constants.

$$ \sum_{n=0}^{N-1} \sum_{m=0}^{N-1} \sum_{l=0}^{L-1} J_0\big( 2(n-m)\pi f_d T_s \big) \exp\left( \frac{-2i(n-m)\pi}{N} \right)\times \exp\left( {-2i\cot(a)b^2\pi(N-l)(n-m)} \right) $$

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    $\begingroup$ Do you know how to use $\LaTeX$? This is very difficult to read! $\endgroup$ – Sharkos Apr 19 '13 at 12:25
  • $\begingroup$ If my formatting is correct, what is $J_0$? Also I tried to stick to the original formula, but I don't see 'k' anywhere, nor any sigma... $\endgroup$ – Jonathan H Apr 19 '13 at 13:00
  • $\begingroup$ $J_0$ is zero order bessel function of the first kind. I have also edited the equasion in the question. $\endgroup$ – zahra Apr 19 '13 at 13:44
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Just make the substitution $p = m-n$ everywhere ($p$ go to $0-n=-n$ to $N-1-n$) :

$$\sum_{n=0}^{N-1}\sum_{p=-n}^{N-1-n}\sum_{l=0}^{L-1}J_0(-2p\pi f_dT_s)\exp(\frac{2ip\pi}{N})\exp(...)$$

the since $n$ no longer in use in the equation, you can enumerate the number of times that you have to enumerate each $p$ :

  • for n = 0, you have p going to 0 to N-1
  • for n = 1, p go from -1 to N-2
  • ...
  • for n = N-1, p go from -N+1 to 0

So if you enumerate the number of times you have to some over each p, you have :

  • p = N-1 : 1 time (for n=0)
  • p = N-2 : 2 times (for n=0 and n=2)
  • ...
  • p = 0 : N times (foreach n)
  • ...
  • p = -N+1 : 1 time (for n=N-1)

It gives you $N-|p|$ sum for a given p and then :

$$\sum_{p=-N+1}^{N-1}\sum_{l=0}^{L-1}(N-|p|)J_0(2p\pi f_dT_s)\exp(\frac{2ip\pi}{N})\exp(...)$$

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  • $\begingroup$ Your summation starts at $2n-1$. $\endgroup$ – gnometorule Apr 19 '13 at 13:50
  • $\begingroup$ @gnometorule which one? $\endgroup$ – Samuel Caillerie Apr 19 '13 at 13:54
  • $\begingroup$ $N-(-N+1) = 2N -1$ $\endgroup$ – gnometorule Apr 19 '13 at 13:57
  • $\begingroup$ @gnometorule I don't see where but be careful of the minus $n$ and capital $N$ (as in the OP question)... $\endgroup$ – Samuel Caillerie Apr 19 '13 at 13:58
  • $\begingroup$ I agree to all up to "...for a given p and then": they all appear 1 to N times. But the first term of your sum counts down from $2N-1$ to $1$. $\endgroup$ – gnometorule Apr 19 '13 at 14:08

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