While understanding the proof of DTFT from Signals and Systems by Oppenheim, I have confusion in understanding few steps.

$$ x'[n]=\sum_ {k=<N>} a_ke^{jk(2\pi/N)n}$$ $$ a_k= \frac{1}{N} \sum_ {n=<N>} x'[n]e^{-jk(2\pi/N)n}$$ Since $x[n]=x'[n]$ over a perid that includes the interval $-N_1 <= n <= N_2$, it is convinent to choose the interval of summation to include this inteval, so that $x'[n]$ can replaced by $x[n]$ in the summation. Therefore, $$ a_k= \frac{1}{N} \sum_ {n=N_1}^{N_2} x[n]e^{-jk(2\pi/N)n} = \frac{1}{N} \sum_ {n=-\infty} ^{+\infty} x[n]e^{-jk(2\pi/N)n}$$ While in the second equality we have used the fact that $x[n]$ is zero outside the interval $-N_1 <= n <= N_2$ enter image description here I have confusion in understanding that how summation from $-N_1 N_2$ changes to $-\infty +\infty$. The statement they mentioned above is so confusioning for me.

  • $\begingroup$ If $x'[n]$ is the periodic continuation of $x[n]$, then the last equality is wrong. In the sum from $-\infty$ to $\infty$ the (finite-duration) signal $x[n]$ must be used, not the periodic signal. $\endgroup$ – Matt L. Nov 9 '15 at 10:12
  • $\begingroup$ @MattL. I have corrected that mistake.. $\endgroup$ – Aadnan Farooq A Nov 9 '15 at 10:14
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    $\begingroup$ Yes, but now it's obvious, since $x[n]$ is zero outside the interval $[N_1,N_2]$ you can sum over as many terms as you like because they're all zero. $\endgroup$ – Matt L. Nov 9 '15 at 10:17
  • $\begingroup$ but first we have $x[n]$ which was aperiodic and then we have $x'[n]$ which was periodic and making period to infinity so that it should be same as $x[n]$ that $-\infty to +\infty$ is because of that? $\endgroup$ – Aadnan Farooq A Nov 9 '15 at 10:20
  • $\begingroup$ @MattL.is it correct? $\endgroup$ – Aadnan Farooq A Nov 9 '15 at 11:55

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