$$\hat{x}[k] = \frac{1}{2\pi}\int_{-W}^{W}X(e^{j\omega})e^{j\omega k}d\omega,\label{ift}\tag1$$


$$X(e^{j\omega}) = \sum_{n=-\infty}^{+\infty} x[n]e^{-j\omega n}\label{dft}\tag2$$


$$d[k] = x[k] - \hat{x}[k]\text.\label{error}\tag3$$

Show that energy of $d[k]$ goes to zero as $W\to \pi$. That is, $$\lim_{W \to \pi} \sum_{k = -\infty}^{+\infty}|d[k]|^2 = 0\text.$$

My Attempt

Main problem for me is removing magnitude in some way. Writing $|d[k]|^2 = d[k]d^{*}[k]$ complicates equations. Also if we use Parseval's identity, the magnitude is still present. I don't know if we are allowed to move limit inside the summation or not. Even assuming it's possible, I couldn't found a way for calculating $\lim_{W \to \pi} |d[k]|^2$.

  • $\begingroup$ can I rename your $e[n]$ to $d[n]$ to avoid confusion with $e^\cdot$? $\endgroup$ – Marcus Müller Jun 22 '20 at 17:13
  • $\begingroup$ @MarcusMüller Yes, surely. $\endgroup$ – S.H.W Jun 22 '20 at 17:14

HINT: The error energy can be written as


Now compute the integral


and show that $\lim_{W\to\pi}I(k-n,W)=\delta[k-n]$. This reduces the second sum in $(1)$ to a single element and the result follows.

  • $\begingroup$ Thanks. How it's possible to justify moving limit inside summation and then inside magnitude and again summation? $\endgroup$ – S.H.W Jun 22 '20 at 18:02
  • 1
    $\begingroup$ @S.H.W: If a function is continuous you can move the limit inside. $\endgroup$ – Matt L. Jun 22 '20 at 19:32
  • $\begingroup$ That's right but what about moving limit inside summation? $\endgroup$ – S.H.W Jun 22 '20 at 19:33
  • $\begingroup$ @S.H.W: No difference, the same argument still holds. $\endgroup$ – Matt L. Jun 22 '20 at 19:40
  • 1
    $\begingroup$ You generally have to assume convergence of the sums, but note that in this case we don't actually evaluate a limit but we just evaluate a function value: we can just set $W=\pi$. $\endgroup$ – Matt L. Jun 22 '20 at 21:00

[Apologizing: most of the following is already covered by Matt L.]

Expression $$X(e^{j\omega}) = \sum_{n=-\infty}^{+\infty} x[n]e^{-j\omega n}$$ is not valid in general, because we do not know whether it exists. One (classical) sufficient condition for the DTFT (discrete-time Fourier transform) is that the sequence $x[n]$ is summable. In other words, let us suppose that:

$$\sum_{n=-\infty}^{+\infty} |x[n]| < \infty\,.$$

Then, one can deduce that $X(e^{j\omega})$ exists, is $2\pi$ periodic, continuous (and bounded, but we won't use that). Since $X(e^{j\omega})$ is continuous, the following integral exists:

$$\hat{x}[k] = \frac{1}{2\pi}\int_{-W}^{W}X(e^{j\omega})e^{j\omega k}\mathrm{d}\omega,$$

which we can write as:

$$\hat{x}[k] = \frac{1}{2\pi}\int_{-W}^{W}\left(\sum_{n=-\infty}^{+\infty} x[n]e^{-j\omega n}\right)e^{j\omega k}\mathrm{d}\omega,$$


$$\hat{x}[k] = \frac{1}{2\pi}\int_{-W}^{W}\left(\sum_{n=-\infty}^{+\infty} x[n]e^{-j\omega n}e^{j\omega k}\right)\mathrm{d}\omega.$$

With the $\ell_1$ absolute summability stated at the beginning, one can apply summation/integral inversion, also known as Fubini-Tonelli theorems, see also: SE.Maths: When can a sum and integral be interchanged?

Then rewrite $$x[k] = \frac{1}{2\pi}\left(\int_{-W}^{W}x[k]\mathrm{d}\omega+\int_{-\pi}^{-W}x[k]\mathrm{d}\omega+\int_{W}^{\pi}x[k]\mathrm{d}\omega\right)$$

The two last terms vanish to zero, and the first one can be combined with $\hat{x}[k]$ in the difference. Then, a discrete or periodic cardinal sine appears.

Note : this may hold in weaker conditions than $\ell_1$, but honeslty, I do not know them precisely.


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