# Showing error energy goes to zero

Let

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

where

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

Also,

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

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 $$\lim_{W \to \pi} |d[k]|^2$$.

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

HINT: The error energy can be written as

\begin{align}\sum_k\big|d[k]\big|^2&=\sum_k\left|x[k]-\frac{1}{2\pi}\int_{-W}^W\sum_nx[n]e^{-jn\omega}e^{jk\omega}d\omega\right|^2\\&=\sum_k\left|x[k]-\sum_nx[n]\frac{1}{2\pi}\int_{-W}^We^{j(k-n)\omega}d\omega\right|^2\tag{1}\end{align}

Now compute the integral

$$I(k-n,W)=\frac{1}{2\pi}\int_{-W}^We^{j(k-n)\omega}d\omega\tag{2}$$

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.

• Thanks. How it's possible to justify moving limit inside summation and then inside magnitude and again summation? – S.H.W Jun 22 at 18:02
• @S.H.W: If a function is continuous you can move the limit inside. – Matt L. Jun 22 at 19:32
• That's right but what about moving limit inside summation? – S.H.W Jun 22 at 19:33
• @S.H.W: No difference, the same argument still holds. – Matt L. Jun 22 at 19:40
• 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 $W=\pi$. – Matt L. Jun 22 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:

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

or

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

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.