Why integrate over $2\pi$ in inverse DTFT?

Question

In DTFT of a signal, the spectrum of a sequence is periodic with period $2\pi$ and all the information needed for derivation of the original signal from its spectrum is contained in $\pi <\omega <\pi$.

But , why do they integrate over $2\pi$ and not from $-\infty$ to $+\infty$ , as in continuous Fourier transform in inverse DTFT? After all, when you derive a transform of a signal, the inverse of it is not arbitrary and to recover the original signal, you can't integrate over an other interval because all the information is contained in there.

Answer 1

There are two reasons:

1. As you say, the sequence is periodic so all of the information is contained in the $-\pi < w <\pi$ region. Thus, integrating over that region tells you everything that there is to be learned from the inverse transform.
2. Integrating from $-\infty < w< \infty$ would only give answers of $0$, $\infty$, or $-\infty$. The reason for that is that integrating over $-\pi < w <\pi$ will generally give a finite answer which can be something negative (e.g. -4), 0, or something positive (e.g. 2.5). If you integrate over $-\infty < w< \infty$ each $2\pi$ region will give the exact same answer, so the negative results will accumulate to $-\infty$, the zeros will sum to $0$, and the positive results will sum to $\infty$. (Note that the results are often complex, but that doesn't change the primary point. It just increases the dimensionality of the infinities.)

Answer 2

To answer your question, we need to understand DTFT as change of basis of the vector $x[n]$. Lets first agree that DTFT of an infinite length sequence $x[n]$ will be defined only if the DTFT sum converges which means :

$X(e^{jw}) = \sum^{\infty}_{-\infty} x[n].e^{-j\omega n}$ must converge and be finite.

This happens for all square summable sequences, i.e. all infinite length sequences with finite energy. Such sequences live in $l_{2}(\mathbb Z)$.

Now, look at the DTFT formula as inner product of two vectors : $x[n]$ and $e^{j\omega n}$ for $\omega \in \mathbb R $. Therefore, we can write :

$$X(e^{j\omega}) = <x, e^{j\omega n}>$$

What this means is $X(e^{j\omega})$ is nothing but component of $x[n]$ along the basis vector $e^{j\omega n}$. It can be proved that the basis vectors $e^{j\omega n}$ are $2\pi$-Periodic $\forall$ $\omega$ and, are orthogonal.

Basically, the point is DTFT operator maps $l_{2} (\mathbb Z)$ onto $L_{2}([-\pi, \pi])$, which is a space of $2\pi$-periodic functions.

Inverse DTFT is just representing $x[n]$ as sum of the basis vectors $e^{j\omega n}$ weighted by their corresponding components $X(e^{j\omega})$. Since, basis vectors are defined for all $\omega \in \mathbb R$, hence the sum will become integral.

Now the question comes, why to integrate in a period of $2\pi$ and not $\pi$ or $3\pi$ or $m\pi$.

Because when we substitute $\sum^{\infty}_{-\infty}x[m].e^{-j\omega m}$ in place of $X(e^{j\omega})$ in the Inverse DTFT integral, we will get the following :

$$\frac{1}{2\pi} \sum^{\infty}_{-\infty} x[m]. \int_{-\pi}^{\pi}e^{-j\omega [m-n]} d\omega, \forall m,n \in \mathbb I$$. Evaluate this integral to get the following :

$$\int_{-\pi}^{\pi} e^{-j\omega [m-n]}d\omega = 2\pi \frac{sin(\pi[m-n])}{\pi[m-n]} = 2\pi \delta[m-n]$$.

This $2\pi\delta[m-n]$ will only be obtained if you integrate on any $2\pi$ interval not otherwise. And, now we can use the nice sifting property of $\delta[k]$ function to pick $x[n]$ because, $\delta[m-n]$ will only be 1 when $m=n$ and 0 otherwise.

Continuing the Inverse DTFT evaluation :

$$\frac{1}{2\pi} \sum^{\infty}_{-\infty} x[m].2\pi\delta[m-n] = \frac{2\pi}{2\pi} .x[n] = x[n]$$.

Integrate over any other interval and you lose orthogonality of the basis vectors and you won't get $x[n]$ back. for example, if you integrated from $-2\pi$ to $2\pi$, you would have gotten $4\pi \frac{sin(2\pi [m-n])}{2\pi [m-n]}$ which will be 0 always. And you can verify yourself that integrating over an interval of $3\pi$ will give you something like $\frac{-Sin(\pi [m-n]/2)}{[m-n]}$ which will be 1 and -1 for many combinations of m and n and hence won't let you use the nice sifting property of $\delta[k]$ to get $x[n]$ back.

Answer 3

Mathematically , you have to integrate over $2\pi$: $$X(\omega) = \sum_{n=-\infty}^{\infty} x[n] \,e^{-i \omega n}$$

$$x[n]= \frac{1}{2 \pi}\int_{2\pi} X(\omega)\cdot e^{i \omega n} d\omega= \frac{1}{2 \pi}\int_{2\pi} \sum_{k=-\infty}^{\infty} x[k] \,e^{-i \omega k}\cdot e^{i \omega n} d\omega $$ because of orthogonality of sinusoidals , this will be zero for all $k$s except $k=n$: $$\to x[n]= \frac{1}{2 \pi}\int_{2\pi}x[n]d\omega $$

In the case of a continuous Fourier transform you'll arrive at a delta function and you have to integrate over $(-\infty,\infty)$.