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

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.

• Doesn't your first sentence give the answer to your question? – Matt L. May 21 '13 at 15:00
• This transform is somehow derived from the continuous transform,and the definition for inverse of it isn't arbitrary at all in CFT.If you don't integrate over all real axis, you will have wrong answer for $x$. Because this inversion formula must give you your original function,from which you have obtained the spectrum. – Zorich May 21 '13 at 15:10
• Yes, but there's a difference between continuous and discrete-time transforms, as you have correctly pointed out. You can see the inverse transform in the discrete-time case simply as computing the Fourier series coefficients of the periodic spectrum. – Matt L. May 21 '13 at 15:32
• Short answer: for the same reason that when calculating a Fourier series you only integrate over one period of the waveform instead of over the entire real line (as you do in the Fourier transform). It's due to the periodicity of the input (in your case, the periodic DTFT spectrum). – Jason R May 21 '13 at 15:36

## 3 Answers

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.)

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 :

$$X(e^{j\omega}) = $$

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.

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 $(-\infty,\infty)$.