# 2.12 Normalized dtft as an operator

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## Normalized dtft as an operator

Note that by taking the DTFT of a sequence we get a function defined on $\left[-\pi ,\pi \right]$ . In vector space notation we can view the DTFT as an operator (transformation). In this context it is useful to consider the normalizedDTFT

$\mathcal{F}\left(x\right):=X\left({e}^{j\omega }\right)=\frac{1}{\sqrt{2\pi }}\sum _{n=-\infty }^{\infty }x\left[n\right]{e}^{-j\omega n}.$

One can show that the summation converges for any $x\in {\ell }_{2}\left(\pi \right)$ , and yields a function $X\left({e}^{j\omega }\right)\in {L}_{2}\left[-\pi ,\pi \right]$ . Thus,

$\mathcal{F}:{\ell }_{2}\left(\mathbb{Z}\right)\to {L}_{2}\left[-\pi ,\pi \right]$

can be viewed as a linear operator!

Note: It is not at all obvious that $\mathcal{F}$ can be defined for all $x\in {\ell }_{2}\left(\mathbb{Z}\right)$ . To show this, one can first argue that if $x\in {\ell }_{1}\left(\mathbb{Z}\right)$ , then

$\begin{array}{cc}\hfill \left|X,\left(,{e}^{j\omega },\right)\right|& \le \left|\frac{1}{\sqrt{2\pi }},\sum _{n=-\infty }^{\infty },x,\left[n\right],{e}^{-j\omega n}\right|\hfill \\ & \le \frac{1}{\sqrt{2\pi }}\sum _{n=-\infty }^{\infty }\left|x,\left[,n,\right]\right|\left|{e}^{-jwn}\right|\hfill \\ & =\frac{1}{\sqrt{2\pi }}\sum _{n=-\infty }^{\infty }\left|x,\left[,n,\right]\right|<\infty \hfill \end{array}$

For an $x\in {\ell }_{2}\left(\mathbb{Z}\right)\setminus {\ell }_{1}\left(\mathbb{Z}\right)$ , one must show that it is always possible to construct a sequence ${x}_{k}\in {\ell }_{2}\left(\mathbb{Z}\right)\cap {\ell }_{1}\left(\mathbb{Z}\right)$ such that

$\underset{k\to \infty }{lim}{\parallel {x}_{k}-x\parallel }_{2}=0.$

This means $\left\{{x}_{k}\right\}$ is a Cauchy sequence, so that since ${\ell }_{2}\left(\mathbb{Z}\right)$ is a Hilbert space, the limit exists (and is $x$ ). In this case

$X\left({e}^{j\omega }\right)=\underset{k\to \infty }{lim}{X}_{k}\left({e}^{j\omega }\right).$

So for any $x\in {\ell }_{2}\left(\mathbb{Z}\right)$ , we can define $\mathcal{F}\left(x\right)=X\left({e}^{j\omega }\right)$ , where $X\left({e}^{j\omega }\right)\in {L}_{2}\left[-\pi ,\pi \right]$ .

Can we always get the original $x$ back? Yes, the DTFT is invertible

${\mathcal{F}}^{-1}\left(X\right)=\frac{1}{\sqrt{2\pi }}{\int }_{-\pi }^{\pi }X\left({e}^{j\omega }\right)·{e}^{j\omega n}d\omega$

To verify that ${\mathcal{F}}^{-1}\left(\mathcal{F}\left(x\right)\right)=x$ , observe that

$\begin{array}{cc}\hfill \frac{1}{\sqrt{2\pi }}{\int }_{-\pi }^{\pi }\left(\frac{1}{\sqrt{2\pi }},\sum _{k=-\infty }^{\infty },x,\left[k\right],{e}^{-j\omega k}\right){e}^{j\omega n}d\omega & =\frac{1}{2\pi }\sum _{k=-\infty }^{\infty }x\left[k\right]{\int }_{-\pi }^{\pi }{e}^{-j\omega \left(k-n\right)}d\omega \hfill \\ & =\frac{1}{2\pi }\sum _{k=-\infty }^{\infty }x\left[k\right]·2\pi \delta \left[n-k\right]\hfill \\ & =x\left[n\right]\hfill \end{array}$

One can also show that for any $X\in {L}_{2}\left[-\pi ,\pi \right]$ , $\mathcal{F}\left({\mathcal{F}}^{-1}\left(X\right)\right)=X$ .

Operators that satisfy this property are called unitary operators or unitary transformations . Unitary operators are nice! In fact, if $A=X\to Y$ is a unitary operator between two Hilbert spaces, then one can show that

$⟨{x}_{1},{x}_{2}⟩=⟨A{x}_{1},A{x}_{2}⟩\phantom{\rule{1.em}{0ex}}\forall \phantom{\rule{4pt}{0ex}}{x}_{1},{x}_{2}\in X,$

i.e., unitary operators obey Plancherel's and Parseval's theorems!

#### Questions & Answers

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