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Note that by taking the DTFT of a sequence we get a function defined on $[-\pi ,\pi ]$ . In vector space notation we can view the DTFT as an operator (transformation). In this context it is useful to consider the normalizedDTFT
One can show that the summation converges for any $x\in {\ell}_{2}\left(\pi \right)$ , and yields a function $X\left({e}^{j\omega}\right)\in {L}_{2}[-\pi ,\pi ]$ . Thus,
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
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
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
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 $X\left({e}^{j\omega}\right)\in {L}_{2}[-\pi ,\pi ]$ .
Can we always get the original $x$ back? Yes, the DTFT is invertible
To verify that ${\mathcal{F}}^{-1}\left(\mathcal{F}\left(x\right)\right)=x$ , observe that
One can also show that for any $X\in {L}_{2}[-\pi ,\pi ]$ , $\mathcal{F}\left({\mathcal{F}}^{-1}\left(X\right)\right)=X$ .
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
i.e., unitary operators obey Plancherel's and Parseval's theorems!
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