Why is the DTFT (Discrete Time Fourier Transform) unique to each input?

Question

As the title implies. I know the DFT of a signal is unique due to the matrix, but can anyone give a solid explanation as to why the DTFT is unique for each signal input?

Thanks for your time!

Answer 1

We can view the DTFT as a limit of the DFT of an increasingly long signal, i.e. as the signal length $N \to \infty$, it makes sense to replace the fundamental frequency $2\pi/N$ with a continuous variable $\omega$ instead, and the DFT sum becomes an integral. So if we accept that the DFT is unique, then the DTFT is too.

Alternatively, we can view the DTFT as a change of basis; Michael's response above gives some details on this.

Answer 2

The DFT, DTFT, and FT are all unitary operators on Hilbert spaces (on appropriate domains and up to suitable normalization). They take one orthonormal basis to another. In the case of DTFT, this is clear:

$$ \delta_n \mapsto e^{- 2 \pi n \omega} $$

where $\delta_n$ is the sequence that is $1$ at $n$ and zero elsewhere. Such operators are trivially one-to-one.