2.9b.1 Discrete-Time Fourier Transform

Discrete-Time Fourier Transform (DTFT)

The Discrete-Time Fourier Transform is a mathematical tool used in signal processing to analyze the frequency content of non-periodic discrete-time signals into a continuous functions of frequency, providing a complete representation of the signal’s frequency spectrum, such that
$x[n]=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }X(\omega )e^{j\omega n}d\omega ⟺X(\omega )={\sum }_{n=-\infty }^{\infty }x[n]e^{-j\omega n}.$

Intuition

We can easily take the analysis equation of a discrete-time Fourier transform just as
$X(\omega )={\sum }_{n=-\infty }^{\infty }x[n]e^{-j\omega n},$
since this is simply the frequency response of $x[n]$. The intuition behind this equivalence is that both $X(\omega )$ and $H(\omega )$ identically describe how $x[n]$ behaves in the frequency domain.

However, the question is to find the synthesis equation. To proceed, we can equate each of the basis vectors to ${\varphi }_n(\omega )=e^{-j\omega n}$ and find an orthogonal basis by imagining $X(\omega )$ as a weighted linear combination of the basis vectors. We know that $X(\omega )$ is periodic over $2\pi$ (and therefore the DTFT is periodic in frequency). How do we define an inner product in a periodic frequency space? We proceed by defining an inner product of any two $2\pi$-periodic frequency vectors $F,G$, such that the inner product is integrated over a period of $2\pi$:
$⟨F,G⟩={\int }_{-\pi }^{\pi }F(\omega )G\text{textasteriskcentered}(\omega )d\omega .$
Then,
$⟨H,{\varphi }_k⟩={\sum }_{n=-\infty }^{\infty }x[n]⟨{\varphi }_n,{\varphi }_k⟩=x[k]⟨{\varphi }_k,{\varphi }_k⟩⟺x[n]=\frac{⟨H,{\varphi }_k⟩}{⟨{\varphi }_k,{\varphi }_k⟩}.$
Since $⟨{\varphi }_k,{\varphi }_k⟩=2\pi$ (the Fourier basis of $X(\omega )$ is $2\pi$-periodic), then
$x[n]=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }X(\omega ){\varphi }_n\text{textasteriskcentered}(\omega )d\omega =\frac{1}{2\pi }{\int }_{-\pi }^{\pi }X(\omega )e^{j\omega n}d\omega .$

Examples

DTFT Analysis of $x[n]=\delta [n]$

The DTFT of $x[n]=\delta [n]$ can be written as
$X(\omega )={\sum }_{n=-\infty }^{\infty }\delta [n]e^{-j\omega n}=1.$
Therefore, the Fourier transform $x[n]=\delta [n]$ is $X(\omega )=1\forall \omega$.

DTFT Synthesis of $X(\omega )=\delta (\omega )$

The DTFT synthesis of $X(\omega )=\delta (\omega )$ can be written as
$x[n]=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\delta (\omega )e^{j\omega n}d\omega =\frac{1}{2\pi }.$
Therefore, the Fourier synthesis results in $x[n]=\frac{1}{2\pi }\forall n$.