Solution:

The Discrete Time Fourier Transform (DTFT) is a member of the Fourier transform the family that operates on aperiodic, discrete signals.

The best way to understand the DTFT is how it relates to the DFT. To start, imagine that you acquire an N sample signal, and want to find its frequency spectrum.

By using the DFT, the signal can be decomposed into sine and cosine waves, with frequencies equally spaced between zero and one-half of the sampling rate.

As discussed in the last chapter, padding the time domain signal with zeros makes the period of the time domain longer, as well as makes the spacing between samples in the frequency domain narrower.

As N approaches infinity, the time domain becomes aperiodic, and the frequency domain becomes a continuous signal.

This is the DTFT, the Fourier transform that relates an aperiodic, discrete signal, with a periodic, continuous frequency spectrum.

The mathematics of the DTFT can be understood by starting with the synthesis and analysis equations

$ \begin{aligned}\\ x[n] &=\frac{1}{2 \pi} \int_{2 \pi} \times(\Omega) e^{j \Omega n} d \Omega \\\\ x(\Omega) &=\sum_{n=-\infty}^{+\infty} \times[n] e^{-j \Omega n} \\\\ & x[n] \stackrel{F}{\leftrightarrow} \times x(\Omega) \\\\ \times(\Omega) &=\operatorname{Re}\{\times(\Omega)\}+j \operatorname{Im}\{\times(\Omega)\} \\\\ &=|X(\Omega)| e^jX(\Omega)\\ \end{aligned}\\ $

The spectrum of the DTFT is continuous, so either f or ω can be used.

The common choice is ω, because it makes the equations shorter by eliminating the always-present factor of 2π.

Remember, when ω is used, the frequency spectrum extends from 0 to π, which corresponds to DC to one-half of the sampling rate.

To make things even more complicated, many authors use Ω (an upper-case omega) to represent this frequency in the DTFT, rather than ω (a lower-case omega.