For the discrete-time Fourier transform, we can express any discrete-time, energy (aperiodic) signal as a continuous summation (i.e., an integral) of complex exponentials.
Discrete-time Fourier transform analysis, i.e., the [forward] Discrete-time Fourier transform , is defined by $$X(\omega) = \sum_{n = -\infty}^{\infty} x[n] e^{-j \omega n } $$
Discrete-time Fourier transform synthesis, i.e., the inverse Discrete-time Fourier transform , is defined by $$x[n] = \frac{1}{2 \pi} \int_{2 \pi} X(\omega) e^{j \omega n } d \omega$$
One of the important properties of the DTFT is that its frequency domain is periodic with a periodicity of $2 \pi$. That means that for every DTFT representation, $$X(\omega) = X(\omega + 2 \pi k)$$ for any integer $k$. Typically, we study $X(\omega)$ for frequencies in one period from $-\pi$ to $\pi$, where $0$ is the lowest frequency (before periodicity) and $-\pi$ and $\pi$ are the highest negative and positive freqencies.
This periodicity occurs because all discrete sine and cosine (and complex exponential, by extension) frequencies are always periodic with a period of $2 \pi$.
This property is also why the inverse DTFT is only integrated for a period of $2 \pi$. This periodicity will be important to consider later since it affects how we design filters and play a significant role in our discussion on signal resampling.
The DTFT of a discrete-time, aperiodic signal $x[n]$ is guaranteed to exist if $$\sum_{n=-\infty}^{\infty} |x[n]| < \infty \; .$$ Note that if the signal $x[n]$ is a energy signal, it will satisfy this property.
Caveat: Some periodic signals (e.g., sines and cosines) can have a DTFT through the DTFT's connection to the Discret-Time Fourier Series and through the use of the Dirac delta function, even though periodic signals fail to satify the above property.
Discrete-time systems can be characterized by linear, constant-coefficient difference equations. That is a summation of shifted and weighted inputs $x[n]$ will equal a summation of shifted and weighted outputs $y[n]$ such that $$ \sum_{k=0}^{N} a_k y[n-k] = \sum_{k=0}^{N} b_k x[n-k]$$ If we take the DTFT of both sides of this equation, we get that $$ \begin{eqnarray*} \sum_{k=0}^{N} a_k Y(\omega) e^{-j \omega k} &=& \sum_{k=0}^{N} b_k X(\omega) e^{-j \omega k} \\ Y(\omega) \sum_{k=0}^{N} a_k e^{-j \omega k} &=& X(\omega) \sum_{k=0}^{N} b_k e^{-j \omega k} \\ H(\omega) = \frac{Y(\omega)}{X(\omega)} &=& \frac{\sum_{k=0}^{N} b_k e^{-j \omega k}}{\sum_{k=0}^{N} a_k e^{-j \omega k}} \end{eqnarray*}$$ Therefore, we can represent a system by difference equations.
Below is a table of common discrete-time Fourier transforms.
$$ \begin{align} \hline & x[n] &\qquad & X(\omega) &\qquad & \textrm{condition} \\ \hline & \displaystyle a^n u[n] &\qquad & \displaystyle \frac{1}{1 - a e^{-j \omega}} &\qquad & |a| < 1 \\ \\[.05em] & \displaystyle (n+1) a^n u[n] &\qquad & \displaystyle \frac{1}{(1-a e^{-j \omega})^2} &\qquad & |a| < 1 \\ \\[.05em] & \displaystyle \frac{(n+r-1)!}{n! (r-1)!} a^n u[n] &\qquad & \displaystyle \frac{1}{(1-a e^{-j \omega})^r} &\qquad & |a| < 1 \\ \\[.05em] & \displaystyle \delta[n] &\qquad & \displaystyle 1 \\ \\[.05em] & \displaystyle \delta[n - n_0] &\qquad & \displaystyle e^{-j \omega n_0} \\ \\[.05em] & \displaystyle x[n] = 1 &\qquad & \displaystyle 2 \pi \sum_{k = -\infty}^{\infty} \delta(\omega - 2 \pi k) \\ \\[.05em] & \displaystyle u[n] &\qquad & \displaystyle \frac{1}{1-e^{-j \omega}} + \sum_{k=-\infty}^{\infty} \pi \delta(\omega - 2 \pi k) \\ \\[.05em] & \displaystyle e^{j \omega_0 n} &\qquad & \displaystyle 2 \pi \sum_{k = -\infty}^{\infty} \delta(\omega - \omega_0 - 2 \pi k) \\ \\[.05em] & \displaystyle \cos(\omega_0 n) &\qquad & \displaystyle \pi \sum_{k = -\infty}^{\infty} \{ \delta(\omega - \omega_0 - 2 \pi k) + \delta(\omega + \omega_0 - 2 \pi k) \} \\ \\[.05em] & \displaystyle \sin(\omega_0 n) &\qquad & \displaystyle \frac{\pi}{j} \sum_{k = -\infty}^{\infty} \{ \delta(\omega - \omega_0 - 2 \pi k) - \delta(\omega + \omega_0 - 2 \pi k) \} \\ \\[.05em] & \displaystyle \sum_{k=-\infty}^{\infty} \delta[n - kN] &\qquad & \displaystyle \frac{2 \pi}{N} \sum_{k=-\infty}^{\infty} \delta \left(\omega - \frac{2 \pi k}{N} \right) \\ \\[.05em] & \displaystyle x[n] = \left\{ \begin{array}{ccc} 1 & , & |n| \leq N \\ 0 & , & |n| > N \\ \end{array} \right. &\qquad & \displaystyle \frac{\sin(\omega(N + 1/2))}{\sin(\omega / 2)} \\ \\[.05em] & \displaystyle \frac{\sin(W n)}{\pi n}= \frac{W}{\pi} \textrm{sinc} \left( W n \right) &\qquad & \begin{array}{c} X(\omega) = \left\{ \begin{array}{ccc} 1 & , & 0 \leq |\omega| \leq W \\ 0 & , & W < |\omega| \leq \pi \end{array} \right. \\ X(\omega) \textrm{ is periodic with period } 2 \pi \end{array} \\ \hline & \end{align} $$