The DTFT and inverse DTFT are defined as follows:
$X(\omega) ~=~ \sum_{n=-\infty}^{\infty} x[n]\, e^{-j \omega n}, ~~~~~~ -\pi \leq \omega \lt \pi$
$x[n]~=~ \int_{-\pi}^\pi X(\omega)\, e^{j\omega n} \, \frac{d\omega}{2\pi} , ~~~~ \infty\lt n\lt\infty $
Let's work out some examples of DTFT and inverse DTFT calculations
Impulse response of an ideal lowpass filter
We'll start with an ideal lowpass filter. From its frequency response we can see that it blocks all incoming frequencies having a magnitude greater than $|\omega_c|$:
$H(\omega) = \begin{cases}
1&-\omega_c \leq \omega \leq \omega_c \\
0&{\sf otherwise} \\
\end{cases}$
CAPTION.
Dtft of a moving average system
Consider now a moving average system, where the system output at a time $n$ is the average of $M$ input values, symmetric about $n$. The impulse response of such a system is a pulse function:
$p[n]= \begin{cases}
1&-M\leq n \leq M \\
0&{\sf otherwise} \\
\end{cases}$
A symmetrical pulse, with $M=3$. Suppose we would like to find the frequency response of this system. We merely need to take the DTFT of the impulse response:
$P(\omega) ~=~ \sum_{n=-\infty}^{\infty} p[n]\, e^{-j \omega n}\\
~=~ \sum_{n=-M}^M e^{-j \omega n}\\~=~ \sum_{n=-M}^M \left( e^{-j \omega} \right)^n\\
~=~ \frac{ e^{j \omega M} - e^{-j \omega (M+1)}}{1-e^{-j \omega}}\\~=~ frac{ e^{j \omega M} - e^{-j \omega (M+1)}}{1-e^{-j \omega}}\\
~=~ \frac{ e^{-j\omega/2} \left( e^{j \omega \frac{2M+1}{2}} - e^{-j \omega \frac{2M+1}{2}} \right)}{ e^{-j\omega/2}\left( e^{j\omega/2}-e^{-j \omega/2} \right)} \\
~=~ \frac{ 2j \sin\left(\omega \frac{2M+1}{2}\right) }{ 2j \sin\left(\frac{\omega}{2}\right) }\\~=~\frac{ \sin\left(\frac{2M+1}{2}\,\omega \right) }{ \sin\left(\frac{\omega}{2}\right) }
$This frequency correspondence to the time-domain pulse is known as a digital sinc:
A digital sinc function. The digital sinc function is similar to a regular sinc function, except that rather than slowly decaying over time, it is $2\pi$ periodic.
When we compare this to the first example, the impulse response of an ideal lowpass filter, we see an interesting correspondence. Pulses in one domain (either time or frequency) have sinc-like representations in the alternate domain.
Dtft of a one-sided exponential
Having found the frequency response of a moving average system, let's now do the same for a recursive average system. The input/output relationship of such a system can be expressed as $y[n]=x[n]+\alpha y[n-1],$ where $|\alpha|\lt 1$. The impulse response for such a system is $h[n]=\alpha^n u[n]$:
The impulse response of a recursive average system. We'll now take the DTFT of this impulse response:
$H(\omega) ~=~ \sum_{n=-\infty}^{\infty} h[n]\, e^{-j \omega n} ~=~ \sum_{n=0}^\infty \alpha^n\, e^{-j \omega n}
~=~ \sum_{n=0}^\infty ( \alpha\, e^{-j \omega})^n ~=~ \frac{1}{1 - \alpha\, e^{-j \omega} }$
The magnitude of the frequency response of a one-sided exponential.
