\begin{example}

1) Find the DTFT of $x(n) = \alpha^n u(n)$ where $\abs{\alpha} \lt 1$. 

\begin{align*}
X(\omega) &= \sum \limits_{n=-\infty}^{\infty}\alpha^n u(n)e^{-i\omega n} \\
X(\omega) &= \sum \limits_{n=0}^{\infty}\alpha^n e^{-i\omega n} \\
X(\omega) &= \frac{1}{1-\alpha e^{-i\omega}} \\
\end{align*}

If $\alpha \gt 0$, this is a low-pass filter:

If $\alpha \lt 0$, this is a high-pass filter:

What about if $\alpha \in \C$? Lets consider an example where $\abs{\alpha} \lt 1$, and $\alpha$ is in the $\pi/4$ direction. First, analyze the vectors on the unit circle:

% \img{images/unitcircles/complexalpha/unit.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-3,-3)(9,3)
    
% y-axis
\rput(0.3,2.5){ $\Im \mit{m}$ }
\psline{->}(0,-2.5)(0,2.5)

% x-axis
\rput(2.5,0.3){ $\Re \mit{e}$ }
\psline{->}(-2.5,0)(2.5,0)

% quadrants
\rput(0.3,2.2){ $\pi/2$ }
\rput(0.3,-2.5){ $\pi$ }
\rput(-2.5,0.3){ $3\pi/2$ }
\rput(2.2,0.3){ $0$ }
	
		
% the circle
\pscircle(0,0){2}
	

 % new vector 
\rput(1,1.2){$\alpha$}
\psline[linewidth=1.5 pt]{->}(0,0)(1.2,1.2)

 % new vector 
\rput(1.55,1.45){$e^{i\omega}-\alpha$}
\psline[linewidth=2.5 pt, linecolor=blue]{->}(1.2,1.2)(1.73205124983367,0.999999233974511)

 % new vector 
\rput(1.53205124983367,0.65){$e^{i\omega}$}
\psline[linewidth=1.5 pt]{->}(0,0)(1.73205124983367,0.999999233974511)

% y-axis
\rput(6.3,2.5){ $\Im \mit{m}$ }
\psline{->}(6,-2.5)(6,2.5)

% x-axis
\rput(8.5,0.3){ $\Re \mit{e}$ }
\psline{->}(3.5,0)(8.5,0)

% quadrants
\rput(6.3,2.2){ $\pi/2$ }
\rput(6.3,-2.5){ $\pi$ }
\rput(3.5,0.3){ $3\pi/2$ }
\rput(8.2,0.3){ $0$ }
	
% the circle
\pscircle(6,0){2}

 % new vector 
\rput(7,0.65){$\alpha$}
\psline[linewidth=1.5 pt]{->}(6,0)(7.2,1.2)

 % new vector 
\rput(5,0.5){$e^{i\omega}-\alpha$}
\psline[linewidth=2.5 pt, linecolor=blue]{->}(7.2,1.2)(4.12061274158933,-0.684034745406704)

 % new vector 
\rput(3.92061274158933,-0.684034745406704){$e^{i\omega}$}
\psline[linewidth=1.5 pt]{->}(6,0)(4.12061274158933,-0.684034745406704)

\end{pspicture}
\end{center}

Notice that the denominator, $e^{i\omega} - \alpha$ is at a maximum at $-3\pi/4$, giving the magnitude response a minimum. The same vector has its minimum at $\pi/4$, giving the magnitude response a maximum at $\pi/4$.

% \img{images/magnituderesp/complexalpha/mag.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-1,-1)(3,3)
    
% \fileplot[linewidth=1.5pt]{xomega.dat}
% function name
\rput(3.9269908125,2.79932218491997){$X(\omega)$}

% x-axis
\psline(-4.3269908125, 0)(4.3269908125, 0)
\rput(4.3269908125,-0.3){$\omega$}

% begin x-axis labels:
\psline(-3.9269908125, -0.1)(-3.9269908125, 0.1)
\rput(-3.9269908125, -0.3){$-\frac{5\pi}{4}$}
\psline(-3.14159265, -0.1)(-3.14159265, 0.1)
\rput(-3.14159265, -0.3){$-\pi$}
\psline(-2.3561944875, -0.1)(-2.3561944875, 0.1)
\rput(-2.3561944875, -0.3){$-\frac{3\pi}{4}$}
\psline(-1.570796325, -0.1)(-1.570796325, 0.1)
\rput(-1.570796325, -0.3){$-\frac{\pi}{2}$}
\psline(-0.7853981625, -0.1)(-0.7853981625, 0.1)
\rput(-0.7853981625, -0.3){$-\frac{\pi}{4}$}
\psline(0.7853981625, -0.1)(0.7853981625, 0.1)
\rput(0.7853981625, -0.3){$\frac{\pi}{4}$}
\psline(1.570796325, -0.1)(1.570796325, 0.1)
\rput(1.570796325, -0.3){$\frac{\pi}{2}$}
\psline(2.3561944875, -0.1)(2.3561944875, 0.1)
\rput(2.3561944875, -0.3){$\frac{3\pi}{4}$}
\psline(3.14159265, -0.1)(3.14159265, 0.1)
\rput(3.14159265, -0.3){$\pi$}
\psline(3.9269908125, -0.1)(3.9269908125, 0.1)
\rput(3.9269908125, -0.3){$\frac{5\pi}{4}$}
% end x-axis labels:

% y-axis
\psline(0,-0.4)(0,3.4)
% begin x-axis labels:
\psline(-0.1, 1)(0.1, 1)
\psline(-0.1, 2)(0.1, 2)
\psline(-0.1, 3)(0.1, 3)
% end y-axis labels:

\end{pspicture}
\end{center}

Remember that we only need to graph over on period of $2\pi$, but it demonstrates the periodicity of the magnitude response.

\end{example}

\begin{example}

2) Consider an ideal low-pass filter that kills all frequencies except frequencies in $[-A,A]$. What is the impulse response $h(n)$?


% \includegraphics{images/idealpass/idealpass.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-1,-1)(4,1.5)

%axis
\psline(1,0)(3,0)
% 0 
\psline(2,0.1)(2,-0.1)

% pi and -pi
\psline(1,0.1)(1,-0.1)
\rput(1,-0.25){$-\pi$}
\psline(3,0.1)(3,-0.1)
\rput(3,-0.25){$\pi$}

\rput(3.3,-0.25){$\omega$}
\rput(3,1){$H(\omega)$}
\rput(1.35,1){$(1)$}

\psline(1.5,0)(1.5,1)
\psline(1.5,1)(2.5,1)
\psline(2.5,1)(2.5,0)

\rput(2,-0.25){0}
\rput(1.5,-0.25){-A}
\rput(2.5,-0.25){A}

\end{pspicture}
\end{center}

Use the discrete-time fourier transform.
\begin{align*}
h(n) &= \frac{1}{2\pi} \int_{\langle 2\pi\rangle}H(\omega)e^{i\omega n} d\omega \\
h(n) &= \frac{1}{2\pi} \int_{-A}^{A}e^{i\omega n} d\omega \\
h(n) &= \frac{1}{2\pi} \lr{\left. \frac{e^{i\omega n}}{in}\right|_{-A}^{A}} \\
h(n) &= \frac{1}{2\pi} \lr{\frac{e^{iAn}-e^{-iAn}}{in}} \\
h(n) &= \frac{1}{2\pi} \frac{2i\sin (An)}{in} \\
h(n) &= \frac{1}{\pi} \frac{\sin (An)}{n} \\
\end{align*}

Note that we could have also written this as
\begin{align*}
h(n) &= \frac{1}{2\pi}\left[ \int_{-A}^{A}\cos(\omega n) d\omega + i  \int_{-A}^{A}\sin(\omega n) d\omega\right] \\
h(n) &= \frac{1}{2\pi}\int_{-A}^{A}\cos(\omega n) d\omega  \\
h(n) &= \left. \frac{1}{2\pi} \sin(\omega n) \right|_{-A}^{A} \\
h(n) &= \frac{1}{\pi} \frac{\sin (An)}{n} \\
\end{align*}

Since sine is an odd function, then integrating from $-A$ to $A$ gives us zero, so we are left with integrating the cosine only. What happens to the impulse response $h(n)$ as $A$ approaches $\pi$?

$$ \lim_{A\to\pi} \frac{\sin (An)}{\pi n} =  \frac{\sin (\pi n)}{\pi n}  $$

Now for all $n \neq 0$, $h(n) = 0$, because we are feeding a sine function multiples of $\pi$, and $h(0)=1$ by l'Hopital's rule. This means that as $A$ approaches $\pi$, the impulse response becomes the identity element of convolution: 

$$  \lim_{A\to\pi} \frac{\sin (An)}{\pi n} = \delta(n) $$

Another important thing to note is that this signal cannot exist in reality. This is because signals that are absolutely summable have DTFTs that are continuous functions of $\omega$. That means

$$ \sum \limits_{n=-\infty}^{\infty}\abs{x(n)} \lt \infty $$

Since $\frac{1}{\pi} \frac{\sin (An)}{n}$ decays on the order of $\frac{1}{n}$, it is not absolutley summable. But, the absolute sum of its energy is finite. 

$$ \sum \limits_{n=-\infty}^{\infty}\abs{x(n)}^2 \lt \infty $$

This is because $\sum\lr{\frac{1}{n}}^2$ converges. Functions that are not absolutely summable but are square summable (have finite energy) have DTFT's that are discontinuous. You cannot use the analysis equation, but you can use the synthesis equation.

\end{example}

\begin{example}

3) Find the DTFT of $x(n)=e^{i\omega_0 n}$. 

\begin{align*}
\sum \limits_{n=-\infty}^{\infty}\abs{x(n)} \\ 
\sum \limits_{n=-\infty}^{\infty}\abs{e^{i\omega_0 n}} \\ 
\sum \limits_{n=-\infty}^{\infty}1 = \infty \\
\end{align*}

This is not absolutely summable or square summable. The discontinuity gets worse and $X(\omega)$ has deltas!

We know that this represents a dirac delta in the frequency domain centered at $\omega_0$. This gives us a frequency response of $A\delta(\omega -\omega_0)$ for some constant $A$.


% \includegraphics{images/pulsetrains/freqpulse.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-1,-1)(5,5)

\psline{->}(-1,0)(6,0)
\rput(6,-0.25){$\omega$}
\rput(6,1){$X(\omega)$}
\rput(6,0.5){$\cdots$}
\rput(-1,0.5){$\cdots$}
\rput(2.8,1){$(A)$}

\psline{->}(0,0)(0,1)
\rput(0,-0.25){$\omega_0-2\pi$}

\psline{->}(2.5,0)(2.5,1)
\rput(2.5,-0.25){$\omega_0$}

\psline{->}(5,0)(5,1)
\rput(5,-0.25){$\omega_0+2\pi$}

\end{pspicture}
\end{center}

So we can use the synthesis equation over an interval of $2\pi$:

\begin{align*}
x(n) &= \frac{1}{2\pi} \int_{\langle 2\pi\rangle} X(\omega) e^{i\omega n}d\omega \\
e^{i\omega_0 n} &= \frac{1}{2\pi} \int_{\langle 2\pi\rangle} A \delta(\omega-\omega_0) e^{i\omega n}d\omega \\
e^{i\omega_0 n} &= \frac{A}{2\pi} \int_{-\pi}^{\pi} \delta(\omega-\omega_0) e^{i\omega n}d\omega \\
e^{i\omega_0 n} &= \frac{A}{2\pi} e^{i\omega_0 n} \\
A &= 2\pi \\
\end{align*}

This tells us that $X(\omega) = 2\pi\delta(\omega - \omega_0)$, but we know that $X(\omega + 2\pi k) = X(\omega) $ $ \forall k \in \Z$, so we must write the frequency response in the form:

$$ X(\omega) = \sum \limits_{k=-\infty}^{\infty} 2\pi \delta(\omega-\omega_0 +2\pi k) $$

If we restricted the domain, for example, $-\pi \leq \omega \lt \pi$, then $X(\omega) = 2\pi\delta(\omega-\omega_0)$. 

\end{example}

DTFT Examples