\begin{example}

Consider an impulse response defined by
$$
h(t)=
\begin{cases}
1 & -T\leq t\leq T \\
0 & \mbox{otherwise}
\end{cases}
$$

\begin{enumerate}
\item Find $y(t)$ in terms of the input $x(t)$ and impulse response.
\end{enumerate}

\begin{align*}
y(t) &= \infint x(\tau)h(t-\tau)d\tau \\
y(t) &= \int_{t-T}^{t+T}x(\tau)d\tau \quad \mbox{ since } -T \leq t-\leq T \mbox{ for  } h(t-\tau)\\
\end{align*}

We can solve further if necessary:
\begin{align*}
y(t) &= \int_{t-T}^{t+T}e^{i\omega \tau}d\tau \quad \mbox{ let } x(t) = e^{i\omega t} \\
y(t) &= \left. \frac{e^{i\omega \tau}}{i\omega} \right|_{t-T}^{t+T} \\ 
y(t) &= \frac{e^{i\omega (t+T)}}{i\omega}  - \frac{e^{i\omega (t-T)}}{i\omega} \\ 
y(t) &= \frac{e^{i\omega (t+T)}- e^{i\omega (t-T)}}{i\omega} \\ 
y(t) &= \frac{e^{i\omega t}\lr{e^{i\omega T} - e^{-i\omega T}}}{i\omega} \\ 
y(t) &= \frac{2\sin(\omega T)}{\omega}e^{i\omega t} \\ 
\end{align*}

What is this filter doing? Taking a look at $h(t)$ and the frequency response $ H(\omega) = \infint h(t) e^{-i\omega t}dt $, we can see that this is a centered moving average. That is, at $t$, we sample $x(t)$ from $t-T$ to $t+T$ and integrate the area, we then plot this point on $y(t)$ continuously.
\end{example}

\begin{enumerate}
\item What is the frequency response, $H(\omega)$?
\end{enumerate}

We know that the frequency response is defined by $$ H(\omega) = \infint h(t) e^{-i\omega t} dt $$
Since we are using $h(t)$ instead of $h(t-\tau)$, notice the limits of the integral are not the same as the last part of the example.

\begin{align*}
H(\omega) &= \int_{-T}^{T}e^{-\omega t}dt \\
H(\omega) &= \left. \frac{e^{-i\omega t}}{-i\omega} \right|_{-T}^{T} \\
H(\omega) &= \frac{e^{-i\omega T} - e^{-i\omega (-T)} }{-i\omega} \\ 
H(\omega) &= \frac{e^{i\omega T} - e^{-i\omega T} }{i\omega} \\ 
H(\omega) &= \frac{ 2\sin(\omega T)}{\omega} \\
\end{align*}

We must take note of $H(\omega)$ when $\omega=0$. In this case, it is true that $H(0)=\int_{-T}^{T}dt=2T=\lim_{\omega \to 0}H(\omega)$, which agrees with our frequency response representation. When graphing this function, take note that when omega is negative, it causes a reflection over the $y$-axis so the function is even unlike the sine function normally behaves. 

% \begin{center}
% \includegraphics{images/sinc/ctlti.ps}
% \end{center}

\psset{unit=0.75cm}
\begin{center}
\begin{pspicture}(-13,-2)(13,4)
\psplot[algebraic,linewidth=1.5pt]{-12.56}{12.56}{2*sin(x)/x}
\psplot[algebraic,linestyle=dashed]{-12.56}{-0.5}{abs(2/x)}
\psplot[algebraic,linestyle=dashed]{0.5}{12.56}{abs(2/x)}
\psplot[algebraic,linestyle=dashed]{0.5}{12.56}{-abs(2/x)}
\psplot[algebraic,linestyle=dashed]{-12.56}{-0.5}{-abs(2/x)}
\psaxes[showorigin=false,labels=none, Dx=3.14](0,0)(-12.6,0)(12.6,0)
\rput(5.7, 1){$2T$}
\psline[linestyle=dashed](0,2)(5,2)
\psline{<->}(5,0)(5,2)
\rput(3.14, -0.5){$\frac{\pi}{T}$}
\rput(6.28, -0.5){$\frac{2\pi}{T}$}
\rput(9.42, -0.5){$\frac{3\pi}{T}$}
\rput(12.56, -0.5){$\frac{4\pi}{T}$}
\rput(-3.14, -0.5){$\frac{-\pi}{T}$}
\rput(-6.28, -0.5){$\frac{-2\pi}{T}$}
\rput(-9.42, -0.5){$\frac{-3\pi}{T}$}
\rput(-12.56, -0.5){$\frac{-4\pi}{T}$}
\rput(0, -0.5){$0$}
\rput(5,3 ){$H(\omega) = \frac{2\sin(\omega T)}{\omega}$}

\end{pspicture}
\end{center}

When $\omega T = \pi k$, we have frequencies multiplied by zero every $\pi/T k$, where $k \in \Z$. Thus we have a low-pass filter. So if we have an input signal given by $x(t) = e^{ik\pi t/T}$, we can see that it has a frequency that corresponds to the x-intercepts of the magnitude response (the absolute value of the frequency response), and thus gives an output of 0. If $x(t) = 1$, then we have a frequency of 0, thus $y(t)$ in this case is $2T$.

In general, averaging will produce a low-pass filter. Notice that as the gap between $(-T,T)$ shrinks, the interval between $(-\pi/T,\pi/T)$ grows, and vice versa. 

\begin{example}
We can see that we have the following relationship:
$$ h(t) = \delta(t) \iff H(\omega) = 1  \quad \mbox{ because } \infint \delta(t)e^{-i\omega t}dt = 1$$

What happens when we let $h(t) = \delta(t-T)$?
\begin{align*}
H(\omega) &= \infint \delta(t-T)e^{-i\omega t}dt \\
H(\omega) &= e^{-i\omega T}
\end{align*}

We still have the magnitude at 1, since $\abs{H(\omega)} = 1$. We only introduce a phase, $y(t) = x(t-T)$
\end{example}

\begin{example}
Given a first order equation for the voltage across a circuit that contains a capacitor and a resistor, what is the frequency response?

$$ RC\dot{y}(t) + y(t) = x(t) \quad \mbox{(initial conditions are zero)}  $$

Let $y(t) = H(\omega)e^{i\omega t}, x(t) = e^{i\omega t}$, then $\dot{y}(t) = H(\omega)i \omega e^{i\omega t}$. Now we can solve for the frequency response:
\begin{align*}
RC\dot{y}(t) + y(t) &= x(t) \\
RCH(\omega)i\omega e^{i\omega t} + H(\omega)e^{i\omega t} &= e^{i\omega t} \\
H(\omega) \lr{RCi\omega + 1} &= 1 \\
H(\omega) &= \frac{1}{RCi\omega + 1}
\end{align*}

Then we can re-write this as $ \frac{1/RC}{i\omega + 1/RC}$. This looks similar to $\frac{1}{i\omega + \alpha}$, where $H(\omega) = \frac{1}{i\omega + \alpha} \iff h(t) = e^{-\alpha t}u(t)$, which implies $h(t) = \frac{1}{RC} \cdot e^{-t/RC} \cdot u(t)$. The graph of $h(t)$ has a height of $1/RC$ at 0, and is decreasing towards zero as $t$ increases.

When we consider the magnitude response, never separate the $i\omega$!
$$ \left| H(\omega) \right| = \frac{1/RC}{\abs{i\omega + \lr{1/RC}}} $$

Note that $\lim_{\abs{\omega} \to \infty} \frac{1/RC}{\abs{i\omega + \lr{1/RC}}} = 0$. At $\omega = 0$, $\abs{H(\omega)} = 1$.
\end{example}

\begin{example}
Given the equation of an all-pass filter, 
$$ \dot{y} + ay = \dot{x} - ax $$
what is the magnitude response and phase response?

Let $x(t) = e^{i\omega t}, y(t) = H(\omega)x(t)$
\begin{align*}
\dot{y} + ay &= \dot{x} - ax \\
i\omega H(\omega)e^{i\omega t} + aH(\omega)e^{i\omega t}  &= i\omega e^{i\omega t}  - ae^{i\omega t} \\
i\omega H(\omega) + aH(\omega) &= i\omega - a \\
H(\omega)\lr{i\omega + a} &= i\omega - a \\
H(\omega) &= \frac{i\omega - 1}{i\omega + a} \\
\abs{H(\omega)} &= \frac{\abs{i\omega - 1}}{\abs{i\omega + a}}
\end{align*}
If we analyze these two vectors, they are symmetric with respect to the imaginary axis, thus we have equal magnitudes and the magnitude response is always equal to one. However, the point of the all-pass filter is that the phase response it not so trivial.

\begin{align*}
\angle{  \frac{i\omega - 1}{i\omega + a} } &= \angle{ i\omega - 1 } - \angle {i\omega + a } \\
\angle{  \frac{i\omega - 1}{i\omega + a} } &=  \tan^{-1}\lr{-\omega} -  \tan^{-1}\lr{\frac{\omega}{a}} \\
\end{align*}

\end{example}

Frequency response of CT-LTI Systems