\begin{example}
A signal $x$ is passed through a system with a special amplitude modulation:

$$
x(t) = 
\begin{cases}
-1 & t<0 \\
e^{-t}-1 & t \geq 0
\end{cases}
$$

$$ y(t) = \lr{A + x(t)}\cos(\omega_0 t)  \quad \omega_0 \gt\gt 1, A \gt 0$$

Putting these two together, we can form an expression for $y(t)$:

$$
y(t) = 
\begin{cases}
(A-1)\cos(\omega_0 t) & t\lt 0 \\
(A-1+e^{-t})\cos(\omega_0 t) & t \geq 0
\end{cases}
$$

For $A = 1$, we have nothing on the negative interval $(-\infty,0)$. On the right we have a cosine function bounded by $e^{-t}$ with an amplitude of 1 at the origin.


%\includegraphics{images/diswk7a.ps}
\psset{unit=1cm, algebraic, arrowscale=1.5}
\begin{center}
\begin{pspicture}(-7,-3)(7,3)

\psplot[algebraic,linewidth=1.5pt]{-3.14}{0}{2* cos(x*2)}
\psplot[algebraic,linewidth=1.5pt]{0}{3.14}{(2+Euler^(-x))*cos(x*2)}
\psplot[algebraic,linestyle=dashed]{-3.14}{0}{2}
\psplot[algebraic,linestyle=dashed]{0}{3.14}{2 + Euler^(-x)}
\psplot[algebraic,linestyle=dashed]{0}{3.14}{-2 - Euler^(-x)}
\psplot[algebraic,linestyle=dashed]{-3.14}{0}{-2}
\psaxes[showorigin=false,labels=none, Dx=3.14](0,0)(-3.25,-3.2)(3.25,3.2)

\rput(-0.5, 3){$A$}
\rput(-3.5, 2){$(A-1)$}
\rput(-3.5, -2){$-(A-1)$}
\rput(-0.5, -3){$-A$}

\rput(3,3){$A\gt 1$}

\end{pspicture}
\end{center}

notice we have 0 $\forall t \in (-\infty, 0)$



For $A \gt 1$, we have a normal cosine function from $(-\infty,0)$, with an amplitude ranging from $-(A-1)$ to $(A-1)$. On the positive side of the graph we have a cosine function bounded by $(A-1+e^{-t})$, which gives us an amplitude between $-A$ and $A$ at the origin, and then is tapered down by the bound. The reason that $\omega_0 \gt\gt 1$ is so that we can see more oscillations.

%\includegraphics{images/diswk7b.ps}

\psset{unit=1cm, algebraic, arrowscale=1.5}
\begin{center}
\begin{pspicture}(-7,-3)(7,3)

\psplot[algebraic,linewidth=1.5pt]{-3.14}{0}{2* cos(x*2)}
\psplot[algebraic,linewidth=1.5pt]{0}{3.14}{(2+Euler^(-x))*cos(x*2)}
\psplot[algebraic,linestyle=dashed]{-3.14}{0}{2}
\psplot[algebraic,linestyle=dashed]{0}{3.14}{2 + Euler^(-x)}
\psplot[algebraic,linestyle=dashed]{0}{3.14}{-2 - Euler^(-x)}
\psplot[algebraic,linestyle=dashed]{-3.14}{0}{-2}
\psaxes[showorigin=false,labels=none, Dx=3.14](0,0)(-3.25,-3.2)(3.25,3.2)

\rput(-0.5, 3){$A$}
\rput(-3.5, 2){$(A-1)$}
\rput(-3.5, -2){$-(A-1)$}
\rput(-0.5, -3){$-A$}

\rput(3,3){$A\gt 1$}

\end{pspicture}
\end{center}

Now there is a bounded cosine function in the interval $(-\infty, 0)$

\end{example}

\begin{example}
Consider a square area defined by 
$$
f(t) = \begin{cases}
1 & \mbox{if } t \in [-1,1] \\
0 & \mbox{otherwise}
\end{cases}
$$

What is the graphical representation of the convolution of $f$ with itself? 

It just so happens that we get a triangle whose bounds are from -2 to 2. When $t = -2$, the area starts to grow, then the convolution continues until the growth stops at $t=0$ when the squares are completely aligned. Then the decline is when the square begin "moved" through the other is leaving the overlapped area, thus there is a decline in area overlap.

\end{example}

\begin{example}
Consider a frequency response defined by 
$$ H(\omega) = \frac{i\omega}{i\omega + \alpha} \quad \alpha =0.01 $$ 

We can look at the magnitude response $\left| H(\omega) \right| =  \frac{\abs{i\omega}}{\abs{i\omega + \alpha}}$ by considering what the two vectors $i\omega$ and $i\omega + \alpha$ are behaving like. If we take the limit, we find that $\lim_{\abs{\omega}\to\infty}\abs{H(\omega)}=1$:
\begin{align*}
&\lim_{\abs{\omega}\to\infty}\abs{H(\omega)} \\
&\lim_{\abs{\omega}\to\infty}\frac{\abs{i\omega}}{\abs{i\omega + \alpha}} \\
&\lim_{\abs{\omega}\to\infty}\frac{1}{1 + \frac{\alpha}{i\omega}} = 1
\end{align*}

Even for smaller numbers we have a ratio that is very close to one. Yet, at $\omega = 0$, we have a value of zero. Thus we have a notch filter that kills all DC signals, or low frequencies. If we passed the signal $x(t) = 1 + \cos(20\pi t)$ through this system, we would have an output of $y(t) = \cos(20\pi t)$, since the angle for a real number is 0 which is killed.


How can we find $h(t)$?

\begin{align*}
H(\omega) &= \frac{i\omega}{i\omega + \alpha} \\
H(\omega) &= \frac{i\omega + \alpha - \alpha}{i\omega + \alpha} \\
H(\omega) &= 1 - \frac{\alpha}{i\omega + \alpha}  \\
\end{align*}

Again, we can use pattern matching here, using the pattern
\begin{align*}
H(\omega) = 1 &\iff h(t) = \delta(t) \\
H(\omega) = \frac{\alpha}{i\omega + \alpha} &\iff h(t) = \alpha e^{-\alpha t}u(t)
\end{align*}

Thus, the equation for the impulse response of the system is given by 
$$ h(t) = \delta(t) - \alpha e^{-\alpha t}u(t)  $$

Now, what happens if we pass $u(t)$ through this system? In other words, what is the output signal of $h(t)$ convolved with $u(t)$? Let's denote the output signal as $s(t)$:
\begin{align*}
s(t) &= (h*u)(t) \\
s(t) &= \infint h(\tau)u(t-\tau)d\tau \\
s(t) &= \int_{-\infty}^{t} h(\tau)d\tau \quad \mbox{since } \tau \in (-\infty, t] \Rightarrow u(t-\tau) = 1\\
s(t) &= \int_{-\infty}^{t} \lr{\delta(\tau) - \alpha e^{-\alpha \tau}u(\tau) }d\tau \\
s(t) &= \int_{-\infty}^{t} \delta(\tau) - \int_{-\infty}^{t}\alpha e^{-\alpha \tau}u(\tau) d\tau \\
s(t) &= \int_{-\infty}^{t} \delta(\tau) - \int_{0}^{t}\alpha e^{-\alpha \tau}d\tau \\
s(t) &= u(t) + \left. e^{-\alpha \tau} \right|_{0}^{t} \\
s(t) &= u(t) + \lr{e^{-\alpha t} - 1}u(t) \quad \mbox{ the domain was restricted by } u(t) \\
s(t) &= e^{-\alpha t} u(t) \\
\end{align*}

One thing to note here is that in general you can use the unit step to help you simplify the limits of an integral, but make sure to keep it as a restriction on the final answer. 


\end{example}


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