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Interconnections of LTI Systems 

\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}

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\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

\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}
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\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}

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\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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