\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

\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}


Discussion

Week 7

Given a parallel system, $h$, with two systems $f$ and $g$, determine an expression for $h(t)$ and $H(\omega)$. A parallel system is a system that can be graphically described as a circuit with two systems on parallel sides going in the same direction, connecting with an adder before the output.

%\begin{center}
%\includegraphics{images/parallel/para.ps}
%\end{center}

\begin{center}
\begin{pspicture}(0,-3)(8,3)

\rput(0,0){$x(t)$}
\rput(4,1.5){$f(t)$}
\rput(4,-1.5){$g(t)$}
\rput(8.2,0){$y(t)$}
\rput(1.5,-2){$h(t)$}

\psframe(1,-2.5)(7,2.5)
\psframe(3,1)(5,2)
\psframe(3,-1)(5,-2)

\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(0.5,0)(1.5,0)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(1.5,1.5)(3,1.5)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(1.5,-1.5)(3,-1.5)

\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(6.5,1.5)(6.5,0.25)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(6.5,-1.5)(6.5,-0.25)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(6.75,0)(7.75,0)
\psline[linewidth=1.25 pt](1.5,-1.5)(1.5,1.5)
\psline[linewidth=1.25 pt](5,1.5)(6.5,1.5)
\psline[linewidth=1.25 pt](5,-1.5)(6.5,-1.5)
\psline[linewidth=1.25 pt](6,-1.5)(6.5,-1.5)

\pscircle(6.5,0){0.25}
\psline(6.25,0)(6.75,0)
\psline(6.5,0.5)(6.5,-0.5)

\end{pspicture}
\end{center}

Let $x(t) = \delta(t)$
$$
h(t) = f(t) + g(t)
$$

Let $x(t) = e^{i\omega t}$
\begin{align*}
H(\omega)e^{i\omega t} &= F(\omega)e^{i\omega t} + G(\omega)e^{i\omega t} \\
H(\omega)e^{i\omega t} &= e^{i\omega t}\lr{F(\omega) + G(\omega)} \\
H(\omega) &= F(\omega) + G(\omega) \\
\end{align*}

\begin{example}
Consider the system defined in the frequency domain by the equation $$ H(\omega) = \frac{1 - e^{-i\omega T}}{i\omega + \alpha} $$
Write the system as a parallel system composed of two systems $F$ and $G$. Find $h(t)$ in terms of $f(t)$ and $g(t)$.

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

Here is a hint: 
\begin{align*}
H(\omega) = \frac{1}{i\omega + \alpha} &\iff h(t) = e^{-\alpha t}u(t) \\
h(t) = \delta(t) &\iff H(\omega) = 1
\end{align*}

Lets see if we can find a relationship:

\begin{align*}
H(\omega) &= \frac{1}{i\omega + \alpha} - \frac{e^{-i\omega T}}{i\omega + \alpha}\\
H(\omega) &= F(\omega) - e^{-i\omega T}F(\omega) \\
H(\omega) &= F(\omega) - G(\omega) \\
h(t) &= f(t) - g(t) \\
h(t) &= e^{-\alpha t}u(t) - g(t) \\
h(t) &= e^{-\alpha t}u(t) - f(t-T) \\
\end{align*}

Lets solve for $G(\omega)$ using $g(t)$ = $f(t-T)$:
\begin{align*}
G(\omega) &= \infint g(t) e^{-i\omega t}dt \\
G(\omega) &= \infint f(t-T) e^{-i\omega t}dt \\
G(\omega) &= \infint f(\lambda) e^{-i\omega (\lambda + T)}d\lambda \quad \mbox{let } \lambda = t - T\\
G(\omega) &= e^{-i\omega T}\infint f(\lambda) e^{-i\omega \lambda}d\lambda \\
G(\omega) &= e^{-i\omega T}F(\omega) \\
\end{align*}

Note that if we let $g(t) =  \delta(t-T)$ we get $e^{-i\omega T}$:
\begin{align*}
\infint g(t) e^{-i \omega t} dt = e^{-i\omega T} \iff g(t) = \delta(t-T)
\end{align*}

We can now make the following additions to the statements above:
\begin{align*}
H(\omega) = \frac{1}{i\omega + \alpha} &\iff h(t) = e^{-\alpha t}u(t) \\
h(t) = \delta(t) &\iff H(\omega) = 1 \\
h(t) = \delta(t-T) &\iff H(\omega) = e^{-\omega T} \\
g(t) = f(t-T) &\iff G(\omega) = e^{-i\omega T}F(\omega)
\end{align*}

So finally we have
\begin{align*}
h(t) &= f(t) - g(t) \\
h(t) &= e^{-\alpha t}u(t) - e^{-\alpha(t-T)}u(t-T) \\
\end{align*}

Now we can see that we could have defined the system like this:

% \begin{center}
% \includegraphics{images/parallel/para2.ps}
% \end{center}
\begin{center}
\begin{pspicture}(0,-3)(8,3)

\rput(0,0){$x(t)$}
\rput(4,1.5){$f(t)$}
\rput(3,-1.5){$f(t)$}
\rput(5.25,-1.5){$\delta(t-T)$}
\rput(8.2,0){$y(t)$}
\rput(1.5,-2){$h(t)$}

\psframe(1,-2.5)(7,2.5)
\psframe(3,1)(5,2)
\psframe(2.5,-2)(3.5,-1)
\psframe(4.5,-2)(6,-1)

\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(0.5,0)(1.5,0)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(1.5,1.5)(3,1.5)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(1.5,-1.5)(2.5,-1.5)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(3.5,-1.5)(4.5,-1.5)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(6.5,1.5)(6.5,0.25)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(6.5,-1.5)(6.5,-0.25)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(6.75,0)(7.75,0)
\psline[linewidth=1.25 pt](1.5,-1.5)(1.5,1.5)
\psline[linewidth=1.25 pt](5,1.5)(6.5,1.5)
\psline[linewidth=1.25 pt](6,-1.5)(6.5,-1.5)

\pscircle(6.5,0){0.25}
\psline(6.25,0)(6.75,0)
\psline(6.5,0.5)(6.5,-0.5)

\end{pspicture}
\end{center}

We can also describe the same system as a cascade of $f(t)$ and $g(t)$, where $g(t) = \delta(t) - \delta(t-T)$. 

\end{example}

Parallel

Consider the casading system:

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

\begin{center}
\begin{pspicture}(0,-3)(8,3)

\rput(0,0){$x(t)$}
\rput(8.2,0){$y(t)$}
\rput(2.75,0){$f(t)$}
\rput(5.25,0){$g(t)$}

\psframe(1,-1.5)(7,1.5)
\psframe(2,-0.5)(3.5,0.5)
\psframe(4.5,-0.5)(6,0.5)

\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(0.5,0)(2,0)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(3.5,0)(4.5,0)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(6,0)(7.75,0)

\end{pspicture}
\end{center}

To find $h$, simply let $x$ = $\delta$. When $x$ first enters the system, it goes through $f$ and becomes $f$. Then we pass $f$ through the system $g$. Thus $y$ is the convolution of $f$ and $g$:

$$ h(t) = (f*g)(t) $$

To find $H$, let $x(t) = e^{i \omega t}$. When $x$ first enters the system, it goes through $F$ and becomes $F(\omega)e^{i\omega t}$ since the system is linear. Then the signal enters $G$ and becomes $F(\omega)G(\omega)e^{i\omega t}$. Therefore,

$$H(\omega) = G(\omega)F(\omega)$$

We can take note that convolution in the time domain corresponds to multiplication in the frequency domain:

\begin{align*}
h(t) = (f*g)(t) &\iff H(\omega) = F(\omega)G(\omega) \\ 
\end{align*}

\begin{example}
Consider the system
%$$ x(t) \to \fbox{ $\to \fbox{f(t)} \to \fbox{$\delta(t-T)$}\to$ } \to y  $$

\begin{center}
\begin{pspicture}(0,-3)(8,3)

\rput(0,0){$x(t)$}
\rput(8.2,0){$y(t)$}
\rput(2.75,0){$f(t)$}
\rput(5.25,0){$\delta(t-T)$}

\psframe(1,-1.5)(7,1.5)
\psframe(2,-0.5)(3.5,0.5)
\psframe(4.5,-0.5)(6,0.5)

\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(0.5,0)(2,0)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(3.5,0)(4.5,0)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(6,0)(7.75,0)

\end{pspicture}
\end{center}

Let $x$ be $\delta$. When $x$ first enters the system it becomes $f(t)$. Then  $f(t)$ enters the system given by $\delta(t-T)$, thus we get the convolution 
$$ g(t) = \lr{f(t) * \delta(t-T)} $$

We know can determine $G(\omega)$ because $\delta(t-T)$ in the time domain corresponds to $e^{-i\omega T}$ in the frequency domain:

$$ G(\omega) = F(\omega)e^{-i\omega T}  $$
\end{example}

\begin{example}
Given $f(n)$ and $g(n)$, find $H(\omega)$ and $h(n)$:
\begin{align*}
f(n) &= \sum \limits_{k=0}^{\infty}\alpha^k\delta(n-kN) \\
g(n) &= \delta(n) - \alpha\delta(n-N) \\
\end{align*}

Lets find $G(\omega)$ first, since it is simpler: 
\begin{align*}
G(\omega) &= \sum \limits_{n\in\Z} \lr{\delta(n) - \alpha\delta(n-N)}e^{-\omega n} \\
G(\omega) &=  e^{-i\omega 0} - \alpha e^{-i\omega N} \\
G(\omega) &=  1 - \alpha e^{-i\omega N} \\
\end{align*}

This gives us some insight into $F(\omega)$:
\begin{align*}
F(\omega) &= \sum \limits_{n \in \Z} \lr{\sum \limits_{\ell=0}^{\infty} \alpha^\ell \delta(n-\ell N)}e^{-i\omega n} \\
F(\omega) &= \sum \limits_{n \in \Z} \lr{\delta(n) + \alpha \delta(n-N) + \alpha^2 \delta(n-2N) + \cdots }e^{-i\omega n} \\
F(\omega) &= \sum \limits_{n \in \Z}\delta(n)e^{-i\omega n} + \alpha \sum \limits_{n \in \Z}\delta(n-N)e^{-i\omega n} + \alpha^2 \sum \limits_{n \in \Z}\delta(n-2N)e^{-i\omega n} + \cdots \\
F(\omega) &= 1 + \alpha e^{-i\omega N} + \alpha^2 e^{-i\omega 2N} + \cdots  \\
F(\omega) &= \sum \limits_{k \in \Z_{\oplus}} \alpha^k e^{-i\omega kN} \\
F(\omega) &= \sum \limits_{k \in \Z_{\oplus}} \lr{\alpha e^{-i\omega N}}^k \\
F(\omega) &= \frac{1}{1 - \alpha e^{-i\omega N}}
\end{align*}

Now we can solve for $H(\omega)$:
\begin{align*}
H(\omega) &= F(\omega)G(\omega) \\
H(\omega) &=  \frac{1}{1 - \alpha e^{-i\omega N}} \lr{1 - \alpha e^{-i\omega N}} \\
H(\omega) &= 1
\end{align*}
This implies that $h(t) = \delta(t)$.

\end{example}


Cascade (series)

Feedback systems are usually described in the frequency domain since it is simpler to articulate. Typically we are given a \emph{plant} and a \emph{controller} in a feedback system.

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

\begin{center}
\begin{pspicture}(0,-3)(8,3)

\rput(0,0){$x$}
\rput(7.2,0){$y$}
\rput(3.75,0){$F$}
\rput(3.75,-2){$G$}
\rput(1.5,-2.5){$H$}

\psframe(1,-3)(6,1)
\psframe(3,-0.5)(4.5,0.5)
\psframe(3,-2.5)(4.5,-1.5)

% plus or minus for adder
\rput(2.1,-0.5){$-$} % input from loop
\rput(1.45,0.35){$+$} % input signal

\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(0.25,0)(1.5,0)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(1.75,-2)(1.75,-0.25)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(4.5,0)(7,0)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(5.5,-2)(4.5,-2)
\psline[linewidth=1.25 pt](3,-2)(1.75,-2)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(2,0)(3,0)
\psline[linewidth=1.25 pt](5.5,0)(5.5,-2)

\pscircle(1.75,0){0.25}
\psline(1.5,0)(2,0)
\psline(1.75,0.25)(1.75,-0.25)

\end{pspicture}
\end{center}

Lets trace $x$ through the loop where $F$ is the plant, and $G$ is the controller, and the signal is negated after $G$ before added to the input signal for the loop. Let $x(t) = e^{i\omega t}$, then $y(t) = H(\omega)e^{i\omega t}$. Lets call $q(t)$ the signal that we trace through. This signal goes through the system, and $y(t)$ is fed back into $G$ and becomes $q(t)$:


\begin{align*}
q(t) &= G(\omega)H(\omega)e^{i\omega t} \\
q(t) &= e^{i\omega t} - G(\omega)H(\omega)e^{i\omega t} \quad \mbox{picks up a new input and negation and addition occurs} \\
q(t) &=  e^{i\omega t} \lr{1-G(\omega)H(\omega)} \\
q(t) &=  F(\omega)\lr{1-G(\omega)H(\omega)} e^{i\omega t} \quad \mbox{ goes through $F$}\\
q(t) &= y(t) = H(\omega)e^{i\omega t} =  F(\omega)\lr{1-G(\omega)H(\omega)} e^{i\omega t} \\
\end{align*}

Now after the loop has finished, we can solve for $H(\omega)$:

\begin{align*}
H(\omega)e^{i\omega t} &=  F(\omega)\lr{1-G(\omega)H(\omega)} e^{i\omega t} \\
H(\omega) &=  F(\omega)-G(\omega)F(\omega)H(\omega) \\
H(\omega)+G(\omega)F(\omega)H(\omega) &=  F(\omega) \\
H(\omega)\lr{1+G(\omega)F(\omega)} &=  F(\omega) \\
H(\omega) &=  \frac{F(\omega)}{1+G(\omega)F(\omega)} \\
\end{align*}

This is called \emph{Black's equation}. If we had two positive signs for the adder, the equation would only change slightly:
$$ H(\omega) =  \frac{F(\omega)}{1-G(\omega)F(\omega)} $$ 
To generalize this, we can state that the frequency response to the system is $\frac{\mbox{Forward Gain}}{1 - \mbox{Loop Gain}}$. Note that when in a loop, $F(\omega)$ and $G(\omega)$ are in cascade (a series), therefore the frequency response during the loop (the Loop Gain) is $F(\omega)G(\omega)$. 

7.1 Interconnections of LTI Systems