Consider a feedback system where our plant is the system $F(\omega) = \frac{1}{1-\frac{1}{2}e^{-i\omega}}$, and our control is defined by $G(\omega) = z^{-1}$. Also, before the adder, there is a multiplier of $\alpha$. Note that $z^{-1}$ represents a system such that:

\begin{align*}
q(t) \to &\fbox{$z^{-1}$} \to q(n-1) \\
q(t) \to &\fbox{$D_1$} \to q(n-1) \quad \mbox{basic delay}\\
\end{align*}

So in order to determine what happens, lets pass $e^{i\omega n}$ through $z^{-1}$:
$$ e^{i\omega n}  \to \fbox{$z^{-1}$} \to e^{i\omega (n-1)} = e^{-i\omega} \cdot e^{i\omega n} $$
Therefore we know the frequency response, thus $G(\omega) =  e^{-i\omega}$. Since $\alpha$ is simply multiplied before the system finishes the loop, we can just bundle that with $G(\omega)$, so $\alpha  e^{-i\omega}$ can be our $G(\omega)$.


Now we can simply apply Black's equation, 

\begin{align*}
H(\omega) &= \frac{F(\omega)}{1+G(\omega)F(\omega)} \\
H(\omega) &= \frac{ \frac{1}{1-\frac{1}{2}e^{-i\omega}}}{1-\alpha e^{-i\omega} \frac{1}{1-\frac{1}{2}e^{-i\omega}}} \\
H(\omega) &= \frac{1}{1-(\alpha + \frac{1}{2})e^{-i\omega}} \\
\end{align*}

Notice that we know have the ability to create a low-pass or high-pass filter using this feedback system. Originally, we had only a low pass filter. The system could use $\alpha$ as a parameter to determine whether the system represents a low or high pass filter. We know that $\abs{\alpha + \frac{1}{2}} \lt 1$ in order for the sum $\sum_{\R_+}\lr{\lr{\alpha + \frac{1}{2}}e^{-i\omega}}^n$ to converge (since $h(t) = \lr{\alpha + \frac{1}{2}}^n\cdot u(n)$). This tells us the range for $\alpha$ as a parameter: $\alpha \in (-\frac{3}{2},\frac{1}{2})$ for a high pass filter, and $\alpha \in (0,\frac{1}{2})$ for a low pass filter.

Feedback Systems

An inverse system is a system such that $x \to \fbox{$P(\omega)$} \to \fbox{$P(\omega)^{-1}$} \to y = x$.
%\begin{center}
%\includegraphics{images/inverses/invdiag2.ps}
%\end{center}

\begin{center}
\begin{pspicture}(0,-4)(8,2)

\rput(0,0){$x$}
\rput(7.2,0){$y$}
\rput(3.5,0){$k$} 
\rput(3.75,-2){$P(\omega)$}
\rput(1.5,-2.5){$H$}

% plus or minus for adder
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\rput(1.45,0.35){$+$}

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Given the system in the figure, find the frequency response. Note that the plant is a multiplier. After inspection, we find that 
$$ H(\omega) = \frac{k}{1+kP(\omega)}  $$

Thus, if we let $k$ be large enough, then $kP(\omega) \gt\gt 1$ and $kP(\omega)$ dominates the denominator. Using this information, we can make a good approximation:

$$ H(\omega) = \frac{k}{1+kP(\omega)} \approx  \frac{k}{kP(\omega)} =  \frac{1}{P(\omega)}$$  
This works well as long as $kP(\omega) \gt\gt 1$ works for our frequency range of interest. Lets look at a system where the controller is a multiplier:

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

\begin{center}
\begin{pspicture}(0,-4)(8,2)

\rput(0,0){$x$}
\rput(7.2,0){$y$}
\rput(3.75,0){$P(\omega)$}
\rput(4,-2){$k$}
\rput(1.5,-2.5){$H$}

% plus or minus for adder
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\rput(1.45,0.35){$+$}

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We have a similar situation. If we let $k$ be large enough, then $kP(\omega) \gt\gt 1$ and we have 

$$ H(\omega) =  \frac{P(\omega)}{1+kP(\omega)}  \approx  \frac{P(\omega)}{kP(\omega)} = \frac{1}{k} $$

It is important to note that letting $k$ become large introduces some consequences. We lose magnitude for flatness over a frequency range. $1/k$ is not much gain, but a flat signal, whereas $P(\omega)$ may be a signal with large amplitude but not flat. 


Inverse Systems

\begin{definition}
A system $F$ is causal $\iff$ $\forall T\in\R, \forall x_1,x_2 \in X$ such that $x_1(t) = x_2(t)$ $\forall t \leq T$ produce  $y_1(t) = y_2(t)$ $\forall t \leq T$. In general, we say that a causal system doesn't use future values. A causal system may use present or past values only. In other words, a causal system is \emph{nonanticipitory}. In general, to process information in real-time, a system must be causal.

\end{definition}

\begin{example}
Here is an example of a system defined by its output signal:
$
y(n) = 
\begin{cases}
0 & \mbox{if $n$ is odd} \\
1 & \mbox{if $n$  is even} \\
\end{cases}
$

The system must be causal. The method here is to do a brute force proof.

\begin{proof}
Let $x_1(n),x_2(n) \in X$ be arbitrary. In this case, we have $y_1(n) = y_2(n)$ $\forall n$.
\end{proof}

\end{example}

\begin{example}
Here we have a system given by
$$ \delta(n) \to \fbox{F} \to \delta(n+1) + \delta(n) $$

This is only a partial characterization of $F$, thus, the system can be causal, but doesn't have to be. The method here is to show two examples of why both parts of the previous statement are true.

\begin{enumerate}
\item Let $F$ be defined by $y(n) = \begin{cases} 1 & \mbox{if $n\in\{-1,0\}$} \\ 0 & \mbox{otherwise} \\ \end{cases}$. Then this system is causal, because any two arbitrary functions will have the same output.
\item Let $F$ be defined by $y(n) = x(n) + x(n+1)$. This system is not causal since it ``looks into the future''.
\end{enumerate}

\end{example}

\begin{example}
Here we have a system given by
$$ \delta(n) \to \fbox{G} \to \delta(n+1) + \delta(n) $$
where $G$ is linear. Here, the system cannot be causal. The method here is to compare with another signal. If we pass in the zero signal, then in order for the system to be causal, the outputs should be identical up to $n=0$. But the outputs are not the same at $n=-1$, therefore the system is not causal. We can use this property because of the ZIZO property of linear systems. Another method we can use is to scale the input by $k$. Then $x_1(n) = x_2(n)$ $\forall n \leq -1 \wedge y_1(-1) \neq y_2(-1) \Rightarrow$ the system is not causal.
\end{example}

Let $P$ be the statement ``the output of the system precedes the input''

Let $L$ be the statement ``the system is linear''

Let $C$ be the statement ``the system is causal''

Then given a system that is linear:
\begin{align*}
&L \wedge C \Rightarrow \lnot P \\
&P \Rightarrow \lnot \left( L \wedge C \right) \\
&P \Rightarrow \lnot L \vee \lnot C \\ 
&\lnot C \quad \mbox{since we know $L$ is true}
\end{align*}

\begin{example}
Here we have a system given by
$$ \delta(n) \to \fbox{H} \to \delta(n+1) + \delta(n) $$
where $H$ is time invariant. The system cannot be causal. Here, the method is to compare with a time-shifted version of the input signal. 
\begin{proof}
Let $x_2(n) = x_1(n-1)$. Then $x_1(n) = x_2(n)$ $\forall n \leq -1 \wedge y_1(-1) \neq y_2(-1)$. 
\end{proof}
\end{example}

Let $TI$ be the statement ``the system is time invariant''

Then given $TI$ is true:
\begin{align*}
&TI \wedge C \Rightarrow \lnot P \\
&P \Rightarrow \lnot TI \vee \lnot C\\
&\lnot C \quad \mbox{ since we know $TI$ is true}
\end{align*}

Note that the output preceding the input doesn't preclude causality unless additional information is given, such as time invariance or linearity.


Here are some more examples:
\begin{enumerate}
\item $y(t) = cos(x(t))$ is casual because it only depends on the current time $t$.
\item $y(t) = x(cos(t))$ is not causal, since $y(0)  = x(1)$.
\end{enumerate}

Causality

Given the feedback system in the figure, find $H(\omega)$ and $h(n)$.

%\begin{center}
%\includegraphics{images/feedback/feedbackdis8.ps}
%\end{center}

\begin{center}
\begin{pspicture}(0,-4)(8,2)

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We know from previous examples that
$$f(n) = \alpha^nu(n) \iff F(\omega) = \frac{1}{1-e^{-i\omega}} \mbox{,}$$
and that 
$$g(n) = \alpha \delta(n) \iff G(\omega) = \alpha$$

By Black's formula, we know that $H(\omega) = \frac{F(\omega)}{1+F(\omega)G(\omega)}$. Now we can solve for $H(\omega)$:

\begin{align*}
H(\omega) &= \frac{F(\omega)}{1+F(\omega)G(\omega)} \\
H(\omega) &= \frac{ \frac{1}{1-e^{-i\omega}}}{1+ \frac{1}{1-e^{-i\omega}}\alpha} \\
H(\omega) &= \frac{1}{1 + \alpha - \alpha e^{-i\omega}} \\
H(\omega) &= \frac{1}{1+\alpha} \cdot \frac{1}{1-\frac{\alpha}{1+\alpha}e^{-i\omega}} \quad \mbox{factor out} \frac{1}{1+\alpha} \\
\end{align*}

Now we know that 
\begin{align*}
H(\omega) &= \frac{1}{1+\alpha} \cdot \frac{1}{1-\lr{\frac{\alpha}{1+\alpha}}e^{-i\omega}} \iff h(n) = \frac{1}{1+\alpha} \left( \frac{\alpha}{1+\alpha}\right)^nu(n) 
\end{align*}

We can simplify the expression to $h(t) = \frac{\alpha^n}{\left(1+\alpha\right)^{n+1}}u(n)$.

\begin{example}
Given the system in the figure, find $H(\omega)$.

%\begin{center}
%\includegraphics{images/feedback/feedback2dis8.ps}
%\end{center}

\begin{center}
\begin{pspicture}(-2,-4)(8,2)

\rput(-0.5,0){$x$}
\rput(7.7,0){$y$}
\rput(3.75,0){$F$}
\rput(3.75,-1){$G$}
\rput(0.5,-2.5){$H$}
\rput(2.65,-2){$P$}
\rput(4.85,-2){$Q$}

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% for F
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\psframe(4.25,-2.25)(5.5,-1.75)


% plus or minus for adder
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\rput(1.45,0.35){$+$} % input signal

% plus or minus for adder
\rput(1.1,-0.5){$-$} % input from loop
\rput(0.45,0.35){$+$} % input signal

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% from plant
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% to control
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% from control
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The best approach to this is to interpret this as a two feedback systems recursively. In other words, let the overall system have a plant defined by the inner feedback system, and controller defined the the cascade of the inner feedback system with $P$ and $Q$. More abstractly,
\begin{align*}
Plant(\omega) &= \frac{F(\omega)}{1-F(\omega)G(\omega)} \\
Controller(\omega) &= P(\omega)Q(\omega)Plant(\omega) \\
\end{align*}

Thus we can solve for the overall system 
\begin{align*}
H(\omega) &= \frac{Plant(\omega)}{1+Controller(\omega)} \\
H(\omega) &= \frac{\frac{F(\omega)}{1-F(\omega)G(\omega)}}{1+P(\omega)Q(\omega)\lr{\frac{F(\omega)}{1-F(\omega)G(\omega)}}} \\
H(\omega) &= \frac{F(\omega)}{1-F(\omega)G(\omega)+F(\omega)P(\omega)Q(\omega)}
\end{align*}
\end{example}

Discussion

\begin{problem}
Find the frequency response of the impulse response $x(t)= 
\begin{cases}
1 & \left|t\right|\lt\alpha \\
0 & \mbox{otherwise} \\
\end{cases}
$
\end{problem}

\begin{solution}
\begin{align*}
X(\omega) &= \int_{-\infty}^{\infty} x(t) e^{-i\omega t}dt \\
X(\omega) &= \int_{-\alpha}^{\alpha} e^{-i\omega t}dt \\
X(\omega) &= \left. \frac{e^{-i\omega t}}{-i\omega}\right|_{-\alpha}^{\alpha} \\
X(\omega) &= \frac{e^{-i\omega \alpha} - e^{i\omega \alpha}}{-i \omega} \\
X(\omega) &= \frac{2i\sin(\omega \alpha)}{i \omega} \\
X(\omega) &= \frac{2\sin(\omega \alpha)}{\omega}
\end{align*}
\end{solution}
\begin{problem}
Find the frequency response of the impulse response $w(t)= e^{-\alpha \abs{t}}$.
\end{problem}
\begin{solution}
First break up the impulse response to $w(t) = \begin{cases} e^{-\alpha t} & t \geq 0 \\ e^{\alpha t} & t \lt 0 \end{cases}$


Now we can solve for $W(\omega)$.

\begin{align*}
W(\omega) &= \int_{-\infty}^{\infty}w(t) e^{-i\omega t} dt \\
W(\omega) &= \int_{-\infty}^{0}e^{\alpha t}e^{-i\omega t} dt + \int_{0}^{\infty} e^{-\alpha t}e^{-i\omega t} dt \\
W(\omega) &= \int_{-\infty}^{0}e^{t\left( \alpha-i \omega\right)} dt + \int_{0}^{\infty} e^{-t \left( i \omega + \alpha \right)} \\
W(\omega) &= \left. \frac{e^{t\lr{\alpha-i\omega}}}{\left( \alpha-i \omega\right)} \right|_{-\infty}^{0} - \left. \frac{e^{-t\lr{i\omega+\alpha}}}{-\left( i \omega + \alpha\right)} \right|_{0}^{\infty} \\
W(\omega) &= \frac{e^0 - 0}{\left( \alpha-i \omega\right)} - \frac{0 - 1}{\left(i \omega +\alpha\right)}\\
W(\omega) &= \frac{1}{\left( \alpha-i \omega\right)} + \frac{1}{\left(i \omega +\alpha\right)}\\
W(\omega) &= \frac{\left(i \omega +\alpha\right) + \left( \alpha-i \omega\right)}{\left( \alpha-i \omega\right)\left(i \omega +\alpha\right)}\\
W(\omega) &= \frac{2\alpha}{\left( \alpha^2 + \omega^2\right)}\\
\end{align*}
\end{solution}

\begin{problem}
Find the frequency response of the impulse signal $h(n) = u(n) - u(n-N)$
\end{problem}

\begin{solution}
\begin{align*}
H(\omega) &= \sum \limits_{n\in\Z} \left( u(n) - u(n-N) \right) e^{-i\omega n} \\
H(\omega) &= \sum \limits_{n\in\Z_\oplus}e^{-i\omega n} - \sum \limits_{n=N}^{\infty} e^{-i\omega n} \\
H(\omega) &= \sum \limits_{n\in\Z_\oplus \setminus \lr{N,\infty}  }e^{-i\omega n} \\
H(\omega) &= \sum \limits_{n=0}^{N-1}e^{-i\omega n} \\
H(\omega) &= \frac{ 1- e^{-i\omega N} }{1-e^{-i\omega}} \quad \mbox{since} \sum \limits_{n=k}^{N}\alpha^n = \frac{\alpha^{N+1}-\alpha^k}{\alpha - 1} \\
H(\omega) &= \frac{\lr{e^{-i\omega N/2}} \lr{e^{i\omega N/2} - e^{-i\omega N/2}} }{\lr{{e^{-i\omega/2}}}\lr{e^{i\omega / 2} - e^{-i\omega / 2}}} \quad \mbox{ factor out half of the power} \\
H(\omega) &= e^{-i\omega N/2+i\omega/2} \cdot \frac{2i\sin(\omega N/2)}{2i\sin(\omega/2)} \\
H(\omega) &= e^{-i\omega \lr{N-1}/2} \cdot \frac{\sin(\omega N/2)}{\sin(\omega/2)} \\
\end{align*}

The magnitude response for various values of $N$ are displayed:
%\centering
%\includegraphics{images/reviewGraphs/2.ps}
%\includegraphics{images/reviewGraphs/3.ps}
%\includegraphics{images/reviewGraphs/4.ps}
%\includegraphics{images/reviewGraphs/5.ps}
%\includegraphics[height=2.5in]{images/reviewGraphs/6.ps}
%\includegraphics[height=2.5in]{images/reviewGraphs/7.ps}
%\includegraphics[height=2.5in]{images/reviewGraphs/8.ps}
%\includegraphics[height=2.5in]{images/reviewGraphs/9.ps}

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\begin{center}
\begin{pspicture}(-3.5,-2)(3.5,5)
\slider{2}{10}{N}{$N$}{2}
\psplot[algebraic,plotpoints=1000,linewidth=1.5pt]{-3.14}{3.14}{abs(sin(N*x/2) / sin(x/2))}


\psplot[algebraic,plotpoints=1000,linewidth=0.5pt,linestyle=dashed]{-3.14}{3.14}{abs(sin(N*x/2))}
\psplot[algebraic,plotpoints=1000,linewidth=0.5pt,linestyle=dashed]{-3.14}{3.14}{abs(sin(x/2))}
\psaxes[showorigin=false,labels=none, Dx=3.14](0,0)(-3.25,0)(3.25,5)
\rput(3.14, -0.35){$\pi$}
\rput(-3.14, -0.35){$-\pi$}
\rput(0, -0.35){$0$}
\end{pspicture}
\end{center}

\end{solution}

\begin{problem}
Given the difference formula $y(n) =\alpha y(n-1)=x(n)$, find the frequency response amd impulse reponse, and state whether the system is causal. Then determine the impulse response to $G(\omega) = \frac{H(\omega)}{e^{i\omega N}}$. Then find the difference equation for this system.
\end{problem}

\begin{solution}
By pattern recognition, we know the following
\begin{align*}
H(\omega) &= \frac{1}{1-\alpha e^{-i\omega}} \\
h(n) &= a^nu(n) \\
\end{align*}

Note that this system is casual since two arbitrary input signals with identical inputs up to a given time $N$ will produce identical outputs $\forall N \in \Z$. Lets continue to solve for $G(\omega)$:
\begin{align*}
G(\omega) &= \frac{H(\omega)}{e^{i\omega N}} \\
G(\omega) &= \frac{e^{-i\omega N}}{1-\alpha e^{-i\omega}} \\
\end{align*}

We can infer what the impulse reponse is since multiplication in the frequency domain corresponds to convolution in the time domain:
\begin{align*}
g(n) &=  \delta(n-N) * \alpha^n u(n) \\
g(n) &=  \alpha^{n-N} u(n-N)\\
g(n) &=  h(n-N)\\
\end{align*}

To get the difference equation, we simply cross multiply the equation for $G(\omega)$:
\begin{align*}
G(\omega) &= \frac{e^{-i\omega N}}{1-\alpha e^{-i\omega}} \\
G(\omega)\lr{1-\alpha e^{-i\omega}} &= e^{-i\omega N} \\
G(\omega)e^{i\omega n}\lr{1-\alpha e^{-i\omega}} &= e^{-i\omega N}e^{i\omega n} \\
y(n)-\alpha G(\omega)e^{i\omega \lr{n-1}} &= e^{i\omega \lr{n-N}} \\
y(n)-\alpha y(n-1) &= x(n-N) \\
\end{align*}
\end{solution}

\begin{problem}
The filter represented by the following LCCDE is a low-pass filter. Find the LCCDE for a high-pass filter.
$$ y(n) = \sum \limits_{k=1}^{M}a_ky(n-k) + \sum \limits_{k=0}^{N}b_kx(n-k) $$
\end{problem}

\begin{solution}
\begin{align*}
y(n) &= \sum \limits_{k=1}^{M}a_ky(n-k) + \sum \limits_{k=0}^{N} b_k x(n-k) \\
L(\omega)e^{i\omega n} &= \sum \limits_{k=1}^{M} a_k L(\omega)e^{i\omega\lr{n-k}} + \sum \limits_{k=0}^{N} b_k e^{i \omega \lr{n-k}} \\
L(\omega)&= \sum \limits_{k=1}^{M}a_k L(\omega)e^{-i\omega k} + \sum \limits_{k=0}^{N} b_k e^{-i \omega k} \\
L(\omega)&= \frac{\sum \limits_{k=0}^{N} b_k e^{-i \omega k}}{1-\sum \limits_{k=1}^{M}a_k e^{-i\omega k}} \\
\end{align*}

Now to transform this from a low pass to a high pass, we simply shift the phase by $\pi$:
\begin{align*}
H(\omega) &= L(\omega + \pi) \\
H(\omega)&= \frac{\sum \limits_{k=0}^{N} b_k e^{-i \lr{\omega+\pi} k}}{1-\sum \limits_{k=1}^{M}a_k e^{-i\lr{\omega+\pi} k}} \\
H(\omega)&= \frac{\sum \limits_{k=0}^{N} b_k e^{-i \omega k}\lr{-1}^k }{1-\sum \limits_{k=1}^{M}a_k e^{-i\omega k}\lr{-1}^k} \\
H(\omega)&= \sum \limits_{k=1}^{M}a_k H(\omega) e^{-i\omega k}\lr{-1}^k + \sum \limits_{k=0}^{N} b_k e^{-i \omega k}\lr{-1}^k  \\
y(n) &= \sum \limits_{k=1}^{M}\lr{-1}^k a_k y(n-k) + \sum \limits_{k=0}^{N} \lr{-1}^k b_k x(n-k) \\
\end{align*}
\end{solution}

8 Week 8