Looking at the difference equation:

$$y(n)=\alpha y(n-1)+x(n)$$

First note that we need zero initial conditions for this to be LTI. This is because if we had an initial condition like $y(-1)=k$ for some $k \in \R$. Let $\hat{x}(n) = x(n-N)$. Then if $\hat{y}$ is the response to $\hat{x}$, then both $\hat{x}$ and $\hat{y}$ must satisfy the difference equation $\hat{y}(n) = \alpha \hat{y}(n-1) + \hat{x}(n)$ and the condition that $\hat{y}(-1)=k$. If $\hat{y}=y(n-N)$, then $\hat{y}(-1)=y(-1-N)=k$, but there is no such condition. The system can not be LTI.


The following three equations are interrelated closely. We should be able to traverse between them:
$$ 
h(n)=\alpha^n u(n) \iff H(\omega) = \frac{1}{1-\alpha e^{-i\omega}} \iff y(n)=\alpha y(n-1)+x(n)
$$


Difference Equations

The continuous-time impulse, Dirac Delta:

%\begin{figure}[htp]
%\centering
%\includegraphics{images/dirac/dirac.ps}
%\caption{ The Dirac Delta }
%\label{ fig:dirac }
%\end{figure}

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The (1) in the figure represents the area under the impulse, or the \emph{strength} of the impulse. It is not the height as in discrete-time, as the hieght of the Dirac Delta is infinite.

A definition, but not very useful one, is

$$ \delta(t) = \begin{cases} \infty & t=0 \\ 0 & \mbox{otherwise} \end{cases} $$

Lets look at some others that may prove to be more useful. Consider the unit area, where its minimum $t$ value is $-\Delta/2$, maximum $t$ value is $\Delta/2$, and a height of $1/\Delta$. This can be defined more concretely as 

$$
\delta_\Delta(t)=
\begin{cases}
0 & t<-\frac{\Delta}{2} \\
\frac{1}{\Delta} & -\frac{\Delta}{2}\leq t \leq\frac{\Delta}{2} \\
0 & t\gt\frac{\Delta}{2} \\
\end{cases}
$$

%\begin{figure}[htp]
%\centering
%\includegraphics{images/dirac/dirac-unitarea.ps}
%\caption{ The Unit Area }
%\label{ fig:dirac-unitarea }
%\end{figure}

\begin{center}
\begin{pspicture}(-3,-3)(3,3)
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Then $\delta(t) = \lim_{\Delta \to 0}\delta_\Delta(t)$. The unit area goes to an infinite height, zero width, but its area does not change. 

Lets consider a triangle whose base spans $t$ from $-\Delta$ to $\Delta$, and whose height is $1/\Delta$. This can be defined as
$$
\delta_\Delta(t)=
\begin{cases}
0 & t\lt -\frac{\Delta}{2} \\
\frac{1}{\Delta^2}t+\frac{1}{\Delta} & -\frac{\Delta}{2}\leq t \lt 0 \\
-\frac{1}{\Delta^2}t+\frac{1}{\Delta} & 0\leq t \leq\frac{\Delta}{2} \\
1 & t\gt\frac{\Delta}{2} \\
\end{cases}
$$

%\begin{figure}[htp]
%\centering
%\includegraphics{images/dirac/dirac-triangle.ps}
%\caption{ A triangle with an area of 1 }
%\label{ fig:dirac-triangle }
%\end{figure}

{\bf A triangle with an area of 1}


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Again, $\delta(t) = \lim_{\Delta \to 0}\delta_\Delta(t)$. Lets consider the gaussian probability density function, where $\delta_\Delta(t) = \frac{1}{\sqrt{2\pi\Delta^2}}e^{-t^2/2\Delta^2}$.

%\begin{figure}[htp]
%\centering
%\includegraphics{images/dirac/dirac-gaussian.ps}
%\caption{ Gaussian Probability Density Function }
%\label{ fig:dirac-gaussian }
%\end{figure}



The standard deviation is $\Delta$, the maximum height at 0 is $\frac{1}{\sqrt{2\pi\Delta^2}}$. In this case $\int_{-\infty}^{\infty}\delta_\Delta(t)dt=1$, and again, $\delta(t) = \lim_{\Delta \to 0}\delta_\Delta(t)$.


The truth is, we don't care about what the delta \emph{is}, we care about what it does to other functions when it comes into contact with them.


Take an arbitrary function $f(t)$ and multiply it with the unit area definition $\delta_\Delta(t)$. We would be left with the area underneath $f(t)$ where it comes into contact in the section of the domain in $\pm\Delta/2$. The height is then $f(t)/\Delta$. If $\Delta$ is sufficiently small, then we can represent the height as $f(0)/\Delta$ from $-\Delta/2$ to $\Delta/2$. Thus, the area is equal to $f(0)$.

$$ \int_{-\infty}^{\infty}f(t)\delta_\Delta(t)dt = f(0)$$

Lets take the limit of this integral as $\Delta$ approaches 0:
\begin{align*}
\int_{-\infty}^{\infty}f(t)\delta_\Delta(t)dt &= f(0) \\
\lim_{\Delta \to 0}\int_{-\infty}^{\infty}f(t)\delta_\Delta(t)dt &= f(0) \\
\int_{-\infty}^{\infty}f(t)\lim_{\Delta \to 0}\delta_\Delta(t)dt &= f(0) \\
\int_{-\infty}^{\infty}f(t)\lim_{\Delta \to 0}\delta_\Delta(t)dt &= f(0) \\
\int_{-\infty}^{\infty}f(t)\delta(t)dt &= f(0)
\end{align*}
The value of the function is picked up wherever the impulse is active. This is a sampling property of the Dirac Delta. We can generalize this formula slightly: 

$$ \int_{-\infty}^{\infty}f(t)\delta(t-t_0)dt=f(t_0) $$

The new resulting strength on a graph would be ($f(t_0)$). This is called the \emph{sifting property} of the Dirac Delta.

\begin{problem}
What are the values of these integrals?

\begin{enumerate}
\item $\int_{-\infty}^{\infty}\cos(\omega_0t)\delta(t-T)dT$
\item $\int_{-\infty}^{\infty}\delta(t)e^{i\omega_0 t}dt$
\end{enumerate}

\end{problem}

\begin{solution}
These can be fairly simple to integrate:

\begin{enumerate}
\item $\int_{-\infty}^{\infty}\cos(\omega_0t)\delta(t-T)dT = \cos(\omega_0T)$
\item $\int_{-\infty}^{\infty}\delta(t)e^{i\omega_0 t}dt = 1$
\end{enumerate}
\end{solution}

$$ x(t) = \int_{-\infty}^{\infty}x(\tau)\delta(t-\tau)d\tau $$
$$ \delta(t) \to \fbox{S} \to h(t)  $$
This gives us the continuous-time convolution integral:
$$ y(t) = \int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau $$

Dirac Delta

Looking at the Dirac Delta, we can see some interesting properties:

\begin{align*}
\int_{-\infty}^{\infty}\delta(t-\tau)f(t)dt &= f(\tau) \\
\int_{\tau^-}^{\tau^+}\delta(t-\tau)f(t)dt &= f(\tau) \\
\int_{\tau-\epsilon}^{\tau+\epsilon}\delta(t-\tau)f(t)dt &= f(\tau) \\
\int_{\tau-\epsilon}^{\tau+\epsilon}\delta(t-\tau)dt &= 1 \\
\end{align*}

What happens if we write an integral like this?
$$\int_{-\infty}^{t}\delta(\tau)d\tau$$

In turns out that this is a very important element in characterizing continuous-time LTI systems, known as the \emph{Continuous-Time Unit Step}, which can be defined as

$$ u(t) = \int_{-\infty}^{t}\delta(\tau)d\tau = \begin{cases} 0 & t<0 \\ 1 & t \geq 0 \\ \end{cases}$$

Note, however, that there is a discontinuity at $t=0$, but by convention, we shall include it for $t \geq 0$. The height of the discontinuity happens to be the coefficient in front of $\delta(t)$. We just showed how the CT unit step can be defined in terms of the integral of the Dirac Delta. Notice that then the derivative the the unit step will spawn a DIrac Delta, thus, we can define the Dirac Delta as:

$$ \delta(t) = \dot{u}(t) = \frac{d}{dt}u(t) = \frac{d}{dt}\int_{-\infty}^{t}\delta(\tau)d\tau = \delta(t) $$

If we define $u_\Delta(t)$ as
$$
u_\Delta(t)=
\begin{cases}
0 & t<-\frac{\Delta}{2} \\
\frac{1}{\Delta}t+\frac{1}{2} & -\frac{\Delta}{2}\leq t \leq\frac{\Delta}{2} \\
1 & t>\frac{\Delta}{2} \\
\end{cases}
$$

% \begin{figure}[htp]
% \centering
% \includegraphics{images/dirac/delta-unit-step.ps}
% \caption{ The unit step with the discontinuity from $-\frac{\Delta}{2}$ to $\frac{\Delta}{2}$ }
% \label{ fig:delta-unit-step }
% \end{figure}

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then we can take the derivative and we find that  $\dot{u}_\Delta(t) = \delta_\Delta(t)$, where $\delta_\Delta(t)$ represents the unit area given by 
$$
\dot{u}_\Delta(t) = \delta_\Delta(t)=
\begin{cases}
0 & t<-\frac{\Delta}{2} \\
\frac{1}{\Delta} & -\frac{\Delta}{2}\leq t \leq\frac{\Delta}{2} \\
0 & t>\frac{\Delta}{2} \\
\end{cases}
$$

% \begin{figure}[htp]
% \centering
% \includegraphics{images/dirac/dirac-unitarea.ps}
% \caption{ The unit area }
% \label{ fig:dirac-unit }
% \end{figure}


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If we take the limit, 
$$ \lim_{\Delta \to 0} \dot{u}_\Delta(t) = \delta(t) = \dot{u}(t) $$
we can clearly see that the derivative relates the CT unit step and the Dirac Delta. What about the discrete-time counter parts? We can write the Kronecker Delta as $\delta(n) = u(n) - u(n-1)$, the difference of a unit step and a shifted version of a unit step. We can write the unit step in discrete-time in two ways:
\begin{align*}
u(n) &= \sum \limits_{k=0}^{\infty}\delta(n-k) \\
u(n) &= \sum \limits_{\ell=-\infty}^{n}\delta(\ell) \quad \mbox{substitute } \ell = n-k
\end{align*}
We can verify that this works because when $n<0$ we get $u(n)=0$, and for $n \geq 0$, we have $u(n)=1$.


Dirac Delta and Unit Step

Lets consider a continuous-time LTI system where $x(t) =\delta(t)$, then we have 
\begin{align*}
x \to &\fbox{H} \to y \\
\delta(t) \to &\fbox{H} \to h(t) \\
\end{align*}

where $h(t)$ represents our impulse response. Note that if we mention ``impulse response'', then it is implied that a given system is LTI. We can demonstrate this property here: 

$$ \alpha\delta(t-T) \to \fbox{H} \to \alpha h(t-T) $$

We know from the sifting property, we can write any signal as the integral of the product of itself and the Dirac Delta:

$$ x(t) = \int_{-\infty}^{\infty}x(\tau)\delta(t-\tau)d\tau$$

This looks very similar to the DT counterpart, where we write a signal as a linear combination of impulses:

\begin{align*}
x(n) &= \sum \limits_{m=-\infty}^{\infty}x(m)\delta(n-m) \\
y(n) &= \sum \limits_{m=-\infty}^{\infty}x(m)h(n-m) = (x*h)(n)\\
\end{align*}

Just as we wrote signals in DT as sums of weighted delta functions, in CT we can write the convolution integral:

\begin{nicebox}
$$y(t) = \int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau = (x*h)(t) $$
\end{nicebox}

The convolution operator still has its commutativity property, which we can demonstrate by a substitution:

\begin{align*}
y(t) &= \int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau = (x*h)(t)\\
y(t) &= \int_{\infty}^{-\infty}x(t-\mu)h(\mu)\left(-d\mu\right) \quad \mbox{let } \mu=t-\tau, \mbox{limits have switched}\\
y(t) &= \int_{-\infty}^{\infty}h(\mu)x(t-\mu)d\mu \quad \mbox{absorb negative by switching limits again }\\
y(t) &= \int_{-\infty}^{\infty}h(\mu)x(t-\mu)d\mu = (h*x)(t)\\
\end{align*}

LTI Systems and their impulse responses

There exists a set of functions that physicists refer to as \emph{Strongly-Peaked Functions}, which are narrow and very tall. Consider Newton's Second Law, $F=Ma$, he actually wrote it like this: $F=\frac{d}{dt}(Mv)$, where $P=Mv$ is the momentum, which implies that $F = \frac{dP}{dt}$. Then the change in momentum can be written as
$$ \int_{-\epsilon}^{\epsilon}F(\tau)d\tau = P(\epsilon) - P(-\epsilon) $$
Consider a Pinto and a Mazeradi. If we look at the graph of their forces over time, we would see that there is far more area under the curve representing the Mazeradi over the same distance. From a book "Intro to Mechanics" written by Kleppner and Kolenkow in 1973, gives an example with a rubber ball: Consider a rubber ball whose mass is $0.2kg$, velocity at impact is $8\frac{m}{s}$, and duration of contact is $2\epsilon=10^{-3}s$. We find that $P(-\epsilon)=-1.6kg\frac{m}{s}$, $P(\epsilon)=1.6kg\frac{m}{s}$, and $ P(\epsilon) - P(-\epsilon) = 3.2kg\frac{m}{s}$. This tells us that 


$$ 3.2 =  \int_{-\epsilon}^{\epsilon}F(\tau)d\tau \mbox{,}$$ 

and the force graph represents a strongly-peaked function. Take the average height, and call this $F_{\mbox{avg}}$. Then 
\begin{align*}
F_{\mbox{avg}} \Delta t &= 3.2 \\
F_{\mbox{avg}} &= 3.2/\Delta t = 3200 N = 720lbs
\end{align*}

In other words, 
$$\lim_{\Delta \to 0} F = \infty$$ 

Strongly-Peaked Functions

Lets pass a pure frequency in continuous-time through the convolution integral:

\begin{align*}
y(t) &= \int_{-\infty}^{\infty}h(\tau)x(t-\tau)d\tau \\
y(t) &= \int_{-\infty}^{\infty}h(\tau)e^{i \omega (t-\tau)}d\tau \\
y(t) &= \int_{-\infty}^{\infty}h(\tau)e^{i \omega t}e^{-i \omega \tau}d\tau \\
y(t) &= \left( \int_{-\infty}^{\infty}h(\tau)e^{-i \omega \tau}d\tau\right)e^{i \omega t} \\
y(t) &= H(\omega)e^{i \omega t} \\
\end{align*}

In CT (and DT), an LTI system cannot create new frequencies. The golden rule still holds:

\begin{align*}
x & \to \fbox{h} \to y \\
e^{i\omega t} & \to \fbox{h} \to H(\omega)e^{i\omega t} 
\end{align*}

Thus, \emph{the continuous-time frequency response} is defined as $$H(\omega) = \int_{-\infty}^{\infty}h(\tau)e^{-i\omega \tau}d\tau$$

Some important distinctions between the DT and CT frequency response:

\begin{enumerate}
\item In discrete-time, we have the $2\pi$ ambiguity:
$$H(\omega + 2\pi)=\sum \limits_{n\in Z} h(n) e^{-i\omega n} e^{-i2\pi n} = \sum \limits_{n\in Z} h(n) e^{-i\omega n} = H(\omega)$$
because $n$ is an element of $\Z$.
\item In continuous-time, we have a different story:
$$ H(\omega +2\pi) = \int_{-\infty}^{\infty}h(t)e^{-i\omega t}e^{-i2\pi t}dt \neq H(\omega)$$
because $t$ is an element of $\R$.
\end{enumerate}

\begin{example}

Given an impulse response $h(t) = e^{-\alpha t} u(t)$, where $\alpha \gt 0$, graph the magnitude response and solve for the frequency response.

\begin{align*}
h(t) &= e^{-\alpha t} u(t)\mbox{} \quad \alpha>0 \\
H(\omega) &= \int_{-\infty}^{\infty} h(t)x(t)dt \\
H(\omega) &= \int_{-\infty}^{\infty} \left( e^{-\alpha t} u(t) \right)e^{i \omega t}dt \\
H(\omega) &= \int_{0}^{\infty} e^{-\left(i \omega + \alpha\right)t}dt \\
H(\omega) &= \left. \frac{e^{-\left(i \omega + \alpha\right)t}}{-\left(i \omega + \alpha\right)} \right|_{0}^{\infty} \\
\end{align*}

Since $\alpha \gt 0$, as $t$ approaches $\infty$, $e^{-\left( i\omega + \alpha\right)t} $ approaches $0$.
\begin{align*}
H(\omega) &= \lim_{t \to \infty} \left( \frac{e^{-\left(i \omega + \alpha\right)t}}{-\left(i \omega + \alpha\right)}\right) - \left( \frac{e^{0}}{-\left(i \omega + \alpha\right)} \right) \\
H(\omega) &= 0 - \left( \frac{1}{-\left(i \omega + \alpha\right)} \right) \\
H(\omega) &= \frac{1}{i \omega + \alpha} \\
\end{align*}

\end{example}

A good way to graph this is to write the equation as $\frac{1}{i\omega - (-a)}$, and analyze the graph of $i \omega - (-a)$

CT Frequency Response

Week 6

If we look at the low-pass filter $H(\omega) = \frac{1}{1-\alpha e^{-i\omega}}$, we find that we have a filter whose magnitude response looks like

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

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\rput(3.6,0.3){$\frac{1}{1+\alpha}$}

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\rput(-3.14, -0.35){$-\pi$}
\rput(0, -0.35){$0$}

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Lets define a new frequency response to be $H_N(\omega) = \frac{1}{1-\alpha e^{-i\omega N}}$. Notice that this corresponds to the difference equation $y(n) - \alpha y(n-N) = x(n)$:

\begin{align*}
H(\omega) &= \frac{1}{1-\alpha e^{-i\omega N}} \\
H(\omega) - \alpha H(\omega) e^{-i\omega N} &= 1 \\
H(\omega) e^{i\omega n} - \alpha H(\omega)e^{i\omega n} e^{-i\omega N} &= e^{i\omega n} \\
y(n) - \alpha y(n-N) &= x(n) \\
\end{align*}

To graph the magnitude response for a particular value of $N$, we can analyze the vectors representing the numerator and denominator. But, we can also perform a simple trick, since we know the graph of $H_1(\omega)$:

%\begin{center}
%\includegraphics{images/comb/comb1.ps}
%\end{center}

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\rput(3.6,2.3){$\frac{1}{1-\alpha}$}
\rput(3.6,0.3){$\frac{1}{1+\alpha}$}

\rput(0,3){$N=1$}

\rput(3.14, -0.35){$\pi$}
\rput(1.62, -0.35){$\pi/2$}
\rput(-1.62, -0.35){$-\pi/2$}
\rput(-3.14, -0.35){$-\pi$}
\rput(0, -0.35){$0$}

\end{pspicture}
\end{center}

Now let $H_2(\omega) = H_1(2\omega)$, then we need to divide by 2 on our axes:

%\begin{center}
%\includegraphics{images/comb/comb2.ps}
%\end{center}

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\begin{pspicture}(-3.5,-1.25)(3.5,3)
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\rput(3.6,0.3){$\frac{1}{1+\alpha}$}

\rput(0,3){$N=2$}

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\rput(1.62, -0.35){$\pi/2$}
\rput(-1.62, -0.35){$-\pi/2$}
\rput(-3.14, -0.35){$-\pi$}
\rput(0, -0.35){$0$}

\end{pspicture}
\end{center}

If you notice there will always be $N$ peaks and $N$ valleys. In general, we can use the first graph to find the graph of the magnitude response for all value of $N$. The higher the $N$, the more oscillations.

%\begin{center}
%\includegraphics{images/comb/comb4.ps}
%\end{center}

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\begin{center}
\begin{pspicture}(-3.5,-1.25)(3.5,3)
\psplot[algebraic,linewidth=1.5pt,plotpoints=1000]{-3.14}{3.14}{cos(8*x/2)+1.3}
\psaxes[showorigin=false,labels=none, Dx=1.62](0,0)(-3.25,0)(3.25,2.5)

\psline[linestyle=dashed](-3.14,0.3)(3.14,0.3)
\psline[linestyle=dashed](-3.14,2.3)(3.14,2.3)
\rput(3.6,2.3){$\frac{1}{1-\alpha}$}
\rput(3.6,0.3){$\frac{1}{1+\alpha}$}

\rput(0,3){$N=4$}

\rput(3.14, -0.35){$\pi$}
\rput(1.62, -0.35){$\pi/2$}
\rput(-1.62, -0.35){$-\pi/2$}
\rput(-3.14, -0.35){$-\pi$}
\rput(0, -0.35){$0$}

\end{pspicture}
\end{center}


\psset{unit=1cm}

\begin{center}
\begin{pspicture}(-3.5,-1)(3.75,3.5)

\slider{1}{8}{n}{$2N$}{4}
%\rput(0,3){$N=4$}

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\rput(3.6,0.3){$\frac{1}{1+\alpha}$}


\rput(3.14, -0.35){$\pi$}
\rput(1.62, -0.35){$\pi/2$}
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\rput(-3.14, -0.35){$-\pi$}
\rput(0, -0.35){$0$}

\end{pspicture}
\end{center}


Comb Filter

\begin{problem}
Graph the magnitude response and determine the first difference equation for the output signal $y(n)$ if the frequency response is given by:


$$
H(\omega) = \frac{ \left( e^{i\omega} - e^{i\frac{\pi}{4}}\right) \left(e^{i\omega} - e^{-i\frac{\pi}{4}}\right)}{ \left( e^{i\omega} - \alpha e^{i\frac{\pi}{4}}\right) \left(e^{i\omega} - \alpha e^{-i\frac{\pi}{4}}\right)   }
$$
\end{problem}

\begin{solution}
We can break this problem down to its four component vectors and find that it is a \emph{notch filter} that zeros out frequencies at $\pm \pi/4$ and is approximately 1 everywhere else.

To determine the difference equation, lets simplify the expression:

\begin{align*}
H(\omega) &= \frac{ \left( e^{i\omega} - e^{i\frac{\pi}{4}}\right) \left(e^{i\omega} - e^{-i\frac{\pi}{4}}\right)}{ \left( e^{i\omega} - \alpha e^{i\frac{\pi}{4}}\right) \left(e^{i\omega} - \alpha e^{-i\frac{\pi}{4}}\right)   } \\
H(\omega) &= \frac {e^{i2\omega} - e^{i\omega}e^{-i\pi/4}-e^{i\omega}e^{i\pi/4}+1}{e^{i2\omega} - \alpha e^{i\omega}e^{-i\pi/4}-\alpha e^{i\omega}e^{i\pi/4}+\alpha^2} \\
H(\omega) &= \frac {e^{i2\omega} - e^{i\omega} \left( e^{-i\pi/4}+e^{i\pi/4} \right) +1}{e^{i2\omega} - \alpha e^{i\omega} \left( e^{-i\pi/4}+e^{i\pi/4}\right)+\alpha^2} \\
H(\omega) &= \frac {e^{i2\omega} - e^{i\omega} \left( 2\cos(\frac{\pi}{4}) \right) +1}{e^{i2\omega} - \alpha e^{i\omega} \left( 2\cos(\frac{\pi}{4})\right)+\alpha^2} \\
H(\omega) &= \frac {e^{i2\omega} - e^{i\omega} \left( \sqrt{2} \right) +1}{e^{i2\omega} - \alpha e^{i\omega} \left( \sqrt{2}\right)+\alpha^2} \\
H(\omega) &= \frac {1 - \sqrt{2}e^{-i\omega} + e^{-i2\omega}}{1 - \alpha \sqrt{2}e^{-i\omega}+\alpha^2e^{-i2\omega}} \mbox{ multiply by } e^{-2\omega} \mbox{ on top and bottom}\\
\end{align*}

Now we have a much simpler form:
$$H(\omega) = \frac {1 - \sqrt{2}e^{-i\omega} + e^{-i2\omega}}{1 - \alpha \sqrt{2}e^{-i\omega}+\alpha^2e^{-i2\omega}} $$

Lets cross multiply (imagine having a 1 beneath $H(\omega)$:

\begin{align*}
H(\omega) - \alpha\sqrt{2}H(\omega)e^{-i\omega} + \alpha^2H(\omega)e^{-i2\omega} &= 1 - \sqrt{2}e^{-i\omega} + e^{-i2\omega} \\
H(\omega)e^{i\omega n} - \alpha\sqrt{2}H(\omega)e^{i\omega(n-1)} + \alpha^2H(\omega)e^{i\omega(n-2)} &= e^{i\omega n}  - \sqrt{2}e^{i\omega(n-1)} + e^{i\omega(n-2)} \\
y(n) - \alpha\sqrt{2}y(n-1) + \alpha^2y(n-2) &= x(n)  - \sqrt{2}x(n-1) + x(n-2)
\end{align*}

\end{solution}

6.1 Discussion