\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}

%\psset{xunit=1, yunit=1, algebraic, arrowscale=1.5}
\begin{center}
\begin{pspicture}(-3.5,-2)(3.5,3)
\psplot[algebraic,plotpoints=1000,linewidth=1.5pt]{-3.14}{3.14}{abs(sin(2*x/2) / sin(x/2))}
\psplot[algebraic,plotpoints=1000,linewidth=0.5pt, linestyle=dashed]{-3.14}{3.14}{abs(sin(2*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)(3.25,2)
\rput(3.14, -0.35){$\pi$}
\rput(-3.14, -0.35){$-\pi$}
\rput(0, -0.35){$0$}
\rput(1.5, 2){$N=2$}

\end{pspicture}
\end{center}

\begin{center}
\begin{pspicture}(-3.5,-2)(3.5,3)
\psplot[algebraic,plotpoints=1000,linewidth=1.5pt]{-3.14}{3.14}{abs(sin(3*x/2) / sin(x/2))}
\psplot[algebraic,plotpoints=1000,linewidth=0.5pt,linestyle=dashed]{-3.14}{3.14}{abs(sin(3*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)(3.25,3)
\rput(3.14, -0.35){$\pi$}
\rput(-3.14, -0.35){$-\pi$}
\rput(0, -0.35){$0$}
\rput(1.5, 3){$N=3$}

\end{pspicture}
\end{center}


\begin{center}
\begin{pspicture}(-3.5,-2)(3.5,5)
\psplot[algebraic,plotpoints=1000,linewidth=1.5pt]{-3.14}{3.14}{abs(sin(4*x/2) / sin(x/2))}
\psplot[algebraic,plotpoints=1000,linewidth=0.5pt,linestyle=dashed]{-3.14}{3.14}{abs(sin(4*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)(3.25,4)
\rput(3.14, -0.35){$\pi$}
\rput(-3.14, -0.35){$-\pi$}
\rput(0, -0.35){$0$}
\rput(1.5, 4){$N=4$}

\end{pspicture}
\end{center}

\begin{center}
\begin{pspicture}(-3.5,-2)(3.5,5)
\psplot[algebraic,plotpoints=1000,linewidth=1.5pt]{-3.14}{3.14}{abs(sin(5*x/2) / sin(x/2))}
\psplot[algebraic,plotpoints=1000,linewidth=0.5pt,linestyle=dashed]{-3.14}{3.14}{abs(sin(5*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){$0$}
\rput(1.5, 5){$N=5$}
\end{pspicture}
\end{center}

\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}

Review