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\documentclass[12pt,a4paper,fleqn]{article}
\usepackage{../styles/defsDSPcourse}
\title{COM-303 - Signal Processing for Communications}
\author{Solutions for Homework \#4}
\date{}
\begin{document}
\maketitle
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{solution}{Implementing the DFT}
Set $W = e^{-j2\pi/5}$; the DFT matrix for a 5-point DFT is
\[
\mathbf{W} =
\begin{bmatrix}
1 & 1 & 1 & 1 & 1\\
1 & W & W^2 & W^3 & W^4 \\
1 & W^2 & W^4 & W^6 & W^8 \\
1 & W^3 & W^6 & W^9 & W^{12} \\
1 & W^4 & W^8 & W^{12} & W^{16}
\end{bmatrix}
=
\begin{bmatrix}
1 & 1 & 1 & 1 & 1\\
1 & W & W^2 & W^3 & W^4 \\
1 & W^2 & W^4 & W & W^3 \\
1 & W^3 & W & W^4 & W^2 \\
1 & W^4 & W^3 & W^2 & W
\end{bmatrix}
\]
where we have exploited the fact that $W^n = W^{(n \mod 5)}$. The elements of the matrix that we need to compute are $W$ to $W^4$. We have:
\begin{align*}
W &= C - jS \\
W^2 &= (C-jS)^2 = (C^2 - S^2) - 2jCS \\
W^3 &= e^{-j6\pi/5} = e^{j4\pi/5} = (W^2)^* = (C^2 - S^2) + 2jCS \\
W^4 &= e^{-j8\pi/5} = e^{j2\pi/5} = W^* = C + jS
\end{align*}
With this, given a real-valued input vector $\mathbf{x} = \begin{bmatrix} x_0 & x_1 & x_2 & x_3 & x_4 \end{bmatrix}^T$, we have
\begin{align*}
X_0 &= x_0 + x_1 + x_2 + x_3 + x_4 \\
X_1 &= x_0 + (x_2+x_3)(C^2-S^2) + (x_1 + x_4)C - j[(x_1 - x_4)S + 2(x_2-x_3)CS] \\
X_2 &= x_0 + (x_1+x_4)(C^2-S^2) + (x_2 + x_3)C - j[(x_3 - x_2)S + 2(x_1-x_4)CS] \\
X_3 &= X_2^* \\
X_4 &= X_1^*
\end{align*}
\end{solution}
\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{solution}{DTFTs}
\begin{enumerate}
\item by linearity:
\begin{align*}
X(e^{j\omega}) &= \frac{1}{1 - \frac{1}{2}e^{-j\omega}} - \sum_{n=1}^{\infty}4^{-n}e^{-j\omega n} \\
&= \frac{1}{1 - \frac{1}{2}e^{-j\omega}} - \sum_{n=0}^{\infty}4^{-n}e^{-j\omega n} + 1 \\
&= \frac{1}{1 - \frac{1}{2}e^{-j\omega}} - \frac{1}{1 - \frac{1}{4}e^{-j\omega}} + 1
\end{align*}
\item using the modulation theorem:
\begin{align*}
X(e^{j\omega}) &= \frac{1}{2}\frac{1}{1-ae^{-j(\omega-\omega_0)}} + \frac{1}{2}\frac{1}{1-ae^{-j(\omega+\omega_0)}} \\
&= \frac{1 - a\cos\omega_0 e^{-j\omega}}{1-2a\cos \omega_0 e^{-j\omega} + a^2e^{-j2\omega}}
\end{align*}
\end{enumerate}
\end{solution}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{solution}{DTFT}
Set $t[n] = a^n\,u[n]$; then
\[
x[n] = t[n] + t[-n] - \delta[n].
\]
Since $T(e^{j\omega}) = (1-ae^{-j\omega})^{-1}$, it is
\begin{align*}
X(e^{j\omega}) &= \frac{1}{1-ae^{-j\omega}} + \frac{1}{1-ae^{j\omega}} - 1 \\
&= \frac{1-a^2}{1 - 2a\cos\omega + a^2}
\end{align*}
Here is a plot for $a=0.99$:
\begin{center}
\def\a{0.99 }
\begin{dspPlot}[height=3cm,width=10cm,xtype=freq,xout=true,yticks=50]{-1,1}{0,250}
\dspFunc{x 180 mul cos -2 mul \a mul 1 add \a \a mul add 1 \a \a mul sub exch div }
\end{dspPlot}
\end{center}
\end{solution}
\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{solution}{DTFT}
The sequence $t[n] = (1+\cos\pi n)/2$ is equal to 1 for $n$ even and to 0 for $n$ odd. With this we can write
\begin{align*}
y[n] &= x[n]\,t[n] + t[n+1] \\
&= \frac{1}{2}\left( x[n] + x[n]\cos\pi n + 1 + \cos\pi(n+1) \right) \\
&= \frac{1}{2}\left( x[n] + x[n]\cos\pi n + 1 - \cos\pi n) \right)
\end{align*}
Now recall that the (generalized) DTFT of a cosine with frequency $\omega = \pi$ is
\begin{align*}
\mbox{DTFT}\{\cos\pi n\} &= \frac{1}{2}\tilde{\delta}(\omega + \pi) + \frac{1}{2}\tilde{\delta}(\omega - \pi) \\
&= \tilde{\delta}(\omega - \pi)
\end{align*}
since $\tilde{\delta}(\omega)$ is $2\pi$-periodic. Therefore, by linearity,
\[
Y(e^{j\omega}) = \frac{1}{2}[X(e^{j\omega}) + X(e^{j(\omega - \pi)}) + \tilde{\delta}(\omega) - \tilde{\delta}(\omega-\pi)]
\]
\end{solution}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{solution}{Modulation}
\begin{enumerate}
\item The modulated signal is
\[
y[n]=\Re\{x[n]e^{j\omega_c n}\} = x[n]\cos(\omega_c n)
\]
so that its DTFT is simply:
\[
Y(e^{j\omega})=\frac{1}{2} \left[X \left( e^{j(\omega-\omega_c)} \right) + X \left( e^{j(\omega+\omega_c)} \right) \right].
\]
\item The baseband signal has spectral support $[\frac{-\omega_b}{2},\frac{\omega_b}{2}]$. First note that the maximum frequency of the modulated signal is $\omega_c + \frac{\omega_b}{2}$. To avoid overlap with the first repetition of the spectrum, we must guarantee that:
\[
\omega_c + \frac{\omega_b}{2} < \pi,
\]
which limit the maximum carrier frequency to: $\omega_c < \pi - \frac{\omega_b}{2}$
\item The sampling frequency is $F_s=48000$ Hz and the signal bandwidth is 8~KHz and therefore
\[
\omega_b=8000\mbox{ Hz}\frac{2\pi}{F_s}=\frac{\pi}{3}.
\]
Applying the result of previous exercise:
\[
\omega_c < \pi - \frac{\omega_b}{2} = \frac{5\pi}{6},
\]
and converting back into Hz,
\[
f_c = \omega_c \frac{F_s}{2\pi} < \frac{5\pi}{6}\frac{48000\mbox{Hz}}{2\pi} = 20\mbox{KHz}.
\]
\end{enumerate}
\end{solution}
\end{document}

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