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\documentclass[12pt,a4paper,fleqn]{article}
\usepackage{../styles/defsDSPcourse}
\title{COM-303 - Signal Processing for Communications}
\author{Solutions for Homework \#12}
\date{}
\begin{document}
\maketitle
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{solution}{Channel capacity}
Using Shannon's capacity formula, we can derive the upper limit on the channel's througput as $C=W\log_2(1+\mbox{SNR})=39.949$ Mbps.
\end{solution}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{solution}{QAM error rate}
The correct answers are (b) and (d).
\end{solution}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{solution}{QAM demodulation}
Remember that the input signal is $s[n] = b_r[n]\cos(\omega_c n) - b_i[n]\sin(\omega_c n)$, and the goal is to retrieve the baseband signals $b_r[n]$ and $b_i[n]$.
\begin{enumerate}
\item For $m_1[n]=\cos(\omega_c n)$, $m_2[n]=\sin(\omega_c n)$ we can use the "double angle" trigonometric formulas to obtain
\begin{align*}
s[n]m_1[n] &= \frac{1}{2}\left[b_r[n] + b_r[n]\cos(2\omega_c n) -b_i[n]\sin(2\omega_c n)\right] \\
s[n]m_2[n] &= \frac{1}{2}\left[-b_i[n] + b_r[n]\sin(2\omega_c n) - b_i[n] \cos(2\omega_c n) \right];
\end{align*}
all modulated terms are at double the carrier frequency so that, by filtering out frequencies over $\omega_c$, we can retrieve the basesband signals.
\item For $m_1[n]=1+\cos\omega_cn$, $m_2[n]=1+\sin\omega_cn$ we have
\begin{align*}
s[n]m_1[n] &= \frac{b_r[n]}{2}+\frac{1}{2}(b_r[n] \cos(2\omega_c n) - b_i[n]\sin(2\omega_c n)) \\
&+ b_r[n]\cos(\omega_c n) - b_i[n] \sin(\omega_c n) \\
s[n]m_2[n] &= \frac{b_i[n]}{2} + \frac{1}{2}(b_r[n]\sin(2\omega_c n) - b_i[n]\cos(2\omega_c n)) \\
&+ b_r[n]\cos(\omega_c n) - b_i[n]\sin(\omega_c n);
\end{align*}
here we have modulated terms at the carrier frequency but, by filtering out frequencies over $\omega_0$ we can still retrieve the basesband signals.
\item For $m_1[n]=\cos\frac{\omega_cn}{2} \cos\frac{3\omega_cn}{2}$, $m_2[n]=\sin\frac{\omega_cn}{2} \cos\frac{3\omega_cn}{2}$,we have
\begin{align*}
s[n]m_1[n] &= \frac{b_r[n]}{4} + \frac{b_r[n]}{4}( \cos(\omega_c n) + \cos(2\omega_c n) + \cos(3\omega_c n)) \\
&- \frac{b_i[n]}{4}( -\sin (\omega_c n) + \sin(2\omega_c n) + \sin(3\omega_c n)) \\
s[n]m_2[n] &= \frac{b_i[n]}{4} + \frac{b_r[n]}{4}( \sin(\omega_c n) - \sin(2\omega_c n) + \sin(3\omega_c n)) \\
&+\frac{b_i[n]}{4}( \cos (\omega_c n) - \cos(2\omega_c n)-\cos(3\omega_c n));
\end{align*}
here we have modulated terms at the 1x, 2x, and 3x the carrier frequency but, again, by filtering out frequencies over $\omega_0$ we can still retrieve the basesband signals.
\end{enumerate}
\end{solution}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{solution}{ADSL}
\begin{enumerate}
\item Throughput is proportional to the SNR, therefore the channel with the largest possible throughput is $\mbox{CH}_7$ while the channels with the lowest throughput are $\mbox{CH}_2$ and $\mbox{CH}_5$.
\item For each channel, we need to first isolate the operational quadrant on the QAM error rate chart, and then select the curve with the highest number of bits per symbol that intersects the quadrant. The upper bound for the quadrant is given by the max probability of error: we can first draw a horizontal line on the SNR chart, intersecting the ordinate at $P_{\it err}=10^{-6}$ and the operational area of the SNR chart will be \textit{below} this line. Then, for each sub-channel we can draw a vertical line at the subchannel SNR: the area to its \textit{left} will completely identify the operational quadrant. Finally, once we have determined the number of bits per symbol $M$, the resulting throughput will be equal to $MF_s\omega_b$ where $\omega_b$ is the bandwidth of the subchannel, i.e. $\pi/8$.
\begin{itemize}
\item $\mbox{CH}_3$ has a max SNR of $24$dB so that both the $M=2$ and $M=4$ curves intersect the quadrant. We can choose $M=4$ so that the final throughput is $500$Kbps.
\item $\mbox{CH}_4$ has a max SNR of $32$dB; we can use $M=6$ for a max throughput of 750Kbps
\item $\mbox{CH}_7$ has a max SNR of $32$dB; we can use $M=10$ for a max throughput of 1.25Mbps
\end{itemize}
\end{enumerate}
\end{solution}
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\begin{solution}{Bandwidth constraint}
\begin{enumerate}
\item In general, the Baud rate is equal to the available bandwidth so we have $250\cdot 10^6$ symbols/s. $\cal{A}$ has 32 symbols, so we need $\log_2 32 = 5 $ bits/symbol. Hence the throughput is $250\cdot 10^6 \cdot 5 = 1.25\cdot 10^9$ bits/s
\item The transmitter's sampling rate will need to satisfy $F_s > 2F_{max}$, as per the sampling theorem. With that, we should choose $F_s$ to be an integer multiple of the bandwidth, $F_s = KW$. Among the possibilities listed in the exercise, only (b), $K=12$ for $F_s = 2.4$GHz, is thus a valid choice.
\end{enumerate}
\end{solution}
\begin{solution}{Power constraint}
The channel bandwidth is 3 kHz, so the Baud rate is $W = 3000$ symbols/s; in other to find the total bit rate we need to determine the maximum amount of bits per symbol that we can send given the power constraint (i.e. the max SNR) and the accepted probability of error. We know tha for QAM signaling $P_{err} = e^{-3\cdot2^{-(M+1)}\cdot \mathrm{SNR}}$. This gives us $M \approx 6$, and so the total bitrate $18$Kbit/sec.
\end{solution}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{solution}{Gain and probability of error}
Assume each received symbol is affected by a noise sample $\eta[n] \in \mathcal{U}[-100, 100]$. Then
\begin{eqnarray*}
P_{err} &=& P[|\eta[n]| >G \mid \text{n}^{th} \text{bit} \neq \pm 31G ]\cdot P[\text{n}^{th} \text{bit} \neq \pm 31G] \nonumber\\
&&+ P[\eta[n]>G\mid \text{n}^{th} \text{bit} = -31G] \cdot P[\text{n}^{th} \text{bit} = -31G]\nonumber\\
&&+ P[\eta[n]<-G\mid \text{n}^{th} \text{bit} = 31G] \cdot P[\text{n}^{th} \text{bit} = 31G]\nonumber\\
&=&P[|\eta[n]| >G]\cdot \frac{30}{32} + P[\eta[n] >G]\cdot \frac{1}{32} + P[\eta[n] <-G]\cdot \frac{1}{32}\nonumber\\
&=&\frac{30}{32}\cdot 2 \cdot P[\eta[n]>G] + \frac{2}{32} \cdot P[\eta[n]>G]
\end{eqnarray*}
We know that $P_{err} = 10^{-2}$, so $P[\eta[n]>G] = 10^{-2}\cdot \frac{32}{62} = 0.0051$, and $G = 99$.
\end{solution}
\end{document}
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