Added tasks for Exercise 5

3 files changed, 184 insertions, 6 deletions

diff --git a/chapter06/content_ch06.tex b/chapter06/content_ch06.tex
index 23ad171..e50260c 100644
--- a/chapter06/content_ch06.tex
+++ b/chapter06/content_ch06.tex
@@ -186,8 +186,8 @@ However, the filter is implemented in the time-domain.
 	\caption{Block diagram of an example \acs{IIR} filter}
 	\label{fig:ch06:iir_filt}
 \end{figure}%
-\nomenclature[Bm]{\begin{circuitikz}[baseline={(current bounding box.center)}]\draw (0,0) to[twoport,t=$z^{-1}$,>] (2,0);\end{circuitikz}}{Delay element}%
-\nomenclature[Bm]{\begin{circuitikz}[baseline={(current bounding box.center)}]\node[adder](){};\end{circuitikz}}{Adder}
+\nomenclature[Bd]{\begin{circuitikz}[baseline={(current bounding box.center)}]\draw (0,0) to[twoport,t=$z^{-1}$,>] (2,0);\end{circuitikz}}{Delay element}%
+\nomenclature[Ba]{\begin{circuitikz}[baseline={(current bounding box.center)}]\node[adder](){};\end{circuitikz}}{Adder}
 
 Figure \ref{fig:ch06:iir_filt} shows an example filter. The block diagram has following digital components:
 \begin{itemize}
@@ -251,7 +251,7 @@ A stable filter has always a value-limited impulse response (\ac{BIBO} stable).
 	\end{equation}
 	\item The conditions for \ac{BIBO} stability is that all poles are located \underline{within the unit circle}.
 	\begin{equation}
-		\left|\underline{z}_{\infty,l}\right| < 1 \qquad \forall \; 0 \leq l \leq Q
+		\left|\underline{z}_{\infty,l}\right| < 1 \qquad \forall \; 0 \leq l \leq P
 	\end{equation}
 \end{itemize}
 
@@ -270,7 +270,7 @@ A digital filter without the feed-back path will not have any problems with stab
 		\underline{H}(\underline{z}) = \sum\limits_{i=0}^{P} \underline{b}_i \underline{z}^{-i}
 	\end{equation}
 	\item The number of feed-back filter taps is $Q = 0$.
-	\item The filter does not have any poles. \textbf{The filter will always be \ac{BIBO} stable.}
+	\item All poles of the filter are $\underline{z}_{\infty,l} = 0 \quad \forall \; 0 \leq l \leq P$. \textbf{The filter will always be \ac{BIBO} stable.}
 \end{itemize}
 
 \begin{figure}[H]
@@ -324,6 +324,8 @@ There is another simple explanation for the \ac{BIBO} stability.
 	Digital filters without a feed-back branch will always have a finite-length impulse response. They are called \index{finite impulse response filter} \textbf{\acf{FIR} filters}. \ac{FIR} filters are always \ac{BIBO} stable.
 \end{definition}
 
+As a drawback, \ac{FIR} filters require higher orders than an equivalent \ac{IIR} filter. This increases the complexity of its implementation.
+
 \begin{example}{Gliding average filter}
 	The formula of the average of a series of $N$ values is:
 	\begin{equation}
@@ -360,8 +362,41 @@ There is another simple explanation for the \ac{BIBO} stability.
 
 Both \ac{IIR} and \ac{FIR} are causal. Their impulse response is $\underline{h}[n] = 0 \quad \forall \; n < 0$.
 
+\section{Digital Mixer}
+
 \section{Resampling}
 
+\begin{figure}[H]
+	\centering
+	\begin{tikzpicture}
+		\node[block,draw,align=center](High){High sampling rate};
+		\node[block,draw,align=center,right=3cm of High](Low){Low sampling rate};
+		
+		\draw[-latex] ([xshift=5mm] High.north east) -- node[midway,above,align=center]{Down-sampling\\ (Decimation)} ([xshift=-5mm] Low.north west);
+		\draw[-latex] ([xshift=-5mm] Low.south west) -- node[midway,below,align=center]{Up-sampling\\ (Interpolation)} ([xshift=5mm] High.south east);
+	\end{tikzpicture}
+	\caption{Relation between down-sampling (decimation) and up-sampling (interpolation).}
+\end{figure}
+
+\begin{figure}[H]
+	\centering
+	\begin{circuitikz}
+		\node[block,draw,minimum height=3cm](Data){Data\\ Processing};
+		
+		\draw ([shift={(-4cm,1cm)}] Data.west) node[left,align=right]{Input} to[adc,>,o-] ++(2cm,0) to[twoport,t=$\downarrow N$,>] ([yshift=1cm] Data.west) node[inputarrow]{};
+		\draw ([yshift=-1cm] Data.west) to[twoport,t=$\uparrow M$,>] ++(-2cm,0) to[dac,>] ++(-2cm,0) node[inputarrow,rotate=180]{} node[left,align=right]{Output};
+	\end{circuitikz}
+	\caption{A system with a down-sampler (decimation factor $N$) and up-sampler (interpolation factor $M$)}
+\end{figure}%
+\nomenclature[Bd]{\begin{circuitikz}[baseline={(current bounding box.center)}]\draw (0,0) to[twoport,t=$\downarrow N$,>] (2,0);\end{circuitikz}}{Down-sampler (decimation factor $N$)}%
+\nomenclature[Bu]{\begin{circuitikz}[baseline={(current bounding box.center)}]\draw (0,0) to[twoport,t=$\uparrow M$,>] (2,0);\end{circuitikz}}{Up-sampler (interpolation factor $M$)}%
+
+\textbf{Why resampling?}
+\begin{itemize}
+	\item Signals at lower sampling rates require less computation time and memory (software), or lower hardware complexity (less logic gates). The power consumption is reduced.
+	\item The \ac{ADC} can be operated at maximum sampling rate. The signal is oversampled. Down-sampling provides processing gain and enhances the receiver performance.
+\end{itemize}
+
 \todo{Downsampling, Decimation}
 
 \todo{Upsampling, Interpolation}
@@ -370,8 +405,6 @@ Both \ac{IIR} and \ac{FIR} are causal. Their impulse response is $\underline{h}[
 
 \todo{CIC filter}
 
-\section{Digital Mixer}
-
 \section{Fast Fourier Transform}
 
 \todo{FFT}
diff --git a/exercise05/exercise05.tex b/exercise05/exercise05.tex
index 0591f3c..0c1ae31 100644
--- a/exercise05/exercise05.tex
+++ b/exercise05/exercise05.tex
@@ -12,6 +12,137 @@
 \addcontentsline{toc}{section}{Exercise 5}
 \section*{Exercise 5}
 
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\begin{question}[subtitle={Mixers}]
+	\begin{tasks}
+		\task
+		Is the mixer a linear device like filters and amplifiers?
+		\task
+		What is the difference between unbalanced and balanced mixers?
+		\task
+		Why do mixers need a non-linear component?
+	\end{tasks}
+\end{question}
+
+\begin{solution}
+	\begin{tasks}
+	\end{tasks}
+\end{solution}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\begin{question}[subtitle={Mirror frequencies}]
+	This is simplified block diagram of a receiver.
+	\begin{figure}[H]
+		\centering
+		\begin{adjustbox}{scale=0.8}
+			\begin{circuitikz}
+				\node[mixer](Mixer){};
+				\node[oscillator, below=1cm of Mixer](LO){};
+				\node[adcshape, right=2cm of Mixer](ADC){};
+				\node[block, draw, right=1cm of ADC](Baseband){Digital signal\\ processing};
+				
+				\draw (LO.south) node[below,align=center,yshift=-5mm]{LO};
+				\draw (Mixer.north) node[above,align=center,yshift=3mm]{Mixer};
+				
+				\draw (Mixer.west) -- ++(-1cm,0) node[rxantenna,xscale=-1]{};
+				
+				\draw[-latex] (LO.north) -- (Mixer.south);
+				\draw[-latex] (Mixer.east) to[lowpass] (ADC.west);
+				\draw[-latex] (ADC.east) -- (Baseband.west);
+			\end{circuitikz}
+		\end{adjustbox}
+	\end{figure}
+	A signal of \SI{868}{MHz} should be received. The baseband is not zero-IF. The signal shall be mixed to \SI{1}{MHz} centre frequency.
+
+	\begin{tasks}
+		\task
+		How much is the minimum ADC sampling rate?
+		\task
+		To which frequencies can the LO be tuned to?
+		\task
+		The \SI{868}{MHz}-band is shared with lots of other users. Which important piece is missing in the receiver signal chain?
+		\task
+		An IQ demodulator is used instead of the single mixer. Sketch the spectrum of the complex-valued baseband signal for both possible LO frequencies!
+	\end{tasks}
+\end{question}
+
+\begin{solution}
+	\begin{tasks}
+	\end{tasks}
+\end{solution}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\begin{question}[subtitle={Constellation diagrams}]
+	Draw a constellation diagram of:
+	\begin{tasks}
+		\task
+		ASK (with 2 steps)
+		\task
+		BPSK
+		\task
+		QPSK
+		\task
+		16-QAM
+	\end{tasks}
+\end{question}
+
+\begin{solution}
+	\begin{tasks}
+	\end{tasks}
+\end{solution}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\begin{question}[subtitle={Constellation diagrams}]
+	A QPSK modulator has the following mapping and symbol constellation:
+	\begin{table}[H]
+		\centering
+		\begin{tabular}{|l|l|l|}
+			\hline
+			Data & Symbol & Phasor \\
+			\hline
+			\hline
+			$(00)_2$ & 0 & $\SI{2}{mV} \cdot e^{j 0}$ \\
+			\hline
+			$(01)_2$ & 1 & $\SI{2}{mV} \cdot e^{j \frac{\pi}{2}}$ \\
+			\hline
+			$(10)_2$ & 2 & $\SI{2}{mV} \cdot e^{j \pi}$ \\
+			\hline
+			$(11)_2$ & 3 & $\SI{2}{mV} \cdot e^{j \frac{3 \pi}{2}}$ \\
+			\hline
+		\end{tabular}
+	\end{table}
+	The carrier is:
+	\begin{equation}
+		x_C(t) = \SI{2}{mV} \cdot \cos\left(2\pi \cdot \SI{50}{MHz} \cdot t\right)
+	\end{equation}
+	The symbol rate is $\SI{25}{MHz}$. After the DAC, an ideal low-pass filter with $\SI{25}{MHz}$ cut-off frequency is applied.
+	
+	\begin{tasks}
+		\task
+		How much is the transmission bandwidth?
+		\task
+		How many bits can be encoded per QPSK symbol? How many symbols are required to encode one byte (8 bits)?
+		\task
+		Draw the constellation diagram!
+		\task
+		The data byte $(2E)_{16}$ shall be transmitted. Give the sequence of phasors representing the data byte!
+		\task
+		Describe the problem with inter-symbol interference!
+		\task
+		Plot the I and Q baseband signals! Plot the RF signal after IQ modulation!
+		\task
+		The following phasors are received at the receiver:
+		\begin{equation}
+			[\SI{1.5}{mV} e^{j \SI{120}{\degree}}, \SI{1.5}{mV} e^{j \SI{300}{\degree}}, \SI{1.5}{mV} e^{j \SI{30}{\degree}}, \SI{1.5}{mV} e^{j \SI{210}{\degree}}]
+		\end{equation}
+		What would the decoded data be? What is the matter?
+	\end{tasks}
+\end{question}
+
+\begin{solution}
+	\begin{tasks}
+	\end{tasks}
+\end{solution}
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 %\begin{question}[subtitle={Decibel}]
diff --git a/exercise06/exercise06.tex b/exercise06/exercise06.tex
index 169b3f6..954357e 100644
--- a/exercise06/exercise06.tex
+++ b/exercise06/exercise06.tex
@@ -14,6 +14,20 @@
 
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\begin{question}[subtitle={Decibel}]
+	Proof mathematically that all poles of the FIR filter are $0$!
+	
+	\begin{tasks}
+	\end{tasks}
+\end{question}
+
+\begin{solution}
+	\begin{tasks}
+	\end{tasks}
+\end{solution}
+
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 %\begin{question}[subtitle={Decibel}]
 %	\begin{tasks}
 %	\end{tasks}