From afed1ed6b62aa9e0c347e608d57d3ae9118e80ca Mon Sep 17 00:00:00 2001 From: Philipp Le Date: Sun, 7 Jun 2020 23:55:33 +0200 Subject: Added tasks for Exercise 5 --- chapter06/content_ch06.tex | 45 +++++++++++++--- exercise05/exercise05.tex | 131 +++++++++++++++++++++++++++++++++++++++++++++ exercise06/exercise06.tex | 14 +++++ 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} -- cgit v1.1