diff options
Diffstat (limited to 'exercise04')
| -rw-r--r-- | exercise04/exercise04.tex | 126 | ||||
| -rw-r--r-- | exercise04/helpers/ex_4_2.py | 63 |
2 files changed, 179 insertions, 10 deletions
diff --git a/exercise04/exercise04.tex b/exercise04/exercise04.tex index 27551d4..44dea36 100644 --- a/exercise04/exercise04.tex +++ b/exercise04/exercise04.tex @@ -15,7 +15,7 @@ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{question}[subtitle={Sampling Periodic Signals}] \begin{equation*} - u(t) = \SI{2}{V} \cos\left(2\pi \SI{2}{MHz} t + \SI{60}{\degree}\right) + u(t) = \SI{2}{V} \cos\left(2\pi \cdot \SI{2}{MHz} \cdot t + \SI{60}{\degree}\right) \end{equation*} The signal is sampled with a sampling period of $T_S = \SI{125}{\nano\second}$. The first sample taken is $u(t = 0)$. @@ -54,15 +54,121 @@ \end{solution} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -%\begin{question}[subtitle={Sampling Non-Periodic Signals}] -% \begin{tasks} -% \end{tasks} -%\end{question} -% -%\begin{solution} -% \begin{tasks} -% \end{tasks} -%\end{solution} +\begin{question}[subtitle={Sampling Non-Periodic Signals}] + The signal $x[n]$ is given in the time domain. + \begin{table}[H] + \centering + \begin{tabular}{|l|r|r|r|r|r|r|r|r|} + \hline + $n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ + \hline + \hline + $x[n]$ & 0.5 & 1 & 0 & 0.5 & -0.5 & -1 & -0.5 & -0.75 \\ + \hline + \end{tabular} + \end{table} + + \begin{tasks} + \task + The signal is windowed with $N = 4$ starting at $x[n = 0]$. A hamming window with $M = 2$ is applied. Calculate the values of $\underline{x}_W[n]$! + + Hamming window: + \begin{equation*} + w[n] = \begin{cases}0.54 - 0.46 \cos\left(\frac{2 \pi n}{M}\right), &\quad 0 \leq n \leq M,\\ 0, &\quad \text{otherwise}.\end{cases} + \end{equation*} + + \task + Calculate the discrete Fourier transform of the windowed signal! + + \task + The signal has been sampled with $T_S = \SI{1}{ms}$. What frequency values do the $k$ represent? + \end{tasks} +\end{question} + +\begin{solution} + \begin{tasks} + \task + At first the signal is truncated. Only the first $4$ samples are considered. + + The window function is: + \begin{equation*} + w[n] = \begin{cases}0.54 - 0.46 \cos\left(\frac{2 \pi n}{M}\right), &\quad 0 \leq n \leq M,\\ 0, &\quad \text{otherwise}.\end{cases} + \end{equation*} + + The signal is then multiplied with the window: + \begin{equation*} + x_w[n] = x[n] \cdot w[n] + \end{equation*} + + \begin{table}[H] + \centering + \begin{tabular}{|l|r|r|r|r|} + \hline + $n$ & 0 & 1 & 2 & 3 \\ + \hline + \hline + $w[n]$ & 0.08 & 1.0 & 0.08 & 0 \\ + \hline + $x_w[n]$ & 0.02 & 1.0 & -0.04 & 0 \\ + \hline + \end{tabular} + \end{table} + + \task + The signal is periodically repeated. + + The DFT is calculated over $N = 4$. + \begin{equation*} + \underline{X}_w[k] = \mathcal{F}_{\text{DFT}}\left\{\underline{x}[n]\right\} = \sum\limits_{n \in N} \underline{x}[n] \cdot e^{-j 2\pi \frac{k}{N} n} + \end{equation*} + + \begin{table}[H] + \centering + \begin{tabular}{|l|r|r|r|r|} + \hline + $k$ & 0 & 1 & 2 & 3 \\ + \hline + $k$ (alternate) & 0 & 1 & -2 & -1 \\ + \hline + \hline + $\underline{X}_w[k]$ & $0.98$ & $(0.06-1j)$ & $-1.02$ & $(0.06+1j)$
\\ + \hline + $|\underline{X}_w[k]|$ & $0.98$ & $1.00$ & $1.02$ & $1.00$
\\ + \hline + $\arg\left(\underline{X}_w[k]\right)$ & $0$ & $-1.51 \approx -\pi$ & $3.14 \approx 2\pi$ & $1.51 \approx \pi$ \\ + \hline + \end{tabular} + \end{table} + + \task + \begin{equation*} + \begin{split} + \phi[k] &= 2 \pi \frac{k}{N} \\ + \omega[k] &= \frac{\phi[k]}{T_S} \\ + f[k] &= \frac{\omega[k]}{2 \pi} \\ + \end{split} + \end{equation*} + + \begin{table}[H] + \centering + \begin{tabular}{|l|r|r|r|r|} + \hline + $k$ & 0 & 1 & 2 & 3 \\ + \hline + $k$ (alternate) & 0 & 1 & -2 & -1 \\ + \hline + \hline + $\phi[k]$ & $0$ & $1.57 \approx \pi$ & $3.14 \approx 2 \pi \equiv -2\pi$ & $4.71 \approx 3 \pi \equiv -\pi$ \\ + \hline + $\omega[k]$ & $\SI{0}{s^{-1}}$ & $\SI{1570.8}{s^{-1}}$ & $\SI{3141.6}{s^{-1}}$ & $\SI{4712.4}{s^{-1}}$
\\ + \hline + \hline + $f[k]$ & $\SI{0}{Hz}$ & $\SI{250}{Hz}$ & $\SI{500}{Hz} \equiv \SI{-500}{Hz}$ & $\SI{750}{Hz} \equiv \SI{-250}{Hz}$ \\ + \hline + \end{tabular} + \end{table} + \end{tasks} +\end{solution} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{question}[subtitle={Quantization}] diff --git a/exercise04/helpers/ex_4_2.py b/exercise04/helpers/ex_4_2.py new file mode 100644 index 0000000..a57907b --- /dev/null +++ b/exercise04/helpers/ex_4_2.py @@ -0,0 +1,63 @@ +#!/usr/bin/env python3 + +# SPDX-License-Identifier: BSD-3-Clause +# +# Copyright 2020 Philipp Le +# +# Redistribution and use in source and binary forms, with or without +# modification, are permitted provided that the following conditions +# are met: +# +# 1. Redistributions of source code must retain the above copyright +# notice, this list of conditions and the following disclaimer. +# +# 2. Redistributions in binary form must reproduce the above copyright +# notice, this list of conditions and the following disclaimer in the +# documentation and/or other materials provided with the distribution. +# +# 3. Neither the name of the copyright holder nor the names of its +# contributors may be used to endorse or promote products derived from +# this software without specific prior written permission. +# +# THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS" +# AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE +# IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE +# ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE +# LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR +# CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF +# SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS +# INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN +# CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) +# ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF +# THE POSSIBILITY OF SUCH DAMAGE. + + +import numpy + +def latex_table(ar): + return " & ".join(["$"+str(it)+"$" for it in ar]) + +x = numpy.array([0.25, 1, -0.5, 0.5, -0.5, -1, -0.5, -0.75]) +print("$x[n]$ & "+latex_table(x)) + +w = 0.54 - 0.46 * numpy.cos(2*numpy.pi*numpy.array([0,1,2])/2) +w = numpy.concatenate([w, numpy.zeros(1)]) +print("$w[n]$ & "+latex_table(w)) + +x_w = x[0:4]*w +print("$x_w[n]$ & "+latex_table(x_w)) + +Xft = numpy.fft.fft(x_w) +Xft_abs = numpy.abs(Xft) +Xft_phase = numpy.angle(Xft) +print("$\\underline{X}_w[k]$ & "+latex_table(Xft)) +print("$|\\underline{X}_w[k]|$ & "+latex_table(Xft_abs)) +print("$\\arg\\left(\\underline{X}_w[k]\\right)$ & "+latex_table(Xft_phase)) + +phi=2*numpy.pi*numpy.array([0,1,2,3])/4 +omega=phi/1e-3 +f=omega/(2*numpy.pi) +print("$\\phi[k]$ & "+latex_table(phi)) +print("$\\omega[k]$ & "+latex_table(omega)) +print("$f[k]$ & "+latex_table(f)) + |