blob: 44dea360bc44530009d37900469bd7ba97289a0e (
plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
|
% SPDX-License-Identifier: CC-BY-SA-4.0
%
% Copyright (c) 2020 Philipp Le
%
% Except where otherwise noted, this work is licensed under a
% Creative Commons Attribution-ShareAlike 4.0 License.
%
% Please find the full copy of the licence at:
% https://creativecommons.org/licenses/by-sa/4.0/legalcode
\phantomsection
\addcontentsline{toc}{section}{Exercise 4}
\section*{Exercise 4}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{question}[subtitle={Sampling Periodic Signals}]
\begin{equation*}
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 $u(t = 0)$.
\begin{tasks}
\task
Plot the function from $t = 0$ to $t = \SI{1}{\micro\second}$!
\task
Calculate the samples $n = 0 \dots 8$!
\task
What is the DTFT of the signal?
Hints:
\begin{equation*}
\begin{split}
x[n] = e^{-j a n} &= \underline{X}_{\frac{2\pi}{T_S}}\left(e^{-j T_S \omega}\right) = 2 \pi \cdot \delta \left(\omega + a\right) \\
\cos\left(b\right) &= \frac{1}{2} \left(e^{j b} + e^{-j b}\right)
\end{split}
\end{equation*}
\task
Can the DFT directly applied to the signal? If yes, determine the smallest $N$ and give the values of all $\underline{U}[k]$!
\task
What is the longest possible sampling period? What must be considered at this sampling period?
\task
Now, the sampling period is changed to $T_S = \SI{0.5}{\micro\second}$. There is no anti-aliasing filter. The reconstruction filter is an ideal low-pass filter with a cut-off frequency of \SI{50}{kHz}. Give the reconstructed output function in the time domain! Give an explanation in the frequency domain!
\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}]
The signal of task 1b) is now quantized. The quantizer has $8$ discrete values. These values are equally distributed between \SI{-2}{V} and \SI{2}{V}. Prior to sampling, the original time-continuous signal passed through an ideal low-pass filter with a cut-off frequency of \SI{4}{MHz}.
\begin{tasks}
\task
Define a mapping from the value-continuous samples to the value-discrete samples!
\task
The value-discrete samples are now pulse-code modulated. How many bits are required?
\task
Determine the quantization error for each value-discrete sample! How much is the signal-to-noise ratio?
\task
3 bits are a very poor resolution. How many bits are appropriate for the quantizer to obtain the best signal-to-noise ratio? Effects of the window filter are neglected. Assume that the signal has passed through a processing chain with a total gain of \SI{25}{dB} and noise figure of \SI{12}{dB} prior to quantization. The input of the quantizer has an impedance of \SI{50}{\ohm}. % 14 bits
\end{tasks}
\end{question}
\begin{solution}
\begin{tasks}
\end{tasks}
\end{solution}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{question}[subtitle={Decibel}]
% \begin{tasks}
% \end{tasks}
%\end{question}
%
%\begin{solution}
% \begin{tasks}
% \end{tasks}
%\end{solution}
|
