Difference between revisions of "Aufgaben:Exercise 3.2Z: Sinc-Squared Spectrum with Diracs"
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*Further information on this topic can be found in the learning video [[Kontinuierliche_und_diskrete_Spektren_(Lernvideo)|Kontinuierliche und diskrete Spektren]]. | *Further information on this topic can be found in the learning video [[Kontinuierliche_und_diskrete_Spektren_(Lernvideo)|Kontinuierliche und diskrete Spektren]]. | ||
| − | *It can be assumed as known that a triangular pulse $y(t)$ symmetrical about $t = 0$ with the amplitude ${A}$ and the absolute duration $2T$ $($i.e.: he signal values are unequal to$0 | + | *It can be assumed as known that a triangular pulse $y(t)$ symmetrical about $t = 0$ with the amplitude ${A}$ and the absolute duration $2T$ $($i.e.: he signal values are unequal to $ 0 $ only between $–T$ and $+T$ ) has the following spectral function: |
:$$Y( f ) = A \cdot T \cdot {\rm si}^2 ( \pi f T ).$$ | :$$Y( f ) = A \cdot T \cdot {\rm si}^2 ( \pi f T ).$$ | ||
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<quiz display=simple> | <quiz display=simple> | ||
| − | { | + | {What are the values of the parameters ${A}$ (amplitude) and ${T}$ (one-sided duration) of the triangular signal component $x_1(t)$? |
|type="{}"} | |type="{}"} | ||
$A\ = \ $ { 2 3% } $\text{V}$ | $A\ = \ $ { 2 3% } $\text{V}$ | ||
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| − | { | + | {What is the DC component ${B}$ of the signal? |
|type="{}"} | |type="{}"} | ||
$B\ = \ $ { -1.03--0.97 } $\text{V}$ | $B\ = \ $ { -1.03--0.97 } $\text{V}$ | ||
| − | { | + | {What is the amplitude $C$ of the periodic component of $x(t)$? |
|type="{}"} | |type="{}"} | ||
$C\ = \ $ { 1 3% } $\text{V}$ | $C\ = \ $ { 1 3% } $\text{V}$ | ||
| − | { | + | {What are the maximum and minimum values of the signal $x(t)$? |
|type="{}"} | |type="{}"} | ||
$x_\text{max}\ = \ $ { 2 3% } $\text{V}$ | $x_\text{max}\ = \ $ { 2 3% } $\text{V}$ | ||
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</quiz> | </quiz> | ||
| − | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
[[File:P_ID498__Sig_Z_3_2_a_neu.png|right|frame|Fläche des Dreieckimpulses]] | [[File:P_ID498__Sig_Z_3_2_a_neu.png|right|frame|Fläche des Dreieckimpulses]] | ||
Revision as of 20:52, 20 January 2021
$\rm si$-Quadrat-Spektrum mit Diracs
The sketched spectrum ${X(f)}$ of a time signal ${x(t)}$ is composed of
- a continuous component $X_1(f)$,
- plus three dirac-shaped spectral lines.
The continuous component with $f_0 = 200\, \text{kHz}$ and $X_0 = 10^{–5} \text{ V/Hz}$is as follows:
- $$X_1( f ) = X_0 \cdot {\mathop{\rm si}\nolimits} ^2 ( {\pi {f}/{f_0}} ),\quad {\rm where is}\quad {\mathop{\rm si}\nolimits} (x) = {\sin (x)}/{x}.$$
The spectral line at $f = 0$ has the weight $–\hspace{-0.08cm}1\,\text{V}$. In addition, there are two lines at frequencies $\pm f_0$, both with weight $0.5\,\text{V}$.
Hints:
- This exercise belongs to the chapter Fourier Transform and Its Inverse.
- Further information on this topic can be found in the learning video Kontinuierliche und diskrete Spektren.
- It can be assumed as known that a triangular pulse $y(t)$ symmetrical about $t = 0$ with the amplitude ${A}$ and the absolute duration $2T$ $($i.e.: he signal values are unequal to $ 0 $ only between $–T$ and $+T$ ) has the following spectral function:
- $$Y( f ) = A \cdot T \cdot {\rm si}^2 ( \pi f T ).$$
Question
Solution
Fläche des Dreieckimpulses
(1) Die einseitige Dauer des symmetrischen Dreieckimpulses beträgt $T = 1/f_0\hspace{0.15 cm}\underline{ = 5 \,{\rm µ s}}$.
- Der Spektralwert $X_0 = X_1(f = 0)$ gibt die Impulsfläche von $x_1(t)$ an.
- Diese ist gleich ${A} \cdot {T}$. Daraus folgt:
- $$A = \frac{X_0 }{T} = \frac{ 10^{-5}\rm V/Hz }{5 \cdot 10^{-6}{\rm s}}\hspace{0.15 cm}\underline{= 2\;{\rm V}}.$$
(2) Der Gleichsignalanteil ist durch das Diracgewicht bei $f = 0$ gegeben. Man erhält ${B} \hspace{0.15 cm}\underline{= -1 \,\text{V}}$.
(3) Die beiden Spektrallinien bei $\pm f_0$ ergeben zusammen ein Cosinussignal mit der Amplitude ${C} \hspace{0.15 cm}\underline{= 1 \text{V}}$.
(4) Der Maximalwert tritt zum Zeitpunkt ${t} = 0$ auf (hier sind Dreieckimpuls und Cosinussignal maximal):
- $$x_{\text{max}} = A + B + C \hspace{0.15 cm}\underline{= +2 \text{V}}.$$
- Die minimalen Werte von ${x(t)}$ ergeben sich dann, wenn der Dreieckimpuls abgeklungen ist und die Cosinusfunktion den Wert $–\hspace{-0.08 cm}1 \,\text{V}$ liefert:
- $$x_\text{min} = {B} - {C}\hspace{0.15 cm}\underline{ = -2\, \text{V}}.$$
