Aufgaben:Exercise 2.1Z: Sum Signal: Difference between revisions

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''Hinweis:''  
''Hints:''  
*Die Aufgabe gehört zum Kapitel  [[Signal_Representation/General_Description|Allgemeine Beschreibung periodischer Signale]].
*The exercise belongs to the chapter  [[Signal_Representation/General_Description|General description of periodic signals].
*Mit dem interaktiven Applet  [[Applets:Periodendauer_periodischer_Signale|Periodendauer periodischer Signale]]  lässt sich die resultierende Periodendauer zweier harmonischer Schwingungen ermitteln.  
*With the interactive applet  [[Applets:Periodendauer_periodischer_Signale|Periodendauer periodischer Signale]]  the resulting period duration of two harmonic oscillations can be determined.  
   
   




===Fragebogen===
===Questions===


<quiz display=simple>
<quiz display=simple>
{Wie groß ist Periodendauer&nbsp; $T_x$&nbsp; und Grundfrequenz&nbsp; $f_x$&nbsp; des Signals&nbsp; ${x(t)}$?
{What is the period duration&nbsp; $T_x$&nbsp; and the base frequency&nbsp; $f_x$&nbsp; of the signal&nbsp; ${x(t)}$?
|type="{}"}
|type="{}"}
$f_x\ = \ $  { 1 3% } &nbsp; $\text{kHz}$
$f_x\ = \ $  { 1 3% } &nbsp; $\text{kHz}$




{Wie groß ist Periodendauer&nbsp; $T_y$&nbsp; und Grundfrequenz&nbsp; $f_y$&nbsp; des Signals&nbsp; ${y(t)}$?
{What is the period duration&nbsp; $T_y$&nbsp; and the base frequency&nbsp; $f_y$&nbsp; of the signal&nbsp; ${y(t)}$?
|type="{}"}
|type="{}"}
$f_y\ = \ $ { 0.4 3% } &nbsp; $\text{kHz}$
$f_y\ = \ $ { 0.4 3% } &nbsp; $\text{kHz}$




{Bestimmen Sie die Grundfrequenz&nbsp; $f_s$&nbsp; sowie die Periodendauer&nbsp; $T_s$&nbsp; des Summensignals&nbsp; ${s(t)}$&nbsp; und überprüfen Sie das Ergebnis anhand der Skizze.
{Determine the base frequency&nbsp; $f_s$&nbsp; as well as the period duration&nbsp; $T_s$&nbsp; of the sum signal&nbsp; ${s(t)}$&nbsp; and verify your results with the help of the sketched signal.
|type="{}"}
|type="{}"}
$T_s\ = \ $ { 5 3% } &nbsp; $\text{ms}$
$T_s\ = \ $ { 5 3% } &nbsp; $\text{ms}$




{Welche Periodendauer&nbsp; $T_d$&nbsp; weist das Differenzsignal&nbsp; ${d(t)}$&nbsp; auf?
{What is the period duration&nbsp; $T_d$&nbsp; of the difference signal&nbsp; ${d(t)}$&nbsp;?
|type="{}"}
|type="{}"}
$T_d\ = \ $ { 5 3% } &nbsp; $\text{ms}$
$T_d\ = \ $ { 5 3% } &nbsp; $\text{ms}$




{Welche Periodendauer&nbsp; $T_w$&nbsp; besitzt das Signal&nbsp; ${w(t)} = {u(t)} + v(t)$?
{What is the period duration&nbsp; $T_w$&nbsp; of the signal&nbsp; ${w(t)} = {u(t)} + v(t)$?
|type="{}"}
|type="{}"}
$T_w\ = \ $ { 500 3% } &nbsp; $\text{ms}$
$T_w\ = \ $ { 500 3% } &nbsp; $\text{ms}$
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</quiz>
</quiz>


===Musterlösung===
===Solution===
{{ML-Kopf}}
{{ML-Kopf}}
'''(1)'''&nbsp;  Für das Rechtecksignal gilt&nbsp; $T_x = 1 \,\text{ms}$  &nbsp; &rArr; &nbsp; $f_x \hspace{0.15cm}\underline{= 1 \, \text{kHz}}$.
'''(1)'''&nbsp;  Für das Rechtecksignal gilt&nbsp; $T_x = 1 \,\text{ms}$  &nbsp; &rArr; &nbsp; $f_x \hspace{0.15cm}\underline{= 1 \, \text{kHz}}$.

Revision as of 01:44, 27 December 2020

Rechtecksignal, Dreiecksignal und Summensignal

The adjacent diagram shows the two periodic signals  ${x(t)}$  and  ${y(t)}$ , from which the sum signal  ${s(t)}$  – sketched in the lower picture – and the difference signal  ${d(t)}$  are formed.

Furthermore, in this task we consider the signa  ${w(t)}$, which results from the sum of the two periodic signals  ${u(t)}$  und  $v(t)$ . The base frequencies of the signals are

  • $f_u = 998 \,\text{Hz},$
  • $f_v = 1002 \,\text{Hz}.$

That is all we know about the signals  ${u(t)}$  and  $v(t)$ .




Hints:

  • The exercise belongs to the chapter  [[Signal_Representation/General_Description|General description of periodic signals].
  • With the interactive applet  Periodendauer periodischer Signale  the resulting period duration of two harmonic oscillations can be determined.


Questions

1 What is the period duration  $T_x$  and the base frequency  $f_x$  of the signal  ${x(t)}$?

$f_x\ = \ $   $\text{kHz}$

2 What is the period duration  $T_y$  and the base frequency  $f_y$  of the signal  ${y(t)}$?

$f_y\ = \ $   $\text{kHz}$

3 Determine the base frequency  $f_s$  as well as the period duration  $T_s$  of the sum signal  ${s(t)}$  and verify your results with the help of the sketched signal.

$T_s\ = \ $   $\text{ms}$

4 What is the period duration  $T_d$  of the difference signal  ${d(t)}$ ?

$T_d\ = \ $   $\text{ms}$

5 What is the period duration  $T_w$  of the signal  ${w(t)} = {u(t)} + v(t)$?

$T_w\ = \ $   $\text{ms}$


Solution

(1)  Für das Rechtecksignal gilt  $T_x = 1 \,\text{ms}$   ⇒   $f_x \hspace{0.15cm}\underline{= 1 \, \text{kHz}}$.


(2)  Für das Dreiecksignal gilt  $T_y = 2.5 \,\text{ms}$  und  $f_y \hspace{0.15cm}\underline{= 0.4\, \text{kHz}}$.


(3)  Die Grundfrequenz  $f_s$  des Summensignals  $s(t)$  ist der größte gemeinsame Teiler von  $f_x = 1 \,\text{kHz}$  und  $f_y = 0.4 \,\text{kHz}$.

  • Daraus folgt  $f_s = 200 \,\text{Hz}$  und die Periodendauer  $T_s\hspace{0.15cm}\underline{ = 5 \,\text{ms}}$, wie auch aus der grafischen Darstellung des Signals  ${s(t)}$  auf der Angabenseite hervorgeht.


Differenzsignal $d(t) = x(t) - y(t)$

(4)  Die Periodendauer  $T_d$  ändert sich gegenüber der Periodendauer  $T_s$  nicht, wenn das Signal  ${y(t)}$  nicht addiert, sondern subtrahiert wird:     $T_d = T_s \hspace{0.15cm}\underline{= 5\, \text{ms}}$.


(5)  Der größte gemeinsame Teiler von  $f_u = 998 \,\text{Hz}$  und  $f_{v} = 1002 \,\text{Hz}$  ist  $f_w = 2 \,\text{Hz}$.

  • Der Kehrwert hiervon ergibt die Periodendauer $T_w \hspace{0.15cm}\underline{= 500 \,\text{ms}}$.