Aufgaben:Exercise 2.1: Rectification: Difference between revisions

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{{quiz-Header|Buchseite=Signal Representation/General Description
{{quiz-Header|Buchseite=Signal Representation/General Description
}}
}}
[[File:P_ID239__Sig_A_2_1.png|250px|right|frame|Periodisches Dreiecksignal]]
[[File:P_ID239__Sig_A_2_1.png|250px|right|frame|Periodic triangular signal]]
The graph shows the periodic signal  $x(t)$. If  $x(t)$ is applied to the input of a non-linearity with the characteristic curve
The graph shows the periodic signal  $x(t)$.  If  $x(t)$ is applied to the input of a non-linearity with the characteristic curve


:$$y=g(x)=\left\{ {x \; \rm for\; \it x \geq \rm 0, \atop {\rm 0 \;\;\; \rm else,}}\right.$$
:$$y=g(x)=\left\{ {x \; \rm for\; \it x \geq \rm 0, \atop {\rm 0 \;\;\; \rm else,}}\right.$$


the signall  $y(t)$ is obtained at the output. A second non-linear characteristic
the signal  $y(t)$ is obtained at the output.  A second non-linear characteristic
   
   
:$$z=h(x)=|x|$$
:$$z=h(x)=|x|$$
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''Hint:''  
''Note:''  
*This exercise belongs to the chapter  [[Signal_Representation/General_Description|General description of periodic signals]].
*This exercise belongs to the chapter  [[Signal_Representation/General_Description|General description of periodic signals]].
   
   
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{What is the period  $T_0$  of the signal  $y(t)$?
{What is the period duration  $T_0$  of the signal  $y(t)$?
|type="{}"}
|type="{}"}
$T_0 \ = \ $  { 2 3% }   $\text{ms}$
$T_0 \ = \ $  { 2 3% }   $\text{ms}$




{What is the base angular frequency  $\omega_0$  of the signal  $z(t)$?
{What is the basic circular frequency  $\omega_0$  of the signal  $z(t)$?
|type="{}"}
|type="{}"}
$\omega_0 \ = \ $  { 6283 3% }   $\text{1/s}$
$\omega_0 \ = \ $  { 6283 3% }   $\text{1/s}$
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===Solution===
===Solution===
{{ML-Kopf}}
{{ML-Kopf}}
'''(1)'''&nbsp;  Richtig sind die <u>Lösungsvorschläge 1 und 4</u>:
'''(1)'''&nbsp;  Correct are the <u>solutions 1 and 4</u>:
*Die nichtlineare Kennlinie&nbsp; $y = g(x)$&nbsp; beschreibt einen Einweggleichrichter.  
*The non-linear characteristic&nbsp; $y = g(x)$&nbsp; describes a half-wave rectifier.  
*$z = h(x) = |x|$&nbsp; beschreibt einen Zweiweggleichrichter.
*$z = h(x) = |x|$&nbsp; describes a full-wave rectifier.
    
    


'''(2)'''&nbsp;  Die Periodendauer des gegebenen Signals&nbsp; $x(t)$&nbsp; beträgt&nbsp; $T_0 = 2\,\text{ms}$. Der Kehrwert hiervon ergibt die Grundfrequenz&nbsp; $f_0  \hspace{0.1cm}\underline{ = 500\,\text{Hz}}$.
'''(2)'''&nbsp;  The period duration&nbsp; $x(t)$&nbsp; is&nbsp; $T_0 = 2\,\text{ms}$. The inverse magnitudes to the base frequency&nbsp; $f_0  \hspace{0.1cm}\underline{ = 500\,\text{Hz}}$.




'''(3)'''&nbsp;  Die Einweggleichrichtung ändert nichts an der Periodendauer, siehe  linke Skizze. Somit gilt weiterhin&nbsp; $T_0 \hspace{0.1cm}\underline{= 2\,\text{ms}}$.
'''(3)'''&nbsp;  The half-wave rectification does not change the duration of the period, see the left graph:&nbsp; $T_0 \hspace{0.1cm}\underline{= 2\,\text{ms}}$.


[[File:P_ID262__Sig_A_2_1_a.png|center|frame|Periodische Dreiecksignale]]
[[File:P_ID262__Sig_A_2_1_a.png|center|frame|Periodic triangular signals]]


'''(4)'''&nbsp;  Das Signal&nbsp; $z(t)$&nbsp; nach der Doppelweggleichrichtung hat dagegen die doppelte Frequenz (siehe rechte Darstellung). Hier gelten folgende Werte:
'''(4)'''&nbsp;  After full-wave rectification, the signal&nbsp; $z(t)$&nbsp; has double the frequency (see right graph). The following values apply here:
:$$T_0 = 1\,\text{ms}, \hspace{0.5cm}  f_0 = 1\,\text{kHz}, \hspace{0.5cm}  \omega_0 \hspace{0.1cm}\underline{= 6283\,\text{1/s}}.$$
:$$T_0 = 1\,\text{ms}, \hspace{0.5cm}  f_0 = 1\,\text{kHz}, \hspace{0.5cm}  \omega_0 \hspace{0.1cm}\underline{= 6283\,\text{1/s}}.$$
{{ML-Fuß}}
{{ML-Fuß}}


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[[Category:Exercises for Signal Representation|^2.1 General Description about Periodic Signals^]]
[[Category:Signal Representation: Exercises|^2.1 Description of Periodic Signals^]]
[[de:Aufgaben:Aufgabe 2.1: Gleichrichtung]]

Latest revision as of 17:53, 16 March 2026

Periodic triangular signal

The graph shows the periodic signal  $x(t)$.  If  $x(t)$ is applied to the input of a non-linearity with the characteristic curve

$$y=g(x)=\left\{ {x \; \rm for\; \it x \geq \rm 0, \atop {\rm 0 \;\;\; \rm else,}}\right.$$

the signal  $y(t)$ is obtained at the output.  A second non-linear characteristic

$$z=h(x)=|x|$$

delivers the signal  $z(t)$.




Note:



Questions

1 Which of the following statements are true?

$y = g(x)$  describes a half-wave rectifier.
$y = g(x)$  describes a full-wave rectifier.
$z = h(x)$  describes a half-wave rectifier.
$z = h(x)$  describes a full-wave rectifier.

2 What is the base frequency $f_0$  of the signal  $x(t)$?

$f_0 \ = \ $   $\text{Hz}$

3 What is the period duration  $T_0$  of the signal  $y(t)$?

$T_0 \ = \ $   $\text{ms}$

4 What is the basic circular frequency  $\omega_0$  of the signal  $z(t)$?

$\omega_0 \ = \ $   $\text{1/s}$


Solution

Solution

(1)  Correct are the solutions 1 and 4:

  • The non-linear characteristic  $y = g(x)$  describes a half-wave rectifier.
  • $z = h(x) = |x|$  describes a full-wave rectifier.


(2)  The period duration  $x(t)$  is  $T_0 = 2\,\text{ms}$. The inverse magnitudes to the base frequency  $f_0 \hspace{0.1cm}\underline{ = 500\,\text{Hz}}$.


(3)  The half-wave rectification does not change the duration of the period, see the left graph:  $T_0 \hspace{0.1cm}\underline{= 2\,\text{ms}}$.

Periodic triangular signals

(4)  After full-wave rectification, the signal  $z(t)$  has double the frequency (see right graph). The following values apply here:

$$T_0 = 1\,\text{ms}, \hspace{0.5cm} f_0 = 1\,\text{kHz}, \hspace{0.5cm} \omega_0 \hspace{0.1cm}\underline{= 6283\,\text{1/s}}.$$

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