Difference between revisions of "Aufgaben:Exercise 1.2: Signal Classification"

From LNTwww
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- <math>x_3(t)</math>.
 
- <math>x_3(t)</math>.
  
{Berechnen Sie die auf den Einheitswiderstand&nbsp; $R = 1\ Ω$&nbsp; bezogene Energie&nbsp; <math>E_2</math>&nbsp; des Signals&nbsp; <math>x_2(t)</math>. <br>Wie groß ist die Leistung&nbsp; <math>P_2</math>&nbsp; dieses Signals?
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{Calculate the energy&nbsp; $R = 1\ Ω$&nbsp; related to the unit resistance&nbsp; <math>E_2</math>&nbsp; of the signal&nbsp; <math>x_2(t)</math>. <br>What is the power&nbsp; <math>P_2</math>&nbsp; of this signal?
 
|type="{}"}
 
|type="{}"}
 
<math>E_2 \ = \ </math>{ 0.5 5% }  $\ \cdot 10^{-3}\,\text{V}^2\text{s}$
 
<math>E_2 \ = \ </math>{ 0.5 5% }  $\ \cdot 10^{-3}\,\text{V}^2\text{s}$
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'''(1)'''&nbsp;  The <u>solutions 1 and 3</u> are applicable:
 
'''(1)'''&nbsp;  The <u>solutions 1 and 3</u> are applicable:
*Alle Signale können in analytischer Form vollständig beschrieben werden; sie sind deshalb auch deterministisch.  
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*All signals can be described completely in analytical form; therefore they are also deterministic.  
*Alle Signale sind außerdem für alle Zeiten&nbsp; $t$&nbsp; eindeutig definiert, nicht nur zu gewissen Zeitpunkten. Deshalb handelt es sich stets um zeitkontinuierliche Signale.
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*All signals are also clearly defined for all times&nbsp; $t$&nbsp; not only at certain times. Therefore, they are always time-continuous signals.
*Die Signalamplituden von&nbsp; <math>x_2(t)</math>&nbsp; und&nbsp; <math>x_3(t)</math>&nbsp; können alle beliebigen Werte zwischen&nbsp; $0$&nbsp; und&nbsp; $1\,\text{V}$&nbsp; annehmen; sie sind deshalb wertkontinuierlich.  
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*The signal amplitudes of&nbsp; <math>x_2(t)</math>&nbsp; and&nbsp; <math>x_3(t)</math>&nbsp; can take any values between&nbsp; $0$&nbsp; and&nbsp; $1\,\text{V}$&nbsp; they are therefore continuous in value.  
*Dagegen sind beim Signal&nbsp; <math>x_1(t)</math>&nbsp; nur die zwei Signalwerte&nbsp; $0$&nbsp; und&nbsp; $1\,\text{V}$&nbsp; möglich; es liegt ein wertdiskretes Signal vor.  
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*On the other hand, with the signal&nbsp; <math>x_1(t)</math>&nbsp; only the two signal values&nbsp; $0$&nbsp; and&nbsp; $1\,\text{V}$&nbsp; are possible; a discrete-valued signal is present.  
  
  
  
'''(2)'''&nbsp;  Richtig sind die <u>Lösungsvorschläge 1 und 2</u>:
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'''(2)'''&nbsp;  Correct are the <u>solutions 1 and 2</u>:
*Ein Signal bezeichnet man als kausal, wenn es für Zeiten&nbsp; $t < 0$&nbsp; nicht existiert bzw. identisch Null ist. Dies gilt für die Signale&nbsp; <math>x_1(t)</math>&nbsp; und&nbsp; <math>x_2(t)</math>.  
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*A signal is called causal if for times&nbsp; $t < 0$&nbsp; it does not exist or is identically zero. This applies to the signals&nbsp; <math>x_1(t)</math>&nbsp; and&nbsp; <math>x_2(t)</math>.  
*Dagegen gehört&nbsp; <math>x_3(t)</math>&nbsp; zur Klasse der akausalen Signale.
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*In contrast,&nbsp; <math>x_3(t)</math>&nbsp; belongs to the class of non-causal signals.
  
  
  
'''(3)'''&nbsp;  Nach der allgemeinen Definition gilt:
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'''(3)'''&nbsp;  According to the general definition:
  
 
::<math>E_2=\lim_{T_{\rm M}\to\infty}\int^{T_{\rm M}/2}_{-T_{\rm M}/2}x^2_2(t)\,\hspace{0.1cm}{\rm d}t.</math>
 
::<math>E_2=\lim_{T_{\rm M}\to\infty}\int^{T_{\rm M}/2}_{-T_{\rm M}/2}x^2_2(t)\,\hspace{0.1cm}{\rm d}t.</math>
  
Im vorliegenden Fall ist die untere Integrationsgrenze Null und die obere Integrationsgrenze&nbsp; $+\infty$. Man erhält:
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In this case, the lower integration limit is zero and the upper integration limit&nbsp; $+\infty$. You get:
 
   
 
   
 
::<math>E_2=\int^\infty_0 (1{\rm V})^2\cdot{\rm e}^{-2t/(1\rm ms)}\,\hspace{0.1cm}{\rm d}t = 5 \cdot 10^{-4}\hspace{0.1cm} \rm V^2s  \hspace{0.15cm}\underline{= 0.5 \cdot 10^{-3}\hspace{0.1cm} \rm V^2s}.  </math>  
 
::<math>E_2=\int^\infty_0 (1{\rm V})^2\cdot{\rm e}^{-2t/(1\rm ms)}\,\hspace{0.1cm}{\rm d}t = 5 \cdot 10^{-4}\hspace{0.1cm} \rm V^2s  \hspace{0.15cm}\underline{= 0.5 \cdot 10^{-3}\hspace{0.1cm} \rm V^2s}.  </math>  
  
Bei endlicher Energie ist die zugehörige Leistung stets verschwindend klein. Daraus folgt&nbsp; $P_2\hspace{0.15cm}\underline{ = 0}$.
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With finite energy, the associated power is always negligible. From this follows&nbsp; $P_2\hspace{0.15cm}\underline{ = 0}$.
  
  
  
'''(4)'''&nbsp;  Richtig sind die <u>Lösungsvorschläge 2 und 3</u>:
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'''(4)'''&nbsp;  Correct are the <u>solutions 2 and 3</u>:
*Wie bereits in der letzten Teilaufgabe berechnet wurde, besitzt&nbsp; <math>x_2(t)</math>&nbsp; eine endliche Energie:&nbsp;  
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*As already calculated in the last subtask,&nbsp; <math>x_2(t)</math>&nbsp; has a finite energy:&nbsp;  
:$$E_2= 0.5 \cdot 10^{-3}\hspace{0.1cm} {\rm V^2s}. $$  
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:$$E_2= 0.5 \cdot 10^{-3}\hspace{0.1cm} {\rm V^2s}. $$  
  
*Die Energie des Signals&nbsp; <math>x_3(t)</math>&nbsp; ist doppelt so groß, da nun der Zeitbereich&nbsp; $t < 0$&nbsp; den gleichen Beitrag liefert wie der Zeitbereich&nbsp; $t > 0$. Also ist
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*The energy of the signal&nbsp; <math>x_3(t)</math>&nbsp; is twice as large, since now the time domain&nbsp; $t < 0$&nbsp; makes the same contribution as the time domain&nbsp; $t > 0$. So
 
:$$E_3= 10^{-3}\hspace{0.1cm} {\rm V^2s}.$$  
 
:$$E_3= 10^{-3}\hspace{0.1cm} {\rm V^2s}.$$  
  
*Beim Signal&nbsp; <math>x_1(t)</math>&nbsp; divergiert das Energieintegral:&nbsp; $E_1 \rightarrow \infty$. Dieses Signal weist eine endliche Leistung auf &nbsp; &rArr; &nbsp; $P_1= 0.5 \hspace{0.1cm} {\rm V}^2$.
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*At signal&nbsp; <math>x_1(t)</math>&nbsp; the energy integral diverges:&nbsp; $E_1 \rightarrow \infty$. This signal has a finite power &nbsp; &rArr; &nbsp; $P_1= 0.5 \hspace{0.1cm} {\rm V}^2$.
*Das Ergebnis berücksichtigt auch, dass das Signal&nbsp; <math>x_1(t)</math>&nbsp; in der Hälfte der Zeit&nbsp; $(t < 0)$&nbsp; identisch Null ist.
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*The result also takes into account that the signal&nbsp; <math>x_1(t)</math>&nbsp; in half the time&nbsp; $(t < 0)$&nbsp; is identical to zero.
* Das Signal&nbsp; <math>x_1(t)</math>&nbsp; ist dementsprechend&nbsp; <u>leistungsbegrenzt</u>.  
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* The signal&nbsp; <math>x_1(t)</math>&nbsp; is accordingly&nbsp; <u>power limited</u>.  
  
 
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Revision as of 19:04, 13 August 2020

predetermined characteristics

Three signal curves are shown on the Right:

  • The blue signal  \(x_1(t)\)  is switched on at time  $t = 0$  and has at   $t > 0$  the value  $1\,\text{V}$.
  • The blue signal  \(x_2(t)\)  is for  $t < 0$  equals zero, jumps at  $t = 0$  to  $1\,\text{V}$  and then falls down with the time constant  $1\,\text{ms}$ . For  $t > 0$  the following applies:
\[x_2(t) = 1\,\text{V} \cdot {\rm e}^{- {t}/(1\,\text{ms})}.\]
  • Correspondingly, the signal shown in green applies to all times  $t$:
\[x_3(t) = 1\,\text{V} \cdot {\rm e}^{- {|\hspace{0.05cm}t\hspace{0.05cm}|}/(1\,\text{ms})}.\]

You will now classify these three signals according to the following criteria:

  • deterministic or stochastic,
  • causal or acausal,
  • energy limited or power limited,
  • value-continuous or value-discrete,
  • time-continuous or time-discrete.



Notes:


Questions

1

Which of the following statements are true?

All signals considered here are deterministic.
All signals considered here are of stochastic nature.
The signals are always continuous in time.
They are always signals of continuous value.

2

Which signals are causal according to the definition in the theory part?

\(x_1(t)\),
\(x_2(t)\),
\(x_3(t)\).

3

Calculate the energy  $R = 1\ Ω$  related to the unit resistance  \(E_2\)  of the signal  \(x_2(t)\).
What is the power  \(P_2\)  of this signal?

\(E_2 \ = \ \)

$\ \cdot 10^{-3}\,\text{V}^2\text{s}$
\(P_2 \ = \ \)

$\ \cdot \text{Vs}$

4

Which of the signals have a finite energy?

\(x_1(t)\),
\(x_2(t)\),
\(x_3(t)\).


Solutions

(1)  The solutions 1 and 3 are applicable:

  • All signals can be described completely in analytical form; therefore they are also deterministic.
  • All signals are also clearly defined for all times  $t$  not only at certain times. Therefore, they are always time-continuous signals.
  • The signal amplitudes of  \(x_2(t)\)  and  \(x_3(t)\)  can take any values between  $0$  and  $1\,\text{V}$  they are therefore continuous in value.
  • On the other hand, with the signal  \(x_1(t)\)  only the two signal values  $0$  and  $1\,\text{V}$  are possible; a discrete-valued signal is present.


(2)  Correct are the solutions 1 and 2:

  • A signal is called causal if for times  $t < 0$  it does not exist or is identically zero. This applies to the signals  \(x_1(t)\)  and  \(x_2(t)\).
  • In contrast,  \(x_3(t)\)  belongs to the class of non-causal signals.


(3)  According to the general definition:

\[E_2=\lim_{T_{\rm M}\to\infty}\int^{T_{\rm M}/2}_{-T_{\rm M}/2}x^2_2(t)\,\hspace{0.1cm}{\rm d}t.\]

In this case, the lower integration limit is zero and the upper integration limit  $+\infty$. You get:

\[E_2=\int^\infty_0 (1{\rm V})^2\cdot{\rm e}^{-2t/(1\rm ms)}\,\hspace{0.1cm}{\rm d}t = 5 \cdot 10^{-4}\hspace{0.1cm} \rm V^2s \hspace{0.15cm}\underline{= 0.5 \cdot 10^{-3}\hspace{0.1cm} \rm V^2s}. \]

With finite energy, the associated power is always negligible. From this follows  $P_2\hspace{0.15cm}\underline{ = 0}$.


(4)  Correct are the solutions 2 and 3:

  • As already calculated in the last subtask,  \(x_2(t)\)  has a finite energy: 
$$E_2= 0.5 \cdot 10^{-3}\hspace{0.1cm} {\rm V^2s}. $$
  • The energy of the signal  \(x_3(t)\)  is twice as large, since now the time domain  $t < 0$  makes the same contribution as the time domain  $t > 0$. So
$$E_3= 10^{-3}\hspace{0.1cm} {\rm V^2s}.$$
  • At signal  \(x_1(t)\)  the energy integral diverges:  $E_1 \rightarrow \infty$. This signal has a finite power   ⇒   $P_1= 0.5 \hspace{0.1cm} {\rm V}^2$.
  • The result also takes into account that the signal  \(x_1(t)\)  in half the time  $(t < 0)$  is identical to zero.
  • The signal  \(x_1(t)\)  is accordingly  power limited.