Difference between revisions of "Aufgaben:Exercise 4.09Z: Periodic ACF"

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{{quiz-Header|Buchseite=Stochastische Signaltheorie/Autokorrelationsfunktion (AKF)
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{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Auto-Correlation_Function
 
}}
 
}}
  
[[File:P_ID380__Sto_Z_4_9.png|right|frame|Periodisches mehrstufiges Rechtecksignal]]
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[[File:P_ID380__Sto_Z_4_9.png|right|frame|Periodic multilevel rectangular signal]]
Wir betrachten in dieser Aufgabe einen periodischen und gleichzeitig ergodischen stochastischen Prozess  $\{x_i(t)\}$, der durch die dargestellte Musterfunktion  $x(t)$  vollständig charakterisiert ist.
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We consider in this exercise a periodic and simultaneously ergodic stochastic process  $\{x_i(t)\}$,  which is fully characterized by the presented pattern function  $x(t)$.
  
Weitere Mustersignale des Zufallsprozesses  $\{x_i(t)\}$  erhält man durch Verschiebung um unterschiedlich große Verzögerungen  $\tau_i$, wobei  $\tau_i$  als gleichverteilt zwischen  $0$  und der Periodendauer  $T_0$  angenommen wird.
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Further pattern signals of the random process  $\{x_i(t)\}$  are obtained by shifting by  different delays  $\tau_i$,  where  $\tau_i$  is assumed to be uniformly distributed between  $0$  and the period  $T_0$.
  
  
 
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'''Hint:'''  The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Auto-Correlation_Function|Auto-Correlation Function]].
 
 
 
 
 
 
 
 
''Hinweis:''  
 
*Die Aufgabe gehört zum  Kapitel  [[Theory_of_Stochastic_Signals/Autokorrelationsfunktion_(AKF)|Autokorrelationsfunktion]].
 
 
   
 
   
  
  
===Fragebogen===
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===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Ermitteln Sie die Periodendauer&nbsp; $T_0$, normiert auf die in der Skizze definierte Zeitdauer&nbsp; $T$.
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{Determine the period duration&nbsp; $T_0$&nbsp; normalized to the period duration&nbsp; $T$&nbsp; defined in the sketch.
 
|type="{}"}
 
|type="{}"}
$T_0/T \ = \ $ { 5 3% }
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$T_0/T \ = \ $ { 5 3% }
  
  
{Wie gro&szlig; ist der Gleichsignalanteil (lineare Mittelwert)&nbsp; $m_x$&nbsp; des beschriebenen Prozesses&nbsp; $\{x_i(t)\}$?
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{What is the size of the DC signal component &nbsp; &rArr; &nbsp; linear mean&nbsp; $m_x$&nbsp; of the described process&nbsp; $\{x_i(t)\}$?
 
|type="{}"}
 
|type="{}"}
 
$m_x \ = \ $ { 0.4 3% } $\ \rm V$
 
$m_x \ = \ $ { 0.4 3% } $\ \rm V$
  
  
{Wie gro&szlig; ist die (auf den Widerstand&nbsp; $1 \hspace{0.05cm} \rm \Omega$&nbsp; bezogene) Prozessleistung?
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{What is the process power&nbsp; (related to the resistor&nbsp; $1 \hspace{0.05cm} \rm \Omega$&nbsp;)?
 
|type="{}"}
 
|type="{}"}
 
$P_x \ = \ $ { 2 3% } $\ \rm V^2$
 
$P_x \ = \ $ { 2 3% } $\ \rm V^2$
  
  
{Berechnen Sie die AKF-Werte f&uuml;r&nbsp; $\tau = T$&nbsp; und&nbsp; $\tau = 2T$.
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{Calculate the ACF values for&nbsp; $\tau = T$&nbsp; and&nbsp; $\tau = 2T$.
 
|type="{}"}
 
|type="{}"}
 
$\varphi_x(\tau = T) \ = \ $ { 0.6 3% } $\ \rm V^2$
 
$\varphi_x(\tau = T) \ = \ $ { 0.6 3% } $\ \rm V^2$
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{Skizzieren Sie den AKF-Verlauf unter Ber&uuml;cksichtigung von Symmetrieen.&nbsp; Welche Werte ergeben sich f&uuml;r&nbsp; $\tau = 3T$&nbsp; und&nbsp; $\tau = 4T$?
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{Sketch the ACF curve taking into account symmetries.&nbsp; What values result for&nbsp; $\tau = 3T$&nbsp; and&nbsp; $\tau = 4T$?
 
|type="{}"}
 
|type="{}"}
 
$\varphi_x(\tau = 3T) \ = \ $ { -1.236--1.164 } $\ \rm V^2$
 
$\varphi_x(\tau = 3T) \ = \ $ { -1.236--1.164 } $\ \rm V^2$
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{Berechnen Sie den Erwartungswert der AKF bez&uuml;glich aller&nbsp; $\tau$-Werte.&nbsp; Interpretieren Sie das Ergebnis.
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{Calculate the expected value of the ACF with respect to all&nbsp; $\tau$ values.&nbsp; Interpret the result.
 
|type="{}"}
 
|type="{}"}
 
${\rm E}\big[\varphi_x(\tau)\big]\ = \ $ { 0.16 3% } $\ \rm V^2$
 
${\rm E}\big[\varphi_x(\tau)\big]\ = \ $ { 0.16 3% } $\ \rm V^2$
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</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
[[File:P_ID382__Sto_Z_4_9_d.png|right|frame|Zur AKF&ndash;Berechnung]]
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[[File:P_ID382__Sto_Z_4_9_d.png|right|frame|For the ACF calculation]]
'''(1)'''&nbsp; Die (normierte) Periodendauer betr&auml;gt&nbsp; $T_0/T \hspace{0.15cm}\underline{= 5}.$
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'''(1)'''&nbsp; The&nbsp; (normalized)&nbsp; period duration is&nbsp; $T_0/T \hspace{0.15cm}\underline{= 5}.$
  
  
'''(2)'''&nbsp; Aufgrund der Periodizit&auml;t gen&uuml;gt die Mittelung &uuml;ber eine Periodendauer&nbsp; $T_0$:
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'''(2)'''&nbsp; Due to periodicity,&nbsp; the averaging over a periodic time&nbsp; $T_0$:
:$$m_x = \frac{1}{T_0} \cdot \int_0^{T_0} x(t) \hspace{0.1cm}{\rm d} t = \frac{1}{5 T} \cdot (2\hspace{0.05cm}{\rm V} \cdot 2 T - 2\hspace{0.05cm}{\rm V} \cdot 2 T) \hspace{0.15cm}\underline{= \rm 0.4 \,V}.$$
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:$$m_x = \frac{1}{T_0} \cdot \int_0^{T_0} x(t) \hspace{0.1cm}{\rm d} t = \frac{1}{5 T} \cdot (2\hspace{0.05cm}{\rm V} \cdot 2 T - 2\hspace{0.05cm}{\rm V} \cdot 2 T) \hspace{0.15cm}\underline{= \rm 0.4 \,V}.$$
  
  
'''(3)'''&nbsp; In analoger Weise zur letzten Teilaufgabe  erh&auml;lt man f&uuml;r die mittlere Leistung:
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'''(3)'''&nbsp; In analogy to the last subtask,&nbsp; we obtain for the mean power:
:$$P_x = \frac{2 T}{5 T} \cdot \big[(\rm 2V)^2 +(- \rm 1V)^2 \big]\hspace{0.15cm}\underline{ = \rm 2 \,V^2}.$$
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:$$P_x = \frac{2 T}{5 T} \cdot \big[(\rm 2V)^2 +(- \rm 1V)^2 \big]\hspace{0.15cm}\underline{ = \rm 2 \,V^2}.$$
  
  
'''(4)'''&nbsp; Die nebenstehende Grafik zeigt jeweils im Bereich von&nbsp; $0$&nbsp; bis&nbsp; $T_0 = 5T$  
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'''(4)'''&nbsp; The accompanying graph shows in each case in the range from&nbsp; $0$&nbsp; to&nbsp; $T_0 = 5T$.
*oben das Produkt&nbsp; $x(t) \cdot x(t+T)$,
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:*above the product&nbsp; $x(t) \cdot x(t+T)$,
*unten das Produkt&nbsp; $x(t) \cdot x(t+2T)$.   
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:*down the product&nbsp; $x(t) \cdot x(t+2T)$.   
  
  
Zu beachten ist, dass&nbsp; $x(t+T)$&nbsp; eine Verschiebung des Signals&nbsp; $x(t)$&nbsp; um&nbsp; $T$&nbsp; nach links bedeutet.  
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*Note that&nbsp; $x(t+T)$&nbsp; means a shift of the signal&nbsp; $x(t)$&nbsp; by&nbsp; $T$&nbsp; to the left.  
  
Aus diesen Skizzen folgen die Beziehungen:  
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*From these sketches follow the relations:  
:$$\varphi_x (T)= \rm {1}/{5 } \cdot (\rm 4V^2 + \rm 1V^2 - \rm 2V^2) \hspace{0.15cm}\underline{= \rm 0.6\, V^2},$$
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:$$\varphi_x (T)= \rm {1}/{5 } \cdot (\rm 4V^2 + \rm 1V^2 - \rm 2V^2) \hspace{0.15cm}\underline{= \rm 0.6\, V^2},$$
:$$\varphi_x ( 2 T)= \rm {1}/{5 } \cdot(-\rm 2V^2 \cdot 3) \hspace{0.15cm}\underline{= - \rm 1.2 \,V^2}.$$
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:$$\varphi_x ( 2 T)= \rm {1}/{5 } \cdot(-\rm 2V^2 \cdot 3) \hspace{0.15cm}\underline{= - \rm 1.2 \,V^2}.$$
  
  
'''(5)'''&nbsp; Eine Autokorrelationsfunktion ist stets gerade: &nbsp; $\varphi_x (-\tau)= \varphi_x (\tau)$.&nbsp;  
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[[File:P_ID383__Sto_Z_4_9_e.png|right|frame|Wanted auto-correlation function]]
*Bei periodischen Prozessen ist die AKF zudem ebenfalls periodisch und zwar mit der gleichen Periodendauer&nbsp; $T_0$&nbsp; wie die einzelnen Musterfunktionen. Daraus folgt:
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'''(5)'''&nbsp; An auto-correlation function is always even: &nbsp;  
[[File:P_ID383__Sto_Z_4_9_e.png|right|frame|Gesuchte Autokorrelationsfunktion]]
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:$$\varphi_x (-\tau)= \varphi_x (\tau).$$  
 +
*In addition,&nbsp; for periodic processes,&nbsp; the ACF is also periodic with the same period duration&nbsp; $T_0$&nbsp; as the individual pattern functions.&nbsp; It follows that:
  
:$$\varphi_x ( 0) = \varphi_x (5 T) = \varphi_x (10 T) = \ \text{...} \ = \it P_x = \rm 2 \,V^2,$$
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:$$\varphi_x ( 0) = \varphi_x (5 T) = \varphi_x (10 T) = \ \text{...} \ = \it P_x = \rm 2 \,V^2,$$
:$$\varphi_x (3 T) = \varphi_x (-3 T) =\varphi_x (2 T) = \ \text{...} \ \hspace{0.15cm}\underline{= - \rm 1.2 \,V^2},$$
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:$$\varphi_x (3 T) = \varphi_x (-3 T) =\varphi_x (2 T) = \ \text{...} \ \hspace{0.15cm}\underline{= - \rm 1.2 \,V^2},$$
:$$\varphi_x (4 T) = \varphi_x (-4 T) =\varphi_x ( T) = \ \text{...} \ \hspace{0.15cm}\underline{= \rm 0.6 \,V^2}.$$
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:$$\varphi_x (4 T) = \varphi_x (-4 T) =\varphi_x ( T) = \ \text{...} \ \hspace{0.15cm}\underline{= \rm 0.6 \,V^2}.$$
  
*Die berechneten AKF-Werte k&ouml;nnen durch Geradenabschnitte miteinander verbunden werden, da die Integration &uuml;ber Rechteckfunktionen stets lineare Teilabschnitte ergibt.
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*The calculated ACF values can be connected by straight line sections,&nbsp; since integration over rectangular functions always yields linear subsections.
  
  
  
'''(6)'''&nbsp; Die fünf Intervalle&nbsp; $(0$ bis $T)$,&nbsp; $(T$ bis $2T)$, ... ,&nbsp; $(4T$ bis $5T)$&nbsp; liefern die Beiträge
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'''(6)'''&nbsp; The five intervals&nbsp; $(0$ to $T)$,&nbsp; $(T$ to $2T)$, ... ,&nbsp; $(4T$ to $5T)$&nbsp; provide the contributions.
:$$(+1.3;   -0.3;   -1.2;   -0.3;   +1.3) \cdot \rm V^2.$$  
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:$$(+1.3; -0.3; -1.2; -0.3; +1.3) \cdot \rm V^2.$$  
*Daraus ergibt sich der Erwartungswert (lineare Mittelwert):
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*This gives the expected value&nbsp; (linear mean):
:$${\rm E}\big[\varphi_x(\tau)\big] = 1/5 \cdot (1.3-0.3 -1.2 -0.3 +1.3]\hspace{0.15cm}\underline{= \rm 0.16 \,V^2}.$$
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:$${\rm E}\big[\varphi_x(\tau)\big] = 1/5 \cdot (1.3-0.3 -1.2 -0.3 +1.3]\hspace{0.15cm}\underline{= \rm 0.16 \,V^2}.$$
  
*Dies entspricht dem Quadrat des Mittelwertes&nbsp; $m_x$ &nbsp; &rArr; &nbsp; siehe Teilaufgabe&nbsp; '''(2)'''.
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*This corresponds to the square of the mean&nbsp; $m_x$ &nbsp; &rArr; &nbsp; see subtask&nbsp; '''(2)'''.
  
 
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[[Category:Aufgaben zu Stochastische Signaltheorie|^4.4 Autokorrelationsfunktion (AKF)^]]
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[[Category:Theory of Stochastic Signals: Exercises|^4.4 Auto-Correlation Function^]]

Latest revision as of 17:33, 19 March 2022

Periodic multilevel rectangular signal

We consider in this exercise a periodic and simultaneously ergodic stochastic process  $\{x_i(t)\}$,  which is fully characterized by the presented pattern function  $x(t)$.

Further pattern signals of the random process  $\{x_i(t)\}$  are obtained by shifting by different delays  $\tau_i$,  where  $\tau_i$  is assumed to be uniformly distributed between  $0$  and the period  $T_0$.


Hint:  The exercise belongs to the chapter  Auto-Correlation Function.


Questions

1

Determine the period duration  $T_0$  normalized to the period duration  $T$  defined in the sketch.

$T_0/T \ = \ $

2

What is the size of the DC signal component   ⇒   linear mean  $m_x$  of the described process  $\{x_i(t)\}$?

$m_x \ = \ $

$\ \rm V$

3

What is the process power  (related to the resistor  $1 \hspace{0.05cm} \rm \Omega$ )?

$P_x \ = \ $

$\ \rm V^2$

4

Calculate the ACF values for  $\tau = T$  and  $\tau = 2T$.

$\varphi_x(\tau = T) \ = \ $

$\ \rm V^2$
$\varphi_x(\tau = 2T) \ = \ $

$\ \rm V^2$

5

Sketch the ACF curve taking into account symmetries.  What values result for  $\tau = 3T$  and  $\tau = 4T$?

$\varphi_x(\tau = 3T) \ = \ $

$\ \rm V^2$
$\varphi_x(\tau = 4T)\ = \ $

$\ \rm V^2$

6

Calculate the expected value of the ACF with respect to all  $\tau$ values.  Interpret the result.

${\rm E}\big[\varphi_x(\tau)\big]\ = \ $

$\ \rm V^2$


Solution

For the ACF calculation

(1)  The  (normalized)  period duration is  $T_0/T \hspace{0.15cm}\underline{= 5}.$


(2)  Due to periodicity,  the averaging over a periodic time  $T_0$:

$$m_x = \frac{1}{T_0} \cdot \int_0^{T_0} x(t) \hspace{0.1cm}{\rm d} t = \frac{1}{5 T} \cdot (2\hspace{0.05cm}{\rm V} \cdot 2 T - 2\hspace{0.05cm}{\rm V} \cdot 2 T) \hspace{0.15cm}\underline{= \rm 0.4 \,V}.$$


(3)  In analogy to the last subtask,  we obtain for the mean power:

$$P_x = \frac{2 T}{5 T} \cdot \big[(\rm 2V)^2 +(- \rm 1V)^2 \big]\hspace{0.15cm}\underline{ = \rm 2 \,V^2}.$$


(4)  The accompanying graph shows in each case in the range from  $0$  to  $T_0 = 5T$.

  • above the product  $x(t) \cdot x(t+T)$,
  • down the product  $x(t) \cdot x(t+2T)$.


  • Note that  $x(t+T)$  means a shift of the signal  $x(t)$  by  $T$  to the left.
  • From these sketches follow the relations:
$$\varphi_x (T)= \rm {1}/{5 } \cdot (\rm 4V^2 + \rm 1V^2 - \rm 2V^2) \hspace{0.15cm}\underline{= \rm 0.6\, V^2},$$
$$\varphi_x ( 2 T)= \rm {1}/{5 } \cdot(-\rm 2V^2 \cdot 3) \hspace{0.15cm}\underline{= - \rm 1.2 \,V^2}.$$


Wanted auto-correlation function

(5)  An auto-correlation function is always even:  

$$\varphi_x (-\tau)= \varphi_x (\tau).$$
  • In addition,  for periodic processes,  the ACF is also periodic with the same period duration  $T_0$  as the individual pattern functions.  It follows that:
$$\varphi_x ( 0) = \varphi_x (5 T) = \varphi_x (10 T) = \ \text{...} \ = \it P_x = \rm 2 \,V^2,$$
$$\varphi_x (3 T) = \varphi_x (-3 T) =\varphi_x (2 T) = \ \text{...} \ \hspace{0.15cm}\underline{= - \rm 1.2 \,V^2},$$
$$\varphi_x (4 T) = \varphi_x (-4 T) =\varphi_x ( T) = \ \text{...} \ \hspace{0.15cm}\underline{= \rm 0.6 \,V^2}.$$
  • The calculated ACF values can be connected by straight line sections,  since integration over rectangular functions always yields linear subsections.


(6)  The five intervals  $(0$ to $T)$,  $(T$ to $2T)$, ... ,  $(4T$ to $5T)$  provide the contributions.

$$(+1.3; -0.3; -1.2; -0.3; +1.3) \cdot \rm V^2.$$
  • This gives the expected value  (linear mean):
$${\rm E}\big[\varphi_x(\tau)\big] = 1/5 \cdot (1.3-0.3 -1.2 -0.3 +1.3]\hspace{0.15cm}\underline{= \rm 0.16 \,V^2}.$$
  • This corresponds to the square of the mean  $m_x$   ⇒   see subtask  (2).