Difference between revisions of "Aufgaben:Exercise 2.2: Multi-Level Signals"

From LNTwww
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{What is the variance  $\sigma_x^2$  of the random variable  $x$  in general and für  $M= 5$?
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{What is the variance  $\sigma_x^2$  of the random variable  $x$  in general and for  $M= 5$?
 
|type="{}"}
 
|type="{}"}
 
$\sigma_x^2\ = \ $ { 2 3% }
 
$\sigma_x^2\ = \ $ { 2 3% }
  
  
{Calculate the mean  $m_y$  of the random variable  $y$  für  $M= 5$.
+
{Calculate the mean  $m_y$  of the random variable  $y$  for  $M= 5$.
 
|type="{}"}
 
|type="{}"}
 
$m_y \ = \ $ { 0. } $\ \rm V$
 
$m_y \ = \ $ { 0. } $\ \rm V$
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===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
 
'''(1)'''  Man erhält durch Mittelung über alle möglichen Signalwerte für den linearen Mittelwert:
 
'''(1)'''  Man erhält durch Mittelung über alle möglichen Signalwerte für den linearen Mittelwert:
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{{ML-Fuß}}
 
{{ML-Fuß}}
 +
 +
'''(1)'''  One obtains by averaging over all possible signal values for the linear mean:
 +
:$$m_{\it x}=\rm \sum_{\mu=0}^{\it M-{\rm 1}} \it p_\mu\cdot x_{\mu}=\frac{\rm 1}{\it M} \cdot \sum_{\mu=\rm 0}^{\it M-\rm 1}\mu=\frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M}{\rm 2}=\frac{\it M-\rm 1}{\rm 2}.$$
 +
 +
*In the special case  $M= 5$  the linear mean results in  $m_x \;\underline{= 2}$.
 +
 +
 +
 +
'''(2)'''  Analogously, for the root mean square:
 +
: $$m_{\rm 2\it x}= \rm \sum_{\mu=0}^{\it M -\rm 1}\it p_\mu\cdot x_{\mu}^{\rm 2}=\frac{\rm 1}{\it M}\cdot \sum_{\mu=\rm 0}^{\rm M- 1}\mu^{\rm 2} = \frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M\cdot(\rm 2\it M-\rm 1)}{\rm 6} = \frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}. $$
 +
 +
*In the special case $M= 5$  the root mean square results in  $m_{2x} {=6}$.
 +
*From this, the variance can be calculated using Steiner's theorem:
 +
:$$\sigma_x^{\rm 2}=m_{\rm 2\it x}-m_x^{\rm 2}=\frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}-\frac{(\it M-\rm 1)^{\rm 2}}{\rm 4}=\frac{\it M^{\rm 2}-\rm 1}{\rm 12}.$$
 +
 +
*In the special case  $M= 5$  the result for the variance  $\sigma_x^2 \;\underline{= 2}$.
 +
 +
 +
 +
'''(3)'''  Because of the symmetry of  $y$ , holds independently of  $M$:
 +
:$$m_x \;\underline{= 2}.$$
 +
 +
 +
 +
'''(4)'''  Between  $x(t)$  and  $y(t)$  the following relation holds:
 +
:$$y(t)=\frac{2\cdot y_{\rm 0}}{M-\rm 1}\cdot \big[x(t)-m_x\big].$$
 +
 +
*From this it follows for the variances:
 +
: $$\sigma_y^{\rm 2}=\frac{4\cdot y_{\rm 0}^{\rm 2}}{( M - 1)^{\rm 2}}\cdot \sigma_x^{\rm 2}=\frac{y_{\rm 0}^{\rm 2}\cdot (M^{\rm 2}- 1)}{3\cdot (M- 1)^{\rm 2}}=\frac{y_{\rm 0}^{\rm 2}\cdot ( M+ 1)}{ 3\cdot ( M- 1)}. $$
 +
 +
*In the special case  $M= 5$  this results in:
 +
:$$\it \sigma_y^{\rm 2}= \frac {\it y_{\rm 0}^{\rm 2} \cdot {\rm 6}}{\rm 3 \cdot 4}\hspace{0.15cm} \underline{=\rm2\,V^{2}}.$$
 +
  
  
  
 
[[Category:Theory of Stochastic Signals: Exercises|^2.2 Moments of Discrete Random Variables^]]
 
[[Category:Theory of Stochastic Signals: Exercises|^2.2 Moments of Discrete Random Variables^]]

Revision as of 19:39, 3 December 2021


Two similar multilevel signals

Let the square wave signal  $x(t)$  be dimensionless and can only have the current values  $0, \ 1, \ 2, \ \text{...} \ , \ M-2, \ M-1$  with equal probability. The upper graph shows this signal for the special case  $M = 5$.


The square wave signal  $y(t)$  is also  $M$–stepped, but zero mean and restricted to the range from  $y > -y_0$  to  $y < +y_0$  .


In the graph below you can see the signal  $y(t)$, again for the number of steps  $M = 5$.





Hints:

  • Fur numerical calculations, use  $y_0 = 2\hspace{0.05cm}V$.



Questions

1

What is the linear mean  $m_x$  of the random variable  $x$  for  $M= 5$?

$m_x \ = \ $

2

What is the variance  $\sigma_x^2$  of the random variable  $x$  in general and for  $M= 5$?

$\sigma_x^2\ = \ $

3

Calculate the mean  $m_y$  of the random variable  $y$  for  $M= 5$.

$m_y \ = \ $

$\ \rm V$

4

What is the variance  $\sigma_y^2$  of the random variable  $y$?  Consider the result from  (2).  What is the value again for  $M= 5$?

$\sigma_y^2\ = \ $

$\ \rm V^2$


Solution

(1)  Man erhält durch Mittelung über alle möglichen Signalwerte für den linearen Mittelwert:

$$m_{\it x}=\rm \sum_{\mu=0}^{\it M-{\rm 1}} \it p_\mu\cdot x_{\mu}=\frac{\rm 1}{\it M} \cdot \sum_{\mu=\rm 0}^{\it M-\rm 1}\mu=\frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M}{\rm 2}=\frac{\it M-\rm 1}{\rm 2}.$$
  • Im Sonderfall  $M= 5$  ergibt sich der lineare Mittelwert zu  $m_x \;\underline{= 2}$.


(2)  Analog gilt für den quadratischen Mittelwert:

$$m_{\rm 2\it x}= \rm \sum_{\mu=0}^{\it M -\rm 1}\it p_\mu\cdot x_{\mu}^{\rm 2}=\frac{\rm 1}{\it M}\cdot \sum_{\mu=\rm 0}^{\rm M-1}\mu^{\rm 2} = \frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M\cdot(\rm 2\it M-\rm 1)}{\rm 6} = \frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}.$$
  • Im Sonderfall $M= 5$  ergibt sich der quadratische Mittelwert zu  $m_{2x} {=6}$.
  • Daraus kann die Varianz mit dem Satz von Steiner berechnet werden:
$$\sigma_x^{\rm 2}=m_{\rm 2\it x}-m_x^{\rm 2}=\frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}-\frac{(\it M-\rm 1)^{\rm 2}}{\rm 4}=\frac{\it M^{\rm 2}-\rm 1}{\rm 12}.$$
  • Im Sonderfall  $M= 5$  ergibt sich für die Varianz  $\sigma_x^2 \;\underline{= 2}$.


(3)  Aufgrund der Symmetrie von  $y$  gilt unabhängig von  $M$:

$$m_x \;\underline{= 2}.$$


(4)  Zwischen  $x(t)$  und  $y(t)$  gilt folgender Zusammenhang:

$$y(t)=\frac{2\cdot y_{\rm 0}}{M-\rm 1}\cdot \big[x(t)-m_x\big].$$
  • Daraus folgt für die Varianzen:
$$\sigma_y^{\rm 2}=\frac{4\cdot y_{\rm 0}^{\rm 2}}{( M - 1)^{\rm 2}}\cdot \sigma_x^{\rm 2}=\frac{y_{\rm 0}^{\rm 2}\cdot (M^{\rm 2}-1)}{3\cdot (M- 1)^{\rm 2}}=\frac{y_{\rm 0}^{\rm 2}\cdot ( M+ 1)}{ 3\cdot ( M- 1)}.$$
  • Im Sonderfall  $M= 5$  ergibt sich hierfür:
$$\it \sigma_y^{\rm 2}= \frac {\it y_{\rm 0}^{\rm 2} \cdot {\rm 6}}{\rm 3 \cdot 4}\hspace{0.15cm} \underline{=\rm2\,V^{2}}.$$

(1)  One obtains by averaging over all possible signal values for the linear mean:

$$m_{\it x}=\rm \sum_{\mu=0}^{\it M-{\rm 1}} \it p_\mu\cdot x_{\mu}=\frac{\rm 1}{\it M} \cdot \sum_{\mu=\rm 0}^{\it M-\rm 1}\mu=\frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M}{\rm 2}=\frac{\it M-\rm 1}{\rm 2}.$$
  • In the special case  $M= 5$  the linear mean results in  $m_x \;\underline{= 2}$.


(2)  Analogously, for the root mean square:

$$m_{\rm 2\it x}= \rm \sum_{\mu=0}^{\it M -\rm 1}\it p_\mu\cdot x_{\mu}^{\rm 2}=\frac{\rm 1}{\it M}\cdot \sum_{\mu=\rm 0}^{\rm M- 1}\mu^{\rm 2} = \frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M\cdot(\rm 2\it M-\rm 1)}{\rm 6} = \frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}. $$
  • In the special case $M= 5$  the root mean square results in  $m_{2x} {=6}$.
  • From this, the variance can be calculated using Steiner's theorem:
$$\sigma_x^{\rm 2}=m_{\rm 2\it x}-m_x^{\rm 2}=\frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}-\frac{(\it M-\rm 1)^{\rm 2}}{\rm 4}=\frac{\it M^{\rm 2}-\rm 1}{\rm 12}.$$
  • In the special case  $M= 5$  the result for the variance  $\sigma_x^2 \;\underline{= 2}$.


(3)  Because of the symmetry of  $y$ , holds independently of  $M$:

$$m_x \;\underline{= 2}.$$


(4)  Between  $x(t)$  and  $y(t)$  the following relation holds:

$$y(t)=\frac{2\cdot y_{\rm 0}}{M-\rm 1}\cdot \big[x(t)-m_x\big].$$
  • From this it follows for the variances:
$$\sigma_y^{\rm 2}=\frac{4\cdot y_{\rm 0}^{\rm 2}}{( M - 1)^{\rm 2}}\cdot \sigma_x^{\rm 2}=\frac{y_{\rm 0}^{\rm 2}\cdot (M^{\rm 2}- 1)}{3\cdot (M- 1)^{\rm 2}}=\frac{y_{\rm 0}^{\rm 2}\cdot ( M+ 1)}{ 3\cdot ( M- 1)}. $$
  • In the special case  $M= 5$  this results in:
$$\it \sigma_y^{\rm 2}= \frac {\it y_{\rm 0}^{\rm 2} \cdot {\rm 6}}{\rm 3 \cdot 4}\hspace{0.15cm} \underline{=\rm2\,V^{2}}.$$