Aufgaben:Exercise 3.9Z: Sine Transformation - Revision history

Guenter at 15:09, 17 February 2022

2022-02-17T15:09:10Z

Bene: Text replacement - "(rms)" to "(standard deviation)"

2022-02-17T11:36:56Z

Text replacement - "(rms)" to "(standard deviation)"

Bene: Text replacement - "rms value" to "standard deviation"

2022-02-17T11:11:50Z

Text replacement - "rms value" to "standard deviation"

Bene: Text replacement - "root mean square" to "standard deviation"

2022-02-17T11:00:51Z

Text replacement - "root mean square" to "standard deviation"

Guenter at 16:48, 2 February 2022

2022-02-02T16:48:36Z

Guenter at 15:55, 2 February 2022

2022-02-02T15:55:34Z

Guenter at 15:38, 2 February 2022

2022-02-02T15:38:50Z

Noah at 18:59, 15 January 2022

2022-01-15T18:59:14Z

Noah at 18:57, 15 January 2022

2022-01-15T18:57:34Z

Noah at 12:12, 12 January 2022

2022-01-12T12:12:22Z

@@ Line 50: / Line 50: @@
-{Calculate the the&nbsp; standard deviation&nbsp; (standard deviation)&nbsp; of  the random variable&nbsp; $y$.
+{Calculate the the&nbsp; standard deviation of  the random variable&nbsp; $y$.
 |type="{}"}
 $\sigma_y \ = \ $ { 0.172 3% }
@@ Line 92: / Line 92: @@
 .75}.$$
-*With the result from&nbsp; '''(2)'''&nbsp; it thus follows for the rms&nbsp; ("standard deviation"):
+*With the result from&nbsp; '''(2)'''&nbsp; it thus follows for the standard deviation:
 :$$ \sigma_{y}=\sqrt{\frac{\rm 3}{\rm 4}-\Big(\frac{\rm 8}{\rm 3\cdot\pi}\Big)^{\rm 2}} \hspace{0.15cm}\underline{\approx \rm 0.172}.$$

@@ Line 50: / Line 50: @@
-{Calculate the the&nbsp; standard deviation&nbsp; (rms)&nbsp; of  the random variable&nbsp; $y$.
+{Calculate the the&nbsp; standard deviation&nbsp; (standard deviation)&nbsp; of  the random variable&nbsp; $y$.
 |type="{}"}
 $\sigma_y \ = \ $ { 0.172 3% }

@@ Line 9: / Line 9: @@
 Outside of this,&nbsp; the PDF is identically zero.
-The mean and the rms value of this random variable&nbsp; $x$&nbsp; have already been determined in the&nbsp; [[Aufgaben:Exercise_3.3:_Moments_for_cos²-PDF|Exercise 3.3]]:
+The mean and the standard deviation of this random variable&nbsp; $x$&nbsp; have already been determined in the&nbsp; [[Aufgaben:Exercise_3.3:_Moments_for_cos²-PDF|Exercise 3.3]]:
 :$$m_x = 1,\hspace{0.2cm}\sigma_x = 0.361.$$

@@ Line 50: / Line 50: @@
-{Calculate the the&nbsp; root mean square&nbsp; (rms)&nbsp; of  the random variable&nbsp; $y$.
+{Calculate the the&nbsp; standard deviation&nbsp; (rms)&nbsp; of  the random variable&nbsp; $y$.
 |type="{}"}
 $\sigma_y \ = \ $ { 0.172 3% }
@@ Line 92: / Line 92: @@
 .75}.$$
-*With the result from&nbsp; '''(2)'''&nbsp; it thus follows for the rms&nbsp; ("root mean square"):
+*With the result from&nbsp; '''(2)'''&nbsp; it thus follows for the rms&nbsp; ("standard deviation"):
 :$$ \sigma_{y}=\sqrt{\frac{\rm 3}{\rm 4}-\Big(\frac{\rm 8}{\rm 3\cdot\pi}\Big)^{\rm 2}} \hspace{0.15cm}\underline{\approx \rm 0.172}.$$

@@ Line 69: / Line 69: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Correct are <u>the second and the third suggested solutions</u>:
+'''(1)'''&nbsp; Correct are&nbsp; <u>the second and the third suggested solutions</u>:
 *Because of the range of values of&nbsp; $x$&nbsp; and the given characteristic curve,&nbsp; $y$&nbsp; cannot take values smaller than&nbsp; $0$&nbsp; or larger than&nbsp; $1$&nbsp; respectively.
-*The value&nbsp; $y = 0$&nbsp; cannot occur either, however, since neither&nbsp; $x = 0$&nbsp; nor&nbsp; $x = 2$&nbsp; are possible.
+*The value&nbsp; $y = 0$&nbsp; cannot occur either,&nbsp; however,&nbsp; since neither&nbsp; $x = 0$&nbsp; nor&nbsp; $x = 2$&nbsp; are possible.
-*With these properties, the result is surely &nbsp;$m_y < 1$, i.e., a smaller value than &nbsp;$m_x = 1$&nbsp; (see specification).
+*With these properties,&nbsp; the result is surely &nbsp;$m_y < 1$, i.e., a smaller value than &nbsp;$m_x = 1$&nbsp; (see specification).
-'''(2)'''&nbsp; To solve this task, one could, for example, first determine the PDF&nbsp; $f_y(y)$&nbsp; and calculate &nbsp;$m_y$&nbsp; from it in the usual way.
+'''(2)'''&nbsp; To solve this task,&nbsp; one could,&nbsp; for example,&nbsp; first determine the PDF&nbsp; $f_y(y)$&nbsp; and calculate &nbsp;$m_y$&nbsp; from it in the usual way.
 *The direct way leads to the same result:
 :$$m_y={\rm E}\big[y\big]={\rm E}\big[g(x)\big]=\int_{-\infty}^{+\infty}g(x)\cdot f_x(x)\,{\rm d}x.$$
@@ Line 92: / Line 92: @@
 .75}.$$
-*With the result from&nbsp; '''(2)'''&nbsp; it thus follows for the rms:
+*With the result from&nbsp; '''(2)'''&nbsp; it thus follows for the rms&nbsp; ("root mean square"):
 :$$ \sigma_{y}=\sqrt{\frac{\rm 3}{\rm 4}-\Big(\frac{\rm 8}{\rm 3\cdot\pi}\Big)^{\rm 2}} \hspace{0.15cm}\underline{\approx \rm 0.172}.$$
-'''(4)'''&nbsp; Due to the symmetry of PDF&nbsp; $f_x(x)$&nbsp; and characteristic curve&nbsp; $y =g(x)$&nbsp; um&nbsp; $x = 1$&nbsp; the two domains yield.
+'''(4)'''&nbsp; Due to the symmetry of the PDF&nbsp; $f_x(x)$&nbsp; and the characteristic curve&nbsp; $y =g(x)$&nbsp; um&nbsp; $x = 1$&nbsp; the two domains yield.
 *$0 \le x \le 1$&nbsp; and
 *$1 \le x \le 2$
@@ Line 103: / Line 103: @@
 each give the same contribution for $f_y(y)$.
-*In the first domain, the derivative of the characteristic curve is positive:&nbsp; $g\hspace{0.05cm}'(x)={\rm \pi}/{\rm 2}\cdot \cos({\rm \pi}/{\rm 2}\cdot x).$
+*In the first domain,&nbsp; the derivative of the characteristic curve is positive:&nbsp; $g\hspace{0.05cm}'(x)={\rm \pi}/{\rm 2}\cdot \cos({\rm \pi}/{\rm 2}\cdot x).$
 *The inverse function is:&nbsp; $ x=h(y)={\rm 2}/{\rm \pi}\cdot \arcsin( y).$
-*Taking into account the second contribution by the factor&nbsp; $2$&nbsp; we get&auml;r the searched PDF in the range&nbsp; $0 \le y \le 1$ ):
+*Taking into account the second contribution by the factor&nbsp; $2$&nbsp; we get the searched PDF in the range&nbsp; $0 \le y \le 1$:
 :$$f_y(y)= 2\cdot\frac{\sin^{ 2}({ \pi}/{ 2}\cdot x)}{{ \pi}/{ 2}\cdot \cos({ \pi}/{ 2}\cdot x)}\Big|_{\, x={ 2}/{ \pi}\cdot \arcsin( y)}.$$
-[[File:P_ID138__Sto_Z_3_9_e_neu.png|right|frame|Wanted PDF]]
+[[File:P_ID138__Sto_Z_3_9_e_neu.png|right|frame|PDF after transformation]]
-*outside $f_y(y) is \equiv 0$. This leads to the intermediate result
+*Outside of this range:&nbsp; $f_y(y) \equiv 0$.&nbsp; This leads to the intermediate result
 :$$f_y(y)=\frac{4}{\pi}\cdot \frac{\sin^{2}(\arcsin( y ))}{\sqrt{\rm 1-\sin^{ 2}(\arcsin( y \rm ))}}.$$
 *And because of&nbsp; $\sin\big (\arcsin(y)\big) = y$:
 :$$f_y(y)=\frac{ 4}{\pi}\cdot \frac{ y^{2}}{\sqrt{1- y^{\rm 2}}}.$$
 *At the point&nbsp; $y = 0.6$&nbsp; one obtains the value&nbsp; $f_y(y= 0.6)\hspace{0.15cm}\underline{=0.573}$.
-*On the right, this PDF&nbsp; $f_y(y)$&nbsp; is shown graphically.
+*On the right,&nbsp; this PDF&nbsp; $f_y(y)$&nbsp; is shown graphically.
-'''(5)'''&nbsp; The PDF is infinitely large at the point&nbsp; $y = 1$&nbsp;.
+'''(5)'''&nbsp; The PDF is infinitely large at the point&nbsp; $y = 1$.
 *This is due to the fact that at this point the derivative&nbsp; $g\hspace{0.05cm}'(x)$&nbsp; of the characteristic curve runs horizontally.
-* However, since&nbsp; $y$&nbsp; is a continuous random gr&ouml;&szlig;e, nevertheless&nbsp; ${\rm Pr}(y = 1) \hspace{0.15cm}\underline{= 0}$ holds.
+* However,&nbsp; since&nbsp; $y$&nbsp; is a continuous random quantity,&nbsp; nevertheless&nbsp; ${\rm Pr}(y = 1) \hspace{0.15cm}\underline{= 0}$&nbsp; holds.
 This means:
-*An infinity point in the PDF is not identical to a Dirac function.
+*An infinity point in the PDF is not identical to a Dirac delta function.
-*Or more casually expressed: &nbsp; An infinity point in the PDF is "less" than a Dirac function.
+*Or more casually expressed: &nbsp; An infinity point in the PDF is&nbsp; "less"&nbsp; than a Dirac delta function.
 {{ML-Fuß}}

@@ Line 28: / Line 28: @@
 Hints:
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Exponentially_Distributed_Random_Variables|Exponentially Distributed Random Variables]].
-*In particular, reference is made to the page&nbsp; [[Theory_of_Stochastic_Signals/Exponentially_Distributed_Random_Variables#Transformation_of_random_variables|"Transformation of random variables"]]&nbsp; and the chapter&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments|"Expected Values and Moments"]].
+*In particular, reference is made to the section&nbsp; [[Theory_of_Stochastic_Signals/Exponentially_Distributed_Random_Variables#Transformation_of_random_variables|"Transformation of random variables"]]&nbsp; and the chapter&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments|"Expected Values and Moments"]].
 *Given are the two indefinite integrals:
@@ Line 40: / Line 40: @@
 {Which of the following statements are true?
 |type="[]"}
-- $y$&nbsp; is limited to the value range&nbsp; $0 \le y \le 1$&nbsp; .
+- $y$&nbsp; is limited to the range&nbsp; $0 \le y \le 1$&nbsp; .
-+ $y$&nbsp; is limited to the value range&nbsp; $0 < y \le 1$&nbsp;.
++ $y$&nbsp; is limited to the range&nbsp; $0 < y \le 1$&nbsp;.
 + The mean&nbsp; $m_y$&nbsp; is less than the mean&nbsp; $m_x$.
@@ Line 50: / Line 50: @@
-{Calculate the root mean square of&nbsp; $y$&nbsp; and the rms&nbsp;$\sigma_y$ .
+{Calculate the the&nbsp; root mean square&nbsp; (rms)&nbsp; of  the random variable&nbsp; $y$.
 |type="{}"}
 $\sigma_y \ = \ $ { 0.172 3% }
-{Calculate the PDF $f_y(y)$.&nbsp; Note the symmetry properties.&nbsp; What PDF&ndash;value results for&nbsp; $y = 0.6$&nbsp;?
+{Calculate the PDF $f_y(y)$.&nbsp; Note the symmetry properties.&nbsp; What PDF value results for&nbsp; $y = 0.6$?
 |type="{}"}
 $f_y(y=0.6) \ = \ $ { 0.573 3% }
-{What is the PDF value for&nbsp; $y = 1$?&nbsp; Interpret the result.&nbsp; What is the probability that&nbsp; $y$&nbsp; is exactly equal&nbsp; $1$&nbsp;?
+{What is the PDF value for&nbsp; $y = 1$?&nbsp; Interpret the result.&nbsp; What is the probability that&nbsp; $y$&nbsp; is exactly equal&nbsp; $1$?
 |type="{}"}
 ${\rm Pr}(y=1) \ = \ $ { 0. }

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID137__Sto_Z_3_9.png|right|frame|Input PDF and characteristic curve]]
+[[File:P_ID137__Sto_Z_3_9.png|right|frame|Input PDF, characteristic curve]]
-In this task, we consider a random variable&nbsp; $x$&nbsp; with&nbsp; $\sin^2$& shaped PDF in the range between&nbsp; $x= 0$&nbsp; and&nbsp; $x= 2$:
+In this task,&nbsp; we consider a random variable&nbsp; $x$&nbsp; with sine-square shaped&nbsp; $\rm PDF$&nbsp; in the range between&nbsp; $x= 0$&nbsp; and&nbsp; $x= 2$:
-:$$f_x(x)= \sin^2({\rm\pi}/{\rm 2}\cdot x) \hspace{1cm}\rm for\hspace{0.15cm}{\rm 0\le \it x \le \rm 2} .$$
+:$$f_x(x)= \sin^2({\rm\pi}/{\rm 2}\cdot x) \hspace{1cm}\rm for\hspace{0.25cm}{\rm 0\le \it x \le \rm 2} .$$
-Outside of this, the PDF is identically zero.
+Outside of this,&nbsp; the PDF is identically zero.
-The mean and rms of this random variable&nbsp; $x$&nbsp; have already been determined in the&nbsp; [[Aufgaben:Exercise_3.3:_Moments_for_cos²-PDF|Exercise 3.3]]&nbsp; :
+The mean and the rms value of this random variable&nbsp; $x$&nbsp; have already been determined in the&nbsp; [[Aufgaben:Exercise_3.3:_Moments_for_cos²-PDF|Exercise 3.3]]:
 :$$m_x = 1,\hspace{0.2cm}\sigma_x = 0.361.$$
@@ Line 15: / Line 15: @@
 :$$y= g(x) =\sin({\rm\pi}/{\rm 2}\cdot x).$$
-The figure shows in each case in the range&nbsp; $0 \le x \le 2$:
+The figure shows in each case in the range &nbsp; $0 \le x \le 2$:
 *above the PDF&nbsp; $f_x(x)$,
 *below the nonlinear characteristic&nbsp; $y = g(x)$.
@@ Line 28: / Line 28: @@
 Hints:
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Exponentially_Distributed_Random_Variables|Exponentially Distributed Random Variables]].
-*In particular, reference is made to the page&nbsp; [[Theory_of_Stochastic_Signals/Exponentially_Distributed_Random_Variables#Transformation_of_random_variables|Transformation of random variables]]&nbsp; and the chapter&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments|Expected Values and Moments]].
+*In particular, reference is made to the page&nbsp; [[Theory_of_Stochastic_Signals/Exponentially_Distributed_Random_Variables#Transformation_of_random_variables|"Transformation of random variables"]]&nbsp; and the chapter&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments|"Expected Values and Moments"]].
 *Given are the two indefinite integrals:

@@ Line 89: / Line 89: @@
 *This leads to the result:
-:$$ m_{ 2 y}=\int_{\rm 0}^{\rm 2}\hspace{-0. 15cm}\sin^{\rm 4}({\rm \pi}/{\rm 2}\cdot x)\, {\rm d} x= \frac{\rm 3}{\rm 8}\cdot x-\frac{\rm 1}{\rm 2\cdot\pi}\cdot \sin(\rm \pi\cdot{\it x})+\frac{\rm 1}{\rm 16\cdot\pi}\cdot \sin(\rm 2 \pi\cdot {\it x})\Big|_{\rm 0}^{\rm 2} \hspace{0.15cm}{= \rm
+:$$ m_{ 2 y}=\int_{\rm 0}^{\rm 2}\hspace{-0.15cm}\sin^{\rm 4}({\rm \pi}/{\rm 2}\cdot x)\,{\rm d} x= \frac{\rm 3}{\rm 8}\cdot  x-\frac{\rm 1}{\rm 2\cdot\pi}\cdot \sin(\rm \pi\cdot{\it  x})+\frac{\rm 1}{\rm 16\cdot\pi}\cdot \sin(\rm 2 \pi\cdot {\it  x})\Big|_{\rm 0}^{\rm 2} \hspace{0.15cm}{= \rm
 .75}.$$
@@ Line 109: / Line 109: @@
 *Taking into account the second contribution by the factor&nbsp; $2$&nbsp; we get&auml;r the searched PDF in the range&nbsp; $0 \le y \le 1$ ):
 :$$f_y(y)= 2\cdot\frac{\sin^{ 2}({ \pi}/{ 2}\cdot x)}{{ \pi}/{ 2}\cdot \cos({ \pi}/{ 2}\cdot x)}\Big|_{\, x={ 2}/{ \pi}\cdot \arcsin( y)}.$$
-[[File:P_ID138__Sto_Z_3_9_e_new.png|right|frame|Sought PDF]]
+[[File:P_ID138__Sto_Z_3_9_e_neu.png|right|frame|Sought PDF]]
 *outside $f_y(y) is \equiv 0$. This leads to the intermediate result
 :$$f_y(y)=\frac{4}{\pi}\cdot \frac{\sin^{2}(\arcsin( y ))}{\sqrt{\rm 1-\sin^{ 2}(\arcsin( y \rm ))}}.$$