Aufgaben:Exercise 3.8: Amplification and Limitation - Revision history

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

2022-02-17T11:11:50Z

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

Guenter at 13:37, 2 February 2022

2022-02-02T13:37:49Z

Guenter at 13:22, 2 February 2022

2022-02-02T13:22:06Z

Guenter at 13:14, 2 February 2022

2022-02-02T13:14:26Z

Noah at 18:56, 15 January 2022

2022-01-15T18:56:00Z

Noah at 18:55, 15 January 2022

2022-01-15T18:55:42Z

Noah at 18:52, 15 January 2022

2022-01-15T18:52:14Z

Noah at 19:46, 11 January 2022

2022-01-11T19:46:23Z

Noah at 19:23, 11 January 2022

2022-01-11T19:23:39Z

Noah at 19:21, 11 January 2022

2022-01-11T19:21:26Z

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@@ Line 33: / Line 33: @@
-{Calculate the moments&nbsp; $m_k$&nbsp; of the random variable&nbsp; $x$.&nbsp; Reason that all moments with odd index are zero.&nbsp; How big is the rms value?
+{Calculate the moments&nbsp; $m_k$&nbsp; of the random variable&nbsp; $x$.&nbsp; Reason that all moments with odd index are zero.&nbsp; How big is the standard deviation?
 |type="{}"}
 $\sigma_x \ = \ $ { 0.707 3% }

@@ Line 19: / Line 19: @@
 Hints:
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Exponentially_Distributed_Random_Variables|Exponentially Distributed Random Variable]].
 *Given the following definite integral:
 :$$\int_{0}^{\infty}\it x^n\cdot\rm e^{-\it a \hspace{0.03cm}\cdot \hspace{0.03cm}x}\, d{\it x} =\frac{\it n{\rm !}}{\it a^{n}}.$$

@@ Line 71: / Line 71: @@
 :$$\it F=\rm 2\cdot \it A\int_{\rm 0}^{\infty}\hspace{-0.15cm}\rm e^{\rm -2\it x}\, \rm d \it x=\frac{\rm 2\cdot \it A}{\rm -2}\cdot \rm e^{\rm -2\it x}\Big|_{\rm 0}^{\infty}=\it A.$$
-*Since this area must be equal by definition&nbsp; $F = 1$&nbsp; $\underline{A = 1}$.
+*Since this area must be equal by definition&nbsp; $F = 1$ &nbsp; &rArr; &nbsp; $\underline{A = 1}$.
-'''(2)'''&nbsp; All moments with odd index&nbsp; $k$&nbsp; are equal to zero due to symmetric PDF.
+'''(2)'''&nbsp; All moments with odd index&nbsp; $k$&nbsp; are equal to zero due to the symmetrical PDF.
 *For even&nbsp; $k$&nbsp; the left part of the PDF can be mirrored into the right one and we get:
 :$$\it m_k=\rm 2 \cdot \int_{\rm 0}^{\infty}\hspace{-0.15cm}\it x^{k}\cdot \rm e^{-\rm 2\it x}\,\rm d \it x=\frac{\rm 2\cdot\rm\Gamma(\it k{\rm +}\rm 1)}{\rm 2^{\it k{\rm +}\rm 1}}=\frac{\it k{\rm !}}{\rm 2^{\it k}}.$$
@@ Line 83: / Line 83: @@
-[[File:P_ID131__Sto_A_3_8_e.png|right|frame|PDF after reinforcement and boundary]]
+[[File:P_ID131__Sto_A_3_8_e.png|right|frame|PDF after amplification and boundary]]
 '''(3)'''&nbsp; The fourth-order central moment is&nbsp; $\mu_4 = m_4 = 4!/2^4 = 1.5$.
 *From this follows for the kurtosis:
@@ Line 93: / Line 93: @@
-'''(5)'''&nbsp; Correct are <u>solutions 1 and 3</u>:
+'''(5)'''&nbsp; Correct are the&nbsp; <u>solutions 1 and 3</u>:
 *The PDF&nbsp; $f_y(y)$&nbsp; involves a Dirac delta function at the point&nbsp; $y= 0$&nbsp; with weight&nbsp; ${\rm Pr}(x < 0) = 0.5$.
-*In addition, another Dirac delta function at&nbsp; $y= 1$&nbsp; with weight&nbsp; ${\rm Pr}(x > 0.5) = 0.184$.
+*In addition,&nbsp; another Dirac delta function at&nbsp; $y= 1$&nbsp; with weight&nbsp; ${\rm Pr}(x > 0.5) = 0.184$.
@@ Line 103: / Line 103: @@
 :$$f_y(y)=\frac{f_x(x)}{|g'(x)|}\Bigg|_{x=h(y)}=\frac{\rm e^{-\rm 2\it x}}{\rm 2}\Bigg|_{\it x={\it y}/{\rm 2}}=0.5 \cdot {\rm e^{\it -y}} .$$
-*For&nbsp; $y= 0.5$&nbsp; accordingly, the continuous PDF component is
+*For&nbsp; $y= 0.5$&nbsp; accordingly,&nbsp; the continuous PDF component is
 :$$f_y(y = 0.5)\hspace{0.15cm}\underline{\approx 0.304}.$$
@@ Line 110: / Line 110: @@
 :$$m_y=\frac{1}{\rm 2\rm e} \cdot 1 +\int_{\rm 0}^{\rm 1}\frac{\it y}{\rm 2}\cdot \rm e^{\it -y}\, \rm d \it y=\frac{\rm 1}{\rm 2\rm e}{\rm +}\frac{\rm 1}{\rm 2}-\frac{\rm 1}{\rm e}=\frac{\rm 1}{\rm 2}-\frac{\rm 1}{\rm 2 e}\hspace{0.15cm}\underline{=\rm 0.316}.$$
-*The first term is from the Dirac delta at&nbsp; $y= 1$, the second from the continuous PDF component.
+*The first term is from the Dirac delta at&nbsp; $y= 1$,&nbsp; the second from the continuous PDF component.
 {{ML-Fuß}}

@@ Line 4: / Line 4: @@
 [[File:P_ID130__Sto_A_3_8.png|right|frame|Amplification and limitation<br>of the random variable&nbsp; $x$]]
-We consider a random signal&nbsp; $x(t)$&nbsp; with symmetric probability density function:
+We consider a random signal&nbsp; $x(t)$&nbsp; with symmetric probability density function&nbsp; $\rm (PDF)$:
 :$$f_x(x)=A\cdot \rm e^{\rm -2 \hspace{0.05cm}\cdot \hspace{0.05cm}|\it x|}.$$
-*This signal is applied to the input of a nonlinearity with the characteristic curve (see lower figure):
+*This signal is applied to the input of a nonlinearity with the characteristic curve&nbsp; (see lower figure):
 :$$y=\left\{\begin{array}{*{4}{c}}0  &\rm for\hspace{0.2cm} \it x <\rm 0, \\\rm2\it x   & \rm for\hspace{0.2cm} \rm 0\le \it x\le \rm 0.5,  \\1 & \rm for\hspace{0.2cm}\it x > \rm 0.5\\\end{array}\right.$$
@@ Line 19: / Line 19: @@
 Hints:
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Exponentially_Distributed_Random_Variables|Exponentially Distributed Random Variable]].
 *Given the following definite integral:
 :$$\int_{0}^{\infty}\it x^n\cdot\rm e^{-\it a \hspace{0.03cm}\cdot \hspace{0.03cm}x}\, d{\it x} =\frac{\it n{\rm !}}{\it a^{n}}.$$
@@ Line 33: / Line 32: @@
-{Calculate the moments&nbsp; $m_k$&nbsp; of the random variable&nbsp; $x$.&nbsp; Reason that all moments with odd index are zero. <br>How big is the rms value?
+{Calculate the moments&nbsp; $m_k$&nbsp; of the random variable&nbsp; $x$.&nbsp; Reason that all moments with odd index are zero.&nbsp; How big is the rms value?
 |type="{}"}
 $\sigma_x \ = \ $ { 0.707 3% }
-{What is the value of the kurtosis of the random size $x$?
+{What is the value of the kurtosis of the random variable&nbsp; $x$?
 |type="{}"}
 $K_x \ = \ $ { 6 3% }
@@ Line 60: / Line 59: @@
-{What is the mean of the bounded and amplified random variable $y$?
+{What is the mean of the bounded and amplified random variable&nbsp; $y$?
 |type="{}"}
 $m_y \ = \ $ { 0.316 3% }

@@ Line 104: / Line 104: @@
 :$$f_y(y)=\frac{f_x(x)}{|g'(x)|}\Bigg|_{x=h(y)}=\frac{\rm e^{-\rm 2\it x}}{\rm 2}\Bigg|_{\it x={\it y}/{\rm 2}}=0.5 \cdot {\rm e^{\it -y}} .$$
-*For&nbsp; $y= 0.5$&nbsp; accordingly, the continuous PDF portion is
+*For&nbsp; $y= 0.5$&nbsp; accordingly, the continuous PDF component is
 :$$f_y(y = 0.5)\hspace{0.15cm}\underline{\approx 0.304}.$$

@@ Line 78: / Line 78: @@
 '''(2)'''&nbsp; All moments with odd index&nbsp; $k$&nbsp; are equal to zero due to symmetric PDF.
 *For even&nbsp; $k$&nbsp; the left part of the PDF can be mirrored into the right one and we get:
-:$$\it m_k=\rm 2 \cdot \int_{\rm 0}^{\infty}\hspace{-0. 15cm}\it x^{k}\cdot \rm e^{-\rm 2\it x}\,\rm d \it x=\frac{\rm 2\cdot\rm\Gamma(\it k{\rm +}\rm 1)}{\rm 2^{\it k{\rm +}\rm 1}}=\frac{\it k{\rm !}}{\rm 2^{\it k}}.$$
+:$$\it m_k=\rm 2 \cdot \int_{\rm 0}^{\infty}\hspace{-0.15cm}\it x^{k}\cdot \rm e^{-\rm 2\it x}\,\rm d \it x=\frac{\rm 2\cdot\rm\Gamma(\it k{\rm +}\rm 1)}{\rm 2^{\it k{\rm +}\rm 1}}=\frac{\it k{\rm !}}{\rm 2^{\it k}}.$$
 *From this it follows with&nbsp; $k = 2$&nbsp; considering the mean&nbsp; $m_1 = 0$:
@@ Line 102: / Line 102: @@
 '''(6)'''&nbsp; The signal range&nbsp; $0 \le x \le 0.5$&nbsp; is linearly mapped to the range&nbsp; $0 \le y \le 1$&nbsp; at the output.
 *The derivative of the characteristic curve is constantly equal to $2$&nbsp; (amplification). From this one obtains:
-:$$f_y(y)=\frac{f_x(x)}{|g'(x)|}\Bigg|_{x=h(y)}=\frac{\rm e^{-\rm 2\it x}{\rm 2}\Bigg|_{\it x={\it y}/{\rm 2}}=0.5 \cdot {\rm e^{\it -y}} .$$
+:$$f_y(y)=\frac{f_x(x)}{|g'(x)|}\Bigg|_{x=h(y)}=\frac{\rm e^{-\rm 2\it x}}{\rm 2}\Bigg|_{\it x={\it y}/{\rm 2}}=0.5 \cdot {\rm e^{\it -y}} .$$
 *For&nbsp; $y= 0.5$&nbsp; accordingly, the continuous PDF portion is
@@ Line 108: / Line 108: @@
-'''(7)'''&nbsp; For the mean value of the random gr&ouml;&aerospace;e&nbsp; $y$&nbsp; holds:
+'''(7)'''&nbsp; For the mean value of the random variable&nbsp; $y$&nbsp; holds:
-:$$m_y=\frac{1}{\rm 2\rm e} \cdot 1 +\int_{\rm 0}^{\rm 1}\frac{\it y}{\rm 2}\cdot \rm e^{\it -y}\, \rm d \it y=\frac{\rm 1}{\rm 2\rm e}{\rm +}\frac{\rm 1}{\rm 2}-\frac{\rm 1}{\rm e}=\frac{\rm 1}{\rm 2}-\frac{\rm 1}{\rm 2 e}\hspace{0. 15cm}\underline{=\rm 0.316}.$$
+:$$m_y=\frac{1}{\rm 2\rm e} \cdot 1 +\int_{\rm 0}^{\rm 1}\frac{\it y}{\rm 2}\cdot \rm e^{\it -y}\, \rm d \it y=\frac{\rm 1}{\rm 2\rm e}{\rm +}\frac{\rm 1}{\rm 2}-\frac{\rm 1}{\rm e}=\frac{\rm 1}{\rm 2}-\frac{\rm 1}{\rm 2 e}\hspace{0.15cm}\underline{=\rm 0.316}.$$
-*The first term is from the Dirac delta at&nbsp; $y= 1$, the second from the continuous PDF&ndash;fraction.
+*The first term is from the Dirac delta at&nbsp; $y= 1$, the second from the continuous PDF component.
 {{ML-Fuß}}

@@ Line 8: / Line 8: @@
 *This signal is applied to the input of a nonlinearity with the characteristic curve (see lower figure):
-:$$y=\left\{\begin{array}{*{4}{c}}0 &\rm f\ddot{u}r\hspace{0.2cm} \it x <\rm 0, \\rm2\it x & \rm f\ddot{u}r\hspace{0.2cm} \rm 0\le \it x\le \rm 0.5, \\1 & \rm f\ddot{u}r\hspace{0.2cm}\it x > \rm 0.5\\end{array}\right.$$
+:$$y=\left\{\begin{array}{*{4}{c}}0  &\rm for\hspace{0.2cm} \it x <\rm 0, \\\rm2\it x   & \rm for\hspace{0.2cm} \rm 0\le \it x\le \rm 0.5,  \\1 & \rm for\hspace{0.2cm}\it x > \rm 0.5\\\end{array}\right.$$
 *The characteristic sketched below limits the variable&nbsp; $x(t)$&nbsp; at the input asymmetrically and amplifies it in the linear range.<br><br>