Aufgaben:Exercise 4.09Z: Laplace Distributed Noise - Revision history

Guenter at 15:15, 6 September 2022

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Guenter at 10:12, 23 June 2022

2022-06-23T10:12:17Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.09Z: Laplace-verteiltes Rauschen to Aufgaben:Exercise 4.09Z: Laplace Distributed Noise

2022-06-23T10:04:04Z

Guenter moved page Aufgabe 4.09Z: Laplace-verteiltes Rauschen to Exercise 4.09Z: Laplace Distributed Noise

@@ Line 71: / Line 71: @@
 *Thus,&nbsp; the variance&nbsp; $\sigma^2$&nbsp; is actually equal to the second moment,&nbsp; as already stated in the question:
-[[File:P_ID2045__Dig_Z_4_9c.png|right|frame|Contour lines of the two-dimensional Laplace PDF]]
+[[File:EN_Dig_Z_4_9ML.png|right|frame|Contour lines of the two-dimensional Laplace PDF]]
 :$$\sigma^2 = {\rm E}[n_1^2] = 2 \cdot \frac{a}{2} \cdot \int_{0}^{\infty} x^2 \cdot {\rm e}^{-a \hspace{0.03cm} \cdot \hspace{0.03cm} x} \,{\rm d} x  = a \cdot {2}/{a^3}=
   {2}/{a^2} \hspace{0.05cm}. \hspace{0.2cm}{\rm With}\hspace{0.15cm}a = 1\text{:} \hspace{0.2cm}\hspace{0.1cm}\underline {\sigma^2 = 2 }\hspace{0.05cm}.$$

@@ Line 71: / Line 71: @@
 *Thus,&nbsp; the variance&nbsp; $\sigma^2$&nbsp; is actually equal to the second moment,&nbsp; as already stated in the question:
-[[File:P_ID2045__Dig_Z_4_9c.png|right|frame|Contour lines of the two-dimensional Laplace PDF '''KORREKTUR: zu''']]
+[[File:P_ID2045__Dig_Z_4_9c.png|right|frame|Contour lines of the two-dimensional Laplace PDF]]
 :$$\sigma^2 = {\rm E}[n_1^2] = 2 \cdot \frac{a}{2} \cdot \int_{0}^{\infty} x^2 \cdot {\rm e}^{-a \hspace{0.03cm} \cdot \hspace{0.03cm} x} \,{\rm d} x  = a \cdot {2}/{a^3}=
   {2}/{a^2} \hspace{0.05cm}. \hspace{0.2cm}{\rm With}\hspace{0.15cm}a = 1\text{:} \hspace{0.2cm}\hspace{0.1cm}\underline {\sigma^2 = 2 }\hspace{0.05cm}.$$

@@ Line 73: / Line 73: @@
 [[File:P_ID2045__Dig_Z_4_9c.png|right|frame|Contour lines of the two-dimensional Laplace PDF '''KORREKTUR: zu''']]
 :$$\sigma^2 = {\rm E}[n_1^2] = 2 \cdot \frac{a}{2} \cdot \int_{0}^{\infty} x^2 \cdot {\rm e}^{-a \hspace{0.03cm} \cdot \hspace{0.03cm} x} \,{\rm d} x  = a \cdot {2}/{a^3}=
-  {2}/{a^2} \hspace{0.05cm}. \hspace{0.2cm}{\rm Mit}\hspace{0.15cm}a = 1\text{:} \hspace{0.2cm}\hspace{0.1cm}\underline {\sigma^2 = 2 }\hspace{0.05cm}.$$
+  {2}/{a^2} \hspace{0.05cm}. \hspace{0.2cm}{\rm With}\hspace{0.15cm}a = 1\text{:} \hspace{0.2cm}\hspace{0.1cm}\underline {\sigma^2 = 2 }\hspace{0.05cm}.$$

@@ Line 71: / Line 71: @@
 *Thus,&nbsp; the variance&nbsp; $\sigma^2$&nbsp; is actually equal to the second moment,&nbsp; as already stated in the question:
-[[File:P_ID2045__Dig_Z_4_9c.png|right|frame|Contour lines of the two-dimensional Laplace PDF]]
+[[File:P_ID2045__Dig_Z_4_9c.png|right|frame|Contour lines of the two-dimensional Laplace PDF '''KORREKTUR: zu''']]
 :$$\sigma^2 = {\rm E}[n_1^2] = 2 \cdot \frac{a}{2} \cdot \int_{0}^{\infty} x^2 \cdot {\rm e}^{-a \hspace{0.03cm} \cdot \hspace{0.03cm} x} \,{\rm d} x  = a \cdot {2}/{a^3}=
   {2}/{a^2} \hspace{0.05cm}. \hspace{0.2cm}{\rm Mit}\hspace{0.15cm}a = 1\text{:} \hspace{0.2cm}\hspace{0.1cm}\underline {\sigma^2 = 2 }\hspace{0.05cm}.$$

@@ Line 60: / Line 60: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; <u>Solution 2</u> is correct:
+'''(1)'''&nbsp; <u>Solution 2</u>&nbsp; is correct:
-*The area under the PDF must equal $1$:
+*The area under the PDF must equal&nbsp; $1$:
 :$$\int_{-\infty}^{+\infty}  p_{n_1}(x) \,{\rm d} x  = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
   \int_{0}^{+\infty}  p_{n_1}(x) \,{\rm d} x  = 0.5  \hspace{0.3cm}
@@ Line 68: / Line 68: @@
-'''(2)'''&nbsp; The <u>linear mean</u> is <u>equal to 0</u> due to PDF symmetry.
+'''(2)'''&nbsp; The&nbsp; <u>linear mean</u>&nbsp; is&nbsp; <u>equal to 0</u>&nbsp; due to PDF symmetry.
-Thus, the variance $\sigma^2$ is actually equal to the root mean square, as already stated in the question:
+*Thus,&nbsp; the variance&nbsp; $\sigma^2$&nbsp; is actually equal to the second moment,&nbsp; as already stated in the question:
 :$$\sigma^2 = {\rm E}[n_1^2] = 2 \cdot \frac{a}{2} \cdot \int_{0}^{\infty} x^2 \cdot {\rm e}^{-a \hspace{0.03cm} \cdot \hspace{0.03cm} x} \,{\rm d} x  = a \cdot {2}/{a^3}=
   {2}/{a^2} \hspace{0.05cm}. \hspace{0.2cm}{\rm Mit}\hspace{0.15cm}a = 1\text{:} \hspace{0.2cm}\hspace{0.1cm}\underline {\sigma^2 = 2 }\hspace{0.05cm}.$$
-[[File:P_ID2045__Dig_Z_4_9c.png|right|frame|Contour lines of the 2D Laplace distribution]]
+'''(3)'''&nbsp; <u>Solution 1</u>&nbsp; is correct:
-'''(3)'''&nbsp; <u>Solution 1</u> is correct:
+*In the first quadrant&nbsp; $(x &#8805; 0, y &#8805; 0)$,&nbsp; the magnitude formation can be omitted.&nbsp; Then the two-dimensional PDF is given by:
 :$$\boldsymbol{ p }_{\boldsymbol{ n }}  (x,\hspace{0.15cm} y) = {a^2}/{4} \cdot {\rm e}^{- a \hspace{0.03cm}\cdot \hspace{0.03cm}x} \cdot {\rm e}^{- a \hspace{0.03cm}\cdot \hspace{0.03cm}y }= {a^2}/{4} \cdot {\rm e}^{- a \hspace{0.03cm}\cdot \hspace{0.03cm}(x+y)}\hspace{0.05cm}.$$
-*A contour line with factor $\beta$ versus maximum then has the following shape ($0 < \beta < 1$):
+*A contour line with factor&nbsp; $\beta$&nbsp; versus maximum then has the following shape&nbsp; $(0 < \beta < 1)$:
 :$${\rm e}^{- a \hspace{0.03cm}\cdot \hspace{0.03cm}(x+y)} = \beta
   \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x + y = \frac{{\rm ln}\hspace{0.15cm}1/\beta}{a}
    \hspace{0.05cm}.$$
-*The graph shows the contour lines for $a = 1$ and some values of $\beta$, each giving a square rotated by $45^\circ$ &nbsp;&#8658;&nbsp; the contour lines are thus even.
+*The graph shows the contour lines for&nbsp; $a = 1$&nbsp; and some values of&nbsp; $\beta$,&nbsp; each giving a square rotated by&nbsp; $45^\circ$ &nbsp; &#8658; &nbsp; the contour lines are thus even.
-'''(4)'''&nbsp; The probability event considered here corresponds exactly to the third quadrant of the composite PDF sketched above. Because of symmetry, this probability is:
+'''(4)'''&nbsp; The probability event considered here corresponds exactly to the third quadrant of the composite PDF sketched above.
 :$${\rm Pr}[(n_1 < 0) ∩ (n_2 < 0)]\hspace{0.15cm}\underline {=25\%}.$$
-'''(5)'''&nbsp; For this, the composite PDF can be written:
+'''(5)'''&nbsp; For this,&nbsp; the composite PDF can be written:
 :$${\rm Pr} \left [ (n_1 > 1)\cap (n_2 > 1)\right ] = {1}/{4} \cdot \int_{1}^{\infty} \int_{1}^{\infty}{\rm e}^{-  (x+y)} \,{\rm d} x  \,{\rm d} y == {1}/{2} \cdot \int_{1}^{\infty} {\rm e}^{-  x} \,{\rm d} x  \hspace{0.15cm} \cdot \hspace{0.15cm}
    {1}/{2} \cdot \int_{1}^{\infty} {\rm e}^{-  y} \,{\rm d} y $$
 :$$ \Rightarrow \hspace{0.3cm}{\rm Pr} \left [ (n_1 > 1)\cap (n_2 > 1)\right ] =  \left [ {\rm Pr} (n_1 > 1)\right ] \cdot \left [ {\rm Pr} (n_2 > 1)\right ]\hspace{0.05cm}. $$
-Considered is the statistical independence between $n_1$ and $n_2$ and the equality $p_{\it n1}(x) = p_{\it n2}(y)$.  For $a = 1$ holds:
+*Considered is the statistical independence between&nbsp; $n_1$&nbsp; and&nbsp; $n_2$&nbsp; and the equality&nbsp; $p_{\it n1}(x) = p_{\it n2}(y)$.&nbsp;  For $a = 1$&nbsp; holds:
 :$${\rm Pr} (n_1 > 1) = {1}/{2} \cdot \int_{1}^{\infty} {\rm e}^{-  x} \,{\rm d} x = {1}/({2{\rm e}})\approx 0.184\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm} {\rm Pr} \left [ (n_1 > 1)\cap (n_2 > 1)\right ] = {1}/({4{\rm e}^2)}\hspace{0.1cm}\hspace{0.15cm}\underline {\approx 3.4\%}\hspace{0.05cm}.$$
@@ Line 106: / Line 107: @@
 [[File:P_ID2046__Dig_Z_4_9f.png|right|frame|Division of the integration domain]]
 '''(6)'''&nbsp; The region considered here is highlighted in color in the following graph.
-*However, the regions extend to the right and above to infinity.
+*However,&nbsp; the regions extend to the right and above to infinity.
 *The searched probability results in
 :$${\rm Pr}  [ n_1 \hspace{-0.2cm} \ + \ \hspace{-0.2cm} n_2 > 2 ]  =\frac{1}{4} \cdot \int\limits_{-\infty}^{+\infty} {\rm e}^{-|x|} \int\limits_{2-x}^{\infty}{\rm e}^{-|y|} \,{\rm d} y  \,{\rm d} x = I_1 + I_2 + I_3 + I_4 \hspace{0.05cm}.$$
-Because of the magnitude, a splitting into partial integrals has to be done. Upwards and to the right all areas extend to infinity. Because of the symmetry $I_4 = I_3$ is valid.
+*Because of the magnitude,&nbsp; a splitting into partial integrals has to be done.
 :$$I_1 = {1}/{4} \cdot \int_{2}^{+\infty} \hspace{-0.15cm}{\rm e}^{-x} \int_{0}^{\infty}\hspace{-0.15cm}{\rm e}^{-y} \,{\rm d} y  \,{\rm d} x =
   {1}/{4} \cdot \int_{2}^{+\infty} {\rm e}^{-x}  \,{\rm d} x ={1}/({4{\rm e}^2})\hspace{0.05cm},$$
@@ Line 125: / Line 131: @@
    {1}/({8{\rm e}^2}) = I_3\hspace{0.05cm}.$$
-Thus, the overall result is:
+*Thus,&nbsp; the overall result is:
 :$${\rm Pr} \left [ n_1 + n_2 > 2 \right ] = {\rm e}^{-2} \cdot ({1}/{4} +{1}/{2} +{1}/{8} +{1}/{8})= {\rm e}^{-2} \hspace{0.1cm}\hspace{0.15cm}\underline {\approx 13.5\%}\hspace{0.05cm}.$$
 {{ML-Fuß}}

@@ Line 29: / Line 29: @@
 <quiz display=simple>
 {What is the size of the constant&nbsp; $K$&nbsp; of the one-dimensional PDF?
-|type="[]"}
+|type="()"}
 - $K = 1$.
 + $K = a/2$

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Digital_Signal_Transmission/Approximation_of_the_Error_Probability}}
-[[File:P_ID2042__Dig_Z_4_9.png|right|frame|2D Laplacian PDF]]
+[[File:P_ID2042__Dig_Z_4_9.png|right|frame|Two-dimensional Laplacian PDF]]
 We consider two-dimensional noise&nbsp; $\boldsymbol{n} = (n_1, n_2)$.
-The two noise variables are "<i>independent and identically distributed</i>&nbsp;", abbreviated i.i.d., and each has a Laplace probability density:
+The two noise variables are&nbsp; "independent and identically distributed",&nbsp; abbreviated&nbsp; "i.i.d.",&nbsp; and each has a Laplace probability density function&nbsp; $\rm (PDF)$:
 :$$p_{n_1}(x)  \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  K \cdot {\rm e}^{- a \hspace{0.03cm}\cdot \hspace{0.03cm} |x|}  \hspace{0.05cm},$$
 :$$ p_{n_2}(y)  \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  K \cdot {\rm e}^{- a \hspace{0.03cm}\cdot \hspace{0.03cm} |y|}  \hspace{0.05cm}. $$
-*The 2D probability density function&nbsp; $p_{\it \boldsymbol{n}}(x, y)$&nbsp; is shown in the graph.
+*The two-dimensional probability density function&nbsp; $p_{\it \boldsymbol{n}}(x, y)$&nbsp; is shown in the graph.
-''Notes:''
-* The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability|"Approximation of the Error Probability"]].
+Notes:
-* We would like to refer to the interactive applet &nbsp;[[Applets:2D_Laplace|"2D Laplace"]].&nbsp;
+* The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability|"Approximation of the Error Probability"]].
 * The integral resulting in subtask '''(6)''' must be split into several partial integrals due to the magnitude formation.
-*Furthermore, it holds:
+* We would like to refer to the&nbsp; (German language)&nbsp; interactive SWF applet &nbsp;[[Applets:2D_Laplace|"2D Laplace"]].&nbsp;
-:$$\int_{0}^{\infty} x^2 \cdot {\rm e}^{-a \hspace{0.03cm}\cdot \hspace{0.03cm} x} \,{\rm d} x  = {2}/{a^3} \hspace{0.05cm}.$$
@@ Line 25: / Line 28: @@
 ===Questions===
 <quiz display=simple>
-{What is the size of the constant&nbsp; $K$&nbsp; of the 1D PDF?
+{What is the size of the constant&nbsp; $K$&nbsp; of the one-dimensional PDF?
 |type="[]"}
 - $K = 1$.
@@ Line 31: / Line 34: @@
 - $K = 1/a$.
-{Let&nbsp; $a = 1$. What are the mean&nbsp; ${\rm E}\big[n_i \big]$&nbsp; and variance&nbsp; $\sigma^2 = {\rm E}\big[n_i^2\big]$&nbsp; of the two 1D random variables? &nbsp;$(i = 1, 2)$
+{Let&nbsp; $a = 1$.&nbsp; What are the mean&nbsp; ${\rm E}\big[n_i \big]$&nbsp; and the variance&nbsp; $\sigma^2 = {\rm E}\big[n_i^2\big]$&nbsp; of the two one-dimensional random variables? &nbsp;$(i = 1,\ 2)$.
 |type="{}"}
 ${\rm E}\big[n_i\big] \ = \ $ { 0. }
 ${\rm E}\big[n_i^2\big] \ = \ ${ 2 3% }
-{What is the shape of the contour lines of the 2D PDF in the first quadrant?
+{What is the shape of the contour lines of the two-dimensional PDF in the first quadrant?
-|type="[]"}
+|type="()"}
 + They are straight lines.
 - They are hyperbolas.
 - They are circles.
-{Let further&nbsp; $a = 1$. What is the probability that both&nbsp; $n_1$&nbsp; and&nbsp; $n_2$&nbsp; are negative?
+{Let further&nbsp; $a = 1$.&nbsp; What is the probability that both&nbsp; $n_1$&nbsp; and&nbsp; $n_2$&nbsp; are negative?
 |type="{}"}
 ${\rm Pr}\big[(n_1 < 0) ∩ (n_2 < 0)\big]\ = \ $ { 25 3% } $\ \%$
@@ Line 50: / Line 53: @@
 ${\rm Pr}\big[(n_1 > 1) ∩ (n_2 > 1)\big]\ = \ $ { 3.4 3% } $\ \%$
-{What is the probability that the sum is&nbsp; $n_1 + n_2 > 2$?&nbsp;
+{What is the probability that the sum&nbsp; $n_1 + n_2 > 2$?&nbsp;
 |type="{}"}
 $ {\rm Pr}\big[n_1 + n_2 > 2)\big] \ = \ $ { 13.5 3% } $\ \%$

← Older revision		Revision as of 10:12, 23 June 2022
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