Aufgaben:Exercise 1.3: Rayleigh Fading - Versionsgeschichte

Guenter am 11. Mai 2020 um 13:06 Uhr

2020-05-11T13:06:51Z

Javier am 25. März 2020 um 16:52 Uhr

2020-03-25T16:52:01Z

Javier am 25. März 2020 um 16:51 Uhr

2020-03-25T16:51:25Z

Javier am 25. März 2020 um 16:50 Uhr

2020-03-25T16:50:31Z

Javier am 25. März 2020 um 16:49 Uhr

2020-03-25T16:49:56Z

Javier am 25. März 2020 um 16:46 Uhr

2020-03-25T16:46:51Z

Javier am 25. März 2020 um 16:45 Uhr

2020-03-25T16:45:55Z

Javier am 25. März 2020 um 16:44 Uhr

2020-03-25T16:44:54Z

Javier am 25. März 2020 um 16:43 Uhr

2020-03-25T16:43:01Z

Javier am 25. März 2020 um 16:41 Uhr

2020-03-25T16:41:53Z

@@ Zeile 91: / Zeile 91: @@
 $$f_p(p) = \frac{1}{2\sigma^2} \cdot {\rm exp} [ -\frac{p}{2\sigma^2}]
   \hspace{0.05cm}.$$
-[[File:P_ID2112__Mob_A_1_3c.png|right|frame|PDF and probability regions ]]
+[[File:EN_Mob_A_1_3_c.png|right|frame|PDF and probability regions ]]
 *It follows:

@@ Zeile 107: / Zeile 107: @@
-'''(4)''&nbsp; From $10 \cdot {\rm lg} \ p_1 = \ &ndash;3 \ \ \rm dB$ follows $p_1 = 0.5$. The upper limit of the integration range results from the condition $10 \cdot {\rm lg} \ p_2 = +3 \ \ \rm dB$ to $p_2 = 2$.
+'''(4)''&nbsp; From $10 \cdot {\rm lg} \ p_1 = \ &ndash;3 \ \ \rm dB$ follows $p_1 = 0.5$. The upper limit of the integration range results from the condition $10 \cdot {\rm lg} \ p_2 = +3 \ \ \rm dB$, so $p_2 = 2$.
 *This gives, according to the above graph:
 $${\rm Pr}(-3\,\,{\rm dB}\le 10 \cdot {\rm lg}\hspace{0.15cm}p(t) \le +3\,\,{\rm dB}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}   \int_{0.5}^{2}f_p(p)\hspace{0.15cm}{\rm d}p =

@@ Zeile 107: / Zeile 107: @@
-'''(4)''&nbsp; From $10 \cdot {\rm lg} \ p_1 = \ &ndash;3 \ \ \rm dB$ follows $p_1 = 0.5$ and the upper limit of the integration range results from the condition $10 \cdot {\rm lg} \ p_2 = +3 \ \ \rm dB$ to $p_2 = 2$.
+'''(4)''&nbsp; From $10 \cdot {\rm lg} \ p_1 = \ &ndash;3 \ \ \rm dB$ follows $p_1 = 0.5$. The upper limit of the integration range results from the condition $10 \cdot {\rm lg} \ p_2 = +3 \ \ \rm dB$ to $p_2 = 2$.
-*This gives you, according to the above graphic:
+*This gives, according to the above graph:
 $${\rm Pr}(-3\,\,{\rm dB}\le 10 \cdot {\rm lg}\hspace{0.15cm}p(t) \le +3\,\,{\rm dB}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}   \int_{0.5}^{2}f_p(p)\hspace{0.15cm}{\rm d}p =
    \left [ - {\rm e}^{ -{p}/(2\sigma^2)}\hspace{0.15cm} \right ]_{0.5}^{2} ={\rm e}^{-0.25}- {\rm e}^{-1} \approx 0.779 - 0.368 \hspace{0.15cm} \underline{ = 0.411} \hspace{0.05cm}.$$

@@ Zeile 91: / Zeile 91: @@
 $$f_p(p) = \frac{1}{2\sigma^2} \cdot {\rm exp} [ -\frac{p}{2\sigma^2}]
   \hspace{0.05cm}.$$
-[[File:P_ID2112__Mob_A_1_3c.png|right|frame|WDF and probability regions ]]
+[[File:P_ID2112__Mob_A_1_3c.png|right|frame|PDF and probability regions ]]
 *It follows:

@@ Zeile 70: / Zeile 70: @@
 ===Sample solution===
 {{ML-Kopf}}
-'''(1)''&nbsp; From ${\rm Max}[a(t)] = 2$ follows directly:
+'''(1)'''&nbsp; From ${\rm Max}[a(t)] = 2$ follows directly:
-$${\rm Max} \left [ 20 \cdot {\rm lg}\hspace{0.15cm}a(t) \right ] = 20 \cdot {\rm lg}\hspace{0.15cm}(2) \hspace{0.15cm} \underline{\approx 6\,\,{\rm dB}}
+:$${\rm Max} \left [ 20 \cdot {\rm lg}\hspace{0.15cm}a(t) \right ] = 20 \cdot {\rm lg}\hspace{0.15cm}(2) \hspace{0.15cm} \underline{\approx 6\,\,{\rm dB}}
    \hspace{0.05cm}.$$
 '''(2)'''&nbsp; The maximum value of the square $p(t) = a(t)^2$ is
-$$${\rm Max} \left [ p(t) \right ] = {\rm Max} \left [ a(t)^2 \right ] \hspace{0.15cm} \underline{\4}
+$${\rm Max} \left [ p(t) \right ] = {\rm Max} \left [ a(t)^2 \right ] \hspace{0.15cm} \underline{=4}
    \hspace{0.05cm}.$$
-*The logarithmic representation of the square of the amount $p(t)$ is identical to the logarithmic representation of the amount $a(t)$. Since $p(t)$ is a power quantity
+*The logarithmic representation of the squared magnitude $p(t)$ is identical to the logarithmic representation of the amount $a(t)$. Since $p(t)$ is a power quantity
-$$10 \cdot {\rm lg}\hspace{0.15cm} p(t) = 10 \cdot {\rm lg}\hspace{0.15cm}a(t)^2 = 20 \cdot {\rm lg}\hspace{0.15cm} a(t)
+:$${\rm Max} \left [  p(t) \right ] = {\rm Max} \left [  a(t)^2 \right ]  \hspace{0.15cm} \underline{= 4}
    \hspace{0.05cm}.$$
@@ Zeile 88: / Zeile 88: @@
 '''(3)'''&nbsp; The condition $a(t) &#8804; 1$ is equivalent to the requirement $p(t) = a(t)^2 &#8804; 1$.
-*The absolute square is known to be exponentially distributed, and for $p &#8805; 0$ applies accordingly:
+*The absolute square is known to be exponentially distributed, and for $p &#8805; 0$ we have
 $$f_p(p) = \frac{1}{2\sigma^2} \cdot {\rm exp} [ -\frac{p}{2\sigma^2}]
   \hspace{0.05cm}.$$
@@ Zeile 100: / Zeile 100: @@
   \underline{\sigma = 1} \hspace{0.05cm}.$$
-The graphic shows
+The graph shows
-* left the probability&nbsp; ${\rm Pr}(p(t) &#8804; 1)$,
+* on the left side the probability&nbsp; ${\rm Pr}(p(t) &#8804; 1)$,
-* right the probability&nbsp; ${\rm Pr}(0.5 \le p(t) &#8804; 2)$.
+* on the right side the probability&nbsp; ${\rm Pr}(0.5 \le p(t) &#8804; 2)$.

@@ Zeile 63: / Zeile 63: @@
 $\sigma \ = \ $ { 1 3% }
-{What is the probability that the logarithmic Rayleigh coefficient &nbsp; &#8658; &nbsp; $10 \cdot {\rm lg} \ p(t)$&nbsp; is in the range between &nbsp; $\text{-3 dB}$&nbsp; and&nbsp; $\text{+3 dB}$?
+{What is the probability that the logarithmic Rayleigh coefficient &nbsp;  $10 \cdot {\rm lg} \ p(t)$&nbsp; is between &nbsp; $\text{-3 dB}$&nbsp; and&nbsp; $\text{+3 dB}$?
 |type="{}"}
 ${\rm Pr}(|10 \cdot {\rm lg} \ p(t)| < 3 \ \rm dB) \ = \ $ { 0.411 3% }

@@ Zeile 56: / Zeile 56: @@
 {What is the maximum value of&nbsp; $p(t) = |z(t)|^2$&nbsp; both in linear and logarithmic representation?
 |type="{}"}
-${\rm Max}\big[p(t)\big] \ = \ $ {\ $ 4 3% }
+${\rm Max}\big[p(t)\big] \ = \  $ { 4 3% }
-${\rm Max}\big [10 \cdot {\rm lg} \ p(t)\big] \ = \ $ { 6 3% } $ \ \rm dB$
+${\rm Max}\big[10 \cdot {\rm lg} \ p(t)\big] \ = \  $ { 6 3% } $ \ \rm dB$
-{Let &nbsp; ${\rm Pr}\big[a(t) &#8804; 1\big] = $0.394 Determine the Rayleigh&ndash;parameter&nbsp; $\sigma$.
+{Let &nbsp; ${\rm Pr}\big[a(t) &#8804; 1\big] = $0.394 Determine the Rayleigh parameter&nbsp; $\sigma$.
 |type="{}"}
 $\sigma \ = \ $ { 1 3% }

@@ Zeile 50: / Zeile 50: @@
 <quiz display=simple>
-{For the entire range, the amount function&nbsp; $a(t) &#8804; 2$ applies. What is the maximum value for the logarithmic quantity in this range?
+{For the entire range, we have&nbsp; $a(t) &#8804; 2$. What is the maximum value of the logarithmic quantity in this range?
 |type="{}"}
 ${\rm Max}\big [20 \cdot {\rm lg} \ {a(t)}\big] \ = \ $ { 6 3% } $\ \rm dB$
-{What is the maximum value for&nbsp; $p(t) = |z(t)|^2$&nbsp; both in linear and logarithmic representation?
+{What is the maximum value of&nbsp; $p(t) = |z(t)|^2$&nbsp; both in linear and logarithmic representation?
 |type="{}"}
 ${\rm Max}\big[p(t)\big] \ = \ $ {\ $ 4 3% }

@@ Zeile 37: / Zeile 37: @@
   \hspace{0.05cm}.$$
-In the lower graph, the value range between&nbsp; $\text{-3 dB}$&nbsp; and&nbsp; $\text{+3 dB}$&nbsp; regarding the logarithmic Rayleigh&ndash;Size&nbsp; $20 \cdot {\rm lg} \ a(t)$ is highlighted in green. The subtask '''(4)''' refers to this.
+In the lower graph, the value range between&nbsp; $\text{-3 dB}$&nbsp; and&nbsp; $\text{+3 dB}$&nbsp; of the logarithmic Rayleigh coefficient&nbsp; $20 \cdot {\rm lg} \ a(t)$ is highlighted in green. The subtask '''(4)''' refers to this.
 ''Notes:''
-* The task belongs to chapter&nbsp; [[Mobile_Communications/Probability Density_of_Rayleigh%E2%80%93Fadings|Probability Density of Rayleigh&ndash;Fadings]]&nbsp; of this book.
+* This task belongs to chapter&nbsp; [[Mobile_Kommunikation/Wahrscheinlichkeitsdichte_des_Rayleigh%E2%80%93Fadings|Wahrscheinlichkeitsdichte des Rayleigh&ndash;Fadings]]&nbsp; of this book.
-* A similar topic is treated with a different approach in chapter&nbsp; [[Stochastic_Signal Theory/Weitere_Verteilungen|Weitere Verteilungen]]&nbsp; of the book &bdquo;Stochastic Signal Theory&rdquo;.
+* A similar topic is treated with a different approach in chapter&nbsp; [[Stochastische_Signaltheorie/Weitere_Verteilungen|Weitere Verteilungen]]&nbsp; of the book &bdquo;Stochastic Signal Theory&rdquo;.
-* To check your results you can use the interactive applet&nbsp; [[Applets:WDF_VTF|WDF, VTF and Moments]]&nbsp; of the book &bdquo;Stochastic Signal Theory&rdquo;.
+* To check your results, you can use the interactive applet&nbsp; [[Applets:WDF_VTF|WDF, VTF and Moments]]&nbsp; of the book &bdquo;Stochastic Signal Theory&rdquo;.