Aufgaben:Exercise 1.2: Lognormal Channel Model: Unterschied zwischen den Versionen
Aus LNTwww
Zeile 6: | Zeile 6: | ||
Thus the total path loss can be described by the following equation: | Thus the total path loss can be described by the following equation: | ||
− | $$V_{\rm P} = V_{\rm 0} + V_{\rm S} \hspace{0.05cm}.$$ | + | :$$V_{\rm P} = V_{\rm 0} + V_{\rm S} \hspace{0.05cm}.$$ |
*$V_0$ takes into account the distance-dependent path loss which is assumed to be constant: $V_0 = 80 \ \rm dB$ . | *$V_0$ takes into account the distance-dependent path loss which is assumed to be constant: $V_0 = 80 \ \rm dB$ . | ||
*The loss $V_{\rm S}$ is due to shadowing caused by the lognormal–distribution with the probability density function (PDF) | *The loss $V_{\rm S}$ is due to shadowing caused by the lognormal–distribution with the probability density function (PDF) | ||
− | :$$f_{ | + | :$$f_{V_{\rm S}}(V_{\rm S}) = \frac {1}{ \sqrt{2 \pi }\cdot \sigma_{\rm S}} \cdot {\rm e }^{ - { (V_{\rm S}\hspace{0.05cm}- \hspace{0.05cm}m_{\rm S})^2}/(2 \hspace{0.05cm}\cdot \hspace{0.05cm}\sigma_{\rm S}^2) }$$ |
: see diagram. The following numerical values apply: | : see diagram. The following numerical values apply: | ||
− | $$m_{\rm S} = 20\,\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm} \sigma_{\rm S} = 10\,\,{\rm dB}\hspace{0.15cm}{\rm or }\hspace{0.15cm}\sigma_{\rm S} = 0\,\,{\rm dB}\hspace{0.15cm}{\rm (subtask\hspace{0.15cm} 2)}\hspace{0.05cm}.$$ | + | :$$m_{\rm S} = 20\,\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm} \sigma_{\rm S} = 10\,\,{\rm dB}\hspace{0.15cm}{\rm or }\hspace{0.15cm}\sigma_{\rm S} = 0\,\,{\rm dB}\hspace{0.15cm}{\rm (subtask\hspace{0.15cm} 2)}\hspace{0.05cm}.$$ |
Also make the following simple assumptions: | Also make the following simple assumptions: | ||
Zeile 25: | Zeile 25: | ||
Notes:'' | Notes:'' | ||
− | * This task belongs to the chapter [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3% | + | * This task belongs to the chapter [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung_X|Distanzabhängige Dämpfung und Abschattung]]. |
* You can use the following (rough) approximations for the complementary Gaussian error integral: | * You can use the following (rough) approximations for the complementary Gaussian error integral: | ||
:$${\rm Q}(1) \approx 0.16\hspace{0.05cm},\hspace{0.2cm} {\rm Q}(2) \approx 0.02\hspace{0.05cm},\hspace{0.2cm} | :$${\rm Q}(1) \approx 0.16\hspace{0.05cm},\hspace{0.2cm} {\rm Q}(2) \approx 0.02\hspace{0.05cm},\hspace{0.2cm} | ||
{\rm Q}(3) \approx 10^{-3}\hspace{0.05cm}.$$ | {\rm Q}(3) \approx 10^{-3}\hspace{0.05cm}.$$ | ||
− | * Or use the interaction module provided by $\rm LNTwww$ [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen_(neues_Applet)|Komplementäre Gaußsche Fehlerfunktionen]]. | + | * Or use the interaction module provided by $\rm LNTwww$ [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen_(neues_Applet)_X|Komplementäre Gaußsche Fehlerfunktionen]]. |
Zeile 59: | Zeile 59: | ||
'''(1)''' The correct answer is <u>YES</u>: | '''(1)''' The correct answer is <u>YES</u>: | ||
*From the $\rm dB$–value $V_0 = 80 \ \rm dB$ follows the absolute (linear) value $K_0 = 10^8$. Thus the received power is | *From the $\rm dB$–value $V_0 = 80 \ \rm dB$ follows the absolute (linear) value $K_0 = 10^8$. Thus the received power is | ||
− | $$P_{\rm E} = P_{\rm S}/K_0 = 10 \ {\rm W}/10^8 = 100 \ {\rm nW} > 10 \ \ \rm pW.$$ | + | :$$P_{\rm E} = P_{\rm S}/K_0 = 10 \ {\rm W}/10^8 = 100 \ {\rm nW} > 10 \ \ \rm pW.$$ |
*You can also solve this problem directly with the logarithmic quantities: | *You can also solve this problem directly with the logarithmic quantities: | ||
Zeile 86: | Zeile 86: | ||
*The probability density $f_{\rm VS}(V_{\rm S})$ of the path loss due to shadowing (Longnormal–Fading) is shown here. | *The probability density $f_{\rm VS}(V_{\rm S})$ of the path loss due to shadowing (Longnormal–Fading) is shown here. | ||
− | *The probability that the system will fail is marked in red | + | *The probability that the system will fail is marked in red. |
<br clear=all> | <br clear=all> | ||
'''(4)''' From the availability probability $99.9 \%$ follows the failure probability $10^{\rm –3} \approx \ {\rm Q}(3)$. | '''(4)''' From the availability probability $99.9 \%$ follows the failure probability $10^{\rm –3} \approx \ {\rm Q}(3)$. | ||
*If the distance-dependent path loss $V_0$ is reduced by $10 \ \ \rm dB$ to $\underline {70 \ \rm dB}$, a failure will only occur when $V_{\rm S} ≥ 50 \ \ \rm dB$. | *If the distance-dependent path loss $V_0$ is reduced by $10 \ \ \rm dB$ to $\underline {70 \ \rm dB}$, a failure will only occur when $V_{\rm S} ≥ 50 \ \ \rm dB$. | ||
*This would achieve exactly the required reliability, as the following calculation shows: | *This would achieve exactly the required reliability, as the following calculation shows: | ||
− | $${\rm Pr}({\rm "System\hspace{0.15cm} does\hspace{0.15cm} not\hspace{0.15cm} work\hspace{0.15cm}"})= | + | :$${\rm Pr}({\rm "System\hspace{0.15cm} does\hspace{0.15cm} not\hspace{0.15cm} work\hspace{0.15cm}"})= |
{\rm Q}\left ( \frac{120-70-20}{10}\right ) | {\rm Q}\left ( \frac{120-70-20}{10}\right ) | ||
= {\rm Q}(3) \approx 0.001 \hspace{0.05cm}.$$ | = {\rm Q}(3) \approx 0.001 \hspace{0.05cm}.$$ |
Version vom 13. Mai 2020, 12:56 Uhr
We consider a mobile radio cell in an urban area and a vehicle that is approximately at a fixed distance $d_0$ from the base station. For example, it moves on an arc around the base station.
Thus the total path loss can be described by the following equation:
- $$V_{\rm P} = V_{\rm 0} + V_{\rm S} \hspace{0.05cm}.$$
- $V_0$ takes into account the distance-dependent path loss which is assumed to be constant: $V_0 = 80 \ \rm dB$ .
- The loss $V_{\rm S}$ is due to shadowing caused by the lognormal–distribution with the probability density function (PDF)
- $$f_{V_{\rm S}}(V_{\rm S}) = \frac {1}{ \sqrt{2 \pi }\cdot \sigma_{\rm S}} \cdot {\rm e }^{ - { (V_{\rm S}\hspace{0.05cm}- \hspace{0.05cm}m_{\rm S})^2}/(2 \hspace{0.05cm}\cdot \hspace{0.05cm}\sigma_{\rm S}^2) }$$
- see diagram. The following numerical values apply:
- $$m_{\rm S} = 20\,\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm} \sigma_{\rm S} = 10\,\,{\rm dB}\hspace{0.15cm}{\rm or }\hspace{0.15cm}\sigma_{\rm S} = 0\,\,{\rm dB}\hspace{0.15cm}{\rm (subtask\hspace{0.15cm} 2)}\hspace{0.05cm}.$$
Also make the following simple assumptions:
- The transmit power is $P_{\rm S} = 10 \ \rm W$ (or $40 \ \rm dBm$).
- The receive power should be at least $P_{\rm E} = 10 \ \rm pW$ (or $–80 \ \rm dBm$)
Notes:
- This task belongs to the chapter Distanzabhängige Dämpfung und Abschattung.
- You can use the following (rough) approximations for the complementary Gaussian error integral:
- $${\rm Q}(1) \approx 0.16\hspace{0.05cm},\hspace{0.2cm} {\rm Q}(2) \approx 0.02\hspace{0.05cm},\hspace{0.2cm} {\rm Q}(3) \approx 10^{-3}\hspace{0.05cm}.$$
- Or use the interaction module provided by $\rm LNTwww$ Komplementäre Gaußsche Fehlerfunktionen.
Questionnaire
Sample solution
(1) The correct answer is YES:
- From the $\rm dB$–value $V_0 = 80 \ \rm dB$ follows the absolute (linear) value $K_0 = 10^8$. Thus the received power is
- $$P_{\rm E} = P_{\rm S}/K_0 = 10 \ {\rm W}/10^8 = 100 \ {\rm nW} > 10 \ \ \rm pW.$$
- You can also solve this problem directly with the logarithmic quantities:
- $$10 \cdot {\rm lg}\hspace{0.15cm} \frac{P_{\rm E}}{1\,\,{\rm mW}} = 10 \cdot {\rm lg}\hspace{0.15cm} \frac{P_{\rm S}}{1\,\,{\rm mW}} - V_0 = 40\,{\rm dBm} -80\,\,{\rm dB} = -40\,\,{\rm dBm} \hspace{0.05cm}.$$
- Only the limit value $–80 \ \rm dBm$ is required.
(2) Lognormal–Fading with $\sigma_{\rm S} = 0 \ \rm dB$ is equivalent to a constant receive power $P_{\rm E}$.
- Compared to the subtask '(1) this is $m_{\rm S} = 20 \ \ \rm dB$ smaller ⇒ $P_{\rm E} = \ –60 \ \ \rm dBm$.
- But it is still greater than the specified limit value ($–80 \ \rm dBm$).
- It follows: The system is (almost) 100% functional. „Almost” because with a Gaussian random quantity there is always a (small) residual uncertainty.
(3) The receive power is too low (less than $–80 \ \rm dBm$) if the power loss due to the lognormal–term is $40 \ \rm dB$ or more.
- The variable portion $V_{\rm S}$ must therefore not be greater than $20 \ \rm dB$.
- So it follows:
- $${\rm Pr}({\rm "System\hspace{0.15cm}does\hspace{0.15cm}not\hspace{0.15cm}work"})= {\rm Q}\left ( \frac{20\,\,{\rm dB}}{\sigma_{\rm S} = 10\,{\rm dB}}\right ) = {\rm Q}(2) \approx 0.02\hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm Pr}({\rm "System\hspace{0.15cm}works"})= 1- 0.02 \hspace{0.15cm} \underline{\approx 98\,\%}\hspace{0.05cm}.$$
The graphic illustrates the result.
- The probability density $f_{\rm VS}(V_{\rm S})$ of the path loss due to shadowing (Longnormal–Fading) is shown here.
- The probability that the system will fail is marked in red.
(4) From the availability probability $99.9 \%$ follows the failure probability $10^{\rm –3} \approx \ {\rm Q}(3)$.
- If the distance-dependent path loss $V_0$ is reduced by $10 \ \ \rm dB$ to $\underline {70 \ \rm dB}$, a failure will only occur when $V_{\rm S} ≥ 50 \ \ \rm dB$.
- This would achieve exactly the required reliability, as the following calculation shows:
- $${\rm Pr}({\rm "System\hspace{0.15cm} does\hspace{0.15cm} not\hspace{0.15cm} work\hspace{0.15cm}"})= {\rm Q}\left ( \frac{120-70-20}{10}\right ) = {\rm Q}(3) \approx 0.001 \hspace{0.05cm}.$$