Difference between revisions of "Aufgaben:Exercise 1.2: Lognormal Channel Model"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Mobile_Communications/Distance_Dependent_Attenuation_and_Shading |
}} | }} | ||
− | [[File: | + | [[File:EN_Mob_A1_2.png|right|frame|PDF of lognormal fading]] |
− | 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. | + | 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: | 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 | + | *$V_0$ 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 | + | *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) },$$ |
− | : | + | : see diagram. The following numerical values apply: |
− | $$m_{\rm S} = 20\,\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm} \sigma_{\rm S} = 10\,\,{\rm dB}\hspace{0. | + | :$$m_{\rm S} = 20\,\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm} \sigma_{\rm S} = 10\,\,{\rm dB}\hspace{0.25cm}{\rm or }\hspace{0.25cm}\sigma_{\rm S} = 0\,\,{\rm dB}\hspace{0.15cm}{\rm (subtask\hspace{0.15cm} 2)}\hspace{0.05cm}.$$ |
Also make the following simple assumptions: | Also make the following simple assumptions: | ||
− | * The transmit power is $P_{\rm S} = 10 \ \rm W$ (or | + | * The transmit power is $P_{\rm S} = 10 \ \rm W$ $\text{(or } 40 \ \rm dBm$). |
− | * The | + | * The received power should be at least $P_{\rm E} = 10 \ \rm pW$ $\text{(or } -80 \ \rm dBm$) |
Line 25: | Line 25: | ||
Notes:'' | Notes:'' | ||
− | * | + | * This task belongs to the chapter [[Mobile_Communications/Distance_dependent_attenuation_and_shading|Distance dependent attenuation and shading]]. |
* 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}.$$ |
− | * Or use the interaction module | + | * Or use the interaction module [[Applets:Complementary_Gaussian_Error_Functions|Complementary Gaussian Error Functions]] provided by $\rm LNTwww$. |
− | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
− | {Would $P_{\rm E}$ | + | {Would $P_{\rm E}$ be sufficient if the loss $V_S$ due to shadowing is not present? |
|type="()"} | |type="()"} | ||
+ Yes, | + Yes, | ||
- No. | - No. | ||
− | The | + | {The parameters of the lognormal distribution are $m_{\rm S} = 20 \, \rm dB$ and $\sigma_{\rm S} = 0 \, \rm dB$. What percentage of the time does the system work? |
|type="{}"} | |type="{}"} | ||
${\rm Pr(System \ works)} \ = \ $ { 100 3% } $\ \%$ | ${\rm Pr(System \ works)} \ = \ $ { 100 3% } $\ \%$ | ||
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${\rm Pr(System \ works)}\ = \ $ { 98 3% } $\ \%$ | ${\rm Pr(System \ works)}\ = \ $ { 98 3% } $\ \%$ | ||
− | {How big can $V_0$ be | + | {How big can $V_0$ be at most, so that the reliability of $99.9\%$ is reached? |
|type="{}"} | |type="{}"} | ||
$V_0 \ = \ $ { 70 3% } $\ \ \rm dB$ | $V_0 \ = \ $ { 70 3% } $\ \ \rm dB$ | ||
− | </quiz | + | </quiz> |
− | === | + | ===Solution=== |
− | {{ML | + | {{ML-Kopf}} |
− | '''(1)''' | + | '''(1)''' The correct answer is <u>YES</u>: |
− | *From the $\rm dB$–value $V_0 = 80 \ \rm dB$ follows the absolute (linear) value $K_0 = 10^8$. Thus the received power is | + | *From the $\rm dB$–value $V_0 = 80 \ \rm dB$ follows the absolute (linear) value $K_0 = 10^8$. Thus the received power is |
− | $$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.$$ |
*You can also solve this problem directly with the logarithmic quantities: | *You can also solve this problem directly with the logarithmic quantities: | ||
− | $10 \cdot {\rm lg}\hspace{0.15cm} \frac{P_{\rm E | + | :$$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}.$$ | \hspace{0.05cm}.$$ | ||
− | *Only the limit value | + | *Only the limit value $-80 \ \rm dBm$ is required. |
− | '''(2)''' Lognormal& | + | '''(2)''' Lognormal fading with $\sigma_{\rm S} = 0 \ \rm dB$ is equivalent to a constant received power $P_{\rm E}$. |
− | *Compared to the subtask '''(1)'' this is $m_{\rm S} = 20 \ \ \rm dB$ smaller ⇒ $P_{\rm E} = \ –60 \ \ \rm dBm$. | + | *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 | + | *But it is still greater than the specified limit value $(-80 \ \rm dBm)$. |
− | *It follows: The system is (almost) <u>100% functional</u>. & | + | *It follows: The system is (almost) <u>100% functional</u>. "Almost" because with a Gaussian random quantity there is always a (small) residual uncertainty. |
− | '''(3)''' The | + | '''(3)''' The received 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 | + | [[File:EN_Mob_A_1_2c.png|right|frame|Loss due to lognormal fading]] |
+ | |||
+ | *The distance-dependent path loss $V_{\rm S}$ must therefore not be greater than $20 \ \rm dB$. | ||
*So it follows: | *So it follows: | ||
− | $${\rm Pr}({\rm "System\hspace{0.15cm} does not | + | :$${\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 | + | = {\rm Q}(2) \approx 0.02$$ |
− | \Rightarrow \hspace{0.3cm}{\rm Pr}({\rm "System\hspace{0.15cm} works"})= 1- 0.02 \hspace{0.15cm} \underline{\approx 98\,\%}\hspace{0.05cm}.$$ | + | :$$\Rightarrow \hspace{0.3cm}{\rm Pr}({\rm "System\hspace{0.15cm}works"})= 1- 0.02 \hspace{0.15cm} \underline{\approx 98\,\%}\hspace{0.05cm}.$$ |
− | |||
The graphic illustrates the result. | The graphic illustrates the result. | ||
− | *The probability density $f_{\rm VS}(V_{\rm S})$ of the path loss due to shadowing (Longnormal& | + | *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 $V_{\rm S} ≥ 50 \ \ \rm dB$. | |
− | '''(4)''' From the availability probability $99.9 \%$ follows the | ||
− | *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: | ||
− | $${\rm Pr}({\rm "System\hspace{0.15cm} does not work\hspace{0.15cm}"})= | + | :$${\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}.$$ | = {\rm Q}(3) \approx 0.001 \hspace{0.05cm}.$$ | ||
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− | [[Category: | + | [[Category:Mobile Communications: Exercises|^1.1 Distance-Dependent Attenuation^]] |
Latest revision as of 15:58, 2 July 2021
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:
- $$m_{\rm S} = 20\,\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm} \sigma_{\rm S} = 10\,\,{\rm dB}\hspace{0.25cm}{\rm or }\hspace{0.25cm}\sigma_{\rm S} = 0\,\,{\rm dB}\hspace{0.15cm}{\rm (subtask\hspace{0.15cm} 2)}\hspace{0.05cm}.$$
Also make the following simple assumptions:
- The transmit power is $P_{\rm S} = 10 \ \rm W$ $\text{(or } 40 \ \rm dBm$).
- The received power should be at least $P_{\rm E} = 10 \ \rm pW$ $\text{(or } -80 \ \rm dBm$)
Notes:
- This task belongs to the chapter Distance dependent attenuation and shading.
- 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 Complementary Gaussian Error Functions provided by $\rm LNTwww$.
Questions
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 received 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 received 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 distance-dependent path loss $V_{\rm S}$ must therefore not be greater than $20 \ \rm dB$.
- So it follows:
- $${\rm Pr}({\rm "System\hspace{0.15cm}does\hspace{0.15cm}not\hspace{0.15cm}work"})= {\rm Q}\left ( \frac{20\,\,{\rm dB}}{\sigma_{\rm S} = 10\,{\rm dB}}\right ) = {\rm Q}(2) \approx 0.02$$
- $$\Rightarrow \hspace{0.3cm}{\rm Pr}({\rm "System\hspace{0.15cm}works"})= 1- 0.02 \hspace{0.15cm} \underline{\approx 98\,\%}\hspace{0.05cm}.$$
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:
- $${\rm Pr}({\rm "System\hspace{0.15cm} does\hspace{0.15cm} not\hspace{0.15cm} work\hspace{0.15cm}"})= {\rm Q}\left ( \frac{120-70-20}{10}\right ) = {\rm Q}(3) \approx 0.001 \hspace{0.05cm}.$$