Aufgaben:Exercise 1.2: Lognormal Channel Model: Unterschied zwischen den Versionen

Aus LNTwww
Wechseln zu:Navigation, Suche
Zeile 2: Zeile 2:
 
}}
 
}}
  
[[Datei:P_ID2122__Mob_A_1_2.png|right|frame|PDF of lognormal fading]]
+
[[Datei: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.  
  
Zeile 82: Zeile 82:
 
   = {\rm Q}(2) \approx 0.02\hspace{0.3cm}
 
   = {\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}.$$
 
\Rightarrow \hspace{0.3cm}{\rm Pr}({\rm "System\hspace{0.15cm}works"})= 1-  0.02 \hspace{0.15cm} \underline{\approx 98\,\%}\hspace{0.05cm}.$$
[[File:P_ID2187__Mob_A_1_2c_v1.png|right|frame|loss due to lognormal fading]]
+
[[File:EN_Mob_A_1_2c.png|right|frame|loss due to lognormal fading]]
 
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–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.  

Version vom 11. Mai 2020, 13:53 Uhr

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.

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)
$$f_{V{\rm S}}(V_{\rm S}) = \frac {1}{ \sqrt{2 \pi }\cdot \sigma_{\rm S}} \cdot {\rm exp } \left [ - \frac{ (V_{\rm S}- m_{\rm S})^2}{2 \cdot \sigma_{\rm S}^2} \right ] \hspace{0.05cm}$$
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:

  • You can use the following (rough) approximations for the complementary Gaussian error integral:
$${\rm Q}(1) \approx 0.16\hspace{0.05cm},\hspace{0.2cm} {\rm Q}(2) \approx 0.02\hspace{0.05cm},\hspace{0.2cm} {\rm Q}(3) \approx 10^{-3}\hspace{0.05cm}.$$


Questionnaire

1

Would  $P_{\rm E}$  without consideration of the lognormal–fading be sufficient?

Yes,
No.

2

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?

${\rm Pr(System \ works)} \ = \ $

$\ \%$

3

What is the probability with  $m_{\rm S} = 20 \ \ \rm dB$  and  $\sigma_{\rm S} = 10 \ \ \rm dB$?

${\rm Pr(System \ works)}\ = \ $

$\ \%$

4

How big can  $V_0$  be at most, so that the reliability of   $99.9\%$  is reached?

$V_0 \ = \ $

$\ \ \rm dB$


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:
$${\rm Pr}({\rm "System\hspace{0.15cm}does\hspace{0.15cm}not\hspace{0.15cm}work"})= {\rm Q}\left ( \frac{20\,\,{\rm dB}}{\sigma_{\rm S} = 10\,{\rm dB}}\right ) = {\rm Q}(2) \approx 0.02\hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm Pr}({\rm "System\hspace{0.15cm}works"})= 1- 0.02 \hspace{0.15cm} \underline{\approx 98\,\%}\hspace{0.05cm}.$$
loss due to lognormal fading

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:

$${\rm Pr}({\rm "System\hspace{0.15cm} does\hspace{0.15cm} not\hspace{0.15cm} work\hspace{0.15cm}"})= {\rm Q}\left ( \frac{120-70-20}{10}\right ) = {\rm Q}(3) \approx 0.001 \hspace{0.05cm}.$$