View source for Aufgaben:Exercise 1.2: Lognormal Channel Model

{{quiz-Header|Buchseite=Mobile Kommunikation/Distanzabhängige Dämpfung und Abschattung
}}

[[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&nbsp; $d_0$&nbsp; from the base station.&nbsp; For example, it moves on an arc around the base station. 

Thus the total path loss can be described by the following equation:
:$$V_{\rm P} = V_{\rm 0} + V_{\rm S}  \hspace{0.05cm}.$$

*$V_0$&nbsp; takes into account the distance-dependent path loss which is assumed to be constant:&nbsp; $V_0 = 80 \ \rm dB$&nbsp;. 
*The loss&nbsp; $V_{\rm S}$&nbsp; 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&nbsp; $P_{\rm S} = 10 \ \rm W$&nbsp; $\text{(or  } 40 \ \rm dBm$).
* The received power should be at least&nbsp; $P_{\rm E} = 10 \ \rm pW$&nbsp; $\text{(or  } -80 \ \rm dBm$)






Notes:'' 
* This task belongs to the chapter&nbsp; [[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:
:$${\rm Q}(1) \approx 0.16\hspace{0.05cm},\hspace{0.2cm} {\rm Q}(2) \approx 0.02\hspace{0.05cm},\hspace{0.2cm}
 {\rm Q}(3) \approx 10^{-3}\hspace{0.05cm}.$$
* Or use the interaction module&nbsp; [[Applets:Complementary_Gaussian_Error_Functions|Complementary Gaussian Error Functions]]&nbsp; provided by&nbsp; $\rm LNTwww$.



===Questions===

<quiz display=simple>
{Would&nbsp; $P_{\rm E}$&nbsp;  be sufficient without consideration of the lognormal&ndash;fading?
|type="()"}
+ Yes,
- No.

{The parameters of the lognormal distribution are&nbsp; $m_{\rm S} = 20 \, \rm dB$&nbsp; and&nbsp; $\sigma_{\rm S} = 0 \, \rm dB$.&nbsp; What percentage of the time does the system work?
|type="{}"}
${\rm Pr(System \ works)} \ = \ $ { 100 3% } $\ \%$

{What is the probability with&nbsp; $m_{\rm S} = 20 \ \ \rm dB$&nbsp; and&nbsp; $\sigma_{\rm S} = 10 \ \ \rm dB$?
|type="{}"}
${\rm Pr(System \ works)}\ = \ $ { 98 3% } $\ \%$

{How big can&nbsp; $V_0$&nbsp; be at most, so that the reliability of&nbsp; $99.9\%$&nbsp; is reached?
|type="{}"}
$V_0 \ = \ $ { 70 3% } $\ \ \rm dB$
</quiz>

===Solutions===
{{ML-Kopf}}
'''(1)'''&nbsp; The correct answer is <u>YES</u>:
*From the&nbsp; $\rm dB$&ndash;value&nbsp; $V_0 = 80 \ \rm dB$&nbsp; follows the absolute (linear) value&nbsp; $K_0 = 10^8$.&nbsp; 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&nbsp; $-80 \ \rm dBm$&nbsp; is required.



'''(2)'''&nbsp; Lognormal fading with&nbsp; $\sigma_{\rm S} = 0 \ \rm dB$&nbsp; is equivalent to a constant received power&nbsp; $P_{\rm E}$. 
*Compared to the subtask&nbsp; '''(1)'''&nbsp; this is&nbsp; $m_{\rm S} = 20 \ \ \rm dB$&nbsp; smaller &nbsp; &#8658; &nbsp; $P_{\rm E} = \ &ndash;60 \ \ \rm dBm$. 
*But it is still greater than the specified limit value&nbsp; $(-80 \ \rm dBm)$. 
*It follows: &nbsp; The system is (almost)&nbsp; <u>100% functional</u>.&nbsp; "Almost" because with a Gaussian random quantity there is always a (small) residual uncertainty.



'''(3)'''&nbsp; The received power is too low&nbsp; $($less than $&ndash;80 \ \rm dBm)$&nbsp; if the power loss due to the lognormal&ndash;term is&nbsp; $40 \ \rm dB$&nbsp; or more.
[[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&nbsp; $20 \ \rm dB$. 
*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&nbsp; $f_{\rm VS}(V_{\rm S})$&nbsp; of the path loss due to shadowing&nbsp; (Longnormal Fading)&nbsp; is shown here. 

*The probability that the system will fail is marked in red.



'''(4)'''&nbsp; From the availability probability&nbsp; $99.9 \%$&nbsp; follows the failure probability&nbsp; $10^{\rm -3} \approx \ {\rm Q}(3)$. 
*If the distance-dependent path loss&nbsp; $V_0$&nbsp; is reduced by&nbsp; $10 \ \ \rm dB$&nbsp; to&nbsp; $\underline {70 \ \rm dB}$, a failure will only occur when&nbsp; $V_{\rm S} &#8805; 50 \ \ \rm dB$. 
*This would achieve exactly the required reliability, as the following calculation shows: 
:$${\rm Pr}({\rm "System\hspace{0.15cm} does\hspace{0.15cm} not\hspace{0.15cm} work\hspace{0.15cm}"})=
  {\rm Q}\left ( \frac{120-70-20}{10}\right ) 
  = {\rm Q}(3) \approx 0.001 \hspace{0.05cm}.$$

{{ML-Fuß}}






[[Category:Exercises for Mobile Communications|^1.1 Distance-Dependent Attenuation^]]