Exercise 1.2: Lognormal Channel Model

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)

$$f_{V_{\rm S}}(V_{\rm S}) = \frac {1}{ \sqrt{2 \pi }\cdot \sigma_{\rm S}} \cdot {\rm e }^{ - { (V_{\rm S}\hspace{0.05cm}- \hspace{0.05cm}m_{\rm S})^2}/(2 \hspace{0.05cm}\cdot \hspace{0.05cm}\sigma_{\rm S}^2) },$$

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 receive 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 $\rm LNTwww$.

Questionnaire

	Yes,
	No.

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

$\ \%$

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

$\ \%$

$V_0 \ = \ $

$\ \ \rm dB$

Sample solution

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:

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