# Exercise 1.2: Lognormal Channel Model

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:

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

### Questions

1

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

 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 \ = \$ $\ \ \rm dB$

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