Quelltext der Seite Aufgaben:Exercise 1.1Z: Simple Path Loss Model

{{quiz-Header|Buchseite=Mobile Kommunikation/Distanzabhängige Dämpfung und Abschattung
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[[Datei:P_ID2121__Mob_Z_1_1.png|right|frame|Path loss model]]
Radio transmission with line-of-sight can be described by the so-called path loss model, which is given by the following equations:
$$V_{\rm P}(d) = V_{\rm 0} + \gamma \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (d/d_0)\hspace{0.05cm},$$
$$V_{\rm 0} = \gamma \cdot 10\,{\rm dB}  \cdot {\rm lg} \hspace{0.1cm} \frac{4 \cdot \pi \cdot d_0}{\lambda} \hspace{0.05cm}.$$

The graphic shows the path loss&nbsp; $V_{\rm P}(d)$&nbsp; in&nbsp; $\rm dB$. The abscissa&nbsp; $d$&nbsp; is also displayed logarithmically. 

In the above equation, the following parameters are used:
* the distance&nbsp; $d$&nbsp; of transmitter and receiver,
* the reference distance&nbsp; $d_0 = 1 \ \rm m$,
* the path loss exponent&nbsp; $\gamma$,
* the wavelength&nbsp; $\lambda$&nbsp; of the electromagnetic wave.


Two scenarios are shown &nbsp;$\rm (A)$&nbsp; and &nbsp;$\rm (B)$&nbsp; with the same path loss at distance&nbsp; $d_0 = 1 \ \rm m$:
$$V_{\rm 0} = V_{\rm P}(d = d_0) = 20\,{\rm dB}  \hspace{0.05cm}.$$

One of these two scenarios describes the so-called <i>free space attenuation</i>, characterized by the path loss exponent&nbsp; $\gamma = 2$. However, the equation for the free space attenuation only applies in the <i>far-field</i>, i.e. when the distance&nbsp; $d$&nbsp; between transmitter and receiver is greater than the <i>Fraunhofer distance</i>;
$$d_{\rm F} = {2 D^2}/{\lambda} \hspace{0.05cm}.$$

Here,&nbsp; $D$&nbsp; is the largest physical dimension of the transmitting antenna. With an&nbsp; $\lambda/2$&ndash;antenna, the Fraunhofer distance has a simple expression:
$$d_{\rm F} = \frac{2 \cdot (\lambda/2)^2}{\lambda} = {\lambda}/{2}\hspace{0.05cm}.$$






''Notes:'' 
* This task belongs to the chapter&nbsp; [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung|Distanzabhängige Dämpfung und Abschattung]].
* The speed of light is&nbsp; $c = 3 \cdot 10^8 \ {\rm m/s}$.
 




===Questionnaire===

<quiz display=simple>
{Which path loss exponents apply to the scenarios &nbsp;$\rm (A)$&nbsp; and &nbsp;$\rm (B)$?
|type="{}"}
$\gamma_{\rm A}\ = \ $ { 2 3% }
$\gamma_{\rm B} \ = \ $ { 2.5 3% }

{Which scenario describes free-space attenuation?
|type="()"}
+ Scenario &nbsp;$\rm (A)$,
- Scenario &nbsp;$\rm (B)$.

{Which signal frequencies are the basis for the scenarios &nbsp;$\rm (A)$&nbsp; and &nbsp;$\rm (B)$&nbsp;?
|type="{}"}
$f_{\rm A} \ = \ $ { 240 3% } $\ \ \rm MHz$
$f_{\rm B} \ = \ $ { 151.4 3% } $\ \ \rm MHz$

{Does the free-space scenario apply to all distances between&nbsp; $1 \ \rm m$&nbsp; and&nbsp; $10 \ \rm km$?
|type="()"}
+ Yes,
- No.
</quiz>

===Sample solution===
{{ML-Kopf}}
'''(1)'''&nbsp; The (simplest) path loss equation is
$$V_{\rm P}(d) = V_{\rm 0} + \gamma \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (d/d_0)\hspace{0.05cm}.$$

*In scenario (A), the decay per decade (for example, between $d_0 = 1 \ \rm m$ and $d = 10 \ \rm m$) is exactly $20 \ \rm dB$ and in scenario (B) $25 \ \rm dB$. 
*It follows:
$$\gamma_{\rm A} \hspace{0.15cm} \underline{= 2}\hspace{0.05cm},\hspace{0.2cm}\gamma_{\rm B} \hspace{0.15cm} \underline{= 2.5}\hspace{0.05cm}.$$



'''(2)'''&nbsp; <u>Solution 1</u> is correct, since the free space attenuation is characterized by the path loss exponent $\gamma = 2$.



'''(3)'''&nbsp; The path loss at $d_0 = 1 \ \rm m$ is in both cases $V_0 = 20 \ \rm dB$. For scenario (A) the same applies:
$$10 \cdot {\rm lg}\hspace{0.1cm} \left [ \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm A}}\right ]^2 = 20\,{\rm dB} \hspace{0.2cm} \Rightarrow \hspace{0.2cm}
 \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm A}} = 10 \hspace{0.2cm} \Rightarrow \hspace{0.2cm}
 \lambda_{\rm A} = 4 \pi \cdot 0.1\,{\rm m} = 1,257\,{\rm m}
  \hspace{0.05cm}.$$

*The frequency $f_{\rm A}$ is related to the wavelength $\lambda_{\rm A}$ over the speed of light $c$:
$$f_{\rm A} = \frac{c}{\lambda_{\rm A}} = \frac{3 \cdot 10^8\,{\rm m/s}}}{1.257\,{\rm m}} = 2.39 \cdot 10^8\,{\rm Hz}
 \hspace{0.15cm} \underline{\approx 240 \,\,{\rm MHz}}
  \hspace{0.05cm}.$$

*On the other hand, the scenario (B)
$$10 \cdot {\rm lg}\hspace{0.1cm} \left [ \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm B}}\right ]^{2.5} = 20\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 25 \cdot {\rm lg}\hspace{0.1cm} \left [ \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm B}}\right ] = 20\,{\rm dB}$$
$$\Rightarrow \hspace{0.3cm}  \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm B}} = 10^{0.8} \approx 6.31
  \hspace {0.3cm} \Rightarrow \hspace{0.3cm}
 {\lambda_{\rm B}} = \frac{10}{6.31} \cdot {\lambda_{\rm A}}\hspace{0.3cm}
\Rightarrow \hspace{0.3cm}
 {f_{\rm B}}} = \frac{6.31}{10} \cdot {f_{\rm A}}} = 0.631 \cdot 240 \,{\rm MHz}\hspace{0.15cm} \underline {\approx 151.4 \,\,{\rm MHz}}
  \hspace{0.05cm}.$$


'''(4)'''&nbsp; <u>first suggested solution</u> is correct: 
*In free space&ndash;scenario (A) the Fraunhofer&ndash;distance&nbsp; $d_{\rm F} = \lambda_{\rm A}/2 \approx 63 \ \rm cm$. Thus, the following always applies&nbsp; $d > d_{\rm F}$. 
*Also in scenario (B) is because of&nbsp; $\lambda_{\rm B} \approx 2 \ \rm m$&nbsp; or &nbsp;$d_{\rm F} \approx 1 \ \rm m$&nbsp; the entire displayed course correct.
{{ML-Fuß}}



[[Category:Exercises for Mobile Communications|^1.1 Distance-dependent attenuation^]]