Difference between revisions of "Aufgaben:Exercise 1.1: Dual Slope Loss Model"
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To simulate path loss in an urban environment, the asymptotic dual-slope model is often used, which is shown as a red curve in the diagram. This simple model is characterized by two linear sections separated by the so-called breakpoint $\rm (BP)$: | To simulate path loss in an urban environment, the asymptotic dual-slope model is often used, which is shown as a red curve in the diagram. This simple model is characterized by two linear sections separated by the so-called breakpoint $\rm (BP)$: | ||
* For $d \le d_{\rm BP}$ and the exponent $\gamma_0$ we have: | * For $d \le d_{\rm BP}$ and the exponent $\gamma_0$ we have: | ||
| − | :$$ | + | :$$L_{\rm P}(d) = L_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_0})\hspace{0.05cm}.$$ |
* For $d > d_{\rm BP}$ we must apply the path loss exponent $\gamma_1$ where $\gamma_1 > \gamma_0$: | * For $d > d_{\rm BP}$ we must apply the path loss exponent $\gamma_1$ where $\gamma_1 > \gamma_0$: | ||
| − | :$$ | + | :$$L_{\rm P}(d) = L_{\rm BP} + \gamma_1 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_{\rm BP}})\hspace{0.05cm}.$$ |
In these equations, the variables are: | In these equations, the variables are: | ||
| − | * $ | + | * $L_0$ is the path loss (in dB) at $d_0$ (normalization distance). |
| − | * $ | + | * $L_{\rm BP}$ is the path loss (in dB) at $d=d_{\rm BP}$ ("Breakpoint"). |
The graph applies to the model parameters | The graph applies to the model parameters | ||
:$$d_0 = 1\,{\rm m}\hspace{0.05cm},\hspace{0.2cm}d_{\rm BP} = 100\,{\rm m}\hspace{0.05cm},\hspace{0.2cm} | :$$d_0 = 1\,{\rm m}\hspace{0.05cm},\hspace{0.2cm}d_{\rm BP} = 100\,{\rm m}\hspace{0.05cm},\hspace{0.2cm} | ||
| − | + | L_0 = 10\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm}\gamma_0 = 2 \hspace{0.05cm},\hspace{0.2cm}\gamma_1 = 4 \hspace{0.3cm} | |
\Rightarrow \hspace{0.3cm} | \Rightarrow \hspace{0.3cm} | ||
| − | + | L_{\rm BP} = 50\,{\rm dB}\hspace{0.05cm}.$$ | |
In the questions, this piece-wise defined profile is called $\rm A$. | In the questions, this piece-wise defined profile is called $\rm A$. | ||
The second curve is the profile $\rm B$ given by the following equation: | The second curve is the profile $\rm B$ given by the following equation: | ||
| − | :$$ | + | :$$L_{\rm P}(d) = L_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ( {d}/{d_0} \right ) |
+ (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left (1 + {d}/{d_{\rm BP}} \right )\hspace{0.05cm}.$$ | + (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left (1 + {d}/{d_{\rm BP}} \right )\hspace{0.05cm}.$$ | ||
With this dual model, the entire distance course can be written in closed form, and the received power depends on the distance $d$ according to the following equation: | With this dual model, the entire distance course can be written in closed form, and the received power depends on the distance $d$ according to the following equation: | ||
| − | :$$P_{\rm | + | :$$P_{\rm t}(d) = \frac{P_{\rm t} \cdot G_{\rm t} \cdot G_{\rm r} /L_{\rm tot}}{L_{\rm P}(d)} |
| − | \hspace{0.05cm},\hspace{0.2cm} | + | \hspace{0.05cm},\hspace{0.2cm}L_{\rm P}(d) = 10^{L_{\rm P}(d)/10} |
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
| − | Here, all parameters are in natural units (not in dB). The transmit power is assumed to be $P_{\rm | + | Here, all parameters are in natural units (not in dB). The transmit power is assumed to be $P_{\rm t} = 5 \ \rm W$ . The other quantities have the following meanings and values: |
| − | * $10 \cdot \lg \ G_{\rm | + | * $10 \cdot \lg \ G_{\rm t} = 17 \ \rm dB$ (gain of the transmit antenna), |
| − | * $10 \cdot \lg \ G_{\rm | + | * $10 \cdot \lg \ G_{\rm r} = -3 \ \ \rm dB$ (gain of receiving antenna – so actually a loss), |
| − | * $10 \cdot \lg \ | + | * $10 \cdot \lg \ L_{\rm tot} = 4 \ \ \rm dB$ (loss through feeds). |
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*This task belongs to the chapter [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung_|Distanzabhängige Dämpfung und Abschattung]]. | *This task belongs to the chapter [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung_|Distanzabhängige Dämpfung und Abschattung]]. | ||
*If the profile $\rm B$ were | *If the profile $\rm B$ were | ||
| − | :$$ | + | :$$L_{\rm P}(d) = L_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} \left ( {d}/{d_0} \right ) |
+ (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right )$$ | + (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right )$$ | ||
:then profile $\rm A$ and profile $\rm B$ for $d ≥ d_{\rm BP}$ would be identical. | :then profile $\rm A$ and profile $\rm B$ for $d ≥ d_{\rm BP}$ would be identical. | ||
*In this case, however, profile $\rm B$ would be above profile $\rm A$ for $(d < d_{\rm BP})$ , suggesting clearly too good conditions. | *In this case, however, profile $\rm B$ would be above profile $\rm A$ for $(d < d_{\rm BP})$ , suggesting clearly too good conditions. | ||
*For example, $d = d_0 = 1 \ \ \rm m$ with the given numerical values gives a result that is $40 \ \ \rm dB$ too good: | *For example, $d = d_0 = 1 \ \ \rm m$ with the given numerical values gives a result that is $40 \ \ \rm dB$ too good: | ||
| − | :$$ | + | :$$L_{\rm P}(d) = L_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ( {d}/{d_0} \right ) + (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right ) =10\,{\rm dB} + 2 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({1}/{100} \right ) |
= -30\,{\rm dB} \hspace{0.05cm}. $$ | = -30\,{\rm dB} \hspace{0.05cm}. $$ | ||
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{How large is the path loss $($in $\rm dB)$ after $d= 100 \ \rm m$ according to profile $\rm A$? | {How large is the path loss $($in $\rm dB)$ after $d= 100 \ \rm m$ according to profile $\rm A$? | ||
|type="{}"} | |type="{}"} | ||
| − | $ | + | $L_{\rm P}(d = 100 \ \rm m) \ = \ $ { 50 3% } $\ \rm dB$ |
{How large is the path loss $($in $\rm dB)$ after $d= 100 \ \rm m$ according to profile $\rm B$? | {How large is the path loss $($in $\rm dB)$ after $d= 100 \ \rm m$ according to profile $\rm B$? | ||
|type="{}"} | |type="{}"} | ||
| − | $ | + | $L_{\rm P}(d = 100 \ \rm m) \ = \ $ { 56 3% } $\ \rm dB$ |
{What is the receive power after $100 \ \ \rm m$ with both profiles? | {What is the receive power after $100 \ \ \rm m$ with both profiles? | ||
|type="{}"} | |type="{}"} | ||
| − | Profile $\text{A:} \hspace{0.2cm} P_{\rm | + | Profile $\text{A:} \hspace{0.2cm} P_{\rm r}(d = 100 \ \rm m) \ = \ $ { 0.5 3% } $\ \ \rm mW$ |
| − | Profile $\text{B:} \hspace{0.2cm} P_{\rm | + | Profile $\text{B:} \hspace{0.2cm} P_{\rm r}(d = 100 \ \rm m) \ = \ $ { 0.125 3% } $\ \ \rm mW$ |
| − | {How big is the deviation $ | + | {How big is the deviation $ΔL_{\rm P}$ between profile $\rm A$ and $\rm B$ at $d = 50 \ \rm m$? |
|type="{}"} | |type="{}"} | ||
| − | $ | + | $ΔL_{\rm P}(d = 50 \ \rm m) \ = \ $ { 3.5 3% } $\ \rm dB$ |
| − | {How big is the deviation $ | + | {How big is the deviation $ΔL_{\rm P}$ between profile $\rm A$ and $\rm B$ at $d = 200 \ \rm m$? |
|type="{}"} | |type="{}"} | ||
| − | $ | + | $ΔL_{\rm P}(d = 200 \ \rm m)\ = \ $ { 3.5 3% } $\ \rm dB$ |
</quiz> | </quiz> | ||
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{{ML-Kopf}} | {{ML-Kopf}} | ||
'''(1)''' You can see directly from the graphic that the profile $\rm (A)$ with the two linear sections at the breakpoint $(d = 100 \ \rm m)$ gives the following result: | '''(1)''' You can see directly from the graphic that the profile $\rm (A)$ with the two linear sections at the breakpoint $(d = 100 \ \rm m)$ gives the following result: | ||
| − | :$$ | + | :$$L_{\rm P}(d = 100\,{\rm m})\hspace{0.15cm} \underline{= 50\,{\rm dB}} \hspace{0.05cm}.$$ |
| − | '''(2)''' With the profile $\rm (B)$ on the other hand, using $ | + | '''(2)''' With the profile $\rm (B)$ on the other hand, using $L_0 = 10 \ \rm dB$, $\gamma_0 = 2$ and $\gamma_1 = 4$: |
| − | :$$ | + | :$$L_{\rm P}(d = 100\,{\rm m}) = 10\,{\rm dB} + 20\,{\rm dB}\cdot {\rm lg} \hspace{0.1cm}(100)+ 20\,{\rm dB}\cdot {\rm lg} \hspace{0.1cm}(2) |
\hspace{0.15cm} \underline{\approx 56\,{\rm dB}} \hspace{0.05cm}.$$ | \hspace{0.15cm} \underline{\approx 56\,{\rm dB}} \hspace{0.05cm}.$$ | ||
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'''(3)''' The antenna gains from the transmitter $(+17 \ \rm dB)$ and receiver $(-3 \ \rm dB)$ and the internal losses of the base station $(+4 \ \rm dB)$ can be combined to | '''(3)''' The antenna gains from the transmitter $(+17 \ \rm dB)$ and receiver $(-3 \ \rm dB)$ and the internal losses of the base station $(+4 \ \rm dB)$ can be combined to | ||
| − | :$$10 \cdot {\rm lg}\hspace{0.1cm} G = 10 \cdot {\rm lg}\hspace{0.1cm} G_{\rm S} + 10 \cdot {\rm lg} \hspace{0.1cm} G_{\rm E} - 10 \cdot {\rm lg}\hspace{0.1cm} | + | :$$10 \cdot {\rm lg}\hspace{0.1cm} G = 10 \cdot {\rm lg}\hspace{0.1cm} G_{\rm S} + 10 \cdot {\rm lg} \hspace{0.1cm} G_{\rm E} - 10 \cdot {\rm lg}\hspace{0.1cm} L_{\rm tot} = 17\,{\rm dB} -3\,{\rm dB} - 4\,{\rm dB} = 10\,{\rm dB} |
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {G = 10} | \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {G = 10} | ||
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
*For the profile $\rm (A)$ the following path loss occurred: | *For the profile $\rm (A)$ the following path loss occurred: | ||
| − | :$$ | + | :$$L_{\rm P}(d = 100\,{\rm m})\hspace{0.05cm} {= 50\,{\rm dB}} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} K_{\rm P} = 10^5 \hspace{0.05cm}.$$ |
:This gives you for the received power after $d = 100\ \rm m$: | :This gives you for the received power after $d = 100\ \rm m$: | ||
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'''(4)''' Below the breakpoint $(d < 100 \ \rm m)$, the deviation is determined by the last summand of profile $\rm (B)$: | '''(4)''' Below the breakpoint $(d < 100 \ \rm m)$, the deviation is determined by the last summand of profile $\rm (B)$: | ||
| − | :$${\rm \Delta} | + | :$${\rm \Delta}L_{\rm P}(d = 50\,{\rm m}) = (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left (1 + {d}/{d_{\rm BP}} \right )= (4-2) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (1.5)\hspace{0.15cm} |
\underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$ | \underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$ | ||
| − | '''(5)''' Here the profile $\rm (A)$ with $ | + | '''(5)''' Here the profile $\rm (A)$ with $L_{\rm BP} = 50 \ \rm dB$ gives: |
| − | :$$ | + | :$$L_{\rm P}(d = 200\,{\rm m}) = 50\,{\rm dB} + 4 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (2)\hspace{0.15cm} |
{\approx 62\,{\rm dB}} \hspace{0.05cm}.$$ | {\approx 62\,{\rm dB}} \hspace{0.05cm}.$$ | ||
*On the other hand, the profile $\rm (B)$ leads to the result: | *On the other hand, the profile $\rm (B)$ leads to the result: | ||
| − | :$$ | + | :$$L_{\rm P}(d = 200\,{\rm m}) = 50\,{\rm dB} + 20\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (200) |
+ 20\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (3) = 10\,{\rm dB} + 46\,{\rm dB} + 9.5\,{\rm dB} | + 20\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (3) = 10\,{\rm dB} + 46\,{\rm dB} + 9.5\,{\rm dB} | ||
\hspace{0.15cm} {\approx 65.5\,{\rm dB}}$$ | \hspace{0.15cm} {\approx 65.5\,{\rm dB}}$$ | ||
| − | :$$\Rightarrow \hspace{0.3cm} {\rm \Delta} | + | :$$\Rightarrow \hspace{0.3cm} {\rm \Delta}L_{\rm P}(d = 200\,{\rm m}) = 65.5\,{\rm dB} - 62\,{\rm dB}\hspace{0.15cm} |
\underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$ | \underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$ | ||
| − | *You can see that $\Delta | + | *You can see that $\Delta L_{\rm P}$ is almost symmetrical to $d = d_{\rm BP}$ if you plot the distance $d$ logarithmically as in the given graph. |
{{ML-Fuß}} | {{ML-Fuß}} | ||
Revision as of 11:23, 27 June 2020
Dual-slope path loss diagram
To simulate path loss in an urban environment, the asymptotic dual-slope model is often used, which is shown as a red curve in the diagram. This simple model is characterized by two linear sections separated by the so-called breakpoint $\rm (BP)$:
- For $d \le d_{\rm BP}$ and the exponent $\gamma_0$ we have:
- $$L_{\rm P}(d) = L_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_0})\hspace{0.05cm}.$$
- For $d > d_{\rm BP}$ we must apply the path loss exponent $\gamma_1$ where $\gamma_1 > \gamma_0$:
- $$L_{\rm P}(d) = L_{\rm BP} + \gamma_1 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_{\rm BP}})\hspace{0.05cm}.$$
In these equations, the variables are:
- $L_0$ is the path loss (in dB) at $d_0$ (normalization distance).
- $L_{\rm BP}$ is the path loss (in dB) at $d=d_{\rm BP}$ ("Breakpoint").
The graph applies to the model parameters
- $$d_0 = 1\,{\rm m}\hspace{0.05cm},\hspace{0.2cm}d_{\rm BP} = 100\,{\rm m}\hspace{0.05cm},\hspace{0.2cm} L_0 = 10\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm}\gamma_0 = 2 \hspace{0.05cm},\hspace{0.2cm}\gamma_1 = 4 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} L_{\rm BP} = 50\,{\rm dB}\hspace{0.05cm}.$$
In the questions, this piece-wise defined profile is called $\rm A$.
The second curve is the profile $\rm B$ given by the following equation:
- $$L_{\rm P}(d) = L_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ( {d}/{d_0} \right ) + (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left (1 + {d}/{d_{\rm BP}} \right )\hspace{0.05cm}.$$
With this dual model, the entire distance course can be written in closed form, and the received power depends on the distance $d$ according to the following equation:
- $$P_{\rm t}(d) = \frac{P_{\rm t} \cdot G_{\rm t} \cdot G_{\rm r} /L_{\rm tot}}{L_{\rm P}(d)} \hspace{0.05cm},\hspace{0.2cm}L_{\rm P}(d) = 10^{L_{\rm P}(d)/10} \hspace{0.05cm}.$$
Here, all parameters are in natural units (not in dB). The transmit power is assumed to be $P_{\rm t} = 5 \ \rm W$ . The other quantities have the following meanings and values:
- $10 \cdot \lg \ G_{\rm t} = 17 \ \rm dB$ (gain of the transmit antenna),
- $10 \cdot \lg \ G_{\rm r} = -3 \ \ \rm dB$ (gain of receiving antenna – so actually a loss),
- $10 \cdot \lg \ L_{\rm tot} = 4 \ \ \rm dB$ (loss through feeds).
Notes:
- This task belongs to the chapter Distanzabhängige Dämpfung und Abschattung.
- If the profile $\rm B$ were
- $$L_{\rm P}(d) = L_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} \left ( {d}/{d_0} \right ) + (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right )$$
- then profile $\rm A$ and profile $\rm B$ for $d ≥ d_{\rm BP}$ would be identical.
- In this case, however, profile $\rm B$ would be above profile $\rm A$ for $(d < d_{\rm BP})$ , suggesting clearly too good conditions.
- For example, $d = d_0 = 1 \ \ \rm m$ with the given numerical values gives a result that is $40 \ \ \rm dB$ too good:
- $$L_{\rm P}(d) = L_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ( {d}/{d_0} \right ) + (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right ) =10\,{\rm dB} + 2 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({1}/{100} \right ) = -30\,{\rm dB} \hspace{0.05cm}. $$
Questionnaire
Sample solution
(1) You can see directly from the graphic that the profile $\rm (A)$ with the two linear sections at the breakpoint $(d = 100 \ \rm m)$ gives the following result:
- $$L_{\rm P}(d = 100\,{\rm m})\hspace{0.15cm} \underline{= 50\,{\rm dB}} \hspace{0.05cm}.$$
(2) With the profile $\rm (B)$ on the other hand, using $L_0 = 10 \ \rm dB$, $\gamma_0 = 2$ and $\gamma_1 = 4$:
- $$L_{\rm P}(d = 100\,{\rm m}) = 10\,{\rm dB} + 20\,{\rm dB}\cdot {\rm lg} \hspace{0.1cm}(100)+ 20\,{\rm dB}\cdot {\rm lg} \hspace{0.1cm}(2) \hspace{0.15cm} \underline{\approx 56\,{\rm dB}} \hspace{0.05cm}.$$
(3) The antenna gains from the transmitter $(+17 \ \rm dB)$ and receiver $(-3 \ \rm dB)$ and the internal losses of the base station $(+4 \ \rm dB)$ can be combined to
- $$10 \cdot {\rm lg}\hspace{0.1cm} G = 10 \cdot {\rm lg}\hspace{0.1cm} G_{\rm S} + 10 \cdot {\rm lg} \hspace{0.1cm} G_{\rm E} - 10 \cdot {\rm lg}\hspace{0.1cm} L_{\rm tot} = 17\,{\rm dB} -3\,{\rm dB} - 4\,{\rm dB} = 10\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {G = 10} \hspace{0.05cm}.$$
- For the profile $\rm (A)$ the following path loss occurred:
- $$L_{\rm P}(d = 100\,{\rm m})\hspace{0.05cm} {= 50\,{\rm dB}} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} K_{\rm P} = 10^5 \hspace{0.05cm}.$$
- This gives you for the received power after $d = 100\ \rm m$:
- $$P_{\rm E}(d = 100\,{\rm m}) = \frac{P_{\rm S} \cdot G}{K_{\rm P}} = \frac{5\,{\,} \cdot 10}{10^5}\hspace{0.15cm} \underline{= 0.5\,{\rm mW}} \hspace{0.05cm}.$$
- For profile $\rm (B)$ the received power is about $4$ times smaller:
- $$P_{\rm E}(d = 100\,{\rm m}) = \frac{5\,{\rm W} \cdot 10}{10^{5.6}}\approx \frac{5\,{\rm W} \cdot 10}{4 \cdot 10^{5}}\hspace{0.15cm} \underline{= 0.125\,{\rm mW}} \hspace{0.05cm}.$$
(4) Below the breakpoint $(d < 100 \ \rm m)$, the deviation is determined by the last summand of profile $\rm (B)$:
- $${\rm \Delta}L_{\rm P}(d = 50\,{\rm m}) = (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left (1 + {d}/{d_{\rm BP}} \right )= (4-2) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (1.5)\hspace{0.15cm} \underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$
(5) Here the profile $\rm (A)$ with $L_{\rm BP} = 50 \ \rm dB$ gives:
- $$L_{\rm P}(d = 200\,{\rm m}) = 50\,{\rm dB} + 4 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (2)\hspace{0.15cm} {\approx 62\,{\rm dB}} \hspace{0.05cm}.$$
- On the other hand, the profile $\rm (B)$ leads to the result:
- $$L_{\rm P}(d = 200\,{\rm m}) = 50\,{\rm dB} + 20\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (200) + 20\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (3) = 10\,{\rm dB} + 46\,{\rm dB} + 9.5\,{\rm dB} \hspace{0.15cm} {\approx 65.5\,{\rm dB}}$$
- $$\Rightarrow \hspace{0.3cm} {\rm \Delta}L_{\rm P}(d = 200\,{\rm m}) = 65.5\,{\rm dB} - 62\,{\rm dB}\hspace{0.15cm} \underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$
- You can see that $\Delta L_{\rm P}$ is almost symmetrical to $d = d_{\rm BP}$ if you plot the distance $d$ logarithmically as in the given graph.
