View source for Aufgaben:Exercise 2.5Z: Multi-Path Scenario


{{quiz-Header|Buchseite=Mobile Kommunikation/Das GWSSUS&ndash;Kanalmodell}}

[[File:EN_Mob_A_2_5Z.png|right|frame|Mobile radio scenario with three paths]]
In&nbsp; [[Aufgaben:Exercise_2.5:_Scatter_Function| Exercise 2.5]], a delay&ndash;Doppler function (or scatter function) was given. From this, we will calculate and interpret the other system functions. The given scatter function&nbsp; $s(\tau_0, f_{\rm D})$&nbsp; was
:$$s(\tau_0, f_{\rm D}) =\frac{1}{\sqrt{2}} \cdot \delta (\tau_0) \cdot \delta (f_{\rm D} - 100\,{\rm Hz}) \ - \ $$
:$$\hspace{1.5cm} \ - \ \hspace{-0.2cm} \frac{1}{2} \cdot \delta (\tau_0 \hspace{-0.05cm}- \hspace{-0.05cm}1\,{\rm \mu s}) \cdot \delta (f_{\rm D} \hspace{-0.05cm}- \hspace{-0.05cm}50\,{\rm Hz}) \ - \frac{1}{2} \cdot \delta (\tau_0 \hspace{-0.05cm}- \hspace{-0.05cm}1\,{\rm \mu s}) \cdot \delta (f_{\rm D}\hspace{-0.05cm} + \hspace{-0.05cm}50\,{\rm Hz}) 
 \hspace{0.05cm}.$$


''Note:'' &nbsp; In our learning tutorial,&nbsp; $s(\tau_0, \hspace{0.05cm} f_{\rm D})$&nbsp; is also identified with&nbsp; $\eta_{\rm VD}(\tau_0, \hspace{0.05cm}f_{\rm D})$&nbsp;. 

Here we have replaced the delay variable&nbsp; $\tau$&nbsp; with&nbsp; $\tau_0$&nbsp;. The new variable&nbsp; $\tau_0$&nbsp; describes the difference between the delay of a path and the delay&nbsp; $\tau_1$&nbsp; of the main path. The main path is thus identified in the above equation by&nbsp; $\tau_0 = 0$&nbsp;.

Now, we try to find a mobile radio scenario in which this scatter function would actually occur. The basic structure is sketched above as a top view, and the following hold:
* A single frequency is transmitted&nbsp; $f_{\rm S} = 2 \ \rm GHz$.
* The mobile receiver&nbsp; $\rm (E)$&nbsp; is represented here by a yellow dot. It is not known whether the vehicle is stationary, moving towards the transmitter&nbsp; $\rm (S)$&nbsp; or moving away from it.
* The signal reaches the receiver via a main path (red) and two secondary paths (blue and green). Reflections from the obstacles cause phase shifts of&nbsp; $\pi$.
* ${\rm S}_2$&nbsp; and&nbsp; ${\rm S}_3$&nbsp; are to be understood here as fictitious transmitters from whose position the angles of incidence&nbsp; $\alpha_2$&nbsp; and&nbsp; $\alpha_3$&nbsp; of the secondary paths can be determined.
* Let the signal frequency be &nbsp; $f_{\rm S}$, the angle of incidence &nbsp; $\alpha$, the velocity&nbsp; $v$&nbsp; and the velocity of light&nbsp; $c = 3 \cdot 10^8 \ \rm m/s$. Then, the Doppler frequency is
:$$f_{\rm D}= {v}/{c} \cdot f_{\rm S} \cdot \cos(\alpha)
 \hspace{0.05cm}.$$
* The damping factors&nbsp; $k_1$,&nbsp; $k_2$&nbsp; and&nbsp; $k_3$&nbsp; are inversely proportional to the path lengths&nbsp; $d_1$,&nbsp; $d_2$&nbsp; and&nbsp; $d_3$. This corresponds to the path loss exponent&nbsp; $\gamma = 2$.
*This means: &nbsp; The signal power decreases quadratically with distance&nbsp; $d$&nbsp; and accordingly the signal amplitude decreases linearly with&nbsp; $d$.







''Notes:''
* This task belongs to chapter&nbsp; [[Mobile_Communications/Das_GWSSUS%E2%80%93Kanalmodell| Das GWSSUS&ndash;Kanalmodell]].
*We focus especially on the &nbsp; [[Mobile_Communications/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung#Gebr.C3.A4uchliches_Pfadverlustmodell| path-loss model]]&nbsp; and the &nbsp; [[Mobile_Communications/Statistische_Bindungen_innerhalb_des_Rayleigh-Prozesses#Dopplerfrequenz_und_deren_Verteilung| Doppler effect]].
 



===Questionnaire===
<quiz display=simple>
{At first, consider only the Dirac function at&nbsp; $\tau = 0$&nbsp; and&nbsp; $f_{\rm D} = 100 \ \rm Hz$. Which statements apply to the receiver?
|type="()"}
- The receiver is standing.
+ The receiver moves directly towards the transmitter.
- The receiver moves away in the opposite direction to the transmitter.

{What is the vehicle speed?
|type="{}"}
$v \ = \ ${ 54 3% } $\ \ \rm km/h$

{Which statements apply to the Dirac at&nbsp; $\tau_0 = 1 \ \ \rm &micro; s$&nbsp; and&nbsp; $f_{\rm D} = +50 \ \ \rm Hz$?
|type="[]"}
+ This Dirac comes from the blue path.
- This Dirac comes from the green path.
- The angle&nbsp; is&nbsp; $30^\circ$.
+ The angle&nbsp; is&nbsp; $60^\circ$.

{What statements apply to the green path?
|type="[]"}
+ We have $\tau_0 = 1 \ \ \rm &micro; s$&nbsp; and&nbsp; $f_{\rm D} = \, &ndash;50 \ \ \rm Hz$.
- The angle&nbsp; $\alpha_3$&nbsp; (see graph) is&nbsp; $60^\circ$.
+ The angle&nbsp; $\alpha_3$&nbsp; is&nbsp; $240^\circ$.

{Which of the following relations hold between the two side paths?
|type="[]"}
+ $d_3 = d_2$.
+ $k_3 = k_2$.
+ $\tau_3 = \tau_2$.

{What is the difference in time&nbsp; $\Delta d = d_2 - d_1$?
|type="{}"}
$\Delta d \ = \ ${ 300 3% } $\ \ \rm m$

{What is the relationship between&nbsp; $d_2$&nbsp; and&nbsp; $d_1$?
|type="{}"}
$d_2/d_1 \ = \ ${ 1,414 3% }

{Find the distances&nbsp; $d_1$&nbsp; and&nbsp; $d_2$&nbsp;.
|type="{}"}
$d_1 \ = \ ${ 724 3% } $\ \rm m$
$d_2 \ = \ ${ 1024 3% } $\ \rm m$
</quiz>

===Sample solution===
{{ML-Kopf}}
'''(1)'''&nbsp; The Doppler frequency is positive for $\tau_0$. This means that the receiver is moving towards the transmitter &nbsp; &#8658; &nbsp; <u>solution 2</u> is correct.


'''(2)'''&nbsp; The equation for the Doppler frequency is
:$$f_{\rm D}= \frac{v}{c} \cdot f_{\rm S} \cdot \cos(\alpha)
 \hspace{0.05cm},$$
 If the angle of incidence is $\alpha=0$, the Doppler frequency is
:$$f_d=\frac{v}{c}\cdot f_S$$

*The speed of the receiver is then
:$$v = \frac{f_{\rm D}}{f_{\rm S}} \cdot c = \frac{10^2\,{\rm Hz}}{2 \cdot 10^9\,{\rm Hz}} \cdot 3 \cdot 10^8\,{\rm m/s} = 15\,{\rm m/s}
  \hspace{0.1cm} \underline {= 54 \,{\rm km/h}} 
  \hspace{0.05cm}.$$


'''(3)'''&nbsp; <u>Solutions 1 and 4</u> are correct:
*The Doppler frequency $f_{\rm D} = 50 \ \rm Hz$ comes from the blue path, because the receiver moves towards the virtual transmitter ${\rm S}_2$ (i.e., towards the reflection point), although not directly. In other words, the movement of the receiver <b>reduces</b> the blue path's length.
*The angle $\alpha_2$ between the direction of movement and the connecting line ${\rm S_2 &ndash; E}$ is $60^\circ$:
:$$\cos(\alpha_2) = \frac{f_{\rm D}}{f_{\rm S}} \cdot \frac{c}{v} = \frac{50 \,{\rm Hz}\cdot 3 \cdot 10^8\,{\rm m/s}}{2 \cdot 10^9\,{\rm Hz}\cdot 15\,{\rm m/s}}  = 0.5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \alpha_2
  \hspace{0.1cm} \underline {= 60^{\circ} } 
  \hspace{0.05cm}.$$


'''(4)'''&nbsp; <u>Statements 1 and 3</u> are correct: 
*From $f_{\rm D} = \, &ndash;50 \ \rm Hz$ follows $\alpha_3 = \alpha_2 &plusmn; \pi$, so $\alpha_3 \ \underline {= 240^\circ}$.



'''(5)'''&nbsp; <u>All statements are correct</u>: 
*The two Dirac functions at $&plusmn; 50 \ \ \rm Hz$ have the same delay. We have $\tau_3 = \tau_2 = \tau_1 + \tau_0$. 
*From the equality of the delays, however, also follows that&nbsp; $d_3 = d_2$. As both paths have the same length, their damping factors are also equal.



'''(6)'''&nbsp; The delay difference is $\tau_0 = 1 \ \rm &micro; s$, as shown in the equation for $s(\tau_0, f_{\rm D})$.
* This gives the difference in length: 
:$$\Delta d = \tau_0 \cdot c = 10^{&ndash;6} {\rm s} \cdot 3 \cdot 10^8 \ \rm m/s \ \ \underline {= 300 \ \ \rm m}.$$



'''(7)'''&nbsp; The path loss exponent was assumed to be $\gamma = 2$ for this task. 
*Then $k_1 = K/d_1$ and $k_2 = K/d_2$. 

*The minus sign takes into account the $180^\circ$ phase rotation on the secondary paths. 
*From the weights of the Dirac functions one can read $k_1 = \sqrt{0.5}$ and $k_2 = -0.5$. From this follows:
:$$\frac{d_2}{d_1}  = \frac{k_1}{-k_2} = \frac{1/\sqrt{2}}{0.5} = \sqrt{2}
  \hspace{0.15cm} \underline {= 1.414} 
  \hspace{0.05cm}.$$

*The constant $K$ is only an auxiliary variable that does not need to be considered further.



'''(8)'''&nbsp; From &nbsp; $d_2/d_1 = 2^{-0.5}$&nbsp; and&nbsp; $\Delta d = d_2 \, - d_1 = 300 \ \rm m$&nbsp; finally follows:
:$$\sqrt{2} \cdot d_1 - d_1 = 300\,{\rm m} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}
  d_1 = \frac{300\,{\rm m}}{\sqrt{2}  - 1}   \hspace{0.15cm} \underline {= 724\,{\rm m}} 
  \hspace{0.3cm} \Rightarrow \hspace{0.3cm} d_2 = \sqrt{2} \cdot d_1 \hspace{0.15cm} \underline {= 1024\,{\rm m}}
  \hspace{0.05cm}. $$
{{ML-Fuß}}


[[Category:Exercises for Mobile Communications|^2.3 The GWSSUS Channel Model^]]