Aufgaben:Exercise 2.5Z: Multi-Path Scenario

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Mobile radio scenario with three paths

In  Exercise 2.5, a delay–Doppler function (or scatter function) was given.  From this, you should calculate and interpret the other system functions.  The given scatter function  $s(\tau_0, f_{\rm D})$  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:   In our learning tutorial,  $s(\tau_0, \hspace{0.05cm} f_{\rm D})$  is also identified with  $\eta_{\rm VD}(\tau_0, \hspace{0.05cm}f_{\rm D})$.

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

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:  $f_{\rm S} = 2 \ \rm GHz$.
• The mobile receiver  $\rm (E)$  is represented here by a yellow dot.  It is not known whether the vehicle is stationary, moving towards the transmitter  $\rm (S)$  or moving away.
• 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  $\pi$.
• ${\rm S}_2$  and  ${\rm S}_3$  are to be understood here as fictitious transmitters from whose position the angles of incidence  $\alpha_2$  and  $\alpha_3$  of the secondary paths can be determined.
• Let the signal frequency be  $f_{\rm S}$,  the angle of incidence  $\alpha$, the velocity  $v$  and the velocity of light  $c = 3 \cdot 10^8 \ \rm m/s$.  Then, the Doppler frequency is

$$f_{\rm D}= {v}/{c} \cdot f_{\rm S} \cdot \cos(\alpha)\hspace{0.05cm}.$$

• The damping factors  $k_1$,  $k_2$  and  $k_3$  are inversely proportional to the path lengths  $d_1$,  $d_2$  and  $d_3$. This corresponds to the path loss exponent  $\gamma = 2$.
• This means:   The signal power decreases quadratically with distance  $d$  and accordingly the signal amplitude decreases linearly with  $d$.

Notes:

• This task belongs to chapter  The GWSSUS Channel Model.
• We also refer to the pages  Common path-loss model  and  Doppler frequency and its distribution.

Questionnaire

1 At first, consider only the Dirac function at  $\tau = 0$  and  $f_{\rm D} = 100 \ \rm Hz$.  Which statements apply to the receiver?

	The receiver is standing.
	The receiver moves directly towards the transmitter.
	The receiver moves away in the opposite direction to the transmitter.

2 What is the vehicle speed?

$v \ = \ $ $\ \ \rm km/h$

3 Which statements apply to the Dirac at  $\tau_0 = 1 \ \ \rm µ s$  and  $f_{\rm D} = +50 \ \ \rm Hz$?

	This Dirac comes from the blue path.
	This Dirac comes from the green path.
	The angle  is  $30^\circ$.
	The angle  is  $60^\circ$.

4 What statements apply to the green path?

	We have $\tau_0 = 1 \ \rm µ s$  and  $f_{\rm D} = -50 \ \rm Hz$.
	The angle  $\alpha_3$  (see graph) is  $60^\circ$.
	The angle  $\alpha_3$  is  $240^\circ$.

5 Which of the following relations hold between the two side paths?

	$d_3 = d_2$.
	$k_3 = k_2$.
	$\tau_3 = \tau_2$.

6 What is the difference  $\Delta d = d_2 - d_1$  in time?

$\Delta d \ = \ $ $\ \ \rm m$

7 What is the relationship between  $d_2$  and  $d_1$?

$d_2/d_1 \ = \ $

8 Find the distances  $d_1$  and  $d_2$ .

$d_1 \ = \ $ $\ \rm m$

$d_2 \ = \ $ $\ \rm m$

Solution

Solution

(1)  The Doppler frequency is positive for  $\tau_0$.  This means that the receiver is moving towards the transmitter   ⇒   solution 2 is correct.

(2)  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  $\alpha=0$, then the Doppler frequency is

$$f_{\rm D}=\frac{v}{c}\cdot f_{\rm S}.$$

• In this case the speed of the receiver is

$$v = \frac{f_{\rm D}}{f_{\rm S}} \cdot c = \frac{10^2\,{\rm Hz}}{2 \cdot 10^9\,{\rm Hz}} \cdot 3 \cdot 10^8\,{\rm m/s} = 15\,{\rm m/s}\hspace{0.1cm} \underline {= 54 \,{\rm km/h}}\hspace{0.05cm}.$$

(3)  Solutions 1 and 4 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 reduces the blue path's length.
• The angle  $\alpha_2$  between the direction of movement and the connecting line  ${\rm S_2 – E}$  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)  Statements 1 and 3 are correct:  From  $f_{\rm D} = \, –50 \ \rm Hz$  follows  $\alpha_3 = \alpha_2 ± \pi$   ⇒   $\alpha_3 \ \underline {= 240^\circ}$.

(5)  All statements are correct:

• The two Dirac functions at  $± 50 \ \ \rm Hz$  have the same delay.  We have  $\tau_3 = \tau_2 = \tau_1 + \tau_0$.
• From the equality of the delays, however, it follows also  $d_3 = d_2$.
• As both paths have the same length, their damping factors are equal, too.

(6)  The delay difference is  $\tau_0 = 1 \ \rm µ s$, as shown in the equation for  $s(\tau_0, f_{\rm D})$.  This gives the difference in length:

$$\Delta d = \tau_0 \cdot c = 10^{–6} {\rm s} \cdot 3 \cdot 10^8 \ \rm m/s \ \ \underline {= 300 \ \ \rm m}.$$

(7)  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 constant  $K$  is only an auxiliary variable that does not need to be considered further.

• 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  $k_2 = -0.5$.  Therefore:

$$\frac{d_2}{d_1} = \frac{k_1}{-k_2} = \frac{1/\sqrt{2}}{0.5} = \sqrt{2}\hspace{0.15cm} \underline {= 1.414}\hspace{0.05cm}.$$

(8)  From   $d_2/d_1 = 2^{-0.5}$  and  $\Delta d = d_2 \, - d_1 = 300 \ \rm m$  finally follows:

$$\sqrt{2} \cdot d_1 - d_1 = 300\,{\rm m} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}d_1 = \frac{300\,{\rm m}}{\sqrt{2} - 1} \hspace{0.15cm} \underline {= 724\,{\rm m}}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} d_2 = \sqrt{2} \cdot d_1 \hspace{0.15cm} \underline {= 1024\,{\rm m}}\hspace{0.05cm}. $$