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  $d$.




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.


What is the vehicle speed?

$v \ = \ $

$\ \ \rm km/h$


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  $60^\circ$.


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  $240^\circ$.


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

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


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

$\Delta d \ = \ $

$\ \ \rm m$


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

$d_2/d_1 \ = \ $


Find the distances  $d_1$  and  $d_2$ .

$d_1 \ = \ $

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

$\ \rm m$


(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:

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