Aufgaben:Exercise 1.6Z: Comparison of Rayleigh and Rice

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In this task Rayleigh–Fading and Rice–Fading are to be compared with each other.

The graphic shows the complex factor  $z(t) = x(t) + {\rm j} \cdot y(t)$  in the complex plane For the low-pass–transmit signal  $s(t) = 1$, which with respect to a bandpass–system corresponds to a cosine oscillation with amplitude  $1$  the low-pass–receive signal  $r(t)$  is identical to  $z(t)$.

• The upper diagram describes  Rayleigh–Fading, where the component signals  $x(t)$  and  $y(t)$  are each Gaussian distributed with the variance  $\sigma^2$. The probability density function of the amount  $a(t) = |z(t)|$  is for  $a ≥ 0$:

$$f_a(a) = \frac{a}{\sigma^2} \cdot {\rm exp} [ -\frac{a^2 }{2\sigma^2}] \hspace{0.05cm}.$ :The square expectation value of  $z(t)$  is  $1$: $${\rm E}\left [ |z(t)|^2 \right ] = 2 \sigma^2 = 1 \hspace{0.3cm}

\Rightarrow \hspace{0.3cm}

   \sigma = {1}/{\sqrt{2} \approx 0.707
  \hspace{0.05cm}.$$

*The lower phase diagram is created at  [[Mobile_Communication/Non-Frequency-Selective_Fading_with_Direct Component#Example_Signal_Verl.C3.A4ufe_bei_Rice.E2.80.93Fading|Rice–Fading]]. Here, too, are  $x(t)$  and  $y(t)$  gaussian distributed with the variance  $\sigma^2$, but now with mean value  $x_0$  and  $y_0$. The WDF with the modified Bessel function  ${\rm I}_0$  for  $a ≥ 0$:
:$$f_a(a) = \frac{a}{\sigma^2} \cdot {\rm exp} \left[ -\frac{a^2 + |z_0|^2}{2\sigma^2}\right] \cdot {\rm I}_0 \left [ \frac{a \cdot |z_0|}{\sigma^2} \right ]\hspace{0.05cm}.$$

:The root mean square now includes the direct component  $z_0 = x_0 + {\rm j} \cdot y_0$
$${\rm E}\left [ |z(t)|^2 \right ] = 2 \cdot \sigma^2 + |z_0|^2 \hspace{0.05cm}.$$

For the system comparison
* is assumed to be constant  ${\rm E}\big[|z(t)|^2\big] = 1$ ,
* the <i>Rice–Fading</i> is based on the preferred direction as shown in the graphic, 
* let the power be split between the direct path  $(|z_0|^2)$  and the scattering paths  $(2\sigma^2)$  in the ratio  $4:1$ .


For the subtasks '''(1)''' to '''(4)''',   $s(t) = 1$, while a BPSK–signal is assumed in the subtasks '''(5)'' or '''(6)'''. The low pass–signal  $s(t)$  thus has a rectangular shape with the possible values  $±1$. The duration of a rectangular pulse is  $T = 10 \ \rm ms$.







''Notes:''
* The task belongs to chapter  [[Mobile_Communication/Non-Frequency Selective_Fading_with_Direct Component| Non-Frequency Selective Fading with Direct Component]].
* The circles drawn in the graphic (violet and green) refer to the subtasks '''(3)'' and '''(4)''.
 



==='"`UNIQ--h-0--QINU`"'Questionnaire===
'"`UNIQ--quiz-00000002-QINU`"'

==='"`UNIQ--h-1--QINU`"'Sample solution===
{[[:Template:ML Head]]
'''(1)'''  Correct is the <u>solution 2</u>: 
*The first suggestion returns a <i>Rayleigh–fading</i>–model. With the last setting the result would be
:$$|z(t)| = {\rm j}  \hspace{0.3cm}\Rightarrow \hspace{0.3cm} r(t) = s(t) \cdot z(t) = {\rm j} \cdot s(t) 
  \hspace{0.05cm}.$$

*If we take into account that we are in the equivalent low-pass range, then a cosine input would produce a minus–sinusoidal output signal $r_{\rm BP}(t)$. 
*On the other hand, solution 2 applies to all possible signals:
$$|z(t)| = x_0 = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} r(t) = s(t)  
  \hspace{0.05cm}.$$


'''(2)'''  For the given <i>Rayleigh–Fading</i> the parameter $\sigma^2 = 0.5$. This results in $z(t)$ for the quadratic mean of the multiplicative factor:
$${\rm E}\left [ |z(t)|^2 \right ] = 2 \sigma^2 = 1 
  \hspace{0.05cm}.$$

The <i>Rice–Fading</i> should have exactly the same performance. That is, it should apply:
$$|z_0|^2 + 2 \sigma^2 = 1 
  \hspace{0.05cm}.$$

Furthermore, it was requested:
* The ratio of the power of the deterministic part ($|z_0|^2$) and the stochastic part ($2\sigma^2$) is $4$. It follows from this:
:$$2 \sigma^2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \sigma^2 = 0.1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}     \sigma = \frac{1}{\sqrt{10}} \hspace{0.25cm} \underline{ \approx 0.316} \hspace{0.05cm},$$
:$$|z_0|^2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.8 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} |z_0| = 0.894
  \hspace{0.05cm}.$$
* The division of $z_0 = x_0 + {\rm j} \cdot y_0$ results from the graph. You can see that $y_0 = x_0$ (center of the cloud in the first quadrant below $45^\circ$):
:$$x_0 = y_0 = \frac{|z_0|}{\sqrt{2}}} = \frac{0.894}{\sqrt{2}} \hspace{0.25cm} \underline{ \approx 0.632} \hspace{0.05cm}.$$


'''(3)'''  The range „$20 \cdot {\rm lg} \, |z(t)| ≤ \, –6 \ \ \rm dB$” is equal to „$|z(t)|≤0.5$”. 
*This area is marked by the purple circle in the graph on the specifications page. 
*You can see from this that the <u>statement 1 is correct</u>, because with Rayleigh there are more points within the violet circle.


[[File:P_ID2136__Mob_Z_1_6c.png|right|frame|Signal Section and WDF $f_a(a)$ for Rayleigh and Rice]]

This graphic shows the result of a system simulation with the program „Mobile radio channel” from the (former) practical course „Simulation of digital transmission systems”. 
*Both from the signal section as well as from the WDF one can see that the blue curve (Rayleigh) has more parts under the violet curve than the red curve (Rice).
<br clear="all">
'''(4)'''  Correct is <u>solution 3</u>: 
*For the Rayleigh distribution the probability
$$${\rm Pr}(20 \cdot {\rm lg}\hspace{0.15cm} a(t) \le 0\,\,{\rm dB}) = {\rm Pr}(a(t) \le 1) = 1 - {\rm e}^{-1} \approx 63\,\,\
  \hspace{0.05cm}.$$

*From the graph you can see that the probability for the riceversion (with the selected parameters) is about the same as the probability of falling below. 
*This result can also be guessed from the complex representation of $z(t)$ on the data page (easier if you already know the result).
* Among other things, because the tip of the Gaussian cloud is still within the green circle.


[[File:P_ID2137__Mob_Z_1_6e.png|right|frame|Complex receive signal $r(t)$ for Rayleigh and Rice]]
'''(5)''''  Correct are the <u>solution proposals 2 and 3</u>:
* At <i>Rice–Fading</i> the point cloud of $z(t)$ is in the first quadrant. If you multiply $z(t)$ by $s(t) = ±1$, you get two point clouds in the first and third quadrant. This does not change the WDF $f_a(a)$ of the amount.
* Even with <i>Rayleigh–Fading</i> the density functions $f_a(a)$ of the amount for $|z(t)|$ and $|r(t)|$ do not show any differences. Since the phase $\phi$ is equally distributed, the final result also has the same point clouds.
*However, if you look at the development of the complex representation of $z(t)$ and $r(t)$ dynamically, there are indeed differences. 
*Larger jumps occur in the complex representation of $r(t)$, whenever there are phase jumps by $±180^\circ$ in the transmitted signal $s(t)$, i.e. when changing symbols. 
*Thus, ${\it \Phi}_z(f_{\rm D})$ and ${\it \Phi}_r(f_{\rm D})$ &ndash are different for Rice–Fading; the latter is wider – and accordingly the corresponding autocorrelation function.

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