Exercise 2.2: Simple Two-Path Channel Model

Zwei äquivalente Modelle
für den Zweiwege-Kanal

Here we consider a two-way–channel for mobile radio according to the adjacent graph, characterized by the model parameters

$$k_1 = 10^{-4}\hspace{0.05cm}, \hspace{0.2cm} \tau_{1} = 10\,{\rm µ s}\hspace{0.05cm}, \hspace{0.2cm}\tau_{2} = 11\,{\rm µ s} \hspace{0.05cm}.$$

Two different numerical values are considered for the damping factor on the secondary path:

$k_2 = 2 \cdot 10^{-5}$ ⇒ subtasks (1) to (4),
$k_2 = 10^{-4}$ ⇒ subtasks (5) and (6).

An equivalent channel model is shown below, with only the part highlighted in green being considered further. This means:

The basic attenuation (path loss) and the basic propagation time are not taken into account here.
The frequency response of this $(k_0, \tau_0$)–model is designated $H_0(f)$ .

An important descriptive parameter of any mobile radio system is the coherence bandwidth $B_{\rm K}$, which is defined in the chapter GWSSUS–Kanalmodell . The coherence bandwidth indicates whether the system can be approximated as non-frequency-selective:

This is justified if the signal bandwidth $B_{\rm S}$ is significantly smaller than the coherence bandwidth $B_{\rm K}$.
Otherwise, the mobile radio system is frequency-selective, which requires a more complicated description.

As a simple approximation for the coherence bandwidth, the reciprocal value of pulse broadening is often used in the literature (marked by an apostrophe in our learning tutorial):

$$B_{\rm K}\hspace{0.01cm}' = \frac{1}{\tau_{\rm max} - \tau_{\rm min}} \hspace{0.05cm}.$$

Notes:

This task belongs to the topic of the chapter Mehrwegeempfang beim Mobilfunk.
For the solution you also need the speed of light $c = 3 \cdot 10^8 \ \rm m/s$.
For $k_2$ only positive values are used here. However, as you may remember,if the secondary path is created by reflection on a wall, a phase change by $\pi$ occurs, resulting in a negative value of $k_2$.

Questionnaire

$d_1 \ = \ $

$\ \ \rm km$

$k_0 \ = \ $

$\tau_0 \ = \ $

$\ \ \rm µ s$

$|H_0(f = 0)| \ = \ $

$|H_0(f = 250 \ \rm kHz)| \ = \ $

$|H_0(f = 500 \ \rm kHz)| \ = \ $

	$f_{\rm S} = 500 \ \rm kHz$,
	$f_{\rm S} = 750 \ \rm kHz$,
	$f_{\rm S} = 1 \ \rm MHz$.

$k_2 = 2 \cdot 10^{-5} \text{:} \ \hspace{0.4cm} B_{\rm K}\hspace{0.01cm}' = \ $

$\ \rm MHz

$k_2 = 10^{-4} \text{:} \ \hspace{0.4cm} B_{\rm K}\hspace{0.01cm}' = \ $

$\ \rm MHz

	For GSM: $(B_{\rm S} = 200 \ \rm kHz)$ the channel is frequency selective.
	For UMTS: $(B_{\rm S} = 5 \ \rm MHz)$ the channel is frequency selective.

Sample solution

Musterlösung

(1) We have $\tau_1 = d_1/c$ ⇒ $ d_1 = \tau_1 \cdot c = 10^{-5} \rm s \cdot 3 \cdot 10^8 \ m/s \ \ \underline {= 3 \ km}$.

(2) The damping factor is $k_0 = k_2/k_1 \ \ \underline {= 0.2}$ and the delay time $\tau_0 = \tau_2 \ – \tau_1 \ \underline {= 1 \ \ \rm µ s}$.

The path loss effective for both paths is thus $k_1 = 10^{-4}$ and the basic delay time is $\tau_1 = 10 \ \ \rm µ s$.

(3) The impulse location is $$h_{\rm 0}(\tau) = \delta(\tau) + k_0 \cdot \delta(\tau - \tau_0) \hspace{0.05cm}.$ By Fourier transformation you get the frequency response :$$H_{\rm 0}(f) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1 + k_0 \cdot {\rm e}^{- {\rm j}\cdot 2 \pi f \tau_0}=1 + k_0 \cdot {\cos}( 2 \pi f \tau_0) + {\rm j}\cdot k_0 \cdot {\sin

}( 2 \pi f \tau_0)
 \hspace{0.05cm},$$

and thus to the following magnitude frequency response:
$$|H_{\rm 0}(f)| = \sqrt{ \left [ 1 + k_0 \cdot {\cos}( 2 \pi f \tau_0)\right ]^2 + k_0^2 \cdot {\sin^2 }( 2 \pi f \tau_0)}\hspace{0.3cm}

\Rightarrow \hspace{0.3cm}|H_{\rm 0}(f = 0)| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1+ k_0 \hspace{0.1cm} \underline {=1.2} \hspace{0.05cm},$$ $$|H_{\rm 0}(f = 250\,{\rm kHz})| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{ \left [ 1 + k_0 \cdot {\cos}( \pi/2)\right ]^2 + k_0^2 \cdot {\sin^2 }( \pi/2)} = \sqrt{1+ k_0^2} \hspace{0.1cm} \underline {\approx 1.02} \hspace{0.05cm},$$ $$|H_{\rm 0}(f = 500\,{\rm kHz})| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{ \left [ 1 + k_0 \cdot {\cos}( \pi)\right ]^2 + k_0^2 \cdot {\sin^2 }( \pi)} = {1- k_0} \hspace{0.1cm} \underline {= 0.8} \hspace{0.05cm}.$$ [[File:P_ID2158__Mob_A_2_2c.png|right|frame|Amount frequency response of a two-way channel]] The graphic (red curve) shows the function $|H_0(f)|$. *The values you are looking for are marked by the yellow dots. *The blue curve refers to the subtask (5) with $k_0 = 1 \ \Rightarrow \ k_2 = k_0 \cdot k_1 = 10^{–4}$. '''(4)''' Correct is the <u>solution 1</u>: *Destructive overlays exist for $|H_0(f)| < 1$, for example for $f = 500 \ \rm kHz$. *On the other hand: :$$|H_{\rm 0}(f = 750\,{\rm kHz})| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} |H_{\rm 0}(f = 250\,{\rm kHz})| \approx 1.02 > 1\hspace{0.05cm},$$ :$$|H_{\rm 0}(f = 1\,{\rm MHz})| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} |H_{\rm 0}(f = 0)| = 1.2 > 1 \hspace{0.05cm}.$ '''(5)''' The difference $\tau_{\rm max} \ – \tau_{\rm min}$ of the delay times in the two paths is equal to $\tau_0 = 1 \ \ \rm µ s$. *So the coherence bandwidth :'"`UNIQ-MathJax35-QINU`"' *The result is independent from $k_2$. It applies to $k_2 = 2 \cdot 10^{-5} \Rightarrow k_0 = 0.2$ and $k_2 = 10^{-4} \Rightarrow k_0 = $1 in the same way. *In the graph this approximation $B_{\rm K}\hspace{0.01cm}'$ is drawn for the coherence bandwidth. '''(6)''' Correct is the <u>solution 2</u>: *The channel is not frequency selective if the coherence bandwidth $B_{\rm K}$ is significantly larger than the signal bandwidth $B_{\rm S}$.

For the given channel, this is true for GSM, but not for UMTS. For UMTS there is a frequency-selective channel.

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