Aufgaben:Exercise 4.2: UMTS Radio Channel Basics

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Path loss,  frequency–selective and time–selective fading Korrektur

UMTS also has quite a few effects leading to degradation that must be taken into account during system planning:

• ${\rm Interference}$:  Since all users are simultaneously served in the same frequency band,  each user is interfered by other users.

• ${\rm Path\:loss}$:  The received power  $P_{\rm E}$  of a radio signal decreases with distance   $d$   by a factor  $d^{- \gamma}$.

• ${\rm Multipath\:propagation}$:  The signal reaches the mobile receiver not only through the direct path,  but through several paths – differently attenuated and differently delayed.

• ${\rm Doppler\:effect}$:  If transmitter and/or receiver move,  frequency shifts can occur depending on speed and the direction  $($Which angle?  Towards each other? Away from each other?$)$.

In the book  "Mobile Communications"  these effects have already been discussed in detail. The diagrams convey only a few pieces of information regarding

• Path loss:  Path loss indicates the decrease in the received power with distance  $d$  from the transmitter.  Above the so-called  "break point"  applies approximately to the received power:

$$\frac{P(d)}{P(d_0)} = \alpha_0 \cdot \left ( {d}/{d_0}\right )^{-4}.$$

According to the upper graph  $\alpha_{0} = 10^{-5}$  $($correspondingly  $50 \ \rm dB)$  and  $d_{0} = 100 \ \rm m$.

• Frequency-selective fading:  The power transfer function  $|H_{\rm K}(f)|^{2}$  at a given time according to the middle graph illustrates frequency-selective fading.  The blue-dashed horizontal line,  on the other hand,  indicates non-frequency-selective fading.

Such frequency-selective fading occurs when the coherence bandwidth  $B_{\rm K}$  is much smaller than the signal bandwidth  $B_{\rm S}$.  Here,  with the  "delay spread"  $T_{\rm V}$   ⇒   difference between the maximum and minimum delay times:

$$B_{\rm K}\approx \frac{1}{T_{\rm V}}= \frac{1}{\tau_{\rm max}- \tau_{\rm min}}.$$

• Time-selective fading:  The bottom graph shows the power transfer function  $|H_{\rm K}(t)|^{2}$  for a fixed frequency  $f_{0}$.  The sketch is to be understood schematically,  because for the time-selective fading considered here exactly the same course was chosen as in the middle diagram for the frequency-selective fading  $($pure convenience of the author$)$.

Here a so-called  "Doppler spread"  $B_{\rm D}$  arises,  defined as the difference between the maximum and the minimum Doppler frequency.  The inverse  $T_{\rm D} = 1/B_{\rm D}$  is called  "coherence time"  or also  "correlation duration".  In UMTS,  time-selective fading occurs whenever  $T_{\rm D} \ll T_{\rm C}$  $($chip duration$)$.

Hints:

• This exercise belongs to the chapter  "General Description of UMTS".

• Reference is made in particular to the sections   "Properties of the UMTS radio channel"   and   "Frequency-selective and time-selective fading".

• For UMTS,  the bandwidth:  $B_{\rm S} = 5 \ \rm MHz$  and the chip duration:  $T_{\rm C} \approx 0.26 \ \rm µ s$.

Questions

1

Starting from the top graph on the information page,  calculate the path loss  $($in  $\rm dB)$  for  $d = \rm 5 \ km$.

${\rm path\ loss} \ = \ $ $\ \rm dB $.

2

What statements are true regarding frequency-selective fading?

	This is caused by multipath reception.
	It is caused by movement of transmitter and/or receiver.
	Different frequencies are attenuated differently.
	An echo at a distance  $1\ \rm µ s$  results in frequency-selective fading.

3

What statements are true regarding time-selective fading?

	This arises due to multipath reception.
	It results from movement of transmitter and/or receiver.
	Different frequencies are attenuated differently.

Solution

Solution

(1)  According to the sketch, the breakpoint is at $d_{0} = 100 \ \rm m$.

• For $d ≤ d_{0}$, the path loss is equal to $\alpha_{0} \cdot (d/d_{0})^{-2}$. For $d = d_{0} = 100 \ \rm m$ holds:

$${\rm path loss} = \alpha_0 = 10^{-5}\hspace{0.5cm}\Rightarrow\hspace{0.5cm}{50\,{\rm dB}}.$$

• Above $d_{0}$, the path loss is equal to $\alpha_{0} \cdot (d/d_{0})^{-4}$. Thus, at $5 \ \rm km$ distance, one obtains:

$${\rm path loss} = 10^{-5}\cdot 50^{-4} = 1.6 \cdot 10^{-12}\hspace{0.5cm}\rightarrow\hspace{0.5cm}\underline{118\,{\rm dB}}.$$

(2)  Correct statements 1, 3, and 4:

• Frequency selective fading is due to multipath reception. This means:
• Different frequency components are delayed and attenuated differently by the channel.
• This results in attenuation and phase distortion.
• Because $\tau_{\rm max} = 1 \ \rm µ s$ (simplifying $\tau_{\rm min} = 0$ is set) further results in

$$B_{\rm K} = \frac{1}{\tau_{\rm max}- \tau_{\rm min}} = 1\,{\rm MHz}\ \ll \ B_{\rm S} \hspace{0.15cm}\underline {= 5\,{\rm MHz}}.$$

(3)  Correct is statement 2.

• Statements 1 and 3, on the other hand, are valid for frequency-selective fading – see subtask (2).