View source for Aufgaben:Exercise 4.2: UMTS Radio Channel Basics


{{quiz-Header|Buchseite=Examples_of_Communication_Systems/General_Description_of_UMTS

}}

[[File:P_ID1931__Bei_A_4_2.png|right|frame|Path loss,&nbsp; frequency&ndash;selective and time&ndash;selective fading '''Korrektur''']]
UMTS also has quite a few effects leading to degradation that must be taken into account during system planning:
*${\rm Interference}$:&nbsp; Since all users are simultaneously served in the same frequency band,&nbsp; each user is interfered by other users.

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

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

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


In the book&nbsp; "[[Mobile Communications]]"&nbsp; these effects have already been discussed in detail. The diagrams convey only a few pieces of information regarding
*<u>Path loss:</u>&nbsp; Path loss indicates the decrease in the received power with distance&nbsp; $d$&nbsp; from the transmitter.&nbsp; Above the so-called&nbsp; "break point"&nbsp; 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&nbsp; $\alpha_{0} = 10^{-5}$&nbsp; $($correspondingly&nbsp; $50 \ \rm dB)$&nbsp; and&nbsp; $d_{0} = 100 \ \rm m$.

*<u>Frequency-selective fading:</u>&nbsp; The power transfer function&nbsp; $|H_{\rm K}(f)|^{2}$&nbsp; at a given time according to the middle graph illustrates frequency-selective fading.&nbsp; The blue-dashed horizontal line,&nbsp; on the other hand,&nbsp; indicates non-frequency-selective fading.
::Such frequency-selective fading occurs when the coherence bandwidth&nbsp; $B_{\rm K}$&nbsp; is much smaller than the signal bandwidth&nbsp; $B_{\rm S}$.&nbsp; Here,&nbsp; with the&nbsp; "delay spread"&nbsp; $T_{\rm V}$ &nbsp; &rArr; &nbsp; difference between the maximum and minimum delay times:
::$$B_{\rm K}\approx \frac{1}{T_{\rm V}}= \frac{1}{\tau_{\rm max}- \tau_{\rm min}}.$$

*<u>Time-selective fading:</u>&nbsp; The bottom graph shows the power transfer function&nbsp; $|H_{\rm K}(t)|^{2}$&nbsp; for a fixed frequency&nbsp; $f_{0}$.&nbsp; The sketch is to be understood schematically,&nbsp; because for the time-selective fading considered here exactly the same course was chosen as in the middle diagram for the frequency-selective fading&nbsp; $($pure convenience of the author$)$.
::Here a so-called&nbsp; "Doppler spread"&nbsp; $B_{\rm D}$&nbsp; arises,&nbsp; defined as the difference between the maximum and the minimum Doppler frequency.&nbsp; The inverse&nbsp; $T_{\rm D} = 1/B_{\rm D}$&nbsp; is called&nbsp; "coherence time"&nbsp; or also&nbsp; "correlation duration".&nbsp; In UMTS,&nbsp; time-selective fading occurs whenever&nbsp; $T_{\rm D} \ll T_{\rm C}$&nbsp; $($chip duration$)$.



<u>Hints:</u> 

*This exercise belongs to the chapter&nbsp; [[Examples_of_Communication_Systems/General_Description_of_UMTS|"General Description of UMTS"]].
 
*Reference is made in particular to the sections &nbsp; [[Examples_of_Communication_Systems/General_Description_of_UMTS#Properties_of_the_UMTS_radio_channel|"Properties of the UMTS radio channel"]] &nbsp; and &nbsp; [[Examples_of_Communication_Systems/General_Description_of_UMTS#Frequency. E2.80.93_and_time.E2.80.93selective_fading|"Frequency- and time-selective fading"]].

*For UMTS,&nbsp; the bandwidth:&nbsp; $B_{\rm S} = 5 \ \rm MHz$&nbsp; and the chip duration:&nbsp; $T_{\rm C} \approx 0.26 \ \rm &micro; s$.
 



===Questions===

<quiz display=simple>

{Starting from the top graph on the information page,&nbsp; calculate the path loss&nbsp; $($in&nbsp; $\rm dB)$&nbsp; for&nbsp; $d = \rm 5 \ km$.
|type="{}"}
${\rm path\ loss} \ = \ $ { 118 3% } $\ \rm dB $.

{What statements are true regarding frequency-selective fading?
|type="[]"}
+ 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&nbsp; $1\ \rm &micro; s$&nbsp; results in frequency-selective fading.

{What statements are true regarding time-selective fading?
|type="[]"}
- This arises due to multipath reception.
+ It results from movement of transmitter and/or receiver.
- Different frequencies are attenuated differently.

</quiz>

===Solution===
{{ML-Kopf}}

'''(1)'''&nbsp; 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)'''&nbsp; Correct <u>statements 1, 3, and 4</u>: 
*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 $\tau_{\rm max} = 1 \ \rm &micro; s$ (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)'''&nbsp; Correct is <u>statement 2</u>. 
*Statements 1 and 3, on the other hand, are valid for frequency-selective fading &ndash; see subtask '''(2)'''.


{{ML-Fuß}}



[[Category:Examples of Communication Systems: Exercises|^4.1 General Description of UMTS
^]]