Difference between revisions of "Aufgaben:Exercise 4.2: UMTS Radio Channel Basics"

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{{quiz-Header|Buchseite=Beispiele von Nachrichtensystemen/Allgemeine Beschreibung von UMTS
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{{quiz-Header|Buchseite=Examples_of_Communication_Systems/General_Description_of_UMTS
  
 
}}
 
}}
  
[[File:P_ID1931__Bei_A_4_2.png|right|frame|Pfadverlust, frequenzselektives und zeitselektives Fading]]
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[[File:EN_Bei_A_4_2_v2.png|right|frame|Path loss,  frequency/time–selective fading ]]
Auch bei UMTS gibt es etliche negative Effekte, die man bei der Systemplanung berücksichtigen muss:
+
UMTS also has quite a few effects leading to degradation that must be taken into account during system planning:
*$\color{red}{\rm Interferenzen}$ durch andere Nutzer, da alle Nutzer gleichzeitig im gleichen Frequenzband versorgt werden.
+
*${\rm Interference}$:  Since all users are simultaneously served in the same frequency band,  each user is interfered by other users.
*$\color{red}{\rm Pfadverlust}$: Die Empfangsleistung $P_{\rm E}$ eines Funksignals nimmt mit der Entfernung $d$ um den Faktor $d^{– \gamma}$ ab.
 
*$\color{red}{\rm Mehrwegeempfang}$: Signal erreicht den mobilen Empfänger nicht nur über den direkten Pfad, sondern auf mehreren Wegen unterschiedlich gedämpft und verschieden verzögert.
 
*$\color{red}{\rm Dopplereffekt}$: Bewegen sich der Sender und/oder der Empfänger, so kann es zu Frequenzverschiebungen der Frequenz kommen abhängig von Geschwindigkeit und Richtung:
 
  
– Welcher Winkel?
+
*${\rm Path\:loss}$:  The received power  $P_{\rm E}$  of a radio signal decreases with distance   $d$   by a factor  $d^{- \gamma}$.
+
 
– Aufeinander zu?
+
*${\rm Multipath\:propagation}$:  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
 +
*<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:
 +
::$$B_{\rm K}\approx \frac{1}{T_{\rm V}}= \frac{1}{\tau_{\rm max}- \tau_{\rm min}}.$$
  
– Voneinander weg?
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*<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$)$.
  
Im Buch „Mobile Kommunikation” wurden diese Effekte bereits im Detail behandelt. Die Diagramme vermitteln nur einige wenige Informationen bezüglich
 
*Pfadverlust (obere Grafik),
 
*frequenzselektives Fading (Mitte),
 
*zeitselektives Fading (untere Grafik).
 
  
  
Der Pfadverlust gibt die Verminderung der Empfangsleistung mit der Entfernung $d$ vom Sender an. Oberhalb des sog. ''Break Points'' gilt für die Empfangsleistung näherungsweise:
+
<u>Hints:</u>
:$$\frac{P(d)}{P(d_0)} = \alpha_0 \cdot \left ( {d}/{d_0}\right )^{-4}.$$
 
Entsprechend der oberen Grafik gilt $\alpha_{0} = 10^{–5}$ (entsprechend $50 \ \rm dB$) und $d_{0} = 100 \ \rm m$.
 
  
Die Leistungsübertragungsfunktion $|H_{\rm K}(f)|^{2}$ zu einem gegebenen Zeitpunkt gemäß der mittleren Grafik verdeutlicht frequenzselektives Fading. Die blau–gestrichelt eingezeichnete Horizontale kennzeichnet nichtfrequenzselektives Fading. Frequenzselektives Fading entsteht, wenn die Kohärenzbandbreite $B_{\rm K}$ sehr viel kleiner als die Signalbandbreite $B_{\rm S}$ ist. Dabei gilt mit der Mehrwegeverbreiterung (englisch: ''Delay Spread'') $T_{\rm V} \Rightarrow$  Differenz zwischen der maximalen und der minimalen Verzögerungszeit:
+
*This exercise belongs to the chapter&nbsp; [[Examples_of_Communication_Systems/General_Description_of_UMTS|"General Description of UMTS"]].
:$$B_{\rm K}\approx \frac{1}{T_{V}}= \frac{1}{\tau_{\rm max}- \tau_{\rm min}}.$$
+
   
Die untere Grafik zeigt schematisch die Leistungsübertragungsfunktion $H_{\rm K}(t)^{2}$ für eine feste Frequenz $f_{0}$. Schematisch deshalb, weil für das hier betrachtete zeitselektive Fading genau der gleiche Verlauf gewählt wurde wie in der mittleren Grafik für das frequenzselektive Fading (reine Bequemlichkeit der Autoren).
+
*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.E2.80.93selective_and_time.E2.80.93selective_fading|"Frequency-selective and time-selective fading"]].
Hier entsteht eine so genannte Dopplerverbreiterung $B_{\rm D}$, definiert als Differenz zwischen der maximalen und der minimalen Dopplerfrequenz. Der Kehrwert $T_{\rm D} = 1/B_{\rm D}$ wird als Kohärenzzeit oder auch als Korrelationsdauer bezeichnet. Bei UMTS tritt immer dann zeitselektives Fading auf, wenn $T_{\rm D} << T_{\rm C}$ (Chipdauer) ist.
 
  
 +
*For UMTS,&nbsp; the bandwidth:&nbsp; $B_{\rm S} = 5 \ \rm MHz$&nbsp; and the chip duration:&nbsp; $T_{\rm C} \approx 0.26 \ \rm &micro; s$.
 +
  
''Hinweis:''
 
  
Die Aufgabe gehört zum Themengebiet von [[Beispiele_von_Nachrichtensystemen/Allgemeine_Beschreibung_von_UMTS|Allgemeine Beschreibung von UMTS]]. Die Bandbreite beträgt bei UMTS $B_{\rm S} = 5 \ \rm MHz$ und die Chipdauer $T_{\rm C} \approx 0.26 \  \rm \mu s$.
+
 
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
  
{Berechnen Sie – ausgehend von der oberen Grafik auf der Angabenseite – den Pfadverlust (in $\rm dB$) für $d = \ \rm 5 km$.
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{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="{}"}
 
|type="{}"}
${\rm Pfadverlust} \ = \ $ { 118 3% } $\ \rm dB $
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${\rm path\ loss} \ = \ $ { 118 3% } $\ \rm dB $.
  
{Welche Aussagen gelten bezüglich des frequenzselektiven Fadings?
+
{What statements are true regarding frequency-selective fading?
 
|type="[]"}
 
|type="[]"}
- Dieses entsteht durch Mehrwegeempfang.
+
+ This is caused by multipath reception.
+ Es entsteht durch Bewegung von Sender und/oder Empfänger.
+
- It is caused by movement of transmitter and/or receiver.
+ Verschiedene Frequenzen werden unterschiedlich gedämpft.
+
+ Different frequencies are attenuated differently.
+ Ein Echo im Abstand 1μs führt zu frequenzselektivem Fading.
+
+ An echo at a distance&nbsp; $1\ \rm &micro; s$&nbsp; results in frequency-selective fading.
  
{Welche Aussagen gelten bezüglich des zeitselektiven Fadings?
+
{What statements are true regarding time-selective fading?
 
|type="[]"}
 
|type="[]"}
- Dieses entsteht durch Mehrwegeempfang.
+
- This arises due to multipath reception.
+ Es entsteht durch Bewegung von Sender und/oder Empfänger.
+
+ It results from movement of transmitter and/or receiver.
- Verschiedene Frequenzen werden unterschiedlich gedämpft.
+
- Different frequencies are attenuated differently.
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
  
'''(1)'''&nbsp; Entsprechend der Skizze liegt der Breakpoint bei $d_{0} = 100 \ \rm m$. Für $d ≤ d_{0}$ ist der Pfadverlust gleich $\alpha_{0} \cdot (d/d_{0})^{–2}$. Für $d = d_{0} = 100 \ \rm m$ gilt:
+
'''(1)'''&nbsp; According to the sketch,&nbsp; the breakpoint is at&nbsp; $d_{0} = 100 \ \rm m$.  
:$${\rm Pfadverlust} = \alpha_0 = 10^{-5}\hspace{0.5cm}\Rightarrow\hspace{0.5cm}{50\,{\rm dB}}.$$
+
*For&nbsp; $d ≤ d_{0}$,&nbsp; the path loss is equal to&nbsp; $\alpha_{0} \cdot (d/d_{0})^{-2}$.&nbsp; For $d = d_{0} = 100 \ \rm m$&nbsp; holds:
Oberhalb von $d_{0}$ ist der Pfadverlust gleich  $\alpha_{0} \cdot ((d/d_{0})^{–4}$. Somit erhält man in $5 \ \rm km$ Entfernung:
+
:$${\rm path\ loss} = \alpha_0 = 10^{-5}\hspace{0.5cm}\Rightarrow\hspace{0.5cm}{50\,{\rm dB}}.$$
:$${\rm Pfadverlust} = 10^{-5}\cdot 50^{-4} = 1.6 \cdot 10^{-12}\hspace{0.5cm}\Rightarrow\hspace{0.5cm}\underline{118\,{\rm dB}}.$$
+
 
 +
*Above&nbsp; $d_{0}$,&nbsp; the path loss is equal to&nbsp; $\alpha_{0} \cdot (d/d_{0})^{-4}$. &nbsp; Thus,&nbsp; at&nbsp; $5 \ \rm km$&nbsp; distance,&nbsp; 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 are the&nbsp; <u>statements 1, 3, and 4</u>:
 +
*Frequency-selective fading is due to multipath reception.&nbsp; This means:
 +
 
 +
*Different frequency components are delayed and attenuated differently by the channel.
 +
 
 +
*This results in attenuation and phase distortion.
 +
 +
*Because&nbsp; $\tau_{\rm max} = 1 \ \rm &micro; s$&nbsp; $($simplifying&nbsp; $\tau_{\rm min} = 0$&nbsp; is set$)$&nbsp; further results in
 +
:$$B_{\rm K} = \frac{1}{\tau_{\rm max}- \tau_{\rm min}} = 1\,{\rm MHz}\ \ll \ B_{\rm S} \hspace{0.15cm}\underline {= 5\,{\rm MHz}}.$$
  
'''(2)'''&nbsp; Richtig sind die <u>Aussagen 1, 3 und 4</u>. Das frequenzselektive Fading ist auf Mehrwegeempfang zurückzuführen: Unterschiedliche Frequenzanteile werden durch den Kanal unterschiedlich verzögert und gedämpft und es entstehen dadurch Dämpfungs– und Phasenverzerrungen. Wegen $\tau_{\rm max} = 1 \ \rm \mu s$ (vereinfachend wird $\tau_{\rm min} = 0$ gesetzt) ergibt sich weiter
 
:$$B_{\rm K} = \frac{1}{\tau_{\rm max}- \tau_{\rm min}} = 1\,{\rm MHz} << B_{\rm S} \hspace{0.15cm}\underline {= 5\,{\rm MHz}}.$$
 
  
'''(3)'''&nbsp; Richtig ist <u>Aussage 2</u>. Die Aussagen 1 und 3 würden für frequenzselektives Fading gelten.
+
'''(3)'''&nbsp; Correct is&nbsp; <u>statement 2</u>.  
 +
*Statements 1 and 3,&nbsp; on the other hand,&nbsp; are valid for frequency-selective fading &ndash; see subtask&nbsp; '''(2)'''.
  
  
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[[Category:Aufgaben zu Beispiele von Nachrichtensystemen|^4.1 Allgemeine Beschreibung von UMTS
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[[Category:Examples of Communication Systems: Exercises|^4.1 General Description of UMTS
 
^]]
 
^]]

Latest revision as of 17:26, 13 February 2023

Path loss,  frequency/time–selective fading

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:

  • 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

(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 are the  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).