Difference between revisions of "Aufgaben:Exercise 1.3: System Comparison at AWGN Channel"

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{{quiz-Header|Buchseite=Modulationsverfahren/Qualitätskriterien
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{{quiz-Header|Buchseite=Modulation_Methods/Quality_Criteria
 
}}
 
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[[File:P_ID960__Mod_A_1_3.png|right|frame|Systemvergleich beim AWGN–Kanal]]
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[[File:P_ID960__Mod_A_1_3.png|right|frame|System comparison at AWGN channel]]
Für den Vergleich verschiedener Modulationsverfahren und Demodulatoren hinsichtlich der Rauschempfindlichkeit gehen wir meist vom so genannten  [[Modulationsverfahren/Qualitätskriterien#Einige_Anmerkungen_zum_AWGN.E2.80.93Kanalmodell|AWGN–Kanal]]  aus und beschreiben folgendes doppelt–logarithmische Diagramm:
+
For the comparison of different modulation and demodulation methods with regard to noise sensitivity,   we usually assume the so-called  [[Modulation_Methods/Quality_Criteria#Some_remarks_on_the_AWGN_channel_model|AWGN channel]]  and present the following double logarithmic diagram:  
*Die Ordinate gibt den Sinken–Störabstand (SNR logarithmiert)  $10 · \lg ρ_v$  in dB an.
+
*The y-axis indicates the   "sink-to-noise ratio"   (logarithmic SNR)   ⇒    $10 · \lg ρ_v$  in dB.
*Auf der Abszisse ist  $10 · \lg ξ$  aufgetragen; für die normierte Leistungskenngröße gilt:
+
* $10 · \lg ξ$  is plotted on the x-axis;   the normalized power parameter   ("performance parameter")  is characterized by:
 
:$$ \xi = \frac{P_{\rm S} \cdot \alpha_{\rm K}^2 }{{N_0} \cdot B_{\rm NF}}\hspace{0.05cm}.$$
 
:$$ \xi = \frac{P_{\rm S} \cdot \alpha_{\rm K}^2 }{{N_0} \cdot B_{\rm NF}}\hspace{0.05cm}.$$
*In  $ξ$  sind also die Sendeleistung  $P_{\rm S}$, der Kanaldämpfungsfaktor  $α_{\rm K}$, die Rauschleistungsdichte  $N_0$  sowie die Bandbreite  $B_{\rm NF}$  des Nachrichtensignals in geeigneter Weise zusammengefasst.
+
*Thus,    the transmission power  $P_{\rm S}$,  the channel attenuation factor $α_{\rm K}$,  the noise power density  $N_0$  and the bandwidth  $B_{\rm NF}$  of the message signal are suitably summarised together in  $ξ$.
* Wenn nicht ausdrücklich etwas anderes angegeben ist, soll in der Aufgabe von folgenden Werten ausgegangen werden:
+
* Unless explicitly stated otherwise,  the following values shall be assumed in the exercise:
 
:$$P_{\rm S}= 5 \;{\rm kW}\hspace{0.05cm}, \hspace{0.2cm} \alpha_{\rm
 
:$$P_{\rm S}= 5 \;{\rm kW}\hspace{0.05cm}, \hspace{0.2cm} \alpha_{\rm
 
K} = 0.001\hspace{0.05cm}, \hspace{0.2cm} {N_0} =
 
K} = 0.001\hspace{0.05cm}, \hspace{0.2cm} {N_0} =
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B_{\rm NF}= 5\; {\rm kHz}\hspace{0.05cm}.$$
 
B_{\rm NF}= 5\; {\rm kHz}\hspace{0.05cm}.$$
  
In der Grafik sind zwei Systeme eingezeichnet, deren $(x, y)$–Verlauf wie folgt beschrieben werden kann:
+
Two systems are plotted in the graph and their   $(x, y)$-curve can be described as follows:
*Das  $\text{System A}$  ist gekennzeichnet durch die folgende Gleichung:
+
*$\text{System A}$  is characterized by the following equation:
 
:$$y = x+1.$$
 
:$$y = x+1.$$
*Entsprechend gilt für das  $\text{System B}$:
+
* $\text{System B}$  is instead characterized by:
 
:$$ y= 6 \cdot \left(1 - {\rm e}^{-x+1} \right)\hspace{0.05cm}.$$
 
:$$ y= 6 \cdot \left(1 - {\rm e}^{-x+1} \right)\hspace{0.05cm}.$$
Die in der Grafik zusätzlich grün eingezeichneten Achsenbeschriftungen haben folgende Bedeutung:
+
The additional axis labels drawn in green have the following meaning:
 
:$$ x = \frac{10 \cdot {\rm lg} \hspace{0.1cm}\xi} {10 \,{\rm dB}}\hspace{0.05cm}, \hspace{0.3cm}y = \frac{10 \cdot {\rm lg} \hspace{0.1cm}\rho_v} {10 \,{\rm dB}}\hspace{0.05cm}.$$
 
:$$ x = \frac{10 \cdot {\rm lg} \hspace{0.1cm}\xi} {10 \,{\rm dB}}\hspace{0.05cm}, \hspace{0.3cm}y = \frac{10 \cdot {\rm lg} \hspace{0.1cm}\rho_v} {10 \,{\rm dB}}\hspace{0.05cm}.$$
So steht  $x = 4$  für  $10 · \lg ξ = 40\text{ dB}$  bzw.  $ξ = 10^4$  und  $y = 5$  steht für  $10 · \lg ρ_v= 50\text{ dB}$ , also  $ρ_v = 10^5$.
+
*Thus  $x = 4$  represents  $10 · \lg ξ = 40\text{ dB}$  or  $ξ = 10^4$   
 +
*and  $y = 5$  represents  $10 · \lg ρ_v= 50\text{ dB}$ , i.e.,  $ρ_v = 10^5$.
  
  
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''Hinweise:''  
+
 
*Die Aufgabe gehört zum  Kapitel  [[Modulationsverfahren/Qualitätskriterien|Qualitätskriterien]].
+
 
*Bezug genommen wird insbesondere auf die Seite   [[Modulationsverfahren/Qualitätskriterien#Untersuchungen_beim_AWGN.E2.80.93Kanal|Untersuchungen beim AWGN-Kanal]].
+
 
*Durch die Angabe der Leistungen in  $\rm W$att sind diese unabhängig vom Bezugswiderstand  $R$.
+
''Hints:''  
 +
*This exercise belongs to the chapter   [[Modulation_Methods/Quality_Criteria|Quality Criteria]].
 +
*Particular reference is made to the page    [[Modulation_Methods/Quality_Criteria#Investigating_at_the_AWGN_channel|Investigating at the AWGN channel]].
 +
*By specifying the powers in watts,  they are independent of the reference resistance  $R$.
 
   
 
   
  
 
   
 
   
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Welcher Sinken–Störabstand (in dB) ergibt sich bei &nbsp;$\text{System A}$&nbsp; mit <br>$P_{\rm S}= 5 \;{\rm kW}$, &nbsp; $\alpha_{\rm
+
{What is the&nbsp; sink signal-to-noise ratio&nbsp;  (in dB)&nbsp; for &nbsp;$\text{System A}$&nbsp; with &nbsp;$P_{\rm S}= 5 \;{\rm kW}$, &nbsp; $\alpha_{\rm
K} = 0.001$, &nbsp; $N_0 = 10^{-10}\;{\rm W}/{\rm Hz}$ und $B_{\rm NF}= 5\; {\rm kHz}$?
+
K} = 0.001$, &nbsp; $N_0 = 10^{-10}\;{\rm W}/{\rm Hz}$, &nbsp; $B_{\rm NF}= 5\; {\rm kHz}$?
 
|type="{}"}
 
|type="{}"}
 
$10 · \lg \hspace{0.05cm}ρ_v \ = \ $ { 50 3% } $\ \text{dB}$
 
$10 · \lg \hspace{0.05cm}ρ_v \ = \ $ { 50 3% } $\ \text{dB}$
  
{Es wird nun &nbsp;$10 · \lg \hspace{0.05cm} ρ_v ≥ 60\text{ dB}$&nbsp; gefordert. Durch welche Maßnahmen (jeweils für sich allein) ist dies zu erreichen?
+
{Now &nbsp;$10 · \lg \hspace{0.05cm} ρ_v ≥ 60\text{ dB}$&nbsp; is required.&nbsp; Which independent measures can be taken to achieve this?
 
|type="[]"}
 
|type="[]"}
- Erhöhung der Sendeleistung von &nbsp;$P_{\rm S}= 5\text{ kW}$&nbsp; auf $10\text{ kW}$&nbsp;.
+
- Increasing the transmission power from &nbsp;$P_{\rm S}= 5\text{ kW}$&nbsp; to $10\text{ kW}$&nbsp;.
+ Erhöhung des Kanaldämpfungsfaktors von &nbsp;$α_{\rm K} = 0.001$&nbsp; auf &nbsp;$0.004$.
+
+ Increasing the channel transmission factor from &nbsp;$α_{\rm K} = 0.001$&nbsp; to &nbsp;$0.004$.
+ Reduzierung der Rauschleistungsdichte auf &nbsp;$N_0=10^{–11 }\text{ W/Hz}$.
+
+ Reducing the noise power density to &nbsp;$N_0=10^{–11 }\text{ W/Hz}$.
- Erhöhung der NF–Bandbreite von &nbsp;$B_{\rm NF}= 5\text{ kHz}$&nbsp; auf &nbsp;$\text{ kHz}$.
+
- Increasing the source signal bandwidth from &nbsp;$B_{\rm NF}= 5\text{ kHz}$&nbsp; to &nbsp;$10\text{ kHz}$.
  
{Welcher Störabstand ergibt sich bei &nbsp;$\text{System B}$&nbsp; mit &nbsp;$10 · \lg ξ = 40\text{ dB}$?
+
{What is the sink signal-to-noise ratio for &nbsp;$\text{System B}$&nbsp; with &nbsp;$10 · \lg ξ = 40\text{ dB}$?
 
|type="{}"}
 
|type="{}"}
 
$10 · \lg \hspace{0.05cm}ρ_v \ = \ $ { 57 3% } $\ \text{dB}$
 
$10 · \lg \hspace{0.05cm}ρ_v \ = \ $ { 57 3% } $\ \text{dB}$
  
{Gefordert wird der Störabstand &nbsp;$10 · \lg ρ_v = 50\text{ dB}$. Welche Sendeleistung &nbsp;$P_{\rm S}$ genügt bei &nbsp;$\text{System B}$, um diese Qualität zu erzielen?
+
{If the required sink signal-to-noise ratio is  &nbsp;$10 · \lg ρ_v = 50\text{ dB}$,&nbsp; what transmission power &nbsp;$P_{\rm S}$ is sufficient to achieve this for &nbsp;$\text{System B}$?
 
|type="{}"}
 
|type="{}"}
 
$P_{\rm S} \ = \ $ { 0.3 3% } $\ \text{ kW }$
 
$P_{\rm S} \ = \ $ { 0.3 3% } $\ \text{ kW }$
  
{Für welchen Wert von &nbsp;$10 · \lg ξ$&nbsp; ist die Verbesserung von &nbsp;$\text{System B}$&nbsp; gegenüber &nbsp;$\text{System A}$&nbsp; am größten?
+
{What value of &nbsp;$10 · \lg ξ$&nbsp; gives the greatest improvement for &nbsp;$\text{System B}$&nbsp; relative to &nbsp;$\text{System A}$&nbsp;?
 
|type="{}"}
 
|type="{}"}
 
$10 · \lg \hspace{0.05cm} ξ \ = \ ${ 27.9 3% } $\ \text{dB}$
 
$10 · \lg \hspace{0.05cm} ξ \ = \ ${ 27.9 3% } $\ \text{dB}$
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</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Die normierte Leistungskenngröße ergibt sich mit diesen Werten zu
+
'''(1)'''&nbsp; The normalized performance parameter is calculated using these values as follows:
 
:$$\xi = \frac{5 \cdot 10^3\,{\rm W}\cdot 10^{-6} }{10^{-10}\,{\rm W}/{\rm Hz} \cdot 5 \cdot 10^3\,{\rm Hz}} = 10^4 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = 40\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x=4 \hspace{0.05cm}.$$
 
:$$\xi = \frac{5 \cdot 10^3\,{\rm W}\cdot 10^{-6} }{10^{-10}\,{\rm W}/{\rm Hz} \cdot 5 \cdot 10^3\,{\rm Hz}} = 10^4 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = 40\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x=4 \hspace{0.05cm}.$$
Damit ergibt sich der Hilfsordinatenwert $y = 5$, was zum Sinken-Störabstand $10 · \lg \hspace{0.05cm} ρ_v\hspace{0.15cm}\underline{ = 50 \ \rm dB}$ führt.
+
*This gives the auxiliary coordinate value&nbsp; $y = 5$,&nbsp; which leads to a sink SNR of &nbsp; $10 · \lg \hspace{0.05cm} ρ_v\hspace{0.15cm}\underline{ = 50 \ \rm dB}$.
  
  
'''(2)'''&nbsp; Richtig sind also die <u>Alternativen 2 und 3</u>:
+
'''(2)'''<u>&nbsp;Answers 2 and 3</u>&nbsp; are correct:
  
Diese Forderung  entspricht gegenüber dem bisher betrachteten System einer Erhöhung des Störabstandes um $10$ dB, so dass auch $10 ·  \lg \hspace{0.05cm}ξ$ um $10$ dB erhöht werden muss:
+
This requirement corresponds to a&nbsp; $10$&nbsp; dB&nbsp; increase in the sink SNR compared to the previous system,&nbsp; so &nbsp;$10 ·  \lg \hspace{0.05cm}ξ$&nbsp; must also be increased by&nbsp;$10$&nbsp; dB:
 
:$$10 \cdot {\rm lg} \hspace{0.1cm}\xi = 50\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \xi=10^5 \hspace{0.05cm}.$$
 
:$$10 \cdot {\rm lg} \hspace{0.1cm}\xi = 50\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \xi=10^5 \hspace{0.05cm}.$$
  
Ein $10$–fach größerer $ξ$–Wert wird erreicht – vorausgesetzt die anderen Parameter bleiben jeweils gleich:
+
A tenfold larger&nbsp; $ξ$&nbsp; value is achieved&nbsp;  (provided all other parameters are held constant in each case)
*durch die Sendeleistung $P_{\rm S} = 50$ kW statt $5$ kW,
+
*by a transmission power of&nbsp; $P_{\rm S} = 50$&nbsp; kW&nbsp; instead of &nbsp; $5$&nbsp; kW,
*durch den Dämpfungsfaktor $α_{\rm K} = 0.00316$ anstelle von $0.001$,
+
*by a channel transmission factor of &nbsp; $α_{\rm K} = 0.00316$&nbsp; instead of&nbsp; $0.001$,
*durch die Rauschleistungsdichte $N_0 = 10^{ –11 }$ W/Hz statt $10^{ –10 }$ W/Hz,
+
*by a noise power density of &nbsp; $N_0 = 10^{ –11 }$&nbsp; W/Hz&nbsp; instead of&nbsp; $10^{ –10 }$&nbsp; W/Hz,
*durch die Bandbreite $B_{\rm NF} = 0.5$ kHz statt $5$ kHz.
+
*by a signal bandwidth of&nbsp; $B_{\rm NF} = 0.5$&nbsp; kHz&nbsp; instead of &nbsp; $5$&nbsp; kHz.
  
  
'''(3)'''&nbsp; Für $10 · \lg \hspace{0.05cm} ξ = 40$ dB ist die Hilfsgröße $x = 4$. Damit ergibt sich für die Hilfsgröße der Ordinate:
+
'''(3)'''&nbsp; For&nbsp; $10 · \lg \hspace{0.05cm} ξ = 40$&nbsp; dB,&nbsp; the auxiliary value is &nbsp; $x = 4$.&nbsp; This gives the auxiliary&nbsp; $y$&ndash;value:
 
:$$y= 6 \cdot \left(1 - {\rm e}^{-3} \right)\approx 5.7 \hspace{0.05cm}.$$
 
:$$y= 6 \cdot \left(1 - {\rm e}^{-3} \right)\approx 5.7 \hspace{0.05cm}.$$
Dies entspricht dem Sinken–Störabstand $10 · \lg \hspace{0.05cm} ρ_v\hspace{0.15cm}\underline{ = 57 \ \rm dB}$, also einer Verbesserung gegenüber dem '''System A''' um $7$ dB.
+
*This corresponds to a sink SNR of &nbsp; $10 · \lg \hspace{0.05cm} ρ_v\hspace{0.15cm}\underline{ = 57 \ \rm dB}$ &nbsp; &rArr; &nbsp; $7$&nbsp; dB improvement over &nbsp;$\text{System A}$.
  
  
'''(4)'''&nbsp; Diese Problemstellung wird durch folgende Gleichung beschrieben:
+
 
 +
'''(4)'''&nbsp; This problem is described by the following equation:
 
:$$ y= 6 \cdot \left(1 - {\rm e}^{-x+1} \right) = 5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm e}^{-x+1} ={1}/{6}\hspace{0.3cm}
 
:$$ y= 6 \cdot \left(1 - {\rm e}^{-x+1} \right) = 5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm e}^{-x+1} ={1}/{6}\hspace{0.3cm}
 
  \Rightarrow \hspace{0.3cm} x = 1+ {\rm ln} \hspace{0.1cm}6 \approx 2.79 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = 27.9\,{\rm dB}\hspace{0.05cm}.$$
 
  \Rightarrow \hspace{0.3cm} x = 1+ {\rm ln} \hspace{0.1cm}6 \approx 2.79 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = 27.9\,{\rm dB}\hspace{0.05cm}.$$
Bei '''System A''' war hierfür $10 · \lg \hspace{0.05cm} \xi = 40$ dB notwendig, was bei den weiter gegebenen Zahlenwerten durch $P_{\rm S} = 5$ kW erreicht wurde. Nun kann die Sendeleistung um etwa $12.1$ dB verringert werden:
+
*For &nbsp;$\text{System A}$&nbsp; $10 · \lg \hspace{0.05cm} \xi = 40$&nbsp; dB is required,&nbsp; which was achieved with &nbsp; $P_{\rm S} = 5$&nbsp; kW and the other numerical values given.&nbsp;
 +
*Now the transmission power can be reduced by about &nbsp; $12.1$&nbsp; dB:
 
:$$ 10 \cdot {\rm lg} \hspace{0.1cm} \frac{P_{\rm S}}{ 5 \;{\rm kW}}= -12.1\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac{P_{\rm S}}{ 5 \;{\rm kW}} = 10^{-1.21}\approx 0.06\hspace{0.05cm}.$$
 
:$$ 10 \cdot {\rm lg} \hspace{0.1cm} \frac{P_{\rm S}}{ 5 \;{\rm kW}}= -12.1\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac{P_{\rm S}}{ 5 \;{\rm kW}} = 10^{-1.21}\approx 0.06\hspace{0.05cm}.$$
Das bedeutet: Bei '''System B''' wird mit nur 6% der Sendeleistung von '''System A''' also mit nur $P_{\rm S} \hspace{0.15cm}\underline{ = 0.3 \ \rm kW}$ – die gleiche Systemqualität erzielt.
+
*This means that in &nbsp;$\text{System B}$&nbsp; the same system quality is achieved with only &nbsp; $6\%$&nbsp; of the transmission power of &nbsp;$\text{System A}$&nbsp; i.e., with only &nbsp; $P_{\rm S} \hspace{0.15cm}\underline{ = 0.3 \ \rm kW}$.
 +
 
  
  
'''(5)'''&nbsp; Wir bezeichnen mit ''V'' (steht für Verbesserung) den größeren Sinken–Störabstand von  '''System B''' gegenüber '''System A''':
+
'''(5)'''&nbsp; The larger sink SNR of &nbsp;$\text{System B}$&nbsp; compared to &nbsp;$\text{System A}$&nbsp; we will denote with &nbsp; $V$&nbsp; (from German&nbsp; "Verbesserung" &nbsp; &rArr; &nbsp; "improvement"):
 
:$$V  =  10 \cdot {\rm lg} \hspace{0.1cm}\rho_v \hspace{0.1cm}{\rm (System\;B)} - 10 \cdot {\rm lg} \hspace{0.1cm}\rho_v \hspace{0.1cm}{\rm (System\;A)}
 
:$$V  =  10 \cdot {\rm lg} \hspace{0.1cm}\rho_v \hspace{0.1cm}{\rm (System\;B)} - 10 \cdot {\rm lg} \hspace{0.1cm}\rho_v \hspace{0.1cm}{\rm (System\;A)}
 
  =  \left[6 \cdot \left(1 - {\rm e}^{-x+1} \right) -x -1 \right] \cdot 10\,{\rm dB}\hspace{0.05cm}.$$
 
  =  \left[6 \cdot \left(1 - {\rm e}^{-x+1} \right) -x -1 \right] \cdot 10\,{\rm dB}\hspace{0.05cm}.$$
Durch Nullsetzen der Ableitung ergibt sich derjenige $x$–Wert, der zur maximalen Verbesserung führt:
+
*Setting the derivative to zero yields the &nbsp;$x$–value that leads to the maximum improvement:  
 
:$$ \frac{{\rm d}V}{{\rm d}x} = 6 \cdot {\rm e}^{-x+1} -1\Rightarrow \hspace{0.3cm} x = 1+ {\rm ln} \hspace{0.1cm}6 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = \hspace{0.15cm}\underline {27.9\,{\rm dB}}\hspace{0.05cm}.$$
 
:$$ \frac{{\rm d}V}{{\rm d}x} = 6 \cdot {\rm e}^{-x+1} -1\Rightarrow \hspace{0.3cm} x = 1+ {\rm ln} \hspace{0.1cm}6 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = \hspace{0.15cm}\underline {27.9\,{\rm dB}}\hspace{0.05cm}.$$
Es ergibt sich also genau der in der Teilaufgabe (4) behandelte Fall mit $10 · \lg ρ_υ = 50$ dB, während der Störabstand bei  '''System A''' nur $37.9$ dB beträgt. Die Verbesserung ist demnach $12.1$ dB.
+
*This results in exactly the case discussed in subtask &nbsp; '''(4)'''&nbsp; with &nbsp; $10 · \lg ρ_υ = 50$&nbsp; dB,&nbsp; while the sink SNR for &nbsp;$\text{System A}$&nbsp; is only&nbsp; $37.9$&nbsp; dB.&nbsp;
 +
*The improvement is therefore&nbsp; $12.1$&nbsp; dB.
  
 
{{ML-Fuß}}
 
{{ML-Fuß}}
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[[Category:Aufgaben zu Modulationsverfahren|^1.2 Qualitätskriterien^]]
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[[Category:Modulation Methods: Exercises|^1.2 Quality Criteria^]]

Latest revision as of 17:54, 23 March 2022

System comparison at AWGN channel

For the comparison of different modulation and demodulation methods with regard to noise sensitivity,  we usually assume the so-called  AWGN channel  and present the following double logarithmic diagram:

  • The y-axis indicates the  "sink-to-noise ratio"  (logarithmic SNR)   ⇒   $10 · \lg ρ_v$  in dB.
  •  $10 · \lg ξ$  is plotted on the x-axis;  the normalized power parameter  ("performance parameter")  is characterized by:
$$ \xi = \frac{P_{\rm S} \cdot \alpha_{\rm K}^2 }{{N_0} \cdot B_{\rm NF}}\hspace{0.05cm}.$$
  • Thus,  the transmission power  $P_{\rm S}$,  the channel attenuation factor $α_{\rm K}$,  the noise power density  $N_0$  and the bandwidth  $B_{\rm NF}$  of the message signal are suitably summarised together in  $ξ$.
  • Unless explicitly stated otherwise,  the following values shall be assumed in the exercise:
$$P_{\rm S}= 5 \;{\rm kW}\hspace{0.05cm}, \hspace{0.2cm} \alpha_{\rm K} = 0.001\hspace{0.05cm}, \hspace{0.2cm} {N_0} = 10^{-10}\;{\rm W}/{\rm Hz}\hspace{0.05cm}, \hspace{0.2cm} B_{\rm NF}= 5\; {\rm kHz}\hspace{0.05cm}.$$

Two systems are plotted in the graph and their   $(x, y)$-curve can be described as follows:

  • $\text{System A}$  is characterized by the following equation:
$$y = x+1.$$
  •  $\text{System B}$  is instead characterized by:
$$ y= 6 \cdot \left(1 - {\rm e}^{-x+1} \right)\hspace{0.05cm}.$$

The additional axis labels drawn in green have the following meaning:

$$ x = \frac{10 \cdot {\rm lg} \hspace{0.1cm}\xi} {10 \,{\rm dB}}\hspace{0.05cm}, \hspace{0.3cm}y = \frac{10 \cdot {\rm lg} \hspace{0.1cm}\rho_v} {10 \,{\rm dB}}\hspace{0.05cm}.$$
  • Thus  $x = 4$  represents  $10 · \lg ξ = 40\text{ dB}$  or  $ξ = 10^4$ 
  • and  $y = 5$  represents  $10 · \lg ρ_v= 50\text{ dB}$ , i.e.,  $ρ_v = 10^5$.





Hints:


Questions

1

What is the  sink signal-to-noise ratio  (in dB)  for  $\text{System A}$  with  $P_{\rm S}= 5 \;{\rm kW}$,   $\alpha_{\rm K} = 0.001$,   $N_0 = 10^{-10}\;{\rm W}/{\rm Hz}$,   $B_{\rm NF}= 5\; {\rm kHz}$?

$10 · \lg \hspace{0.05cm}ρ_v \ = \ $

$\ \text{dB}$

2

Now  $10 · \lg \hspace{0.05cm} ρ_v ≥ 60\text{ dB}$  is required.  Which independent measures can be taken to achieve this?

Increasing the transmission power from  $P_{\rm S}= 5\text{ kW}$  to $10\text{ kW}$ .
Increasing the channel transmission factor from  $α_{\rm K} = 0.001$  to  $0.004$.
Reducing the noise power density to  $N_0=10^{–11 }\text{ W/Hz}$.
Increasing the source signal bandwidth from  $B_{\rm NF}= 5\text{ kHz}$  to  $10\text{ kHz}$.

3

What is the sink signal-to-noise ratio for  $\text{System B}$  with  $10 · \lg ξ = 40\text{ dB}$?

$10 · \lg \hspace{0.05cm}ρ_v \ = \ $

$\ \text{dB}$

4

If the required sink signal-to-noise ratio is  $10 · \lg ρ_v = 50\text{ dB}$,  what transmission power  $P_{\rm S}$ is sufficient to achieve this for  $\text{System B}$?

$P_{\rm S} \ = \ $

$\ \text{ kW }$

5

What value of  $10 · \lg ξ$  gives the greatest improvement for  $\text{System B}$  relative to  $\text{System A}$ ?

$10 · \lg \hspace{0.05cm} ξ \ = \ $

$\ \text{dB}$


Solution

(1)  The normalized performance parameter is calculated using these values as follows:

$$\xi = \frac{5 \cdot 10^3\,{\rm W}\cdot 10^{-6} }{10^{-10}\,{\rm W}/{\rm Hz} \cdot 5 \cdot 10^3\,{\rm Hz}} = 10^4 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = 40\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x=4 \hspace{0.05cm}.$$
  • This gives the auxiliary coordinate value  $y = 5$,  which leads to a sink SNR of   $10 · \lg \hspace{0.05cm} ρ_v\hspace{0.15cm}\underline{ = 50 \ \rm dB}$.


(2) Answers 2 and 3  are correct:

This requirement corresponds to a  $10$  dB  increase in the sink SNR compared to the previous system,  so  $10 · \lg \hspace{0.05cm}ξ$  must also be increased by $10$  dB:

$$10 \cdot {\rm lg} \hspace{0.1cm}\xi = 50\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \xi=10^5 \hspace{0.05cm}.$$

A tenfold larger  $ξ$  value is achieved  (provided all other parameters are held constant in each case)

  • by a transmission power of  $P_{\rm S} = 50$  kW  instead of   $5$  kW,
  • by a channel transmission factor of   $α_{\rm K} = 0.00316$  instead of  $0.001$,
  • by a noise power density of   $N_0 = 10^{ –11 }$  W/Hz  instead of  $10^{ –10 }$  W/Hz,
  • by a signal bandwidth of  $B_{\rm NF} = 0.5$  kHz  instead of   $5$  kHz.


(3)  For  $10 · \lg \hspace{0.05cm} ξ = 40$  dB,  the auxiliary value is   $x = 4$.  This gives the auxiliary  $y$–value:

$$y= 6 \cdot \left(1 - {\rm e}^{-3} \right)\approx 5.7 \hspace{0.05cm}.$$
  • This corresponds to a sink SNR of   $10 · \lg \hspace{0.05cm} ρ_v\hspace{0.15cm}\underline{ = 57 \ \rm dB}$   ⇒   $7$  dB improvement over  $\text{System A}$.


(4)  This problem is described by the following equation:

$$ y= 6 \cdot \left(1 - {\rm e}^{-x+1} \right) = 5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm e}^{-x+1} ={1}/{6}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} x = 1+ {\rm ln} \hspace{0.1cm}6 \approx 2.79 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = 27.9\,{\rm dB}\hspace{0.05cm}.$$
  • For  $\text{System A}$  $10 · \lg \hspace{0.05cm} \xi = 40$  dB is required,  which was achieved with   $P_{\rm S} = 5$  kW and the other numerical values given. 
  • Now the transmission power can be reduced by about   $12.1$  dB:
$$ 10 \cdot {\rm lg} \hspace{0.1cm} \frac{P_{\rm S}}{ 5 \;{\rm kW}}= -12.1\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac{P_{\rm S}}{ 5 \;{\rm kW}} = 10^{-1.21}\approx 0.06\hspace{0.05cm}.$$
  • This means that in  $\text{System B}$  the same system quality is achieved with only   $6\%$  of the transmission power of  $\text{System A}$  – i.e., with only   $P_{\rm S} \hspace{0.15cm}\underline{ = 0.3 \ \rm kW}$.


(5)  The larger sink SNR of  $\text{System B}$  compared to  $\text{System A}$  we will denote with   $V$  (from German  "Verbesserung"   ⇒   "improvement"):

$$V = 10 \cdot {\rm lg} \hspace{0.1cm}\rho_v \hspace{0.1cm}{\rm (System\;B)} - 10 \cdot {\rm lg} \hspace{0.1cm}\rho_v \hspace{0.1cm}{\rm (System\;A)} = \left[6 \cdot \left(1 - {\rm e}^{-x+1} \right) -x -1 \right] \cdot 10\,{\rm dB}\hspace{0.05cm}.$$
  • Setting the derivative to zero yields the  $x$–value that leads to the maximum improvement:
$$ \frac{{\rm d}V}{{\rm d}x} = 6 \cdot {\rm e}^{-x+1} -1\Rightarrow \hspace{0.3cm} x = 1+ {\rm ln} \hspace{0.1cm}6 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = \hspace{0.15cm}\underline {27.9\,{\rm dB}}\hspace{0.05cm}.$$
  • This results in exactly the case discussed in subtask   (4)  with   $10 · \lg ρ_υ = 50$  dB,  while the sink SNR for  $\text{System A}$  is only  $37.9$  dB. 
  • The improvement is therefore  $12.1$  dB.