Aufgaben:Exercise 1.3: System Comparison at AWGN Channel - Revision history

Reed at 16:54, 23 March 2022

2022-03-23T16:54:47Z

Reed at 16:53, 23 March 2022

2022-03-23T16:53:44Z

Reed at 16:52, 23 March 2022

2022-03-23T16:52:49Z

Reed at 10:11, 24 November 2021

2021-11-24T10:11:16Z

Guenter at 12:14, 16 November 2021

2021-11-16T12:14:14Z

Guenter at 11:46, 16 November 2021

2021-11-16T11:46:04Z

Guenter at 11:44, 16 November 2021

2021-11-16T11:44:36Z

Guenter at 11:43, 16 November 2021

2021-11-16T11:43:32Z

Reed: /* Solution */

2021-11-10T15:34:19Z

Solution

Reed at 15:31, 10 November 2021

2021-11-10T15:31:09Z

@@ Line 63: / Line 63: @@
 $P_{\rm S} \ = \ $ { 0.3 3% } $\ \text{ kW }$
-{What value of &nbsp;$10 · \lg ξ$&nbsp; gives the greatest improvement of &nbsp;$\text{System B}$&nbsp; relative to &nbsp;$\text{System A}$&nbsp;?
+{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="{}"}
 $10 · \lg \hspace{0.05cm} ξ \ = \ ${ 27.9 3% } $\ \text{dB}$

@@ Line 18: / Line 18: @@
 *$\text{System A}$&nbsp; is characterized by the following equation:
 :$$y = x+1.$$
-*Accordingly, &nbsp;$\text{System B}$&nbsp; is characterized by:
+*&nbsp;$\text{System B}$&nbsp; 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:

@@ Line 4: / Line 4: @@
 [[File:P_ID960__Mod_A_1_3.png|right|frame|System comparison at AWGN channel]]
-For the comparison of different modulation and demodulation methods with regard to noise sensitivity,&nbsp;  we usually start from the so-called&nbsp;  [[Modulation_Methods/Quality_Criteria#Some_remarks_on_the_AWGN_channel_model|AWGN channel]]&nbsp; and present the following double logarithmic diagram:
+For the comparison of different modulation and demodulation methods with regard to noise sensitivity,&nbsp;  we usually assume the so-called&nbsp;  [[Modulation_Methods/Quality_Criteria#Some_remarks_on_the_AWGN_channel_model|AWGN channel]]&nbsp; and present the following double logarithmic diagram:
 *The y-axis indicates the&nbsp;  "sink-to-noise ratio"&nbsp;  (logarithmic SNR) &nbsp; &rArr; &nbsp;  $10 · \lg ρ_v$&nbsp; in dB.
 *&nbsp;$10 · \lg ξ$&nbsp; is plotted on the x-axis;&nbsp;  the normalized power parameter&nbsp;  ("performance parameter")&nbsp;  is characterized by:

@@ Line 34: / Line 34: @@
 ''Hints:''
-*This exercise belongs to the chapter &nbsp; [[Modulation_Methods/Qualitätskriterien|Quality Criteria]].
+*This exercise belongs to the chapter &nbsp; [[Modulation_Methods/Quality_Criteria|Quality Criteria]].
 *Particular reference is made to the page &nbsp;  [[Modulation_Methods/Quality_Criteria#Investigating_at_the_AWGN_channel|Investigating at the AWGN channel]].
 *By specifying the powers in watts,&nbsp; they are independent of the reference resistance &nbsp;$R$.

@@ Line 72: / Line 72: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The normalized performance parameter is calculated using these values as follows
+'''(1)'''&nbsp; The normalized performance parameter is calculated using these values as follows:
 :$$\xi = \frac{5 \cdot 10^3\,{\rm W}\cdot 10^{-6} }{10^{-10}\,{\rm W}/{\rm Hz} \cdot 5 \cdot 10^3\,{\rm Hz}} = 10^4 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = 40\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x=4 \hspace{0.05cm}.$$
-*This gives the auxiliary coordinate value &nbsp; $y = 5$, which leads to a sink-to-noise ratio of &nbsp; $10 · \lg \hspace{0.05cm} ρ_v\hspace{0.15cm}\underline{ = 50 \ \rm dB}$&nbsp;.
+*This gives the auxiliary coordinate value&nbsp; $y = 5$,&nbsp; which leads to a sink SNR of &nbsp; $10 · \lg \hspace{0.05cm} ρ_v\hspace{0.15cm}\underline{ = 50 \ \rm dB}$.
+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:
 :$$10 \cdot {\rm lg} \hspace{0.1cm}\xi = 50\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \xi=10^5 \hspace{0.05cm}.$$
-A tenfold larger &nbsp; $ξ$ value is achieved – provided all other parameters are held constant in each case –
+A tenfold larger&nbsp; $ξ$&nbsp; value is achieved&nbsp;  (provided all other parameters are held constant in each case)
 *by a transmission power of&nbsp; $P_{\rm S} = 50$&nbsp; kW&nbsp; instead of &nbsp; $5$&nbsp; kW,
 *by a channel transmission factor of &nbsp; $α_{\rm K} = 0.00316$&nbsp; instead of&nbsp; $0.001$,
 *by a noise power density of &nbsp; $N_0 = 10^{ –11 }$&nbsp; W/Hz&nbsp; instead of&nbsp; $10^{ –10 }$&nbsp; W/Hz,
-*by a bandwidth of&nbsp; $B_{\rm NF} = 0.5$&nbsp; kHz&nbsp; instead of &nbsp; $5$&nbsp; kHz.
+*by a signal bandwidth of&nbsp; $B_{\rm NF} = 0.5$&nbsp; kHz&nbsp; instead of &nbsp; $5$&nbsp; kHz.
-'''(3)'''&nbsp; For&nbsp; $10 · \lg \hspace{0.05cm} ξ = 40$&nbsp; dB, the auxiliary value is &nbsp; $x = 4$.&nbsp; This gives the auxiliary y-value:
+'''(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:
 :$$y= 6 \cdot \left(1 - {\rm e}^{-3} \right)\approx 5.7 \hspace{0.05cm}.$$
-*This corresponds to a sink-to-noise ratio of &nbsp; $10 · \lg \hspace{0.05cm} ρ_v\hspace{0.15cm}\underline{ = 57 \ \rm dB}$, a $7$&nbsp; dB improvement over &nbsp;$\text{System A}$&nbsp;.
+*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;$\text{System A}$.
@@ Line 101: / Line 98: @@
 :$$ y= 6 \cdot \left(1 - {\rm e}^{-x+1} \right) = 5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm e}^{-x+1} ={1}/{6}\hspace{0.3cm}
   \Rightarrow \hspace{0.3cm} x = 1+ {\rm ln} \hspace{0.1cm}6 \approx 2.79 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = 27.9\,{\rm dB}\hspace{0.05cm}.$$
-*For &nbsp;$\text{System A}$&nbsp; this required &nbsp; $10 · \lg \hspace{0.05cm} \xi = 40$&nbsp; dB , which was achieved with &nbsp; $P_{\rm S} = 5$&nbsp; kW  and the other numerical values given.&nbsp;
+*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:
 :$$ 10 \cdot {\rm lg} \hspace{0.1cm} \frac{P_{\rm S}}{ 5 \;{\rm kW}}= -12.1\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac{P_{\rm S}}{ 5 \;{\rm kW}} = 10^{-1.21}\approx 0.06\hspace{0.05cm}.$$
@@ Line 108: / Line 105: @@
+'''(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"):
 :$$V  =  10 \cdot {\rm lg} \hspace{0.1cm}\rho_v \hspace{0.1cm}{\rm (System\;B)} - 10 \cdot {\rm lg} \hspace{0.1cm}\rho_v \hspace{0.1cm}{\rm (System\;A)}
   =  \left[6 \cdot \left(1 - {\rm e}^{-x+1} \right) -x -1 \right] \cdot 10\,{\rm dB}\hspace{0.05cm}.$$
 *Setting the derivative to zero yields the &nbsp;$x$–value that leads to the maximum improvement:
 :$$ \frac{{\rm d}V}{{\rm d}x} = 6 \cdot {\rm e}^{-x+1} -1\Rightarrow \hspace{0.3cm} x = 1+ {\rm ln} \hspace{0.1cm}6 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = \hspace{0.15cm}\underline {27.9\,{\rm dB}}\hspace{0.05cm}.$$
-*This results in exactly the case discussed in subtask &nbsp; '''(4)'''&nbsp; with &nbsp; $10 · \lg ρ_υ = 50$&nbsp; dB, while the signal-to-noise ratio for &nbsp;$\text{System A}$&nbsp; is only&nbsp; $37.9$&nbsp; dB.&nbsp;
+*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.

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID960__Mod_A_1_3.png|right|frame|System Comparison at the AWGN Channel]]
+[[File:P_ID960__Mod_A_1_3.png|right|frame|System comparison at AWGN channel]]
-For the comparison of different modulation and demodulation methods with regard to noise sensitivity, we usually start from the so-called [[Modulation_Methods/Qualitätskriterien#Einige_Anmerkungen_zum_AWGN.E2.80.93Kanalmodell|AWGN Channel]]&nbsp; and present the following double logarithmic diagram:
+For the comparison of different modulation and demodulation methods with regard to noise sensitivity,&nbsp;  we usually start from the so-called&nbsp;  [[Modulation_Methods/Qualitätskriterien#Einige_Anmerkungen_zum_AWGN.E2.80.93Kanalmodell|AWGN channel]]&nbsp; and present the following double logarithmic diagram:
-*The y-axis indicates the sinkt-to-noise ratio (logarithmic SNR) &nbsp;$10 · \lg ρ_v$&nbsp; in dB.
+*The y-axis indicates the&nbsp;  "sink-to-noise ratio"&nbsp;  (logarithmic SNR) &nbsp; &rArr; &nbsp;  $10 · \lg ρ_v$&nbsp; in dB.
-*&nbsp;$10 · \lg ξ$&nbsp; is plotted on the x-axis; the normalized power parameter ("performance parameter") is characterized by:
+*&nbsp;$10 · \lg ξ$&nbsp; is plotted on the x-axis;&nbsp;  the normalized power parameter&nbsp;  ("performance parameter")&nbsp;  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 &nbsp;$P_{\rm S}$, the channel attenuation factor&nbsp;$α_{\rm K}$, the noise power density &nbsp;$N_0$&nbsp; and the bandwidth &nbsp;$B_{\rm NF}$&nbsp; of the message signal are suitably summarised together in &nbsp;$ξ$&nbsp;.
+*Thus,&nbsp;   the transmission power &nbsp;$P_{\rm S}$,&nbsp; the channel attenuation factor&nbsp;$α_{\rm K}$,&nbsp; the noise power density &nbsp;$N_0$&nbsp; and the bandwidth &nbsp;$B_{\rm NF}$&nbsp; of the message signal are suitably summarised together in &nbsp;$ξ$.
-* Unless explicitly stated otherwise, the following values shall be assumed in the exercise:
+* Unless explicitly stated otherwise,&nbsp; the following values shall be assumed in the exercise:
 :$$P_{\rm S}= 5 \;{\rm kW}\hspace{0.05cm}, \hspace{0.2cm} \alpha_{\rm
 K} = 0.001\hspace{0.05cm}, \hspace{0.2cm} {N_0} =
@@ Line 18: / Line 18: @@
 *$\text{System A}$&nbsp; is characterized by the following equation:
 :$$y = x+1.$$
-*Accordingly, &nbsp;$\text{System B}$ is characterized by:
+*Accordingly, &nbsp;$\text{System B}$&nbsp; is characterized by:
 :$$ 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 &nbsp;$x = 4$&nbsp; represents &nbsp;$10 · \lg ξ = 40\text{ dB}$&nbsp; or &nbsp;$ξ = 10^4$&nbsp; and &nbsp;$y = 5$&nbsp; represents &nbsp;$10 · \lg ρ_v= 50\text{ dB}$&nbsp;, i.e., &nbsp;$ρ_v = 10^5$.
+*Thus &nbsp;$x = 4$&nbsp; represents &nbsp;$10 · \lg ξ = 40\text{ dB}$&nbsp; or &nbsp;$ξ = 10^4$&nbsp;
@@ Line 34: / Line 35: @@
 ''Hints:''
 *This exercise belongs to the chapter &nbsp; [[Modulation_Methods/Qualitätskriterien|Quality Criteria]].
-*Particular reference is made to the page &nbsp;  [[Modulation_Methods/Qualitätskriterien#Untersuchungen_beim_AWGN.E2.80.93Kanal|Investigating at the AWGN Channel]].
+*Particular reference is made to the page &nbsp;  [[Modulation_Methods/Qualitätskriterien#Untersuchungen_beim_AWGN.E2.80.93Kanal|Investigating at the AWGN channel]].
-*By specifying the powers in watts, they are independent of the reference resistance, 𝑅. &nbsp;$R$.
+*By specifying the powers in watts,&nbsp; they are independent of the reference resistance &nbsp;$R$.
@@ Line 42: / Line 43: @@
 <quiz display=simple>
-{What is the sink-to-noise ratio  (in dB) for &nbsp;$\text{System A}$&nbsp; with &nbsp;$P_{\rm S}= 5 \;{\rm kW}$, &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; $N_0 = 10^{-10}\;{\rm W}/{\rm Hz}$, &nbsp; $B_{\rm NF}= 5\; {\rm kHz}$?
 |type="{}"}
@@ Line 50: / Line 51: @@
 |type="[]"}
 - Increasing the transmission power from &nbsp;$P_{\rm S}= 5\text{ kW}$&nbsp; to $10\text{ kW}$&nbsp;.
-+ Increasing the channel transmission factor from nbsp;$α_{\rm K} = 0.001$&nbsp; to &nbsp;$0.004$.
++ Increasing the channel transmission factor from &nbsp;$α_{\rm K} = 0.001$&nbsp; to &nbsp;$0.004$.
 + Reducing the noise power density to &nbsp;$N_0=10^{–11 }\text{ W/Hz}$.
-- Increasing the AF bandwidth from &nbsp;$B_{\rm NF}= 5\text{ kHz}$&nbsp; to &nbsp;$\text{ kHz}$.
+- Increasing the source signal bandwidth from &nbsp;$B_{\rm NF}= 5\text{ kHz}$&nbsp; to &nbsp;$10\text{ kHz}$.
-{What is the signal-to-noise ratio for &nbsp;$\text{System B}$&nbsp; with &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="{}"}
 $10 · \lg \hspace{0.05cm}ρ_v \ = \ $ { 57 3% } $\ \text{dB}$
-{If the required signal-to-noise ratio is  &nbsp;$10 · \lg ρ_v = 50\text{ dB}$, what transmission power &nbsp;$P_{\rm S}$ is sufficient to achieve this for &nbsp;$\text{System B}$?
+{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="{}"}
 $P_{\rm S} \ = \ $ { 0.3 3% } $\ \text{ kW }$

@@ Line 108: / Line 108: @@
-'''(5)'''&nbsp; Denoting the larger sink-to-noise ratio of Wir bezeichnen mit (steht für "Verbesserung" )&nbsp; den größeren Sinken–Störabstand von  &nbsp;$\text{System B}$&nbsp; compared to  &nbsp;$\text{System A}$&nbsp; we will denote with &nbsp; $V$&nbsp; (from German "Verbesserung" $$\Rightarrow$$ Improvement):
+'''(5)'''&nbsp; The larger sink-to-noise ratio of &nbsp;$\text{System B}$&nbsp; compared to  &nbsp;$\text{System A}$&nbsp; we will denote with &nbsp; $V$&nbsp; (from German "Verbesserung" $\Rightarrow$ Improvement):
 :$$V  =  10 \cdot {\rm lg} \hspace{0.1cm}\rho_v \hspace{0.1cm}{\rm (System\;B)} - 10 \cdot {\rm lg} \hspace{0.1cm}\rho_v \hspace{0.1cm}{\rm (System\;A)}
   =  \left[6 \cdot \left(1 - {\rm e}^{-x+1} \right) -x -1 \right] \cdot 10\,{\rm dB}\hspace{0.05cm}.$$