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

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	*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  $ξ$.		*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:		* Unless explicitly stated otherwise,  the following values shall be assumed in the exercise:
	:$$P_{\rm S}= 5 \;{\rm kW}\hspace{0.05cm}, \hspace{0.2cm} \alpha_{\~~rmK~~} = 0.001\hspace{0.05cm}, \hspace{0.2cm} {N_0} =10^{-10}\;{\rm W}/{\rm Hz}\hspace{0.05cm}, \hspace{0.2cm}B_{\rm NF}= 5\; {\rm kHz}\hspace{0.05cm}.$$		:$$P_{\rm S}= 5 \;{\rm kW}\hspace{0.05cm}, \hspace{0.2cm} \alpha_{\rm K} = 0.001\hspace{0.05cm}, \hspace{0.2cm} {N_0} =10^{-10}\;{\rm W}/{\rm Hz}\hspace{0.05cm}, \hspace{0.2cm}B_{\rm NF}= 5\; {\rm kHz}\hspace{0.05cm}.$$

	Two systems are plotted in the graph and their   $(x, y)$-curve can be described as follows:		Two systems are plotted in the graph and their   $(x, y)$-curve can be described as follows:
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	[[Category:Modulation Methods: Exercises\|^1.2 Quality Criteria^]]		[[Category:Modulation Methods: Exercises\|^1.2 Quality Criteria^]]
	[[de:Aufgaben:1.~~3_Systemvergleich_beim_AWGN–Kanal~~]]		[[de:Aufgaben:Aufgabe 1.3: Systemvergleich beim AWGN–Kanal]]

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	*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  $ξ$.		*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:		* 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_{\rmK} = 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}.$$
	K} = 0.001\hspace{0.05cm}, \hspace{0.2cm} {N_0} =
	10^{-10}\;{\rm W}/{\rm Hz}\hspace{0.05cm}, \hspace{0.2cm}
	B_{\rm NF}= 5\; {\rm kHz}\hspace{0.05cm}.$$

	Two systems are plotted in the graph and their   $(x, y)$-curve can be described as follows:		Two systems are plotted in the graph and their   $(x, y)$-curve can be described as follows:
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	'''(4)'''  This problem is described by the following equation:		'''(4)'''  This problem is described by the following equation:
	:$$ y= 6 \cdot \left(1 - {\rm e}^{-x+1} \right) = 5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm e}^{-x+1} ={1}/{6}\hspace{0.3cm}		:$$ 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}.$$
	\Rightarrow \hspace{0.3cm} x = 1+ {\rm ln} \hspace{0.1cm}6 \approx 2.79 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = 27.9\,{\rm dB}\hspace{0.05cm}.$$
	*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.		*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:		*Now the transmission power can be reduced by about   $12.1$  dB:
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	'''(5)'''  The larger sink SNR of  $\text{System B}$  compared to  $\text{System A}$  we will denote with   $V$  (from German  "Verbesserung"   ⇒   "improvement"):		'''(5)'''  The larger sink SNR of  $\text{System B}$  compared to  $\text{System A}$  we will denote with   $V$  (from German  "Verbesserung"   ⇒   "improvement"):
	:$$V = 10 \cdot {\rm lg} \hspace{0.1cm}\rho_v \hspace{0.1cm}{\rm (System\;B)} - 10 \cdot {\rm lg} \hspace{0.1cm}\rho_v \hspace{0.1cm}{\rm (System\;A)}		:$$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}.$$
	= \left[6 \cdot \left(1 - {\rm e}^{-x+1} \right) -x -1 \right] \cdot 10\,{\rm dB}\hspace{0.05cm}.$$
	*Setting the derivative to zero yields the  $x$–value that leads to the maximum improvement:		*Setting the derivative to zero yields the  $x$–value that leads to the maximum improvement:
	:$$ \frac{{\rm d}V}{{\rm d}x} = 6 \cdot {\rm e}^{-x+1} -1\Rightarrow \hspace{0.3cm} x = 1+ {\rm ln} \hspace{0.1cm}6 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = \hspace{0.15cm}\underline {27.9\,{\rm dB}}\hspace{0.05cm}.$$		:$$ \frac{{\rm d}V}{{\rm d}x} = 6 \cdot {\rm e}^{-x+1} -1\Rightarrow \hspace{0.3cm} x = 1+ {\rm ln} \hspace{0.1cm}6 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = \hspace{0.15cm}\underline {27.9\,{\rm dB}}\hspace{0.05cm}.$$

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	{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  $\text{System B}$?		{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  $\text{System B}$?
	\|type="{}"}		\|type="{}"}
	$P_{\rm S} \ = \ $ { 0.3 3% } $\ \text{ kW }$		$P_{\rm S} \ = \ $ { 0.3 3% } $\ \text{kW}$

	{What value of  $10 · \lg ξ$  gives the greatest improvement for  $\text{System B}$  relative to  $\text{System A}$ ?		{What value of  $10 · \lg ξ$  gives the greatest improvement for  $\text{System B}$  relative to  $\text{System A}$ ?
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	*by a transmission power of  $P_{\rm S} = 50$  kW  instead of   $5$  kW,		*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 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 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.		*by a signal bandwidth of  $B_{\rm NF} = 0.5$  kHz  instead of   $5$  kHz.

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	[[Category:Modulation Methods: Exercises\|^1.2 Quality Criteria^]]		[[Category:Modulation Methods: Exercises\|^1.2 Quality Criteria^]]
			[[de:Aufgaben:1.3_Systemvergleich_beim_AWGN–Kanal]]

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	$P_{\rm S} \ = \ $ { 0.3 3% } $\ \text{ kW }$		$P_{\rm S} \ = \ $ { 0.3 3% } $\ \text{ kW }$

	{What value of  $10 · \lg ξ$  gives the greatest improvement of  $\text{System B}$  relative to  $\text{System A}$ ?		{What value of  $10 · \lg ξ$  gives the greatest improvement for  $\text{System B}$  relative to  $\text{System A}$ ?
	\|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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	*$\text{System A}$  is characterized by the following equation:		*$\text{System A}$  is characterized by the following equation:
	:$$y = x+1.$$		:$$y = x+1.$$
	*~~Accordingly,~~  $\text{System B}$  is characterized by:		* $\text{System B}$  is instead characterized by:
	:$$ y= 6 \cdot \left(1 - {\rm e}^{-x+1} \right)\hspace{0.05cm}.$$		:$$ y= 6 \cdot \left(1 - {\rm e}^{-x+1} \right)\hspace{0.05cm}.$$
	The additional axis labels drawn in green have the following meaning:		The additional axis labels drawn in green have the following meaning:

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	[[File:P_ID960__Mod_A_1_3.png\|right\|frame\|System comparison at 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/Quality_Criteria#Some_remarks_on_the_AWGN_channel_model\|AWGN channel]]  and present the following double logarithmic diagram:		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:
	*The y-axis indicates the  "sink-to-noise ratio"  (logarithmic SNR)   ⇒   $10 · \lg ρ_v$  in dB.		*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:		* $10 · \lg ξ$  is plotted on the x-axis;  the normalized power parameter  ("performance parameter")  is characterized by:

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	''Hints:''		''Hints:''
	*This exercise belongs to the chapter   [[Modulation_Methods/~~Qualitätskriterien~~\|Quality Criteria]].		*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]].		*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  $R$.		*By specifying the powers in watts,  they are independent of the reference resistance  $R$.

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	''Hints:''		''Hints:''
	*This exercise belongs to the chapter   [[Modulation_Methods/Qualitätskriterien\|Quality Criteria]].		*This exercise belongs to the chapter   [[Modulation_Methods/Qualitätskriterien\|Quality Criteria]].
	*Particular reference is made to the page   [[Modulation_Methods/~~Qualitätskriterien~~#~~Untersuchungen_beim_AWGN.E2.80.93Kanal~~\|Investigating at the AWGN channel]].		*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  $R$.		*By specifying the powers in watts,  they are independent of the reference resistance  $R$.

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The normalized performance parameter is calculated using these values as follows		'''(1)'''  The normalized performance parameter is calculated using these values as follows:
	:$$\xi = \frac{5 \cdot 10^3\,{\rm W}\cdot 10^{-6} }{10^{-10}\,{\rm W}/{\rm Hz} \cdot 5 \cdot 10^3\,{\rm Hz}} = 10^4 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = 40\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x=4 \hspace{0.05cm}.$$		:$$\xi = \frac{5 \cdot 10^3\,{\rm W}\cdot 10^{-6} }{10^{-10}\,{\rm W}/{\rm Hz} \cdot 5 \cdot 10^3\,{\rm Hz}} = 10^4 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = 40\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x=4 \hspace{0.05cm}.$$
	*This gives the auxiliary coordinate value   $y = 5$, which leads to a sink~~-to-noise ratio~~ of   $10 · \lg \hspace{0.05cm} ρ_v\hspace{0.15cm}\underline{ = 50 \ \rm dB}$~~ ~~.		*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)'''<u> Answers 2 and 3</u>  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:
	~~'''(2)'''<u> Answers 2 and 3</u> are correct:~~

	This requirement corresponds to a $10$  dB, increase in the ~~signal-to-noise ratio~~ compared to the previous system, so  $10 · \lg \hspace{0.05cm}ξ$  must also be increased by $10$  dB:
	:$$10 \cdot {\rm lg} \hspace{0.1cm}\xi = 50\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \xi=10^5 \hspace{0.05cm}.$$		:$$10 \cdot {\rm lg} \hspace{0.1cm}\xi = 50\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \xi=10^5 \hspace{0.05cm}.$$

	A tenfold larger   $ξ$ value is achieved – provided all other parameters are held constant in each case –		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 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 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 noise power density of   $N_0 = 10^{ –11 }$  W/Hz  instead of  $10^{ –10 }$  W/Hz,
	*by a bandwidth of  $B_{\rm NF} = 0.5$  kHz  instead of   $5$  kHz.		*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:		'''(3)'''  For  $10 · \lg \hspace{0.05cm} ξ = 40$  dB,  the auxiliary value is   $x = 4$.  This gives the auxiliary  $y$–value:
	:$$y= 6 \cdot \left(1 - {\rm e}^{-3} \right)\approx 5.7 \hspace{0.05cm}.$$		:$$y= 6 \cdot \left(1 - {\rm e}^{-3} \right)\approx 5.7 \hspace{0.05cm}.$$
	*This corresponds to a sink~~-to-noise ratio~~ of   $10 · \lg \hspace{0.05cm} ρ_v\hspace{0.15cm}\underline{ = 57 \ \rm dB}$~~, a~~ $7$  dB improvement over  $\text{System A}$~~ ~~.		*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  $\text{System A}$.


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	:$$ y= 6 \cdot \left(1 - {\rm e}^{-x+1} \right) = 5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm e}^{-x+1} ={1}/{6}\hspace{0.3cm}		:$$ 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}.$$		\Rightarrow \hspace{0.3cm} x = 1+ {\rm ln} \hspace{0.1cm}6 \approx 2.79 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = 27.9\,{\rm dB}\hspace{0.05cm}.$$
	*For  $\text{System A}$~~  this required~~   $10 · \lg \hspace{0.05cm} \xi = 40$  dB , which was achieved with   $P_{\rm S} = 5$  kW and the other numerical values given.		*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:		*Now the transmission power can be reduced by about   $12.1$  dB:
	:$$ 10 \cdot {\rm lg} \hspace{0.1cm} \frac{P_{\rm S}}{ 5 \;{\rm kW}}= -12.1\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac{P_{\rm S}}{ 5 \;{\rm kW}} = 10^{-1.21}\approx 0.06\hspace{0.05cm}.$$		:$$ 10 \cdot {\rm lg} \hspace{0.1cm} \frac{P_{\rm S}}{ 5 \;{\rm kW}}= -12.1\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac{P_{\rm S}}{ 5 \;{\rm kW}} = 10^{-1.21}\approx 0.06\hspace{0.05cm}.$$
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			'''(5)'''  The larger sink SNR of  $\text{System B}$  compared to  $\text{System A}$  we will denote with   $V$  (from German  "Verbesserung"   ⇒   "improvement"):
	'''(5)'''  The larger sink~~-to-noise ratio~~ of  $\text{System B}$  compared to  $\text{System A}$  we will denote with   $V$  (from German "Verbesserung" ~~$\Rightarrow$ Improvement~~):
	:$$V = 10 \cdot {\rm lg} \hspace{0.1cm}\rho_v \hspace{0.1cm}{\rm (System\;B)} - 10 \cdot {\rm lg} \hspace{0.1cm}\rho_v \hspace{0.1cm}{\rm (System\;A)}		:$$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}.$$		= \left[6 \cdot \left(1 - {\rm e}^{-x+1} \right) -x -1 \right] \cdot 10\,{\rm dB}\hspace{0.05cm}.$$
	*Setting the derivative to zero yields the  $x$–value that leads to the maximum improvement:		*Setting the derivative to zero yields the  $x$–value that leads to the maximum improvement:
	:$$ \frac{{\rm d}V}{{\rm d}x} = 6 \cdot {\rm e}^{-x+1} -1\Rightarrow \hspace{0.3cm} x = 1+ {\rm ln} \hspace{0.1cm}6 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = \hspace{0.15cm}\underline {27.9\,{\rm dB}}\hspace{0.05cm}.$$		:$$ \frac{{\rm d}V}{{\rm d}x} = 6 \cdot {\rm e}^{-x+1} -1\Rightarrow \hspace{0.3cm} x = 1+ {\rm ln} \hspace{0.1cm}6 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = \hspace{0.15cm}\underline {27.9\,{\rm dB}}\hspace{0.05cm}.$$
	*This results in exactly the case discussed in subtask   '''(4)'''  with   $10 · \lg ρ_υ = 50$  dB, while the ~~signal-to-noise ratio~~ for  $\text{System A}$  is only  $37.9$  dB.		*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.		*The improvement is therefore  $12.1$  dB.