Aufgaben:Exercise 1.7: System Efficiencies - Revision history

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	:$${E_{\rm B}}/{2} = s_0^2 \cdot \frac {T_1}{2}+ s_0^2 \cdot \frac {T-T_1}{2 \cdot3}= s_0^2 \cdot \left [\frac{T}{6} + \frac{T_1}{3}\right ]\hspace{0.3cm}\hspace{0.3cm}\Rightarrow E_{\rm B} = {s_0^2}/{3}\cdot \left (T + 2 \cdot T_1 \right )\hspace{0.05cm}.$$		:$${E_{\rm B}}/{2} = s_0^2 \cdot \frac {T_1}{2}+ s_0^2 \cdot \frac {T-T_1}{2 \cdot3}= s_0^2 \cdot \left [\frac{T}{6} + \frac{T_1}{3}\right ]\hspace{0.3cm}\hspace{0.3cm}\Rightarrow E_{\rm B} = {s_0^2}/{3}\cdot \left (T + 2 \cdot T_1 \right )\hspace{0.05cm}.$$
	*With the given values  ${s_{0}}^{2} = 10 \ \rm mW$  and  $T = 3\ \rm µ s$  we obtain:		*With the given values  ${s_{0}}^{2} = 10 \ \rm mW$  and  $T = 3\ \rm µ s$  we obtain:
	:$$T_1 = 0\text{:} \hspace{0.75cm} {E_{\rm B}} = \ 1/3 \cdot{s_0^2 \cdot T}= 1/3 \cdot {10^{-2}\,{\rm W} \cdot 3 \cdot 10^{-6}\,{\rm s}} \hspace{0.1cm}\underline {= 1 \cdot 10^{-8}\,{\~~rmWs~~}}\hspace{0.05cm},$$		:$$T_1 = 0\text{:} \hspace{0.75cm} {E_{\rm B}} = \ 1/3 \cdot{s_0^2 \cdot T}= 1/3 \cdot {10^{-2}\,{\rm W} \cdot 3 \cdot 10^{-6}\,{\rm s}} \hspace{0.1cm}\underline {= 1 \cdot 10^{-8}\,{\rm Ws}}\hspace{0.05cm},$$
	:$$T_1 = T/2\text{:} \hspace{0.2cm} {E_{\rm B}} = \ 2/3 \cdot{ s_0^2 \cdot T}= \hspace{2.6cm}\text{...} \hspace{1.4cm}\hspace{0.1cm}\underline {= 2 \cdot 10^{-8}\,{\~~rmWs~~}} \hspace{0.05cm},$$		:$$T_1 = T/2\text{:} \hspace{0.2cm} {E_{\rm B}} = \ 2/3 \cdot{ s_0^2 \cdot T}= \hspace{2.6cm}\text{...} \hspace{1.4cm}\hspace{0.1cm}\underline {= 2 \cdot 10^{-8}\,{\rm Ws}} \hspace{0.05cm},$$
	:$$T_1 = T\text{:} \hspace{0.65cm} {E_{\rm B}} = \ { s_0^2 \cdot T}= \hspace{3.65cm}\text{...} \hspace{1.4cm}\hspace{0.1cm}\underline {= 3 \cdot 10^{-8}\,{\~~rmWs~~}} \hspace{0.05cm}.$$		:$$T_1 = T\text{:} \hspace{0.65cm} {E_{\rm B}} = \ { s_0^2 \cdot T}= \hspace{3.65cm}\text{...} \hspace{1.4cm}\hspace{0.1cm}\underline {= 3 \cdot 10^{-8}\,{\rm Ws}} \hspace{0.05cm}.$$


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	'''(3)'''  Under the condition given in question  '''(2)''',  the maximum SNR is obtained:		'''(3)'''  Under the condition given in question  '''(2)''',  the maximum SNR is obtained:
	:$$\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm}\| \hspace{0.05cm}L}}= \frac{2 \cdot E_{\rm B}}{N_0} = \frac{2 \cdot 3 \cdot 10^{-8}\,{\rm Ws}}{3 \cdot 10^{-10}\,{\~~rmW~~/Hz}}\hspace{0.1cm}\underline {= 200} \hspace{0.05cm}.$$		:$$\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm}\| \hspace{0.05cm}L}}= \frac{2 \cdot E_{\rm B}}{N_0} = \frac{2 \cdot 3 \cdot 10^{-8}\,{\rm Ws}}{3 \cdot 10^{-10}\,{\rm W/Hz}}\hspace{0.1cm}\underline {= 200} \hspace{0.05cm}.$$


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	[[Category:Digital Signal Transmission: Exercises\|^1.4 Optimization of Baseband Systems^]]		[[Category:Digital Signal Transmission: Exercises\|^1.4 Optimization of Baseband Systems^]]
	[[de:Aufgaben:1.~~7_Systemwirkungsgrade~~]]		[[de:Aufgaben:Aufgabe 1.7: Systemwirkungsgrade]]

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	[[File:P_ID1294__Dig_A_1_7.png\|right\|frame\|Transmission pulse  "Trapezoid" ]]		[[File:P_ID1294__Dig_A_1_7.png\|right\|frame\|Transmission pulse  "Trapezoid" ]]
	The receiver of a binary transmission system with symbol duration  $T$  consists of an integrator,  which is represented by the impulse response		The receiver of a binary transmission system with symbol duration  $T$  consists of an integrator,  which is represented by the impulse response
	:$$h_{\rm E}(t) = \left\{ \begin{array}{c} 1/T \\		:$$h_{\rm E}(t) = \left\{ \begin{array}{c} 1/T \\0 \\ \end{array} \right.\quad\begin{array}{{1}c} {\rm{for}}\\ {\rm{for}} \\ \end{array}\begin{array}{{20}c}\hspace{0.05cm}\|t\| < T/2 \hspace{0.05cm}, \\\|t\| > T/2\hspace{0.05cm}. \\\end{array}$$
	0 \\ \end{array} \right.\quad
	\begin{array}{*{1}c} {\rm{for}}
	\\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c}
	\hspace{0.05cm}\|t\| < T/2 \hspace{0.05cm}, \\
	\|t\| > T/2\hspace{0.05cm}. \\
	\end{array}$$
	This is followed by a threshold decision with optimal parameters.		This is followed by a threshold decision with optimal parameters.

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	*According to this division,  it can also be written:		*According to this division,  it can also be written:
	:$${E_{\rm B}}/{2} = s_0^2 \cdot T_1\hspace{0.0cm}' + E_2\hspace{0.05cm},\hspace{0.3cm}{\rm with}\hspace{0.3cm}E_{\rm 2} = \\int_{T_1\hspace{0.0cm}'}^{T/2}g_s^2(t) \,{\rm d} t= s_0^2 \cdot \int_{0}^{T_2\hspace{0.0cm}'}\left ( 1 - \frac {t}{T_2\hspace{0.0cm}'}\right )^2 \,{\rm d}t $$		:$${E_{\rm B}}/{2} = s_0^2 \cdot T_1\hspace{0.0cm}' + E_2\hspace{0.05cm},\hspace{0.3cm}{\rm with}\hspace{0.3cm}E_{\rm 2} = \\int_{T_1\hspace{0.0cm}'}^{T/2}g_s^2(t) \,{\rm d} t= s_0^2 \cdot \int_{0}^{T_2\hspace{0.0cm}'}\left ( 1 - \frac {t}{T_2\hspace{0.0cm}'}\right )^2 \,{\rm d}t $$
	:$$\Rightarrow \hspace{0.3cm}E_{\rm 2} = \ s_0^2 \cdot \left [ \int_{0}^{T_2\hspace{0.0cm}'}\,\,{\rm d} t- \frac {2}{T_2\hspace{0.0cm}'} \cdot		:$$\Rightarrow \hspace{0.3cm}E_{\rm 2} = \ s_0^2 \cdot \left [ \int_{0}^{T_2\hspace{0.0cm}'}\,\,{\rm d} t- \frac {2}{T_2\hspace{0.0cm}'} \cdot\int_{0}^{T_2\hspace{0.0cm}'}t \,\,{\rm d} t + \frac {1}{(T_2\hspace{0.0cm}'\hspace{0.02cm})^2} \cdot\int_{0}^{T_2\hspace{0.0cm}'}t^2 \,\,{\rm d} t\right ] = \ s_0^2 \cdot \left [ {T_2\hspace{0.0cm}'} - \frac {2}{T_2\hspace{0.0cm}'} \cdot\frac {(T_2\hspace{0.0cm}'\hspace{0.02cm})^2}{2} + \frac {1}{(T_2\hspace{0.0cm}'\hspace{0.02cm})^2} \cdot\frac {(T_2\hspace{0.0cm}'\hspace{0.02cm})^3}{3}\right ] = s_0^2\cdot\frac {T_2\hspace{0.0cm}'\hspace{0.02cm}}{3}\hspace{0.05cm}.$$
	\int_{0}^{T_2\hspace{0.0cm}'}t \,\,{\rm d} t + \frac {1}{(T_2\hspace{0.0cm}'\hspace{0.02cm})^2} \cdot
	\int_{0}^{T_2\hspace{0.0cm}'}t^2 \,\,{\rm d} t\right ] = \ s_0^2 \cdot \left [ {T_2\hspace{0.0cm}'} - \frac {2}{T_2\hspace{0.0cm}'} \cdot
	\frac {(T_2\hspace{0.0cm}'\hspace{0.02cm})^2}{2} + \frac {1}{(T_2\hspace{0.0cm}'\hspace{0.02cm})^2} \cdot
	\frac {(T_2\hspace{0.0cm}'\hspace{0.02cm})^3}{3}\right ] = s_0^2
	\cdot\frac {T_2\hspace{0.0cm}'\hspace{0.02cm}}{3}
	\hspace{0.05cm}.$$
	*Substituted into the above equation one obtains:		*Substituted into the above equation one obtains:
	:$${E_{\rm B}}/{2} = s_0^2 \cdot \frac {T_1}{2}+ s_0^2 \cdot \frac {T-T_1}{2 \cdot3}= s_0^2 \cdot \left [\frac{T}{6} + \frac{T_1}{3}\right ]\hspace{0.3cm}\hspace{0.3cm}\Rightarrow E_{\rm B} = {s_0^2}/{3}\cdot \left (T + 2 \cdot T_1 \right )\hspace{0.05cm}.$$		:$${E_{\rm B}}/{2} = s_0^2 \cdot \frac {T_1}{2}+ s_0^2 \cdot \frac {T-T_1}{2 \cdot3}= s_0^2 \cdot \left [\frac{T}{6} + \frac{T_1}{3}\right ]\hspace{0.3cm}\hspace{0.3cm}\Rightarrow E_{\rm B} = {s_0^2}/{3}\cdot \left (T + 2 \cdot T_1 \right )\hspace{0.05cm}.$$

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	:$$E_{\rm B} =\int_{-\infty}^{+\infty}g_s^2(t) \,{\rm d} t =2 \cdot \int_{0}^{T_1\hspace{0.0cm}'}g_s^2(t) \,{\rm d}t\hspace{0.2cm}+ \hspace{0.2cm}2 \cdot \int_{T_1\hspace{0.0cm}'}^{T/2}g_s^2(t) \,{\rm d} t\hspace{0.05cm}.$$		:$$E_{\rm B} =\int_{-\infty}^{+\infty}g_s^2(t) \,{\rm d} t =2 \cdot \int_{0}^{T_1\hspace{0.0cm}'}g_s^2(t) \,{\rm d}t\hspace{0.2cm}+ \hspace{0.2cm}2 \cdot \int_{T_1\hspace{0.0cm}'}^{T/2}g_s^2(t) \,{\rm d} t\hspace{0.05cm}.$$
	*According to this division,  it can also be written:		*According to this division,  it can also be written:
	:$${E_{\rm B}}/{2} = s_0^2 \cdot T_1\hspace{0.0cm}' + E_2		:$${E_{\rm B}}/{2} = s_0^2 \cdot T_1\hspace{0.0cm}' + E_2\hspace{0.05cm},\hspace{0.3cm}{\rm with}\hspace{0.3cm}E_{\rm 2} = \\int_{T_1\hspace{0.0cm}'}^{T/2}g_s^2(t) \,{\rm d} t= s_0^2 \cdot \int_{0}^{T_2\hspace{0.0cm}'}\left ( 1 - \frac {t}{T_2\hspace{0.0cm}'}\right )^2 \,{\rm d}t $$
	\hspace{0.05cm},\hspace{0.3cm}{\rm with}\hspace{0.3cm}
	E_{\rm 2} = \
	\int_{T_1\hspace{0.0cm}'}^{T/2}g_s^2(t) \,{\rm d} t
	= s_0^2 \cdot \int_{0}^{T_2\hspace{0.0cm}'}\left ( 1 - \frac {t}{T_2\hspace{0.0cm}'}\right )^2 \,{\rm d}
	t $$
	:$$\Rightarrow \hspace{0.3cm}E_{\rm 2} = \ s_0^2 \cdot \left [ \int_{0}^{T_2\hspace{0.0cm}'}\,\,{\rm d} t- \frac {2}{T_2\hspace{0.0cm}'} \cdot		:$$\Rightarrow \hspace{0.3cm}E_{\rm 2} = \ s_0^2 \cdot \left [ \int_{0}^{T_2\hspace{0.0cm}'}\,\,{\rm d} t- \frac {2}{T_2\hspace{0.0cm}'} \cdot
	\int_{0}^{T_2\hspace{0.0cm}'}t \,\,{\rm d} t + \frac {1}{(T_2\hspace{0.0cm}'\hspace{0.02cm})^2} \cdot		\int_{0}^{T_2\hspace{0.0cm}'}t \,\,{\rm d} t + \frac {1}{(T_2\hspace{0.0cm}'\hspace{0.02cm})^2} \cdot

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	These definitions can be used to specify the system efficiencies:		These definitions can be used to specify the system efficiencies:
	:$$\eta_{\rm L} = \ \frac{\rho_d}{\rho_{d, \hspace{0.08cm}{\rm max \hspace{0.05cm}\|\hspace{0.05cm}		:$$\eta_{\rm L} = \ \frac{\rho_d}{\rho_{d, \hspace{0.08cm}{\rm max \hspace{0.05cm}\|\hspace{0.05cm}L}}}\hspace{0.05cm},$$
	L}}}\hspace{0.05cm},$$		:$$\eta_{\rm A} = \ \frac{\rho_d}{\rho_{d, \hspace{0.08cm}{\rm max\hspace{0.05cm} \|\hspace{0.05cm} A}}} = {1}/{C_{\rm S}^2}\cdot \eta_{\rm L} \hspace{0.05cm}.$$
	:$$\eta_{\rm A} = \ \frac{\rho_d}{\rho_{d, \hspace{0.08cm}{\rm max\hspace{0.05cm} \|
	\hspace{0.05cm} A}}} = {1}/{C_{\rm S}^2}\cdot \eta_{\rm L} \hspace{0.05cm}.$$

	Here,  the  "crest factor"   $C_{\rm S}$  denotes the ratio between the maximum value and the rms value  (root of power)  of the transmitted signal  $s(t)$.		Here,  the  "crest factor"   $C_{\rm S}$  denotes the ratio between the maximum value and the rms value  (root of power)  of the transmitted signal  $s(t)$.
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	'''(1)'''  To simplify the calculations,  we set  $T_1' = T_1/2$  and  $T_2' = (T - T_1)/2$.		'''(1)'''  To simplify the calculations,  we set  $T_1' = T_1/2$  and  $T_2' = (T - T_1)/2$.
	*This gives for the transmitted pulse energy:		*This gives for the transmitted pulse energy:
	:$$E_{\rm B} =		:$$E_{\rm B} =\int_{-\infty}^{+\infty}g_s^2(t) \,{\rm d} t =2 \cdot \int_{0}^{T_1\hspace{0.0cm}'}g_s^2(t) \,{\rm d}t\hspace{0.2cm}+ \hspace{0.2cm}2 \cdot \int_{T_1\hspace{0.0cm}'}^{T/2}g_s^2(t) \,{\rm d} t\hspace{0.05cm}.$$
	\int_{-\infty}^{+\infty}g_s^2(t) \,{\rm d} t =
	2 \cdot \int_{0}^{T_1\hspace{0.0cm}'}g_s^2(t) \,{\rm d}
	t\hspace{0.2cm}+ \hspace{0.2cm}2 \cdot \int_{T_1\hspace{0.0cm}'}^{T/2}g_s^2(t) \,{\rm d} t
	\hspace{0.05cm}.$$
	*According to this division,  it can also be written:		*According to this division,  it can also be written:
	:$${E_{\rm B}}/{2} = s_0^2 \cdot T_1\hspace{0.0cm}' + E_2		:$${E_{\rm B}}/{2} = s_0^2 \cdot T_1\hspace{0.0cm}' + E_2
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	*Substituted into the above equation one obtains:		*Substituted into the above equation one obtains:
	:$${E_{\rm B}}/{2} = s_0^2 \cdot \frac {T_1}{2}+ s_0^2 \cdot \frac {T-T_1}{2 \~~cdot~~		:$${E_{\rm B}}/{2} = s_0^2 \cdot \frac {T_1}{2}+ s_0^2 \cdot \frac {T-T_1}{2 \cdot3}= s_0^2 \cdot \left [\frac{T}{6} + \frac{T_1}{3}\right ]\hspace{0.3cm}\hspace{0.3cm}\Rightarrow E_{\rm B} = {s_0^2}/{3}\cdot \left (T + 2 \cdot T_1 \right )\hspace{0.05cm}.$$
	3}= s_0^2 \cdot \left [\frac{T}{6} + \frac{T_1}{3}\right ]\hspace{0.3cm}
	\hspace{0.3cm}\Rightarrow E_{\rm B} = {s_0^2}/{3}\cdot \left (T + 2 \cdot T_1 \right )\hspace{0.05cm}.$$
	*With the given values  ${s_{0}}^{2} = 10 \ \rm mW$  and  $T = 3\ \rm µ s$  we obtain:		*With the given values  ${s_{0}}^{2} = 10 \ \rm mW$  and  $T = 3\ \rm µ s$  we obtain:
	:$$T_1 = 0\text{:} \hspace{0.75cm} {E_{\rm B}} = \ 1/3 \cdot{s_0^2 \cdot T}= 1/3 \cdot {10^{-2}\,{\rm W} \cdot 3 \cdot 10^{-6}\,{\rm s}} \hspace{0.1cm}\underline {= 1 \cdot 10^{-8}\,{\rm		:$$T_1 = 0\text{:} \hspace{0.75cm} {E_{\rm B}} = \ 1/3 \cdot{s_0^2 \cdot T}= 1/3 \cdot {10^{-2}\,{\rm W} \cdot 3 \cdot 10^{-6}\,{\rm s}} \hspace{0.1cm}\underline {= 1 \cdot 10^{-8}\,{\rmWs}}\hspace{0.05cm},$$
	Ws}}\hspace{0.05cm},$$		:$$T_1 = T/2\text{:} \hspace{0.2cm} {E_{\rm B}} = \ 2/3 \cdot{ s_0^2 \cdot T}= \hspace{2.6cm}\text{...} \hspace{1.4cm}\hspace{0.1cm}\underline {= 2 \cdot 10^{-8}\,{\rmWs}} \hspace{0.05cm},$$
	:$$T_1 = T/2\text{:} \hspace{0.2cm} {E_{\rm B}} = \ 2/3 \cdot{ s_0^2 \cdot T}= \hspace{2.6cm}\text{...} \hspace{1.4cm}\hspace{0.1cm}\underline {= 2 \cdot 10^{-8}\,{\rm		:$$T_1 = T\text{:} \hspace{0.65cm} {E_{\rm B}} = \ { s_0^2 \cdot T}= \hspace{3.65cm}\text{...} \hspace{1.4cm}\hspace{0.1cm}\underline {= 3 \cdot 10^{-8}\,{\rmWs}} \hspace{0.05cm}.$$
	Ws}} \hspace{0.05cm},$$
	:$$T_1 = T\text{:} \hspace{0.65cm} {E_{\rm B}} = \ { s_0^2 \cdot T}= \hspace{3.65cm}\text{...} \hspace{1.4cm}\hspace{0.1cm}\underline {= 3 \cdot 10^{-8}\,{\rm
	Ws}} \hspace{0.05cm}.$$


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	'''(3)'''  Under the condition given in question  '''(2)''',  the maximum SNR is obtained:		'''(3)'''  Under the condition given in question  '''(2)''',  the maximum SNR is obtained:
	:$$\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm}\| \hspace{0.05cm}		:$$\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm}\| \hspace{0.05cm}L}}= \frac{2 \cdot E_{\rm B}}{N_0} = \frac{2 \cdot 3 \cdot 10^{-8}\,{\rm Ws}}{3 \cdot 10^{-10}\,{\rmW/Hz}}\hspace{0.1cm}\underline {= 200} \hspace{0.05cm}.$$
	L}}= \frac{2 \cdot E_{\rm B}}{N_0} = \frac{2 \cdot 3 \cdot 10^{-8}\,{\rm Ws}}{3 \cdot 10^{-10}\,{\rm
	W/Hz}}\hspace{0.1cm}\underline {= 200} \hspace{0.05cm}.$$


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	'''(5)'''  With  $T_1 = T/2$  (trapezoidal transmitted pulses),  we obtain for the signal-to-noise ratio:		'''(5)'''  With  $T_1 = T/2$  (trapezoidal transmitted pulses),  we obtain for the signal-to-noise ratio:
	:$$\rho_d = \frac{g_0^2}{\sigma_d^2}\hspace{0.3cm}{\rm with}\hspace{0.3cm} g_0^2=0.075^2\, {\rm W},\hspace{0.1cm} \sigma_d^2 = \frac{N_0}{2 \cdot T} = 5 \cdot 10^{-5}\,{\rm W}\hspace{0.3cm}		:$$\rho_d = \frac{g_0^2}{\sigma_d^2}\hspace{0.3cm}{\rm with}\hspace{0.3cm} g_0^2=0.075^2\, {\rm W},\hspace{0.1cm} \sigma_d^2 = \frac{N_0}{2 \cdot T} = 5 \cdot 10^{-5}\,{\rm W}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\rho_d = \frac{0.075^2\, {\rm W}}{5 \cdot 10^{-5}\,{\rm W}} = 112.5 \hspace{0.05cm}.$$
	\Rightarrow \hspace{0.3cm}\rho_d = \frac{0.075^2\, {\rm W}}{5 \cdot 10^{-5}\,{\rm W}} = 112.5 \hspace{0.05cm}.$$
	*Thus,  the system efficiency under power limitation with the result of  '''(3)''':		*Thus,  the system efficiency under power limitation with the result of  '''(3)''':
	:$$\eta_{\rm L} = \frac{\rho_d}{\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm} \| \hspace{0.05cm} L}}}= \frac{112.5}{200}\hspace{0.1cm}\underline {= 0.5625 }\hspace{0.05cm}.$$		:$$\eta_{\rm L} = \frac{\rho_d}{\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm} \| \hspace{0.05cm} L}}}= \frac{112.5}{200}\hspace{0.1cm}\underline {= 0.5625 }\hspace{0.05cm}.$$

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	[[Category:Digital Signal Transmission: Exercises\|^1.4 Optimization of Baseband Systems^]]		[[Category:Digital Signal Transmission: Exercises\|^1.4 Optimization of Baseband Systems^]]
			[[de:Aufgaben:1.7_Systemwirkungsgrade]]

Guenter at 16:01, 9 April 2023

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	In the course of this exercise,  the following quantities will be used:		In the course of this exercise,  the following quantities will be used:
	*$\rho_{d,\rm\hspace{0.08cm} max \hspace{0.03cm}\|\hspace{0.03cm} L}$  is the maximum SNR under the constraint ~~of ~~ of  "power limitation".		*$\rho_{d,\rm\hspace{0.08cm} max \hspace{0.03cm}\|\hspace{0.03cm} L}$  is the maximum SNR under the constraint of  "power limitation".
	*$\rho_{d,\rm\hspace{0.08cm} max \hspace{0.03cm}\|\hspace{0.03cm} A}$  is the maximum SNR under the constraint of  "peak limitation"  (or  "amplitude limitation").		*$\rho_{d,\rm\hspace{0.08cm} max \hspace{0.03cm}\|\hspace{0.03cm} A}$  is the maximum SNR under the constraint of  "peak limitation"  (or  "amplitude limitation").

Guenter at 12:17, 4 May 2022

2022-05-04T12:17:45Z

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	$g_{0} \ = \ $ { 0.075 3% } $\ \rm \sqrt{W}$		$g_{0} \ = \ $ { 0.075 3% } $\ \rm \sqrt{W}$

	{Calculate the system efficiency  $\eta_{\rm L}$  when the power is limited   $(T_{1} = T/2)$.		{Calculate the system efficiency  $\eta_{\rm L}$  when the power is limited  $(T_{1} = T/2)$.
	\|type="{}"}		\|type="{}"}
	$\eta_{\rm L} \ = \ $ { 0.5625 3% }		$\eta_{\rm L} \ = \ $ { 0.5625 3% }

	{Calculate the crest factor   $(T_{1} = T/2)$.		{Calculate the crest factor  $(T_{1} = T/2)$.
	\|type="{}"}		\|type="{}"}
	$C_{\rm S} \ = \ $ { 1.225 3% }		$C_{\rm S} \ = \ $ { 1.225 3% }

	{Calculate the system efficiency at peak limitation   $(T_{1} = T/2)$.		{Calculate the system efficiency at peak limitation  $(T_{1} = T/2)$.
	\|type="{}"}		\|type="{}"}
	$\eta_{\rm A} \ = \ $ { 0.375 3% }		$\eta_{\rm A} \ = \ $ { 0.375 3% }
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  To simplify the calculations, we set $T_1' = T_1/2$ and $T_2' = (T – T_1)/2$.		'''(1)'''  To simplify the calculations,  we set  $T_1' = T_1/2$  and  $T_2' = (T - T_1)/2$.
	*This gives for the transmitted pulse energy:		*This gives for the transmitted pulse energy:
	:$$E_{\rm B} =		:$$E_{\rm B} =
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	t\hspace{0.2cm}+ \hspace{0.2cm}2 \cdot \int_{T_1\hspace{0.0cm}'}^{T/2}g_s^2(t) \,{\rm d} t		t\hspace{0.2cm}+ \hspace{0.2cm}2 \cdot \int_{T_1\hspace{0.0cm}'}^{T/2}g_s^2(t) \,{\rm d} t
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	*According to this division, it can also be written:		*According to this division,  it can also be written:
	:$${E_{\rm B}}/{2} = s_0^2 \cdot T_1\hspace{0.0cm}' + E_2		:$${E_{\rm B}}/{2} = s_0^2 \cdot T_1\hspace{0.0cm}' + E_2
	\hspace{0.05cm},\hspace{0.3cm}{\rm with}\hspace{0.3cm}		\hspace{0.05cm},\hspace{0.3cm}{\rm with}\hspace{0.3cm}
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	3}= s_0^2 \cdot \left [\frac{T}{6} + \frac{T_1}{3}\right ]\hspace{0.3cm}		3}= s_0^2 \cdot \left [\frac{T}{6} + \frac{T_1}{3}\right ]\hspace{0.3cm}
	\hspace{0.3cm}\Rightarrow E_{\rm B} = {s_0^2}/{3}\cdot \left (T + 2 \cdot T_1 \right )\hspace{0.05cm}.$$		\hspace{0.3cm}\Rightarrow E_{\rm B} = {s_0^2}/{3}\cdot \left (T + 2 \cdot T_1 \right )\hspace{0.05cm}.$$
	*With the given values ${s_{0}}^{2} = 10 \ \rm mW$ and $T = 3\ \rm µ s$ we obtain:		*With the given values  ${s_{0}}^{2} = 10 \ \rm mW$  and  $T = 3\ \rm µ s$  we obtain:
	:$$T_1 = 0\text{:} \hspace{0.75cm} {E_{\rm B}} = \ 1/3 \cdot{s_0^2 \cdot T}= 1/3 \cdot {10^{-2}\,{\rm W} \cdot 3 \cdot 10^{-6}\,{\rm s}} \hspace{0.1cm}\underline {= 1 \cdot 10^{-8}\,{\rm		:$$T_1 = 0\text{:} \hspace{0.75cm} {E_{\rm B}} = \ 1/3 \cdot{s_0^2 \cdot T}= 1/3 \cdot {10^{-2}\,{\rm W} \cdot 3 \cdot 10^{-6}\,{\rm s}} \hspace{0.1cm}\underline {= 1 \cdot 10^{-8}\,{\rm
	Ws}}\hspace{0.05cm},$$		Ws}}\hspace{0.05cm},$$
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	'''(2)'''  The system efficiency under power limitation is maximum $(\eta_{\rm L} = 1)$, when the basic transmission pulse $g_{s}(t)$ has the same shape as the impulse response $h_{\rm E}(t)$. This is true here for the NRZ transmitted pulse:   $T_1/T \ \underline{= 1}$.		'''(2)'''  The system efficiency under power limitation is maximum  $(\eta_{\rm L} = 1)$,  when the basic transmission pulse  $g_{s}(t)$  has the same shape as the impulse response $h_{\rm E}(t)$.
			*This is true here for the NRZ transmitted pulse:   $T_1/T \ \underline{= 1}$.


	'''(3)'''  Under the condition given in question '''(2)''', the maximum SNR is obtained:
			'''(3)'''  Under the condition given in question  '''(2)''',  the maximum SNR is obtained:
	:$$\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm}\| \hspace{0.05cm}		:$$\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm}\| \hspace{0.05cm}
	L}}= \frac{2 \cdot E_{\rm B}}{N_0} = \frac{2 \cdot 3 \cdot 10^{-8}\,{\rm Ws}}{3 \cdot 10^{-10}\,{\rm		L}}= \frac{2 \cdot E_{\rm B}}{N_0} = \frac{2 \cdot 3 \cdot 10^{-8}\,{\rm Ws}}{3 \cdot 10^{-10}\,{\rm
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	'''(4)'''  In general, $g_{d}(t) = g_{s}(t) ∗ h_{\rm E}(t)$. For $t = 0$, $T_1 = T/2$ gives the trapezoidal area for this:		'''(4)'''  In general,  $g_{d}(t) = g_{s}(t) ∗ h_{\rm E}(t)$.  For  $t = 0$,  $T_1 = T/2$  gives the trapezoidal area for this:
	:$$g_0 = g_d(t=0) = \frac{1}{T} \cdot \int_{-\infty}^{+\infty}g_s(t) \,{\rm d} t = \frac{T + T_1}{2} \cdot s_0 = 0.75 \cdot 0.1 \cdot \sqrt{\rm W} \hspace{0.1cm}\underline {= 0.075 \,\sqrt{\rm W}} \hspace{0.05cm}.$$		:$$g_0 = g_d(t=0) = \frac{1}{T} \cdot \int_{-\infty}^{+\infty}g_s(t) \,{\rm d} t = \frac{T + T_1}{2} \cdot s_0 = 0.75 \cdot 0.1 \cdot \sqrt{\rm W} \hspace{0.1cm}\underline {= 0.075 \,\sqrt{\rm W}} \hspace{0.05cm}.$$


	'''(5)'''  With $T_1 = T/2$ (trapezoidal transmitted pulses), we obtain for the signal-to-noise ratio:		'''(5)'''  With  $T_1 = T/2$  (trapezoidal transmitted pulses),  we obtain for the signal-to-noise ratio:
	:$$\rho_d = \frac{g_0^2}{\sigma_d^2}\hspace{0.3cm}{\rm with}\hspace{0.3cm} g_0^2=0.075^2\, {\rm W},\hspace{0.1cm} \sigma_d^2 = \frac{N_0}{2 \cdot T} = 5 \cdot 10^{-5}\,{\rm W}\hspace{0.3cm}		:$$\rho_d = \frac{g_0^2}{\sigma_d^2}\hspace{0.3cm}{\rm with}\hspace{0.3cm} g_0^2=0.075^2\, {\rm W},\hspace{0.1cm} \sigma_d^2 = \frac{N_0}{2 \cdot T} = 5 \cdot 10^{-5}\,{\rm W}\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm}\rho_d = \frac{0.075^2\, {\rm W}}{5 \cdot 10^{-5}\,{\rm W}} = 112.5 \hspace{0.05cm}.$$		\Rightarrow \hspace{0.3cm}\rho_d = \frac{0.075^2\, {\rm W}}{5 \cdot 10^{-5}\,{\rm W}} = 112.5 \hspace{0.05cm}.$$
	*Thus, the system efficiency under power limitation with the result of '''(3)''':		*Thus,  the system efficiency under power limitation with the result of  '''(3)''':
	:$$\eta_{\rm L} = \frac{\rho_d}{\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm} \| \hspace{0.05cm} L}}}= \frac{112.5}{200}\hspace{0.1cm}\underline {= 0.5625 }\hspace{0.05cm}.$$		:$$\eta_{\rm L} = \frac{\rho_d}{\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm} \| \hspace{0.05cm} L}}}= \frac{112.5}{200}\hspace{0.1cm}\underline {= 0.5625 }\hspace{0.05cm}.$$
	*Due to the mismatch, $\eta_{\rm L} < 1$.		*Due to the mismatch, $\eta_{\rm L} < 1$.


	'''(6)'''  With the maximum value $s_{0}$ and the result of ~~question~~ '''(1)''':
			'''(6)'''  With the maximum value  $s_{0}$  and the result of  '''(1)''':
	:$$s_{\rm eff} = \sqrt{{ E_{\rm B}}/{T}}= \sqrt{{ 2/3 \cdot s_{0}^2}} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}C_{\rm S} ={ s_{\rm 0}}/{s_{\rm eff}}= \sqrt{{ 3}/{2}}\hspace{0.1cm}\underline { \approx 1.225}\hspace{0.05cm}.$$		:$$s_{\rm eff} = \sqrt{{ E_{\rm B}}/{T}}= \sqrt{{ 2/3 \cdot s_{0}^2}} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}C_{\rm S} ={ s_{\rm 0}}/{s_{\rm eff}}= \sqrt{{ 3}/{2}}\hspace{0.1cm}\underline { \approx 1.225}\hspace{0.05cm}.$$


	'''(7)'''  The system efficiency under peak limitation is smaller than that under power limitation, because the non-optimal transmitted signal (too small energy) plays a role here in addition to the mismatch:		'''(7)'''  The system efficiency under peak limitation is smaller than that under power limitation,  <br>because the non-optimal transmitted signal  (too small energy)  plays a role here in addition to the mismatch:
	:$$\eta_{\rm A} = \frac{1}{C_{\rm S}^2}\cdot \eta_{\rm L} = \frac{ 2}{3} \cdot 0.5625 =\hspace{0.1cm}\underline { 0.375} \hspace{0.05cm}.$$		:$$\eta_{\rm A} = \frac{1}{C_{\rm S}^2}\cdot \eta_{\rm L} = \frac{ 2}{3} \cdot 0.5625 =\hspace{0.1cm}\underline { 0.375} \hspace{0.05cm}.$$

Guenter at 12:08, 4 May 2022

2022-05-04T12:08:06Z

← Older revision		Revision as of 14:08, 4 May 2022
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	[[File:P_ID1294__Dig_A_1_7.png\|right\|frame\|~~Basic transmission~~ pulse "~~trapezoid~~" ]]		[[File:P_ID1294__Dig_A_1_7.png\|right\|frame\|Transmission pulse  "Trapezoid" ]]
	The receiver of a binary ~~message~~ transmission system with symbol duration  $T$  consists of an integrator, which is represented by the impulse response		The receiver of a binary transmission system with symbol duration  $T$  consists of an integrator,  which is represented by the impulse response
	:$$h_{\rm E}(t) = \left\{ \begin{array}{c} 1/T \\		:$$h_{\rm E}(t) = \left\{ \begin{array}{c} 1/T \\
	0 \\ \end{array} \right.\quad		0 \\ \end{array} \right.\quad
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	\\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c}		\\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c}
	\hspace{0.05cm}\|t\| < T/2 \hspace{0.05cm}, \\		\hspace{0.05cm}\|t\| < T/2 \hspace{0.05cm}, \\
	\|t\| > T/2 \\		\|t\| > T/2\hspace{0.05cm}. \\
	\end{array}$$		\end{array}$$
	This is followed by a threshold decision with optimal parameters.		This is followed by a threshold decision with optimal parameters.

	The basic transmission pulse  $g_{s}(t)$  according to the diagram is generally trapezoidal and is parameterized by the time  $T_{1}$:		The basic transmission pulse  $g_{s}(t)$  according to the diagram is generally trapezoidal and is parameterized by the time  $T_{1}$:
	*For  $T_{1} = 0$  the result is a triangular pulse, for  $T_{1} = T$  the NRZ rectangle.		*For  $T_{1} = 0$  the result is a triangular pulse,  for  $T_{1} = T$  the NRZ rectangle.
	*The absolute pulse duration  $T_{\rm S}$  is always equal to the symbol duration  $T$, i.e. the ~~distance~~ between two transmission pulses.		*The absolute pulse duration  $T_{\rm S}$  is always equal to the symbol duration  $T$,  i.e. the spacing between two transmission pulses.


	The signal-to-noise power ratio (SNR) before the threshold decision can be calculated as follows, assuming no intersymbol interfering:		The signal-to-noise power ratio  $\rm (SNR)$  before the threshold decision can be calculated as follows,  assuming no intersymbol interfering:
	:$$\rho_d = {g_0^2}/{\sigma_d^2}\hspace{0.05cm}.$$		:$$\rho_d = {g_0^2}/{\sigma_d^2}\hspace{0.05cm}.$$
	Here,  $g_{0} = g_{d}(t = 0)$  is the maximum value of the basic ~~transmitter~~ pulse and		Here,  $g_{0} = g_{d}(t = 0)$  is the maximum value of the basic transmission pulse,  and
	:$$\sigma_d^2 = {N_0}/{2} \cdot \int_{-\infty}^{+\infty}\|h_{\rm E}(t)\|^2 \,{\rm d} t = \frac{N_0}{2 \cdot T}$$		:$$\sigma_d^2 = {N_0}/{2} \cdot \int_{-\infty}^{+\infty}\|h_{\rm E}(t)\|^2 \,{\rm d} t = \frac{N_0}{2 \cdot T}$$
	the noise power after the receiver filter in the presence of AWGN noise at its input.		the noise power after the receiver filter in the presence of AWGN noise at its input.

	In the course of this exercise, the following quantities will be used:		In the course of this exercise,  the following quantities will be used:
	*$\rho_{d,\rm\hspace{0.~~05cm~~} max \| L}$  is the maximum SNR under the constraint of power limitation.		*$\rho_{d,\rm\hspace{0.08cm} max \hspace{0.03cm}\|\hspace{0.03cm} L}$  is the maximum SNR under the constraint of  of  "power limitation".
	*$\rho_{d,\rm\hspace{0.~~05cm~~} max \| A}$  is the maximum SNR under peak limitation (amplitude limitation).		*$\rho_{d,\rm\hspace{0.08cm} max \hspace{0.03cm}\|\hspace{0.03cm} A}$  is the maximum SNR under the constraint of  "peak limitation"  (or  "amplitude limitation").


	These definitions can be used to specify the system efficiencies:		These definitions can be used to specify the system efficiencies:
	:$$\eta_{\rm L} = \ \frac{\rho_d}{\rho_{d, \hspace{0.~~05cm~~}{\rm max \hspace{0.05cm}\|\hspace{0.05cm}		:$$\eta_{\rm L} = \ \frac{\rho_d}{\rho_{d, \hspace{0.08cm}{\rm max \hspace{0.05cm}\|\hspace{0.05cm}
	L}}}\hspace{0.05cm},$$		L}}}\hspace{0.05cm},$$
	:$$\eta_{\rm A} = \ \frac{\rho_d}{\rho_{d, \hspace{0.~~05cm~~}{\rm max\hspace{0.05cm} \|		:$$\eta_{\rm A} = \ \frac{\rho_d}{\rho_{d, \hspace{0.08cm}{\rm max\hspace{0.05cm} \|
	\hspace{0.05cm} A}}} = {1}/{C_{\rm S}^2}\cdot \eta_{\rm L} \hspace{0.05cm}.$$		\hspace{0.05cm} A}}} = {1}/{C_{\rm S}^2}\cdot \eta_{\rm L} \hspace{0.05cm}.$$

	Here, the ''crest factor''  $C_{\rm S}$  denotes the ratio between the maximum value and the rms value (root of power) of the transmitted signal  $s(t)$.		Here,  the  "crest factor"   $C_{\rm S}$  denotes the ratio between the maximum value and the rms value  (root of power)  of the transmitted signal  $s(t)$.


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			Notes:
	''Notes:''		*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Optimization_of_Baseband_Transmission_Systems\|"Optimization of Baseband Transmission Systems"]].
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Optimization_of_Baseband_Transmission_Systems\|Optimization of Baseband Transmission Systems]].

	*Use the following numerical values to solve the exercise:		*Use the following numerical values to solve the exercise:
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	<quiz display=simple>		<quiz display=simple>

	{Calculate the pulse energy  $E_{\rm B}$  as a function of  $T_{1}$. What are the values for  $T_{1} = 0$ ,  $T_{1} = T/2$  and  $T_{1} = T$?		{Calculate the pulse energy  $E_{\rm B}$  as a function of  $T_{1}$.  What are the values for  $T_{1} = 0$,   $T_{1} = T/2$  and  $T_{1} = T$?
	\|type="{}"}		\|type="{}"}
	$T_{1} = 0\text{:} \hspace{0.75cm} E_{\rm B} \ = \ $ { 1 3% } $\ \cdot 10^{-8} \, \rm Ws$		$T_{1} = 0\text{:} \hspace{0.75cm} E_{\rm B} \ = \ $ { 1 3% } $\ \cdot 10^{-8} \, \rm Ws$
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	$T_{1} = T\text{:}\hspace{0.65cm} E_{\rm B} \ = \ $ { 3 3% } $\ \cdot 10^{-8} \, \rm Ws$		$T_{1} = T\text{:}\hspace{0.65cm} E_{\rm B} \ = \ $ { 3 3% } $\ \cdot 10^{-8} \, \rm Ws$

	{What value  $T_{1}$  leads to the maximum possible SNR when power is limited?		{What value  $T_{1}$  leads to the maximum possible SNR when the power is limited?
	\|type="{}"}		\|type="{}"}
	$T_{1}/T \ = \ $ { 1 3% }		$T_{1}/T \ = \ $ { 1 3% }

	{Therefore, what is the maximum SNR with power limitation?		{Therefore,  what is the maximum SNR with power limitation?
	\|type="{}"}		\|type="{}"}
	$\rho_{d,\hspace{0.~~05cm~~}\rm max \hspace{0.05cm}\|\hspace{0.05cm} L} \ = \ $ { 200 3% }		$\rho_{d,\hspace{0.08cm}\rm max \hspace{0.05cm}\|\hspace{0.05cm} L} \ = \ $ { 200 3% }

	{How large is the basic transmitter pulse  $g_{d}(t)$  in pulse center for  $T_{1} = T/2$?		{How large is the basic transmitter pulse  $g_{d}(t)$  in pulse center for  $T_{1} = T/2$?
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	$g_{0} \ = \ $ { 0.075 3% } $\ \rm \sqrt{W}$		$g_{0} \ = \ $ { 0.075 3% } $\ \rm \sqrt{W}$

	{Calculate the system efficiency  $\eta_{\rm L}$  when power is limited  $(T_{1} = T/2)$.		{Calculate the system efficiency  $\eta_{\rm L}$  when the power is limited   $(T_{1} = T/2)$.
	\|type="{}"}		\|type="{}"}
	$\eta_{\rm L} \ = \ $ { 0.5625 3% }		$\eta_{\rm L} \ = \ $ { 0.5625 3% }

	{Calculate the crest factor  $(T_{1} = T/2)$.		{Calculate the crest factor   $(T_{1} = T/2)$.
	\|type="{}"}		\|type="{}"}
	$C_{\rm S} \ = \ $ { 1.225 3% }		$C_{\rm S} \ = \ $ { 1.225 3% }

	{Calculate the system efficiency at peak limitation  $(T_{1} = T/2)$.		{Calculate the system efficiency at peak limitation   $(T_{1} = T/2)$.
	\|type="{}"}		\|type="{}"}
	$\eta_{\rm A} \ = \ $ { 0.375 3% }		$\eta_{\rm A} \ = \ $ { 0.375 3% }

Guenter at 13:58, 19 March 2022

2022-03-19T13:58:45Z

← Older revision		Revision as of 15:58, 19 March 2022
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	[[Category:Digital Signal Transmission: Exercises\|^1.4 ~~Optimierung der Basisbandsysteme~~^]]		[[Category:Digital Signal Transmission: Exercises\|^1.4 Optimization of Baseband Systems^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 1.7: Systemwirkungsgrade to Aufgaben:Exercise 1.7: System Efficiencies

2022-03-19T13:56:08Z

Guenter moved page Aufgaben:Aufgabe 1.7: Systemwirkungsgrade to Aufgaben:Exercise 1.7: System Efficiencies

← Older revision	Revision as of 15:56, 19 March 2022
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