Aufgaben:Exercise 1.7: System Efficiencies - Revision history

Guenter at 16:01, 9 April 2023

2023-04-09T16:01:27Z

Guenter at 12:17, 4 May 2022

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Guenter at 12:08, 4 May 2022

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Guenter at 13:58, 19 March 2022

2022-03-19T13:58:45Z

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

2022-03-19T13:56:08Z

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

Hwang at 14:13, 18 March 2022

2022-03-18T14:13:19Z

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Javier: Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

2021-03-23T12:47:17Z

Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

Javier: Text replacement - "[[Digitalsignalübertragung" to "[[Digital_Signal_Transmission"

2020-07-09T12:46:51Z

Text replacement - "[[Digitalsignalübertragung" to "[[Digital_Signal_Transmission"

@@ Line 27: / Line 27: @@
 In the course of this exercise,&nbsp; the following quantities will be used:
-*$\rho_{d,\rm\hspace{0.08cm} max \hspace{0.03cm}|\hspace{0.03cm} L}$&nbsp; is the maximum SNR under the constraint of&nbsp; of&nbsp; "power limitation".
+*$\rho_{d,\rm\hspace{0.08cm} max \hspace{0.03cm}|\hspace{0.03cm} L}$&nbsp; is the maximum SNR under the constraint  of&nbsp; "power limitation".
 *$\rho_{d,\rm\hspace{0.08cm} max \hspace{0.03cm}|\hspace{0.03cm} A}$&nbsp; is the maximum SNR under the constraint of&nbsp; "peak limitation"&nbsp; (or&nbsp; "amplitude limitation").

@@ Line 73: / Line 73: @@
 $g_{0} \ = \ $ { 0.075 3% } $\ \rm \sqrt{W}$
-{Calculate the system efficiency &nbsp;$\eta_{\rm L}$&nbsp; when the power is limited  &nbsp; $(T_{1} = T/2)$.
+{Calculate the system efficiency &nbsp;$\eta_{\rm L}$&nbsp; when the power is limited  &nbsp;$(T_{1} = T/2)$.
 |type="{}"}
 $\eta_{\rm L} \ = \ $ { 0.5625 3% }
-{Calculate the crest factor &nbsp; $(T_{1} = T/2)$.
+{Calculate the crest factor &nbsp;$(T_{1} = T/2)$.
 |type="{}"}
 $C_{\rm S} \ = \ $ { 1.225 3% }
-{Calculate the system efficiency at peak limitation &nbsp; $(T_{1} = T/2)$.
+{Calculate the system efficiency at peak limitation &nbsp;$(T_{1} = T/2)$.
 |type="{}"}
 $\eta_{\rm A} \ = \ $ { 0.375 3% }
@@ Line 92: / Line 92: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; To simplify the calculations, we set $T_1' = T_1/2$ and $T_2' = (T – T_1)/2$.
+'''(1)'''&nbsp; To simplify the calculations,&nbsp; we set&nbsp; $T_1' = T_1/2$&nbsp; and&nbsp; $T_2' = (T - T_1)/2$.
 *This gives for the transmitted pulse energy:
 :$$E_{\rm B} =
@@ Line 99: / Line 99: @@
   t\hspace{0.2cm}+ \hspace{0.2cm}2 \cdot \int_{T_1\hspace{0.0cm}'}^{T/2}g_s^2(t) \,{\rm d} t
   \hspace{0.05cm}.$$
-*According to this division, it can also be written:
+*According to this division,&nbsp; it can also be written:
 :$${E_{\rm B}}/{2} = s_0^2 \cdot T_1\hspace{0.0cm}' + E_2
   \hspace{0.05cm},\hspace{0.3cm}{\rm with}\hspace{0.3cm}
@@ Line 117: / Line 117: @@
 }= s_0^2  \cdot \left [\frac{T}{6} + \frac{T_1}{3}\right ]\hspace{0.3cm}
 \hspace{0.3cm}\Rightarrow E_{\rm B}   = {s_0^2}/{3}\cdot   \left (T + 2 \cdot T_1 \right )\hspace{0.05cm}.$$
-*With the given values ${s_{0}}^{2} = 10 \ \rm mW$ and $T = 3\ \rm &micro; s$ we obtain:
+*With the given values&nbsp; ${s_{0}}^{2} = 10 \ \rm mW$&nbsp; and&nbsp; $T = 3\ \rm &micro; s$&nbsp; we obtain:
 :$$T_1 = 0\text{:} \hspace{0.75cm} {E_{\rm B}}  = \ 1/3 \cdot{s_0^2 \cdot T}= 1/3 \cdot {10^{-2}\,{\rm W} \cdot 3 \cdot 10^{-6}\,{\rm s}} \hspace{0.1cm}\underline {=  1 \cdot 10^{-8}\,{\rm
   Ws}}\hspace{0.05cm},$$
@@ Line 126: / Line 126: @@
-'''(2)'''&nbsp; 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: &nbsp;  $T_1/T \ \underline{= 1}$.
+'''(2)'''&nbsp; The system efficiency under power limitation is maximum&nbsp; $(\eta_{\rm L} = 1)$,&nbsp; when the basic transmission pulse&nbsp; $g_{s}(t)$&nbsp; has the same shape as the impulse response $h_{\rm E}(t)$.&nbsp;
 '''(3)'''&nbsp; Under the condition given in question '''(2)''', the maximum SNR is obtained:
 :$$\rho_{d, \hspace{0.05cm}{\rm max  \hspace{0.05cm}|  \hspace{0.05cm}
   L}}= \frac{2 \cdot E_{\rm B}}{N_0} = \frac{2 \cdot 3 \cdot 10^{-8}\,{\rm Ws}}{3 \cdot 10^{-10}\,{\rm
@@ Line 135: / Line 137: @@
-'''(4)'''&nbsp; In general, $g_{d}(t) = g_{s}(t) ∗ h_{\rm E}(t)$. For $t = 0$, $T_1 = T/2$ gives the trapezoidal area for this:
+'''(4)'''&nbsp; In general,&nbsp; $g_{d}(t) = g_{s}(t) ∗ h_{\rm E}(t)$.&nbsp; For&nbsp; $t = 0$,&nbsp; $T_1 = T/2$&nbsp; gives the trapezoidal area for this:
 :$$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}.$$
-'''(5)'''&nbsp; With $T_1 = T/2$ (trapezoidal transmitted pulses), we obtain for the signal-to-noise ratio:
+'''(5)'''&nbsp; With&nbsp; $T_1 = T/2$&nbsp; (trapezoidal transmitted pulses),&nbsp; we obtain for the signal-to-noise ratio:
 :$$\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}
 \Rightarrow \hspace{0.3cm}\rho_d = \frac{0.075^2\, {\rm W}}{5 \cdot 10^{-5}\,{\rm W}} = 112.5 \hspace{0.05cm}.$$
-*Thus, the system efficiency under power limitation with the result of '''(3)''':
+*Thus,&nbsp; the system efficiency under power limitation with the result of&nbsp; '''(3)''':
 :$$\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}.$$
 *Due to the mismatch, $\eta_{\rm L} < 1$.
-'''(6)'''&nbsp; With the maximum value $s_{0}$ and the result of question '''(1)''':
 :$$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}.$$
-'''(7)'''&nbsp; The system efficiency under peak limitation is smaller than that under power limitation, because the non-optimal transmitted signal (too small energy) plays a role here in addition to the mismatch:
+'''(7)'''&nbsp; The system efficiency under peak limitation is smaller than that under power limitation,&nbsp; <br>because the non-optimal transmitted signal&nbsp; (too small energy)&nbsp; plays a role here in addition to the mismatch:
 :$$\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}.$$

@@ Line 4: / Line 4: @@
-[[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&nbsp; "Trapezoid" ]]
-The receiver of a binary message transmission system with symbol duration &nbsp;$T$&nbsp; consists of an integrator, which is represented by the impulse response
+The receiver of a binary transmission system with symbol duration &nbsp;$T$&nbsp; consists of an integrator,&nbsp; which is represented by the impulse response
 :$$h_{\rm E}(t)  =   \left\{ \begin{array}{c} 1/T   \\
 \\  \end{array} \right.\quad
@@ Line 11: / Line 11: @@
 \\  {\rm{for}} \\ \end{array}\begin{array}{*{20}c}
 \hspace{0.05cm}|t| < T/2 \hspace{0.05cm}, \\
-  |t| > T/2 \\
+  |t| > T/2\hspace{0.05cm}. \\
 \end{array}$$
 This is followed by a threshold decision with optimal parameters.
 The basic transmission pulse &nbsp;$g_{s}(t)$&nbsp; according to the diagram is generally trapezoidal and is parameterized by the time &nbsp;$T_{1}$:&nbsp;
-*For &nbsp;$T_{1} = 0$&nbsp; the result is a triangular pulse, for &nbsp;$T_{1} = T$&nbsp; the NRZ rectangle.
+*For &nbsp;$T_{1} = 0$&nbsp; the result is a triangular pulse,&nbsp; for &nbsp;$T_{1} = T$&nbsp; the NRZ rectangle.
-*The absolute pulse duration &nbsp;$T_{\rm S}$&nbsp; is always equal to the symbol duration &nbsp;$T$, i.e. the distance between two transmission pulses.
+*The absolute pulse duration &nbsp;$T_{\rm S}$&nbsp; is always equal to the symbol duration &nbsp;$T$,&nbsp; 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&nbsp; $\rm (SNR)$&nbsp; before the threshold decision can be calculated as follows,&nbsp; assuming no intersymbol interfering:
 :$$\rho_d = {g_0^2}/{\sigma_d^2}\hspace{0.05cm}.$$
-Here, &nbsp;$g_{0} = g_{d}(t = 0)$&nbsp; is the maximum value of the basic transmitter pulse and
+Here, &nbsp;$g_{0} = g_{d}(t = 0)$&nbsp; is the maximum value of the basic transmission pulse,&nbsp; and
 :$$\sigma_d^2 = {N_0}/{2} \cdot \int_{-\infty}^{+\infty}|h_{\rm E}(t)|^2 \,{\rm d} t = \frac{N_0}{2 \cdot T}$$
 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,&nbsp; the following quantities will be used:
-*$\rho_{d,\rm\hspace{0.05cm} max | L}$&nbsp; is the maximum SNR under the constraint of power limitation.
+*$\rho_{d,\rm\hspace{0.08cm} max \hspace{0.03cm}|\hspace{0.03cm} L}$&nbsp; is the maximum SNR under the constraint of&nbsp; of&nbsp; "power limitation".
-*$\rho_{d,\rm\hspace{0.05cm} max | A}$&nbsp; is the maximum SNR under peak limitation (amplitude limitation).
+*$\rho_{d,\rm\hspace{0.08cm} max \hspace{0.03cm}|\hspace{0.03cm} A}$&nbsp; is the maximum SNR under the constraint of&nbsp; "peak limitation"&nbsp; (or&nbsp; "amplitude limitation").
 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},$$
-:$$\eta_{\rm A} = \ \frac{\rho_d}{\rho_{d, \hspace{0.05cm}{\rm max\hspace{0.05cm} |
+:$$\eta_{\rm A} = \ \frac{\rho_d}{\rho_{d, \hspace{0.08cm}{\rm max\hspace{0.05cm} |
 \hspace{0.05cm} A}}} = {1}/{C_{\rm S}^2}\cdot \eta_{\rm L} \hspace{0.05cm}.$$
-Here, the ''crest factor'' &nbsp;$C_{\rm S}$&nbsp; denotes the ratio between the maximum value and the rms value (root of power) of the transmitted signal &nbsp;$s(t)$.
+Here,&nbsp; the&nbsp; "crest factor" &nbsp; $C_{\rm S}$&nbsp; denotes the ratio between the maximum value and the rms value&nbsp; (root of power)&nbsp; of the transmitted signal &nbsp;$s(t)$.
@@ Line 43: / Line 43: @@
 Notes:
-''Notes:''
+*The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Optimization_of_Baseband_Transmission_Systems|"Optimization of Baseband Transmission Systems"]].
 *Use the following numerical values to solve the exercise:
@@ Line 56: / Line 55: @@
 <quiz display=simple>
-{Calculate the pulse energy &nbsp;$E_{\rm B}$&nbsp; as a function of &nbsp;$T_{1}$. What are the values for &nbsp;$T_{1} = 0$ , &nbsp;$T_{1} = T/2$&nbsp; and &nbsp;$T_{1} = T$?
+{Calculate the pulse energy &nbsp;$E_{\rm B}$&nbsp; as a function of &nbsp;$T_{1}$.&nbsp; What are the values for &nbsp;$T_{1} = 0$,&nbsp; &nbsp;$T_{1} = T/2$&nbsp; and &nbsp;$T_{1} = T$?
 |type="{}"}
 $T_{1} = 0\text{:} \hspace{0.75cm}  E_{\rm B} \ = \ $ { 1 3% } $\ \cdot 10^{-8} \, \rm Ws$
@@ Line 62: / Line 61: @@
 $T_{1} = T\text{:}\hspace{0.65cm} E_{\rm B} \ = \ $ { 3 3% } $\ \cdot 10^{-8} \, \rm Ws$
-{What value &nbsp;$T_{1}$&nbsp; leads to the maximum possible SNR when power is limited?
+{What value &nbsp;$T_{1}$&nbsp; leads to the maximum possible SNR when the power is limited?
 |type="{}"}
 $T_{1}/T \ = \ $ { 1 3% }
-{Therefore, what is the maximum SNR with power limitation?
+{Therefore,&nbsp; what is the maximum SNR with power limitation?
 |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 &nbsp;$g_{d}(t)$&nbsp; in pulse center for &nbsp;$T_{1} = T/2$?
@@ Line 74: / Line 73: @@
 $g_{0} \ = \ $ { 0.075 3% } $\ \rm \sqrt{W}$
-{Calculate the system efficiency &nbsp;$\eta_{\rm L}$&nbsp; when power is limited  &nbsp;$(T_{1} = T/2)$.
+{Calculate the system efficiency &nbsp;$\eta_{\rm L}$&nbsp; when the power is limited  &nbsp; $(T_{1} = T/2)$.
 |type="{}"}
 $\eta_{\rm L} \ = \ $ { 0.5625 3% }
-{Calculate the crest factor &nbsp;$(T_{1} = T/2)$.
+{Calculate the crest factor &nbsp; $(T_{1} = T/2)$.
 |type="{}"}
 $C_{\rm S} \ = \ $ { 1.225 3% }
-{Calculate the system efficiency at peak limitation &nbsp;$(T_{1} = T/2)$.
+{Calculate the system efficiency at peak limitation &nbsp; $(T_{1} = T/2)$.
 |type="{}"}
 $\eta_{\rm A} \ = \ $ { 0.375 3% }

← Older revision		Revision as of 13:58, 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^]]

@@ Line 4: / Line 4: @@
-[[File:P_ID1294__Dig_A_1_7.png|right|frame|Sendegrundimpuls "Trapez&rdquo; ]]
+[[File:P_ID1294__Dig_A_1_7.png|right|frame|Sendegrundimpuls "Trapez" ]]
 Der Empfänger eines binären Nachrichtenübertragungssystems mit Symboldauer &nbsp;$T$&nbsp; besteht aus einem Integrator, der durch die Impulsantwort
 :$$h_{\rm E}(t)  =   \left\{ \begin{array}{c} 1/T   \\

← Older revision	Revision as of 13:56, 19 March 2022
(No difference)

← Older revision		Revision as of 12:47, 23 March 2021
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−	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^1.4 Optimierung der Basisbandsysteme^]]	+	[[Category:Digital Signal Transmission: Exercises\|^1.4 Optimierung der Basisbandsysteme^]]

@@ Line 45: / Line 45: @@
 ''Hinweise:''
-*Die Aufgabe gehört zum  Kapitel&nbsp;  [[Digitalsignalübertragung/Optimierung_der_Basisbandübertragungssysteme|Optimierung der Basisbandübertragungssysteme]].
+*Die Aufgabe gehört zum  Kapitel&nbsp;  [[Digital_Signal_Transmission/Optimierung_der_Basisbandübertragungssysteme|Optimierung der Basisbandübertragungssysteme]].
 *Verwenden Sie zur Lösung der Aufgabe folgende Zahlenwerte: