Aufgaben:Exercise 2.1: ACF and PSD with Coding - Revision history

Hwang at 14:23, 23 May 2022

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Hwang at 14:22, 23 May 2022

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Guenter at 15:40, 13 May 2022

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Hwang at 15:03, 7 April 2022

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Guenter at 12:48, 4 April 2022

2022-04-04T12:48:51Z

Guenter: Guenter moved page Aufgaben:Aufgabe 2.1: AKF und LDS nach Codierung to Aufgaben:Exercise 2.1: ACF and PSD with Coding

2022-04-04T12:40:38Z

Guenter moved page Aufgabe 2.1: AKF und LDS nach Codierung to Exercise 2.1: ACF and PSD with Coding

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

2021-03-23T12:53:23Z

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

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

2020-07-09T12:52:14Z

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

Guenter at 10:37, 11 February 2019

2019-02-11T10:37:58Z

@@ Line 69: / Line 69: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Since&nbsp; ${\it \Phi}_{a}(f)$&nbsp; as a power-spectral density is always rea&nbsp;l '''KORREKTUR: real'''(plus even and positive,&nbsp; but that does not matter here)&nbsp; and the ACF values&nbsp; $\varphi_{a}(\lambda)$&nbsp; are symmetric about&nbsp; $\lambda = 0$,&nbsp; the given equation can be transformed as follows:
+'''(1)'''&nbsp; Since&nbsp; ${\it \Phi}_{a}(f)$&nbsp; as a power-spectral density is always real &nbsp;(plus even and positive,&nbsp; but that does not matter here)&nbsp; and the ACF values&nbsp; $\varphi_{a}(\lambda)$&nbsp; are symmetric about&nbsp; $\lambda = 0$,&nbsp; the given equation can be transformed as follows:
 :$${\it \Phi}_a(f) = \sum_{\lambda = -\infty}^{+\infty}\varphi_a(\lambda)\cdot {\rm e}^{- {\rm j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \lambda T} = \varphi_a(0) + \sum_{\lambda = 1}^{\infty}2 \cdot \varphi_a(\lambda)\cdot\cos ( 2 \pi f \hspace{0.02cm} \lambda T) \hspace{0.05cm}.$$
 *By comparison with the sketched function

@@ Line 69: / Line 69: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Since&nbsp; ${\it \Phi}_{a}(f)$&nbsp; as a power-spectral density is always rea&nbsp;l (plus even and positive,&nbsp; but that does not matter here)&nbsp; and the ACF values&nbsp; $\varphi_{a}(\lambda)$&nbsp; are symmetric about&nbsp; $\lambda = 0$,&nbsp; the given equation can be transformed as follows:
+'''(1)'''&nbsp; Since&nbsp; ${\it \Phi}_{a}(f)$&nbsp; as a power-spectral density is always rea&nbsp;l '''KORREKTUR: real'''(plus even and positive,&nbsp; but that does not matter here)&nbsp; and the ACF values&nbsp; $\varphi_{a}(\lambda)$&nbsp; are symmetric about&nbsp; $\lambda = 0$,&nbsp; the given equation can be transformed as follows:
 :$${\it \Phi}_a(f) = \sum_{\lambda = -\infty}^{+\infty}\varphi_a(\lambda)\cdot {\rm e}^{- {\rm j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \lambda T} = \varphi_a(0) + \sum_{\lambda = 1}^{\infty}2 \cdot \varphi_a(\lambda)\cdot\cos ( 2 \pi f \hspace{0.02cm} \lambda T) \hspace{0.05cm}.$$
 *By comparison with the sketched function

@@ Line 69: / Line 69: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Since ${\it \Phi}_{a}(f)$ as a power-spectral density is always real (plus even and positive, but that does not matter here) and the ACF values $\varphi_{a}(\lambda)$ are symmetric about $\lambda = 0$, the given equation can be transformed as follows:
+'''(1)'''&nbsp; Since&nbsp; ${\it \Phi}_{a}(f)$&nbsp; as a power-spectral density is always rea&nbsp;l (plus even and positive,&nbsp; but that does not matter here)&nbsp; and the ACF values&nbsp; $\varphi_{a}(\lambda)$&nbsp; are symmetric about&nbsp; $\lambda = 0$,&nbsp; the given equation can be transformed as follows:
 :$${\it \Phi}_a(f) = \sum_{\lambda = -\infty}^{+\infty}\varphi_a(\lambda)\cdot {\rm e}^{- {\rm j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \lambda T} = \varphi_a(0) + \sum_{\lambda = 1}^{\infty}2 \cdot \varphi_a(\lambda)\cdot\cos ( 2 \pi f \hspace{0.02cm} \lambda T) \hspace{0.05cm}.$$
 *By comparison with the sketched function
@@ Line 75: / Line 75: @@
 :one obtains:
 :$${\it \varphi}_a(\lambda = 0)\hspace{0.15cm}\underline { = 0.5}, \hspace{0.2cm} {\it \varphi}_a(\lambda = 2) = {\it \varphi}_a(\lambda = -2) \hspace{0.15cm}\underline {= -0.25} \hspace{0.05cm}.$$
-*All other ACF values result to zero, so also $\varphi_{a}(\lambda = ±1)\hspace{0.15cm}\underline {=0}$.
+*All other ACF values result to zero,&nbsp; so also&nbsp; $\varphi_{a}(\lambda = ±1)\hspace{0.15cm}\underline {=0}$.
-'''(2)'''&nbsp; For the rectangular NRZ basic pulse, due to the limitation of the energy ACF to the range $|\tau| ≤ T$, we obtain:
+'''(2)'''&nbsp; For the rectangular NRZ basic pulse,&nbsp; due to the limitation of the energy ACF to the range $|\tau| ≤ T$,&nbsp; we obtain:
 :$$P_{\rm S} = \varphi_s(\tau = 0) = \frac{1}{T} \cdot \varphi_a(\lambda = 0)\cdot \varphi^{^{\bullet}}_{gs}(\tau = 0)= \frac{1}{T} \cdot \frac{1}{2} \cdot s_0^2 \cdot T = \frac{s_0^2}{2} \hspace{0.15cm}\underline {= 5\,\,{\rm mW}}\hspace{0.05cm}.$$
-'''(3)'''&nbsp; For rectangular spectral function, it is more convenient to calculate the transmit power by integration over the power-spectral density:
+'''(3)'''&nbsp; For rectangular spectral function,&nbsp; it is more convenient to calculate the transmit power by integration over the power-spectral density:
 :$$P_{\rm S}  = \ \int_{-1/(2T)}^{+1/(2T)} {\it \Phi}_s(f) \,{\rm d} f = \frac{1}{T} \cdot \int_{-1/(2T)}^{+1/(2T)} {\it \Phi}_a(f) \cdot {\it \Phi}^{^{\bullet}}_{gs}(f) \,{\rm d} f$$
 :$$\Rightarrow\hspace{0.3cm}P_{\rm S} = \ \frac{1}{T} \cdot \left [ s_0^2 \cdot T^2 \right ] \cdot \int_{-1/(2T)}^{+1/(2T)} \left( {1}/{2} - {1}/{2} \cdot \cos (4 \pi f \hspace{0.02cm} T)\right ) \,{\rm d} f\hspace{0.05cm} = {s_0^2}/{2}\hspace{0.15cm}\underline { = 5\,\,{\rm mW}} .$$
-*Here it is considered that for this exercise the energy PSD $|G_{s}(f)|^{2}$ is given as constant (within the integration interval) and thus can be drawn in front of the integral.
+*Here it is considered that the energy PSD&nbsp; $|G_{s}(f)|^{2}$&nbsp; is given as constant&nbsp; (within the integration interval)&nbsp; and thus can be drawn in front of the integral.
-*In spite of a completely different signal form $s(t)$, the same transmit power results here, since the integral yields the value $1/(2T)$.
+*In spite of a completely different signal form&nbsp; $s(t)$,&nbsp; the same transmit power results here,&nbsp; since the integral yields the value $1/(2T)$.
 *It should be noted that this simple calculation is only possible for the rolloff factor $r = 0$.

@@ Line 7: / Line 7: @@
 [[File:P_ID1308__Dig_A_2_1.png|right|frame|Power-spectral density with coding]]
-We consider the digital signal &nbsp;$s(t)$, using the following descriptive quantities:
+We consider the digital signal &nbsp;$s(t)$,&nbsp;  using the following descriptive quantities:
 *$a_{\nu}$&nbsp; are the amplitude coefficients,
 *$g_{s}(t)$&nbsp; indicates the basic transmission pulse,
-*$T$&nbsp; is the symbol duration (spacing of the pulses).
+*$T$&nbsp; is the symbol duration&nbsp; (spacing of the pulses).
 Then holds:
 :$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g_s ( t - \nu \cdot T) \hspace{0.05cm}.$$
 To characterize the spectral properties resulting from the coding and pulse shaping, one uses, among other things
-*the auto-correlation function (ACF)
+To characterize the spectral properties resulting from the coding and pulse shaping,&nbsp; one uses,&nbsp; among other things
 :$$\varphi_s(\tau) = \sum_{\lambda = -\infty}^{+\infty}{1}/{T} \cdot \varphi_a(\lambda)\cdot \varphi^{^{\bullet}}_{gs}(\tau - \lambda \cdot T)\hspace{0.05cm},$$
-*the power-spectral density  (PSD)
+*the power-spectral density&nbsp;  $\rm (PSD)$
 :$${\it \Phi}_s(f) = {1}/{T} \cdot {\it \Phi}_a(f) \cdot {\it \Phi}^{^{\bullet}}_{gs}(f) \hspace{0.05cm}.$$
-Here, &nbsp;$\varphi_{a}(\lambda)$&nbsp; denotes the discrete auto-correlation function of the amplitude coefficients related to the power-spectral density &nbsp;${\it \Phi}_{a}(f)$&nbsp; via the Fourier transform. Thus, for this holds:
+Here, &nbsp;$\varphi_{a}(\lambda)$&nbsp; denotes the discrete ACF of the amplitude coefficients related to the power-spectral density &nbsp;${\it \Phi}_{a}(f)$&nbsp; via the Fourier transform.&nbsp; Thus,&nbsp; for this holds:
 :$${\it \Phi}_a(f) = \sum_{\lambda = -\infty}^{+\infty}\varphi_a(\lambda)\cdot {\rm e}^{- {\rm j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \lambda T} \hspace{0.05cm}.$$
-Furthermore, the energy ACF and energy spectrum are used in above equations:
+Furthermore,&nbsp; the energy ACF and energy spectrum are used in above equations:
 :$$\varphi^{^{\bullet}}_{gs}(\tau) = \int_{-\infty}^{+\infty} g_s ( t ) \cdot g_s ( t + \tau)\,{\rm d} t \hspace{0.4cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \hspace{0.4cm} {\it \Phi}^{^{\bullet}}_{gs}(f) = |G_s(f)|^2 \hspace{0.05cm}.$$
-In the present exercise, the following function is to be assumed for the power-spectral density of the amplitude coefficients (see graph):
+In the present exercise,&nbsp; the following function is to be assumed for the power-spectral density of the amplitude coefficients&nbsp; (see graph):
 :$${\it \Phi}_a(f) = {1}/{2} - {1}/{2} \cdot \cos (4 \pi f \hspace{0.02cm} T)\hspace{0.05cm}.$$
 The following assumptions are made for the basic transmission pulse:
-*In question '''(2)''', let &nbsp;$g_{s}(t)$&nbsp; be an NRZ rectangular pulse, so that there is a triangular energy ACF confined to the range &nbsp;$|\tau| ≤ T$.&nbsp; The maximum here is
+*In question&nbsp; '''(2)''',&nbsp; let &nbsp;$g_{s}(t)$&nbsp; be an NRZ rectangular pulse,&nbsp; so that there is a triangular energy ACF confined to the range &nbsp;$|\tau| ≤ T$.&nbsp; The maximum value here is
 :$$\varphi^{^{\bullet}}_{gs}(\tau = 0) = s_0^2 \cdot T \hspace{0.05cm}.$$
-*For question '''(3)''', assume a root-Nyquist characteristic with rolloff factor &nbsp;$r = 0$.&nbsp; In this case holds:
+*For question&nbsp; '''(3)''',&nbsp; assume a root-Nyquist characteristic with rolloff factor &nbsp;$r = 0$.&nbsp; In this case holds:
 :$$|G_s(f)|^2 = \left\{ \begin{array}{c} s_0^2 \cdot T^2 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{1}c} {\rm{for}} \\ {\rm{for}} \\ \end{array} \begin{array}{*{20}c} |f| < {1}/({2T}) \hspace{0.05cm}, \\ |f| > {1}/({2T}) \hspace{0.05cm}.\\ \end{array}$$
-*For numerical calculations, always use &nbsp;$s_{0}^{2} = 10 \ \rm mW$.&nbsp;
+*For numerical calculations,&nbsp; use always &nbsp;$s_{0}^{2} = 10 \ \rm mW$.&nbsp;
-''Notes:''
+Notes:
-*The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Basics_of_Coded_Transmission|Basics of Coded Transmission]].
+*The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Basics_of_Coded_Transmission|"Basics of Coded Transmission"]].
-*Consider that the transmit power &nbsp;$P_{\rm S}$&nbsp; is equal to the ACF &nbsp;$\varphi_{s}(\tau)$&nbsp; at the point &nbsp;$\tau = 0$,&nbsp; but can also be calculated as an integral over the PSD &nbsp;$\Phi_{s}(f)$.&nbsp;
+*Consider that the transmit power &nbsp;$P_{\rm S}$&nbsp; is equal to the ACF &nbsp;$\varphi_{s}(\tau)$&nbsp; at the point &nbsp;$\tau = 0$,&nbsp; but can also be calculated as an integral over the PSD &nbsp;${\it \Phi}_{s}(f)$.&nbsp;
 ===Questions===

← Older revision		Revision as of 12:48, 4 April 2022
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−	[[Category:Digital Signal Transmission: Exercises\|^2.1 ~~Codierte Übertragung - Grundlagen~~^]]	+	[[Category:Digital Signal Transmission: Exercises\|^2.1 Basics of Coded Transmission^]]

← Older revision		Revision as of 12:53, 23 March 2021
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−	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^2.1 Codierte Übertragung - Grundlagen^]]	+	[[Category:Digital Signal Transmission: Exercises\|^2.1 Codierte Übertragung - Grundlagen^]]

@@ Line 42: / Line 42: @@
 ''Hinweise:''
-*Die Aufgabe gehört zum  Kapitel&nbsp;  [[Digitalsignalübertragung/Grundlagen_der_codierten_Übertragung|Grundlagen der codierten Übertragung]].
+*Die Aufgabe gehört zum  Kapitel&nbsp;  [[Digital_Signal_Transmission/Grundlagen_der_codierten_Übertragung|Grundlagen der codierten Übertragung]].
 *Berücksichtigen Sie, dass die Sendeleistung &nbsp;$P_{\rm S}$&nbsp; gleich der AKF &nbsp;$\varphi_{s}(\tau)$&nbsp; an der Stelle &nbsp;$\tau = 0$&nbsp; ist, aber auch als Integral über das LDS &nbsp;$\Phi_{s}(f)$&nbsp; berechnet werden kann.

@@ Line 72: / Line 72: @@
 '''(1)'''&nbsp; Da ${\it \Phi}_{a}(f)$ als eine spektrale Leistungsdichte stets reell ist (dazu gerade und positiv, aber das spielt hier keine Rolle) und die AKF–Werte $\varphi_{a}(\lambda)$ symmetrisch um $\lambda = 0$ sind, kann die angegebene Gleichung wie folgt umgewandelt werden:
 :$${\it \Phi}_a(f) = \sum_{\lambda = -\infty}^{+\infty}\varphi_a(\lambda)\cdot {\rm e}^{- {\rm j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \lambda T} = \varphi_a(0) + \sum_{\lambda = 1}^{\infty}2 \cdot \varphi_a(\lambda)\cdot\cos ( 2 \pi f \hspace{0.02cm} \lambda T) \hspace{0.05cm}.$$
-Durch Vergleich mit der skizzierten Funktion
+*Durch Vergleich mit der skizzierten Funktion
 :$${\it \Phi}_a(f) = {1}/{2} - {1}/{2} \cdot \cos (4 \pi f \hspace{0.02cm} T)\hspace{0.05cm}.$$
-erhält man:
+:erhält man:
 :$${\it \varphi}_a(\lambda = 0)\hspace{0.15cm}\underline { = 0.5}, \hspace{0.2cm} {\it \varphi}_a(\lambda = 2) = {\it \varphi}_a(\lambda = -2) \hspace{0.15cm}\underline {= -0.25} \hspace{0.05cm}.$$
-Alle anderen AKF–Werte ergeben sich zu Null, also auch $\varphi_{a}(\lambda = ±1)\hspace{0.15cm}\underline {=0}$.
+*Alle anderen AKF–Werte ergeben sich zu Null, also auch $\varphi_{a}(\lambda = ±1)\hspace{0.15cm}\underline {=0}$.
 '''(2)'''&nbsp; Für den rechteckförmigen NRZ–Grundimpuls ergibt sich aufgrund der Begrenzung der Energie–AKF auf den Bereich $|\tau| ≤ T$:
@@ Line 82: / Line 84: @@
-'''(3)'''&nbsp; Im hier zu betrachtenden Fall (rechteckförmige Spektralfunktion) ist es günstiger, die Sendeleistung durch Integration über das Leistungsdichtespektrum zu berechnen:
+'''(3)'''&nbsp; Bei rechteckförmiger Spektralfunktion ist es günstiger, die Sendeleistung durch Integration über das Leistungsdichtespektrum zu berechnen:
 :$$P_{\rm S}  = \ \int_{-1/(2T)}^{+1/(2T)} {\it \Phi}_s(f) \,{\rm d} f = \frac{1}{T} \cdot \int_{-1/(2T)}^{+1/(2T)} {\it \Phi}_a(f) \cdot {\it \Phi}^{^{\bullet}}_{gs}(f) \,{\rm d} f$$
 :$$\Rightarrow\hspace{0.3cm}P_{\rm S} = \ \frac{1}{T} \cdot \left [ s_0^2 \cdot T^2 \right ] \cdot \int_{-1/(2T)}^{+1/(2T)} \left( {1}/{2} - {1}/{2} \cdot \cos (4 \pi f \hspace{0.02cm} T)\right ) \,{\rm d} f\hspace{0.05cm} = {s_0^2}/{2}\hspace{0.15cm}\underline { = 5\,\,{\rm mW}} .$$
-Hierbei ist berücksichtigt, dass für diese Aufgabe das Energie–LDS $|G_{s}(f)|^{2}$ als konstant vorgegeben ist (innerhalb des Integrationsintervalls) und somit vor das Integral gezogen werden kann.
+*Hierbei ist berücksichtigt, dass für diese Aufgabe das Energie–LDS $|G_{s}(f)|^{2}$ als konstant vorgegeben ist (innerhalb des Integrationsintervalls) und somit vor das Integral gezogen werden kann.
-Trotz völlig anderer Signalform $s(t)$ ergibt sich hier die gleiche Sendeleistung, da das Integral den Wert $1/(2T)$ liefert. Anzumerken ist, dass diese einfache Rechnung nur für den Rolloff-Faktor $r = 0$ möglich ist.
+*Trotz völlig anderer Signalform $s(t)$ ergibt sich hier die gleiche Sendeleistung, da das Integral den Wert $1/(2T)$ liefert.