Aufgaben:Exercise 2.2: Binary Bipolar Rectangles - Revision history

Guenter at 17:02, 13 May 2022

2022-05-13T17:02:01Z

Guenter at 16:35, 13 May 2022

2022-05-13T16:35:05Z

Guenter at 16:34, 13 May 2022

2022-05-13T16:34:43Z

Guenter at 16:34, 13 May 2022

2022-05-13T16:34:27Z

Hwang at 16:48, 7 April 2022

2022-04-07T16:48:21Z

Show changes

Guenter at 12:49, 4 April 2022

2022-04-04T12:49:46Z

Guenter: Guenter moved page Aufgaben:Aufgabe 2.2: Binäre bipolare Rechtecke to Aufgaben:Exercise 2.2: Binary Bipolar Rectangles

2022-04-04T12:46:59Z

Guenter moved page Aufgabe 2.2: Binäre bipolare Rechtecke to Exercise 2.2: Binary Bipolar Rectangles

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

2021-03-23T12:53:44Z

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

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

2020-07-09T12:44:43Z

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

Guenter at 11:07, 11 February 2019

2019-02-11T11:07:07Z

@@ Line 87: / Line 87: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; A digital signal is said to be redundancy-free if
-*the amplitude coefficients do not depend on each other (this was assumed here),
+*the amplitude coefficients do not depend on each other&nbsp; (this was assumed here),
 *all possible amplitude coefficients are equally probable.
-In this sense, $s_{0.5}(t)$ is a redundancy-free signal &nbsp; &rArr; &nbsp;  <u>solution 2</u>.
+In this sense,&nbsp; $s_{0.5}(t)$&nbsp; is a redundancy-free signal &nbsp; &rArr; &nbsp;  <u>solution 2</u>.
-*Thus, here the entropy (the average information content per transmitted binary symbol) is at most equal to the decision content:
+*Thus,&nbsp; here the entropy&nbsp; (the average information content per transmitted binary symbol)&nbsp; is at most equal to the decision content:
-:$$H_{\rm max} = {1}/{2}\cdot {\rm log}_2 (2)+{1}/{2}\cdot {\rm log}_2 (2) = 1 \,\,{\rm bit/binary symbol} \hspace{0.05cm}.$$
+:$$H_{\rm max} = {1}/{2}\cdot {\rm log}_2 (2)+{1}/{2}\cdot {\rm log}_2 (2) = 1 \,\,{\rm bit/binary\ symbol} \hspace{0.05cm}.$$
-*In contrast, the entropies of the other two binary signals are:
+*In contrast,&nbsp; the entropies of the other two binary signals are:
 :$$H  = \  \frac{3}{4}\cdot {\rm log}_2 (\frac{4}{3})+ \frac{1}{4}\cdot {\rm log}_2 (4)
   = \left( \frac{3}{4} + \frac{1}{4}\right)\cdot {\rm log}_2 (4) -
   \frac{3}{4}\cdot{\rm log}_2 (3) =$$
 :$$ \hspace{0.5cm} = \ 2 - \frac{3}{4}\cdot{\rm log}_2 (3) =
-.811 \,\,{\rm bit/binary symbol} \hspace{0.05cm}.$$
+.811 \,\,{\rm bit/binary\ symbol} \hspace{0.05cm}.$$
-*From this, the relative redundancy of these signals is:
+*From this,&nbsp; the relative redundancy of these signals is:
 :$$r = \frac{H_{\rm max} - H}{H_{\rm max}}\hspace{0.15cm} \approx 18.9\%\hspace{0.05cm}.$$
-'''(2)'''&nbsp; The quadratic mean is equal to $m_{2a} = 1$ independent of $p$:
+'''(2)'''&nbsp; The second order moment&nbsp; ("power")&nbsp; is equal to&nbsp; $m_{2a} = 1$&nbsp; independent of&nbsp; $p$:
 :$$m_{2a}={\rm E}[a_\nu^2] = p \cdot (+1)^2 + (1-p)\cdot (-1)^2 \hspace{0.15cm}\underline { = 1 \hspace{0.05cm}}.$$
-'''(3)'''&nbsp; For the linear mean value we get
+'''(3)'''&nbsp; For the first order moment&nbsp; ("linear mean")&nbsp; we get:
 :$$m_{a}={\rm E}[a_\nu] = p \cdot (+1) + (1-p)\cdot (-1) = 2 p -1 \hspace{0.05cm}.$$
 :$$\Rightarrow \hspace{0.3cm} p = 0.75\text{:} \hspace{0.4cm} m_{a}\hspace{0.15cm}\underline {=0.50},\hspace{0.2cm} p = 0.50\text{:} \hspace{0.4cm} m_{a}\hspace{0.15cm}\underline {=0},\hspace{0.2cm} p = 0.25\text{:} \hspace{0.4cm}  m_{a}\hspace{0.15cm}\underline { =-0.50 \hspace{0.05cm}}.$$
-'''(4)'''&nbsp; Using the results from '''(2)''' and '''(4)''', we obtain:
+'''(4)'''&nbsp; Using the results from&nbsp; ( '''(2)'''&nbsp; and&nbsp; '''(4)''',&nbsp; we obtain:
 :$$p = 0.75\text{:} \hspace{0.4cm} \sigma_{a}^2 \hspace{0.15cm}\underline {=0.75},$$
 :$$ p = 0.50\text{:} \hspace{0.4cm}  \sigma_{a}^2\hspace{0.15cm} \underline { =1.00 \hspace{0.05cm}},$$
@@ Line 120: / Line 120: @@
 [[File:P_ID1311__Dig_A_2_2e.png|right|frame|ACF with equal symbol probabilities]]
-'''(5)'''&nbsp; Only the <u>first two statements</u> are correct:
+'''(5)'''&nbsp; Only the&nbsp; <u>first two statements</u>&nbsp; are correct:
-*For $p = 0.5$, $\varphi_{a}(\lambda = 0) = 1$ and $\varphi_{a}(\lambda \neq 0) = 0$. It follows that:
+*For&nbsp; $p = 0.5$, &nbsp;  $\varphi_{a}(\lambda = 0) = 1$&nbsp; and&nbsp; $\varphi_{a}(\lambda \neq 0) = 0$.&nbsp; It follows that:
 :$$\varphi_s(\tau) = \frac{1}{T} \cdot \varphi^{^{\bullet}}_{gs}(\tau )\hspace{0.05cm}.$$
-*This results in a triangular ACF and a ${\rm si}^{2}$–shaped PSD for both the NRZ and RZ basic pulses.
+*This results in a triangular ACF and a&nbsp; ${\rm sinc}^{2}$–shaped PSD for both the NRZ and RZ basic pulses.
-*The area under the PSD is smaller by a factor of $T_{\rm S}/T$ for the RZ pulse than for the NRZ pulse, since the ACF values also differ by this factor at $\tau = 0$.
+*The area under the PSD is smaller by a factor of&nbsp; $T_{\rm S}/T$&nbsp; for the RZ pulse than for the NRZ pulse,&nbsp; <br>since the ACF values also differ by this factor at&nbsp; $\tau = 0$.
 *The PSD is continuous in both cases because the ACF does not contain a DC component or periodic components.
 [[File:P_ID1330__Dig_A_2_2f.png|right|frame|ACF with unequal symbol probabilities]]
-'''(6)'''&nbsp; <u>All statements except the third</u> are correct:
+'''(6)'''&nbsp; <u>All statements except the third</u>&nbsp; are correct:
-*For $p = 0.75$, the ACF $\varphi_{s}(\tau)$ is composed of infinitely many triangular functions, all of which have the same height $s_{0}^{2}/4$ except for the middle triangle around $\tau = 0$.
+*For&nbsp; $p = 0.75$, &nbsp; the ACF $\varphi_{s}(\tau)$ is composed of infinitely many triangular functions,&nbsp; all of which have the same height&nbsp; $s_{0}^{2}/4$&nbsp; except for the middle triangle around&nbsp; $\tau = 0$.
-*According to the sketch, one can combine all these triangle functions into a DC component of height $m_{a}^{2} \cdot  s_{0}^{2} = s_{0}^{2}/4$ and a single triangle around $\tau = 0$ with height $\sigma_{a}^{2} \cdot  s_{0}^{2} = 3/4 · s_{0}^{2}$.
+*According to the sketch,&nbsp; one can combine all these triangle functions into a DC component of height&nbsp; $m_{a}^{2} \cdot  s_{0}^{2} = s_{0}^{2}/4$&nbsp; and a single triangle around&nbsp; $\tau = 0$&nbsp; with height&nbsp; $\sigma_{a}^{2} \cdot  s_{0}^{2} = 3/4 · s_{0}^{2}$.
-*In the PSD, this leads to a continuous ${\rm si}^{2}$–shaped component and a Dirac delta function at $f = 0$. The weight of this Dirac is $s_{0}^{2}/4$.
+*In the PSD,&nbsp; this leads to a continuous&nbsp; ${\rm sinc}^{2}$–shaped component and a Dirac delta function at&nbsp; $f = 0$.&nbsp; The weight of this Dirac is&nbsp; $s_{0}^{2}/4$.
-*For $p = 0.25$ we get the same ACF as with $p = 0.75$, since both the quadratic mean $m_{2a} = 1$ and $m_{a}^{2} = 0.25$ coincide. Thus, of course, the power-spectral densities also match.
+*For&nbsp; $p = 0.25$&nbsp; we get the same ACF as with&nbsp; $p = 0.75$,&nbsp; since both the second order moment&nbsp; $m_{2a} = 1$&nbsp; and&nbsp; $m_{a}^{2} = 0.25$&nbsp; coincide.&nbsp; Thus,&nbsp; of course,&nbsp; the power-spectral densities also match.
 [[File:P_ID1331__Dig_A_2_2g.png|right|frame|ACF for RZ rectangular pulses]]
 '''(7)'''&nbsp; <u>Both proposed solutions are correct:</u>:
-*With the RZ duty cycle $T_{\rm S}/T = 0.5$ the sketched ACF is obtained, which can also be represented by a periodic triangular function of height $s_{0}^{2}/8$ (with red filling) and a single triangular pulse of height $3/8 \cdot s_{0}^{2}$ (green filling).
+*With the RZ duty cycle&nbsp; $T_{\rm S}/T = 0.5$&nbsp; the sketched ACF is obtained,&nbsp; which can also be represented by a periodic triangular function of height&nbsp; $s_{0}^{2}/8$&nbsp; (red filling)&nbsp; and a single triangular pulse of height&nbsp; $3/8 \cdot s_{0}^{2}$&nbsp; (green filling).
-*This non-periodic component leads to a continuous, ${\rm si}^{2}$–shaped PSD with zeros at multiples of $2/T$.
+*This non-periodic component leads to a continuous-valued,&nbsp; ${\rm sinc}^{2}$–shaped PSD with zeros at multiples of&nbsp; $2/T$.
-*The periodic triangular ACF causes Dirac delta functions in the PSD at multiples of $1/T$.
+*The periodic triangular ACF causes Dirac delta functions in the PSD at multiples of&nbsp; $1/T$.
-*However, due to the antimetry of the periodic component, the Dirac delta functions at multiples of $2/T$ each have weight $0$.
+*However,&nbsp; due to the antimetry of the periodic component,&nbsp; the Dirac delta functions at multiples of&nbsp; $2/T$&nbsp; each have weight&nbsp; $0$.
-*The weights of the Dirac delta functions at distance $1/T$ are proportional to the continuous PSD component.
+*The weights of the Dirac delta functions at distance&nbsp; $1/T$&nbsp; are proportional to the continuous PSD component.

← Older revision		Revision as of 16:35, 13 May 2022
Line 26:		Line 26:
	:$$\varphi_s(\tau) = \sum_{\lambda = -\infty}^{+\infty}{1}/{T} \cdot \varphi_a(\lambda)\cdot \varphi^{^{\bullet}}_{g_s}(\tau - \lambda \cdot T)\hspace{0.05cm}.$$		:$$\varphi_s(\tau) = \sum_{\lambda = -\infty}^{+\infty}{1}/{T} \cdot \varphi_a(\lambda)\cdot \varphi^{^{\bullet}}_{g_s}(\tau - \lambda \cdot T)\hspace{0.05cm}.$$
	*The power-spectral density  ${\it \Phi}_{s}(f)$  is the Fourier transform of the ACF  $\varphi_{s}(\tau)$.		*The power-spectral density  ${\it \Phi}_{s}(f)$  is the Fourier transform of the ACF  $\varphi_{s}(\tau)$.
−
−
−

@@ Line 14: / Line 14: @@
 :$${\rm Pr}(a_\nu = -1)  \ = \ 1 - p \hspace{0.05cm}.$$
 The three signal sections shown in the graph are valid for &nbsp;$p = 0.75$, &nbsp;$p = 0.50$&nbsp; and &nbsp;$p = 0.25$.
 Throughout this exercise,&nbsp; reference is made to the following descriptive quantities:

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Digital_Signal_Transmission/Basics_of_Coded_Transmission
 }}
-[[File:P_ID1310__Dig_A_2_2.png|right|frame|Examples of binary bipolar rectangular signals]]
+[[File:P_ID1310__Dig_A_2_2.png|right|frame|Binary bipolar rectangular signals]]
 We assume the following signal:
 :$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g_s ( t - \nu \cdot T) \hspace{0.05cm}.$$
-The basic transmission pulse &nbsp;$g_{s}(t)$&nbsp; is always assumed to be rectangular in this exercise, with the NRZ format (blue signal curves in the graph) as well as the RZ format with duty cycle &nbsp;$T_{\rm S}/T = 0.5$&nbsp; (red signal curves) to be investigated.
+The basic transmission pulse &nbsp;$g_{s}(t)$&nbsp; is always assumed to be rectangular in this exercise,&nbsp; with the NRZ format&nbsp; (blue signal curves in the graph)&nbsp; as well as the RZ format with duty cycle &nbsp;$T_{\rm S}/T = 0.5$&nbsp; (red signal curves)&nbsp; to be investigated.
 The amplitude coefficients have the following properties:
@@ Line 16: / Line 16: @@
-Throughout this exercise, reference is made to the following descriptive quantities:
+Throughout this exercise,&nbsp; reference is made to the following descriptive quantities:
-*$m_{a} = \E\big[a_{\nu}\big]$&nbsp; indicates the linear mean of the amplitude coefficients.
+*$m_{a} = \E\big[a_{\nu}\big]$&nbsp; indicates the linear mean&nbsp; (first order moment)&nbsp; of the amplitude coefficients.
-*$m_{2a} = \E\big[a_{\nu}^{2}\big]$&nbsp; is the quadratic mean.
+*$m_{2a} = \E\big[a_{\nu}^{2}\big]$&nbsp; is the power&nbsp; (second order moment).
-*Thus, the variance &nbsp;$\sigma_{a}^{2} = m_{2a} – m_{a}^{2}$&nbsp; can also be calculated.
+*Thus,&nbsp; the variance &nbsp;$\sigma_{a}^{2} = m_{2a} - m_{a}^{2}$&nbsp; can also be calculated.
-*The discrete ACF of the amplitude coefficients is &nbsp;$\varphi_{a}(\lambda) = \E\big[a_{\nu} \cdot a_{\nu} + \lambda \big]$. It holds here:
+*The discrete ACF of the amplitude coefficients is &nbsp;$\varphi_{a}(\lambda) = \E\big[a_{\nu} \cdot a_{\nu + \lambda} \big]$.&nbsp; It holds here:
 :$$\varphi_a(\lambda) = \left\{ \begin{array}{c} m_2 \\ m_1^2 \\ \end{array} \right.\quad \begin{array}{*{1}c} {\rm{for}}\\ {\rm{for}} \\ \end{array} \begin{array}{*{20}c}\lambda = 0, \\ \lambda \ne 0 \hspace{0.05cm}.\\ \end{array}$$
 *The energy ACF of the basic transmission pulse is:
 :$$\varphi^{^{\bullet}}_{g_s}(\tau) = \left\{ \begin{array}{c} s_0^2 \cdot T_{\rm S} \cdot \left( 1 - {|\tau|}/{T_{\rm S}}\right) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{1}c} {\rm{for}}\\ {\rm{for}} \\ \end{array} \begin{array}{*{20}c}|\tau| \le T_{\rm S} \\ |\tau| \ge T_{\rm S} \hspace{0.05cm}.\\ \end{array}$$
-*Thus, for the total ACF of the transmitted signal, we obtain:
+*Thus,&nbsp; for the total ACF of the transmitted signal,&nbsp; we obtain:
 :$$\varphi_s(\tau) = \sum_{\lambda = -\infty}^{+\infty}{1}/{T} \cdot \varphi_a(\lambda)\cdot \varphi^{^{\bullet}}_{g_s}(\tau - \lambda \cdot T)\hspace{0.05cm}.$$
 *The power-spectral density &nbsp;${\it \Phi}_{s}(f)$&nbsp; is the Fourier transform of the ACF &nbsp;$\varphi_{s}(\tau)$.
@@ Line 33: / Line 33: @@
-''Note:''
+Note: &nbsp; The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Basics_of_Coded_Transmission|"Basics of Coded Transmission"]].
@@ Line 50: / Line 49: @@
-{What is the quadratic mean &nbsp;$m_{2a}$&nbsp; of the amplitude coefficients as a function of &nbsp;$p$?
+{What is the second order moment&nbsp; ("power") &nbsp;$m_{2a}= \E\big[a_{\nu}^{2}\big]$&nbsp; of the amplitude coefficients as a function of &nbsp;$p$?
 |type="{}"}
 $p = 0.75\text{:} \hspace{0.4cm} m_{2a} \ = \ $ { 1 3% }
@@ Line 56: / Line 55: @@
 $p = 0.25\text{:} \hspace{0.4cm} m_{2a} \ = \ $ { 1 3% }
-{Calculate the linear mean &nbsp;$m_{a}$&nbsp; in relation to &nbsp;$p$.
+{Calculate the first order moment&nbsp; ("linear mean") &nbsp;$m_{a}= \E\big[a_{\nu}\big]$&nbsp; in relation to &nbsp;$p$.
 |type="{}"}
 $p = 0.75\text{:} \hspace{0.4cm} m_{a} \ = \ $ { 0.5 3% }
@@ Line 68: / Line 67: @@
 $p = 0.25\text{:} \hspace{0.4cm} \sigma_{a}^{2} \ = \ $ { 0.75 3% }
-{Let &nbsp;$p = 0.5$ hold initially. Sketch the ACF &nbsp;$\varphi_{s}(\tau)$&nbsp; for the NRZ and RZ basic pulses and evaluate the following statements:
+{Let &nbsp;$p = 0.5$&nbsp; hold initially.&nbsp; Sketch the ACF &nbsp;$\varphi_{s}(\tau)$&nbsp; for the NRZ and RZ basic transmission pulses and evaluate the following statements:
 |type="[]"}
 + The ACF is triangular in both cases.
-+ The PSD is &nbsp;${\rm si}^{2}$–shaped in both cases.
++ The PSD is &nbsp;${\rm sinc}^{2}$–shaped in both cases.
 -The PSD area is the same in both cases.
 -In the case of RZ pulses, &nbsp;${\it \Phi}_{s}(f)$&nbsp; involves additional Dirac delta functions.
-{Let $p = 0.75$ now hold. Sketch the ACF for the NRZ basic pulse and evaluate the following statements:
+{Let be&nbsp; $p = 0.75$.&nbsp; Sketch the ACF for the NRZ basic pulse and evaluate the following statements:
 |type="[]"}
 + The ACF consists of a triangle and a DC component.
-+ The PSD consists of a &nbsp;${\rm si}^{2}$ component and a Dirac.
++ The PSD consists of a &nbsp;${\rm sinc}^{2}$ component and a Dirac delta function.
 -The Dirac delta function has the weight &nbsp;$s_{0}^{2}$.
-+With $p = 0.25$, the same power-spectral density is obtained.
++With $p = 0.25$,&nbsp; the same power-spectral density is obtained.
-{Let &nbsp;$p = 0.75$. Sketch the ACF for the RZ basic pulse and evaluate the following statements:
+{Let be &nbsp;$p = 0.75$.&nbsp; Sketch the ACF for the RZ basic pulse and evaluate the following statements:
 |type="[]"}
-+ Again, the PSD contains a &nbsp;${\rm si}^{2}$–shaped component.
++ Again,&nbsp; the PSD contains a &nbsp;${\rm sinc}^{2}$–shaped component.
-+ At the same time, there are still infinitely many Dirac delta lines in the PSD.
++ At the same time,&nbsp; there are still infinitely many Dirac delta lines in the PSD.
 </quiz>

← Older revision		Revision as of 12:49, 4 April 2022
Line 155:		Line 155:


−	[[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
Line 155:		Line 155:


−	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^2.1 Codierte Übertragung - Grundlagen^]]	+	[[Category:Digital Signal Transmission: Exercises\|^2.1 Codierte Übertragung - Grundlagen^]]

← Older revision		Revision as of 12:44, 9 July 2020
Line 34:		Line 34:

	''Hinweis:''		''Hinweis:''
−	*Die Aufgabe gehört zum Kapitel  [[~~Digitalsignalübertragung~~/Grundlagen_der_codierten_Übertragung\|Grundlagen der codierten Übertragung]].	+	*Die Aufgabe gehört zum Kapitel  [[Digital_Signal_Transmission/Grundlagen_der_codierten_Übertragung\|Grundlagen der codierten Übertragung]].

@@ Line 96: / Line 96: @@
-In diesem Sinne ist $s_{0.5}(t)$ ein redundanzfreies Signal &nbsp; &rArr; &nbsp;  <u>Lösungsvorschlag 2</u>. Somit ist hier die Entropie (der mittlere Informationsgehalt pro übertragenem Binärsymbol) maximal gleich dem Entscheidungsgehalt:
+In diesem Sinne ist $s_{0.5}(t)$ ein redundanzfreies Signal &nbsp; &rArr; &nbsp;  <u>Lösungsvorschlag 2</u>.
 :$$H_{\rm max} = {1}/{2}\cdot {\rm log}_2 (2)+{1}/{2}\cdot {\rm log}_2 (2) = 1 \,\,{\rm bit/Bin\ddot{a}rsymbol} \hspace{0.05cm}.$$
-Dagegen gilt für die Entropien der beiden anderen Binärsignale:
+*Dagegen gilt für die Entropien der beiden anderen Binärsignale:
 :$$H  = \  \frac{3}{4}\cdot {\rm log}_2 (\frac{4}{3})+ \frac{1}{4}\cdot {\rm log}_2 (4)
   = \left( \frac{3}{4} + \frac{1}{4}\right)\cdot {\rm log}_2 (4) -
@@ Line 104: / Line 105: @@
 :$$ \hspace{0.5cm} = \ 2 - \frac{3}{4}\cdot{\rm log}_2 (3) =
 .811 \,\,{\rm bit/Bin\ddot{a}rsymbol} \hspace{0.05cm}.$$
-Daraus ergibt sich für die relative Redundanz dieser Signale:
+*Daraus ergibt sich für die relative Redundanz dieser Signale:
 :$$r = \frac{H_{\rm max} - H}{H_{\rm max}}\hspace{0.15cm} \approx 18.9\%\hspace{0.05cm}.$$
 '''(2)'''&nbsp; Der quadratische Mittelwert ist unabhängig von $p$ gleich $m_{2a} = 1$:
 :$$m_{2a}={\rm E}[a_\nu^2] = p \cdot (+1)^2 + (1-p)\cdot (-1)^2 \hspace{0.15cm}\underline { = 1 \hspace{0.05cm}}.$$
 '''(3)'''&nbsp; Für den linearen Mittelwert erhält man
@@ Line 114: / Line 117: @@
 :$$\Rightarrow \hspace{0.3cm} p = 0.75\text{:} \hspace{0.4cm} m_{a}\hspace{0.15cm}\underline {=0.50},\hspace{0.2cm} p = 0.50\text{:} \hspace{0.4cm} m_{a}\hspace{0.15cm}\underline {=0},\hspace{0.2cm} p = 0.25\text{:} \hspace{0.4cm}  m_{a}\hspace{0.15cm}\underline { =-0.50 \hspace{0.05cm}}.$$
 '''(4)'''&nbsp; Mit den Ergebnissen aus (2) und (4) erhält man:
 :$$p = 0.75\text{:} \hspace{0.4cm} \sigma_{a}^2 \hspace{0.15cm}\underline {=0.75},$$
 :$$ p = 0.50\text{:} \hspace{0.4cm}  \sigma_{a}^2\hspace{0.15cm} \underline { =1.00 \hspace{0.05cm}},$$
@@ Line 120: / Line 124: @@
-[[File:P_ID1311__Dig_A_2_2e.png|right|frame|AKF bei gleichwahrscheinlichen Symbolen]]
+[[File:P_ID1311__Dig_A_2_2e.png|right|frame|AKF bei gleichen Symbolwahrscheinlichkeiten]]
 '''(5)'''&nbsp; Richtig sind nur die <u>beiden ersten Aussagen</u>:
 *Für $p = 0.5$ gilt $\varphi_{a}(\lambda = 0) = 1$ und $\varphi_{a}(\lambda \neq 0) = 0$. Daraus folgt:
@@ Line 141: / Line 145: @@
 *Mit dem RZ–Tastverhältnis $T_{\rm S}/T = 0.5$ ergibt sich die skizzierte AKF, die auch durch eine periodische Dreieckfunktion der Höhe $s_{0}^{2}/8$ (mit roter Füllung) und einem einzigen Dreieckimpuls der Höhe $3/8 \cdot s_{0}^{2}$ (grün gefüllt) dargestellt werden kann.
 *Dieser nichtperiodische Anteil führt zu einem kontinuierlichen, ${\rm si}^{2}$–förmigen LDS mit Nullstellen bei Vielfachen von $2/T$.
-*Das periodische Dreiecksignal bewirkt Diracfunktionen bei Vielfachen von $1/T$.
+*Die periodische Dreieck&ndash;AKF bewirkt im LDS Diracfunktionen bei Vielfachen von $1/T$.
-*Aufgrund der Antimetrie des periodischen Anteils besitzen die Diracfunktionen bei Vielfachen von $2/T$ jeweils das Gewicht $0$.
+*Aufgrund der Antimetrie des periodischen Anteils besitzen die Diracfunktionen bei Vielfachen von $2/T$ allerdings jeweils das Gewicht $0$.
 *Die Gewichte der Diracfunktionen im Abstand $1/T$ sind proportional zum kontinuierlichen LDS–Anteil.