Aufgaben:Exercise 2.3: Binary Signal and Quaternary Signal - Revision history

Guenter at 15:16, 3 June 2022

2022-06-03T15:16:22Z

Guenter at 15:06, 3 June 2022

2022-06-03T15:06:15Z

Hwang at 12:42, 17 May 2022

2022-05-17T12:42:19Z

Guenter at 12:32, 16 May 2022

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Guenter at 12:19, 16 May 2022

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Hwang at 12:01, 19 April 2022

2022-04-19T12:01:57Z

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Guenter at 12:56, 4 April 2022

2022-04-04T12:56:14Z

Guenter: Guenter moved page Aufgaben:Aufgabe 2.3: Binärsignal und Quaternärsignal to Aufgaben:Exercise 2.3: Binary Signal and Quaternary Signal

2022-04-04T12:51:03Z

Guenter moved page Aufgabe 2.3: Binärsignal und Quaternärsignal to Exercise 2.3: Binary Signal and Quaternary Signal

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

2021-03-23T12:53:02Z

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

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

2020-07-09T12:44:20Z

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

@@ Line 92: / Line 92: @@
 *The last statement is not true.&nbsp; For the following reasoning,&nbsp; we assume for simplicity that&nbsp; $g_{s}(t)$&nbsp; is symmetric and thus&nbsp; $G_{s}(f)$&nbsp; is real.&nbsp; Then holds:
 :$${\it \Phi}_{s}(f) = {1 }/ { T} \cdot |G_s(f)|^2\hspace{0.3cm}\Rightarrow \hspace{0.3cm}G_s(f) = \sqrt{{ T} \cdot {\it \Phi}_{s}(f)}\hspace{0.4cm} \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ \hspace{0.4cm}g_s(t) \hspace{0.05cm}.$$
-*Due to the square root in the above equation,&nbsp; the basic transmission pulse&nbsp; $g_{s}(t)$&nbsp; is not&nbsp; $\rm sinc^{2}$–shaped in contrast to the basic detection pulse '''KORREKTUR: basic transmitter pulse'''$g_{d}(t)$,&nbsp; which is equal in shape to the energy ACF&nbsp; $\varphi^{^{\bullet}}_{gs}(\tau)$&nbsp; and thus&nbsp; $\rm sinc^{2}$–shaped.&nbsp; At the same time,&nbsp; $\varphi^{^{\bullet}}_{gs}(\tau) = g_{s}(\tau) ∗ g_{s}(–\tau)$&nbsp; holds.
+*Due to the square root in the above equation,&nbsp; the basic transmission pulse&nbsp; $g_{s}(t)$&nbsp; is not&nbsp; $\rm sinc^{2}$–shaped in contrast to the basic detection pulse $g_{d}(t)$,&nbsp; which is equal in shape to the energy ACF&nbsp; $\varphi^{^{\bullet}}_{gs}(\tau)$&nbsp; and thus&nbsp; $\rm sinc^{2}$–shaped.&nbsp; At the same time,&nbsp; $\varphi^{^{\bullet}}_{gs}(\tau) = g_{s}(\tau) ∗ g_{s}(–\tau)$&nbsp; holds.

@@ Line 4: / Line 4: @@
 [[File:P_ID1324__Dig_A_2_3.png|right|frame|ACF and PSD of binary signal&nbsp; $\rm (B)$&nbsp; and quaternary signal&nbsp; $\rm (Q)$]]
-Two redundancy-free transmission systems &nbsp;$\rm B$&nbsp; and &nbsp;$\rm Q$&nbsp; each with bipolar amplitude coefficients &nbsp;$a_{\nu}$&nbsp; are to be compared.&nbsp; Both systems satisfy the first Nyquist condition.&nbsp; According to the root-root splitting,&nbsp; the spectrum &nbsp;$G_{d}(f)$&nbsp; of the basic detection pulse '''KORREKTUR: basic transmitter pulse''' is equal in shape to the power-spectral density &nbsp;${\it \Phi}_{s}(f)$&nbsp; of the transmitted signal.
+Two redundancy-free transmission systems &nbsp;$\rm B$&nbsp; and &nbsp;$\rm Q$&nbsp; each with bipolar amplitude coefficients &nbsp;$a_{\nu}$&nbsp; are to be compared.&nbsp; Both systems satisfy the first Nyquist condition.&nbsp; According to the root-root splitting,&nbsp; the spectrum &nbsp;$G_{d}(f)$&nbsp; of the basic detection pulse  is equal in shape to the power-spectral density &nbsp;${\it \Phi}_{s}(f)$&nbsp; of the transmitted signal.
 The following properties of the two systems are known:

@@ Line 4: / Line 4: @@
 [[File:P_ID1324__Dig_A_2_3.png|right|frame|ACF and PSD of binary signal&nbsp; $\rm (B)$&nbsp; and quaternary signal&nbsp; $\rm (Q)$]]
-Two redundancy-free transmission systems &nbsp;$\rm B$&nbsp; and &nbsp;$\rm Q$&nbsp; each with bipolar amplitude coefficients &nbsp;$a_{\nu}$&nbsp; are to be compared.&nbsp; Both systems satisfy the first Nyquist condition.&nbsp; According to the root-root splitting,&nbsp; the spectrum &nbsp;$G_{d}(f)$&nbsp; of the basic detection pulse is equal in shape to the power-spectral density &nbsp;${\it \Phi}_{s}(f)$&nbsp; of the transmitted signal.
+Two redundancy-free transmission systems &nbsp;$\rm B$&nbsp; and &nbsp;$\rm Q$&nbsp; each with bipolar amplitude coefficients &nbsp;$a_{\nu}$&nbsp; are to be compared.&nbsp; Both systems satisfy the first Nyquist condition.&nbsp; According to the root-root splitting,&nbsp; the spectrum &nbsp;$G_{d}(f)$&nbsp; of the basic detection pulse '''KORREKTUR: basic transmitter pulse''' is equal in shape to the power-spectral density &nbsp;${\it \Phi}_{s}(f)$&nbsp; of the transmitted signal.
 The following properties of the two systems are known:
@@ Line 92: / Line 92: @@
 *The last statement is not true.&nbsp; For the following reasoning,&nbsp; we assume for simplicity that&nbsp; $g_{s}(t)$&nbsp; is symmetric and thus&nbsp; $G_{s}(f)$&nbsp; is real.&nbsp; Then holds:
 :$${\it \Phi}_{s}(f) = {1 }/ { T} \cdot |G_s(f)|^2\hspace{0.3cm}\Rightarrow \hspace{0.3cm}G_s(f) = \sqrt{{ T} \cdot {\it \Phi}_{s}(f)}\hspace{0.4cm} \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ \hspace{0.4cm}g_s(t) \hspace{0.05cm}.$$
-*Due to the square root in the above equation,&nbsp; the basic transmission pulse&nbsp; $g_{s}(t)$&nbsp; is not&nbsp; $\rm sinc^{2}$–shaped in contrast to the basic detection pulse $g_{d}(t)$,&nbsp; which is equal in shape to the energy ACF&nbsp; $\varphi^{^{\bullet}}_{gs}(\tau)$&nbsp; and thus&nbsp; $\rm sinc^{2}$–shaped.&nbsp; At the same time,&nbsp; $\varphi^{^{\bullet}}_{gs}(\tau) = g_{s}(\tau) ∗ g_{s}(–\tau)$&nbsp; holds.
+*Due to the square root in the above equation,&nbsp; the basic transmission pulse&nbsp; $g_{s}(t)$&nbsp; is not&nbsp; $\rm sinc^{2}$–shaped in contrast to the basic detection pulse '''KORREKTUR: basic transmitter pulse'''$g_{d}(t)$,&nbsp; which is equal in shape to the energy ACF&nbsp; $\varphi^{^{\bullet}}_{gs}(\tau)$&nbsp; and thus&nbsp; $\rm sinc^{2}$–shaped.&nbsp; At the same time,&nbsp; $\varphi^{^{\bullet}}_{gs}(\tau) = g_{s}(\tau) ∗ g_{s}(–\tau)$&nbsp; holds.

@@ Line 40: / Line 40: @@
-{What is the power of the transmitted signal&nbsp; $s(t)$&nbsp; of binary system &nbsp;$\rm B$?
+{What is the transmitted power of the binary system &nbsp;$\rm B$?
 |type="{}"}
 $P_{\rm S} \ = \ $ { 200 3% } $\ \rm mW$
@@ Line 60: / Line 60: @@
-{What is the averaged power &nbsp;$P_{\rm S}$&nbsp; of the transmitted signal of the quaternary system  &nbsp;$\rm Q$?
+{What is the transmitted  power &nbsp;$P_{\rm S}$&nbsp; of the quaternary system  &nbsp;$\rm Q$?
 |type="{}"}
 $P_{\rm S} \ = \ $ { 100 3% } $\ \rm mW$
@@ Line 73: / Line 73: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The Nyquist frequency $f_{\rm Nyq} = 100 \ \rm MHz$ can be read from the diagram. From this follows according to the properties of Nyquist systems:
+'''(1)'''&nbsp; The Nyquist frequency&nbsp; $f_{\rm Nyq} = 100 \ \rm MHz$&nbsp; can be read from the diagram.&nbsp; From this follows according to the properties of Nyquist systems:
 :$$f_{\rm Nyq} = \frac{1 } {2 \cdot T} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} T = \frac{1 } {2 \cdot f_{\rm Nyq}} \hspace{0.15cm}\underline{ =5\,{\rm ns}}\hspace{0.05cm}.$$
-'''(2)'''&nbsp; In the binary system, the bit rate is also the information flow and it holds:
+'''(2)'''&nbsp; In the binary system,&nbsp; the bit rate is also the information flow and it holds:
 :$$R_{\rm B} = {1 }/ { T} \hspace{0.15cm}\underline {= 200\,{\rm Mbit/s}}= 2 \cdot f_{\rm Nyq} \cdot{\rm bit}/{\rm Hz}\hspace{0.05cm}.$$
-'''(3)'''&nbsp; The transmitted power is equal to the integral over $\it \Phi_{s}(f)$ and can be calculated as a triangular area:
+'''(3)'''&nbsp; The transmitted power is equal to the integral over&nbsp; ${\it \Phi}_{s}(f)$&nbsp; and can be calculated as a triangular area:
 :$$P_{\rm S} = \ \int_{-\infty}^{+\infty} {\it \Phi}_s(f) \,{\rm d} f = 10^{-9} \frac{\rm W}{\rm Hz} \cdot 200\,\,{\rm MHz} \hspace{0.15cm}\underline { = 200\,\,{\rm mW}}.$$
-'''(4)'''&nbsp; The <u>first two statements</u> are correct:
+'''(4)'''&nbsp; The&nbsp; <u>first two statements</u>&nbsp; are correct:
-*The Fourier inverse transform of the power-spectral density ${\it \Phi}_{s}(f)$ gives the $\rm sinc^{2}$–shaped ACF $\varphi_{s}(\tau)$. In general, the following relationship also holds:
+*The Fourier inverse transform of the power-spectral density&nbsp; ${\it \Phi}_{s}(f)$&nbsp; gives the $\rm sinc^{2}$–shaped ACF&nbsp; $\varphi_{s}(\tau)$.&nbsp; In general,&nbsp; the following relationship also holds:
 :$$ \varphi_s(\tau) = \sum_{\lambda = -\infty}^{+\infty}{1}/{T} \cdot \varphi_a(\lambda)\cdot \varphi^{^{\bullet}}_{gs}(\tau - \lambda \cdot T)\hspace{0.05cm}.$$
-*However, for a redundancy-free binary system, &nbsp;$\varphi_{a}(\lambda = 0) = 1$, while all other discrete ACF values &nbsp;$\varphi_{a}(\lambda \neq 0)$&nbsp; are equal to $0$. Thus, the energy ACF also has a $\rm sinc^{2}$–shaped curve (''note:'' &nbsp; energy ACF and energy PSD are each dotted in this tutorial):
+*However,&nbsp; for a redundancy-free binary system, &nbsp;$\varphi_{a}(\lambda = 0) = 1$,&nbsp; while all other discrete ACF values &nbsp;$\varphi_{a}(\lambda \neq 0)$&nbsp; are equal to $0$.&nbsp; Thus,&nbsp; the energy ACF also has a&nbsp; $\rm sinc^{2}$–shaped curve &nbsp; (note: &nbsp; energy ACF and energy PSD are each dotted in this tutorial):
 :$$\varphi^{^{\bullet}}_{gs}(\tau ) = T \cdot \varphi_s(\tau) \hspace{0.05cm}.$$
-*The last statement is not true. For the following reasoning, we assume for simplicity that $g_{s}(t)$ is symmetric and thus $G_{s}(f)$ is real. Then holds:
+*The last statement is not true.&nbsp; For the following reasoning,&nbsp; we assume for simplicity that&nbsp; $g_{s}(t)$&nbsp; is symmetric and thus&nbsp; $G_{s}(f)$&nbsp; is real.&nbsp; Then holds:
 :$${\it \Phi}_{s}(f) = {1 }/ { T} \cdot |G_s(f)|^2\hspace{0.3cm}\Rightarrow \hspace{0.3cm}G_s(f) = \sqrt{{ T} \cdot {\it \Phi}_{s}(f)}\hspace{0.4cm} \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ \hspace{0.4cm}g_s(t) \hspace{0.05cm}.$$
-*Due to the square root in the above equation, the basic transmission pulse $g_{s}(t)$ is not $\rm sinc^{2}$–shaped in contrast to the basic transmitter pulse $g_{d}(t)$, which is equal in shape to the energy ACF $\varphi^{^{\bullet}}_{gs}(\tau)$ and thus $\rm sinc^{2}$–shaped. At the same time, $\varphi^{^{\bullet}}_{gs}(\tau) = g_{s}(\tau) ∗ g_{s}(–\tau)$ holds.
+*Due to the square root in the above equation,&nbsp; the basic transmission pulse&nbsp; $g_{s}(t)$&nbsp; is not&nbsp; $\rm sinc^{2}$–shaped in contrast to the basic detection pulse $g_{d}(t)$,&nbsp; which is equal in shape to the energy ACF&nbsp; $\varphi^{^{\bullet}}_{gs}(\tau)$&nbsp; and thus&nbsp; $\rm sinc^{2}$–shaped.&nbsp; At the same time,&nbsp; $\varphi^{^{\bullet}}_{gs}(\tau) = g_{s}(\tau) ∗ g_{s}(–\tau)$&nbsp; holds.
-'''(5)'''&nbsp; The ACF $\varphi_{s}(\tau)$ is limited to the range $|\tau| ≤ T$ when the basic transmission pulse is an NRZ rectangle. From the graph, the symbol duration $T \underline{= 10 \ \rm ns}$.
+'''(5)'''&nbsp; The ACF&nbsp; $\varphi_{s}(\tau)$&nbsp; is limited to the range&nbsp; $|\tau| ≤ T$&nbsp; when the basic transmission pulse is an NRZ rectangle.&nbsp; From the graph,&nbsp; the symbol duration&nbsp; $T \underline{= 10 \ \rm ns}$.
-'''(6)'''&nbsp; For the quaternary signal, the information flow is the same as for the binary signal above because of the double symbol duration:
+'''(6)'''&nbsp; For the quaternary signal,&nbsp; the information flow is the same as for the binary signal above because of the double symbol duration:
 :$$R_{\rm B} = {{\rm log_2(4)} }/ { T} \hspace{0.15cm}\underline {= 200\,\,{\rm Mbit/s}}\hspace{0.05cm}.$$
-'''(7)'''&nbsp; The transmitted power is equal to the ACF value at $\tau = 0$ and can be read from the graph:
+'''(7)'''&nbsp; The transmitted power is equal to the ACF value at&nbsp; $\tau = 0$&nbsp; and can be read from the graph:
 :$$P_{\rm S} = \hspace{0.15cm}\underline {100\,\,{\rm mW}}.$$
-'''(8)'''&nbsp; For the redundancy-free quaternary signal with NRZ rectangular pulses, the average transmitted power is:
+'''(8)'''&nbsp; For the redundancy-free quaternary signal with NRZ rectangular pulses,&nbsp; the average transmitted power is:
 :$$P_{\rm S} = {1}/ { 4} \cdot \left [ (-s_0)^2 + (-s_0/3)^2 + (+s_0/3)^2 +(+s_0)^2 \right ] = {5}/ { 9} \cdot s_0^2$$
 :$$\Rightarrow \hspace{0.3cm}s_0^2 = {9}/ {5} \cdot P_{\rm S} = {9}/ {5} \cdot 100\,\,{\rm mW}\hspace{0.15cm}\underline { = 180\,\,{\rm mW}}\hspace{0.05cm}.$$

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1324__Dig_A_2_3.png|right|frame|ACF and PSD of <br>binary signal and quaternary signal]]
+[[File:P_ID1324__Dig_A_2_3.png|right|frame|ACF and PSD of binary signal&nbsp; $\rm (B)$&nbsp; and quaternary signal&nbsp; $\rm (Q)$]]
-Two redundancy-free transmission systems &nbsp;$\rm B$&nbsp; and &nbsp;$\rm Q$&nbsp; each with bipolar amplitude coefficients &nbsp;$a_{\nu}$&nbsp; are to be compared. Both systems satisfy the first Nyquist condition. According to the root-root splitting, the spectrum &nbsp;$G_{d}(f)$&nbsp; of the basic transmitter pulse is equal in shape to the power-spectral density &nbsp;${\it \Phi}_{s}(f)$&nbsp; of the transmitted signal.
+Two redundancy-free transmission systems &nbsp;$\rm B$&nbsp; and &nbsp;$\rm Q$&nbsp; each with bipolar amplitude coefficients &nbsp;$a_{\nu}$&nbsp; are to be compared.&nbsp; Both systems satisfy the first Nyquist condition.&nbsp; According to the root-root splitting,&nbsp; the spectrum &nbsp;$G_{d}(f)$&nbsp; of the basic detection pulse is equal in shape to the power-spectral density &nbsp;${\it \Phi}_{s}(f)$&nbsp; of the transmitted signal.
 The following properties of the two systems are known:
 *From the binary system &nbsp;$\rm B$,&nbsp; the power-spectral density &nbsp;${\it \Phi}_{s}(f)$&nbsp; at the transmitter is known and shown in the graph together with the description parameters.
-''Notes:''
+*Consider that auto-correlation function&nbsp; $\rm (ACF)$&nbsp; and power-spectral density&nbsp; $\rm (PSD)$&nbsp; of a stochastic signal are always related via the Fourier transform.
@@ Line 39: / Line 40: @@
-{What is the power of the transmitted signal of binary system &nbsp;$\rm B$&nbsp;?
+{What is the power of the transmitted signal&nbsp; $s(t)$&nbsp; of binary system &nbsp;$\rm B$?
 |type="{}"}
 $P_{\rm S} \ = \ $ { 200 3% } $\ \rm mW$
@@ Line 46: / Line 47: @@
 |type="[]"}
 + The ACF &nbsp;$\varphi_{s}(\tau)$&nbsp; of the transmitted signal is &nbsp;$\rm sinc^{2}$–shaped.
-+ The energy ACF &nbsp;$\varphi^{^{\bullet}}_{gs}(\tau)$&nbsp; of the basic pulse is &nbsp;$\rm si^{2}$–shaped.
++ The energy ACF &nbsp;$\varphi^{^{\bullet}}_{gs}(\tau)$&nbsp; of the basic transmission pulse is &nbsp;$\rm sinc^{2}$–shaped.
 - The basic transmission pulse &nbsp;$g_{s}(t)$&nbsp; itself is &nbsp;$\rm sinc^{2}$–shaped.
@@ Line 54: / Line 55: @@
-{What is the equivalent bit rate &nbsp;$R_{\rm B}$&nbsp; of the quaternary system &nbsp;$\rm Q$&nbsp;?
+{What is the equivalent bit rate &nbsp;$R_{\rm B}$&nbsp; of the quaternary system &nbsp;$\rm Q$?
 |type="{}"}
 $R_{\rm B} \ = \ $ { 200 3% } $\ \rm Mbit/s$
-{What is the power &nbsp;$P_{\rm S}$&nbsp; of the transmitted signal of the quaternary system  &nbsp;$\rm Q$&nbsp;?
+{What is the averaged power &nbsp;$P_{\rm S}$&nbsp; of the transmitted signal of the quaternary system  &nbsp;$\rm Q$?
 |type="{}"}
 $P_{\rm S} \ = \ $ { 100 3% } $\ \rm mW$

← Older revision		Revision as of 12:56, 4 April 2022
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−	[[Category:Digital Signal Transmission: Exercises\|^2.2 ~~Redundanzfreie Codierung~~^]]	+	[[Category:Digital Signal Transmission: Exercises\|^2.2 Redundancy-Free Coding^]]

← Older revision	Revision as of 12:51, 4 April 2022
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← Older revision		Revision as of 12:53, 23 March 2021
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−	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^2.2 Redundanzfreie Codierung^]]	+	[[Category:Digital Signal Transmission: Exercises\|^2.2 Redundanzfreie Codierung^]]

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 ''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]].
-*Bezug genommen wird auch auf das Kapitel&nbsp; [[Digitalsignalübertragung/Redundanzfreie_Codierung|Redundanzfreie Codierung]].
+*Bezug genommen wird auch auf das Kapitel&nbsp; [[Digital_Signal_Transmission/Redundanzfreie_Codierung|Redundanzfreie Codierung]].
 *Berücksichtigen Sie, dass Autokorrelationsfunktion (AKF) und Leistungsdichtespektrum (LDS) eines stochastischen Signals stets über die Fouriertransformation zusammenhängen.