Aufgaben:Exercise 1.1: Basic Transmission Pulses - Revision history

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	:$$T_{\rm S} / T = 1\hspace{0.05cm},\hspace{0.3cm}\Delta t_{\rm S} / T \hspace{0.1cm}\underline{ = 1 }\hspace{0.05cm}.$$		:$$T_{\rm S} / T = 1\hspace{0.05cm},\hspace{0.3cm}\Delta t_{\rm S} / T \hspace{0.1cm}\underline{ = 1 }\hspace{0.05cm}.$$
	*The basic transmission pulse for the signal  $s_{\rm C}(t)$  is:		*The basic transmission pulse for the signal  $s_{\rm C}(t)$  is:
	:$$g_s(t) = \left\{ \begin{array}{c} s_0 \cdot \cos^2(\pi \cdot \frac{t}{T}) \\		:$$g_s(t) = \left\{ \begin{array}{c} s_0 \cdot \cos^2(\pi \cdot \frac{t}{T}) \\0 \\ \end{array} \right.\quad\begin{array}{{1}c} {\rm{for}}\\ \\ \end{array}\begin{array}{{20}c}-T/2 \le t \le +T/2 \hspace{0.05cm}, \\{\rm else} \hspace{0.05cm}. \\\end{array}$$
	0 \\ \end{array} \right.\quad
	\begin{array}{*{1}c} {\rm{for}}
	\\ \\ \end{array}\begin{array}{*{20}c}
	-T/2 \le t \le +T/2 \hspace{0.05cm}, \\
	{\rm else} \hspace{0.05cm}. \\
	\end{array}$$
	*From the diagram on the information section,  we can see that the following values apply to the cosine–square pulse:		*From the diagram on the information section,  we can see that the following values apply to the cosine–square pulse:
	:$$T_{\rm S} / T = 1\hspace{0.05cm},\hspace{0.3cm}\Delta t_{\rm S} / T \hspace{0.1cm}\underline{ = 0.5} \hspace{0.05cm}.$$		:$$T_{\rm S} / T = 1\hspace{0.05cm},\hspace{0.3cm}\Delta t_{\rm S} / T \hspace{0.1cm}\underline{ = 0.5} \hspace{0.05cm}.$$

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	In particular,  the following characteristics are to be calculated for the respective basic transmission pulses  $g_s(t)$:		In particular,  the following characteristics are to be calculated for the respective basic transmission pulses  $g_s(t)$:
	*the equivalent pulse duration of  $g_s(t)$:		*the equivalent pulse duration of  $g_s(t)$:
	:$$\Delta t_{\rm S} = \frac {\int ^{+\infty} _{-\infty} \hspace{0.15cm} g_s(t)\,{\rm		:$$\Delta t_{\rm S} = \frac {\int ^{+\infty} _{-\infty} \hspace{0.15cm} g_s(t)\,{\rmd}t}{{\rm Max} \hspace{0.05cm}[g_s(t)]} \hspace{0.05cm},$$
	d}t}{{\rm Max} \hspace{0.05cm}[g_s(t)]} \hspace{0.05cm},$$
	*the energy of  $g_s(t)$:		*the energy of  $g_s(t)$:
	:$$E_g = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm		:$$E_g = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rmd}t \hspace{0.05cm},$$
	d}t \hspace{0.05cm},$$
	*the power of the transmitted signal  $s(t)$:		*the power of the transmitted signal  $s(t)$:
	:$$P_{\rm S} = \lim_{T_{\rm M} \to \infty} \frac{1}{+T_{\rm M}} \cdot \int^{+T_{\rm M}/2} _{-T_{\rm M}/2} s^2(t)\,{\rm		:$$P_{\rm S} = \lim_{T_{\rm M} \to \infty} \frac{1}{+T_{\rm M}} \cdot \int^{+T_{\rm M}/2} _{-T_{\rm M}/2} s^2(t)\,{\rmd}t \hspace{0.05cm}.$$
	d}t \hspace{0.05cm}.$$


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	*Reference is made in particular to the section  [[Digital_Signal_Transmission/System_Components_of_a_Baseband_Transmission_System#Characteristics_of_the_digital_transmitter\|"Characteristics of the digital transmitter"]].		*Reference is made in particular to the section  [[Digital_Signal_Transmission/System_Components_of_a_Baseband_Transmission_System#Characteristics_of_the_digital_transmitter\|"Characteristics of the digital transmitter"]].
	*Given is the following indefinite integral:		*Given is the following indefinite integral:
	:$$\int \cos^4(a x)\,{\rm d}x = \frac{3}{8} \cdot x + \frac{1}{4a} \cdot \sin(2 a x)+ \frac{1}{32a} \cdot \sin(4 a		:$$\int \cos^4(a x)\,{\rm d}x = \frac{3}{8} \cdot x + \frac{1}{4a} \cdot \sin(2 a x)+ \frac{1}{32a} \cdot \sin(4 ax)\hspace{0.05cm}.$$
	x)\hspace{0.05cm}.$$


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	'''(3)'''  For the energy of the rectangular pulse holds:		'''(3)'''  For the energy of the rectangular pulse holds:
	:$$E_g = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm		:$$E_g = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rmd}t = s_0^2 \cdot T = 0.5\, {\rm W} \cdot 1\, {\rm µ s} \hspace{0.1cm}\underline{= 0.5 \cdot 10^{-6}\, {\rmWs}}\hspace{0.05cm}.$$
	d}t = s_0^2 \cdot T = 0.5\, {\rm W} \cdot 1\, {\rm µ s} \hspace{0.1cm}\underline{= 0.5 \cdot 10^{-6}\, {\rm
	Ws}}\hspace{0.05cm}.$$


	'''(4)'''  For a bipolar rectangular signal,  the following would apply:		'''(4)'''  For a bipolar rectangular signal,  the following would apply:
	:$$s_{\rm R}^2(t)= s_0^2 = {\rm const.} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} P_s = s_0^2 \cdot		:$$s_{\rm R}^2(t)= s_0^2 = {\rm const.} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} P_s = s_0^2 \cdot\lim_{T_{\rm M} \to \infty} \frac{1}{T_{\rm M}} \cdot \int ^{T_{\rm M}/2} _{-T_{\rm M}/2} \,{\rmd}t = s_0^2 \hspace{0.05cm}.$$
	\lim_{T_{\rm M} \to \infty} \frac{1}{T_{\rm M}} \cdot \int ^{T_{\rm M}/2} _{-T_{\rm M}/2} \,{\rm
	d}t = s_0^2 \hspace{0.05cm}.$$
	*However,  since the signal  $s_{\rm R}(t)$ is  unipolar here,  in half the time $s_{\rm R}(t)= 0$.  Thus,  we get:		*However,  since the signal  $s_{\rm R}(t)$ is  unipolar here,  in half the time $s_{\rm R}(t)= 0$.  Thus,  we get:
	:$$P_{\rm S} = {1}/{2} \cdot s_0^2 \hspace{0.1cm}\underline{= 0.25 \,{\rm		:$$P_{\rm S} = {1}/{2} \cdot s_0^2 \hspace{0.1cm}\underline{= 0.25 \,{\rmW}} \hspace{0.05cm}.$$
	W}} \hspace{0.05cm}.$$


	'''(5)'''  For the energy of the cosine–square pulse holds:		'''(5)'''  For the energy of the cosine–square pulse holds:
	:$$E_g = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm		:$$E_g = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rmd}t = 2 \cdot s_0^2 \cdot \int^{T/2} _{0} \cos^4(\pi \cdot {t}/{T})\,{\rmd}t \hspace{0.05cm}.$$
	d}t = 2 \cdot s_0^2 \cdot \int^{T/2} _{0} \cos^4(\pi \cdot {t}/{T})\,{\rm
	d}t \hspace{0.05cm}.$$
	*Here,  the formula derived in subtask  '''(3)'''  and the symmetry of  $g_s(t)$  about time  $t = 0$  are considered.		*Here,  the formula derived in subtask  '''(3)'''  and the symmetry of  $g_s(t)$  about time  $t = 0$  are considered.
	*The integral is given in the task description,  where $a = π/T$  is to be set:		*The integral is given in the task description,  where $a = π/T$  is to be set:
	:$$E_g = 2 \cdot s_0^2 \cdot \left [ \frac{3}{8} \cdot t + \frac{T}{4\pi} \cdot \sin(2 \pi \frac{t}{T})+ \frac{T}{32\pi} \cdot		:$$E_g = 2 \cdot s_0^2 \cdot \left [ \frac{3}{8} \cdot t + \frac{T}{4\pi} \cdot \sin(2 \pi \frac{t}{T})+ \frac{T}{32\pi} \cdot\sin(4 \pi \frac{t}{T})\right ]_{0}^{T/2}\hspace{0.05cm}.$$
	\sin(4 \pi \frac{t}{T})\right ]_{0}^{T/2}\hspace{0.05cm}.$$
	*The lower bound  $t = 0$  always yields the result  $0$.   With respect to the upper bound,  only the first term yields a result different from  $0$.  Thus:		*The lower bound  $t = 0$  always yields the result  $0$.   With respect to the upper bound,  only the first term yields a result different from  $0$.  Thus:
	:$$E_g = 2 \cdot s_0^2 \cdot \frac{3}{8} \cdot \frac{T}{2} = \frac{3}{8} \cdot 5 \cdot 10^{-7}\, {\rm		:$$E_g = 2 \cdot s_0^2 \cdot \frac{3}{8} \cdot \frac{T}{2} = \frac{3}{8} \cdot 5 \cdot 10^{-7}\, {\rmWs} \hspace{0.1cm}\underline{ = 0.1875 \cdot 10^{-6}\, {\rmWs}}\hspace{0.05cm}.$$
	Ws} \hspace{0.1cm}\underline{ = 0.1875 \cdot 10^{-6}\, {\rm
	Ws}}\hspace{0.05cm}.$$


	'''(6)'''  The following relationship holds for the bipolar signal $s_{\rm C}(t)$:		'''(6)'''  The following relationship holds for the bipolar signal $s_{\rm C}(t)$:
	:$$P_{\rm S} = \frac{ E_g}{T} = \frac{ 1.875 \cdot 10^{-7}\, {\rm		:$$P_{\rm S} = \frac{ E_g}{T} = \frac{ 1.875 \cdot 10^{-7}\, {\rmWs}}{10^{-6}\, {\rm s}}\hspace{0.1cm}\underline{ = 0.1875 \,{\rm W}} \hspace{0.05cm}.$$
	Ws}}{10^{-6}\, {\rm s}}\hspace{0.1cm}\underline{ = 0.1875 \,{\rm W}} \hspace{0.05cm}.$$

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	[[Category:Digital Signal Transmission: Exercises\|^1.1 Baseband System Components^]]		[[Category:Digital Signal Transmission: Exercises\|^1.1 Baseband System Components^]]
			[[de:Aufgaben:1.0_Test]]

Hwang at 13:13, 9 May 2022

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	:$$g_s(t) = \left\{ \begin{array}{c} s_0 \cdot \cos^2(\pi \cdot \frac{t}{T}) \\		:$$g_s(t) = \left\{ \begin{array}{c} s_0 \cdot \cos^2(\pi \cdot \frac{t}{T}) \\
	0 \\ \end{array} \right.\quad		0 \\ \end{array} \right.\quad
	\begin{array}{*{1}c} {\rm{~~f\ddot{u}r~~}}		\begin{array}{*{1}c} {\rm{for}}
	\\ \\ \end{array}\begin{array}{*{20}c}		\\ \\ \end{array}\begin{array}{*{20}c}
	-T/2 \le t \le +T/2 \hspace{0.05cm}, \\		-T/2 \le t \le +T/2 \hspace{0.05cm}, \\

Hwang at 13:12, 9 May 2022

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	\|type="{}"}		\|type="{}"}
	$\text{For the signal}\ \ s_{\rm R}(t) \text{:} \ \ \Delta t_{\rm S}/T \ = \ $ { 1 3% }		$\text{For the signal}\ \ s_{\rm R}(t) \text{:} \ \ \Delta t_{\rm S}/T \ = \ $ { 1 3% }
	$\text{~~for~~ the signal}\ \ s_{\rm C}(t) \text{:} \ \ \Delta t_{\rm S}/T \ = \ $ { 0.5 3% }		$\text{For the signal}\ \ s_{\rm C}(t) \text{:} \ \ \Delta t_{\rm S}/T \ = \ $ { 0.5 3% }

	{What is the energy of the rectangular basic transmission pulse  $g_s(t)$?		{What is the energy of the rectangular basic transmission pulse  $g_s(t)$?

Guenter at 15:37, 27 April 2022

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Solution 2</u> is correct:		'''(1)'''  <u>Solution 2</u>  is correct:
	*In both cases, the transmitted signal can be represented in the following form:		*In both cases,  the transmitted signal can be represented in the following form:
	:$$s(t) = \sum_{(\nu)} a_\nu \cdot g_s ( t - \nu \cdot T)$$		:$$s(t) = \sum_{(\nu)} a_\nu \cdot g_s ( t - \nu \cdot T)$$
	*For the signal $s_{\rm R}(t)$, the amplitude coefficients $a_ν$ are either $0$ or $1$. Thus, a unipolar signal is present.		*For the signal $s_{\rm R}(t)$,  the amplitude coefficients  $a_ν$  are either  $0$  or  $1$.  Thus,  a unipolar signal is present.
	*In contrast, for the bipolar signal $s_{\rm R}(t)$, $a_ν ∈ \{–1, +1\}$ holds.		*In contrast,  for the bipolar signal $s_{\rm R}(t)$   ⇒   $a_ν ∈ \{–1, +1\}$ holds.


	'''(2)'''  The signal $s_{\rm R}(t)$ is NRZ rectangular.		'''(2)'''  The signal  $s_{\rm R}(t)$  is NRZ  ("non-return-to-zero")  rectangular.
	*Accordingly, both the absolute pulse duration $T_{\rm S}$ and the equivalent pulse duration $\Delta t_{\rm S}$ are equal to the symbol duration $T$:		*Accordingly,  both the absolute pulse duration  $T_{\rm S}$  and the equivalent pulse duration  $\Delta t_{\rm S}$  are equal to the symbol duration  $T$:
	:$$T_{\rm S} / T = 1\hspace{0.05cm},\hspace{0.3cm}\Delta t_{\rm S} / T \hspace{0.1cm}\underline{ = 1 }\hspace{0.05cm}.$$		:$$T_{\rm S} / T = 1\hspace{0.05cm},\hspace{0.3cm}\Delta t_{\rm S} / T \hspace{0.1cm}\underline{ = 1 }\hspace{0.05cm}.$$
	*The basic transmission pulse for the signal $s_{\rm C}(t)$ is:		*The basic transmission pulse for the signal  $s_{\rm C}(t)$  is:
	:$$g_s(t) = \left\{ \begin{array}{c} s_0 \cdot \cos^2(\pi \cdot \frac{t}{T}) \\		:$$g_s(t) = \left\{ \begin{array}{c} s_0 \cdot \cos^2(\pi \cdot \frac{t}{T}) \\
	0 \\ \end{array} \right.\quad		0 \\ \end{array} \right.\quad
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	\\ \\ \end{array}\begin{array}{*{20}c}		\\ \\ \end{array}\begin{array}{*{20}c}
	-T/2 \le t \le +T/2 \hspace{0.05cm}, \\		-T/2 \le t \le +T/2 \hspace{0.05cm}, \\
	{\rm ~~sonst~~} \hspace{0.05cm}. \\		{\rm else} \hspace{0.05cm}. \\
	\end{array}$$		\end{array}$$
	*From the diagram on the information section, we can see that the following values apply to the ~~$\cos^2$~~ pulse:		*From the diagram on the information section,  we can see that the following values apply to the cosine–square pulse:
	:$$T_{\rm S} / T = 1\hspace{0.05cm},\hspace{0.3cm}\Delta t_{\rm S} / T \hspace{0.1cm}\underline{ = 0.5} \hspace{0.05cm}.$$		:$$T_{\rm S} / T = 1\hspace{0.05cm},\hspace{0.3cm}\Delta t_{\rm S} / T \hspace{0.1cm}\underline{ = 0.5} \hspace{0.05cm}.$$

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	'''(4)'''  For a bipolar ~~square wave~~ signal, the following would apply:		'''(4)'''  For a bipolar rectangular signal,  the following would apply:
	:$$s_{\rm R}^2(t)= s_0^2 = {\rm const.} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} P_s = s_0^2 \cdot		:$$s_{\rm R}^2(t)= s_0^2 = {\rm const.} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} P_s = s_0^2 \cdot
	\lim_{T_{\rm M} \to \infty} \frac{1}{T_{\rm M}} \cdot \int ^{T_{\rm M}/2} _{-T_{\rm M}/2} \,{\rm		\lim_{T_{\rm M} \to \infty} \frac{1}{T_{\rm M}} \cdot \int ^{T_{\rm M}/2} _{-T_{\rm M}/2} \,{\rm
	d}t = s_0^2 \hspace{0.05cm}.$$		d}t = s_0^2 \hspace{0.05cm}.$$
	*However, since the signal $s_{\rm R}(t)$ is unipolar here, in half the time $s_{\rm R}(t)= 0$. Thus, we get:		*However,  since the signal  $s_{\rm R}(t)$ is  unipolar here,  in half the time $s_{\rm R}(t)= 0$.  Thus,  we get:
	:$$P_{\rm S} = {1}/{2} \cdot s_0^2 \hspace{0.1cm}\underline{= 0.25 \,{\rm		:$$P_{\rm S} = {1}/{2} \cdot s_0^2 \hspace{0.1cm}\underline{= 0.25 \,{\rm
	W}} \hspace{0.05cm}.$$		W}} \hspace{0.05cm}.$$


	'''(5)'''  For the energy of the ~~$\cos^2$~~ pulse holds:		'''(5)'''  For the energy of the cosine–square pulse holds:
	:$$E_g = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm		:$$E_g = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm
	d}t = 2 \cdot s_0^2 \cdot \int^{T/2} _{0} \cos^4(\pi \cdot {t}/{T})\,{\rm		d}t = 2 \cdot s_0^2 \cdot \int^{T/2} _{0} \cos^4(\pi \cdot {t}/{T})\,{\rm
	d}t \hspace{0.05cm}.$$		d}t \hspace{0.05cm}.$$
	*Here, the formula derived in ~~point~~ '''(3)''' and the symmetry of $g_s(t)$ about time $t = 0$ are considered.		*Here,  the formula derived in subtask  '''(3)'''  and the symmetry of  $g_s(t)$  about time  $t = 0$  are considered.
	*The integral is given in the task description, where $a = π/T$ is to be set:		*The integral is given in the task description,  where $a = π/T$  is to be set:
	:$$E_g = 2 \cdot s_0^2 \cdot \left [ \frac{3}{8} \cdot t + \frac{T}{4\pi} \cdot \sin(2 \pi \frac{t}{T})+ \frac{T}{32\pi} \cdot		:$$E_g = 2 \cdot s_0^2 \cdot \left [ \frac{3}{8} \cdot t + \frac{T}{4\pi} \cdot \sin(2 \pi \frac{t}{T})+ \frac{T}{32\pi} \cdot
	\sin(4 \pi \frac{t}{T})\right ]_{0}^{T/2}\hspace{0.05cm}.$$		\sin(4 \pi \frac{t}{T})\right ]_{0}^{T/2}\hspace{0.05cm}.$$
	*The lower bound $t = 0$ always yields the result $0$. With respect to the upper bound, only the first term yields a result different from $0$. Thus:		*The lower bound  $t = 0$  always yields the result  $0$.   With respect to the upper bound,  only the first term yields a result different from  $0$.  Thus:
	:$$E_g = 2 \cdot s_0^2 \cdot \frac{3}{8} \cdot \frac{T}{2} = \frac{3}{8} \cdot 5 \cdot 10^{-7}\, {\rm		:$$E_g = 2 \cdot s_0^2 \cdot \frac{3}{8} \cdot \frac{T}{2} = \frac{3}{8} \cdot 5 \cdot 10^{-7}\, {\rm
	Ws} \hspace{0.1cm}\underline{ = 0.1875 \cdot 10^{-6}\, {\rm		Ws} \hspace{0.1cm}\underline{ = 0.1875 \cdot 10^{-6}\, {\rm

Guenter at 15:21, 27 April 2022

2022-04-27T15:21:41Z

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	Always assume in your calculations that the two possible amplitude coefficients are equally likely and that the distance between adjacent symbols is  $T = 1 \ \rm µ s$.  This corresponds to a bit rate of  $R = 1 \ \rm Mbit/s$.		Always assume in your calculations that the two possible amplitude coefficients are equally likely and that the distance between adjacent symbols is  $T = 1 \ \rm µ s$.  This corresponds to a bit rate of  $R = 1 \ \rm Mbit/s$.

	*The (positive) maximum value of the transmitted signal is the same in both cases		*The  (positive)  maximum value of the transmitted signal is the same in both cases:
	:$$s_0 = \sqrt{0.5\, {\rm W}} \hspace{0.05cm}.$$		:$$s_0 = \sqrt{0.5\, {\rm W}} \hspace{0.05cm}.$$
	*Assuming that the transmitter is terminated with a  $50\ \rm Ω$  resistor, this corresponds to the following voltage value:		*Assuming that the transmitter is terminated with a  $50\ \rm Ω$  resistor,  this corresponds to the following voltage value:
	:$$s_0^2 = 0.5\, {\rm W} \cdot 50\, {\rm \Omega} = 25\, {\rm V}^2 \hspace{0.15cm} \Rightarrow \hspace{0.15cm} s_0 =5\, {\rm V} \hspace{0.05cm}.$$		:$$s_0^2 = 0.5\, {\rm W} \cdot 50\, {\rm \Omega} = 25\, {\rm V}^2 \hspace{0.15cm} \Rightarrow \hspace{0.15cm} s_0 =5\, {\rm V} \hspace{0.05cm}.$$

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	Notes:		Notes:
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/System_Components_of_a_Baseband_Transmission_System\|"System Components of a Baseband Transmission System"]].		*The exercise belongs to the chapter  [[Digital_Signal_Transmission/System_Components_of_a_Baseband_Transmission_System\|"System Components of a Baseband Transmission System"]].
	*Reference is made in particular to the section  [[Digital_Signal_Transmission/System_Components_of_a_Baseband_Transmission_System#Characteristics_of_the_digital_transmitter\|"Characteristics of the digital transmitter"]].		*Reference is made in particular to the section  [[Digital_Signal_Transmission/System_Components_of_a_Baseband_Transmission_System#Characteristics_of_the_digital_transmitter\|"Characteristics of the digital transmitter"]].

	*Given is the following indefinite integral:		*Given is the following indefinite integral:
	:$$\int \cos^4(a x)\,{\rm d}x = \frac{3}{8} \cdot x + \frac{1}{4a} \cdot \sin(2 a x)+ \frac{1}{32a} \cdot \sin(4 a		:$$\int \cos^4(a x)\,{\rm d}x = \frac{3}{8} \cdot x + \frac{1}{4a} \cdot \sin(2 a x)+ \frac{1}{32a} \cdot \sin(4 a

Guenter at 15:18, 27 April 2022

2022-04-27T15:18:37Z

← Older revision		Revision as of 17:18, 27 April 2022
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	[[File:P_ID1256__Dig_A_1_1.png\|right\|frame\|Considered basic transmission pulses]]		[[File:P_ID1256__Dig_A_1_1.png\|right\|frame\|Considered basic transmission pulses]]
	In this exercise, we examine the two transmitted signals  $s_{\rm R}(t)$  and  $s_{\rm C}(t)$  with ~~square-wave and~~ &~~nbsp~~;~~$\cos^2$~~ basic transmission pulse, ~~respectively,~~ shown in the diagram. In particular, the following characteristics are to be calculated for the respective pulses  $g_s(t)$:		In this exercise,  we examine the two transmitted signals  $s_{\rm R}(t)$  and  $s_{\rm C}(t)$  with rectangular resp. cosine–square basic transmission pulse,  shown in the diagram.

			In particular,  the following characteristics are to be calculated for the respective basic transmission pulses  $g_s(t)$:
	*the equivalent pulse duration of  $g_s(t)$:		*the equivalent pulse duration of  $g_s(t)$:
	:$$\Delta t_{\rm S} = \frac {\int ^{+\infty} _{-\infty} \hspace{0.15cm} g_s(t)\,{\rm		:$$\Delta t_{\rm S} = \frac {\int ^{+\infty} _{-\infty} \hspace{0.15cm} g_s(t)\,{\rm
	d}t}{{\rm Max} \hspace{0.05cm}[g_s(t)]} \hspace{0.05cm},$$		d}t}{{\rm Max} \hspace{0.05cm}[g_s(t)]} \hspace{0.05cm},$$
	*the energy of ~~the basic transmission pulse~~  $g_s(t)$:		*the energy of  $g_s(t)$:
	:$$E_g = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm		:$$E_g = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm
	d}t \hspace{0.05cm},$$		d}t \hspace{0.05cm},$$
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	:$$s_0 = \sqrt{0.5\, {\rm W}} \hspace{0.05cm}.$$		:$$s_0 = \sqrt{0.5\, {\rm W}} \hspace{0.05cm}.$$
	*Assuming that the transmitter is terminated with a  $50\ \rm Ω$  resistor, this corresponds to the following voltage value:		*Assuming that the transmitter is terminated with a  $50\ \rm Ω$  resistor, this corresponds to the following voltage value:
	:$$s_0 = \~~sqrt~~{0.5\, {\rm W}} \hspace{0.05cm}.$$		:$$s_0^2 = 0.5\, {\rm W} \cdot 50\, {\rm \Omega} = 25\, {\rm V}^2 \hspace{0.15cm} \Rightarrow \hspace{0.15cm} s_0 =5\, {\rm V} \hspace{0.05cm}.$$



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	''Notes:''		Notes:
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/System_Components_of_a_Baseband_Transmission_System\|System Components of a Baseband Transmission System]].		*The exercise belongs to the chapter  [[Digital_Signal_Transmission/System_Components_of_a_Baseband_Transmission_System\|"System Components of a Baseband Transmission System"]].
	*Reference is made in particular to the section  [[Digital_Signal_Transmission/System_Components_of_a_Baseband_Transmission_System#Characteristics_of_the_digital_transmitter\|Characteristics of the digital transmitter]].		*Reference is made in particular to the section  [[Digital_Signal_Transmission/System_Components_of_a_Baseband_Transmission_System#Characteristics_of_the_digital_transmitter\|"Characteristics of the digital transmitter"]].

	*Given is the following indefinite integral:		*Given is the following indefinite integral:
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	{What is the equivalent pulse duration  $\Delta t_{\rm S}$, normalized to the symbol duration  $T$?		{What is the equivalent pulse duration  $\Delta t_{\rm S}$,  normalized to the symbol duration  $T$?
	\|type="{}"}		\|type="{}"}
	$\text{~~for~~ the signal}\ \ s_{\rm R}(t) \text{:} \ \ \Delta t_{\rm S}/T \ = \ $ { 1 3% }		$\text{For the signal}\ \ s_{\rm R}(t) \text{:} \ \ \Delta t_{\rm S}/T \ = \ $ { 1 3% }
	$\text{for the signal}\ \ s_{\rm C}(t) \text{:} \ \ \Delta t_{\rm S}/T \ = \ $ { 0.5 3% }		$\text{for the signal}\ \ s_{\rm C}(t) \text{:} \ \ \Delta t_{\rm S}/T \ = \ $ { 0.5 3% }

	{What is the energy of the rectangular basic transmission pulse??		{What is the energy of the rectangular basic transmission pulse  $g_s(t)$?
	\|type="{}"}		\|type="{}"}
	$E_g \ = \ $ { 0.5 } $\ \cdot 10^{-6}\ \rm Ws$		$E_g \ = \ $ { 0.5 } $\ \cdot 10^{-6}\ \rm Ws$
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	$P_{\rm S} \ = \ $ { 0.25 3% } $\ \rm W$		$P_{\rm S} \ = \ $ { 0.25 3% } $\ \rm W$

	{What is the energy of the  $~~\cos^2~~$ ~~basic transmission pulse~~?		{What is the energy of the cosine–square basic transmission pulse  $g_s(t)$?
	\|type="{}"}		\|type="{}"}
	$E_g \ = \ $ { 0.1875 3% } $\ \cdot 10^{-6}\ \rm Ws$		$E_g \ = \ $ { 0.1875 3% } $\ \cdot 10^{-6}\ \rm Ws$

Guenter: Guenter moved page Aufgaben:Exercise 1.1: Basic Transmitter Pulses to Aufgaben:Exercise 1.1: Basic Transmission Pulses over redirect

2022-04-27T11:44:25Z

Guenter moved page Aufgaben:Exercise 1.1: Basic Transmitter Pulses to Aufgaben:Exercise 1.1: Basic Transmission Pulses over redirect

← Older revision	Revision as of 13:44, 27 April 2022
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← Older revision		Revision as of 17:56, 16 March 2026
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	In particular,  the following characteristics are to be calculated for the respective basic transmission pulses  $g_s(t)$:		In particular,  the following characteristics are to be calculated for the respective basic transmission pulses  $g_s(t)$:
	*the equivalent pulse duration of  $g_s(t)$:		*the equivalent pulse duration of  $g_s(t)$:
	:$$\Delta t_{\rm S} = \frac {\int ^{+\infty} _{-\infty} \hspace{0.15cm} g_s(t)\,{\~~rmd~~}t}{{\rm Max} \hspace{0.05cm}[g_s(t)]} \hspace{0.05cm},$$		:$$\Delta t_{\rm S} = \frac {\int ^{+\infty} _{-\infty} \hspace{0.15cm} g_s(t)\,{\rm d}t}{{\rm Max} \hspace{0.05cm}[g_s(t)]} \hspace{0.05cm},$$
	*the energy of  $g_s(t)$:		*the energy of  $g_s(t)$:
	:$$E_g = \int^{+\infty} _{-\infty} g_s^2(t)\,{\~~rmd~~}t \hspace{0.05cm},$$		:$$E_g = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm d}t \hspace{0.05cm},$$
	*the power of the transmitted signal  $s(t)$:		*the power of the transmitted signal  $s(t)$:
	:$$P_{\rm S} = \lim_{T_{\rm M} \to \infty} \frac{1}{+T_{\rm M}} \cdot \int^{+T_{\rm M}/2} _{-T_{\rm M}/2} s^2(t)\,{\~~rmd~~}t \hspace{0.05cm}.$$		:$$P_{\rm S} = \lim_{T_{\rm M} \to \infty} \frac{1}{+T_{\rm M}} \cdot \int^{+T_{\rm M}/2} _{-T_{\rm M}/2} s^2(t)\,{\rm d}t \hspace{0.05cm}.$$


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	'''(3)'''  For the energy of the rectangular pulse holds:		'''(3)'''  For the energy of the rectangular pulse holds:
	:$$E_g = \int^{+\infty} _{-\infty} g_s^2(t)\,{\~~rmd~~}t = s_0^2 \cdot T = 0.5\, {\rm W} \cdot 1\, {\rm µ s} \hspace{0.1cm}\underline{= 0.5 \cdot 10^{-6}\, {\~~rmWs~~}}\hspace{0.05cm}.$$		:$$E_g = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm d}t = s_0^2 \cdot T = 0.5\, {\rm W} \cdot 1\, {\rm µ s} \hspace{0.1cm}\underline{= 0.5 \cdot 10^{-6}\, {\rm Ws}}\hspace{0.05cm}.$$


	'''(4)'''  For a bipolar rectangular signal,  the following would apply:		'''(4)'''  For a bipolar rectangular signal,  the following would apply:
	:$$s_{\rm R}^2(t)= s_0^2 = {\rm const.} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} P_s = s_0^2 \cdot\lim_{T_{\rm M} \to \infty} \frac{1}{T_{\rm M}} \cdot \int ^{T_{\rm M}/2} _{-T_{\rm M}/2} \,{\~~rmd~~}t = s_0^2 \hspace{0.05cm}.$$		:$$s_{\rm R}^2(t)= s_0^2 = {\rm const.} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} P_s = s_0^2 \cdot\lim_{T_{\rm M} \to \infty} \frac{1}{T_{\rm M}} \cdot \int ^{T_{\rm M}/2} _{-T_{\rm M}/2} \,{\rm d}t = s_0^2 \hspace{0.05cm}.$$
	*However,  since the signal  $s_{\rm R}(t)$ is  unipolar here,  in half the time $s_{\rm R}(t)= 0$.  Thus,  we get:		*However,  since the signal  $s_{\rm R}(t)$ is  unipolar here,  in half the time $s_{\rm R}(t)= 0$.  Thus,  we get:
	:$$P_{\rm S} = {1}/{2} \cdot s_0^2 \hspace{0.1cm}\underline{= 0.25 \,{\~~rmW~~}} \hspace{0.05cm}.$$		:$$P_{\rm S} = {1}/{2} \cdot s_0^2 \hspace{0.1cm}\underline{= 0.25 \,{\rm W}} \hspace{0.05cm}.$$


	'''(5)'''  For the energy of the cosine–square pulse holds:		'''(5)'''  For the energy of the cosine–square pulse holds:
	:$$E_g = \int^{+\infty} _{-\infty} g_s^2(t)\,{\~~rmd~~}t = 2 \cdot s_0^2 \cdot \int^{T/2} _{0} \cos^4(\pi \cdot {t}/{T})\,{\~~rmd~~}t \hspace{0.05cm}.$$		:$$E_g = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm d}t = 2 \cdot s_0^2 \cdot \int^{T/2} _{0} \cos^4(\pi \cdot {t}/{T})\,{\rm d}t \hspace{0.05cm}.$$
	*Here,  the formula derived in subtask  '''(3)'''  and the symmetry of  $g_s(t)$  about time  $t = 0$  are considered.		*Here,  the formula derived in subtask  '''(3)'''  and the symmetry of  $g_s(t)$  about time  $t = 0$  are considered.
	*The integral is given in the task description,  where $a = π/T$  is to be set:		*The integral is given in the task description,  where $a = π/T$  is to be set:
	:$$E_g = 2 \cdot s_0^2 \cdot \left [ \frac{3}{8} \cdot t + \frac{T}{4\pi} \cdot \sin(2 \pi \frac{t}{T})+ \frac{T}{32\pi} \cdot\sin(4 \pi \frac{t}{T})\right ]_{0}^{T/2}\hspace{0.05cm}.$$		:$$E_g = 2 \cdot s_0^2 \cdot \left [ \frac{3}{8} \cdot t + \frac{T}{4\pi} \cdot \sin(2 \pi \frac{t}{T})+ \frac{T}{32\pi} \cdot\sin(4 \pi \frac{t}{T})\right ]_{0}^{T/2}\hspace{0.05cm}.$$
	*The lower bound  $t = 0$  always yields the result  $0$.   With respect to the upper bound,  only the first term yields a result different from  $0$.  Thus:		*The lower bound  $t = 0$  always yields the result  $0$.   With respect to the upper bound,  only the first term yields a result different from  $0$.  Thus:
	:$$E_g = 2 \cdot s_0^2 \cdot \frac{3}{8} \cdot \frac{T}{2} = \frac{3}{8} \cdot 5 \cdot 10^{-7}\, {\~~rmWs~~} \hspace{0.1cm}\underline{ = 0.1875 \cdot 10^{-6}\, {\~~rmWs~~}}\hspace{0.05cm}.$$		:$$E_g = 2 \cdot s_0^2 \cdot \frac{3}{8} \cdot \frac{T}{2} = \frac{3}{8} \cdot 5 \cdot 10^{-7}\, {\rm Ws} \hspace{0.1cm}\underline{ = 0.1875 \cdot 10^{-6}\, {\rm Ws}}\hspace{0.05cm}.$$


	'''(6)'''  The following relationship holds for the bipolar signal $s_{\rm C}(t)$:		'''(6)'''  The following relationship holds for the bipolar signal $s_{\rm C}(t)$:
	:$$P_{\rm S} = \frac{ E_g}{T} = \frac{ 1.875 \cdot 10^{-7}\, {\~~rmWs~~}}{10^{-6}\, {\rm s}}\hspace{0.1cm}\underline{ = 0.1875 \,{\rm W}} \hspace{0.05cm}.$$		:$$P_{\rm S} = \frac{ E_g}{T} = \frac{ 1.875 \cdot 10^{-7}\, {\rm Ws}}{10^{-6}\, {\rm s}}\hspace{0.1cm}\underline{ = 0.1875 \,{\rm W}} \hspace{0.05cm}.$$


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	[[Category:Digital Signal Transmission: Exercises\|^1.1 Baseband System Components^]]		[[Category:Digital Signal Transmission: Exercises\|^1.1 Baseband System Components^]]
	[[de:Aufgaben:1.~~0_Test~~]]		[[de:Aufgaben:Aufgabe 1.1: Sendegrundimpulse]]