Aufgaben:Exercise 3.2: Laplace Transform - Revision history

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 * In the sample solution,&nbsp; we use of the two comparable functions&nbsp; ${\rm si}(x)$&nbsp; and&nbsp; ${\rm sinc}(x)$&nbsp; the former.
 *The following definite integrals are given:
-:$$\int_{0}^{
+:$$\int_{0}^{\infty}{  {\rm e}^{-p x} \cdot \cos(qx)}\hspace{0.1cm}{\rmd}x = \frac{p}{p^2 + q^2}\hspace{0.05cm} , \hspace{1.0cm}\int_{0}^{\infty}{  {\rm e}^{-p x} \cdot \sin(qx)}\hspace{0.1cm}{\rmd}x = \frac{q}{p^2 + q^2}\hspace{0.05cm} , $$:$$\int_{0}^{\infty}{  {\rm e}^{-p x} \cdot \frac{\sin(qx)}{x}}\hspace{0.1cm}{\rmd}x = {\rm arctan}\hspace{0.15cm}\frac{q}{p}\hspace{0.05cm} ,  \hspace{0.6cm}\int_{A}^{B}{  \frac{1}{x}}\hspace{0.1cm}{\rmd}x = {\rm ln}\hspace{0.15cm}\frac{B}{A}\hspace{0.05cm} .$$
@@ Line 97: / Line 84: @@
 '''(1)'''&nbsp; <u>Suggested solution 2</u>&nbsp; is correct:
 *According to the Laplace definition, the following holds with the given equations:
-:$$X_{\rm L}(p) =    \int_{0}^{
+:$$X_{\rm L}(p) =    \int_{0}^{\infty}{ x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\rmd}t = \int\limits_{0}^{\infty}{ {\rm cos} (\omega_0 \cdot T) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\rmd}t = \frac{p}{p^2 + \omega_0^2}\hspace{0.05cm} .$$
 *Suggestion 3 is ruled out since&nbsp; $X_{\rm L}(p)$&nbsp; must have the unit "second" (integral over time) while&nbsp; $p$&nbsp; and&nbsp; $\omega_0$&nbsp; each have the unit "1/s".
@@ Line 130: / Line 110: @@
 *The real part of&nbsp; $Z(f)$&nbsp; thus has the same rectangular shape as&nbsp; $S(f)$,&nbsp; but it is only half as high:
-:$${\rm Re}\{ Z(f)\}= \left\{ \begin{array}{c} T/2 \\
+:$${\rm Re}\{ Z(f)\}= \left\{ \begin{array}{c} T/2 \\0   \end{array} \right.\begin{array}{c}   {\rm{f\ddot{u}r}}  \\ {\rm{f\ddot{u}r}}\end{array}\begin{array}{*{20}c}{  |f|< 1/(2T)\hspace{0.05cm},}  \\{ |f|> 1/(2T)\hspace{0.05cm},}\end{array}\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\underline{{\rm suggestion \hspace{0.15cm} 1}}.$$

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← Older revision		Revision as of 15:48, 16 March 2026
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	[[File:P_ID1763__LZI_A_3_2.png\|right\|frame\|Three causal time functions]]		[[File:P_ID1763__LZI_A_3_2.png\|right\|frame\|Three causal time functions]]
	Causal signals and systems are usually described by means of the Laplace transformation.  If  $x(t)$  is identical to zero for all times  $t < 0$,  then the Laplace transform is:		Causal signals and systems are usually described by means of the Laplace transformation.  If  $x(t)$  is identical to zero for all times  $t < 0$,  then the Laplace transform is:
	:$$X_{\rm L}(p) = \int_{0}^{		:$$X_{\rm L}(p) = \int_{0}^{\infty}{ x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\rmd}t\hspace{0.05cm}\hspace{0.05cm} .$$
	\infty}
	{ x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\rm
	d}t\hspace{0.05cm}\hspace{0.05cm} .$$

	In this exercise, the Laplace transforms of the causal signals shown in the graph are to be determined.  The following equations are valid in each case only for  $t \ge 0$.  For negative times, all signals are identical to zero.		In this exercise, the Laplace transforms of the causal signals shown in the graph are to be determined.  The following equations are valid in each case only for  $t \ge 0$.  For negative times, all signals are identical to zero.
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	The following equation can'''<u>not</u>''' be used to calculate the spectral function since  $z(t)$  is not energy-limited just as the signals  $x(t)$  and  $y(t)$ :		The following equation can'''<u>not</u>''' be used to calculate the spectral function since  $z(t)$  is not energy-limited just as the signals  $x(t)$  and  $y(t)$ :
	:$$Z(f) = Z_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it		:$$Z(f) = Z_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \itf}} .$$
	f}} .$$

	Rather, the fact that  $z(t) = s(t) \cdot \gamma(t)$  holds is to be considered where  $s(t)$  denotes the conventional symmetric  $\rm si$–function here:		Rather, the fact that  $z(t) = s(t) \cdot \gamma(t)$  holds is to be considered where  $s(t)$  denotes the conventional symmetric  $\rm si$–function here:
	:$$s(t) = {\rm si} (\pi \cdot {t}/{T}) \quad		:$$s(t) = {\rm si} (\pi \cdot {t}/{T}) \quad\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad S(f).$$
	\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad S(f).$$

	$S(f)$  is a rectangular function symmetric about  $f = 0$  with height  $T$  and width  $1/T$.		$S(f)$  is a rectangular function symmetric about  $f = 0$  with height  $T$  and width  $1/T$.

	The Fourier transform of the step function  $\gamma(t)$  is:		The Fourier transform of the step function  $\gamma(t)$  is:
	:$$\gamma(t) \quad		:$$\gamma(t) \quad\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad {\it \Gamma}(f) = {1}/{2}\cdot \delta (f) + \frac{1}{{\rm j} \cdot 2\pi f}\hspace{0.05cm}.$$
	\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad {\it \Gamma}(f) = {1}/{2}
	\cdot \delta (f) + \frac{1}{{\rm j} \cdot 2\pi f}\hspace{0.05cm}.$$


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	\infty}		\infty}
	{ {\rm e}^{-p x} \cdot \sin(qx)}\hspace{0.1cm}{\rm		{ {\rm e}^{-p x} \cdot \sin(qx)}\hspace{0.1cm}{\rm
	d}x = \frac{q}{p^2 + q^2}\hspace{0.05cm} , $$		d}x = \frac{q}{p^2 + q^2}\hspace{0.05cm} , $$:$$\int_{0}^{
	:$$\int_{0}^{
	\infty}		\infty}
	{ {\rm e}^{-p x} \cdot \frac{\sin(qx)}{x}}\hspace{0.1cm}{\rm		{ {\rm e}^{-p x} \cdot \frac{\sin(qx)}{x}}\hspace{0.1cm}{\rm
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	'''(2)'''  <u>Suggested solution 1</u>  is correct:		'''(2)'''  <u>Suggested solution 1</u>  is correct:
	*Here, the following holds using the same approach as in the subtask  '''(1)''':		*Here, the following holds using the same approach as in the subtask  '''(1)''':
	:$$Y_{\rm L}(p) = \int_{0}^{		:$$Y_{\rm L}(p) = \int_{0}^{\infty}{ {\rm sin} (\omega_0 \cdot T) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\rmd}t = \frac{\omega_0}{p^2 + \omega_0^2}\hspace{0.05cm} .$$
	\infty}
	{ {\rm sin} (\omega_0 \cdot T) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\rm
	d}t = \frac{\omega_0}{p^2 + \omega_0^2}
	\hspace{0.05cm} .$$


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	'''(3)'''  <u>Suggested solution 3</u>  is correct:		'''(3)'''  <u>Suggested solution 3</u>  is correct:
	*The  $p$–transfer function of the causal  $\rm si$–function is as follows considering the integral given above:		*The  $p$–transfer function of the causal  $\rm si$–function is as follows considering the integral given above:
	:$$Z_{\rm L}(p) = \int_{0}^{		:$$Z_{\rm L}(p) = \int_{0}^{\infty}{ \frac{\sin(\pi \cdot t/T)}{\pi \cdot t/T} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\rmd}t = \frac{T}{\pi} \cdot {\rm arctan} \; \frac{\pi}{p\cdot T}\hspace{0.05cm} .$$
	\infty}
	{ \frac{\sin(\pi \cdot t/T)}{\pi \cdot t/T} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\rm
	d}t = \frac{T}{\pi} \cdot {\rm arctan} \; \frac{\pi}{p\cdot T}
	\hspace{0.05cm} .$$
	*Suggestion 1 only applies to the Fourier transform of the non-causal  $\rm si$–function.		*Suggestion 1 only applies to the Fourier transform of the non-causal  $\rm si$–function.
	*Since here the argument of the  $\rm arctan$  function is dimensional, suggestion 2 cannot be true for this reason alone.		*Since here the argument of the  $\rm arctan$  function is dimensional, suggestion 2 cannot be true for this reason alone.
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	'''(4)'''  <u>Suggested solution 1</u>  is correct:		'''(4)'''  <u>Suggested solution 1</u>  is correct:
	*The following arises as a result from  $z(t) = s(t) \cdot \gamma(t)$  with the convolution theorem:		*The following arises as a result from  $z(t) = s(t) \cdot \gamma(t)$  with the convolution theorem:
	:$$Z(f) = S(f) \star {\it \Gamma}(f) = {1}/{2}		:$$Z(f) = S(f) \star {\it \Gamma}(f) = {1}/{2}\cdot S(f) \star \delta (f) + S(f) \star \frac{1}{{\rm j} \cdot2\pi f}\hspace{0.05cm}.$$
	\cdot S(f) \star \delta (f) + S(f) \star \frac{1}{{\rm j} \~~cdot~~
	2\pi f}\hspace{0.05cm}.$$
	*Since  $S(f)$  is real, the real part of  $Z(f)$  is obtained as the first term of this equation:		*Since  $S(f)$  is real, the real part of  $Z(f)$  is obtained as the first term of this equation:
	:$${\rm Re}[ Z(f)] = {1}/{2}		:$${\rm Re}[ Z(f)] = {1}/{2}\cdot S(f) \star \delta (f) = {1}/{2} \cdot S(f)\hspace{0.05cm}.$$
	\cdot S(f) \star \delta (f) = {1}/{2} \cdot S(f)
	\hspace{0.05cm}.$$

	*The real part of  $Z(f)$  thus has the same rectangular shape as  $S(f)$,  but it is only half as high:		*The real part of  $Z(f)$  thus has the same rectangular shape as  $S(f)$,  but it is only half as high:
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	'''(5)'''  <u>Suggested solution 2</u>  is correct:		'''(5)'''  <u>Suggested solution 2</u>  is correct:
	*With the result of the last subtask,  it follows for the imaginary part:		*With the result of the last subtask,  it follows for the imaginary part:
	:$${\rm Im}\{ Z(f)\} = S(f) \star \frac{(-1)}{{\rm j} \~~cdot~~		:$${\rm Im}\{ Z(f)\} = S(f) \star \frac{(-1)}{{\rm j} \cdot2\pi f} \hspace{0.05cm}.$$
	2\pi f} \hspace{0.05cm}.$$

	*This convolution integral yields the following for sufficiently large frequencies  $f \ge 1/(2T)$:		*This convolution integral yields the following for sufficiently large frequencies  $f \ge 1/(2T)$:
	:$${\rm Im}\{ Z(f)\} = -T \cdot \int_{f- 1/(2T)}^{		:$${\rm Im}\{ Z(f)\} = -T \cdot \int_{f- 1/(2T)}^{f+ 1/(2T)} { \frac{1}{2\pi x}}\hspace{0.1cm}{\rmd}x = \frac{T}{2\pi } \cdot {\rm ln}\hspace{0.15cm}\left \|\frac{f- 1/(2T)}{f+ 1/(2T)}\right \|\hspace{0.05cm}\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\underline{{\rm suggestion \hspace{0.15cm} 2}}.$$
	f+ 1/(2T)} { \frac{1}{2\pi x}}\hspace{0.1cm}{\rm
	d}x = \frac{T}{2\pi } \cdot {\rm ln}\hspace{0.15cm}\left \|\frac{f- 1/(2T)}{f+ 1/(2T)}\right \|
	\hspace{0.05cm}
	\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\underline{{\rm suggestion \hspace{0.15cm} 2}}.$$

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← Older revision		Revision as of 14:32, 16 March 2026
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	[[Category:Linear and Time-Invariant Systems: Exercises\|^3.2 Laplace Transform and p-Transfer Function^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^3.2 Laplace Transform and p-Transfer Function^]]
			[[de:Aufgaben:3.2_Laplace-Transformation]]

Guenter at 11:44, 13 October 2021

2021-10-13T11:44:17Z

← Older revision		Revision as of 13:44, 13 October 2021
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Suggested solution 2</u> is correct:		'''(1)'''  <u>Suggested solution 2</u>  is correct:
	*According to the Laplace~~–~~definition, the following holds with the given equations:		*According to the Laplace definition, the following holds with the given equations:
	:$$X_{\rm L}(p) = \int_{0}^{		:$$X_{\rm L}(p) = \int_{0}^{
	\infty}		\infty}
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	'''(2)'''  <u>Suggested solution 1</u> is correct:		'''(2)'''  <u>Suggested solution 1</u>  is correct:
	*Here, the following holds using the same approach as in the subtask  '''(1)''':		*Here, the following holds using the same approach as in the subtask  '''(1)''':
	:$$Y_{\rm L}(p) = \int_{0}^{		:$$Y_{\rm L}(p) = \int_{0}^{
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	'''(3)'''  <u>Suggested solution 3</u> is correct:		'''(3)'''  <u>Suggested solution 3</u>  is correct:
	*The  $p$–transfer function of the causal  $\rm si$–function is as follows considering the integral given above:		*The  $p$–transfer function of the causal  $\rm si$–function is as follows considering the integral given above:
	:$$Z_{\rm L}(p) = \int_{0}^{		:$$Z_{\rm L}(p) = \int_{0}^{
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	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$
	*Suggestion 1 only applies to the Fourier transform of the non-causal  $\rm si$–function.		*Suggestion 1 only applies to the Fourier transform of the non-causal  $\rm si$–function.
	*Since here the argument of the ~~arctangent~~&~~ndash~~;function is dimensional, suggestion 2 cannot be true for this reason alone.		*Since here the argument of the  $\rm arctan$  function is dimensional, suggestion 2 cannot be true for this reason alone.



	'''(4)'''  <u>Suggested solution 1</u> is correct:		'''(4)'''  <u>Suggested solution 1</u>  is correct:
	*The following arises as a result from  $z(t) = s(t) \cdot \gamma(t)$  with the convolution theorem:		*The following arises as a result from  $z(t) = s(t) \cdot \gamma(t)$  with the convolution theorem:
	:$$Z(f) = S(f) \star {\it \Gamma}(f) = {1}/{2}		:$$Z(f) = S(f) \star {\it \Gamma}(f) = {1}/{2}
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*The real part of  $Z(f)$  thus has the same rectangular shape as  $S(f)$, but it is only half as high:		*The real part of  $Z(f)$  thus has the same rectangular shape as  $S(f)$,  but it is only half as high:
	:$${\rm Re}\{ Z(f)\}= \left\{ \begin{array}{c} T/2 \\		:$${\rm Re}\{ Z(f)\}= \left\{ \begin{array}{c} T/2 \\
	0 \end{array} \right.		0 \end{array} \right.
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	'''(5)'''  <u>Suggested solution 2</u> is correct:		'''(5)'''  <u>Suggested solution 2</u>  is correct:
	*With the result of the last subtask, it follows for the imaginary part:		*With the result of the last subtask,  it follows for the imaginary part:
	:$${\rm Im}\{ Z(f)\} = S(f) \star \frac{(-1)}{{\rm j} \cdot		:$${\rm Im}\{ Z(f)\} = S(f) \star \frac{(-1)}{{\rm j} \cdot
	2\pi f} \hspace{0.05cm}.$$		2\pi f} \hspace{0.05cm}.$$

	*This convolution integral yields the following for sufficiently large frequencies  $f \ge 1/(2T)$~~ ~~:		*This convolution integral yields the following for sufficiently large frequencies  $f \ge 1/(2T)$:
	:$${\rm Im}\{ Z(f)\} = -T \cdot \int_{f- 1/(2T)}^{		:$${\rm Im}\{ Z(f)\} = -T \cdot \int_{f- 1/(2T)}^{
	f+ 1/(2T)} { \frac{1}{2\pi x}}\hspace{0.1cm}{\rm		f+ 1/(2T)} { \frac{1}{2\pi x}}\hspace{0.1cm}{\rm

Guenter at 11:38, 13 October 2021

2021-10-13T11:38:08Z

← Older revision		Revision as of 13:38, 13 October 2021
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	[[File:P_ID1763__LZI_A_3_2.png\|right\|frame\|Three causal time functions]]		[[File:P_ID1763__LZI_A_3_2.png\|right\|frame\|Three causal time functions]]
	Causal signals and systems are usually described by means of the Laplace transformation. If  $x(t)$  is identical to zero for all times  $t < 0$,  then the Laplace transform is:		Causal signals and systems are usually described by means of the Laplace transformation.  If  $x(t)$  is identical to zero for all times  $t < 0$,  then the Laplace transform is:
	:$$X_{\rm L}(p) = \int_{0}^{		:$$X_{\rm L}(p) = \int_{0}^{
	\infty}		\infty}
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	*$\sin(t)/t$–signal with equivalent zero-crossings at a distance of  $T$:		*$\sin(t)/t$–signal with equivalent zero-crossings at a distance of  $T$:
	:$$z(t) = {\rm si} (\pi \cdot {t}/{T})\hspace{0.~~2cm~~}{\rm with}\hspace{0.~~2cm~~}{\rm si}(x)= {\rm sin}(x)/x		:$$z(t) = {\rm si} (\pi \cdot {t}/{T})= {\rm sinc} ({t}/{T})\hspace{0.4cm}{\rm with}\hspace{0.4cm}{\rm si}(x)= {\rm sin}(x)/x ={\rm sinc}(x)/\pi \hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	The following equation can<u>not</u> be used to calculate the spectral function since  $z(t)$  is not energy-limited just as the signals  $x(t)$  and  $y(t)$ :		The following equation can'''<u>not</u>''' be used to calculate the spectral function since  $z(t)$  is not energy-limited just as the signals  $x(t)$  and  $y(t)$ :
	:$$Z(f) = Z_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it		:$$Z(f) = Z_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it
	f}} .$$		f}} .$$
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	\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad {\it \Gamma}(f) = {1}/{2}		\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad {\it \Gamma}(f) = {1}/{2}
	\cdot \delta (f) + \frac{1}{{\rm j} \cdot 2\pi f}\hspace{0.05cm}.$$		\cdot \delta (f) + \frac{1}{{\rm j} \cdot 2\pi f}\hspace{0.05cm}.$$








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	''Please note:''		''Please note:''
	*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function\|Laplace Transform and p-Transfer Function]].		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function\|Laplace Transform and p-Transfer Function]].
			* In the sample solution,  we use of the two comparable functions  ${\rm si}(x)$  and  ${\rm sinc}(x)$  the former.
	*The following definite integrals are given:		*The following definite integrals are given:
	:$$\int_{0}^{		:$$\int_{0}^{
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	{Compute the Laplace transform  $Z_{\rm L}(p)$  of the causal  $\rm si$–function  $z(t)$.  What is the correct solution?		{Compute the Laplace transform  $Z_{\rm L}(p)$  of the causal  $\rm si$–function  $z(t)$.  What is the correct solution?
	\|type="()"}		\|type="()"}
	- $Z_{\rm L}(p)$ has a rectangular shape.		- $Z_{\rm L}(p)$  has a rectangular shape.
	- $Z_{\rm L}(p) = \arctan (1/p)$.		- $Z_{\rm L}(p) = \arctan (1/p)$.
	+ $Z_{\rm L}(p) = T/\pi \cdot \arctan (\pi/(pT))$.		+ $Z_{\rm L}(p) = T/\pi \cdot \arctan (\pi/(pT))$.
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	{Compute the real part of the spectrum  $Z(f)$.  Which statements are true?		{Compute the real part of the spectrum  $Z(f)$.  Which statements are true?
	\|type="[]"}		\|type="[]"}
	+ ${\rm Re}\big[Z(f)\big]$ has a rectangular shape.		+ ${\rm Re}\big[Z(f)\big]$  has a rectangular shape.
	- ${\rm Re}\big[Z(f)\big]$ is proportional to $\ln\; \big\|(f \cdot T -0.5)/(f \cdot T +0.5)\big\|.$		- ${\rm Re}\big[Z(f)\big]$  is proportional to $\ln\; \big\|(f \cdot T -0.5)/(f \cdot T +0.5)\big\|.$

Oezer at 15:13, 10 October 2021

2021-10-10T15:13:28Z

← Older revision		Revision as of 17:13, 10 October 2021
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	{ \|f\|> 1/(2T)\hspace{0.05cm},}		{ \|f\|> 1/(2T)\hspace{0.05cm},}
	\end{array}		\end{array}
	\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\underline{{\rm ~~Vorschlag~~ \hspace{0.15cm} 1}}.$$		\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\underline{{\rm suggestion \hspace{0.15cm} 1}}.$$

Oezer at 15:07, 10 October 2021

2021-10-10T15:07:20Z

← Older revision		Revision as of 17:07, 10 October 2021
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	Rather, the fact that  $z(t) = s(t) \cdot \gamma(t)$  holds is to be considered where  $s(t)$  denotes the conventional symmetric  $\rm si$–function here:		Rather, the fact that  $z(t) = s(t) \cdot \gamma(t)$  holds is to be considered where  $s(t)$  denotes the conventional symmetric  $\rm si$–function here:
	:$$s(t) = {\rm si} (\pi \cdot {t}/{T}) \quad		:$$s(t) = {\rm si} (\pi \cdot {t}/{T}) \quad
	\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad S(f)$$.		\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad S(f).$$

	$S(f)$  is a rectangular function symmetric about  $f = 0$  with height  $T$  and width  $1/T$.		$S(f)$  is a rectangular function symmetric about  $f = 0$  with height  $T$  and width  $1/T$.

Oezer at 15:06, 10 October 2021

2021-10-10T15:06:58Z

← Older revision		Revision as of 17:06, 10 October 2021
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	Rather, the fact that  $z(t) = s(t) \cdot \gamma(t)$  holds is to be considered where  $s(t)$  denotes the conventional symmetric  $\rm si$–function here:		Rather, the fact that  $z(t) = s(t) \cdot \gamma(t)$  holds is to be considered where  $s(t)$  denotes the conventional symmetric  $\rm si$–function here:
	:$$s(t) = {\rm si} (\pi \cdot {t}/{T}) \quad		:$$s(t) = {\rm si} (\pi \cdot {t}/{T}) \quad
	\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad S(f)$$		\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad S(f)$$.

	$S(f)$  is a rectangular function symmetric about  $f = 0$  with height  $T$  and width  $1/T$.		$S(f)$  is a rectangular function symmetric about  $f = 0$  with height  $T$  and width  $1/T$.

Oezer at 15:03, 10 October 2021

2021-10-10T15:03:59Z

← Older revision		Revision as of 17:03, 10 October 2021
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	f}} .$$		f}} .$$

	Rather, the fact that  $z(t) = s(t) \cdot \gamma(t)$  ~~is valid~~ is to be considered where  $s(t)$  denotes the conventional symmetric  $\rm si$–function here:		Rather, the fact that  $z(t) = s(t) \cdot \gamma(t)$  holds is to be considered where  $s(t)$  denotes the conventional symmetric  $\rm si$–function here:
	:$$s(t) = {\rm si} (\pi \cdot {t}/{T}) \quad		:$$s(t) = {\rm si} (\pi \cdot {t}/{T}) \quad
	\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad S(f)$$		\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad S(f)$$
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	{~~Berechnen Sie den Realteil des Spektrums~~  $Z(f)$.  ~~Welche Aussagen treffen zu~~?		{Compute the real part of the spectrum  $Z(f)$.  Which statements are true?
	\|type="[]"}		\|type="[]"}
	+ ${\rm Re}\big[Z(f)\big]$ ~~hat einen rechteckförmigen Verlauf~~.		+ ${\rm Re}\big[Z(f)\big]$ has a rectangular shape.
	- ${\rm Re}\big[Z(f)\big]$ ~~ist~~ proportional zu $\ln\; \big\|(f \cdot T -0.5)/(f \cdot T +0.5)\big\|.$		- ${\rm Re}\big[Z(f)\big]$ is proportional to $\ln\; \big\|(f \cdot T -0.5)/(f \cdot T +0.5)\big\|.$


	{~~Berechnen Sie den Imaginärteil von~~  $Z(f)$.  ~~Welche Aussagen treffen zu~~?		{Compute the imaginary part of  $Z(f)$.  Which statements are true?
	\|type="[]"}		\|type="[]"}
	- ${\rm Im}\big[Z(f)\big]$  ~~hat einen rechteckförmigen Verlauf~~.		- ${\rm Im}\big[Z(f)\big]$  has a rectangular shape.
	+ ${\rm Im}\big[Z(f)\big]$  ~~ist~~ proportional zu  $\ln\; \big\|(f \cdot T -0.5)/(f \cdot T +0.5)\big\|.$		+ ${\rm Im}\big[Z(f)\big]$  is proportional to  $\ln\; \big\|(f \cdot T -0.5)/(f \cdot T +0.5)\big\|.$


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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Richtig ist der~~ <u>~~Vorschlag~~ 2</u>:		'''(1)'''  <u>Suggested solution 2</u> is correct:
	*~~Entsprechend der~~ Laplace–~~Definition gilt mit den vorgegebenen Gleichungen~~:		*According to the Laplace–definition, the following holds with the given equations:
	:$$X_{\rm L}(p) = \int_{0}^{		:$$X_{\rm L}(p) = \int_{0}^{
	\infty}		\infty}
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	d}t = \frac{p}{p^2 + \omega_0^2}		d}t = \frac{p}{p^2 + \omega_0^2}
	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$
	*~~Der Vorschlag~~ 3 ~~scheidet aus, da~~  $X_{\rm L}(p)$  ~~die Einheit~~ "~~Sekunde~~" ~~aufweisen muss~~ (~~Integral über die Zeit~~)~~, während~~  $p$  ~~und~~  $\omega_0$  ~~jeweils die Einheit~~ "1/s" ~~besitzen~~.		*Suggestion 3 is ruled out since  $X_{\rm L}(p)$  must have the unit "second" (integral over time) while  $p$  and  $\omega_0$  each have the unit "1/s".



	'''(2)'''  ~~Richtig ist der~~ <u>~~Lösungsvorschlag~~ 1</u>:		'''(2)'''  <u>Suggested solution 1</u> is correct:
	*~~Hier gilt bei gleicher Vorgehensweise wie~~ in ~~der Teilaufgabe~~  '''(1)''':		*Here, the following holds using the same approach as in the subtask  '''(1)''':
	:$$Y_{\rm L}(p) = \int_{0}^{		:$$Y_{\rm L}(p) = \int_{0}^{
	\infty}		\infty}
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	'''(3)'''  ~~Richtig ist der~~ <u>~~Lösungsvorschlag~~ 3</u>:		'''(3)'''  <u>Suggested solution 3</u> is correct:
	*~~Die~~  $p$–~~Übertragungsfunktion der kausalen~~  $\rm si$–~~Funktion lautet mit dem vorne angegebenen Integral~~:		*The  $p$–transfer function of the causal  $\rm si$–function is as follows considering the integral given above:
	:$$Z_{\rm L}(p) = \int_{0}^{		:$$Z_{\rm L}(p) = \int_{0}^{
	\infty}		\infty}
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	d}t = \frac{T}{\pi} \cdot {\rm arctan} \; \frac{\pi}{p\cdot T}		d}t = \frac{T}{\pi} \cdot {\rm arctan} \; \frac{\pi}{p\cdot T}
	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$
	*~~Der Vorschlag~~ 1 ~~gilt nur für die Fouriertransformierte der akausalen~~  $\rm si$–~~Funktion~~.		*Suggestion 1 only applies to the Fourier transform of the non-causal  $\rm si$–function.
	*~~Der Vorschlag 2 kann schon allein deshalb nicht stimmen, da hier das Argument der Arcustangens~~–~~Funktion dimensionsbehaftet ist~~.		*Since here the argument of the arctangent–function is dimensional, suggestion 2 cannot be true for this reason alone.



	'''(4)'''  ~~Richtig ist der~~ <u>~~Lösungsvorschlag~~ 1</u>:		'''(4)'''  <u>Suggested solution 1</u> is correct:
	*~~Aus~~  $z(t) = s(t) \cdot \gamma(t)$  ~~folgt mit dem Faltungssatz~~:		*The following arises as a result from  $z(t) = s(t) \cdot \gamma(t)$  with the convolution theorem:
	:$$Z(f) = S(f) \star {\it \Gamma}(f) = {1}/{2}		:$$Z(f) = S(f) \star {\it \Gamma}(f) = {1}/{2}
	\cdot S(f) \star \delta (f) + S(f) \star \frac{1}{{\rm j} \cdot		\cdot S(f) \star \delta (f) + S(f) \star \frac{1}{{\rm j} \cdot
	2\pi f}\hspace{0.05cm}.$$		2\pi f}\hspace{0.05cm}.$$
	*Da  $S(f)$  ~~reell ist~~, ~~ergibt sich der Realteil von~~  $Z(f)$  ~~als der erste Term dieser Gleichung~~:		*Since  $S(f)$  is real, the real part of  $Z(f)$  is obtained as the first term of this equation:
	:$${\rm Re}[ Z(f)] = {1}/{2}		:$${\rm Re}[ Z(f)] = {1}/{2}
	\cdot S(f) \star \delta (f) = {1}/{2} \cdot S(f)		\cdot S(f) \star \delta (f) = {1}/{2} \cdot S(f)
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*~~Der Realteil von~~  $Z(f)$  ~~hat somit die gleiche Rechteckform wie~~  $S(f)$, ~~ist aber nur halb so hoch~~:		*The real part of  $Z(f)$  thus has the same rectangular shape as  $S(f)$, but it is only half as high:
	:$${\rm Re}\{ Z(f)\}= \left\{ \begin{array}{c} T/2 \\		:$${\rm Re}\{ Z(f)\}= \left\{ \begin{array}{c} T/2 \\
	0 \end{array} \right.		0 \end{array} \right.
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	'''(5)'''  ~~Richtig ist der~~ <u>~~Lösungsvorschlag~~ 2</u>:		'''(5)'''  <u>Suggested solution 2</u> is correct:
	*~~Mit dem Ergebnis der letzten Teilaufgabe folgt für den Imaginärteil~~:		*With the result of the last subtask, it follows for the imaginary part:
	:$${\rm Im}\{ Z(f)\} = S(f) \star \frac{(-1)}{{\rm j} \cdot		:$${\rm Im}\{ Z(f)\} = S(f) \star \frac{(-1)}{{\rm j} \cdot
	2\pi f} \hspace{0.05cm}.$$		2\pi f} \hspace{0.05cm}.$$

	*~~Für hinreichend große Frequenzen~~  $f \ge 1/(2T)$  ~~liefert dieses Faltungsintegral~~:		*This convolution integral yields the following for sufficiently large frequencies  $f \ge 1/(2T)$ :
	:$${\rm Im}\{ Z(f)\} = -T \cdot \int_{f- 1/(2T)}^{		:$${\rm Im}\{ Z(f)\} = -T \cdot \int_{f- 1/(2T)}^{
	f+ 1/(2T)} { \frac{1}{2\pi x}}\hspace{0.1cm}{\rm		f+ 1/(2T)} { \frac{1}{2\pi x}}\hspace{0.1cm}{\rm
	d}x = \frac{T}{2\pi } \cdot {\rm ln}\hspace{0.15cm}\left \|\frac{f- 1/(2T)}{f+ 1/(2T)}\right \|		d}x = \frac{T}{2\pi } \cdot {\rm ln}\hspace{0.15cm}\left \|\frac{f- 1/(2T)}{f+ 1/(2T)}\right \|
	\hspace{0.05cm}		\hspace{0.05cm}
	\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\underline{{\rm ~~Vorschlag~~ \hspace{0.15cm} 2}}.$$		\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\underline{{\rm suggestion \hspace{0.15cm} 2}}.$$

← Older revision		Revision as of 17:57, 16 March 2026
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	[[File:P_ID1763__LZI_A_3_2.png\|right\|frame\|Three causal time functions]]		[[File:P_ID1763__LZI_A_3_2.png\|right\|frame\|Three causal time functions]]
	Causal signals and systems are usually described by means of the Laplace transformation.  If  $x(t)$  is identical to zero for all times  $t < 0$,  then the Laplace transform is:		Causal signals and systems are usually described by means of the Laplace transformation.  If  $x(t)$  is identical to zero for all times  $t < 0$,  then the Laplace transform is:
	:$$X_{\rm L}(p) = \int_{0}^{\infty}{ x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\~~rmd~~}t\hspace{0.05cm}\hspace{0.05cm} .$$		:$$X_{\rm L}(p) = \int_{0}^{\infty}{ x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\rm d}t\hspace{0.05cm}\hspace{0.05cm} .$$

	In this exercise, the Laplace transforms of the causal signals shown in the graph are to be determined.  The following equations are valid in each case only for  $t \ge 0$.  For negative times, all signals are identical to zero.		In this exercise, the Laplace transforms of the causal signals shown in the graph are to be determined.  The following equations are valid in each case only for  $t \ge 0$.  For negative times, all signals are identical to zero.
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	The following equation can'''<u>not</u>''' be used to calculate the spectral function since  $z(t)$  is not energy-limited just as the signals  $x(t)$  and  $y(t)$ :		The following equation can'''<u>not</u>''' be used to calculate the spectral function since  $z(t)$  is not energy-limited just as the signals  $x(t)$  and  $y(t)$ :
	:$$Z(f) = Z_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \~~itf~~}} .$$		:$$Z(f) = Z_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it f}} .$$

	Rather, the fact that  $z(t) = s(t) \cdot \gamma(t)$  holds is to be considered where  $s(t)$  denotes the conventional symmetric  $\rm si$–function here:		Rather, the fact that  $z(t) = s(t) \cdot \gamma(t)$  holds is to be considered where  $s(t)$  denotes the conventional symmetric  $\rm si$–function here:
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	* In the sample solution,  we use of the two comparable functions  ${\rm si}(x)$  and  ${\rm sinc}(x)$  the former.		* In the sample solution,  we use of the two comparable functions  ${\rm si}(x)$  and  ${\rm sinc}(x)$  the former.
	*The following definite integrals are given:		*The following definite integrals are given:
	:$$\int_{0}^{\infty}{ {\rm e}^{-p x} \cdot \cos(qx)}\hspace{0.1cm}{\~~rmd~~}x = \frac{p}{p^2 + q^2}\hspace{0.05cm} , \hspace{1.0cm}\int_{0}^{\infty}{ {\rm e}^{-p x} \cdot \sin(qx)}\hspace{0.1cm}{\~~rmd~~}x = \frac{q}{p^2 + q^2}\hspace{0.05cm} , $$:$$\int_{0}^{\infty}{ {\rm e}^{-p x} \cdot \frac{\sin(qx)}{x}}\hspace{0.1cm}{\~~rmd~~}x = {\rm arctan}\hspace{0.15cm}\frac{q}{p}\hspace{0.05cm} , \hspace{0.6cm}\int_{A}^{B}{ \frac{1}{x}}\hspace{0.1cm}{\~~rmd~~}x = {\rm ln}\hspace{0.15cm}\frac{B}{A}\hspace{0.05cm} .$$		:$$\int_{0}^{\infty}{ {\rm e}^{-p x} \cdot \cos(qx)}\hspace{0.1cm}{\rm d}x = \frac{p}{p^2 + q^2}\hspace{0.05cm} , \hspace{1.0cm}\int_{0}^{\infty}{ {\rm e}^{-p x} \cdot \sin(qx)}\hspace{0.1cm}{\rm d}x = \frac{q}{p^2 + q^2}\hspace{0.05cm} , $$:$$\int_{0}^{\infty}{ {\rm e}^{-p x} \cdot \frac{\sin(qx)}{x}}\hspace{0.1cm}{\rm d}x = {\rm arctan}\hspace{0.15cm}\frac{q}{p}\hspace{0.05cm} , \hspace{0.6cm}\int_{A}^{B}{ \frac{1}{x}}\hspace{0.1cm}{\rm d}x = {\rm ln}\hspace{0.15cm}\frac{B}{A}\hspace{0.05cm} .$$


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	'''(1)'''  <u>Suggested solution 2</u>  is correct:		'''(1)'''  <u>Suggested solution 2</u>  is correct:
	*According to the Laplace definition, the following holds with the given equations:		*According to the Laplace definition, the following holds with the given equations:
	:$$X_{\rm L}(p) = \int_{0}^{\infty}{ x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\~~rmd~~}t = \int\limits_{0}^{\infty}{ {\rm cos} (\omega_0 \cdot T) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\~~rmd~~}t = \frac{p}{p^2 + \omega_0^2}\hspace{0.05cm} .$$		:$$X_{\rm L}(p) = \int_{0}^{\infty}{ x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\rm d}t = \int\limits_{0}^{\infty}{ {\rm cos} (\omega_0 \cdot T) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\rm d}t = \frac{p}{p^2 + \omega_0^2}\hspace{0.05cm} .$$
	*Suggestion 3 is ruled out since  $X_{\rm L}(p)$  must have the unit "second" (integral over time) while  $p$  and  $\omega_0$  each have the unit "1/s".		*Suggestion 3 is ruled out since  $X_{\rm L}(p)$  must have the unit "second" (integral over time) while  $p$  and  $\omega_0$  each have the unit "1/s".

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	'''(2)'''  <u>Suggested solution 1</u>  is correct:		'''(2)'''  <u>Suggested solution 1</u>  is correct:
	*Here, the following holds using the same approach as in the subtask  '''(1)''':		*Here, the following holds using the same approach as in the subtask  '''(1)''':
	:$$Y_{\rm L}(p) = \int_{0}^{\infty}{ {\rm sin} (\omega_0 \cdot T) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\~~rmd~~}t = \frac{\omega_0}{p^2 + \omega_0^2}\hspace{0.05cm} .$$		:$$Y_{\rm L}(p) = \int_{0}^{\infty}{ {\rm sin} (\omega_0 \cdot T) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\rm d}t = \frac{\omega_0}{p^2 + \omega_0^2}\hspace{0.05cm} .$$


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	'''(3)'''  <u>Suggested solution 3</u>  is correct:		'''(3)'''  <u>Suggested solution 3</u>  is correct:
	*The  $p$–transfer function of the causal  $\rm si$–function is as follows considering the integral given above:		*The  $p$–transfer function of the causal  $\rm si$–function is as follows considering the integral given above:
	:$$Z_{\rm L}(p) = \int_{0}^{\infty}{ \frac{\sin(\pi \cdot t/T)}{\pi \cdot t/T} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\~~rmd~~}t = \frac{T}{\pi} \cdot {\rm arctan} \; \frac{\pi}{p\cdot T}\hspace{0.05cm} .$$		:$$Z_{\rm L}(p) = \int_{0}^{\infty}{ \frac{\sin(\pi \cdot t/T)}{\pi \cdot t/T} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\rm d}t = \frac{T}{\pi} \cdot {\rm arctan} \; \frac{\pi}{p\cdot T}\hspace{0.05cm} .$$
	*Suggestion 1 only applies to the Fourier transform of the non-causal  $\rm si$–function.		*Suggestion 1 only applies to the Fourier transform of the non-causal  $\rm si$–function.
	*Since here the argument of the  $\rm arctan$  function is dimensional, suggestion 2 cannot be true for this reason alone.		*Since here the argument of the  $\rm arctan$  function is dimensional, suggestion 2 cannot be true for this reason alone.
Line 119:		Line 119:

	*This convolution integral yields the following for sufficiently large frequencies  $f \ge 1/(2T)$:		*This convolution integral yields the following for sufficiently large frequencies  $f \ge 1/(2T)$:
	:$${\rm Im}\{ Z(f)\} = -T \cdot \int_{f- 1/(2T)}^{f+ 1/(2T)} { \frac{1}{2\pi x}}\hspace{0.1cm}{\~~rmd~~}x = \frac{T}{2\pi } \cdot {\rm ln}\hspace{0.15cm}\left \|\frac{f- 1/(2T)}{f+ 1/(2T)}\right \|\hspace{0.05cm}\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\underline{{\rm suggestion \hspace{0.15cm} 2}}.$$		:$${\rm Im}\{ Z(f)\} = -T \cdot \int_{f- 1/(2T)}^{f+ 1/(2T)} { \frac{1}{2\pi x}}\hspace{0.1cm}{\rm d}x = \frac{T}{2\pi } \cdot {\rm ln}\hspace{0.15cm}\left \|\frac{f- 1/(2T)}{f+ 1/(2T)}\right \|\hspace{0.05cm}\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\underline{{\rm suggestion \hspace{0.15cm} 2}}.$$


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	[[Category:Linear and Time-Invariant Systems: Exercises\|^3.2 Laplace Transform and p-Transfer Function^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^3.2 Laplace Transform and p-Transfer Function^]]
	[[de:Aufgaben:3.~~2_Laplace~~-Transformation]]		[[de:Aufgaben:Aufgabe 3.2: Laplace-Transformation]]