Aufgaben:Exercise 3.4: Attenuation and Phase Response - Revision history

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	*The conventional transfer function  (the frequency response)  is obtained via the substitution  $p = {\rm j} \cdot 2 \pi f$:		*The conventional transfer function  (the frequency response)  is obtained via the substitution  $p = {\rm j} \cdot 2 \pi f$:
	:$$H(f)= K \cdot \frac {{\rm j \hspace{0.05cm}2\pi \it		:$$H(f)= K \cdot \frac {{\rm j \hspace{0.05cm}2\pi \itf} - p_{\rm o }}{({\rm j \hspace{0.05cm}2\pi \itf} - p_{\rm x 1})({\rm j \hspace{0.05cm}2\pi \itf} - p_{\rm x 2})}= {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}\hspace{0.05cm} .$$*In the limiting case  $f → \infty$,  the following is obtained for the magnitude,  attenuation and phase::$$\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} H(f)= \frac{K}{{\rm j \hspace{0.05cm}2\pi \itf}}\hspace{0.15cm}\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} \|H(f)\|\hspace{0.15cm}\underline {= 0} \hspace{0.05cm}\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} a(f)= \infty,\hspace{0.1cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} b(f)\underline {={\pi}/{2}\hspace{0.1cm}(+90^\circ)}\hspace{0.01cm}.$$
	f} - p_{\rm o }}
	{({\rm j \hspace{0.05cm}2\pi \it
	f} - p_{\rm x 1})({\rm j \hspace{0.05cm}2\pi \it
	f} - p_{\rm x 2})}
	= {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}
	\hspace{0.05cm} .$$*In the limiting case  $f → \infty$,  the following is obtained for the magnitude,  attenuation and phase::$$\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} H(f)= \frac{K}{{\rm j \hspace{0.05cm}2\pi \it
	f}}\hspace{0.15cm}\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} \|H(f)\|\hspace{0.15cm}\underline {= 0} \hspace{0.05cm}
	\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} a(f)= \infty,\hspace{0.1cm}
	\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} b(f)\underline {=
	{\pi}/{2}\hspace{0.1cm}(+90^\circ)}
	\hspace{0.01cm}.$$

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	f} - p_{\rm x 2})}		f} - p_{\rm x 2})}
	= {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}		= {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}
	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$*In the limiting case  $f → \infty$,  the following is obtained for the magnitude,  attenuation and phase::$$\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} H(f)= \frac{K}{{\rm j \hspace{0.05cm}2\pi \it

	*In the limiting case  $f → \infty$,  the following is obtained for the magnitude,  attenuation and phase:
	:$$\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} H(f)= \frac{K}{{\rm j \hspace{0.05cm}2\pi \it
	f}}\hspace{0.15cm}\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} \|H(f)\|\hspace{0.15cm}\underline {= 0} \hspace{0.05cm}		f}}\hspace{0.15cm}\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} \|H(f)\|\hspace{0.15cm}\underline {= 0} \hspace{0.05cm}
	\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} a(f)= \infty,\hspace{0.1cm}		\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} a(f)= \infty,\hspace{0.1cm}
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	'''(2)'''  From the general equation in subtask  '''(1)'''  the following is obtained with the limit process  $f &#8594 0$:		'''(2)'''  From the general equation in subtask  '''(1)'''  the following is obtained with the limit process  $f &#8594 0$:
	[[File:P_ID1769__LZI_A_3_4_d_neu.png\|right\|frame\|Magnitude  (⇒ Betrag)  $\|H(f)\|$,  <br>attenuation  (⇒ Dämpfung)  $a(f)$  and phase  $b(f)$]]		[[File:P_ID1769__LZI_A_3_4_d_neu.png\|right\|frame\|Magnitude  (⇒ Betrag)  $\|H(f)\|$,  <br>attenuation  (⇒ Dämpfung)  $a(f)$  and phase  $b(f)$]]
	:$$\|H(f=0)\|= -\frac {K \cdot p_{\rm o }}		:$$\|H(f=0)\|= -\frac {K \cdot p_{\rm o }}{p_{\rm x 1}\cdot p_{\rm x 2}}= \frac {5 \cdot 1}{(-3 + 3{\rm j})\cdot (-3 + 3{\rm j})}= \frac {5 }{18}\hspace{0.15cm}\underline {\approx0.278}\hspace{0.05cm} ,$$
	{p_{\rm x 1}\cdot p_{\rm x 2}}
	= \frac {5 \cdot 1}{
	(-3 + 3{\rm j})\cdot (-3 + 3{\rm j})}= \frac {5 }{18}\hspace{0.15cm}\underline {\~~approx~~
	0.278}
	\hspace{0.05cm} ,$$
	:$$a(f=0)=- {\rm ln} \hspace{0.1cm}\hspace{0.15cm}\underline { \|H(f=0)\|= 1.281\,{\rm Np }}\hspace{0.05cm} .$$		:$$a(f=0)=- {\rm ln} \hspace{0.1cm}\hspace{0.15cm}\underline { \|H(f=0)\|= 1.281\,{\rm Np }}\hspace{0.05cm} .$$

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	:$$R_{\rm x1} = -3 - 1 \cdot {\rmj}, \hspace{0.2cm}\|R_{\rm x1}\| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{3^2 + 1^2}= 3.162,\hspace{0.5cm}{\rm ln} \hspace{0.1cm}\|R_{\rm x1}\|\hspace{0.25cm} = \hspace{0.2cm}1.151\,{\rm Np }\hspace{0.05cm},$$		:$$R_{\rm x1} = -3 - 1 \cdot {\rmj}, \hspace{0.2cm}\|R_{\rm x1}\| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{3^2 + 1^2}= 3.162,\hspace{0.5cm}{\rm ln} \hspace{0.1cm}\|R_{\rm x1}\|\hspace{0.25cm} = \hspace{0.2cm}1.151\,{\rm Np }\hspace{0.05cm},$$
	:$$ R_{\rm x2} = -3 - 7 \cdot{\rm j},\hspace{0.2cm}\|R_{\rm x2}\| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{3^2 + 7^2}= 7.616,\hspace{0.5cm}{\rm ln} \hspace{0.1cm}\|R_{\rm x2}\|\hspace{0.25cm} = \hspace{0.2cm}2.030\,{\rm Np }\hspace{0.05cm}.$$		:$$ R_{\rm x2} = -3 - 7 \cdot{\rm j},\hspace{0.2cm}\|R_{\rm x2}\| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{3^2 + 7^2}= 7.616,\hspace{0.5cm}{\rm ln} \hspace{0.1cm}\|R_{\rm x2}\|\hspace{0.25cm} = \hspace{0.2cm}2.030\,{\rm Np }\hspace{0.05cm}.$$
	$$\Rightarrow \hspace{0.3cm}a(f = \frac{4}{2\pi})=		$$\Rightarrow \hspace{0.3cm}a(f = \frac{4}{2\pi})=-{\rm ln} \hspace{0.1cm} 5+ 1.151+ 2.030-1.417\hspace{0.15cm}\underline{=0.155\,{\rm Np }}\hspace{0.05cm}.$$
	-{\rm ln} \hspace{0.1cm} 5
	+ 1.151+ 2.030-
	1.417\hspace{0.15cm}\underline{=0.155\,{\rm Np }}
	\hspace{0.05cm}.$$

	This is equivalent to   $0.155\ {\rm Np} \cdot 8.686 \ {\rm dB/Np} \hspace{0.15cm} \underline{= 1.346 \ {\rm dB}}$.		This is equivalent to   $0.155\ {\rm Np} \cdot 8.686 \ {\rm dB/Np} \hspace{0.15cm} \underline{= 1.346 \ {\rm dB}}$.

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	Thus,  the  $p$–transfer function is:		Thus,  the  $p$–transfer function is:
	:$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }}		:$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }}{(p - p_{\rm x 1})(p - p_{\rm x 2})}\hspace{0.05cm} .$$
	{(p - p_{\rm x 1})(p - p_{\rm x 2})}
	\hspace{0.05cm} .$$

	Considering the substitution  $p = {\rm j} \cdot 2 \pi f$,  the conventional transfer function can be specified,  which is also called  "frequency response":		Considering the substitution  $p = {\rm j} \cdot 2 \pi f$,  the conventional transfer function can be specified,  which is also called  "frequency response":
	:$$H(f) = H_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it		:$$H(f) = H_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \itf}} = {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}\hspace{0.05cm}.$$
	f}} = {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}
	\hspace{0.05cm}.$$

	From this equation it can also be seen the relationship between		From this equation it can also be seen the relationship between
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	The attenuation and phase values can be determined as follows for a frequency  $f$  indirectly specified by the point  $p = {\rm j} \cdot 2 \pi f$ :		The attenuation and phase values can be determined as follows for a frequency  $f$  indirectly specified by the point  $p = {\rm j} \cdot 2 \pi f$ :
	:$$a(f)\hspace{0.15cm}{\rm in}\hspace{0.15cm}{\rm Np} \hspace{0.25cm} = \hspace{0.2cm} -{\rm ln} \hspace{0.1cm} K		:$$a(f)\hspace{0.15cm}{\rm in}\hspace{0.15cm}{\rm Np} \hspace{0.25cm} = \hspace{0.2cm} -{\rm ln} \hspace{0.1cm} K+ {\rm ln} \hspace{0.1cm} \|R_{\rm x1}\|+{\rm ln} \hspace{0.1cm} \|R_{\rm x1}\|- {\rm ln} \hspace{0.1cm} \|R_{{\rm o} }\|\hspace{0.05cm},$$
	+ {\rm ln} \hspace{0.1cm} \|R_{\rm x1}\|+{\rm ln} \hspace{0.1cm} \|R_{\rm x1}\|- {\rm ln} \hspace{0.1cm} \|R_{{\rm o} }\|\hspace{0.05cm}		:$$ b(f)\hspace{0.15cm}{\rm in}\hspace{0.15cm}{\rm rad} \hspace{0.25cm} = \hspace{0.2cm}\phi_{\rm x1}+ \phi_{\rm x2}-\phi_{\rm o}\hspace{0.05cm} .$$
	,$$
	:$$ b(f)\hspace{0.15cm}{\rm in}\hspace{0.15cm}{\rm rad} \hspace{0.25cm} = \hspace{0.2cm}
	\phi_{\rm x1}+ \phi_{\rm x2}-\phi_{\rm o}
	\hspace{0.05cm} .$$

	The corresponding magnitudes  $\|R_{\rm o}\|$,  $\|R_{\rm x1}\|$  and  $\|R_{\rm x2}\|$ as well as the angles  $\phi_{\rm o}$,  $\phi_{\rm x1}$  and  $\phi_{\rm x2}$  can be taken from the graph .		The corresponding magnitudes  $\|R_{\rm o}\|$,  $\|R_{\rm x1}\|$  and  $\|R_{\rm x2}\|$ as well as the angles  $\phi_{\rm o}$,  $\phi_{\rm x1}$  and  $\phi_{\rm x2}$  can be taken from the graph .
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The  $p$–transfer function is:		'''(1)'''  The  $p$–transfer function is:
	:$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }}		:$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }}{(p - p_{\rm x 1})(p - p_{\rm x 2})}\hspace{0.05cm} .$$
	{(p - p_{\rm x 1})(p - p_{\rm x 2})}
	\hspace{0.05cm} .$$

	*The conventional transfer function  (the frequency response)  is obtained via the substitution  $p = {\rm j} \cdot 2 \pi f$:		*The conventional transfer function  (the frequency response)  is obtained via the substitution  $p = {\rm j} \cdot 2 \pi f$:
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	0.278}		0.278}
	\hspace{0.05cm} ,$$		\hspace{0.05cm} ,$$
	:$$a(f=0)=- {\rm ln} \hspace{0.1cm}\hspace{0.15cm}\underline { \|H(f=0)\|= 1.281\,{\rm Np }}		:$$a(f=0)=- {\rm ln} \hspace{0.1cm}\hspace{0.15cm}\underline { \|H(f=0)\|= 1.281\,{\rm Np }}\hspace{0.05cm} .$$
	\hspace{0.05cm} .$$

	The screen capture of the Flash–module  "Causal Systems"  summarizes the results of this exercise:		The screen capture of the Flash–module  "Causal Systems"  summarizes the results of this exercise:
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	'''(3)'''  According to the detailed description in the  [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Graphical_determination_of_attenuation_and_phase\|theory part]],  the following holds for the attenuation function:		'''(3)'''  According to the detailed description in the  [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Graphical_determination_of_attenuation_and_phase\|theory part]],  the following holds for the attenuation function:

	:$$a(f)= -{\rm ln} \hspace{0.1cm} K		:$$a(f)= -{\rm ln} \hspace{0.1cm} K+ {\rm ln} \hspace{0.1cm} \|R_{\rm x1}\|+{\rm ln} \hspace{0.1cm} \|R_{\rm x2}\|- {\rm ln} \hspace{0.1cm} \|R_{{\rm o} }\|\hspace{0.05cm} .$$
	+ {\rm ln} \hspace{0.1cm} \|R_{\rm x1}\|+{\rm ln} \hspace{0.1cm} \|R_{\rm x2}\|- {\rm ln} \hspace{0.1cm} \|R_{{\rm o} }\|\hspace{0.05cm} .$$

	*Furthermore,  the additional unit  "neper"  $\rm (Np)$  must be taken into account.		*Furthermore,  the additional unit  "neper"  $\rm (Np)$  must be taken into account.
	*The attenuation at  $f = 2/\pi$  is searched-for.  For this, we set  $p = {\rm j} \cdot 2 \pi f = 4$  and determine the following distances:		*The attenuation at  $f = 2/\pi$  is searched-for.  For this, we set  $p = {\rm j} \cdot 2 \pi f = 4$  and determine the following distances:
	:$$R_{\rm o} = 1 - 4 \cdot {\rm j}, \hspace{0.2cm}\|R_{\rm o}\| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{1^2 + 4^2}= 4.123, \hspace{1.15cm}		:$$R_{\rm o} = 1 - 4 \cdot {\rm j}, \hspace{0.2cm}\|R_{\rm o}\| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{1^2 + 4^2}= 4.123, \hspace{1.15cm}{\rm ln} \hspace{0.1cm}\|R_{\rm o}\|\hspace{0.25cm} = \hspace{0.2cm}1.417\,{\rm Np }\hspace{0.05cm},$$
	{\rm ln} \hspace{0.1cm}\|R_{\rm o}\|		:$$R_{\rm x1} = -3 - 1 \cdot {\rmj}, \hspace{0.2cm}\|R_{\rm x1}\| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{3^2 + 1^2}= 3.162,\hspace{0.5cm}{\rm ln} \hspace{0.1cm}\|R_{\rm x1}\|\hspace{0.25cm} = \hspace{0.2cm}1.151\,{\rm Np }\hspace{0.05cm},$$
	\hspace{0.25cm} = \hspace{0.2cm}1.417\,{\rm Np }\hspace{0.05cm},$$		:$$ R_{\rm x2} = -3 - 7 \cdot{\rm j},\hspace{0.2cm}\|R_{\rm x2}\| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{3^2 + 7^2}= 7.616,\hspace{0.5cm}{\rm ln} \hspace{0.1cm}\|R_{\rm x2}\|\hspace{0.25cm} = \hspace{0.2cm}2.030\,{\rm Np }\hspace{0.05cm}.$$
	:$$R_{\rm x1} = -3 - 1 \cdot {\rm
	j}, \hspace{0.2cm}\|R_{\rm x1}\| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{3^2 + 1^2}= 3.162,\hspace{0.5cm}
	{\rm ln} \hspace{0.1cm}\|R_{\rm x1}\|
	\hspace{0.25cm} = \hspace{0.2cm}1.151\,{\rm Np }\hspace{0.05cm},$$
	:$$ R_{\rm x2} = -3 - 7 \cdot{\rm j},
	\hspace{0.2cm}\|R_{\rm x2}\| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{3^2 + 7^2}= 7.616,\hspace{0.5cm}
	{\rm ln} \hspace{0.1cm}\|R_{\rm x2}\|
	\hspace{0.25cm} = \hspace{0.2cm}2.030\,{\rm Np }\hspace{0.05cm}.$$
	$$\Rightarrow \hspace{0.3cm}a(f = \frac{4}{2\pi})=		$$\Rightarrow \hspace{0.3cm}a(f = \frac{4}{2\pi})=
	-{\rm ln} \hspace{0.1cm} 5		-{\rm ln} \hspace{0.1cm} 5
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	'''(4)'''  The following holds for the phase function according to the description in the theory section due to  $K > 0$ :		'''(4)'''  The following holds for the phase function according to the description in the theory section due to  $K > 0$ :
	:$$b(f ={2}/\pi) = \phi_{\rm x1} + \phi_{\rm x2}-\phi_{\rm o}\hspace{0.05cm},$$		:$$b(f ={2}/\pi) = \phi_{\rm x1} + \phi_{\rm x2}-\phi_{\rm o}\hspace{0.05cm},$$
	:$$\phi_{\rm x1} ={\rm arctan}\hspace{0.15cm}(1/3) =		:$$\phi_{\rm x1} ={\rm arctan}\hspace{0.15cm}(1/3) =18.4^\circ\hspace{0.05cm}, \hspace{0.5cm}\phi_{\rm x2} = {\rm arctan}\hspace{0.15cm}(7/3) =66.8^\circ\hspace{0.05cm},\hspace{0.5cm} \phi_{\rm o} = {\rm arctan}\hspace{0.15cm}(-1/4) =180^\circ - 76^\circ = 104^\circ $$
	18.4^\circ\hspace{0.05cm}, \hspace{0.5cm}\phi_{\rm x2} = {\rm arctan}\hspace{0.15cm}(7/3) =		:$$ \Rightarrow \hspace{0.3cm}b(f ={2}/\pi) =18.4^\circ + 66.8^\circ - 104^\circ \hspace{0.15cm} \underline{= -18.8^\circ}\hspace{0.05cm}.$$
	66.8^\circ\hspace{0.05cm},\hspace{0.5cm} \phi_{\rm o} = {\rm arctan}\hspace{0.15cm}(-1/4) =
	180^\circ - 76^\circ = 104^\circ $$
	:$$ \Rightarrow \hspace{0.3cm}b(f ={2}/\pi) =
	18.4^\circ + 66.8^\circ - 104^\circ \hspace{0.15cm} \underline{= -18.8^\circ}
	\hspace{0.05cm}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

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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:3.4_Dämpfungs-_und_Phasenverlauf]]

Guenter at 15:15, 14 October 2021

2021-10-14T15:15:47Z

← Older revision		Revision as of 17:15, 14 October 2021
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	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$

	*The conventional transfer function (the frequency response) is obtained via the substitution  $p = {\rm j} \cdot 2 \pi f$:		*The conventional transfer function  (the frequency response)  is obtained via the substitution  $p = {\rm j} \cdot 2 \pi f$:
	:$$H(f)= K \cdot \frac {{\rm j \hspace{0.05cm}2\pi \it		:$$H(f)= K \cdot \frac {{\rm j \hspace{0.05cm}2\pi \it
	f} - p_{\rm o }}		f} - p_{\rm o }}
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	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$

	*In the limiting case  $f → \infty$,  the following is obtained for the magnitude, attenuation and phase:		*In the limiting case  $f → \infty$,  the following is obtained for the magnitude,  attenuation and phase:
	:$$\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} H(f)= \frac{K}{{\rm j \hspace{0.05cm}2\pi \it		:$$\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} H(f)= \frac{K}{{\rm j \hspace{0.05cm}2\pi \it
	f}}\hspace{0.15cm}\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} \|H(f)\|\hspace{0.15cm}\underline {= 0} \hspace{0.05cm}		f}}\hspace{0.15cm}\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} \|H(f)\|\hspace{0.15cm}\underline {= 0} \hspace{0.05cm}
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	'''(2)'''  From the general equation in subtask  '''(1)'''  the following is obtained with the limit process  $f &#8594 0$:		'''(2)'''  From the general equation in subtask  '''(1)'''  the following is obtained with the limit process  $f &#8594 0$:
	[[File:P_ID1769__LZI_A_3_4_d_neu.png\|right\|frame\|$\|H(f)\|$,  $a(f)$  and  $b(f)$]]		[[File:P_ID1769__LZI_A_3_4_d_neu.png\|right\|frame\|Magnitude  (⇒ Betrag)  $\|H(f)\|$,  <br>attenuation  (⇒ Dämpfung)  $a(f)$  and phase  $b(f)$]]
	:$$\|H(f=0)\|= -\frac {K \cdot p_{\rm o }}		:$$\|H(f=0)\|= -\frac {K \cdot p_{\rm o }}
	{p_{\rm x 1}\cdot p_{\rm x 2}}		{p_{\rm x 1}\cdot p_{\rm x 2}}
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	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$

	The screen capture of the Flash–module "Causal Systems" summarizes the results of this exercise:		The screen capture of the Flash–module  "Causal Systems"  summarizes the results of this exercise:
	:*middle axis (blue):   magnitude $\|H(f)\|$,		:*middle axis (blue):   magnitude $\|H(f)\|$,   ⇒   here labeled with  $\|Y(f)\|$,
	:*left axis (red):   attenuation $a(f)$,		:*left axis (red):   attenuation  $a(f)$,
	:*right axis (green):   phase $b(f)$.		:*right axis (green):   phase  $b(f)$.
	:*black point:   values for $2\pi f = 4.$		:*black point:   values for  $2\pi f = 4.$
	~~<br clear=all>~~

	[[File:P_ID2843__LZI_A_3_4.png\|right\|frame\|Pole-zero diagram and some auxiliary quantities]]		[[File:P_ID2843__LZI_A_3_4.png\|right\|frame\|Pole-zero diagram and some auxiliary quantities]]
	'''(3)'''  According to the detailed description in the  [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Graphical_determination_of_attenuation_and_phase\|theory part]],  the following holds for the attenuation function:		'''(3)'''  According to the detailed description in the  [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Graphical_determination_of_attenuation_and_phase\|theory part]],  the following holds for the attenuation function:
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	+ {\rm ln} \hspace{0.1cm} \|R_{\rm x1}\|+{\rm ln} \hspace{0.1cm} \|R_{\rm x2}\|- {\rm ln} \hspace{0.1cm} \|R_{{\rm o} }\|\hspace{0.05cm} .$$		+ {\rm ln} \hspace{0.1cm} \|R_{\rm x1}\|+{\rm ln} \hspace{0.1cm} \|R_{\rm x2}\|- {\rm ln} \hspace{0.1cm} \|R_{{\rm o} }\|\hspace{0.05cm} .$$

	*~~The~~ additional unit "neper" $\rm (Np)$ ~~is further to~~ be ~~considered~~.		*Furthermore,  the additional unit  "neper"  $\rm (Np)$  must be taken into account.
			*The attenuation at  $f = 2/\pi$  is searched-for.  For this, we set  $p = {\rm j} \cdot 2 \pi f = 4$  and determine the following distances:
	*The attenuation at  $f = 2/\pi$ is searched-for.  For this, we set  $p = {\rm j} \cdot 2 \pi f = 4$  and determine the following distances:
	:$$R_{\rm o} = 1 - 4 \cdot {\rm j}, \hspace{0.2cm}\|R_{\rm o}\| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{1^2 + 4^2}= 4.123, \hspace{1.15cm}		:$$R_{\rm o} = 1 - 4 \cdot {\rm j}, \hspace{0.2cm}\|R_{\rm o}\| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{1^2 + 4^2}= 4.123, \hspace{1.15cm}
	{\rm ln} \hspace{0.1cm}\|R_{\rm o}\|		{\rm ln} \hspace{0.1cm}\|R_{\rm o}\|
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	This is equivalent to  $0.155\ {\rm Np} \cdot 8.686 \ {\rm dB/Np} \hspace{0.15cm} \underline{= 1.346 \ {\rm dB}}$.		This is equivalent to   $0.155\ {\rm Np} \cdot 8.686 \ {\rm dB/Np} \hspace{0.15cm} \underline{= 1.346 \ {\rm dB}}$.
	~~<br clear=all>~~

	'''(4)'''  The following holds for the phase function according to the description in the theory section due to  $K > 0$ :		'''(4)'''  The following holds for the phase function according to the description in the theory section due to  $K > 0$ :
	:$$b(f ={2}/\pi) = \phi_{\rm x1} + \phi_{\rm x2}-\phi_{\rm o}\hspace{0.05cm},$$		:$$b(f ={2}/\pi) = \phi_{\rm x1} + \phi_{\rm x2}-\phi_{\rm o}\hspace{0.05cm},$$
	:$$\phi_{\rm x1} ={\rm arctan}\hspace{0.15cm}(1/3) =		:$$\phi_{\rm x1} ={\rm arctan}\hspace{0.15cm}(1/3) =
	18.4^\circ\hspace{0.05cm}, \hspace{0.~~2cm~~}\phi_{\rm x2} = {\rm arctan}\hspace{0.15cm}(7/3) =		18.4^\circ\hspace{0.05cm}, \hspace{0.5cm}\phi_{\rm x2} = {\rm arctan}\hspace{0.15cm}(7/3) =
	66.8^\circ\hspace{0.05cm},$$		66.8^\circ\hspace{0.05cm},\hspace{0.5cm} \phi_{\rm o} = {\rm arctan}\hspace{0.15cm}(-1/4) =
	~~:$$~~ \phi_{\rm o} = {\rm arctan}\hspace{0.15cm}(-1/4) =		180^\circ - 76^\circ = 104^\circ $$
	180^\circ - 76^\circ = 104^\circ ~~\hspace{0.3cm}~~		:$$ \Rightarrow \hspace{0.3cm}b(f ={2}/\pi) =
	\Rightarrow \hspace{0.3cm}b(f ={2}/\pi) =
	18.4^\circ + 66.8^\circ - 104^\circ \hspace{0.15cm} \underline{= -18.8^\circ}		18.4^\circ + 66.8^\circ - 104^\circ \hspace{0.15cm} \underline{= -18.8^\circ}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

Guenter at 14:54, 14 October 2021

2021-10-14T14:54:21Z

← Older revision		Revision as of 16:54, 14 October 2021
Line 8:		Line 8:
	:$$ p_{\rm o}= 1,\hspace{0,2cm}p_{\rm x1}= -3 + 3{\rm j},\hspace{0,2cm}p_{\rm x2}= -3 - 3{\rm j}\hspace{0.05cm} .$$		:$$ p_{\rm o}= 1,\hspace{0,2cm}p_{\rm x1}= -3 + 3{\rm j},\hspace{0,2cm}p_{\rm x2}= -3 - 3{\rm j}\hspace{0.05cm} .$$

	Thus, the  $p$–transfer function is:		Thus,  the  $p$–transfer function is:
	:$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }}		:$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }}
	{(p - p_{\rm x 1})(p - p_{\rm x 2})}		{(p - p_{\rm x 1})(p - p_{\rm x 2})}
	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$

	Considering the substitution  $p = {\rm j} \cdot 2 \pi f$,  the conventional transfer function can be specified, which is also called frequency response:		Considering the substitution  $p = {\rm j} \cdot 2 \pi f$,  the conventional transfer function can be specified,  which is also called  "frequency response":
	:$$H(f) = H_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it		:$$H(f) = H_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it
	f}} = {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}		f}} = {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	From this equation, the relationship between		From this equation it can also be seen the relationship between
	*the transfer function  $H(f)$,		*the transfer function  $H(f)$,
	*the attenuation function  $a(f)$  and		*the attenuation function  $a(f)$  and
	*the phase function  $b(f)$		*the phase function  $b(f)$.
	~~can also be seen~~.


	The attenuation and phase values can be determined as follows for a frequency  $f$  indirectly specified by the point $p = {\rm j} \cdot 2 \pi f$ :		The attenuation and phase values can be determined as follows for a frequency  $f$  indirectly specified by the point  $p = {\rm j} \cdot 2 \pi f$ :
	:$$a(f)\hspace{0.15cm}{\rm in}\hspace{0.15cm}{\rm Np} \hspace{0.25cm} = \hspace{0.2cm} -{\rm ln} \hspace{0.1cm} K		:$$a(f)\hspace{0.15cm}{\rm in}\hspace{0.15cm}{\rm Np} \hspace{0.25cm} = \hspace{0.2cm} -{\rm ln} \hspace{0.1cm} K
	+ {\rm ln} \hspace{0.1cm} \|R_{\rm x1}\|+{\rm ln} \hspace{0.1cm} \|R_{\rm x1}\|- {\rm ln} \hspace{0.1cm} \|R_{{\rm o} }\|\hspace{0.05cm}		+ {\rm ln} \hspace{0.1cm} \|R_{\rm x1}\|+{\rm ln} \hspace{0.1cm} \|R_{\rm x1}\|- {\rm ln} \hspace{0.1cm} \|R_{{\rm o} }\|\hspace{0.05cm}
Line 33:		Line 32:
	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$

	The corresponding magnitudes  $\|R_{\rm o}\|$,  $\|R_{\rm x1}\|$  and  $\|R_{\rm x1}\|$ as well as the angles  $\phi_{\rm o}$,  $\phi_{\rm x1}$  and  $\phi_{\rm x2}$  can be taken from the graph .		The corresponding magnitudes  $\|R_{\rm o}\|$,  $\|R_{\rm x1}\|$  and  $\|R_{\rm x2}\|$ as well as the angles  $\phi_{\rm o}$,  $\phi_{\rm x1}$  and  $\phi_{\rm x2}$  can be taken from the graph .



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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]].

Line 61:		Line 59:


	{Compute the attenuation value at  $f =4/(2 \pi)$  in neper  $\rm(Np)$  and decibel  $\rm(dB)$ according to the described approach.		{Compute the attenuation value at  $f =4/(2 \pi)$  in neper  $\rm(Np)$  and decibel  $\rm(dB)$  according to the described approach.
	\|type="{}"}		\|type="{}"}
	$a(f = 2/ \pi)\ = \ $ { 0.155 3% } $\ \rm Np$		$a(f = 2/ \pi)\ = \ $ { 0.155 3% } $\ \rm Np$

Oezer at 16:44, 12 October 2021

2021-10-12T16:44:01Z

← Older revision		Revision as of 18:44, 12 October 2021
Line 119:		Line 119:
	<br clear=all>		<br clear=all>
	[[File:P_ID2843__LZI_A_3_4.png\|right\|frame\|Pole-zero diagram and some auxiliary quantities]]		[[File:P_ID2843__LZI_A_3_4.png\|right\|frame\|Pole-zero diagram and some auxiliary quantities]]
	'''(3)'''  According to the detailed description in the  [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Graphical_determination_of_attenuation_and_phase\|theory part]]  the following holds for the attenuation function:		'''(3)'''  According to the detailed description in the  [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Graphical_determination_of_attenuation_and_phase\|theory part]],  the following holds for the attenuation function:

	:$$a(f)= -{\rm ln} \hspace{0.1cm} K		:$$a(f)= -{\rm ln} \hspace{0.1cm} K

Oezer at 16:42, 12 October 2021

2021-10-12T16:42:58Z

← Older revision		Revision as of 18:42, 12 October 2021
Line 91:		Line 91:
	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$

	*In the limiting case  $f → \infty$  the following is obtained for the magnitude, attenuation and phase:		*In the limiting case  $f → \infty$,  the following is obtained for the magnitude, attenuation and phase:
	:$$\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} H(f)= \frac{K}{{\rm j \hspace{0.05cm}2\pi \it		:$$\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} H(f)= \frac{K}{{\rm j \hspace{0.05cm}2\pi \it
	f}}\hspace{0.15cm}\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} \|H(f)\|\hspace{0.15cm}\underline {= 0} \hspace{0.05cm}		f}}\hspace{0.15cm}\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} \|H(f)\|\hspace{0.15cm}\underline {= 0} \hspace{0.05cm}

Oezer at 16:41, 12 October 2021

2021-10-12T16:41:11Z

← Older revision		Revision as of 18:41, 12 October 2021
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	This is equivalent to  $0.155\ {\rm Np} \cdot 8.686 \ {\rm dB/Np} \hspace{0.15cm} \underline{= 1.346 \ {\rm dB}}$.		This is equivalent to  $0.155\ {\rm Np} \cdot 8.686 \ {\rm dB/Np} \hspace{0.15cm} \underline{= 1.346 \ {\rm dB}}$.
	<br clear=all>		<br clear=all>
	'''(4)'''  ~~Nach der Beschreibung im Theorieteil gilt wegen~~  $K > 0$  ~~für die Phasenfunktion~~:		'''(4)'''  The following holds for the phase function according to the description in the theory section due to  $K > 0$ :
	:$$b(f ={2}/\pi) = \phi_{\rm x1} + \phi_{\rm x2}-\phi_{\rm o}\hspace{0.05cm},$$		:$$b(f ={2}/\pi) = \phi_{\rm x1} + \phi_{\rm x2}-\phi_{\rm o}\hspace{0.05cm},$$
	:$$\phi_{\rm x1} ={\rm arctan}\hspace{0.15cm}(1/3) =		:$$\phi_{\rm x1} ={\rm arctan}\hspace{0.15cm}(1/3) =

← Older revision		Revision as of 17:57, 16 March 2026
Line 12:		Line 12:

	Considering the substitution  $p = {\rm j} \cdot 2 \pi f$,  the conventional transfer function can be specified,  which is also called  "frequency response":		Considering the substitution  $p = {\rm j} \cdot 2 \pi f$,  the conventional transfer function can be specified,  which is also called  "frequency response":
	:$$H(f) = H_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \~~itf~~}} = {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}\hspace{0.05cm}.$$		:$$H(f) = H_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it f}} = {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}\hspace{0.05cm}.$$

	From this equation it can also be seen the relationship between		From this equation it can also be seen the relationship between
Line 71:		Line 71:

	*The conventional transfer function  (the frequency response)  is obtained via the substitution  $p = {\rm j} \cdot 2 \pi f$:		*The conventional transfer function  (the frequency response)  is obtained via the substitution  $p = {\rm j} \cdot 2 \pi f$:
	:$$H(f)= K \cdot \frac {{\rm j \hspace{0.05cm}2\pi \~~itf~~} - p_{\rm o }}{({\rm j \hspace{0.05cm}2\pi \~~itf~~} - p_{\rm x 1})({\rm j \hspace{0.05cm}2\pi \~~itf~~} - p_{\rm x 2})}= {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}\hspace{0.05cm} .$$*In the limiting case  $f → \infty$,  the following is obtained for the magnitude,  attenuation and phase::$$\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} H(f)= \frac{K}{{\rm j \hspace{0.05cm}2\pi \~~itf~~}}\hspace{0.15cm}\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} \|H(f)\|\hspace{0.15cm}\underline {= 0} \hspace{0.05cm}\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} a(f)= \infty,\hspace{0.1cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} b(f)\underline {={\pi}/{2}\hspace{0.1cm}(+90^\circ)}\hspace{0.01cm}.$$		:$$H(f)= K \cdot \frac {{\rm j \hspace{0.05cm}2\pi \it f} - p_{\rm o }}{({\rm j \hspace{0.05cm}2\pi \it f} - p_{\rm x 1})({\rm j \hspace{0.05cm}2\pi \it f} - p_{\rm x 2})}= {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}\hspace{0.05cm} .$$*In the limiting case  $f → \infty$,  the following is obtained for the magnitude,  attenuation and phase::$$\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} H(f)= \frac{K}{{\rm j \hspace{0.05cm}2\pi \it f}}\hspace{0.15cm}\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} \|H(f)\|\hspace{0.15cm}\underline {= 0} \hspace{0.05cm}\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} a(f)= \infty,\hspace{0.1cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} b(f)\underline {={\pi}/{2}\hspace{0.1cm}(+90^\circ)}\hspace{0.01cm}.$$


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	*The attenuation at  $f = 2/\pi$  is searched-for.  For this, we set  $p = {\rm j} \cdot 2 \pi f = 4$  and determine the following distances:		*The attenuation at  $f = 2/\pi$  is searched-for.  For this, we set  $p = {\rm j} \cdot 2 \pi f = 4$  and determine the following distances:
	:$$R_{\rm o} = 1 - 4 \cdot {\rm j}, \hspace{0.2cm}\|R_{\rm o}\| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{1^2 + 4^2}= 4.123, \hspace{1.15cm}{\rm ln} \hspace{0.1cm}\|R_{\rm o}\|\hspace{0.25cm} = \hspace{0.2cm}1.417\,{\rm Np }\hspace{0.05cm},$$		:$$R_{\rm o} = 1 - 4 \cdot {\rm j}, \hspace{0.2cm}\|R_{\rm o}\| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{1^2 + 4^2}= 4.123, \hspace{1.15cm}{\rm ln} \hspace{0.1cm}\|R_{\rm o}\|\hspace{0.25cm} = \hspace{0.2cm}1.417\,{\rm Np }\hspace{0.05cm},$$
	:$$R_{\rm x1} = -3 - 1 \cdot {\~~rmj~~}, \hspace{0.2cm}\|R_{\rm x1}\| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{3^2 + 1^2}= 3.162,\hspace{0.5cm}{\rm ln} \hspace{0.1cm}\|R_{\rm x1}\|\hspace{0.25cm} = \hspace{0.2cm}1.151\,{\rm Np }\hspace{0.05cm},$$		:$$R_{\rm x1} = -3 - 1 \cdot {\rm j}, \hspace{0.2cm}\|R_{\rm x1}\| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{3^2 + 1^2}= 3.162,\hspace{0.5cm}{\rm ln} \hspace{0.1cm}\|R_{\rm x1}\|\hspace{0.25cm} = \hspace{0.2cm}1.151\,{\rm Np }\hspace{0.05cm},$$
	:$$ R_{\rm x2} = -3 - 7 \cdot{\rm j},\hspace{0.2cm}\|R_{\rm x2}\| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{3^2 + 7^2}= 7.616,\hspace{0.5cm}{\rm ln} \hspace{0.1cm}\|R_{\rm x2}\|\hspace{0.25cm} = \hspace{0.2cm}2.030\,{\rm Np }\hspace{0.05cm}.$$		:$$ R_{\rm x2} = -3 - 7 \cdot{\rm j},\hspace{0.2cm}\|R_{\rm x2}\| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{3^2 + 7^2}= 7.616,\hspace{0.5cm}{\rm ln} \hspace{0.1cm}\|R_{\rm x2}\|\hspace{0.25cm} = \hspace{0.2cm}2.030\,{\rm Np }\hspace{0.05cm}.$$
	$$\Rightarrow \hspace{0.3cm}a(f = \frac{4}{2\pi})=-{\rm ln} \hspace{0.1cm} 5+ 1.151+ 2.030-1.417\hspace{0.15cm}\underline{=0.155\,{\rm Np }}\hspace{0.05cm}.$$		$$\Rightarrow \hspace{0.3cm}a(f = \frac{4}{2\pi})=-{\rm ln} \hspace{0.1cm} 5+ 1.151+ 2.030-1.417\hspace{0.15cm}\underline{=0.155\,{\rm Np }}\hspace{0.05cm}.$$
Line 111:		Line 111:

	[[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.~~4_Dämpfungs~~-~~_und_Phasenverlauf~~]]		[[de:Aufgaben:Aufgabe 3.4: Dämpfungs- und Phasenverlauf]]