Aufgaben:Exercise 3.6Z: Two Imaginary Poles - Revision history

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	[[Category:Linear and Time-Invariant Systems: Exercises\|^3.3 Inverse Laplace Transform^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^3.3 Inverse Laplace Transform^]]
	[[de:~~Zusatzaufgaben~~:3.~~6_Zwei_imaginäre_Pole~~]]		[[de:Aufgaben:Aufgabe 3.6Z: Zwei imaginäre Pole]]

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	*The following is obtained for signal  $x(t)$  for positive times by applying the residue theorem:		*The following is obtained for signal  $x(t)$  for positive times by applying the residue theorem:
	:$$x_1(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm}t}\}=\frac {p} { p+{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p\hspace{0.05cm}\cdot \hspace{0.05cm}t}\bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=\frac{1}{2} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}\hspace{0.05cm} ,$$		:$$x_1(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm}t}\}=\frac {p} { p+{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p\hspace{0.05cm}\cdot \hspace{0.05cm}t}\bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=\frac{1}{2} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}\hspace{0.05cm} ,$$
	:$$ x_2(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}		:$$ x_2(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\}=\frac {p} { p-{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p\hspace{0.05cm}\cdot \hspace{0.05cm}t}\bigg \|_{p \hspace{0.05cm}= -{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=\frac{1}{2} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}\hspace{0.05cm} .$$:$$\Rightarrow \hspace{0.3cm} x(t) = x_1(t) + x_2(t) ={1}/{2} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pit}+{\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pit}\right ] = \cos(2\pi t)\hspace{0.05cm} .$$
	\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\}=
	\frac {p} { p-{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
	\hspace{0.05cm}\cdot \hspace{0.05cm}t}
	\bigg \|_{p \hspace{0.05cm}= -{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
	\frac{1}{2} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}
	\hspace{0.05cm} .$$:$$\Rightarrow \hspace{0.3cm} x(t) = x_1(t) + x_2(t) =
	{1}/{2} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
	t}+{\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
	t}\right ] = \cos(2\pi t)
	\hspace{0.05cm} .$$

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	[[File:P_ID1786__LZI_Z_3_6.png\|right\|frame\|Two imaginary poles <br>and one zero ]]		[[File:P_ID1786__LZI_Z_3_6.png\|right\|frame\|Two imaginary poles <br>and one zero ]]
	In this exercise,  we consider a causal signal  $x(t)$  with the Laplace transform		In this exercise,  we consider a causal signal  $x(t)$  with the Laplace transform
	:$$X_{\rm L}(p) =		:$$X_{\rm L}(p) =\frac { p} { p^2 + 4 \pi^2}=\frac { p} { (p-{\rm j} \cdot 2\pi)(p+{\rm j} \cdot 2\pi)}\hspace{0.05cm}$$
	\frac { p} { p^2 + 4 \pi^2}=
	\frac { p} { (p-{\rm j} \cdot 2\pi)(p+{\rm j} \cdot 2\pi)}
	\hspace{0.05cm}$$
	corresponding to the graph  (one red zero and two green poles).		corresponding to the graph  (one red zero and two green poles).

	*In contrast,  the signal  $y(t)$  has the Laplace spectral function		*In contrast,  the signal  $y(t)$  has the Laplace spectral function
	:$$Y_{\rm L}(p) =		:$$Y_{\rm L}(p) =\frac { 1} { p^2 + 4 \pi^2}\hspace{0.05cm}.$$
	\frac { 1} { p^2 + 4 \pi^2}
	\hspace{0.05cm}.$$
	:Thus,  the red zero does not belong to  $Y_{\rm L}(p)$.		:Thus,  the red zero does not belong to  $Y_{\rm L}(p)$.

	*Finally, the signal  $z(t)$  with the Laplace tansform		*Finally, the signal  $z(t)$  with the Laplace tansform
	:$$Z_{\rm L}(p) =		:$$Z_{\rm L}(p) =\frac { p} { (p-{\rm j} \cdot \beta)(p+{\rm j} \cdot \beta)}\hspace{0.05cm}$$
	\frac { p} { (p-{\rm j} \cdot \beta)(p+{\rm j} \cdot \beta)}
	\hspace{0.05cm}$$
	:is considered, in particular the limiting case for  $\beta → 0$.		:is considered, in particular the limiting case for  $\beta → 0$.

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	*A result  $t = 1$  is thus to be interpreted as  $t = T$  with  $T = 1 \ \rm µ s$ .		*A result  $t = 1$  is thus to be interpreted as  $t = T$  with  $T = 1 \ \rm µ s$ .
	*The  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem\|residue theorem]]  is as follows using the example of the function  $X_{\rm L}(p)$  with two simple poles at  $ \pm {\rm j} \cdot \beta$:		*The  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem\|residue theorem]]  is as follows using the example of the function  $X_{\rm L}(p)$  with two simple poles at  $ \pm {\rm j} \cdot \beta$:
	:$$x(t) = X_{\rm L}(p) \cdot (p - {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p\hspace{0.05cm} \cdot		:$$x(t) = X_{\rm L}(p) \cdot (p - {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p\hspace{0.05cm} \cdot\hspace{0.05cm}t} \Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it\beta}}+X_{\rm L}(p) \cdot (p + {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p\hspace{0.05cm} \cdot\hspace{0.05cm}t} \Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{-\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it\beta}}\hspace{0.05cm}.$$
	\hspace{0.05cm}t} \Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
	\beta}}+X_{\rm L}(p) \cdot (p + {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p
	\hspace{0.05cm} \cdot\hspace{0.05cm}t} \Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{-\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
	\beta}}
	\hspace{0.05cm}.$$


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	'''(1)'''  The  <u>suggested solutions 1, 3 and 4</u>  are correct:		'''(1)'''  The  <u>suggested solutions 1, 3 and 4</u>  are correct:
	*The following is obtained for signal  $x(t)$  for positive times by applying the residue theorem:		*The following is obtained for signal  $x(t)$  for positive times by applying the residue theorem:
	:$$x_1(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}		:$$x_1(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm}t}\}=\frac {p} { p+{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p\hspace{0.05cm}\cdot \hspace{0.05cm}t}\bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=\frac{1}{2} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}\hspace{0.05cm} ,$$
	\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm}t}\}=
	\frac {p} { p+{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
	\hspace{0.05cm}\cdot \hspace{0.05cm}t}
	\bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
	\frac{1}{2} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}\hspace{0.05cm} ,$$
	:$$ x_2(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}		:$$ x_2(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
	\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\}=		\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\}=
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	\bigg \|_{p \hspace{0.05cm}= -{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=		\bigg \|_{p \hspace{0.05cm}= -{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
	\frac{1}{2} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}		\frac{1}{2} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}
	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$:$$\Rightarrow \hspace{0.3cm} x(t) = x_1(t) + x_2(t) =
	:$$\Rightarrow \hspace{0.3cm} x(t) = x_1(t) + x_2(t) =
	{1}/{2} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi		{1}/{2} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
	t}+{\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi		t}+{\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
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	*However,  the integration theorem can also be used.		*However,  the integration theorem can also be used.
	*This says among other things that multiplication by  $1/p$  in the spectral domain corresponds to integration in the time domain:		*This says among other things that multiplication by  $1/p$  in the spectral domain corresponds to integration in the time domain:
	:$$Y_{\rm L}(p) = {1}/{p} \cdot X_{\rm L}(p) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} t \ge 0:\quad y(t) = \int_{-\infty}^t \cos(2\pi		:$$Y_{\rm L}(p) = {1}/{p} \cdot X_{\rm L}(p) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} t \ge 0:\quad y(t) = \int_{-\infty}^t \cos(2\pi\tau)\,\,{\rm d}\tau = {1}/({2\pi}) \cdot \sin(2\pi t)\hspace{0.05cm} .$$
	\tau)\,\,{\rm d}\tau = {1}/({2\pi}) \cdot \sin(2\pi t)
	\hspace{0.05cm} .$$

	Please note:  The causal cosine signal  $x(t)$  and the causal sine signal  $y(t)$  are shown on the information page of  [[Aufgaben:Exercise_3.6:_Transient_Behavior\|Exercise 3.6]]  as  $c_{\rm K}(t)$  and  $s_{\rm K}(t)$,  respectively.		Please note:  The causal cosine signal  $x(t)$  and the causal sine signal  $y(t)$  are shown on the information page of  [[Aufgaben:Exercise_3.6:_Transient_Behavior\|Exercise 3.6]]  as  $c_{\rm K}(t)$  and  $s_{\rm K}(t)$,  respectively.
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	*The limit process for  $\beta → 0$  thus results in the step function  $\gamma(t)$.		*The limit process for  $\beta → 0$  thus results in the step function  $\gamma(t)$.
	*The same result is obtained by consideration in the spectral domain:		*The same result is obtained by consideration in the spectral domain:
	:$$Z_{\rm L}(p) = \lim_{\beta \hspace{0.05cm} \rightarrow \hspace{0.05cm} 0}\hspace{0.1cm}\frac{p}{p^2 + \beta^2} = {1}/{p}		:$$Z_{\rm L}(p) = \lim_{\beta \hspace{0.05cm} \rightarrow \hspace{0.05cm} 0}\hspace{0.1cm}\frac{p}{p^2 + \beta^2} = {1}/{p}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}z(t) = \gamma(t)\hspace{0.05cm} .$$
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
	z(t) = \gamma(t)
	\hspace{0.05cm} .$$
	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Linear and Time-Invariant Systems: Exercises\|^3.3 Inverse Laplace Transform^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^3.3 Inverse Laplace Transform^]]
			[[de:Zusatzaufgaben:3.6_Zwei_imaginäre_Pole]]

Guenter at 16:33, 9 December 2021

2021-12-09T16:33:43Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The <u>suggested solutions 1, 3 and 4</u> are correct:		'''(1)'''  The  <u>suggested solutions 1, 3 and 4</u>  are correct:
	*The following is obtained for signal  $x(t)$  for positive times by applying the residue theorem:		*The following is obtained for signal  $x(t)$  for positive times by applying the residue theorem:
	:$$x_1(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}		:$$x_1(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
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	'''(2)'''  The <u>suggested solutions 2 and 4</u> are correct:		'''(2)'''  The  <u>suggested solutions 2 and 4</u>  are correct:
	*In principle, this subtask could be solved in the same way as subtask  '''(1)'''.		*In principle,  this subtask could be solved in the same way as subtask  '''(1)'''.
	*However, the integration theorem can also be used.		*However,  the integration theorem can also be used.
	*This says among other things that multiplication by  $1/p$  in the spectral domain corresponds to integration in the time domain:		*This says among other things that multiplication by  $1/p$  in the spectral domain corresponds to integration in the time domain:
	:$$Y_{\rm L}(p) = {1}/{p} \cdot X_{\rm L}(p) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} t \ge 0:\quad y(t) = \int_{-\infty}^t \cos(2\pi		:$$Y_{\rm L}(p) = {1}/{p} \cdot X_{\rm L}(p) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} t \ge 0:\quad y(t) = \int_{-\infty}^t \cos(2\pi
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	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$

	''Please note:~~''  ~~   The causal cosine signal  $x(t)$  and the causal sine signal  $y(t)$  are shown on the information page of  [[Aufgaben:Exercise_3.6:_Transient_Behavior\|Exercise 3.6]]  as  $c_{\rm K}(t)$  and  $s_{\rm K}(t)$,  respectively.		Please note:  The causal cosine signal  $x(t)$  and the causal sine signal  $y(t)$  are shown on the information page of  [[Aufgaben:Exercise_3.6:_Transient_Behavior\|Exercise 3.6]]  as  $c_{\rm K}(t)$  and  $s_{\rm K}(t)$,  respectively.



	'''(3)'''  The <u>suggested solutions 1 and 3</u> are correct:		'''(3)'''  The  <u>suggested solutions 1 and 3</u>  are correct:
	*A comparison with the computation of  $x(t)$  shows that  $z(t) = \cos (\beta \cdot t)$  holds for  $t \ge 0$  and  $z(t) = 0$  for  $t < 0$~~ ~~.		*A comparison with the computation of  $x(t)$  shows that  $z(t) = \cos (\beta \cdot t)$  holds for  $t \ge 0$  and  $z(t) = 0$  for  $t < 0$.
	*The limit process for  $\beta → 0$  thus results in the step function  $\gamma(t)$.		*The limit process for  $\beta → 0$  thus results in the step function  $\gamma(t)$.
	*The same result is obtained by consideration in the spectral domain:		*The same result is obtained by consideration in the spectral domain:

Guenter at 16:29, 9 December 2021

2021-12-09T16:29:01Z

← Older revision		Revision as of 18:29, 9 December 2021
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	}}		}}

	[[File:P_ID1786__LZI_Z_3_6.png\|right\|frame\|Two imaginary poles and one zero ]]		[[File:P_ID1786__LZI_Z_3_6.png\|right\|frame\|Two imaginary poles <br>and one zero ]]
	In this exercise, we consider a causal signal  $x(t)$  with the Laplace transform		In this exercise,  we consider a causal signal  $x(t)$  with the Laplace transform
	:$$X_{\rm L}(p) =		:$$X_{\rm L}(p) =
	\frac { p} { p^2 + 4 \pi^2}=		\frac { p} { p^2 + 4 \pi^2}=
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	corresponding to the graph  (one red zero and two green poles).		corresponding to the graph  (one red zero and two green poles).

	In contrast, the signal  $y(t)$  has the Laplace spectral function		*In contrast,  the signal  $y(t)$  has the Laplace spectral function
	:$$Y_{\rm L}(p) =		:$$Y_{\rm L}(p) =
	\frac { 1} { p^2 + 4 \pi^2}		\frac { 1} { p^2 + 4 \pi^2}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	Thus, the red zero does not belong to  $Y_{\rm L}(p)$.		:Thus,  the red zero does not belong to  $Y_{\rm L}(p)$.

	Finally, the signal  $z(t)$  with the Laplace tansform		*Finally, the signal  $z(t)$  with the Laplace tansform
	:$$Z_{\rm L}(p) =		:$$Z_{\rm L}(p) =
	\frac { p} { (p-{\rm j} \cdot \beta)(p+{\rm j} \cdot \beta)}		\frac { p} { (p-{\rm j} \cdot \beta)(p+{\rm j} \cdot \beta)}
	\hspace{0.05cm}$$		\hspace{0.05cm}$$
	is considered, in particular the limiting case for  $\beta → 0$.		:is considered, in particular the limiting case for  $\beta → 0$.


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			Please note:


	''Please note:''
	*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform\|Inverse Laplace Transform]].		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform\|Inverse Laplace Transform]].
	*The frequency variable  $p$  is normalized such that time  $t$  is in microseconds after applying the residue theorem.		*The frequency variable  $p$   is normalized such that time  $t$  is in microseconds after applying the residue theorem.
	*A result  $t = 1$  is thus to be interpreted as  $t = T$  with  $T = 1 \ \rm µ s$ .		*A result  $t = 1$  is thus to be interpreted as  $t = T$  with  $T = 1 \ \rm µ s$ .
	*The  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem\|residue theorem]]  is as follows using the example of the function  $X_{\rm L}(p)$  with two simple poles at  $ \pm {\rm j} \cdot \beta$:		*The  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem\|residue theorem]]  is as follows using the example of the function  $X_{\rm L}(p)$  with two simple poles at  $ \pm {\rm j} \cdot \beta$:
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	<quiz display=simple>		<quiz display=simple>
	{Compute the signal  $x(t)$. Which of the following statements are correct?		{Compute the signal  $x(t)$.  Which of the following statements are correct?
	\|type="[]"}		\|type="[]"}
	+ $x(t)$  is a causal cosine signal.		+ $x(t)$  is a causal cosine signal.
	- $x(t)$  is a causal sinusoidal signal.		- $x(t)$  is a causal sinusoidal signal.
	+ The amplitude of  $x(t)$  is  $1$.		+ The amplitude of  $x(t)$  is  $1$.
	+ The period of $x(t)$ is  $T = 1 \ \rm µ s$.		+ The period of  $x(t)$  is  $T = 1 \ \rm µ s$.

Oezer at 08:15, 25 October 2021

2021-10-25T08:15:01Z

← Older revision		Revision as of 10:15, 25 October 2021
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	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$

	''Please note:''     The causal cosine signal  $x(t)$  and the causal sine signal  $y(t)$  are shown on the information page of  [[Aufgaben:Exercise_3.6:_Transient_Behavior\|Exercise 3.6]]  ~~als~~  $c_{\rm K}(t)$  and  $s_{\rm K}(t)$,  respectively.		''Please note:''     The causal cosine signal  $x(t)$  and the causal sine signal  $y(t)$  are shown on the information page of  [[Aufgaben:Exercise_3.6:_Transient_Behavior\|Exercise 3.6]]  as  $c_{\rm K}(t)$  and  $s_{\rm K}(t)$,  respectively.

Oezer at 08:14, 25 October 2021

2021-10-25T08:14:17Z

← Older revision		Revision as of 10:14, 25 October 2021
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	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$

	''Please note'':     The causal cosine signal  $x(t)$  and the causal sine signal  $y(t)$  are shown on the information page of  [[Aufgaben:Exercise_3.6:_Transient_Behavior\|Exercise 3.6]]  als  $c_{\rm K}(t)$  and  $s_{\rm K}(t)$,  respectively.		''Please note:''     The causal cosine signal  $x(t)$  and the causal sine signal  $y(t)$  are shown on the information page of  [[Aufgaben:Exercise_3.6:_Transient_Behavior\|Exercise 3.6]]  als  $c_{\rm K}(t)$  and  $s_{\rm K}(t)$,  respectively.

Oezer at 08:12, 25 October 2021

2021-10-25T08:12:44Z

← Older revision		Revision as of 10:12, 25 October 2021
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	+ $x(t)$  is a causal cosine signal.		+ $x(t)$  is a causal cosine signal.
	- $x(t)$  is a causal sinusoidal signal.		- $x(t)$  is a causal sinusoidal signal.
	+ The amplitude ~~ofn~~  $x(t)$  is  $1$.		+ The amplitude of  $x(t)$  is  $1$.
	+ The period of $x(t)$ is  $T = 1 \ \rm µ s$.		+ The period of $x(t)$ is  $T = 1 \ \rm µ s$.

Oezer at 08:11, 25 October 2021

2021-10-25T08:11:12Z

← Older revision		Revision as of 10:11, 25 October 2021
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	'''(3)'''  The <u>suggested solutions 1 and 3</u> are correct:		'''(3)'''  The <u>suggested solutions 1 and 3</u> are correct:
	*A comparison with the computation of  $x(t)$  shows that  $z(t) = \cos (\beta \cdot t)$  holds for  $t \ge 0$  and  $z(t) = 0$  ~~holds~~ for  $t < 0$ .		*A comparison with the computation of  $x(t)$  shows that  $z(t) = \cos (\beta \cdot t)$  holds for  $t \ge 0$  and  $z(t) = 0$  for  $t < 0$ .
	*The limit process for  $\beta → 0$  thus results in the step function  $\gamma(t)$.		*The limit process for  $\beta → 0$  thus results in the step function  $\gamma(t)$.
	*The same result is obtained by consideration in the spectral domain:		*The same result is obtained by consideration in the spectral domain: