Aufgaben:Exercise 3.2Z: Laplace and Fourier - Revision history

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← Older revision		Revision as of 15:48, 16 March 2026
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	[[File:P_ID1764__LZI_Z_3_2.png\|right\|frame\|Four causal time signals]]		[[File:P_ID1764__LZI_Z_3_2.png\|right\|frame\|Four causal time signals]]
	The Fourier transformation can be applied to any deterministic signal  $x(t)$.  Then, the following holds for the spectral function:		The Fourier transformation can be applied to any deterministic signal  $x(t)$.  Then, the following holds for the spectral function:
	:$$X(f) = \int_{-\infty}^{		:$$X(f) = \int_{-\infty}^{+\infty}{ x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi f t}}\hspace{0.1cm}{\rmd}t\hspace{0.05cm}\hspace{0.05cm} .$$
	+\infty}
	{ x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi f t}}\hspace{0.1cm}{\rm
	d}t\hspace{0.05cm}\hspace{0.05cm} .$$

	For power-limited signals  – characteristics:   infinite energy –  $X(f)$  also includes distributions  (Dirac delta functions).		For power-limited signals  – characteristics:   infinite energy –  $X(f)$  also includes distributions  (Dirac delta functions).

	For all causal signals  (and only for these),  the Laplace transformation is also applicable beside the Fourier transformation:		For all causal signals  (and only for these),  the Laplace transformation is also applicable beside the Fourier transformation:
	:$$X_{\rm L}(p) = \int_{0}^{		:$$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 the diagram you can see several causal time functions that will be covered in this exercise:		In the diagram you can see several causal time functions that will be covered in this exercise:
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	* For example,  the  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem\|shifting theorem]]  in Laplace or Fourier representation is:		* For example,  the  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem\|shifting theorem]]  in Laplace or Fourier representation is:
	:$$x(t- \tau) \quad		:$$x(t- \tau) \quad\circ\!\!-\!\!\!-^{\hspace{-0.25cm}\rmL}\!\!\!-\!\!\hspace{-0.05cm}\bullet\quad X_{\rm L}(p)\cdot {\rme}^{-p \hspace{0.05cm} \cdot \hspace{0.05cm}\tau}\hspace{0.05cm} ,$$
	\circ\!\!-\!\!\!-^{\hspace{-0.25cm}\rm		:$$x(t- \tau) \quad\circ\!\!-\!\!\!-^{\hspace{-0.05cm}}\!\!\!-\!\!\hspace{-0.05cm}\bullet\quadX(f)\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2\pi f \hspace{0.05cm} \cdot \hspace{0.05cm}\tau}\hspace{0.05cm} .$$
	L}\!\!\!-\!\!\hspace{-0.05cm}\bullet\quad X_{\rm L}(p)\cdot {\rm
	e}^{-p \hspace{0.05cm} \cdot \hspace{0.05cm}\tau}\hspace{0.05cm} ,$$
	:$$x(t- \tau) \quad
	\circ\!\!-\!\!\!-^{\hspace{-0.05cm}}\!\!\!-\!\!\hspace{-0.05cm}\bullet\~~quad~~
	X(f)\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2\pi f \hspace{0.05cm} \cdot \hspace{0.05cm}\tau}\hspace{0.05cm} .$$

	* In contrast,  there are differences in the  [[Signal_Representation/Fourier_Transform_Theorems#Integration_Theorem\|integration theorem]] :		* In contrast,  there are differences in the  [[Signal_Representation/Fourier_Transform_Theorems#Integration_Theorem\|integration theorem]] :
	:$$\int {x(\tau)} \hspace{0.1cm}{\rm		:$$\int {x(\tau)} \hspace{0.1cm}{\rmd}\tau \quad\circ\!\!-\!\!\!-^{\hspace{-0.25cm}\rmL}\!\!\!-\!\!\hspace{-0.05cm}\bullet\quad X_{\rm L}(p)\cdot\frac{1}{p}\hspace{0.05cm} ,$$
	d}\tau \quad		:$$\int {x(\tau)} \hspace{0.1cm}{\rmd}\tau \quad\circ\!\!-\!\!\!-^{\hspace{-0.05cm}}\!\!\!-\!\!\hspace{-0.05cm}\bullet\quadX(f)\cdot \left [ {1}/{2} \cdot{\rm \delta } (f) +\frac{1}{{\rm j} \cdot 2\pi f} \right ] \hspace{0.05cm} .$$
	\circ\!\!-\!\!\!-^{\hspace{-0.25cm}\rm
	L}\!\!\!-\!\!\hspace{-0.05cm}\bullet\quad X_{\rm L}(p)\cdot
	\frac{1}{p}\hspace{0.05cm} ,$$
	:$$\int {x(\tau)} \hspace{0.1cm}{\rm
	d}\tau \quad
	\circ\!\!-\!\!\!-^{\hspace{-0.05cm}}\!\!\!-\!\!\hspace{-0.05cm}\bullet\~~quad~~
	X(f)\cdot \left [ {1}/{2} \cdot{\rm \delta } (f) +
	\frac{1}{{\rm j} \cdot 2\pi f} \right ] \hspace{0.05cm} .$$


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	'''(1)'''  <u>Suggested solutions 1 and 3</u>  are correct:		'''(1)'''  <u>Suggested solutions 1 and 3</u>  are correct:
	*Considering that the Dirac delta function is non-zero only at  $t= 0$  and that the integral over the Dirac delta function yields the value  $1$  as long as the integration interval includes the time  $t= 0$,  the following is obtained:		*Considering that the Dirac delta function is non-zero only at  $t= 0$  and that the integral over the Dirac delta function yields the value  $1$  as long as the integration interval includes the time  $t= 0$,  the following is obtained:
	:$$A(f) = 1, \hspace{0.2cm}A_{\rm		:$$A(f) = 1, \hspace{0.2cm}A_{\rmL}(p) = 1 \hspace{0.05cm} .$$
	L}(p) = 1 \hspace{0.05cm} .$$


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	'''(2)'''  <u>Suggested solutions 1 and 3</u>  are correct again:		'''(2)'''  <u>Suggested solutions 1 and 3</u>  are correct again:
	*The step function  $b(t) = \gamma(t)$  is the integral over the Dirac delta function  $a(t) = \delta(t)$   ⇒   the  [[Signal_Representation/Fourier_Transform_Theorems#Integration_Theorem\|integration theorem]]  can be applied:		*The step function  $b(t) = \gamma(t)$  is the integral over the Dirac delta function  $a(t) = \delta(t)$   ⇒   the  [[Signal_Representation/Fourier_Transform_Theorems#Integration_Theorem\|integration theorem]]  can be applied:
	:$$b(t) = \int_{-\infty}^t {a(\tau)} \hspace{0.1cm}{\rm		:$$b(t) = \int_{-\infty}^t {a(\tau)} \hspace{0.1cm}{\rmd}\tau \hspace{0.3cm}\Rightarrow \hspace{0.3cm} B_{\rm L}(p) =A_{\rm L}(p)\cdot{1}/{p} = {1}/{p}\hspace{0.05cm} ,$$
	d}\tau \hspace{0.3cm}\Rightarrow \hspace{0.3cm} B_{\rm L}(p) =A_{\rm L}(p)\cdot		:$$B(f) = A(f)\cdot \left [ {1}/{2} \cdot{\rm \delta } (f) +\frac{1}{{\rm j} \cdot 2\pi f} \right ] = {1}/{2} \cdot{\rm\delta } (f) + \frac{1}{{\rm j} \cdot 2\pi f}\hspace{0.05cm} .$$
	{1}/{p} = {1}/{p}\hspace{0.05cm} ,$$
	:$$B(f) = A(f)\cdot \left [ {1}/{2} \cdot{\rm \delta } (f) +
	\frac{1}{{\rm j} \cdot 2\pi f} \right ] = {1}/{2} \cdot{\rm
	\delta } (f) + \frac{1}{{\rm j} \cdot 2\pi f}\hspace{0.05cm} .$$


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	'''(3)'''  <u>Suggested solutions 2 and 3</u>  are correct:		'''(3)'''  <u>Suggested solutions 2 and 3</u>  are correct:
	*Since the  (causal)  rectangular function can be represented as the difference of two step functions,  the  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem\|shifting theorem]]  yields:		*Since the  (causal)  rectangular function can be represented as the difference of two step functions,  the  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem\|shifting theorem]]  yields:
	:$$c(t)= b(t) - b(t-T) \hspace{0.3cm}		:$$c(t)= b(t) - b(t-T) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} C_{\rm L}(p) =B_{\rm L}(p)- B_{\rm L}(p)\cdot {\rm e}^{-p \hspace{0.05cm} \cdot \hspace{0.05cm}T} = {1}/{p} \cdot \big [ 1- {\rm e}^{-p \hspace{0.05cm} \cdot \hspace{0.05cm}T} \big ]\hspace{0.05cm} .$$
	\Rightarrow \hspace{0.3cm} C_{\rm L}(p) =B_{\rm L}(p)- B_{\rm L}(p)
	\cdot {\rm e}^{-p \hspace{0.05cm} \cdot \hspace{0.05cm}T} = {1}/{p} \cdot \big [ 1- {\rm e}^{-p \hspace{0.05cm} \cdot \hspace{0.05cm}T} \big ]
	\hspace{0.05cm} .$$

	*The following holds for the  [[Signal_Representation/Fourier_Transform_and_its_Inverse#The_first_Fourier_integral\|Fourier spectrum]]  since the rectangular function has finite energy:		*The following holds for the  [[Signal_Representation/Fourier_Transform_and_its_Inverse#The_first_Fourier_integral\|Fourier spectrum]]  since the rectangular function has finite energy:
	:$$C(f) = C_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it		:$$C(f) = C_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \itf}} = \frac{1}{{\rm j} \cdot 2\pi f} \cdot \big [ 1- {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\pi f T} \big ]\hspace{0.05cm}.$$
	f}} = \frac{1}{{\rm j} \cdot 2\pi f} \cdot \big [ 1- {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\pi f T} \big ]
	\hspace{0.05cm}.$$
	*The following can also be written for this using some trigonometric transformations:		*The following can also be written for this using some trigonometric transformations:
	:$$C(f) = T \cdot {\rm si} (2 \pi f{T})+ {\rm j} \cdot \frac{{\rm cos} (2 \pi f{T})-1}{2\pi f}		:$$C(f) = T \cdot {\rm si} (2 \pi f{T})+ {\rm j} \cdot \frac{{\rm cos} (2 \pi f{T})-1}{2\pi f}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$



	'''(4)'''  <u>Suggested solution 1</u>  is correct because the following holds:		'''(4)'''  <u>Suggested solution 1</u>  is correct because the following holds:
	:$$d(t) = \frac{1}{T} \cdot \int\limits_{-\infty}^t {c(\tau)} \hspace{0.1cm}{\rm		:$$d(t) = \frac{1}{T} \cdot \int\limits_{-\infty}^t {c(\tau)} \hspace{0.1cm}{\rmd}\tau \hspace{0.3cm}\Rightarrow \hspace{0.3cm} D_{\rm L}(p) =C_{\rm L}(p)\cdot\frac{1}{p \cdot T} = \frac{1- {\rm e}^{-p \hspace{0.05cm}\cdot \hspace{0.05cm}T}}{p^2 \cdotT}\hspace{0.05cm} .$$
	d}\tau \hspace{0.3cm}\Rightarrow \hspace{0.3cm} D_{\rm L}(p) =C_{\rm L}(p)\cdot
	\frac{1}{p \cdot T} = \frac{1- {\rm e}^{-p \hspace{0.05cm}\cdot \hspace{0.05cm}T}}{p^2 \~~cdot~~
	T}\hspace{0.05cm} .$$
	*Since  $d(t)$  extends to infinity,  the simple relation between  $D_{\rm L}(p)$  and  $D(f)$  according to proposed solution 3 is not valid.		*Since  $d(t)$  extends to infinity,  the simple relation between  $D_{\rm L}(p)$  and  $D(f)$  according to proposed solution 3 is not valid.
	*$D(f)$  rather also includes a Dirac delta function at frequency  $f = 0$.		*$D(f)$  rather also includes a Dirac delta function at frequency  $f = 0$.

Maintenance script: Add German interlanguage link

2026-03-16T12:29:11Z

Add German interlanguage link

← Older revision		Revision as of 14:29, 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:Zusatzaufgaben:3.2_Laplace_und_Fourier]]

Guenter at 12:20, 13 October 2021

2021-10-13T12:20:13Z

← Older revision		Revision as of 14:20, 13 October 2021
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Suggested solutions 1 and 3</u> are correct:		'''(1)'''  <u>Suggested solutions 1 and 3</u>  are correct:
	*Considering that the Dirac function is non-zero only at  $t= 0$  and that the integral over the Dirac yields the value  $1$  as long as the integration interval includes the time  $t= 0$,  the following is obtained:		*Considering that the Dirac delta function is non-zero only at  $t= 0$  and that the integral over the Dirac delta function yields the value  $1$  as long as the integration interval includes the time  $t= 0$,  the following is obtained:
	:$$A(f) = 1, \hspace{0.2cm}A_{\rm		:$$A(f) = 1, \hspace{0.2cm}A_{\rm
	L}(p) = 1 \hspace{0.05cm} .$$		L}(p) = 1 \hspace{0.05cm} .$$
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	'''(2)'''  <u>Suggested solutions 1 and 3</u> are correct again:		'''(2)'''  <u>Suggested solutions 1 and 3</u>  are correct again:
	*The step function  $b(t) = \gamma(t)$  is the integral over the Dirac function  $a(t) = \delta(t)$   ⇒   the  [[Signal_Representation/Fourier_Transform_Theorems#Integration_Theorem\|integration theorem]]  can be applied:		*The step function  $b(t) = \gamma(t)$  is the integral over the Dirac delta function  $a(t) = \delta(t)$   ⇒   the  [[Signal_Representation/Fourier_Transform_Theorems#Integration_Theorem\|integration theorem]]  can be applied:
	:$$b(t) = \int_{-\infty}^t {a(\tau)} \hspace{0.1cm}{\rm		:$$b(t) = \int_{-\infty}^t {a(\tau)} \hspace{0.1cm}{\rm
	d}\tau \hspace{0.3cm}\Rightarrow \hspace{0.3cm} B_{\rm L}(p) =A_{\rm L}(p)\cdot		d}\tau \hspace{0.3cm}\Rightarrow \hspace{0.3cm} B_{\rm L}(p) =A_{\rm L}(p)\cdot
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	'''(3)'''  <u>Suggested solutions 2 and 3</u> are correct:		'''(3)'''  <u>Suggested solutions 2 and 3</u>  are correct:
	*Since the (causal) rectangular function can be represented as the difference of two step functions, the  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem\|shifting theorem]] yields:		*Since the  (causal)  rectangular function can be represented as the difference of two step functions,  the  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem\|shifting theorem]]  yields:
	:$$c(t)= b(t) - b(t-T) \hspace{0.3cm}		:$$c(t)= b(t) - b(t-T) \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} C_{\rm L}(p) =B_{\rm L}(p)- B_{\rm L}(p)		\Rightarrow \hspace{0.3cm} C_{\rm L}(p) =B_{\rm L}(p)- B_{\rm L}(p)
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	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$

	*The following holds for the  [[Signal_Representation/Fourier_Transform_and_its_Inverse#The_first_Fourier_integral\|Fourier spectrum]] since the rectangular function has finite energy:		*The following holds for the  [[Signal_Representation/Fourier_Transform_and_its_Inverse#The_first_Fourier_integral\|Fourier spectrum]]  since the rectangular function has finite energy:
	:$$C(f) = C_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it		:$$C(f) = C_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it
	f}} = \frac{1}{{\rm j} \cdot 2\pi f} \cdot \big [ 1- {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\pi f T} \big ]		f}} = \frac{1}{{\rm j} \cdot 2\pi f} \cdot \big [ 1- {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\pi f T} \big ]
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	'''(4)'''  <u>Suggested solution 1</u> is correct because the following holds:		'''(4)'''  <u>Suggested solution 1</u>  is correct because the following holds:
	:$$d(t) = \frac{1}{T} \cdot \int\limits_{-\infty}^t {c(\tau)} \hspace{0.1cm}{\rm		:$$d(t) = \frac{1}{T} \cdot \int\limits_{-\infty}^t {c(\tau)} \hspace{0.1cm}{\rm
	d}\tau \hspace{0.3cm}\Rightarrow \hspace{0.3cm} D_{\rm L}(p) =C_{\rm L}(p)\cdot		d}\tau \hspace{0.3cm}\Rightarrow \hspace{0.3cm} D_{\rm L}(p) =C_{\rm L}(p)\cdot
	\frac{1}{p \cdot T} = \frac{1- {\rm e}^{-p \hspace{0.05cm}\cdot \hspace{0.05cm}T}}{p^2 \cdot		\frac{1}{p \cdot T} = \frac{1- {\rm e}^{-p \hspace{0.05cm}\cdot \hspace{0.05cm}T}}{p^2 \cdot
	T}\hspace{0.05cm} .$$		T}\hspace{0.05cm} .$$
	*Since  $d(t)$  extends to infinity, the simple relation between  $D_{\rm L}(p)$  and  $D(f)$  according to proposed solution 3 is not valid.		*Since  $d(t)$  extends to infinity,  the simple relation between  $D_{\rm L}(p)$  and  $D(f)$  according to proposed solution 3 is not valid.
	*$D(f)$  rather also includes a Dirac function at frequency  $f = 0$.		*$D(f)$  rather also includes a Dirac delta function at frequency  $f = 0$.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 12:14, 13 October 2021

2021-10-13T12:14:59Z

← Older revision		Revision as of 14:14, 13 October 2021
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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]].
	*The ~~educational~~ video   [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)\|Fourier transform theorems]]  might be helpful.		*The  (German language)  learning video   [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)\|"Gesetzmäßigkeiten der Fouriertransformation"]]  ⇒   "Fourier transform theorems"  might be helpful.


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	- $C_{\rm L}(p) = {\rm si}(pT)$.		- $C_{\rm L}(p) = {\rm si}(pT)$.
	+ $C_{\rm L}(p) = \big [1-{\rm e}^{-p\hspace{0.05cm} \cdot \hspace{0.05cm}T} \big ]/p$.		+ $C_{\rm L}(p) = \big [1-{\rm e}^{-p\hspace{0.05cm} \cdot \hspace{0.05cm}T} \big ]/p$.
	+ $C(f) = C_{\rm L}(p)$  ~~mit~~  $p = 2 \pi f$.		+ $C(f) = C_{\rm L}(p)$  with  $p = 2 \pi f$.


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	+ $D_{\rm L}(p) = \big[1-{\rm e}^{-p\hspace{0.05cm} \cdot \hspace{0.05cm}T}\big]/(p^2T)$.		+ $D_{\rm L}(p) = \big[1-{\rm e}^{-p\hspace{0.05cm} \cdot \hspace{0.05cm}T}\big]/(p^2T)$.
	- $D_{\rm L}(p) = 1-{\rm e}^{-p\hspace{0.05cm} \cdot \hspace{0.05cm}T}$.		- $D_{\rm L}(p) = 1-{\rm e}^{-p\hspace{0.05cm} \cdot \hspace{0.05cm}T}$.
	- $D(f) = D_{\rm L}(p)$  ~~mit~~  $p = 2 \pi f$.		- $D(f) = D_{\rm L}(p)$  with  $p = 2 \pi f$.

Guenter at 12:09, 13 October 2021

2021-10-13T12:09:23Z

← Older revision		Revision as of 14:09, 13 October 2021
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	[[File:P_ID1764__LZI_Z_3_2.png\|right\|frame\|Four causal time signals]]		[[File:P_ID1764__LZI_Z_3_2.png\|right\|frame\|Four causal time signals]]
	The Fourier transformation can be applied to any deterministic signal  $x(t)$ . Then, the following holds for the spectral function:		The Fourier transformation can be applied to any deterministic signal  $x(t)$.  Then, the following holds for the spectral function:
	:$$X(f) = \int_{-\infty}^{		:$$X(f) = \int_{-\infty}^{
	+\infty}		+\infty}
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	d}t\hspace{0.05cm}\hspace{0.05cm} .$$		d}t\hspace{0.05cm}\hspace{0.05cm} .$$

	For power-limited signals – characteristics:   infinite energy – $X(f)$  also includes distributions (Dirac functions).		For power-limited signals  – characteristics:   infinite energy –  $X(f)$  also includes distributions  (Dirac delta functions).

	For all causal signals (and only for these), the Laplace transformation is also applicable beside the Fourier transformation:		For all causal signals  (and only for these),  the Laplace transformation is also applicable beside the Fourier transformation:
	:$$X_{\rm L}(p) = \int_{0}^{		:$$X_{\rm L}(p) = \int_{0}^{
	\infty}		\infty}
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	In the diagram you can see several causal time functions that will be covered in this exercise:		In the diagram you can see several causal time functions that will be covered in this exercise:

	* the Dirac function  $a(t)$,		* the Dirac delta function  $a(t)$,

	* the step function  $b(t)$,		* the unit step function  $b(t)$,

	* the rectangular function  $c(t)$,		* the rectangular function  $c(t)$,
Line 29:		Line 29:


	The  [[Signal_Representation/Fourier_Transform_Theorems\|Fourier transform theorems]]  usually (though not always) also apply to the Laplace transformation where  $p ={\rm j} \cdot 2 \pi f$  is to be set:		The  [[Signal_Representation/Fourier_Transform_Theorems\|Fourier transform theorems]]  usually  (though not always)  also apply to the Laplace transformation where  $p ={\rm j} \cdot 2 \pi f$  is to be set:

	* For example, the  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem\|shifting theorem]]  in Laplace or Fourier representation is:		* For example,  the  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem\|shifting theorem]]  in Laplace or Fourier representation is:
	:$$x(t- \tau) \quad		:$$x(t- \tau) \quad
	\circ\!\!-\!\!\!-^{\hspace{-0.25cm}\rm		\circ\!\!-\!\!\!-^{\hspace{-0.25cm}\rm
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	X(f)\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2\pi f \hspace{0.05cm} \cdot \hspace{0.05cm}\tau}\hspace{0.05cm} .$$		X(f)\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2\pi f \hspace{0.05cm} \cdot \hspace{0.05cm}\tau}\hspace{0.05cm} .$$

	* In contrast, there are differences in the  [[Signal_Representation/Fourier_Transform_Theorems#Integration_Theorem\|integration theorem]] :		* In contrast,  there are differences in the  [[Signal_Representation/Fourier_Transform_Theorems#Integration_Theorem\|integration theorem]] :
	:$$\int {x(\tau)} \hspace{0.1cm}{\rm		:$$\int {x(\tau)} \hspace{0.1cm}{\rm
	d}\tau \quad		d}\tau \quad
Line 57:		Line 57:


			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]].
	*The educational video   [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)\|Fourier transform theorems]]  might be helpful.		*The educational video   [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)\|Fourier transform theorems]]  might be helpful.

Oezer at 09:31, 11 October 2021

2021-10-11T09:31:11Z

← Older revision		Revision as of 11:31, 11 October 2021
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	The  [[Signal_Representation/Fourier_Transform_Theorems\|Fourier transform theorems]]  ~~gelten meist~~ (~~allerdings nicht immer~~) ~~auch für die~~ Laplace~~–Transformation, wobei~~  $p ={\rm j} \cdot 2 \pi f$  ~~zu setzen ist~~:		The  [[Signal_Representation/Fourier_Transform_Theorems\|Fourier transform theorems]]  usually (though not always) also apply to the Laplace transformation where  $p ={\rm j} \cdot 2 \pi f$  is to be set:

	* ~~Zum Beispiel lautet der~~  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem\|shifting theorem]]  in Laplace~~– bzw.~~ Fourier~~–Darstellung~~:		* For example, the  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem\|shifting theorem]]  in Laplace or Fourier representation is:
	:$$x(t- \tau) \quad		:$$x(t- \tau) \quad
	\circ\!\!-\!\!\!-^{\hspace{-0.25cm}\rm		\circ\!\!-\!\!\!-^{\hspace{-0.25cm}\rm
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	X(f)\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2\pi f \hspace{0.05cm} \cdot \hspace{0.05cm}\tau}\hspace{0.05cm} .$$		X(f)\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2\pi f \hspace{0.05cm} \cdot \hspace{0.05cm}\tau}\hspace{0.05cm} .$$

	* ~~Dagegen ergeben sich beim~~  [[Signal_Representation/Fourier_Transform_Theorems#Integration_Theorem\|integration theorem]]  ~~Unterschiede~~:		* In contrast, there are differences in the  [[Signal_Representation/Fourier_Transform_Theorems#Integration_Theorem\|integration theorem]] :
	:$$\int {x(\tau)} \hspace{0.1cm}{\rm		:$$\int {x(\tau)} \hspace{0.1cm}{\rm
	d}\tau \quad		d}\tau \quad
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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]].
	*~~Das Lernvideo~~   [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)\|Fourier transform theorems]]  ~~könnte hilfreich sein~~ .		*The educational video   [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)\|Fourier transform theorems]]  might be helpful.


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	<quiz display=simple>		<quiz display=simple>
	{~~Wie lauten die Spektraltransformationen des Signals~~  $a(t) = \delta(t)$?		{What are the spectral transformations of the signal  $a(t) = \delta(t)$?
	\|type="[]"}		\|type="[]"}
	+ $A_{\rm L}(p) = 1$.		+ $A_{\rm L}(p) = 1$.
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	{~~Wie lauten die Spektraltransformationen der Sprungfunktion~~  $b(t) = \gamma(t)$?		{What are the spectral transformations of the step function  $b(t) = \gamma(t)$?
	\|type="[]"}		\|type="[]"}
	+ $B_{\rm L}(p) = 1/p$.		+ $B_{\rm L}(p) = 1/p$.
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	{~~Wie lauten die Spektraltransformationen der Rechteckfunktion~~  $c(t)$?		{What are the spectral transformations of the rectangular function  $c(t)$?
	\|type="[]"}		\|type="[]"}
	- $C_{\rm L}(p) = {\rm si}(pT)$.		- $C_{\rm L}(p) = {\rm si}(pT)$.
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	{~~Wie lauten die Spektraltransformationen der Rampenfunktion~~  $d(t)$?		{What are the spectral transformations of the ramp function  $d(t)$?
	\|type="[]"}		\|type="[]"}
	+ $D_{\rm L}(p) = \big[1-{\rm e}^{-p\hspace{0.05cm} \cdot \hspace{0.05cm}T}\big]/(p^2T)$.		+ $D_{\rm L}(p) = \big[1-{\rm e}^{-p\hspace{0.05cm} \cdot \hspace{0.05cm}T}\big]/(p^2T)$.
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Richtig sind die~~ <u>~~Lösungsvorschläge~~ 1 ~~und~~ 3</u>:		'''(1)'''  <u>Suggested solutions 1 and 3</u> are correct:
	*~~Berücksichtigt man, dass die Diracfunktion nur bei~~  $t= 0$  ~~ungleich Null ist und das Integral über den~~ Dirac ~~den Wert~~  $1$  ~~liefert, solange das Integrationsintervall den Zeitpunkt~~  $t= 0$  ~~einschließt, so erhält man~~:		*Considering that the Dirac function is non-zero only at  $t= 0$  and that the integral over the Dirac yields the value  $1$  as long as the integration interval includes the time  $t= 0$,  the following is obtained:
	:$$A(f) = 1, \hspace{0.2cm}A_{\rm		:$$A(f) = 1, \hspace{0.2cm}A_{\rm
	L}(p) = 1 \hspace{0.05cm} .$$		L}(p) = 1 \hspace{0.05cm} .$$
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	'''(2)'''  ~~Richtig sind wieder die~~ <u>~~Lösungsvorschläge~~ 1 ~~und~~ 3</u>:		'''(2)'''  <u>Suggested solutions 1 and 3</u> are correct again:
	*~~Die Sprungfunktion~~  $b(t) = \gamma(t)$  ~~ist das Integral über die Diracfunktion~~  $a(t) = \delta(t)$   ⇒   ~~man kann den~~  [[Signal_Representation/Fourier_Transform_Theorems#Integration_Theorem\|integration theorem]]  ~~anwenden~~:		*The step function  $b(t) = \gamma(t)$  is the integral over the Dirac function  $a(t) = \delta(t)$   ⇒   the  [[Signal_Representation/Fourier_Transform_Theorems#Integration_Theorem\|integration theorem]]  can be applied:
	:$$b(t) = \int_{-\infty}^t {a(\tau)} \hspace{0.1cm}{\rm		:$$b(t) = \int_{-\infty}^t {a(\tau)} \hspace{0.1cm}{\rm
	d}\tau \hspace{0.3cm}\Rightarrow \hspace{0.3cm} B_{\rm L}(p) =A_{\rm L}(p)\cdot		d}\tau \hspace{0.3cm}\Rightarrow \hspace{0.3cm} B_{\rm L}(p) =A_{\rm L}(p)\cdot
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	'''(3)'''  ~~Richtig sind die~~ <u>~~Lösungsvorschläge~~ 2 ~~und~~ 3</u>:		'''(3)'''  <u>Suggested solutions 2 and 3</u> are correct:
	*~~Nachdem die~~ (~~kausale~~) ~~Rechteckfunktion als Differenz zweier Sprungfunktionen dargestellt werden kann~~, ~~erhält man mit dem~~  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem\|shifting theorem]]:		*Since the (causal) rectangular function can be represented as the difference of two step functions, the  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem\|shifting theorem]] yields:
	:$$c(t)= b(t) - b(t-T) \hspace{0.3cm}		:$$c(t)= b(t) - b(t-T) \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} C_{\rm L}(p) =B_{\rm L}(p)- B_{\rm L}(p)		\Rightarrow \hspace{0.3cm} C_{\rm L}(p) =B_{\rm L}(p)- B_{\rm L}(p)
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	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$

	*~~Da die Rechteckfunktion eine endliche Energie besitzt, gilt für das~~  [[Signal_Representation/Fourier_Transform_and_its_Inverse#The_first_Fourier_integral\|Fourier spectrum]]:		*The following holds for the  [[Signal_Representation/Fourier_Transform_and_its_Inverse#The_first_Fourier_integral\|Fourier spectrum]] since the rectangular function has finite energy:
	:$$C(f) = C_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it		:$$C(f) = C_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it
	f}} = \frac{1}{{\rm j} \cdot 2\pi f} \cdot \big [ 1- {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\pi f T} \big ]		f}} = \frac{1}{{\rm j} \cdot 2\pi f} \cdot \big [ 1- {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\pi f T} \big ]
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	*~~Nach einigen trigonometrischen Umformungen kann hierfür auch geschrieben werden~~:		*The following can also be written for this using some trigonometric transformations:
	:$$C(f) = T \cdot {\rm si} (2 \pi f{T})+ {\rm j} \cdot \frac{{\rm cos} (2 \pi f{T})-1}{2\pi f}		:$$C(f) = T \cdot {\rm si} (2 \pi f{T})+ {\rm j} \cdot \frac{{\rm cos} (2 \pi f{T})-1}{2\pi f}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
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	'''(4)'''  ~~Richtig ist der~~ <u>~~Lösungsvorschlag~~ 1</u>~~, da gilt~~:		'''(4)'''  <u>Suggested solution 1</u> is correct because the following holds:
	:$$d(t) = \frac{1}{T} \cdot \int\limits_{-\infty}^t {c(\tau)} \hspace{0.1cm}{\rm		:$$d(t) = \frac{1}{T} \cdot \int\limits_{-\infty}^t {c(\tau)} \hspace{0.1cm}{\rm
	d}\tau \hspace{0.3cm}\Rightarrow \hspace{0.3cm} D_{\rm L}(p) =C_{\rm L}(p)\cdot		d}\tau \hspace{0.3cm}\Rightarrow \hspace{0.3cm} D_{\rm L}(p) =C_{\rm L}(p)\cdot
	\frac{1}{p \cdot T} = \frac{1- {\rm e}^{-p \hspace{0.05cm}\cdot \hspace{0.05cm}T}}{p^2 \cdot		\frac{1}{p \cdot T} = \frac{1- {\rm e}^{-p \hspace{0.05cm}\cdot \hspace{0.05cm}T}}{p^2 \cdot
	T}\hspace{0.05cm} .$$		T}\hspace{0.05cm} .$$
	*~~Da sich~~  $d(t)$  ~~bis ins Unendliche erstreckt~~, ~~ist der einfache Zusammenhang zwischen~~  $D_{\rm L}(p)$  ~~und~~  $D(f)$  ~~gemäß Lösungsvorschlag~~ 3 ~~nicht gegeben~~.		*Since  $d(t)$  extends to infinity, the simple relation between  $D_{\rm L}(p)$  and  $D(f)$  according to proposed solution 3 is not valid.
	*$D(f)$  ~~beinhaltet vielmehr auch eine Diracfunktion bei der Frequenz~~  $f = 0$.		*$D(f)$  rather also includes a Dirac function at frequency  $f = 0$.
	{{ML-Fuß}}		{{ML-Fuß}}

Oezer at 22:53, 10 October 2021

2021-10-10T22:53:28Z

← Older revision		Revision as of 00:53, 11 October 2021
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	For power-limited signals – characteristics:   infinite energy – $X(f)$  also includes distributions (Dirac functions).		For power-limited signals – characteristics:   infinite energy – $X(f)$  also includes distributions (Dirac functions).

	For all causal signals (and only for these), the Laplace transformation is also applicable ~~besides~~ the Fourier transformation:		For all causal signals (and only for these), the Laplace transformation is also applicable beside the Fourier transformation:
	:$$X_{\rm L}(p) = \int_{0}^{		:$$X_{\rm L}(p) = \int_{0}^{
	\infty}		\infty}

Oezer at 22:52, 10 October 2021

2021-10-10T22:52:00Z

← Older revision		Revision as of 00:52, 11 October 2021
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	For power-limited signals – characteristics:   infinite energy – $X(f)$  also includes distributions (Dirac functions).		For power-limited signals – characteristics:   infinite energy – $X(f)$  also includes distributions (Dirac functions).

	For all causal signals (and only for these), the Laplace transformation is applicable:		For all causal signals (and only for these), the Laplace transformation is also applicable besides the Fourier transformation:
	:$$X_{\rm L}(p) = \int_{0}^{		:$$X_{\rm L}(p) = \int_{0}^{
	\infty}		\infty}

Oezer at 22:49, 10 October 2021

2021-10-10T22:49:35Z

← Older revision		Revision as of 00:49, 11 October 2021
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	For power-limited signals – characteristics:   infinite energy – $X(f)$  also includes distributions (Dirac functions).		For power-limited signals – characteristics:   infinite energy – $X(f)$  also includes distributions (Dirac functions).

	~~Bei allen kausalen Signalen~~ (~~und nur bei diesen~~) ~~ist daneben auch die~~ Laplace~~-Transformation anwendbar~~:		For all causal signals (and only for these), the Laplace transformation is applicable:
	:$$X_{\rm L}(p) = \int_{0}^{		:$$X_{\rm L}(p) = \int_{0}^{
	\infty}		\infty}

← Older revision		Revision as of 17:56, 16 March 2026
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	[[File:P_ID1764__LZI_Z_3_2.png\|right\|frame\|Four causal time signals]]		[[File:P_ID1764__LZI_Z_3_2.png\|right\|frame\|Four causal time signals]]
	The Fourier transformation can be applied to any deterministic signal  $x(t)$.  Then, the following holds for the spectral function:		The Fourier transformation can be applied to any deterministic signal  $x(t)$.  Then, the following holds for the spectral function:
	:$$X(f) = \int_{-\infty}^{+\infty}{ x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi f t}}\hspace{0.1cm}{\~~rmd~~}t\hspace{0.05cm}\hspace{0.05cm} .$$		:$$X(f) = \int_{-\infty}^{+\infty}{ x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi f t}}\hspace{0.1cm}{\rm d}t\hspace{0.05cm}\hspace{0.05cm} .$$

	For power-limited signals  – characteristics:   infinite energy –  $X(f)$  also includes distributions  (Dirac delta functions).		For power-limited signals  – characteristics:   infinite energy –  $X(f)$  also includes distributions  (Dirac delta functions).

	For all causal signals  (and only for these),  the Laplace transformation is also applicable beside the Fourier transformation:		For all causal signals  (and only for these),  the Laplace transformation is also applicable beside the Fourier transformation:
	:$$X_{\rm L}(p) = \int_{0}^{\infty}{ x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\~~rmd~~}t\hspace{0.05cm}\hspace{0.05cm} .$$		:$$X_{\rm L}(p) = \int_{0}^{\infty}{ x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\rm d}t\hspace{0.05cm}\hspace{0.05cm} .$$

	In the diagram you can see several causal time functions that will be covered in this exercise:		In the diagram you can see several causal time functions that will be covered in this exercise:
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	The  [[Signal_Representation/Fourier_Transform_Theorems\|Fourier transform theorems]]  usually  (though not always)  also apply to the Laplace transformation where  $p ={\rm j} \cdot 2 \pi f$  is to be set:		The  [[Signal_Representation/Fourier_Transform_Theorems\|Fourier transform theorems]]  usually  (though not always)  also apply to the Laplace transformation where  $p ={\rm j} \cdot 2 \pi f$  is to be set:

	* For example,  the  [[Signal_Representation/~~Fourier_Transform_Theorems~~#Shifting_Theorem\|shifting theorem]]  in Laplace or Fourier representation is:		* For example,  the  [[Signal_Representation/The_Fourier_Transform_Theorems#Shifting_Theorem\|shifting theorem]]  in Laplace or Fourier representation is:
	:$$x(t- \tau) \quad\circ\!\!-\!\!\!-^{\hspace{-0.25cm}\~~rmL~~}\!\!\!-\!\!\hspace{-0.05cm}\bullet\quad X_{\rm L}(p)\cdot {\~~rme~~}^{-p \hspace{0.05cm} \cdot \hspace{0.05cm}\tau}\hspace{0.05cm} ,$$		:$$x(t- \tau) \quad\circ\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\hspace{-0.05cm}\bullet\quad X_{\rm L}(p)\cdot {\rm e}^{-p \hspace{0.05cm} \cdot \hspace{0.05cm}\tau}\hspace{0.05cm} ,$$
	:$$x(t- \tau) \quad\circ\!\!-\!\!\!-^{\hspace{-0.05cm}}\!\!\!-\!\!\hspace{-0.05cm}\bullet\quadX(f)\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2\pi f \hspace{0.05cm} \cdot \hspace{0.05cm}\tau}\hspace{0.05cm} .$$		:$$x(t- \tau) \quad\circ\!\!-\!\!\!-^{\hspace{-0.05cm}}\!\!\!-\!\!\hspace{-0.05cm}\bullet\quadX(f)\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2\pi f \hspace{0.05cm} \cdot \hspace{0.05cm}\tau}\hspace{0.05cm} .$$

	* In contrast,  there are differences in the  [[Signal_Representation/~~Fourier_Transform_Theorems~~#Integration_Theorem\|integration theorem]] :		* In contrast,  there are differences in the  [[Signal_Representation/The_Fourier_Transform_Theorems#Integration_Theorem\|integration theorem]] :
	:$$\int {x(\tau)} \hspace{0.1cm}{\~~rmd~~}\tau \quad\circ\!\!-\!\!\!-^{\hspace{-0.25cm}\~~rmL~~}\!\!\!-\!\!\hspace{-0.05cm}\bullet\quad X_{\rm L}(p)\cdot\frac{1}{p}\hspace{0.05cm} ,$$		:$$\int {x(\tau)} \hspace{0.1cm}{\rm d}\tau \quad\circ\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\hspace{-0.05cm}\bullet\quad X_{\rm L}(p)\cdot\frac{1}{p}\hspace{0.05cm} ,$$
	:$$\int {x(\tau)} \hspace{0.1cm}{\~~rmd~~}\tau \quad\circ\!\!-\!\!\!-^{\hspace{-0.05cm}}\!\!\!-\!\!\hspace{-0.05cm}\bullet\quadX(f)\cdot \left [ {1}/{2} \cdot{\rm \delta } (f) +\frac{1}{{\rm j} \cdot 2\pi f} \right ] \hspace{0.05cm} .$$		:$$\int {x(\tau)} \hspace{0.1cm}{\rm d}\tau \quad\circ\!\!-\!\!\!-^{\hspace{-0.05cm}}\!\!\!-\!\!\hspace{-0.05cm}\bullet\quadX(f)\cdot \left [ {1}/{2} \cdot{\rm \delta } (f) +\frac{1}{{\rm j} \cdot 2\pi f} \right ] \hspace{0.05cm} .$$


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	'''(1)'''  <u>Suggested solutions 1 and 3</u>  are correct:		'''(1)'''  <u>Suggested solutions 1 and 3</u>  are correct:
	*Considering that the Dirac delta function is non-zero only at  $t= 0$  and that the integral over the Dirac delta function yields the value  $1$  as long as the integration interval includes the time  $t= 0$,  the following is obtained:		*Considering that the Dirac delta function is non-zero only at  $t= 0$  and that the integral over the Dirac delta function yields the value  $1$  as long as the integration interval includes the time  $t= 0$,  the following is obtained:
	:$$A(f) = 1, \hspace{0.2cm}A_{\~~rmL~~}(p) = 1 \hspace{0.05cm} .$$		:$$A(f) = 1, \hspace{0.2cm}A_{\rm L}(p) = 1 \hspace{0.05cm} .$$



	'''(2)'''  <u>Suggested solutions 1 and 3</u>  are correct again:		'''(2)'''  <u>Suggested solutions 1 and 3</u>  are correct again:
	*The step function  $b(t) = \gamma(t)$  is the integral over the Dirac delta function  $a(t) = \delta(t)$   ⇒   the  [[Signal_Representation/~~Fourier_Transform_Theorems~~#Integration_Theorem\|integration theorem]]  can be applied:		*The step function  $b(t) = \gamma(t)$  is the integral over the Dirac delta function  $a(t) = \delta(t)$   ⇒   the  [[Signal_Representation/The_Fourier_Transform_Theorems#Integration_Theorem\|integration theorem]]  can be applied:
	:$$b(t) = \int_{-\infty}^t {a(\tau)} \hspace{0.1cm}{\~~rmd~~}\tau \hspace{0.3cm}\Rightarrow \hspace{0.3cm} B_{\rm L}(p) =A_{\rm L}(p)\cdot{1}/{p} = {1}/{p}\hspace{0.05cm} ,$$		:$$b(t) = \int_{-\infty}^t {a(\tau)} \hspace{0.1cm}{\rm d}\tau \hspace{0.3cm}\Rightarrow \hspace{0.3cm} B_{\rm L}(p) =A_{\rm L}(p)\cdot{1}/{p} = {1}/{p}\hspace{0.05cm} ,$$
	:$$B(f) = A(f)\cdot \left [ {1}/{2} \cdot{\rm \delta } (f) +\frac{1}{{\rm j} \cdot 2\pi f} \right ] = {1}/{2} \cdot{\rm\delta } (f) + \frac{1}{{\rm j} \cdot 2\pi f}\hspace{0.05cm} .$$		:$$B(f) = A(f)\cdot \left [ {1}/{2} \cdot{\rm \delta } (f) +\frac{1}{{\rm j} \cdot 2\pi f} \right ] = {1}/{2} \cdot{\rm\delta } (f) + \frac{1}{{\rm j} \cdot 2\pi f}\hspace{0.05cm} .$$

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	'''(3)'''  <u>Suggested solutions 2 and 3</u>  are correct:		'''(3)'''  <u>Suggested solutions 2 and 3</u>  are correct:
	*Since the  (causal)  rectangular function can be represented as the difference of two step functions,  the  [[Signal_Representation/~~Fourier_Transform_Theorems~~#Shifting_Theorem\|shifting theorem]]  yields:		*Since the  (causal)  rectangular function can be represented as the difference of two step functions,  the  [[Signal_Representation/The_Fourier_Transform_Theorems#Shifting_Theorem\|shifting theorem]]  yields:
	:$$c(t)= b(t) - b(t-T) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} C_{\rm L}(p) =B_{\rm L}(p)- B_{\rm L}(p)\cdot {\rm e}^{-p \hspace{0.05cm} \cdot \hspace{0.05cm}T} = {1}/{p} \cdot \big [ 1- {\rm e}^{-p \hspace{0.05cm} \cdot \hspace{0.05cm}T} \big ]\hspace{0.05cm} .$$		:$$c(t)= b(t) - b(t-T) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} C_{\rm L}(p) =B_{\rm L}(p)- B_{\rm L}(p)\cdot {\rm e}^{-p \hspace{0.05cm} \cdot \hspace{0.05cm}T} = {1}/{p} \cdot \big [ 1- {\rm e}^{-p \hspace{0.05cm} \cdot \hspace{0.05cm}T} \big ]\hspace{0.05cm} .$$

	*The following holds for the  [[Signal_Representation/~~Fourier_Transform_and_its_Inverse~~#The_first_Fourier_integral\|Fourier spectrum]]  since the rectangular function has finite energy:		*The following holds for the  [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#The_first_Fourier_integral\|Fourier spectrum]]  since the rectangular function has finite energy:
	:$$C(f) = C_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \~~itf~~}} = \frac{1}{{\rm j} \cdot 2\pi f} \cdot \big [ 1- {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\pi f T} \big ]\hspace{0.05cm}.$$		:$$C(f) = C_{\rm L}(p)\Bigg \|_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it f}} = \frac{1}{{\rm j} \cdot 2\pi f} \cdot \big [ 1- {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\pi f T} \big ]\hspace{0.05cm}.$$
	*The following can also be written for this using some trigonometric transformations:		*The following can also be written for this using some trigonometric transformations:
	:$$C(f) = T \cdot {\rm si} (2 \pi f{T})+ {\rm j} \cdot \frac{{\rm cos} (2 \pi f{T})-1}{2\pi f}\hspace{0.05cm}.$$		:$$C(f) = T \cdot {\rm si} (2 \pi f{T})+ {\rm j} \cdot \frac{{\rm cos} (2 \pi f{T})-1}{2\pi f}\hspace{0.05cm}.$$
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	'''(4)'''  <u>Suggested solution 1</u>  is correct because the following holds:		'''(4)'''  <u>Suggested solution 1</u>  is correct because the following holds:
	:$$d(t) = \frac{1}{T} \cdot \int\limits_{-\infty}^t {c(\tau)} \hspace{0.1cm}{\~~rmd~~}\tau \hspace{0.3cm}\Rightarrow \hspace{0.3cm} D_{\rm L}(p) =C_{\rm L}(p)\cdot\frac{1}{p \cdot T} = \frac{1- {\rm e}^{-p \hspace{0.05cm}\cdot \hspace{0.05cm}T}}{p^2 \cdotT}\hspace{0.05cm} .$$		:$$d(t) = \frac{1}{T} \cdot \int\limits_{-\infty}^t {c(\tau)} \hspace{0.1cm}{\rm d}\tau \hspace{0.3cm}\Rightarrow \hspace{0.3cm} D_{\rm L}(p) =C_{\rm L}(p)\cdot\frac{1}{p \cdot T} = \frac{1- {\rm e}^{-p \hspace{0.05cm}\cdot \hspace{0.05cm}T}}{p^2 \cdotT}\hspace{0.05cm} .$$
	*Since  $d(t)$  extends to infinity,  the simple relation between  $D_{\rm L}(p)$  and  $D(f)$  according to proposed solution 3 is not valid.		*Since  $d(t)$  extends to infinity,  the simple relation between  $D_{\rm L}(p)$  and  $D(f)$  according to proposed solution 3 is not valid.
	*$D(f)$  rather also includes a Dirac delta function at frequency  $f = 0$.		*$D(f)$  rather also includes a Dirac delta function at frequency  $f = 0$.
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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:~~Zusatzaufgaben~~:3.~~2_Laplace_und_Fourier~~]]		[[de:Aufgaben:Aufgabe 3.2Z: Laplace und Fourier]]