Aufgaben:Exercise 3.5Z: Integration of Dirac Functions - Revision history

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	Assume the time signal  ${x(t)}$  according to the middle sketch, which is composed of three Dirac pulses at  $–T$,  $0$  and  $+T$  with the pulse weights  ${AT}$,  $-2{AT}$  und  ${AT}$ .		Assume the time signal  ${x(t)}$  according to the middle sketch, which is composed of three Dirac pulses at  $–T$,  $0$  and  $+T$  with the pulse weights  ${AT}$,  $-2{AT}$  und  ${AT}$ .

	The spectral function  ${X(f)}$  can be given directly by applying the  [[Signal_Representation/~~Fourier_Transform_Theorems~~#Duality_Theorem\|Duality Theorem]]  if one takes into account that the time function belonging to  ${U(f)}$  is as follows (see lower sketch):		The spectral function  ${X(f)}$  can be given directly by applying the  [[Signal_Representation/The_Fourier_Transform_Theorems#Duality_Theorem\|Duality Theorem]]  if one takes into account that the time function belonging to  ${U(f)}$  is as follows (see lower sketch):
	:$$u( t ) = - 2A + 2A \cdot \cos ( {2{\rm{\pi }}f_0 t} ).$$		:$$u( t ) = - 2A + 2A \cdot \cos ( {2{\rm{\pi }}f_0 t} ).$$

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	*The following relationship exists between  ${x(t)}$  and  ${y(t)}$ :		*The following relationship exists between  ${x(t)}$  and  ${y(t)}$ :
	:$$y( t ) = \frac{1}{T} \cdot \hspace{-0.1cm} \int_{ - \infty }^{\hspace{0.05cm}t} {x( \tau )}\, {\rm d}\tau .$$		:$$y( t ) = \frac{1}{T} \cdot \hspace{-0.1cm} \int_{ - \infty }^{\hspace{0.05cm}t} {x( \tau )}\, {\rm d}\tau .$$
	*The  [[Signal_Representation/~~Fourier_Transform_Laws~~#Integration_Theorem\|Integration Theorem]]  reads in a correspondingly adapted form:		*The  [[Signal_Representation/The_Fourier_Transform_Theorems#Integration_Theorem\|Integration Theorem]]  reads in a correspondingly adapted form:
	:$$\frac{1}{T}\cdot \hspace{-0.1cm} \int_{ - \infty }^{\hspace{0.05cm}t} {x( \tau )}\,\, {\rm d}\tau\ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ \ X( f ) \cdot \left( {\frac{1}{{{\rm{j}}\cdot 2{\rm{\pi }\cdot }fT}} + \frac{1}{2T}\cdot {\rm \delta} ( f )} \right).$$		:$$\frac{1}{T}\cdot \hspace{-0.1cm} \int_{ - \infty }^{\hspace{0.05cm}t} {x( \tau )}\,\, {\rm d}\tau\ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ \ X( f ) \cdot \left( {\frac{1}{{{\rm{j}}\cdot 2{\rm{\pi }\cdot }fT}} + \frac{1}{2T}\cdot {\rm \delta} ( f )} \right).$$

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	[[Category:Signal Representation: Exercises\|^3.3 Fourier Transform Theorems^]]		[[Category:Signal Representation: Exercises\|^3.3 Fourier Transform Theorems^]]
	[[de:Aufgaben:3.~~5Z_Integration_von_Diracfunktionen~~]]		[[de:Aufgaben:Aufgabe 3.5Z: Integration von Diracfunktionen]]

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	:*At frequency  $f = 0$  the signal  ${x(t)}$  has no spectral components  ⇒   ${X(f = 0)} \;\underline{= 0}$.		:*At frequency  $f = 0$  the signal  ${x(t)}$  has no spectral components  ⇒   ${X(f = 0)} \;\underline{= 0}$.
	:*For  $f = 1 \,\text{kHz}$  – and also  $f \cdot T = 0.5$  –   on the other hand:		:*For  $f = 1 \,\text{kHz}$  – and also  $f \cdot T = 0.5$  –   on the other hand:
	:$$X( f = 1\;{\rm{kHz}} ) = - 4 \cdot A \cdot T = -2 \cdot 10^{ - 3} \;{\rm{V/Hz}}\; \Rightarrow \;		:$$X( f = 1\;{\rm{kHz}} ) = - 4 \cdot A \cdot T = -2 \cdot 10^{ - 3} \;{\rm{V/Hz}}\; \Rightarrow \;\|X( {f = 1\;{\rm{kHz}}} )\| \hspace{0.15 cm}\underline{= 2 \;{\rm{mV/Hz}}}{\rm{.}}$$
	\|X( {f = 1\;{\rm{kHz}}} )\| \hspace{0.15 cm}\underline{= 2 \;{\rm{mV/Hz}}}{\rm{.}}$$

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	[[Category:Signal Representation: Exercises\|^3.3 Fourier Transform Theorems^]]		[[Category:Signal Representation: Exercises\|^3.3 Fourier Transform Theorems^]]
			[[de:Aufgaben:3.5Z_Integration_von_Diracfunktionen]]

Guenter at 13:21, 24 May 2021

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	[[File:P_ID515__Sig_Z_3_5_neu.png\|right\|frame\|Integration ~~von Diracfunktionen~~ ]]		[[File:P_ID515__Sig_Z_3_5_neu.png\|right\|frame\|Integration of Dirac functions ]]
	Like in  [[Aufgaben:3.5_Differentiation_eines_Dreicksignals\|Exercise 3.5]]  the spectrum  ${Y(f)}$  of the signal		Like in  [[Aufgaben:3.5_Differentiation_eines_Dreicksignals\|Exercise 3.5]]  the spectrum  ${Y(f)}$  of the signal
	:$$y( t ) = \left\{ \begin{array}{c} A \\ - A \\ 0 \\ \end{array} \right.\quad \begin{array}{{20}c} {{\rm{for}}} \\ {{\rm{for}}} \\ {\rm{else.}} \\ \end{array}\;\begin{array}{{20}c} { - T \le t < 0,} \\ {0 < t \le T,} \\ {} \\\end{array}$$		:$$y( t ) = \left\{ \begin{array}{c} A \\ - A \\ 0 \\ \end{array} \right.\quad \begin{array}{{20}c} {{\rm{for}}} \\ {{\rm{for}}} \\ {\rm{else.}} \\ \end{array}\;\begin{array}{{20}c} { - T \le t < 0,} \\ {0 < t \le T,} \\ {} \\\end{array}$$

Guenter at 12:48, 27 April 2021

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Guenter at 12:40, 27 April 2021

2021-04-27T12:40:24Z

Javier: Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

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	[[Category:~~Exercises for~~ Signal Representation\|^3.3 Fourier Transform Theorems^]]		[[Category:Signal Representation: Exercises\|^3.3 Fourier Transform Theorems^]]

Noah at 17:04, 27 January 2021

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	can be determined. Again,  $A = 1 \,\text{V}$  and  $T = 0.5 \,\text{ms}$ apply.		can be determined. Again,  $A = 1 \,\text{V}$  and  $T = 0.5 \,\text{ms}$ apply.

	~~Assum ethe~~ time signal  ${x(t)}$  according to the middle sketch, which is composed of three Dirac pulses at  $–T$,  $0$  and  $+T$  with the pulse weights  ${AT}$,  $-2{AT}$  und  ${AT}$ .		Assume the time signal  ${x(t)}$  according to the middle sketch, which is composed of three Dirac pulses at  $–T$,  $0$  and  $+T$  with the pulse weights  ${AT}$,  $-2{AT}$  und  ${AT}$ .

	The spectral function  ${X(f)}$  can be given directly by applying the  [[Signal_Representation/Fourier_Transform_Laws#Vertauschungssatz\|Vertauschungssatz]]  if one takes into account that the time function belonging to  ${U(f)}$  is as follows (see lower sketch):		The spectral function  ${X(f)}$  can be given directly by applying the  [[Signal_Representation/Fourier_Transform_Laws#Vertauschungssatz\|Vertauschungssatz]]  if one takes into account that the time function belonging to  ${U(f)}$  is as follows (see lower sketch):
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	''Hints:''		''Hints:''
	*This exercise belongs to the ~~chpater~~  [[Signal_Representation/Fourier_Transform_Laws\|Fourier Transform Laws]].		*This exercise belongs to the chapter  [[Signal_Representation/Fourier_Transform_Laws\|Fourier Transform Laws]].
	*All of these laws are illustrated with examples in the learning video  [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)\|Gesetzmäßigkeiten der Fouriertransformation]] .		*All of these laws are illustrated with examples in the learning video  [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)\|Gesetzmäßigkeiten der Fouriertransformation]] .
	The following relationship exists between  ${x(t)}$  and  ${y(t)}$ :		The following relationship exists between  ${x(t)}$  and  ${y(t)}$ :

Noah at 17:02, 27 January 2021

2021-01-27T17:02:44Z

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	[[File:P_ID515__Sig_Z_3_5_neu.png\|right\|frame\|Integration von Diracfunktionen ]]		[[File:P_ID515__Sig_Z_3_5_neu.png\|right\|frame\|Integration von Diracfunktionen ]]
	~~Wie~~ in ~~der~~  [[Aufgaben:3.5_Differentiation_eines_Dreicksignals\|~~Aufgabe~~ 3.5]]  ~~soll das Spektrum~~  ${Y(f)}$  ~~des Signals~~		Like in  [[Aufgaben:3.5_Differentiation_eines_Dreicksignals\|Task 3.5]]  the spectrum  ${Y(f)}$  of the signal
	:$$y( t ) = \left\{ \begin{array}{c} A \\ - A \\ 0 \\ \end{array} \right.\quad \begin{array}{{20}c} {{\rm{f \ddot{u}r}}} \\ {{\rm{f\ddot{u} r}}} \\ {\rm{~~sonst~~.}} \\ \end{array}\;\begin{array}{{20}c} { - T \le t < 0,} \\ {0 < t \le T,} \\ {} \\\end{array}$$		:$$y( t ) = \left\{ \begin{array}{c} A \\ - A \\ 0 \\ \end{array} \right.\quad \begin{array}{{20}c} {{\rm{f \ddot{u}r}}} \\ {{\rm{f\ddot{u} r}}} \\ {\rm{else.}} \\ \end{array}\;\begin{array}{{20}c} { - T \le t < 0,} \\ {0 < t \le T,} \\ {} \\\end{array}$$
	~~ermittelt werden~~. ~~Es gelte wieder~~  $A = 1 \,\text{V}$  ~~und~~  $T = 0.5 \,\text{ms}$.		can be determined. Again,  $A = 1 \,\text{V}$  and  $T = 0.5 \,\text{ms}$ apply.

	~~Ausgegangen wird vom Zeitsignal~~  ${x(t)}$  ~~gemäß der mittleren Skizze~~, ~~das sich aus drei Diracimpulsen bei~~  $–T$,  $0$  ~~und~~  $+T$  ~~mit den Impulsgewichte~~  ${AT}$,  $-2{AT}$  und  ${AT}$  ~~zusammensetzt~~.		Assum ethe time signal  ${x(t)}$  according to the middle sketch, which is composed of three Dirac pulses at  $–T$,  $0$  and  $+T$  with the pulse weights  ${AT}$,  $-2{AT}$  und  ${AT}$ .

	~~Die Spektralfunktion~~  ${X(f)}$  ~~kann durch Anwendung des~~  [[Signal_Representation/Fourier_Transform_Laws#Vertauschungssatz\|~~Vertauschungssatzes~~]]  ~~direkt angegeben werden, wenn man berücksichtigt, dass die zu~~  ${U(f)}$  ~~gehörige Zeitfunktion wie folgt lautet~~ (~~siehe untere Skizze~~):		The spectral function  ${X(f)}$  can be given directly by applying the  [[Signal_Representation/Fourier_Transform_Laws#Vertauschungssatz\|Vertauschungssatz]]  if one takes into account that the time function belonging to  ${U(f)}$  is as follows (see lower sketch):
	:$$u( t ) = - 2A + 2A \cdot \cos ( {2{\rm{\pi }}f_0 t} ).$$		:$$u( t ) = - 2A + 2A \cdot \cos ( {2{\rm{\pi }}f_0 t} ).$$

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	''~~Hinweise~~:''		''Hints:''
	*~~Die Aufgabe gehört zum Kapitel~~  [[Signal_Representation/Fourier_Transform_Laws\|~~Gesetzmäßigkeiten der Fouriertransformation~~]].		*This exercise belongs to the chpater  [[Signal_Representation/Fourier_Transform_Laws\|Fourier Transform Laws]].
	*~~Alle diese Gesetzmäßigkeiten werden im Lernvideo~~  [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)\|Gesetzmäßigkeiten der Fouriertransformation]]  ~~an Beispielen verdeutlicht~~.		*All of these laws are illustrated with examples in the learning video  [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)\|Gesetzmäßigkeiten der Fouriertransformation]] .
	*Zwischen  ${x(t)}$  ~~und~~  ${y(t)}$  ~~besteht folgender Zusammenhang~~:		The following relationship exists between  ${x(t)}$  and  ${y(t)}$ :
	:$$y( t ) = \frac{1}{T} \cdot \hspace{-0.1cm} \int_{ - \infty }^{\hspace{0.05cm}t} {x( \tau )}\, {\rm d}\tau .$$		:$$y( t ) = \frac{1}{T} \cdot \hspace{-0.1cm} \int_{ - \infty }^{\hspace{0.05cm}t} {x( \tau )}\, {\rm d}\tau .$$
	*~~Der~~  [[Signal_Representation/Fourier_Transform_Laws#~~Integrationssatz~~\|~~Integrationssatz~~]]  ~~lautet~~ in ~~entsprechend angepasster Form~~:		*The  [[Signal_Representation/Fourier_Transform_Laws#Integration_Theorem\|Integration_Theorem]]  reads in a correspondingly adapted form:
	:$$\frac{1}{T}\cdot \hspace{-0.1cm} \int_{ - \infty }^{\hspace{0.05cm}t} {x( \tau )}\,\, {\rm d}\tau\ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ \ X( f ) \cdot \left( {\frac{1}{{{\rm{j}}\cdot 2{\rm{\pi }\cdot }fT}} + \frac{1}{2T}\cdot {\rm \delta} ( f )} \right).$$		:$$\frac{1}{T}\cdot \hspace{-0.1cm} \int_{ - \infty }^{\hspace{0.05cm}t} {x( \tau )}\,\, {\rm d}\tau\ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ \ X( f ) \cdot \left( {\frac{1}{{{\rm{j}}\cdot 2{\rm{\pi }\cdot }fT}} + \frac{1}{2T}\cdot {\rm \delta} ( f )} \right).$$



	===~~Fragebogen~~===		===Questions===

	<quiz display=simple>		<quiz display=simple>
	{~~Berechnen Sie die Spektralfunktion~~  ${X(f)}$. ~~Wie groß ist deren Betrag bei den Frequenzen~~  $f = 0$  ~~und~~  $f = 1\, \text{kHz}$?		{Calculate the spectral function  ${X(f)}$. What is its magnitude at the frequencies  $f = 0$  and  $f = 1\, \text{kHz}$?
	\|type="{}"}		\|type="{}"}
	$\|{X(f = 0)}\| \ = \ $ { 0. }  $\text{mV/Hz}$		$\|{X(f = 0)}\| \ = \ $ { 0. }  $\text{mV/Hz}$
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	{~~Berechnen Sie die Spektralfunktion~~  ${Y(f)}$. ~~Welche Werte ergeben sich bei den Frequenzen~~  $f = 0$  ~~und~~  $f = 1\, \text{kHz}$?		{Calculate the spectral function  ${Y(f)}$. What values result at the frequencies  $f = 0$  and  $f = 1\, \text{kHz}$?
	\|type="{}"}		\|type="{}"}
	$\|{Y(f = 0)}\|\ = \ $ { 0. }  $\text{mV/Hz}$		$\|{Y(f = 0)}\|\ = \ $ { 0. }  $\text{mV/Hz}$
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	</quiz>		</quiz>

	===~~Musterlösung~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Im Angabenteil zur Aufgabe finden Sie die Fourierkorrespondenz zwischen~~  ${u(t)}$  ~~und~~  ${U(f)}$.		'''(1)'''  In the description of the task you will find the Fourier correspondence between  ${u(t)}$  and  ${U(f)}$.
	*~~Da sowohl die Zeitfunktionen~~  ${u(t)}$  ~~und~~  ${x(t)}$  ~~als auch die dazugehörigen Spektren~~  ${U(f)}$  ~~und~~  ${X(f)}$  ~~gerade und reell sind~~, ~~kann man~~  ${X(f)}$  ~~durch Anwendung des Vertauschungssatzes leicht berechnen~~:		*Since both the time functions  ${u(t)}$  and  ${x(t)}$  and the corresponding spectra  ${U(f)}$  and  ${X(f)}$  are even and real,  ${X(f)}$  can be easily calculated by applying the "Vertauschungssatz":
	:$$X( f ) = - 2 \cdot A \cdot T + 2 \cdot A \cdot T \cdot \cos \left( {{\rm{2\pi }}fT} \right).$$		:$$X( f ) = - 2 \cdot A \cdot T + 2 \cdot A \cdot T \cdot \cos \left( {{\rm{2\pi }}fT} \right).$$
	*~~Wegen der Beziehung~~  $\sin^{2}(\alpha) = (1 – \cos(\alpha))/2$  ~~kann hierfür auch geschrieben werden~~:		*Because of the relation  $\sin^{2}(\alpha) = (1 – \cos(\alpha))/2$  it can also be written for this:
	:$$X( f ) = - 4 \cdot A \cdot T \cdot \sin ^2 ( {{\rm{\pi }}fT} ).$$		:$$X( f ) = - 4 \cdot A \cdot T \cdot \sin ^2 ( {{\rm{\pi }}fT} ).$$
	:*~~Bei der Frequenz~~  $f = 0$  ~~hat~~  ${x(t)}$  ~~keine Spektralanteile~~   ⇒   ${X(f = 0)} \;\underline{= 0}$.		:*At frequency  $f = 0$   ${x(t)}$  has no spectral components  ⇒   ${X(f = 0)} \;\underline{= 0}$.
	:*~~Für~~  $f = 1 \,\text{kHz}$  – also  $f \cdot T = 0.5$  –   ~~gilt dagegen~~:		:*For  $f = 1 \,\text{kHz}$  – also  $f \cdot T = 0.5$  –   on the other hand:
	:$$X( f = 1\;{\rm{kHz}} ) = - 4 \cdot A \cdot T = -2 \cdot 10^{ - 3} \;{\rm{V/Hz}}\; \Rightarrow \;		:$$X( f = 1\;{\rm{kHz}} ) = - 4 \cdot A \cdot T = -2 \cdot 10^{ - 3} \;{\rm{V/Hz}}\; \Rightarrow \;
	\|X( {f = 1\;{\rm{kHz}}} )\| \hspace{0.15 cm}\underline{= 2 \;{\rm{mV/Hz}}}{\rm{.}}$$		\|X( {f = 1\;{\rm{kHz}}} )\| \hspace{0.15 cm}\underline{= 2 \;{\rm{mV/Hz}}}{\rm{.}}$$
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	'''(2)'''  ~~Das Spektrum~~  ${Y(f)}$  ~~kann aus~~  ${X(f)}$  ~~durch Anwendung des Integrationssatzes ermittelt werden~~.		'''(2)'''  The spectrum  ${Y(f)}$  can be determined from  ${X(f)}$  by applying the integration theorem.
	*~~Wegen~~  ${X(f = 0)} = 0$  ~~muss die Diracfunktion bei der Frequenz~~  $f = 0$  ~~nicht berücksichtigt werden und man erhält~~:		*Because of  ${X(f = 0)} = 0$  the Dirac function does not have to be taken into account at the frequency  $f = 0$  and one obtains:
	:$$Y( f ) = \frac{X( f )}{{{\rm{j}} \cdot 2{\rm{\pi }}fT}} = \frac{{ - 4 \cdot A \cdot T \cdot \sin ^2 ( {{\rm{\pi }}fT} )}}{{{\rm{j}}\cdot 2{\rm{\pi }}fT}} = 2{\rm{j}} \cdot A \cdot T \cdot \frac{{\sin ^2 ( {{\rm{\pi }}fT} )}}{{{\rm{\pi }}fT}}.$$		:$$Y( f ) = \frac{X( f )}{{{\rm{j}} \cdot 2{\rm{\pi }}fT}} = \frac{{ - 4 \cdot A \cdot T \cdot \sin ^2 ( {{\rm{\pi }}fT} )}}{{{\rm{j}}\cdot 2{\rm{\pi }}fT}} = 2{\rm{j}} \cdot A \cdot T \cdot \frac{{\sin ^2 ( {{\rm{\pi }}fT} )}}{{{\rm{\pi }}fT}}.$$
	*~~Es ergibt sich selbstverständlich das gleiche Ergebnis wie~~ in ~~der~~  [[Aufgaben:3.5_Differentiation_eines_Dreicksignals\|~~Aufgabe~~ 3.5]]:		*Of course, the result is the same as in  [[Aufgaben:3.5_Differentiation_eines_Dreicksignals\|Task 3.5]]:
	:*~~Bei der Frequenz~~  $f = 0$  ~~hat auch~~  ${y(t)}$  ~~keine Spektralanteile~~   ⇒   ${Y(f = 0)} \;\underline{= 0}$.		:*At frequency  $f = 0$   ${y(t)}$  also has no spectral components  ⇒   ${Y(f = 0)} \;\underline{= 0}$.
	:*~~Für~~  $f = 1\,\text{kHz} \ \ (f \cdot T = 0.5)$  ~~erhält man gegenüber~~  $~~X(f)~~$  ~~einen um den Faktor~~  $~~\pi~~$  ~~kleineren Wert~~:		:*For  $f = 1\,\text{kHz} \ \ (f \cdot T = 0.5)$  one obtains a value smaller by a factor  $\pi$  compared to  $X(f)$ :
	:$$\|Y( {f = 1\;{\rm{kHz}}} )\| = \frac{4 \cdot A \cdot T}{\rm{\pi }} \hspace{0.15 cm}\underline{= {\rm{0}}{\rm{.636}} \;{\rm{mV/Hz}}}{\rm{.}}$$		:$$\|Y( {f = 1\;{\rm{kHz}}} )\| = \frac{4 \cdot A \cdot T}{\rm{\pi }} \hspace{0.15 cm}\underline{= {\rm{0}}{\rm{.636}} \;{\rm{mV/Hz}}}{\rm{.}}$$
	{{ML-Fuß}}		{{ML-Fuß}}

Oezdemir: Oezdemir moved page Aufgaben:Aufgabe 3.5Z: Integration von Diracfunktionen to Aufgaben:Exercise 3.5Z: Integration of Dirac Functions

2020-11-02T15:29:11Z

Oezdemir moved page Aufgaben:Aufgabe 3.5Z: Integration von Diracfunktionen to Aufgaben:Exercise 3.5Z: Integration of Dirac Functions

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  In the description ~~of the task~~ you will find the Fourier correspondence between  ${u(t)}$  and  ${U(f)}$.		'''(1)'''  In the task description you will find the Fourier correspondence between  ${u(t)}$  and  ${U(f)}$.
	*Since both the time functions  ${u(t)}$  and  ${x(t)}$  and the corresponding spectra  ${U(f)}$  and  ${X(f)}$  are even and real,  ${X(f)}$  can be easily calculated by applying the "~~Vertauschungssatz~~":		*Since both the time functions  ${u(t)}$  and  ${x(t)}$  and the corresponding spectra  ${U(f)}$  and  ${X(f)}$  are even and real,  ${X(f)}$  can be easily calculated by applying the "Duality Theorem":
	:$$X( f ) = - 2 \cdot A \cdot T + 2 \cdot A \cdot T \cdot \cos \left( {{\rm{2\pi }}fT} \right).$$		:$$X( f ) = - 2 \cdot A \cdot T + 2 \cdot A \cdot T \cdot \cos \left( {{\rm{2\pi }}fT} \right).$$
	*Because of the relation  $\sin^{2}(\alpha) = (1 – \cos(\alpha))/2$  it can also be written for this:		*Because of the relation  $\sin^{2}(\alpha) = (1 – \cos(\alpha))/2$  it can also be written for this:
	:$$X( f ) = - 4 \cdot A \cdot T \cdot \sin ^2 ( {{\rm{\pi }}fT} ).$$		:$$X( f ) = - 4 \cdot A \cdot T \cdot \sin ^2 ( {{\rm{\pi }}fT} ).$$
	:*At frequency  $f = 0$   ${x(t)}$  has no spectral components  ⇒   ${X(f = 0)} \;\underline{= 0}$.		:*At frequency  $f = 0$  the signal  ${x(t)}$  has no spectral components  ⇒   ${X(f = 0)} \;\underline{= 0}$.
	:*For  $f = 1 \,\text{kHz}$  – also  $f \cdot T = 0.5$  –   on the other hand:		:*For  $f = 1 \,\text{kHz}$  – and also  $f \cdot T = 0.5$  –   on the other hand:
	:$$X( f = 1\;{\rm{kHz}} ) = - 4 \cdot A \cdot T = -2 \cdot 10^{ - 3} \;{\rm{V/Hz}}\; \Rightarrow \;		:$$X( f = 1\;{\rm{kHz}} ) = - 4 \cdot A \cdot T = -2 \cdot 10^{ - 3} \;{\rm{V/Hz}}\; \Rightarrow \;
	\|X( {f = 1\;{\rm{kHz}}} )\| \hspace{0.15 cm}\underline{= 2 \;{\rm{mV/Hz}}}{\rm{.}}$$		\|X( {f = 1\;{\rm{kHz}}} )\| \hspace{0.15 cm}\underline{= 2 \;{\rm{mV/Hz}}}{\rm{.}}$$
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	'''(2)'''  The spectrum  ${Y(f)}$  can be determined from  ${X(f)}$  by applying the ~~integration theorem~~.		'''(2)'''  The spectrum  ${Y(f)}$  can be determined from  ${X(f)}$  by applying the  "Integration Theorem".
	*Because of  ${X(f = 0)} = 0$  the Dirac function does not have to be taken into account at the frequency  $f = 0$  and one obtains:		*Because of  ${X(f = 0)} = 0$  the Dirac function does not have to be taken into account at the frequency  $f = 0$  and one obtains:
	:$$Y( f ) = \frac{X( f )}{{{\rm{j}} \cdot 2{\rm{\pi }}fT}} = \frac{{ - 4 \cdot A \cdot T \cdot \sin ^2 ( {{\rm{\pi }}fT} )}}{{{\rm{j}}\cdot 2{\rm{\pi }}fT}} = 2{\rm{j}} \cdot A \cdot T \cdot \frac{{\sin ^2 ( {{\rm{\pi }}fT} )}}{{{\rm{\pi }}fT}}.$$		:$$Y( f ) = \frac{X( f )}{{{\rm{j}} \cdot 2{\rm{\pi }}fT}} = \frac{{ - 4 \cdot A \cdot T \cdot \sin ^2 ( {{\rm{\pi }}fT} )}}{{{\rm{j}}\cdot 2{\rm{\pi }}fT}} = 2{\rm{j}} \cdot A \cdot T \cdot \frac{{\sin ^2 ( {{\rm{\pi }}fT} )}}{{{\rm{\pi }}fT}}.$$
	*Of course, the result is the same as in  [[Aufgaben:3.5_Differentiation_eines_Dreicksignals\|~~Task~~ 3.5]]:		*Of course, the result is the same as in  [[Aufgaben:3.5_Differentiation_eines_Dreicksignals\|Exercise 3.5]]:
	:*At frequency  $f = 0$   ${y(t)}$  also has no spectral components  ⇒   ${Y(f = 0)} \;\underline{= 0}$.		:*At frequency  $f = 0$  The signal  ${y(t)}$  also has no spectral components  ⇒   ${Y(f = 0)} \;\underline{= 0}$.
	:*For  $f = 1\,\text{kHz} \ \ (f \cdot T = 0.5)$  one obtains a value smaller by a factor  $\pi$  compared to  $X(f)$~~ ~~:		:*For  $f = 1\,\text{kHz} \ \ (f \cdot T = 0.5)$  one obtains a value smaller by a factor  $\pi$  compared to  $X(f)$:
	:$$\|Y( {f = 1\;{\rm{kHz}}} )\| = \frac{4 \cdot A \cdot T}{\rm{\pi }} \hspace{0.15 cm}\underline{= {\rm{0}}{\rm{.636}} \;{\rm{mV/Hz}}}{\rm{.}}$$		:$$\|Y( {f = 1\;{\rm{kHz}}} )\| = \frac{4 \cdot A \cdot T}{\rm{\pi }} \hspace{0.15 cm}\underline{= {\rm{0}}{\rm{.636}} \;{\rm{mV/Hz}}}{\rm{.}}$$
	{{ML-Fuß}}		{{ML-Fuß}}

← Older revision		Revision as of 14:40, 27 April 2021
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	[[File:P_ID515__Sig_Z_3_5_neu.png\|right\|frame\|Integration von Diracfunktionen ]]		[[File:P_ID515__Sig_Z_3_5_neu.png\|right\|frame\|Integration von Diracfunktionen ]]
	Like in  [[Aufgaben:3.5_Differentiation_eines_Dreicksignals\|~~Task~~ 3.5]]  the spectrum  ${Y(f)}$  of the signal		Like in  [[Aufgaben:3.5_Differentiation_eines_Dreicksignals\|Exercise 3.5]]  the spectrum  ${Y(f)}$  of the signal
	:$$y( t ) = \left\{ \begin{array}{c} A \\ - A \\ 0 \\ \end{array} \right.\quad \begin{array}{{20}c} {{\rm{~~f \ddot{u}r~~}}} \\ {{\rm{~~f\ddot{u} r~~}}} \\ {\rm{else.}} \\ \end{array}\;\begin{array}{{20}c} { - T \le t < 0,} \\ {0 < t \le T,} \\ {} \\\end{array}$$		:$$y( t ) = \left\{ \begin{array}{c} A \\ - A \\ 0 \\ \end{array} \right.\quad \begin{array}{{20}c} {{\rm{for}}} \\ {{\rm{for}}} \\ {\rm{else.}} \\ \end{array}\;\begin{array}{{20}c} { - T \le t < 0,} \\ {0 < t \le T,} \\ {} \\\end{array}$$
	can be determined. Again,  $A = 1 \,\text{V}$  and  $T = 0.5 \,\text{ms}$ apply.		can be determined.  Again,  $A = 1 \,\text{V}$  and  $T = 0.5 \,\text{ms}$  apply.

	Assume the time signal  ${x(t)}$  according to the middle sketch, which is composed of three Dirac pulses at  $–T$,  $0$  and  $+T$  with the pulse weights  ${AT}$,  $-2{AT}$  und  ${AT}$ .		Assume the time signal  ${x(t)}$  according to the middle sketch, which is composed of three Dirac pulses at  $–T$,  $0$  and  $+T$  with the pulse weights  ${AT}$,  $-2{AT}$  und  ${AT}$ .

	The spectral function  ${X(f)}$  can be given directly by applying the  [[Signal_Representation/~~Fourier_Transform_Laws~~#~~Vertauschungssatz~~\|~~Vertauschungssatz~~]]  if one takes into account that the time function belonging to  ${U(f)}$  is as follows (see lower sketch):		The spectral function  ${X(f)}$  can be given directly by applying the  [[Signal_Representation/Fourier_Transform_Theorems#Duality_Theorem\|Duality Theorem]]  if one takes into account that the time function belonging to  ${U(f)}$  is as follows (see lower sketch):
	:$$u( t ) = - 2A + 2A \cdot \cos ( {2{\rm{\pi }}f_0 t} ).$$		:$$u( t ) = - 2A + 2A \cdot \cos ( {2{\rm{\pi }}f_0 t} ).$$




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	''Hints:''		''Hints:''
	*This exercise belongs to the chapter  [[Signal_Representation/~~Fourier_Transform_Laws~~\|Fourier Transform ~~Laws~~]].		*This exercise belongs to the chapter  [[Signal_Representation/Fourier_Transform_Theorems\|Fourier Transform Theorems]].
	*All ~~of these~~ laws are illustrated with examples in the learning video  [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)\|Gesetzmäßigkeiten der Fouriertransformation]] .		*All the laws presented here are illustrated with examples in the (German language) learning video<br>      [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)\|Gesetzmäßigkeiten der Fouriertransformation]]   ⇒   "Regularities to the Fourier transform".
	The following relationship exists between  ${x(t)}$  and  ${y(t)}$ :		*The following relationship exists between  ${x(t)}$  and  ${y(t)}$ :
	:$$y( t ) = \frac{1}{T} \cdot \hspace{-0.1cm} \int_{ - \infty }^{\hspace{0.05cm}t} {x( \tau )}\, {\rm d}\tau .$$		:$$y( t ) = \frac{1}{T} \cdot \hspace{-0.1cm} \int_{ - \infty }^{\hspace{0.05cm}t} {x( \tau )}\, {\rm d}\tau .$$
	*The  [[Signal_Representation/Fourier_Transform_Laws#Integration_Theorem\|~~Integration_Theorem~~]]  reads in a correspondingly adapted form:		*The  [[Signal_Representation/Fourier_Transform_Laws#Integration_Theorem\|Integration Theorem]]  reads in a correspondingly adapted form:
	:$$\frac{1}{T}\cdot \hspace{-0.1cm} \int_{ - \infty }^{\hspace{0.05cm}t} {x( \tau )}\,\, {\rm d}\tau\ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ \ X( f ) \cdot \left( {\frac{1}{{{\rm{j}}\cdot 2{\rm{\pi }\cdot }fT}} + \frac{1}{2T}\cdot {\rm \delta} ( f )} \right).$$		:$$\frac{1}{T}\cdot \hspace{-0.1cm} \int_{ - \infty }^{\hspace{0.05cm}t} {x( \tau )}\,\, {\rm d}\tau\ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ \ X( f ) \cdot \left( {\frac{1}{{{\rm{j}}\cdot 2{\rm{\pi }\cdot }fT}} + \frac{1}{2T}\cdot {\rm \delta} ( f )} \right).$$

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	<quiz display=simple>		<quiz display=simple>
	{Calculate the spectral function  ${X(f)}$. What is its magnitude at the frequencies  $f = 0$  and  $f = 1\, \text{kHz}$?		{Calculate the spectral function  ${X(f)}$.  What is its magnitude at the frequencies  $f = 0$  and  $f = 1\, \text{kHz}$?
	\|type="{}"}		\|type="{}"}
	$\|{X(f = 0)}\| \ = \ $ { 0. }  $\text{mV/Hz}$		$\|{X(f = 0)}\| \ = \ $ { 0. }  $\text{mV/Hz}$
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	{Calculate the spectral function  ${Y(f)}$. What values result at the frequencies  $f = 0$  and  $f = 1\, \text{kHz}$?		{Calculate the spectral function  ${Y(f)}$.  What values result at the frequencies  $f = 0$  and  $f = 1\, \text{kHz}$?
	\|type="{}"}		\|type="{}"}
	$\|{Y(f = 0)}\|\ = \ $ { 0. }  $\text{mV/Hz}$		$\|{Y(f = 0)}\|\ = \ $ { 0. }  $\text{mV/Hz}$