Aufgaben:Exercise 3.3: From the Signal to the Spectrum - Revision history

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	*According to the result of subtask  '''(2)'''  the zeros occur at the distance  $f_0 = 1/T$ .		*According to the result of subtask  '''(2)'''  the zeros occur at the distance  $f_0 = 1/T$ .
	*With  $f_0 = 1/(2T) = f = 10 \;{\rm{kHz}}$  the real part is  $0$, but not the imaginary part.		*With  $f_0 = 1/(2T) = f = 10 \;{\rm{kHz}}$  the real part is  $0$, but not the imaginary part.
	*With the arguments  $f \cdot T = 0.5,\ 1.5,\ 2.5,\hspace{0.05cm}\text{ ... }$  the sine function is in each case equal in magnitude to  $1$, and it holds:		*With the arguments  $f \cdot T = 0.5,\ 1.5,\ 2.5,\hspace{0.05cm}\text{...}$  the sine function is in each case equal in magnitude to  $1$, and it holds:

	:$$\left\| {X( f )} \right\| = \frac{A}{ {{\rm{\pi }}\left\| f \right\|}} = X_{\rm S} ( f ).$$		:$$\left\| {X( f )} \right\| = \frac{A}{ {{\rm{\pi }}\left\| f \right\|}} = X_{\rm S} ( f ).$$

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			[[de:Aufgaben:3.3_Vom_Signal_zum_Spektrum]]

Guenter at 09:36, 23 April 2021

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Guenter at 09:08, 23 April 2021

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	''Hints:''		''Hints:''
	*This task belongs to the chapter  [[Signal_Representation/~~Special_Cases_of_Impulse_Signals~~\|Special Cases of ~~Impulse Signals~~]].		*This task belongs to the chapter  [[Signal_Representation/Special_Cases_of_Pulses\|Special Cases of Pulses]].

	*The following trigonometric transformations are given:		*The following trigonometric transformations are given:

Guenter at 09:07, 23 April 2021

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Guenter: Guenter moved page Aufgaben:Exercise 3.3: From The Signal to the Spectrum to Aufgaben:Exercise 3.3: From the Signal to the Spectrum

2021-04-23T08:59:18Z

Guenter moved page Aufgaben:Exercise 3.3: From The Signal to the Spectrum to Aufgaben:Exercise 3.3: From the Signal to the Spectrum

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Guenter at 08:57, 23 April 2021

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Noah at 21:14, 20 January 2021

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	:$$\left\| {X( {f = 20\;{\rm{kHz}}} )} \right\| = \frac{ {A \cdot T}}{ {\rm{\pi }}} \cdot \sin ( {\rm{\pi }} ) \hspace{0.15 cm}\underline{= 0}.$$		:$$\left\| {X( {f = 20\;{\rm{kHz}}} )} \right\| = \frac{ {A \cdot T}}{ {\rm{\pi }}} \cdot \sin ( {\rm{\pi }} ) \hspace{0.15 cm}\underline{= 0}.$$

	*When calculating the value for  $f = 0$  the quotient  $\text{0 divided by 0}$. By applying  [https://de.wikipedia.org/wiki/~~Regel_von_de_l~~%E2%80%~~99Hospital l~~'~~Hospitalschen Regel~~]  the limiting value can be calculated:		*When calculating the value for  $f = 0$  the quotient  $\text{0 divided by 0}$. By applying  [https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule L'Hospitals rule]  the limiting value can be calculated:

	:$$\mathop {\lim }\limits_{x \to 0 } \frac{ {\sin \left( x \right)}}{x} = 1.$$		:$$\mathop {\lim }\limits_{x \to 0 } \frac{ {\sin \left( x \right)}}{x} = 1.$$

	*~~Daraus folgt~~:		*From this follows:

	:$$\left\| {X( {f = 0} )} \right\| = A \cdot T \hspace{0.15 cm}\underline{= 0.1 \;{\rm{mV/Hz}}}{\rm{.}}$$		:$$\left\| {X( {f = 0} )} \right\| = A \cdot T \hspace{0.15 cm}\underline{= 0.1 \;{\rm{mV/Hz}}}{\rm{.}}$$

	*~~Dieses Ergebnis ist einsichtig~~, ~~da nach dem ersten Fourierintegral der Spektralwert bei~~  $f = 0$  ~~genau der Fläche unter der Zeitfunktion entspricht~~.		*This result is obvious because, according to the first Fourier integral, the spectral value at  $f = 0$  corresponds exactly to the area under the time function.



	[[File:P_ID563__Sig_A_3_3_c.png\|right\|frame\|Betragsspektrum des Rechteckimpulses]]		[[File:P_ID563__Sig_A_3_3_c.png\|right\|frame\|Betragsspektrum des Rechteckimpulses]]
	'''(3)'''  ~~Richtig sind die~~ <u>~~Lösungsvorschläge~~ 1 ~~und~~ 3</u>:		'''(3)'''  Proposed<u> solutions 1 and 3</u> are correct::
	*~~Entsprechend dem Ergebnis zur Teilaufgabe~~  '''(2)'''  ~~treten die Nullstellen im Abstand~~  $f_0 = 1/T$  ~~auf~~.		*According to the result of subtask  '''(2)'''  the zeros occur at the distance  $f_0 = 1/T$ .
	*~~Bei~~  $f_0 = 1/(2T) = f = 10 \;{\rm{kHz}}$  ~~ist zwar der Realteil~~  $0$, ~~aber nicht der Imaginärteil~~.		*With  $f_0 = 1/(2T) = f = 10 \;{\rm{kHz}}$  the real part is  $0$, but not the imaginary part.
	*~~Bei den Argumenten~~  $f \cdot T = 0.5, 1.5, 2.5,\hspace{0.05cm}\text{ ... }$  ~~ist die Sinusfunktion jeweils betragsmäßig gleich~~  $1$, ~~und es gilt~~:		*With the arguments  $f \cdot T = 0.5, 1.5, 2.5,\hspace{0.05cm}\text{ ... }$  the sine function is in each case equal in magnitude to  $1$, and it holds:

	:$$\left\| {X( f )} \right\| = \frac{A}{ {{\rm{\pi }}\left\| f \right\|}} = X_{\rm S} ( f ).$$		:$$\left\| {X( f )} \right\| = \frac{A}{ {{\rm{\pi }}\left\| f \right\|}} = X_{\rm S} ( f ).$$

	*~~Bei anderen Frequenzen dient~~  $X_{\rm S}(f)$  als obere Schranke, ~~das heißt~~, ~~es gilt stets~~  $\|Xf)\| \leq X_{\rm S}(f)$.		*At other frequencies,  $X_{\rm S}(f)$  als obere Schranke, serves as an upper bound, i.e.  $\|Xf)\| \leq X_{\rm S}(f)$ always applies.
	*In ~~der Skizze ist diese Schranke zusätzlich zu~~  $\|X(f)\|$  ~~als violette Kurve eingezeichnet~~.		*In the sketch, this bound is drawn as a violet curve in addition to  $\|X(f)\|$ .



	'''(4)'''  ~~Nach der Definition auf der Angabenseite kann man die Phasenfunktion wie folgt berechnen~~:		'''(4)'''  According to the definition on the information page, one can calculate the phase function as follows:

	:$$\varphi ( f ) = - \arctan \frac{ { {\mathop{\rm Im}\nolimits} ( f )}}{ { {\mathop{\rm Re}\nolimits} ( f )}}.$$		:$$\varphi ( f ) = - \arctan \frac{ { {\mathop{\rm Im}\nolimits} ( f )}}{ { {\mathop{\rm Re}\nolimits} ( f )}}.$$

	*~~Mit den Ergebnissen aus Teilaufgabe~~  '''(1)'''  ~~gilt somit~~:		*With the results from subtask  '''(1)'''  the following thus applies:

	:$$\varphi ( f ) = \arctan \left( {\frac{ {1 - \cos ( {\omega T} )}}{ {\sin ( {\omega T} )}}} \right).$$		:$$\varphi ( f ) = \arctan \left( {\frac{ {1 - \cos ( {\omega T} )}}{ {\sin ( {\omega T} )}}} \right).$$

	*~~Das Argument dieser Funktion ist entsprechend der Angabe gleich~~  $\tan(\omega T/2) = \tan(\pi fT)$. ~~Daraus folgt ein mit der Frequenz linear ansteigender Verlauf~~:		*The argument of this function is equal to  $\tan(\omega T/2) = \tan(\pi fT)$ according to the specification. From this follows a linearly increasing course with the frequency:

	:$$\varphi ( f ) = \arctan \left( {\tan ( { {\rm{\pi }}fT} )} \right) = {\rm{\pi }}fT.$$		:$$\varphi ( f ) = \arctan \left( {\tan ( { {\rm{\pi }}fT} )} \right) = {\rm{\pi }}fT.$$

	*~~Mit~~  $f = 10\,\text{kHz}$  ~~und~~  $T = 50\,\text{µs}$  ~~erhält man daraus den Phasenwinkel~~  $\pi /2$  ~~entsprechend~~  $\underline{90^{\circ}}$ .		*With  $f = 10\,\text{kHz}$  and  $T = 50\,\text{µs}$  one obtains from this the phase angle  $\pi /2$  corresponding to  $\underline{90^{\circ}}$ .
	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Exercises for Signal Representation\|^3.2 Special Cases of Impulse Signals^]]		[[Category:Exercises for Signal Representation\|^3.2 Special Cases of Impulse Signals^]]

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	''Hints:''		''Hints:''
	*This task belongs to the chapter  [[Signal_Representation/Special_Cases_of_Pulses\|Special Cases of Pulses]].		*This task belongs to the chapter  [[Signal_Representation/Special_Cases_of_Pulses\|Special Cases of Pulses]].
			*Use one of the functions  $\text{si}(x) = \sin(x)/x$  or  $\text{sinc}(x) = \sin(\pi x)/(\pi x)$.
	*The following trigonometric transformations are given:		*The following trigonometric transformations are given:

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	:$${\mathop{\rm Im}\nolimits} [ {X( f)} ] = \frac{A}{\omega } \cdot \left( {\cos ( {\omega T}) - 1} \right) = - \frac{A}{\omega } \cdot\left( {1 - \cos ( {\omega T} )} \right).$$		:$${\mathop{\rm Im}\nolimits} [ {X( f)} ] = \frac{A}{\omega } \cdot \left( {\cos ( {\omega T}) - 1} \right) = - \frac{A}{\omega } \cdot\left( {1 - \cos ( {\omega T} )} \right).$$

	*For the frequency  $f = 1/(2T) = 10\, \text{kHz}$   ⇒   $\omega \cdot T = \pi$  we get::		*For the frequency  $f = 1/(2T) = 10\, \text{kHz}$   ⇒   $\omega \cdot T = \pi$  we get:

	:$${\mathop{\rm Re}\nolimits}[{X( {f = 10 \;{\rm{kHz}}} )}] = \frac{A}{ {2{\rm{\pi }}f}} \cdot \sin ( {\rm{\pi }} ) \hspace{0.15 cm}\underline{= 0},$$		:$${\mathop{\rm Re}\nolimits}[{X( {f = 10 \;{\rm{kHz}}} )}] = \frac{A}{ {2{\rm{\pi }}f}} \cdot \sin ( {\rm{\pi }} ) \hspace{0.15 cm}\underline{= 0},$$
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	:$$\left\| {X( f )} \right\| = \left\| \frac{A}{ {\rm\pi }f} \cdot \sin ( {\rm \pi }fT ) \right\| = A \cdot T \cdot \left\| \frac{\sin ( {\rm\pi }fT )}{ {\rm \pi}fT} \right\|.$$		:$$\left\| {X( f )} \right\| = \left\| \frac{A}{ {\rm\pi }f} \cdot \sin ( {\rm \pi }fT ) \right\| = A \cdot T \cdot \left\| \frac{\sin ( {\rm\pi }fT )}{ {\rm \pi}fT} \right\|.$$

	*With the ~~shortcut~~  $\text{si}(x) = \sin(x)/x$  the ~~result is~~:		*With the shortcuts  $\text{si}(x) = \sin(x)/x$  or  $\text{sinc}(x) = \sin(\pi x)/(\pi x)$  the results are:

	:$$\left\| {X( f)} \right\| = A \cdot T \cdot\left\|{\rm si} ( { {\rm{\pi }}fT} ) \right\|.$$		:$$\left\| {X( f)} \right\| = A \cdot T \cdot\left\|{\rm si} ( { {\rm{\pi }}fT} ) \right\|,$$
			:$$\left\| {X( f)} \right\| = A \cdot T \cdot\left\|{\rm sinc} ( fT ) \right\|.$$

	*The spectral value at the frequency  $f = 1/T = \text{20 kHz}$  is given by		*The spectral value at the frequency  $f = 1/T = \text{20 kHz}$  is given by
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	:$$\left\| {X( {f = 20\;{\rm{kHz}}} )} \right\| = \frac{ {A \cdot T}}{ {\rm{\pi }}} \cdot \sin ( {\rm{\pi }} ) \hspace{0.15 cm}\underline{= 0}.$$		:$$\left\| {X( {f = 20\;{\rm{kHz}}} )} \right\| = \frac{ {A \cdot T}}{ {\rm{\pi }}} \cdot \sin ( {\rm{\pi }} ) \hspace{0.15 cm}\underline{= 0}.$$

	*When calculating the value for  $f = 0$  the quotient  $\text{0 divided by 0}$. By applying  [https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule L'Hospitals rule]  the limiting value can be calculated:		*When calculating the value for  $f = 0$  the quotient is  $\text{"0 divided by 0"}$.  By applying  [https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule L'Hospitals rule]  the limiting value can be calculated:

	:$$\mathop {\lim }\limits_{x \to 0 } \frac{ {\sin \left( x \right)}}{x} = 1.$$		:$$\mathop {\lim }\limits_{x \to 0 } \frac{ {\sin \left( x \right)}}{x} = 1.$$
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	[[File:P_ID563__Sig_A_3_3_c.png\|right\|frame\|~~Betragsspektrum des Rechteckimpulses~~]]		[[File:P_ID563__Sig_A_3_3_c.png\|right\|frame\|Magnitude spectrum of the rectangular pulse]]
	'''(3)'''  ~~Proposed~~<u> solutions 1 and 3</u> are correct::		'''(3)'''  The <u>proposed solutions 1 and 3</u> are correct:
	*According to the result of subtask  '''(2)'''  the zeros occur at the distance  $f_0 = 1/T$ .		*According to the result of subtask  '''(2)'''  the zeros occur at the distance  $f_0 = 1/T$ .
	*With  $f_0 = 1/(2T) = f = 10 \;{\rm{kHz}}$  the real part is  $0$, but not the imaginary part.		*With  $f_0 = 1/(2T) = f = 10 \;{\rm{kHz}}$  the real part is  $0$, but not the imaginary part.
	*With the arguments  $f \cdot T = 0.5, 1.5, 2.5,\hspace{0.05cm}\text{ ... }$  the sine function is in each case equal in magnitude to  $1$, and it holds:		*With the arguments  $f \cdot T = 0.5,\ 1.5,\ 2.5,\hspace{0.05cm}\text{ ... }$  the sine function is in each case equal in magnitude to  $1$, and it holds:

	:$$\left\| {X( f )} \right\| = \frac{A}{ {{\rm{\pi }}\left\| f \right\|}} = X_{\rm S} ( f ).$$		:$$\left\| {X( f )} \right\| = \frac{A}{ {{\rm{\pi }}\left\| f \right\|}} = X_{\rm S} ( f ).$$

	*At other frequencies,  $X_{\rm S}(f)$  ~~als obere Schranke,~~ serves as an upper bound, i.e.  $\|Xf)\| \leq X_{\rm S}(f)$ always applies.		*At other frequencies,  $X_{\rm S}(f)$  serves as an upper bound, i.e.  $\|Xf)\| \leq X_{\rm S}(f)$  always applies.
	*In the sketch, this bound is drawn as a violet curve in addition to  $\|X(f)\|$~~ ~~.		*In the sketch, this bound is drawn as a violet curve in addition to  $\|X(f)\|$.


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	:$$\varphi ( f ) = \arctan \left( {\frac{ {1 - \cos ( {\omega T} )}}{ {\sin ( {\omega T} )}}} \right).$$		:$$\varphi ( f ) = \arctan \left( {\frac{ {1 - \cos ( {\omega T} )}}{ {\sin ( {\omega T} )}}} \right).$$

	*The argument of this function is equal to  $\tan(\omega T/2) = \tan(\pi fT)$ according to the specification. From this follows a linearly increasing course with ~~the~~ frequency:		*The argument of this function is equal to  $\tan(\omega T/2) = \tan(\pi fT)$  according to the specification.  From this follows a linearly increasing course with frequency:

	:$$\varphi ( f ) = \arctan \left( {\tan ( { {\rm{\pi }}fT} )} \right) = {\rm{\pi }}fT.$$		:$$\varphi ( f ) = \arctan \left( {\tan ( { {\rm{\pi }}fT} )} \right) = {\rm{\pi }}fT.$$

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	}}		}}

	[[File:EN_Sig_A_3_3.png\|250px\|right\|frame\|~~Rechteckimpuls und zugehöriges Spektrum~~]]		[[File:EN_Sig_A_3_3.png\|250px\|right\|frame\|Rectangular pulse and its spectrum]]

	A rectangular pulse  $x(t)$  with a duration of  $T = 50\,\text{µs}$  and ~~the~~ height of  $A = 2\,\text{V}$ is considered. At the ~~jump~~ points at  $t = 0$  and  $t = T$  the signal value is  $A/2$, in each case, but this has no influence on the solution of the task.		A rectangular pulse  $x(t)$  with duration  $T = 50\,\text{µs}$  and height  $A = 2\,\text{V}$ is considered.  At the jumping points at  $t = 0$  and  $t = T$  the signal value is  $A/2$  in each case, but this has no influence on the solution of the task.

	In the ~~graphic below~~, the corresponding spectral function is sketched qualitatively according to magnitude and phase. It is valid:		In the lower graph, the corresponding spectral function is sketched qualitatively according to magnitude and phase.  It is valid:

	:$$X( f ) = \left\| {X( f )} \right\| \cdot {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} \varphi ( f )} .$$		:$$X( f ) = \left\| {X( f )} \right\| \cdot {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} \varphi ( f )} .$$
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	''Hints:''		''Hints:''
	*This ~~exercise~~ belongs to the chapter  [[Signal_Representation/Special_Cases_of_Impulse_Signals\|Special Cases of Impulse Signals]].		*This task belongs to the chapter  [[Signal_Representation/Special_Cases_of_Impulse_Signals\|Special Cases of Impulse Signals]].

	*The following trigonometric transformations are ~~also~~ given:		*The following trigonometric transformations are given:

	:$$\sin ^2( \alpha ) = {1}/{2} \cdot \big( {1 - \cos ( {2\alpha } )} \big),\hspace{0.5cm} \tan( {\alpha /2} ) = \frac{ {1 - \cos ( \alpha )}}{ {\sin ( \alpha )}}.$$		:$$\sin ^2( \alpha ) = {1}/{2} \cdot \big( {1 - \cos ( {2\alpha } )} \big),\hspace{0.5cm} \tan( {\alpha /2} ) = \frac{ {1 - \cos ( \alpha )}}{ {\sin ( \alpha )}}.$$
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	<quiz display=simple>		<quiz display=simple>
	{Calculate the spectral function  $X(f)$. in general. What value results at the frequency  $f=10\, \text{kHz}$?		{Calculate the spectral function  $X(f)$  in general.  What value results at the frequency  $f=10\, \text{kHz}$?
	\|type="{}"}		\|type="{}"}
	${\rm Re}\big[X(f=10 \,\text{kHz})\big] \ = \ $ { 0. }   $\text{mV/Hz}$		${\rm Re}\big[X(f=10 \,\text{kHz})\big] \ = \ $ { 0. }   $\text{mV/Hz}$
	${\rm Im}\big[X(f=10\, \text{kHz})\big]\ = \ $ { -0.064--0.06 }   $\text{mV/Hz}$		${\rm Im}\big[X(f=10\, \text{kHz})\big]\ = \ $ { -0.064--0.06 }   $\text{mV/Hz}$

	{Calculate the magnitude function  $\|X(f)\|$  in general. What values result for the frequencies  $f = 0$  and  $f=20 \,\text{kHz}$?		{Calculate the magnitude function  $\|X(f)\|$  in general.  What values result for the frequencies  $f = 0$  and  $f=20 \,\text{kHz}$?
	\|type="{}"}		\|type="{}"}
	$\|X(f=0)\|\ = \ $ { 0.1 3% }   $\text{mV/Hz}$		$\|X(f=0)\|\ = \ $ { 0.1 3% }   $\text{mV/Hz}$
	$\|X(f=20\, \text{kHz})\|\ = \ $ { 0. }   $\text{mV/Hz}$		$\|X(f=20\, \text{kHz})\|\ = \ $ { 0. }   $\text{mV/Hz}$

	{Which of the following statements are true regarding  $\|X(f)\|$~~ ~~?		{Which of the following statements are true regarding  $\|X(f)\|$?
	\|type="[]"}		\|type="[]"}
	+ $\|X(f)\|$  has zeros at multiples of  $f_0 = 1/T$.		+ $\|X(f)\|$  has zeros at multiples of  $f_0 = 1/T$.
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	+ In the middle between two zeros  $\|X(f)\| = \|A/(\pi f)\|$ is true.		+ In the middle between two zeros  $\|X(f)\| = \|A/(\pi f)\|$ is true.

	{Calculate the phase function  $\varphi (f)$. What phase angle (in degrees) results at the frequency  $f=10\, \text{kHz}$?		{Calculate the phase function  $\varphi (f)$.  What phase angle (in degrees) results at the frequency  $f=10\, \text{kHz}$?
	\|type="{}"}		\|type="{}"}
	$\varphi (f=10\, \text{kHz})\ = \ $ { 90 1% }   $\text{deg}$		$\varphi (f=10\, \text{kHz})\ = \ $ { 90 1% }   $\text{deg}$