Aufgaben:Exercise 4.3Z: Hilbert Transformator - Revision history

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	'''(1)'''  For the spectral function at the model output holds:		'''(1)'''  For the spectral function at the model output holds:
	:$$X_{\rm +}(f)= \left(1 + {\rm j}\cdot H_{\rm HT}(f)\right) \~~cdot~~		:$$X_{\rm +}(f)= \left(1 + {\rm j}\cdot H_{\rm HT}(f)\right) \cdotX(f).$$
	X(f).$$
	*A comparison with the given relation		*A comparison with the given relation
	:$$X_{\rm +}(f)= \left(1 + {\rm		:$$X_{\rm +}(f)= \left(1 + {\rmsign}(f)\right) \cdot X(f)$$
	~~sign~~}(f)\right) \cdot X(f)$$
	:shows that  $H_{\rm HT}(f) = - {\rm j} \cdot \sign(f)$.		:shows that  $H_{\rm HT}(f) = - {\rm j} \cdot \sign(f)$.

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	'''(2)'''  From the spectral function		'''(2)'''  From the spectral function
	:$$X_1(f) = {A}/{2}\cdot\delta (f + f_{0})+		:$$X_1(f) = {A}/{2}\cdot\delta (f + f_{0})+{A}/{2}\cdot\delta (f - f_{0}).$$
	{A}/{2}\cdot\delta (f - f_{0}).$$
	:becomes according to the Hilbert transformer:		:becomes according to the Hilbert transformer:
	:$$Y_1(f) = {\rm j}\cdot {A}/{2}\cdot\delta (f + f_{0})-{\rm		:$$Y_1(f) = {\rm j}\cdot {A}/{2}\cdot\delta (f + f_{0})-{\rmj}\cdot {A}/{2}\cdot\delta (f - f_{0}).$$
	j}\cdot {A}/{2}\cdot\delta (f - f_{0}).$$
	*Thus the signal at the output of the Hilbert transformer is:		*Thus the signal at the output of the Hilbert transformer is:
	:$$y_1(t) = A \cdot {\sin} ( 2 \pi f_0 t ) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}y_1(t=0)\hspace{0.15 cm}\underline{ =0}.$$		:$$y_1(t) = A \cdot {\sin} ( 2 \pi f_0 t ) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}y_1(t=0)\hspace{0.15 cm}\underline{ =0}.$$
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	'''(3)'''  Now the spectral functions at the input and output of the Hilbert transformer are:		'''(3)'''  Now the spectral functions at the input and output of the Hilbert transformer are:
	:$$X_2(f) = {\rm j}\cdot {A}/{2}\cdot\delta (f + f_{0})-{\rm		:$$X_2(f) = {\rm j}\cdot {A}/{2}\cdot\delta (f + f_{0})-{\rmj}\cdot {A}/{2}\cdot\delta (f - f_{0}),$$
	j}\cdot {A}/{2}\cdot\delta (f - f_{0}),$$		:$$Y_2(f) = -{A}/{2}\cdot\delta (f + f_{0})-{A}/{2}\cdot\delta (f - f_{0}).$$
	:$$Y_2(f) = -{A}/{2}\cdot\delta (f + f_{0})-
	{A}/{2}\cdot\delta (f - f_{0}).$$
	*It follows that  $y_2(t) = - A \cdot \cos(2\pi f_0 t)$  and  $y_2(t = 0)\; \underline{= -\hspace{-0.08cm}1 \,\text{V}}$.		*It follows that  $y_2(t) = - A \cdot \cos(2\pi f_0 t)$  and  $y_2(t = 0)\; \underline{= -\hspace{-0.08cm}1 \,\text{V}}$.

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	'''(4)'''  This input signal can also be represented as follows:		'''(4)'''  This input signal can also be represented as follows:
	:$$x_3(t) = A \cdot {\cos} ( 2 \pi f_0 t -		:$$x_3(t) = A \cdot {\cos} ( 2 \pi f_0 t -2 \pi \cdot {\rm 10 \hspace{0.05cm} kHz}\cdot {\rm 0.0125 \hspace{0.05cm} ms}) =A \cdot {\cos} ( 2 \pi f_0 t - \pi/4)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}y_3(t) = A \cdot {\cos} ( 2 \pi f_0 t - 3\pi/4).$$
	2 \pi \cdot {\rm 10 \hspace{0.05cm} kHz}\cdot {\rm 0.0125 \hspace{0.05cm} ms}) =
	A \cdot {\cos} ( 2 \pi f_0 t - \pi/4)\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm}y_3(t) = A \cdot {\cos} ( 2 \pi f_0 t - 3\pi/4).$$
	*The signal phase is thus  $\varphi = \pi /4$.		*The signal phase is thus  $\varphi = \pi /4$.
	*The Hilbert transformer delays this by  $\varphi_{\rm HT} \; \underline{= 90^\circ} \; (\pi /2)$.		*The Hilbert transformer delays this by  $\varphi_{\rm HT} \; \underline{= 90^\circ} \; (\pi /2)$.
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	'''(5)'''  The spectral function of the signal  $x_3(t)$  is:		'''(5)'''  The spectral function of the signal  $x_3(t)$  is:
	:$$X_3(f) = {A_0}/{2} \cdot {\rm e}^{{\rm j} \varphi}\cdot\delta		:$$X_3(f) = {A_0}/{2} \cdot {\rm e}^{{\rm j} \varphi}\cdot\delta(f + f_{\rm 0}) + {A_0}/{2} \cdot {\rm e}^{-{\rm j}\varphi}\cdot\delta (f - f_{\rm 0}) .$$
	(f + f_{\rm 0}) + {A_0}/{2} \cdot {\rm e}^{-{\rm j}
	\varphi}\cdot\delta (f - f_{\rm 0}) .$$
	*For the analytical signal, the first component disappears and the component at  $+f_0$  is doubled:		*For the analytical signal, the first component disappears and the component at  $+f_0$  is doubled:

	:$$X_{3+}(f) = {A_0} \cdot {\rm e}^{-{\rm j} \varphi}\cdot\delta (f		:$$X_{3+}(f) = {A_0} \cdot {\rm e}^{-{\rm j} \varphi}\cdot\delta (f- f_{\rm 0}) .$$
	- f_{\rm 0}) .$$
	*By applying the  "Shifting Theorem" , the associated time function with  $\varphi = \pi /4$ is:		*By applying the  "Shifting Theorem" , the associated time function with  $\varphi = \pi /4$ is:
	:$$x_{3+}(t) = A_0 \cdot {\rm e}^{{\rm j}( 2 \pi f_{\rm 0} t		:$$x_{3+}(t) = A_0 \cdot {\rm e}^{{\rm j}( 2 \pi f_{\rm 0} t\hspace{0.05cm}-\hspace{0.05cm} \varphi)}.$$
	\hspace{0.05cm}-\hspace{0.05cm} \varphi)}.$$
	*Specifically, for time  $t = 0$:		*Specifically, for time  $t = 0$:
	:$$x_{3+}(t = 0) = A_0 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}		:$$x_{3+}(t = 0) = A_0 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\varphi} = A_0 \cdot{\cos} ( 45^\circ)-{\rm j}\cdot A_0\cdot{\sin} ( 45^\circ)= \hspace{0.15 cm}\underline{{\rm 0.707 \hspace{0.05cm} V}-{\rmj}\cdot {\rm 0.707 \hspace{0.05cm} V}}.$$
	\varphi} = A_0 \cdot{\cos} ( 45^\circ)-{\rm j}\cdot A_0
	\cdot{\sin} ( 45^\circ)= \hspace{0.15 cm}\underline{{\rm 0.707 \hspace{0.05cm} V}-{\rm
	j}\cdot {\rm 0.707 \hspace{0.05cm} V}}.$$


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	*To get from  $x(t)$  to  $x_+(t)$,  just replace the cosine function with the complex exponential function.		*To get from  $x(t)$  to  $x_+(t)$,  just replace the cosine function with the complex exponential function.
	*For example, the following applies to a harmonic oscillation:		*For example, the following applies to a harmonic oscillation:
	:$$x(t) = A \cdot {\cos} ( 2 \pi f_0 t		:$$x(t) = A \cdot {\cos} ( 2 \pi f_0 t-\hspace{0.05cm} \varphi) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x_{+}(t) = A \cdot {\rm e}^{{\rm j}( 2 \pi f_{\rm 0} t\hspace{0.05cm}-\hspace{0.05cm} \varphi)}.$$
	-\hspace{0.05cm} \varphi) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x_{+}(t) = A \cdot {\rm e}^{{\rm j}( 2 \pi f_{\rm 0} t
	\hspace{0.05cm}-\hspace{0.05cm} \varphi)}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Signal Representation: Exercises\|^4.2 Analytical Signal and its Spectral Function^]]		[[Category:Signal Representation: Exercises\|^4.2 Analytical Signal and its Spectral Function^]]
			[[de:Aufgaben:4.3Z_Hilbert-Transformator]]

Guenter at 14:13, 24 May 2021

2021-05-24T14:13:04Z

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	[[File:P_ID717__Sig_Z_4_3_neu.png\|right\|frame\|Hilbert ~~transformer~~ ]]		[[File:P_ID717__Sig_Z_4_3_neu.png\|right\|frame\|Hilbert transformator ]]
	The diagram describes a model of how, at least mentally,		The diagram describes a model of how, at least mentally,
	*the analytical signal  $x_{+}(t)$  can be generated,		*the analytical signal  $x_{+}(t)$  can be generated,

Guenter at 14:40, 6 May 2021

2021-05-06T14:40:13Z

Guenter at 14:29, 6 May 2021

2021-05-06T14:29:24Z

← Older revision		Revision as of 16:29, 6 May 2021
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	[[File:P_ID717__Sig_Z_4_3_neu.png\|right\|frame\|Hilbert transformer ]]		[[File:P_ID717__Sig_Z_4_3_neu.png\|right\|frame\|Hilbert transformer ]]
	The diagram describes a model of how, at least mentally		The diagram describes a model of how, at least mentally,
	*~~from~~ the ~~real band-pass~~ signal  $x(t)$		*the analytical signal  $x_{+}(t)$  can be generated,
	*the ~~analytical~~ signal  $~~x_{+}~~(t)$		*from the real band-pass signal  $x(t)$.


			The lower branch contains the so-called  "Hilbert transformer"  with the frequency response  $H_{\rm HT}(f)$.

	~~can be generated.~~		Its output signal  $y(t)$  is multiplied by the imaginary unit  $\rm j$  and added to the signal  $x(t)$ :

	~~The lower branch contains the so-called &bdquo;Hilbert transformer” with the frequency response  $H_{\rm HT}(f)$.~~ Its output signal  $y(t)$  is multiplied by the imaginary unit  $\rm j$  and added to the signal  $x(t)$ :
	:$$x_{\rm +}(t)= x(t) + {\rm j}\cdot y(t) .$$		:$$x_{\rm +}(t)= x(t) + {\rm j}\cdot y(t) .$$
	As test signals are used, each with  $A = 1 \, \text{V}$  and  $f_0 = 10 \, \text{kHz}$:		As test signals are used, each with  $A = 1 \, \text{V}$  and  $f_0 = 10 \, \text{kHz}$:
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	:$$x_2(t) = A \cdot {\sin} ( 2 \pi f_0 t ),$$		:$$x_2(t) = A \cdot {\sin} ( 2 \pi f_0 t ),$$
	:$$x_3(t) = A \cdot {\cos} \big( 2 \pi f_0 (t - \tau) \big) \hspace{0.3cm}{\rm with}\hspace{0.3cm}\tau = 12.5 \hspace{0.1cm}{\rm µ s}.$$		:$$x_3(t) = A \cdot {\cos} \big( 2 \pi f_0 (t - \tau) \big) \hspace{0.3cm}{\rm with}\hspace{0.3cm}\tau = 12.5 \hspace{0.1cm}{\rm µ s}.$$





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	''Hints:''		''Hints:''
	*This exercise belongs to the ~~task~~  [[Signal_Representation/Analytical_Signal_and_Its_Spectral_Function\|Analytical Signal and ~~Its~~ Spectral Function]].		*This exercise belongs to the chapter  [[Signal_Representation/Analytical_Signal_and_Its_Spectral_Function\|Analytical Signal and its Spectral Function]].

	*The following applies to the spectral function of the analytical signal:		*The following applies to the spectral function of the analytical signal:
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	<quiz display=simple>		<quiz display=simple>
	{Calculate the frequency response  $H_{HT}(f)$  of the Hilbert transformer. Which value is valid for the frequency  $f_0 = 10 \text{ kHz}$?		{Calculate the frequency response  $H_{\rm HT}(f)$  of the Hilbert transformer.  Which value is valid for the frequency  $f_0 = 10 \text{ kHz}$?
	\|type="{}"}		\|type="{}"}
	$\text{Re}[H_{\rm HT}(f = f_0)]\ = \ $ { 0. }		$\text{Re}[H_{\rm HT}(f = f_0)]\ = \ $ { 0. }
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	{What is the Hilbert transform  $y_1(t)$  for the input signal  $x_1(t)$? In particular, what value results at  $t = 0$?		{What is the Hilbert transform  $y_1(t)$  for the input signal  $x_1(t)$?  In particular, what value results at  $t = 0$?
	\|type="{}"}		\|type="{}"}
	$y_1(t = 0)\ = \ $ { 0. }  $\rm V$		$y_1(t = 0)\ = \ $ { 0. }  $\rm V$


	{What is the Hilbert transform  $y_2(t)$  for the input signal  $x_2(t)$? Which value results in particular at  $t = 0$?		{What is the Hilbert transform  $y_2(t)$  for the input signal  $x_2(t)$?  Which value results in particular at  $t = 0$?
	\|type="{}"}		\|type="{}"}
	$y_2(t = 0)\ = \ $ { -1.03--0.97 }  $\rm V$		$y_2(t = 0)\ = \ $ { -1.03--0.97 }  $\rm V$


	{What is the Hilbert transform  $y_3(t)$  for the input signal  $x_3(t)$? What value results for  $t=0$? ~~<br>~~What is the phase delay  $\varphi_{\rm HT}$  of the Hilbert transformer?		{What is the Hilbert transform  $y_3(t)$  for the input signal  $x_3(t)$?  What value results for  $t=0$?  What is the phase delay  $\varphi_{\rm HT}$  of the Hilbert transformer?
	\|type="{}"}		\|type="{}"}
	$\varphi_{\rm HT}\ = \ $ { 90 3% }  $\text{Grad}$		$\varphi_{\rm HT}\ = \ $ { 90 3% }  $\text{Grad}$
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	{What is the analytical signal associated with  $x_3(t)$  ~~gehörige analytische Signal?~~ What are the values of the real and imaginary parts of this complex signal at time  $t = 0$?		{What is the analytical signal associated with  $x_3(t)$?  What are the values of the real and imaginary parts of this complex signal at time  $t = 0$?
	\|type="{}"}		\|type="{}"}
	$\text{Re}[x_{3+}(t = 0)]\ = \ $ { 0.707 3% }  $\text{V}$		$\text{Re}[x_{3+}(t = 0)]\ = \ $ { 0.707 3% }  $\text{V}$

Guenter at 14:10, 6 May 2021

2021-05-06T14:10:16Z

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	[[Category:Signal Representation: Exercises\|^4.2 Analytical Signal and ~~Its~~ Spectral Function^]]		[[Category:Signal Representation: Exercises\|^4.2 Analytical Signal and its Spectral Function^]]

Javier: Text replacement - "bandpass" to "band-pass"

2021-05-03T15:12:59Z

Text replacement - "bandpass" to "band-pass"

← Older revision		Revision as of 17:12, 3 May 2021
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	[[File:P_ID717__Sig_Z_4_3_neu.png\|right\|frame\|Hilbert transformer ]]		[[File:P_ID717__Sig_Z_4_3_neu.png\|right\|frame\|Hilbert transformer ]]
	The diagram describes a model of how, at least mentally		The diagram describes a model of how, at least mentally
	*from the real ~~bandpass~~ signal  $x(t)$		*from the real band-pass signal  $x(t)$
	*the analytical signal  $x_{+}(t)$		*the analytical signal  $x_{+}(t)$

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

2021-03-23T12:39:50Z

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

← Older revision		Revision as of 14:39, 23 March 2021
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	[[Category:~~Exercises for~~ Signal Representation\|^4.2 Analytical Signal and Its Spectral Function^]]		[[Category:Signal Representation: Exercises\|^4.2 Analytical Signal and Its Spectral Function^]]

Noah at 20:11, 5 February 2021

2021-02-05T20:11:02Z

Show changes

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	:$$X_{\rm +}(f)= \left(1 + {\rm j}\cdot H_{\rm HT}(f)\right) \cdotX(f).$$		:$$X_{\rm +}(f)= \left(1 + {\rm j}\cdot H_{\rm HT}(f)\right) \cdotX(f).$$
	*A comparison with the given relation		*A comparison with the given relation
	:$$X_{\rm +}(f)= \left(1 + {\~~rmsign~~}(f)\right) \cdot X(f)$$		:$$X_{\rm +}(f)= \left(1 + {\rm sign}(f)\right) \cdot X(f)$$
	:shows that  $H_{\rm HT}(f) = - {\rm j} \cdot \sign(f)$.		:shows that  $H_{\rm HT}(f) = - {\rm j} \cdot \sign(f)$.

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	:$$X_1(f) = {A}/{2}\cdot\delta (f + f_{0})+{A}/{2}\cdot\delta (f - f_{0}).$$		:$$X_1(f) = {A}/{2}\cdot\delta (f + f_{0})+{A}/{2}\cdot\delta (f - f_{0}).$$
	:becomes according to the Hilbert transformer:		:becomes according to the Hilbert transformer:
	:$$Y_1(f) = {\rm j}\cdot {A}/{2}\cdot\delta (f + f_{0})-{\~~rmj~~}\cdot {A}/{2}\cdot\delta (f - f_{0}).$$		:$$Y_1(f) = {\rm j}\cdot {A}/{2}\cdot\delta (f + f_{0})-{\rm j}\cdot {A}/{2}\cdot\delta (f - f_{0}).$$
	*Thus the signal at the output of the Hilbert transformer is:		*Thus the signal at the output of the Hilbert transformer is:
	:$$y_1(t) = A \cdot {\sin} ( 2 \pi f_0 t ) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}y_1(t=0)\hspace{0.15 cm}\underline{ =0}.$$		:$$y_1(t) = A \cdot {\sin} ( 2 \pi f_0 t ) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}y_1(t=0)\hspace{0.15 cm}\underline{ =0}.$$
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	'''(3)'''  Now the spectral functions at the input and output of the Hilbert transformer are:		'''(3)'''  Now the spectral functions at the input and output of the Hilbert transformer are:
	:$$X_2(f) = {\rm j}\cdot {A}/{2}\cdot\delta (f + f_{0})-{\~~rmj~~}\cdot {A}/{2}\cdot\delta (f - f_{0}),$$		:$$X_2(f) = {\rm j}\cdot {A}/{2}\cdot\delta (f + f_{0})-{\rm j}\cdot {A}/{2}\cdot\delta (f - f_{0}),$$
	:$$Y_2(f) = -{A}/{2}\cdot\delta (f + f_{0})-{A}/{2}\cdot\delta (f - f_{0}).$$		:$$Y_2(f) = -{A}/{2}\cdot\delta (f + f_{0})-{A}/{2}\cdot\delta (f - f_{0}).$$
	*It follows that  $y_2(t) = - A \cdot \cos(2\pi f_0 t)$  and  $y_2(t = 0)\; \underline{= -\hspace{-0.08cm}1 \,\text{V}}$.		*It follows that  $y_2(t) = - A \cdot \cos(2\pi f_0 t)$  and  $y_2(t = 0)\; \underline{= -\hspace{-0.08cm}1 \,\text{V}}$.
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	:$$x_{3+}(t) = A_0 \cdot {\rm e}^{{\rm j}( 2 \pi f_{\rm 0} t\hspace{0.05cm}-\hspace{0.05cm} \varphi)}.$$		:$$x_{3+}(t) = A_0 \cdot {\rm e}^{{\rm j}( 2 \pi f_{\rm 0} t\hspace{0.05cm}-\hspace{0.05cm} \varphi)}.$$
	*Specifically, for time  $t = 0$:		*Specifically, for time  $t = 0$:
	:$$x_{3+}(t = 0) = A_0 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\varphi} = A_0 \cdot{\cos} ( 45^\circ)-{\rm j}\cdot A_0\cdot{\sin} ( 45^\circ)= \hspace{0.15 cm}\underline{{\rm 0.707 \hspace{0.05cm} V}-{\~~rmj~~}\cdot {\rm 0.707 \hspace{0.05cm} V}}.$$		:$$x_{3+}(t = 0) = A_0 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\varphi} = A_0 \cdot{\cos} ( 45^\circ)-{\rm j}\cdot A_0\cdot{\sin} ( 45^\circ)= \hspace{0.15 cm}\underline{{\rm 0.707 \hspace{0.05cm} V}-{\rm j}\cdot {\rm 0.707 \hspace{0.05cm} V}}.$$


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	[[Category:Signal Representation: Exercises\|^4.2 Analytical Signal and its Spectral Function^]]		[[Category:Signal Representation: Exercises\|^4.2 Analytical Signal and its Spectral Function^]]
	[[de:Aufgaben:4.~~3Z_Hilbert~~-Transformator]]		[[de:Aufgaben:Aufgabe 4.3Z: Hilbert-Transformator]]

← Older revision		Revision as of 16:40, 6 May 2021
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	:$$X_{\rm +}(f)= \left(1 + {\rm		:$$X_{\rm +}(f)= \left(1 + {\rm
	sign}(f)\right) \cdot X(f)$$		sign}(f)\right) \cdot X(f)$$
	:shows that  $H_{\rm HT}(f) = - {\rm j} \cdot \sign(f)$~~  ist~~.		:shows that  $H_{\rm HT}(f) = - {\rm j} \cdot \sign(f)$.

	*Thus, the real part we are looking for is  ${\rm Re}[X_{\rm +}(f)]\hspace{0.15cm}\underline{=0}$  and the imaginary part is equal to  ${\rm Im}[X_{\rm +}(f)]\hspace{0.15cm}\underline{=-1}$.		*Thus, the real part we are looking for is  ${\rm Re}[X_{\rm +}(f)]\hspace{0.15cm}\underline{=0}$  and the imaginary part is equal to  ${\rm Im}[X_{\rm +}(f)]\hspace{0.15cm}\underline{=-1}$.
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	:$$Y_2(f) = -{A}/{2}\cdot\delta (f + f_{0})-		:$$Y_2(f) = -{A}/{2}\cdot\delta (f + f_{0})-
	{A}/{2}\cdot\delta (f - f_{0}).$$		{A}/{2}\cdot\delta (f - f_{0}).$$
	*It follows that $y_2(t) = - A \cdot \cos(2\pi f_0 t)$ ~~und~~ $y_2(t = 0)\; \underline{= -\hspace{-0.08cm}1 \,\text{V}}$.		*It follows that  $y_2(t) = - A \cdot \cos(2\pi f_0 t)$  and  $y_2(t = 0)\; \underline{= -\hspace{-0.08cm}1 \,\text{V}}$.


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	\Rightarrow \hspace{0.3cm}y_3(t) = A \cdot {\cos} ( 2 \pi f_0 t - 3\pi/4).$$		\Rightarrow \hspace{0.3cm}y_3(t) = A \cdot {\cos} ( 2 \pi f_0 t - 3\pi/4).$$
	*The signal phase is thus  $\varphi = \pi /4$.		*The signal phase is thus  $\varphi = \pi /4$.
	*The Hilbert transformer delays this by  $\varphi_{\rm HT} \; \underline{= 90^\circ} \; (\pi /2)$~~  verzögert~~.		*The Hilbert transformer delays this by  $\varphi_{\rm HT} \; \underline{= 90^\circ} \; (\pi /2)$.
	*Therefore, the output signal  $y_3(t) = A \cdot \cos(2\pi f_0 t -3 \pi /4)$  and the signal value at time  $t = 0$  is  $A \cdot \cos(135^\circ) \; \underline{= -0.707 \,\text{V}}$.		*Therefore, the output signal  $y_3(t) = A \cdot \cos(2\pi f_0 t -3 \pi /4)$  and the signal value at time  $t = 0$  is  $A \cdot \cos(135^\circ) \; \underline{= -0.707 \,\text{V}}$.

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	:$$X_{3+}(f) = {A_0} \cdot {\rm e}^{-{\rm j} \varphi}\cdot\delta (f		:$$X_{3+}(f) = {A_0} \cdot {\rm e}^{-{\rm j} \varphi}\cdot\delta (f
	- f_{\rm 0}) .$$		- f_{\rm 0}) .$$
	*By applying the  ~~[[Signal_Representation/Fourier_Transform_Laws#Verschiebungssatz\|Verschiebungssatzes]]~~ , the associated time function with  $\varphi = \pi /4$ is ~~thus:~~:		*By applying the  "Shifting Theorem" , the associated time function with  $\varphi = \pi /4$ is:
	:$$x_{3+}(t) = A_0 \cdot {\rm e}^{{\rm j}( 2 \pi f_{\rm 0} t		:$$x_{3+}(t) = A_0 \cdot {\rm e}^{{\rm j}( 2 \pi f_{\rm 0} t
	\hspace{0.05cm}-\hspace{0.05cm} \varphi)}.$$		\hspace{0.05cm}-\hspace{0.05cm} \varphi)}.$$
Line 125:		Line 125:
	\cdot{\sin} ( 45^\circ)= \hspace{0.15 cm}\underline{{\rm 0.707 \hspace{0.05cm} V}-{\rm		\cdot{\sin} ( 45^\circ)= \hspace{0.15 cm}\underline{{\rm 0.707 \hspace{0.05cm} V}-{\rm
	j}\cdot {\rm 0.707 \hspace{0.05cm} V}}.$$		j}\cdot {\rm 0.707 \hspace{0.05cm} V}}.$$


	''Hint'':		''Hint'':
	*To get from  $x(t)$  to  $x_+(t)$  , just replace the cosine function with the complex exponential function.		*To get from  $x(t)$  to  $x_+(t)$,  just replace the cosine function with the complex exponential function.
	*For example, the following applies to a harmonic oscillation:		*For example, the following applies to a harmonic oscillation:
	:$$x(t) = A \cdot {\cos} ( 2 \pi f_0 t		:$$x(t) = A \cdot {\cos} ( 2 \pi f_0 t