Aufgaben:Exercise 1.6Z: Interpretation of the Frequency Response - Revision history

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	'''(5)'''  The result of subtask  '''(3)'''  – considering the symmetry – is for  $f_i = f_1$:		'''(5)'''  The result of subtask  '''(3)'''  – considering the symmetry – is for  $f_i = f_1$:
	:$$A_1= \frac{2A_x}{\Delta t} \cdot \int\limits_{ 0 }^{ \Delta t /2 } {\rm cos}(2\pi f_1 \tau )\hspace{0.1cm}{\~~rmd~~}\tau = \frac{2A_x}{2\pi f_1 \cdot \Delta t} \cdot {\rm sin}(2\pi f_1 \frac{\Delta t}{2})= A_x \cdot {\rm si}(\pi f_1 \Delta t ).$$		:$$A_1= \frac{2A_x}{\Delta t} \cdot \int\limits_{ 0 }^{ \Delta t /2 } {\rm cos}(2\pi f_1 \tau )\hspace{0.1cm}{\rm d}\tau = \frac{2A_x}{2\pi f_1 \cdot \Delta t} \cdot {\rm sin}(2\pi f_1 \frac{\Delta t}{2})= A_x \cdot {\rm si}(\pi f_1 \Delta t ).$$
	*Taking   $f_1 · Δt = 0.5$ into account the result is:		*Taking   $f_1 · Δt = 0.5$ into account the result is:
	:$$A_1 = A_x \cdot {\rm si}(\frac{\pi}{2} ) = \frac{2A_x}{\pi} \hspace{0.15cm}\underline{= 0.637\,{\rm V}}.$$		:$$A_1 = A_x \cdot {\rm si}(\frac{\pi}{2} ) = \frac{2A_x}{\pi} \hspace{0.15cm}\underline{= 0.637\,{\rm V}}.$$
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	[[Category:Linear and Time-Invariant Systems: Exercises\|^1.3 Some Low-Pass Functions in Systems Theory^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^1.3 Some Low-Pass Functions in Systems Theory^]]
	[[de:1.~~6Z_Interpretation_von_H(f)~~]]		[[de:Aufgaben:Aufgabe 1.6Z: Interpretation der Übertragungsfunktion]]

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	'''(5)'''  The result of subtask  '''(3)'''  – considering the symmetry – is for  $f_i = f_1$:		'''(5)'''  The result of subtask  '''(3)'''  – considering the symmetry – is for  $f_i = f_1$:
	:$$A_1= \frac{2A_x}{\Delta t} \cdot \int\limits_{ 0 }^{ \Delta t /2 } {\rm cos}(2\pi f_1 \tau )\hspace{0.1cm}{\rm		:$$A_1= \frac{2A_x}{\Delta t} \cdot \int\limits_{ 0 }^{ \Delta t /2 } {\rm cos}(2\pi f_1 \tau )\hspace{0.1cm}{\rmd}\tau = \frac{2A_x}{2\pi f_1 \cdot \Delta t} \cdot {\rm sin}(2\pi f_1 \frac{\Delta t}{2})= A_x \cdot {\rm si}(\pi f_1 \Delta t ).$$
	d}\tau = \frac{2A_x}{2\pi f_1 \cdot \Delta t} \cdot {\rm sin}(2\pi f_1 \frac{\Delta t}{2}
	)= A_x \cdot {\rm si}(\pi f_1 \Delta t ).$$
	*Taking   $f_1 · Δt = 0.5$ into account the result is:		*Taking   $f_1 · Δt = 0.5$ into account the result is:
	:$$A_1 = A_x \cdot {\rm si}(\frac{\pi}{2} ) = \frac{2A_x}{\pi} \hspace{0.15cm}\underline{= 0.637\,{\rm V}}.$$		:$$A_1 = A_x \cdot {\rm si}(\frac{\pi}{2} ) = \frac{2A_x}{\pi} \hspace{0.15cm}\underline{= 0.637\,{\rm V}}.$$

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	[[Category:Linear and Time-Invariant Systems: Exercises\|^1.3 Some Low-Pass Functions in Systems Theory^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^1.3 Some Low-Pass Functions in Systems Theory^]]
			[[de:1.6Z_Interpretation_von_H(f)]]

Oezer at 17:33, 7 September 2021

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	d}\tau = \frac{2A_x}{2\pi f_1 \cdot \Delta t} \cdot {\rm sin}(2\pi f_1 \frac{\Delta t}{2}		d}\tau = \frac{2A_x}{2\pi f_1 \cdot \Delta t} \cdot {\rm sin}(2\pi f_1 \frac{\Delta t}{2}
	)= A_x \cdot {\rm si}(\pi f_1 \Delta t ).$$		)= A_x \cdot {\rm si}(\pi f_1 \Delta t ).$$
	*Taking   $f_1 · Δt = 0.5$ into account ~~thus~~ the result is:		*Taking   $f_1 · Δt = 0.5$ into account the result is:
	:$$A_1 = A_x \cdot {\rm si}(\frac{\pi}{2} ) = \frac{2A_x}{\pi} \hspace{0.15cm}\underline{= 0.637\,{\rm V}}.$$		:$$A_1 = A_x \cdot {\rm si}(\frac{\pi}{2} ) = \frac{2A_x}{\pi} \hspace{0.15cm}\underline{= 0.637\,{\rm V}}.$$
	*Correspondingly, the following is obtained using  $f_3 · Δt = 1.5$:		*Correspondingly, the following is obtained using  $f_3 · Δt = 1.5$:

Oezer at 17:32, 7 September 2021

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	*For the direct (DC) signal  $x_0(t) = A_x$ , set  $f_i = 0$  and one obtains  $A_0 = A_x \ \rm \underline{ = \ 1 \hspace{0.05cm} V}$.		*For the direct (DC) signal  $x_0(t) = A_x$ , set  $f_i = 0$  and one obtains  $A_0 = A_x \ \rm \underline{ = \ 1 \hspace{0.05cm} V}$.
	*In contrast to this, for the cosine frequencies  $f_2 = 2 \ \rm kHz$  and  $f_4 = 4 \ \rm kHz$  the integral vanishes in each case because then it is integrated over one and two periods, respectively:   $A_2 \ \rm \underline{ = \hspace{0.05cm} 0}$  und  $A_4 \hspace{0.05cm} \rm \underline{ = \ 0}$.		*In contrast to this, for the cosine frequencies  $f_2 = 2 \ \rm kHz$  and  $f_4 = 4 \ \rm kHz$  the integral vanishes in each case because then it is integrated over one and two periods, respectively:   $A_2 \ \rm \underline{ = \hspace{0.05cm} 0}$  und  $A_4 \hspace{0.05cm} \rm \underline{ = \ 0}$.
	*In the frequency domain, the ~~case~~ which are dealt with here correspond to:		*In the frequency domain, the cases which are dealt with here correspond to:
	:$$H(f=0) = 1, \hspace{0.3cm}H(f=\Delta f) = 0, \hspace{0.3cm}H(f=2\Delta f) = 0.$$		:$$H(f=0) = 1, \hspace{0.3cm}H(f=\Delta f) = 0, \hspace{0.3cm}H(f=2\Delta f) = 0.$$

Oezer at 17:31, 7 September 2021

2021-09-07T17:31:40Z

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	'''(4)'''  <u>Approaches 2, 3 and 5</u> are correct:		'''(4)'''  <u>Approaches 2, 3 and 5</u> are correct:
	*For the direct (DC) signal  $x_0(t) = A_x$  set  $f_i = 0$  and one obtains  $A_0 = A_x \ \rm \underline{ = \ 1 \hspace{0.05cm} V}$.		*For the direct (DC) signal  $x_0(t) = A_x$ , set  $f_i = 0$  and one obtains  $A_0 = A_x \ \rm \underline{ = \ 1 \hspace{0.05cm} V}$.
	*In contrast to this, for the cosine frequencies  $f_2 = 2 \ \rm kHz$  and  $f_4 = 4 \ \rm kHz$  the integral vanishes in each case because then it is integrated over one and two periods, respectively:   $A_2 \ \rm \underline{ = \hspace{0.05cm} 0}$  und  $A_4 \hspace{0.05cm} \rm \underline{ = \ 0}$.		*In contrast to this, for the cosine frequencies  $f_2 = 2 \ \rm kHz$  and  $f_4 = 4 \ \rm kHz$  the integral vanishes in each case because then it is integrated over one and two periods, respectively:   $A_2 \ \rm \underline{ = \hspace{0.05cm} 0}$  und  $A_4 \hspace{0.05cm} \rm \underline{ = \ 0}$.
	*In the frequency domain, the case which are dealt with here correspond to:		*In the frequency domain, the case which are dealt with here correspond to:

Oezer at 17:29, 7 September 2021

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	''Please note:''		''Please note:''
	*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory\|Some Low-Pass Functions in Systems Theory]].		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory\|Some Low-Pass Functions in Systems Theory]].
	*Contrary to the usual definition of the amplitude, the "$A_i$" may well be negative. Then, this corresponds to the function ~~„minus~~-~~cosine”~~.		*Contrary to the usual definition of the amplitude, the "$A_i$" may well be negative. Then, this corresponds to the function "minus-cosine".

Oezer at 17:29, 7 September 2021

2021-09-07T17:29:07Z

← Older revision		Revision as of 19:29, 7 September 2021
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	''Please note:''		''Please note:''
	*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory\|Some Low-Pass Functions in Systems Theory]].		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory\|Some Low-Pass Functions in Systems Theory]].
	*Contrary to the usual definition of the amplitude, the "$A_i$" may well be negative. ~~This~~ corresponds ~~then~~ to the function „minus-cosine”.		*Contrary to the usual definition of the amplitude, the "$A_i$" may well be negative. Then, this corresponds to the function „minus-cosine”.

Oezer at 17:27, 7 September 2021

2021-09-07T17:27:54Z

← Older revision		Revision as of 19:27, 7 September 2021
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	''Please note:''		''Please note:''
	*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory\| Some Low-Pass Functions in Systems Theory]].		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory\|Some Low-Pass Functions in Systems Theory]].
	*Contrary to the usual definition of the amplitude, the "$A_i$" may well be negative. This corresponds then to the function „minus-cosine”.		*Contrary to the usual definition of the amplitude, the "$A_i$" may well be negative. This corresponds then to the function „minus-cosine”.

Oezer at 17:26, 7 September 2021

2021-09-07T17:26:18Z

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	The task is meant to investigate the influence of a low-pass filter  $H(f)$  on cosinusoidal signals of the form		The task is meant to investigate the influence of a low-pass filter  $H(f)$  on cosinusoidal signals of the form
	:$$x_i(t) = A_x \cdot {\rm cos}(2\pi f_i t ).$$		:$$x_i(t) = A_x \cdot {\rm cos}(2\pi f_i t ).$$
	In the graph you can see the signals  $x_i(t)$, where the index $i$  indicates the frequency in $\rm kHz$ . So,  $x_2(t)$  describes a  $2 \hspace{0.09cm} \rm kHz$–signal.		In the graph you can see the signals  $x_i(t)$ where the index $i$  indicates the frequency in $\rm kHz$ . So,  $x_2(t)$  describes a  $2 \hspace{0.09cm} \rm kHz$–signal.

	The signal amplitude in each case is $A_x = 1 \hspace{0.05cm} \rm V$. The direct (DC) signal  $x_0(t)$  is to be interpreted as a limiting case of a cosine signal with frequeny  $f_0 =0$.		The signal amplitude in each case is $A_x = 1 \hspace{0.05cm} \rm V$. The direct (DC) signal  $x_0(t)$  is to be interpreted as a limiting case of a cosine signal with frequeny  $f_0 =0$.