Aufgaben:Exercise 1.2Z: Measurement of the Frequency Response - Revision history

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	'''(3)'''  With  $A_x = 2 \text{ V}$  and  $\varphi_x = 90^\circ$  (sine function)  the following is obtained for  $f_0 = f_3 =3 \text{ kHz}$:		'''(3)'''  With  $A_x = 2 \text{ V}$  and  $\varphi_x = 90^\circ$  (sine function)  the following is obtained for  $f_0 = f_3 =3 \text{ kHz}$:
	:$$H_{\rm B} (f_3) = \frac{A_y}{A_x} \cdot {\rm e}^{ -{\rm j}(\varphi_x - \varphi_y)} = \frac{1\hspace{0.05cm}{\~~rmV~~}}{2\hspace{0.05cm}{\rm V}} \cdot {\rm e}^{ -{\rm j} (90^{\circ} -90^{\circ})} = 0.5.$$		:$$H_{\rm B} (f_3) = \frac{A_y}{A_x} \cdot {\rm e}^{ -{\rm j}(\varphi_x - \varphi_y)} = \frac{1\hspace{0.05cm}{\rm V}}{2\hspace{0.05cm}{\rm V}} \cdot {\rm e}^{ -{\rm j} (90^{\circ} -90^{\circ})} = 0.5.$$
	Thus, for  $f_0 = f_3 = 3 \text{ kHz}$  the values		Thus, for  $f_0 = f_3 = 3 \text{ kHz}$  the values
	*$a_{\rm B} (f_3)\rm \underline{\: ≈ \: 0.693 \: Np}$ and		*$a_{\rm B} (f_3)\rm \underline{\: ≈ \: 0.693 \: Np}$ and
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	'''(4)'''  Analogously, the frequency response for  $f_0 = f_2 =2 \text{ kHz}$  can be determined:		'''(4)'''  Analogously, the frequency response for  $f_0 = f_2 =2 \text{ kHz}$  can be determined:
	:$$H_{\rm B} ( f_2) = \frac{0.8\hspace{0.05cm}{\~~rmV~~}}{2\hspace{0.05cm}{\rm V}} \cdot {\rm e}^{ -{\rm j} (90^{\circ} -70^{\circ})} = 0.4\cdot {\rm e}^{ -{\rm j} 20^{\circ}}.$$		:$$H_{\rm B} ( f_2) = \frac{0.8\hspace{0.05cm}{\rm V}}{2\hspace{0.05cm}{\rm V}} \cdot {\rm e}^{ -{\rm j} (90^{\circ} -70^{\circ})} = 0.4\cdot {\rm e}^{ -{\rm j} 20^{\circ}}.$$
	Hence, for  $f_0 = f_2 = 2 \ \text{ kHz}$:		Hence, for  $f_0 = f_2 = 2 \ \text{ kHz}$:
	*$a_{\rm B}(f_2) \rm \underline{\: ≈ \: 0.916 \: Np}$,		*$a_{\rm B}(f_2) \rm \underline{\: ≈ \: 0.916 \: Np}$,
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	[[Category:Linear and Time-Invariant Systems: Exercises\|^1.1 System Description in Frequency Domain^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^1.1 System Description in Frequency Domain^]]
	[[de:Aufgaben:1.~~2Z_Messung_der_Übertragungsfunktion~~]]		[[de:Aufgaben:Aufgabe 1.2Z: Messung der Übertragungsfunktion]]

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	*The magnitude spectrum at the output of filter  $\rm A$  with frequency  $f_0 = 1 \ \text{kHz}$  is:		*The magnitude spectrum at the output of filter  $\rm A$  with frequency  $f_0 = 1 \ \text{kHz}$  is:
	:$$\|Y_{\rm A} (f)\| = 1.6\hspace{0.05cm}{\rm V} \cdot {\rm \delta } (f		:$$\|Y_{\rm A} (f)\| = 1.6\hspace{0.05cm}{\rm V} \cdot {\rm \delta } (f\pm f_0) + 0.4\hspace{0.05cm}{\rm V} \cdot {\rm \delta } (f \pm 3 f_0) .$$
	\pm f_0) + 0.4\hspace{0.05cm}{\rm V} \cdot {\rm \delta } (f \pm 3 f_0) .$$
	*For another filter  $\rm B$  the output signal is always a harmonic oscillation with the (single) frequency  $f_0$.  For the frequencies  $f_0$  given in the table the amplitudes  $A_y(f_0)$  and the phases  $φ_y(f_0)$  are measured.  Here, the following holds:		*For another filter  $\rm B$  the output signal is always a harmonic oscillation with the (single) frequency  $f_0$.  For the frequencies  $f_0$  given in the table the amplitudes  $A_y(f_0)$  and the phases  $φ_y(f_0)$  are measured.  Here, the following holds:
	:$$Y_{\rm B} (f) = {A_y}/{2} \cdot {\rm e}^{ {\rm j} \varphi_y}		:$$Y_{\rm B} (f) = {A_y}/{2} \cdot {\rm e}^{ {\rm j} \varphi_y}\cdot {\rm \delta } (f + f_0) + {A_y}/{2} \cdot {\rm e}^{-{\rm j} \varphi_y} \cdot {\rm \delta } (f - f_0).$$
	\cdot {\rm \delta } (f + f_0) + {A_y}/{2} \cdot {\rm e}^{
	-{\rm j} \varphi_y} \cdot {\rm \delta } (f - f_0).$$

	In the exercise, filter  $\rm B$  should be given in the form:$$H_{\rm B}(f) = {\rm e}^{-a_{\rm B}(f)}\cdot {\rm e}^{-{\rm j}		In the exercise, filter  $\rm B$  should be given in the form:$$H_{\rm B}(f) = {\rm e}^{-a_{\rm B}(f)}\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} b_{\rm B}(f)}.$$
	\hspace{0.05cm} \cdot \hspace{0.05cm} b_{\rm B}(f)}.$$

	Here,		Here,
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	'''(3)'''  With  $A_x = 2 \text{ V}$  and  $\varphi_x = 90^\circ$  (sine function)  the following is obtained for  $f_0 = f_3 =3 \text{ kHz}$:		'''(3)'''  With  $A_x = 2 \text{ V}$  and  $\varphi_x = 90^\circ$  (sine function)  the following is obtained for  $f_0 = f_3 =3 \text{ kHz}$:
	:$$H_{\rm B} (f_3) = \frac{A_y}{A_x} \cdot {\rm e}^{ -{\rm j}		:$$H_{\rm B} (f_3) = \frac{A_y}{A_x} \cdot {\rm e}^{ -{\rm j}(\varphi_x - \varphi_y)} = \frac{1\hspace{0.05cm}{\rmV}}{2\hspace{0.05cm}{\rm V}} \cdot {\rm e}^{ -{\rm j} (90^{\circ} -90^{\circ})} = 0.5.$$
	(\varphi_x - \varphi_y)} = \frac{1\hspace{0.05cm}{\rm
	V}}{2\hspace{0.05cm}{\rm V}} \cdot {\rm e}^{ -{\rm j} (90^{\circ} -
	90^{\circ})} = 0.5.$$
	Thus, for  $f_0 = f_3 = 3 \text{ kHz}$  the values		Thus, for  $f_0 = f_3 = 3 \text{ kHz}$  the values
	*$a_{\rm B} (f_3)\rm \underline{\: ≈ \: 0.693 \: Np}$ and		*$a_{\rm B} (f_3)\rm \underline{\: ≈ \: 0.693 \: Np}$ and
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	'''(4)'''  Analogously, the frequency response for  $f_0 = f_2 =2 \text{ kHz}$  can be determined:		'''(4)'''  Analogously, the frequency response for  $f_0 = f_2 =2 \text{ kHz}$  can be determined:
	:$$H_{\rm B} ( f_2) = \frac{0.8\hspace{0.05cm}{\rm		:$$H_{\rm B} ( f_2) = \frac{0.8\hspace{0.05cm}{\rmV}}{2\hspace{0.05cm}{\rm V}} \cdot {\rm e}^{ -{\rm j} (90^{\circ} -70^{\circ})} = 0.4\cdot {\rm e}^{ -{\rm j} 20^{\circ}}.$$
	V}}{2\hspace{0.05cm}{\rm V}} \cdot {\rm e}^{ -{\rm j} (90^{\circ} -
	70^{\circ})} = 0.4\cdot {\rm e}^{ -{\rm j} 20^{\circ}}.$$
	Hence, for  $f_0 = f_2 = 2 \ \text{ kHz}$:		Hence, for  $f_0 = f_2 = 2 \ \text{ kHz}$:
	*$a_{\rm B}(f_2) \rm \underline{\: ≈ \: 0.916 \: Np}$,		*$a_{\rm B}(f_2) \rm \underline{\: ≈ \: 0.916 \: Np}$,

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	[[Category:Linear and Time-Invariant Systems: Exercises\|^1.1 System Description in Frequency Domain^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^1.1 System Description in Frequency Domain^]]
			[[de:Aufgaben:1.2Z_Messung_der_Übertragungsfunktion]]

Guenter at 13:30, 7 October 2021

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	[[File:~~P_ID788__LZI_Z_1_2~~.png \|right\|Measured signal amplitudes <br>and phases for filter  $\rm B$\|frame]]		[[File:EN_LZI_Z_1_2.png\|right\|Measured signal amplitudes <br>and phases for filter  $\rm B$\|frame]]
	For the metrological determination of the filter frequency response a sinusoidal input signal with an amplitude of  $2 \hspace{0.05cm} \text{V}$  and given frequency  $f_0$  is applied.  The output signal  $y(t)$  or its spectrum  $Y(f)$  are then determined according to magnitude and phase.		For the metrological determination of the filter frequency response a sinusoidal input signal with an amplitude of  $2 \hspace{0.05cm} \text{V}$  and given frequency  $f_0$  is applied.  The output signal  $y(t)$  or its spectrum  $Y(f)$  are then determined according to magnitude and phase.

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	\|type="{}"}		\|type="{}"}
	$a_{\rm B}(f_0 = \: \rm 3 \: kHz) \ = \ $ { 0.693 5% }  $\text{Np}$		$a_{\rm B}(f_0 = \: \rm 3 \: kHz) \ = \ $ { 0.693 5% }  $\text{Np}$
	$b_{\rm B}(f_0 = \: \rm 3 \: kHz) \ =\ $ { 0. }  $\text{~~Grad~~}$		$b_{\rm B}(f_0 = \: \rm 3 \: kHz) \ =\ $ { 0. }  $\text{degree}$


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	\|type="{}"}		\|type="{}"}
	$a_{\rm B}(f_0 = \: \rm 2 \: kHz) \ = \ $ { 0.916 5% }  $\text{Np}$		$a_{\rm B}(f_0 = \: \rm 2 \: kHz) \ = \ $ { 0.916 5% }  $\text{Np}$
	$b_{\rm B}(f_0 = \: \rm 2 \: kHz) \ =\ $ { 20 2% }  $\text{~~Grad~~}$		$b_{\rm B}(f_0 = \: \rm 2 \: kHz) \ =\ $ { 20 2% }  $\text{degree}$


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	Thus, for  $f_0 = f_3 = 3 \text{ kHz}$  the values		Thus, for  $f_0 = f_3 = 3 \text{ kHz}$  the values
	*$a_{\rm B} (f_3)\rm \underline{\: ≈ \: 0.693 \: Np}$ and		*$a_{\rm B} (f_3)\rm \underline{\: ≈ \: 0.693 \: Np}$ and
	*$b_{\rm B}(f_3) \rm \underline{\: = \: 0 \: (~~Grad~~)}$ are determined.		*$b_{\rm B}(f_3) \rm \underline{\: = \: 0 \: (degree)}$ are determined.

Guenter at 15:57, 11 July 2021

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	Here,		Here,
	*$a_{\rm B}(f)$  denotes the damping curve, and		*$a_{\rm B}(f_0)$  denotes the damping curve, and
	*$b_{\rm B}(f)$  the phase response.		*$b_{\rm B}(f_0)$  the phase response.


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	</quiz>		</quiz>

	===~~Sample solution~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Approaches 2 und 3</u> are correct:		'''(1)'''  <u>Approaches 2 und 3</u> are correct:
	*For an ~~LTI–system~~,   $Y(f) = X(f) · H(f)$ holds.		*For an LTI system,   $Y(f) = X(f) · H(f)$ holds.
	*Therefore, it is not possible for a component with  $3 f_0$  to be present in the output signal if such a one is missing in the input signal.		*Therefore, it is not possible for a component with  $3 f_0$  to be present in the output signal if such a one is missing in the input signal.
	*This means:   There is no ~~LTI–system~~ on hand and accordingly no frequency response can be specified.		*This means:   There is no LTI system on hand and accordingly no frequency response can be specified.


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	'''(3)'''  With  $A_x = 2 \text{ V}$  and  $\varphi_x = 90^\circ$  (sine function) the following is obtained for  $f_0 = f_3 =3 \text{ kHz}$:		'''(3)'''  With  $A_x = 2 \text{ V}$  and  $\varphi_x = 90^\circ$  (sine function)  the following is obtained for  $f_0 = f_3 =3 \text{ kHz}$:
	:$$H_{\rm B} (f_3) = \frac{A_y}{A_x} \cdot {\rm e}^{ -{\rm j}		:$$H_{\rm B} (f_3) = \frac{A_y}{A_x} \cdot {\rm e}^{ -{\rm j}
	(\varphi_x - \varphi_y)} = \frac{1\hspace{0.05cm}{\rm		(\varphi_x - \varphi_y)} = \frac{1\hspace{0.05cm}{\rm
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	70^{\circ})} = 0.4\cdot {\rm e}^{ -{\rm j} 20^{\circ}}.$$		70^{\circ})} = 0.4\cdot {\rm e}^{ -{\rm j} 20^{\circ}}.$$
	Hence, for  $f_0 = f_2 = 2 \ \text{ kHz}$:		Hence, for  $f_0 = f_2 = 2 \ \text{ kHz}$:
	*$a_{\rm B}(f_2) \rm \underline{\: ≈ \: 0.916 \: Np}$ ~~and~~		*$a_{\rm B}(f_2) \rm \underline{\: ≈ \: 0.916 \: Np}$,
	* $b_{\rm B}(f_2) \rm \underline{\: = \: 20°}$ ~~are obtained~~.		* $b_{\rm B}(f_2) \rm \underline{\: = \: 20°}$.

Guenter at 15:47, 11 July 2021

2021-07-11T15:47:12Z

Oezer at 01:07, 29 June 2021

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	V}}{2\hspace{0.05cm}{\rm V}} \cdot {\rm e}^{ -{\rm j} (90^{\circ} -		V}}{2\hspace{0.05cm}{\rm V}} \cdot {\rm e}^{ -{\rm j} (90^{\circ} -
	70^{\circ})} = 0.4\cdot {\rm e}^{ -{\rm j} 20^{\circ}}.$$		70^{\circ})} = 0.4\cdot {\rm e}^{ -{\rm j} 20^{\circ}}.$$
	~~Hene~~, for  $f_0 = f_2 = 2 \ \text{ kHz}$:		Hence, for  $f_0 = f_2 = 2 \ \text{ kHz}$:
	*$a_{\rm B}(f_2) \rm \underline{\: ≈ \: 0.916 \: Np}$ and		*$a_{\rm B}(f_2) \rm \underline{\: ≈ \: 0.916 \: Np}$ and
	* $b_{\rm B}(f_2) \rm \underline{\: = \: 20°}$ are obtained.		* $b_{\rm B}(f_2) \rm \underline{\: = \: 20°}$ are obtained.

Oezer at 01:06, 29 June 2021

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	Thus, for  $f_0 = f_3 = 3 \text{ kHz}$  the values		Thus, for  $f_0 = f_3 = 3 \text{ kHz}$  the values
	*$a_{\rm B} (f_3)\rm \underline{\: ≈ \: 0.693 \: Np}$ and		*$a_{\rm B} (f_3)\rm \underline{\: ≈ \: 0.693 \: Np}$ and
	*$b_{\rm B}(f_3) \rm \underline{\: = \: 0 \: (Grad)}$ are ~~obtained~~.		*$b_{\rm B}(f_3) \rm \underline{\: = \: 0 \: (Grad)}$ are determined.

Oezer at 01:01, 29 June 2021

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	===Sample solution===		===Sample solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Approaches 2 und 3</u> are correct.:		'''(1)'''  <u>Approaches 2 und 3</u> are correct:
	*For an LTI–system,   $Y(f) = X(f) · H(f)$ holds.		*For an LTI–system,   $Y(f) = X(f) · H(f)$ holds.
	*Therefore, it is not possible for a component with  $3 f_0$  to be present in the output signal if such a one is missing in the input signal.		*Therefore, it is not possible for a component with  $3 f_0$  to be present in the output signal if such a one is missing in the input signal.
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	'''(2)'''  <u>Approach 3</u> is correct.:		'''(2)'''  <u>Approach 3</u> is correct:
	*Based on the given numerical values for  $A_y(f_0)$  filter  $\rm B$  can be assumed to be a <u>band-pass filter</u>.		*Based on the given numerical values for  $A_y(f_0)$  filter  $\rm B$  can be assumed to be a <u>band-pass filter</u>.

Oezer at 01:00, 29 June 2021

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	{What is the damping and phase value ~~for~~ for  $f_0 = 2 \ \text{kHz}$?		{What is the damping and phase value for  $f_0 = 2 \ \text{kHz}$?
	\|type="{}"}		\|type="{}"}
	$a_{\rm B}(f_0 = \: \rm 2 \: kHz) \ = \ $ { 0.916 5% }  $\text{Np}$		$a_{\rm B}(f_0 = \: \rm 2 \: kHz) \ = \ $ { 0.916 5% }  $\text{Np}$

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	[[File:P_ID788__LZI_Z_1_2.png \|right\|Measured signal amplitudes <br>and phases for filter  $\rm B$\|frame]]		[[File:P_ID788__LZI_Z_1_2.png \|right\|Measured signal amplitudes <br>and phases for filter  $\rm B$\|frame]]
	For the metrological determination of the filter frequency response a sinusoidal input signal with an amplitude of  $2 \hspace{0.05cm} \text{V}$  and given frequency  $f_0$  is applied. The output signal  $y(t)$  or its spectrum  $Y(f)$  are then determined according to magnitude and phase.		For the metrological determination of the filter frequency response a sinusoidal input signal with an amplitude of  $2 \hspace{0.05cm} \text{V}$  and given frequency  $f_0$  is applied.  The output signal  $y(t)$  or its spectrum  $Y(f)$  are then determined according to magnitude and phase.

	*The magnitude spectrum at the output of filter  $\rm A$  with frequency  $f_0 = 1 \ \text{kHz}$ is:		*The magnitude spectrum at the output of filter  $\rm A$  with frequency  $f_0 = 1 \ \text{kHz}$  is:
	:$$\|Y_{\rm A} (f)\| = 1.6\hspace{0.05cm}{\rm V} \cdot {\rm \delta } (f		:$$\|Y_{\rm A} (f)\| = 1.6\hspace{0.05cm}{\rm V} \cdot {\rm \delta } (f
	\pm f_0) + 0.4\hspace{0.05cm}{\rm V} \cdot {\rm \delta } (f \pm 3 f_0) .$$		\pm f_0) + 0.4\hspace{0.05cm}{\rm V} \cdot {\rm \delta } (f \pm 3 f_0) .$$
	*For another filter  $\rm B$  ~~on the other hand,~~ the output signal is always a harmonic oscillation with the (single) frequency  $f_0$. For the frequencies  $f_0$  given in the table the amplitudes  $A_y(f_0)$  and the phases  $φ_y(f_0)$  are measured. Here, the following holds:		*For another filter  $\rm B$  the output signal is always a harmonic oscillation with the (single) frequency  $f_0$.  For the frequencies  $f_0$  given in the table the amplitudes  $A_y(f_0)$  and the phases  $φ_y(f_0)$  are measured.  Here, the following holds:
	:$$Y_{\rm B} (f) = {A_y}/{2} \cdot {\rm e}^{ {\rm j} \varphi_y}		:$$Y_{\rm B} (f) = {A_y}/{2} \cdot {\rm e}^{ {\rm j} \varphi_y}
	\cdot {\rm \delta } (f + f_0) + {A_y}/{2} \cdot {\rm e}^{		\cdot {\rm \delta } (f + f_0) + {A_y}/{2} \cdot {\rm e}^{
	-{\rm j} \varphi_y} \cdot {\rm \delta } (f - f_0).$$		-{\rm j} \varphi_y} \cdot {\rm \delta } (f - f_0).$$

	In the ~~task~~, filter  $\rm B$  should be given in the form:$$H_{\rm B}(f) = {\rm e}^{-a_{\rm B}(f)}\cdot {\rm e}^{-{\rm j}		In the exercise, filter  $\rm B$  should be given in the form:$$H_{\rm B}(f) = {\rm e}^{-a_{\rm B}(f)}\cdot {\rm e}^{-{\rm j}
	\hspace{0.05cm} \cdot \hspace{0.05cm} b_{\rm B}(f)}.$$		\hspace{0.05cm} \cdot \hspace{0.05cm} b_{\rm B}(f)}.$$

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	\|type="[]"}		\|type="[]"}
	- The following holds:   $\|H(f)\| = 0.8$.		- The following holds:   $\|H(f)\| = 0.8$.
	+ Filter  $\rm A$  does not represent an ~~LTI–system~~.		+ Filter  $\rm A$  does not represent an LTI system.
	+ The specification of a frequency response is not possible.		+ The specification of a frequency response is not possible.

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	{Determine the damping ~~value~~ and the phase for filter  $\rm B$  and  $f_0 = 3 \ \text{kHz}$.		{Determine the damping and the phase value for filter  $\rm B$  and  $f_0 = 3 \ \text{kHz}$.
	\|type="{}"}		\|type="{}"}
	$a_{\rm B}(f_0 = \: \rm 3 \: kHz) \ = \ $ { 0.693 5% }  $\text{Np}$		$a_{\rm B}(f_0 = \: \rm 3 \: kHz) \ = \ $ { 0.693 5% }  $\text{Np}$