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

Guenter at 13:30, 7 October 2021

2021-10-07T13:30:00Z

Guenter at 15:57, 11 July 2021

2021-07-11T15:57:35Z

Guenter at 15:47, 11 July 2021

2021-07-11T15:47:12Z

Oezer at 01:07, 29 June 2021

2021-06-29T01:07:13Z

Oezer at 01:06, 29 June 2021

2021-06-29T01:06:26Z

Oezer at 01:01, 29 June 2021

2021-06-29T01:01:23Z

Oezer at 01:00, 29 June 2021

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Oezer at 00:57, 29 June 2021

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Oezer at 00:56, 29 June 2021

2021-06-29T00:56:16Z

Oezer at 00:36, 29 June 2021

2021-06-29T00:36:58Z

@@ Line 3: / Line 3: @@
-[[File:P_ID788__LZI_Z_1_2.png |right|Measured signal amplitudes <br>and phases for filter&nbsp; $\rm B$|frame]]
+[[File:EN_LZI_Z_1_2.png|right|Measured signal amplitudes <br>and phases for filter&nbsp; $\rm B$|frame]]
 For the metrological determination of the filter frequency response a sinusoidal input signal with an amplitude of&nbsp; $2 \hspace{0.05cm} \text{V}$&nbsp; and given frequency&nbsp; $f_0$&nbsp; is applied.&nbsp; The output signal&nbsp; $y(t)$&nbsp; or its spectrum&nbsp; $Y(f)$&nbsp; are then determined according to magnitude and phase.
@@ Line 54: / Line 54: @@
 |type="{}"}
 $a_{\rm B}(f_0 = \: \rm 3 \: kHz) \ = \ $ { 0.693 5%  } &nbsp;$\text{Np}$
-$b_{\rm B}(f_0 = \: \rm 3 \: kHz) \ =\ $ { 0. } &nbsp;$\text{Grad}$
+$b_{\rm B}(f_0 = \: \rm 3 \: kHz) \ =\ $ { 0. } &nbsp;$\text{degree}$
@@ Line 61: / Line 61: @@
 |type="{}"}
 $a_{\rm B}(f_0 = \: \rm 2 \: kHz) \ = \ $ { 0.916 5%  } &nbsp;$\text{Np}$
-$b_{\rm B}(f_0 = \: \rm 2 \: kHz) \ =\ $ { 20 2%  } &nbsp;$\text{Grad}$
+$b_{\rm B}(f_0 = \: \rm 2 \: kHz) \ =\ $ { 20 2%  } &nbsp;$\text{degree}$
@@ Line 88: / Line 88: @@
 Thus, for&nbsp; $f_0 =  f_3 = 3  \text{ kHz}$&nbsp; the values
 *$a_{\rm B} (f_3)\rm \underline{\: ≈ \: 0.693 \: Np}$ and
-*$b_{\rm B}(f_3) \rm \underline{\: = \: 0 \: (Grad)}$ are determined.
+*$b_{\rm B}(f_3) \rm \underline{\: = \: 0 \: (degree)}$ are determined.

@@ Line 18: / Line 18: @@
 Here,
-*$a_{\rm B}(f)$&nbsp; denotes the damping curve, and
+*$a_{\rm B}(f_0)$&nbsp; denotes the damping curve, and
-*$b_{\rm B}(f)$&nbsp; the phase response.
+*$b_{\rm B}(f_0)$&nbsp; the phase response.
@@ Line 67: / Line 67: @@
 </quiz>
-===Sample solution===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; <u>Approaches 2 und 3</u> are correct:
-*For an LTI–system, &nbsp; $Y(f) = X(f) · H(f)$ holds.
+*For an LTI system, &nbsp; $Y(f) = X(f) · H(f)$ holds.
 *Therefore, it is not possible for a component with&nbsp; $3 f_0$&nbsp; to be present in the output signal if such a one is missing in the input signal.
-*This means: &nbsp; There is no LTI–system on hand and accordingly no frequency response can be specified.
+*This means: &nbsp; There is no LTI system on hand and accordingly no frequency response can be specified.
@@ Line 81: / Line 81: @@
-'''(3)'''&nbsp; With&nbsp; $A_x = 2  \text{ V}$&nbsp;  and&nbsp; $\varphi_x = 90^\circ$&nbsp;  (sine function) the following is obtained for&nbsp; $f_0 = f_3 =3 \text{ kHz}$:
+'''(3)'''&nbsp; With&nbsp; $A_x = 2  \text{ V}$&nbsp;  and&nbsp; $\varphi_x = 90^\circ$&nbsp;  (sine function)&nbsp; the following is obtained for&nbsp; $f_0 = f_3 =3 \text{ kHz}$:
 :$$H_{\rm B} (f_3) = \frac{A_y}{A_x} \cdot {\rm e}^{ -{\rm j}
 (\varphi_x - \varphi_y)} =  \frac{1\hspace{0.05cm}{\rm
@@ Line 96: / Line 96: @@
 ^{\circ})} = 0.4\cdot {\rm e}^{ -{\rm j} 20^{\circ}}.$$
 Hence, for&nbsp; $f_0 =  f_2 = 2 \ \text{ kHz}$:
-*$a_{\rm B}(f_2) \rm \underline{\: ≈ \: 0.916 \: Np}$ and
+*$a_{\rm B}(f_2) \rm \underline{\: ≈ \: 0.916 \: Np}$,
-* $b_{\rm B}(f_2) \rm \underline{\: = \: 20°}$ are obtained.
+* $b_{\rm B}(f_2) \rm \underline{\: = \: 20°}$.

@@ Line 4: / Line 4: @@
 [[File:P_ID788__LZI_Z_1_2.png |right|Measured signal amplitudes <br>and phases for filter&nbsp; $\rm B$|frame]]
-For the metrological determination of the filter frequency response a sinusoidal input signal with an amplitude of&nbsp; $2 \hspace{0.05cm} \text{V}$&nbsp; and given frequency&nbsp; $f_0$&nbsp; is applied. The output signal&nbsp; $y(t)$&nbsp; or its spectrum&nbsp; $Y(f)$&nbsp; 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&nbsp; $2 \hspace{0.05cm} \text{V}$&nbsp; and given frequency&nbsp; $f_0$&nbsp; is applied.&nbsp; The output signal&nbsp; $y(t)$&nbsp; or its spectrum&nbsp; $Y(f)$&nbsp; are then determined according to magnitude and phase.
-*The magnitude spectrum at the output of filter&nbsp; $\rm A$&nbsp; with frequency&nbsp; $f_0 = 1 \ \text{kHz}$ is:
+*The magnitude spectrum at the output of filter&nbsp; $\rm A$&nbsp; with frequency&nbsp; $f_0 = 1 \ \text{kHz}$&nbsp; is:
 :$$|Y_{\rm A} (f)| = 1.6\hspace{0.05cm}{\rm V} \cdot {\rm \delta } (f
 \pm f_0) + 0.4\hspace{0.05cm}{\rm V} \cdot {\rm \delta }  (f \pm 3 f_0) .$$
-*For another filter&nbsp; $\rm B$&nbsp; on the other hand, the output signal is always a harmonic oscillation with the (single) frequency&nbsp; $f_0$. For the frequencies&nbsp; $f_0$&nbsp; given in the table the amplitudes&nbsp; $A_y(f_0)$&nbsp; and the phases&nbsp; $φ_y(f_0)$&nbsp; are measured. Here, the following holds:
+*For another filter&nbsp; $\rm B$&nbsp; the output signal is always a harmonic oscillation with the (single) frequency&nbsp; $f_0$.&nbsp; For the frequencies&nbsp; $f_0$&nbsp; given in the table the amplitudes&nbsp; $A_y(f_0)$&nbsp; and the phases&nbsp; $φ_y(f_0)$&nbsp; are measured.&nbsp; Here, the following holds:
 :$$Y_{\rm B} (f) = {A_y}/{2} \cdot {\rm e}^{ {\rm j} \varphi_y}
 \cdot {\rm \delta } (f + f_0) +  {A_y}/{2} \cdot {\rm e}^{
 -{\rm j} \varphi_y} \cdot {\rm \delta } (f - f_0).$$
-In the task, filter&nbsp; $\rm B$ &nbsp;should be given in the form:$$H_{\rm B}(f) =  {\rm e}^{-a_{\rm B}(f)}\cdot {\rm e}^{-{\rm j}
+In the exercise, filter&nbsp; $\rm B$ &nbsp;should be given in the form:$$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)}.$$
@@ Line 37: / Line 37: @@
 |type="[]"}
 - The following holds: &nbsp; $|H(f)| = 0.8$.
-+ Filter&nbsp; $\rm A$&nbsp; does not represent an LTI–system.
++ Filter&nbsp; $\rm A$&nbsp; does not represent an LTI system.
 + The specification of a frequency response is not possible.
@@ Line 51: / Line 51: @@
-{Determine the damping value and the phase for filter&nbsp; $\rm B$&nbsp; and&nbsp; $f_0 = 3 \ \text{kHz}$.
+{Determine the damping and the phase value for filter&nbsp; $\rm B$&nbsp; and&nbsp; $f_0 = 3 \ \text{kHz}$.
 |type="{}"}
 $a_{\rm B}(f_0 = \: \rm 3 \: kHz) \ = \ $ { 0.693 5%  } &nbsp;$\text{Np}$

@@ Line 95: / Line 95: @@
 V}}{2\hspace{0.05cm}{\rm V}} \cdot {\rm e}^{ -{\rm j} (90^{\circ} -
 ^{\circ})} = 0.4\cdot {\rm e}^{ -{\rm j} 20^{\circ}}.$$
-Hene, for&nbsp; $f_0 =  f_2 = 2 \ \text{ kHz}$:
+Hence, for&nbsp; $f_0 =  f_2 = 2 \ \text{ kHz}$:
 *$a_{\rm B}(f_2) \rm \underline{\: ≈ \: 0.916 \: Np}$ and
 * $b_{\rm B}(f_2) \rm \underline{\: = \: 20°}$ are obtained.

@@ Line 88: / Line 88: @@
 Thus, for&nbsp; $f_0 =  f_3 = 3  \text{ kHz}$&nbsp; the values
 *$a_{\rm B} (f_3)\rm \underline{\: ≈ \: 0.693 \: Np}$ and
-*$b_{\rm B}(f_3) \rm \underline{\: = \: 0 \: (Grad)}$ are obtained.
+*$b_{\rm B}(f_3) \rm \underline{\: = \: 0 \: (Grad)}$ are determined.

@@ Line 69: / Line 69: @@
 ===Sample solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; <u>Approaches 2 und 3</u> are correct.:
+'''(1)'''&nbsp; <u>Approaches 2 und 3</u> are correct:
 *For an LTI–system, &nbsp; $Y(f) = X(f) · H(f)$ holds.
 *Therefore, it is not possible for a component with&nbsp; $3 f_0$&nbsp; to be present in the output signal if such a one is missing in the input signal.
@@ Line 76: / Line 76: @@
-'''(2)'''&nbsp; <u>Approach 3</u> is correct.:
+'''(2)'''&nbsp; <u>Approach 3</u> is correct:
 *Based on the given numerical values for&nbsp; $A_y(f_0)$&nbsp; filter&nbsp; $\rm B$&nbsp; can be assumed to be a <u>band-pass filter</u>.

@@ Line 58: / Line 58: @@
-{What is the damping and phase value for for&nbsp; $f_0 =  2 \ \text{kHz}$?
+{What is the damping and phase value for&nbsp; $f_0 =  2 \ \text{kHz}$?
 |type="{}"}
 $a_{\rm B}(f_0 = \: \rm 2 \: kHz) \ = \ $ { 0.916 5%  } &nbsp;$\text{Np}$