Difference between revisions of "Aufgaben:Exercise 1.2Z: Measurement of the Frequency Response"

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[[File:P_ID788__LZI_Z_1_2.png |right|Measured signal amplitudes <br>and phases for filter&nbsp; $\rm B$|frame]]
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[[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. 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
 
:$$|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&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}
 
:$$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&nbsp; $\rm B$ &nbsp;should be given in the form:$$H_{\rm B}(f) =  {\rm e}^{-a_{\rm B}(f)}\cdot {\rm e}^{-{\rm j}
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In the exercise, filter&nbsp; $\rm B$ &nbsp;should be given in the form:$$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)}.$$
 
\hspace{0.05cm} \cdot \hspace{0.05cm} b_{\rm B}(f)}.$$
  
 
Here,  
 
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.  
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*$b_{\rm B}(f_0)$&nbsp; the phase response.  
  
  
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|type="[]"}
 
|type="[]"}
 
- The following holds: &nbsp; $|H(f)| = 0.8$.
 
- 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.  
 
+ The specification of a frequency response is not possible.  
  
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{Determine the damping value and the phase for filter&nbsp; $\rm B$&nbsp; and&nbsp; $f_0 = 3 \ \text{kHz}$.  
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{Determine the damping and the phase value for filter&nbsp; $\rm B$&nbsp; and&nbsp; $f_0 = 3 \ \text{kHz}$.  
 
|type="{}"}
 
|type="{}"}
 
$a_{\rm B}(f_0 = \: \rm 3 \: kHz) \ = \ $ { 0.693 5%  } &nbsp;$\text{Np}$
 
$a_{\rm B}(f_0 = \: \rm 3 \: kHz) \ = \ $ { 0.693 5%  } &nbsp;$\text{Np}$
$b_{\rm B}(f_0 = \: \rm 3 \: kHz) \ =\ $ { 0. } &nbsp;$\text{Grad}$
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$b_{\rm B}(f_0 = \: \rm 3 \: kHz) \ =\ $ { 0. } &nbsp;$\text{degree}$
  
  
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|type="{}"}
 
|type="{}"}
 
$a_{\rm B}(f_0 = \: \rm 2 \: kHz) \ = \ $ { 0.916 5%  } &nbsp;$\text{Np}$
 
$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;$\text{Grad}$
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$b_{\rm B}(f_0 = \: \rm 2 \: kHz) \ =\ $ { 20 2%  } &nbsp;$\text{degree}$
  
  
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</quiz>
 
</quiz>
  
===Sample solution===
+
===Solution===
 
{{ML-Kopf}}
 
{{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.  
+
*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.  
 
*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.  
  
  
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'''(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}
 
:$$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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Thus, for&nbsp; $f_0 =  f_3 = 3  \text{ kHz}$&nbsp; the values  
 
Thus, for&nbsp; $f_0 =  f_3 = 3  \text{ kHz}$&nbsp; 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.  
  
  
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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&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
+
*$a_{\rm B}(f_2) \rm \underline{\: ≈ \: 0.916 \: Np}$,
* $b_{\rm B}(f_2) \rm \underline{\: = \: 20°}$ are obtained.  
+
* $b_{\rm B}(f_2) \rm \underline{\: = \: 20°}$.  
  
  

Latest revision as of 14:30, 7 October 2021


Measured signal amplitudes
and phases for filter  $\rm B$

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:
$$|Y_{\rm A} (f)| = 1.6\hspace{0.05cm}{\rm V} \cdot {\rm \delta } (f \pm f_0) + 0.4\hspace{0.05cm}{\rm V} \cdot {\rm \delta } (f \pm 3 f_0) .$$
  • For another filter  $\rm B$  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:
$$Y_{\rm B} (f) = {A_y}/{2} \cdot {\rm e}^{ {\rm j} \varphi_y} \cdot {\rm \delta } (f + f_0) + {A_y}/{2} \cdot {\rm e}^{ -{\rm j} \varphi_y} \cdot {\rm \delta } (f - f_0).$$

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

Here,

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




Please note:


Questions

1

Which of the statements are true regarding filter  $\rm A$ ?

The following holds:   $|H(f)| = 0.8$.
Filter  $\rm A$  does not represent an LTI system.
The specification of a frequency response is not possible.

2

Which of the statements are true regarding filter  $\rm B$ ?

Filter  $\rm B$  is a low-pass filter.
Filter  $\rm B$  is a high-pass filter.
Filter  $\rm B$  is a band-pass filter.
Filter  $\rm B$  is a band-stop filter.

3

Determine the damping and the phase value for filter  $\rm B$  and  $f_0 = 3 \ \text{kHz}$.

$a_{\rm B}(f_0 = \: \rm 3 \: kHz) \ = \ $

 $\text{Np}$
$b_{\rm B}(f_0 = \: \rm 3 \: kHz) \ =\ $

 $\text{degree}$

4

What is the damping and phase value for  $f_0 = 2 \ \text{kHz}$?

$a_{\rm B}(f_0 = \: \rm 2 \: kHz) \ = \ $

 $\text{Np}$
$b_{\rm B}(f_0 = \: \rm 2 \: kHz) \ =\ $

 $\text{degree}$


Solution

(1)  Approaches 2 und 3 are correct:

  • 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.
  • This means:   There is no LTI system on hand and accordingly no frequency response can be specified.


(2)  Approach 3 is correct:

  • Based on the given numerical values for  $A_y(f_0)$  filter  $\rm B$  can be assumed to be a band-pass filter.


(3)  With  $A_x = 2 \text{ V}$  and  $\varphi_x = 90^\circ$  (sine function)  the following is obtained for  $f_0 = f_3 =3 \text{ kHz}$:

$$H_{\rm B} (f_3) = \frac{A_y}{A_x} \cdot {\rm e}^{ -{\rm j} (\varphi_x - \varphi_y)} = \frac{1\hspace{0.05cm}{\rm V}}{2\hspace{0.05cm}{\rm V}} \cdot {\rm e}^{ -{\rm j} (90^{\circ} - 90^{\circ})} = 0.5.$$

Thus, for  $f_0 = f_3 = 3 \text{ kHz}$  the values

  • $a_{\rm B} (f_3)\rm \underline{\: ≈ \: 0.693 \: Np}$ and
  • $b_{\rm B}(f_3) \rm \underline{\: = \: 0 \: (degree)}$ are determined.


(4)  Analogously, the frequency response for  $f_0 = f_2 =2 \text{ kHz}$  can be determined:

$$H_{\rm B} ( f_2) = \frac{0.8\hspace{0.05cm}{\rm V}}{2\hspace{0.05cm}{\rm V}} \cdot {\rm e}^{ -{\rm j} (90^{\circ} - 70^{\circ})} = 0.4\cdot {\rm e}^{ -{\rm j} 20^{\circ}}.$$

Hence, for  $f_0 = f_2 = 2 \ \text{ kHz}$:

  • $a_{\rm B}(f_2) \rm \underline{\: ≈ \: 0.916 \: Np}$,
  • $b_{\rm B}(f_2) \rm \underline{\: = \: 20°}$.


For  $f_0 = -f_2 =-\hspace{-0.01cm}2 \text{ kHz}$  the same damping value applies. However, the phase has the opposite sign. So,   $b_{\rm B}(–f_2) = \ –\hspace{-0.01cm}20^{\circ}.$