Difference between revisions of "Aufgaben:Exercise 1.1: Simple Filter Functions"

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{{quiz-Header|Buchseite=Lineare zeitinvariante Systeme/Systembeschreibung im Frequenzbereich}}
+
{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain}}
  
[[File:P_ID781__LZI_A_1_1.png|Considered quadripoles|right|frame]]
+
[[File:EN_LZI_A_1_1.png|Considered two-port networks|right|frame]]
 
A filter with the frequency response
 
A filter with the frequency response
:$$H_{\rm TP}(f) = \frac{1}{1+ {\rm j}\cdot f/f_0}$$
+
:$$H_{\rm LP}(f) = \frac{1}{1+ {\rm j}\cdot f/f_0}$$
is described as a low-pass filter of first order. Out of it, a first order high-pass filter can be designed according to the following rule:
+
is described as a low-pass filter of first order.  Out of it, a first order high-pass filter can be designed according to the following rule:
:$$H_{\rm HP}(f) = 1- H_{\rm TP}(f) .$$
+
:$$H_{\rm HP}(f) = 1- H_{\rm LP}(f) .$$
  
 
In both cases  $f_0$  represents the so-called  $\text{3 dB}$–cutoff frequency.
 
In both cases  $f_0$  represents the so-called  $\text{3 dB}$–cutoff frequency.
  
The figure shows two quadripoles  $\rm A$  and  $\rm B$. The task is to clarify which of the two quadripoles has a low-pass characteristic and which has a high-pass characteristic.
+
The figure shows two two-port networks  $\rm A$  and  $\rm B$.  The task is to clarify which of the two networks has a low-pass characteristic and which has a high-pass characteristic.
  
 
The components of circuit  $\rm A$  are given as follows:
 
The components of circuit  $\rm A$  are given as follows:
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''Please note:''  
 
''Please note:''  
*The task belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Systembeschreibung_im_Frequenzbereich | Systembeschreibung im Frequenzbereich]].
+
*The task belongs to the chapter  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain |System Description in Frequency Domain]].
 
*For the subtask  '''(4)'''  cosine-shaped input signals are assumed. The frequency  $f_x$  is variable, the power is   $P_x = 10\,{\rm mW}.$
 
*For the subtask  '''(4)'''  cosine-shaped input signals are assumed. The frequency  $f_x$  is variable, the power is   $P_x = 10\,{\rm mW}.$
 
   
 
   
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<quiz display=simple>
 
<quiz display=simple>
{Compute the frequency response&nbsp; $H_{\rm A}(f)$&nbsp; of the quadripole&nbsp; $\rm A$&nbsp; and check the following statements.
+
{Compute the frequency response&nbsp; $H_{\rm A}(f)$&nbsp; of the two-port network&nbsp; $\rm A$&nbsp; and check the following statements.
 
|type="()"}
 
|type="()"}
+ Quadripole&nbsp; $\rm A$&nbsp; is a low-pass filter.
+
+ Two-port network&nbsp; $\rm A$&nbsp; is a low-pass filter.
- Quadripole&nbsp; $\rm A$&nbsp; is a high-pass filter.
+
- Two-port network&nbsp; $\rm A$&nbsp; is a high-pass filter.
  
  
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{Compute the amplitude response&nbsp; $|H_{\rm B}(f)|$&nbsp; of the quadripole&nbsp; $\rm B$&nbsp;  with the components&nbsp; $R$&nbsp; and&nbsp; $L$&nbsp; using the reference frequency&nbsp; $f_0 = R/(2πL)$. <br>What are the values for&nbsp; $f = 0$,&nbsp; $f = f_0$, $f = 2f_0$&nbsp; and for&nbsp; $f → ∞$?
+
{Compute the amplitude response&nbsp; $|H_{\rm B}(f)|$&nbsp; of the two-port network&nbsp; $\rm B$&nbsp;  with the components&nbsp; $R$&nbsp; and&nbsp; $L$&nbsp; using the reference frequency&nbsp; $f_0 = R/(2πL)$. <br>What are the values for&nbsp; $f = 0$,&nbsp; $f = f_0$, $f = 2f_0$&nbsp; and for&nbsp; $f → ∞$?
 
|type="{}"}
 
|type="{}"}
 
$|H_{\rm B}(f = 0)|\ = \ $ { 0 1% }
 
$|H_{\rm B}(f = 0)|\ = \ $ { 0 1% }
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*Because of&nbsp; $H_{\rm A}(f = 0) = 1$&nbsp; this cannot be a high-pass filter; rather it is a low-pass filter.  
 
*Because of&nbsp; $H_{\rm A}(f = 0) = 1$&nbsp; this cannot be a high-pass filter; rather it is a low-pass filter.  
 
*At low frequencies, the reactance of the capacitance is very large and&nbsp; $y_{\rm A}(t) ≈ x_{\rm A}(t)$ applies.  
 
*At low frequencies, the reactance of the capacitance is very large and&nbsp; $y_{\rm A}(t) ≈ x_{\rm A}(t)$ applies.  
*In contrast to this, at very high frequencies, the capacitor acts like a short circuit and &nbsp; $y_{\rm A}(t) ≈ 0$ holds.
+
*In contrast to this, at very high frequencies, the capacitor acts like a short circuit and &nbsp; $y_{\rm A}(t) ≈ 0$&nbsp; holds.
  
  
  
  
'''(2)'''&nbsp; By comparing coefficients between&nbsp; $H_{\rm TP}(f)$&nbsp; given in the statement of the task and&nbsp; $H_{\rm A}(f)$&nbsp; according to subtask&nbsp; '''(1)'''&nbsp; the following is obtained:
+
'''(2)'''&nbsp; By comparing the coefficients between&nbsp; $H_{\rm LP}(f)$&nbsp; given in the statement of the task and&nbsp; $H_{\rm A}(f)$&nbsp; according to subtask&nbsp; '''(1)'''&nbsp; the following is obtained:
 
:$$f_0 = \frac{1}{2\pi \cdot R \cdot C} = \frac{1}{2\pi \cdot{\rm
 
:$$f_0 = \frac{1}{2\pi \cdot R \cdot C} = \frac{1}{2\pi \cdot{\rm
 
50\hspace{0.05cm} \Omega}\cdot {\rm 637 \cdot 10^{-9}\hspace{0.05cm} s/\Omega}}\hspace{0.15cm}\underline{\approx 5 \, {\rm kHz}}.$$
 
50\hspace{0.05cm} \Omega}\cdot {\rm 637 \cdot 10^{-9}\hspace{0.05cm} s/\Omega}}\hspace{0.15cm}\underline{\approx 5 \, {\rm kHz}}.$$
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'''(3)'''&nbsp; The amplitude response is:
 
'''(3)'''&nbsp; The amplitude response is:
 
:$$|H_{\rm A}(f)| = \frac{1}{\sqrt{1+ (f/f_0)^2}}.$$
 
:$$|H_{\rm A}(f)| = \frac{1}{\sqrt{1+ (f/f_0)^2}}.$$
*For&nbsp; $f = f_0$&nbsp; the numerical value&nbsp; $1/\sqrt{2}\hspace{0.1cm} \underline{≈ 0.707}$ is obtained, and   
+
*For&nbsp; $f = f_0$&nbsp; the numerical value&nbsp; $1/\sqrt{2}\hspace{0.1cm} \underline{≈ 0.707}$&nbsp; is obtained, and   
*for&nbsp; $f = 2f_0$&nbsp; approximately the value&nbsp; $1/\sqrt{5}\hspace{0.1cm} \underline{≈ 0.477}$.
+
*for&nbsp; $f = 2f_0$&nbsp; approximately the value&nbsp; $1/\sqrt{5}\hspace{0.1cm} \underline{≈ 0.447}$.
  
  
  
'''(4)'''&nbsp; Die Ausgangsleistung kann nach folgender Gleichung berechnet werden:
+
'''(4)'''&nbsp; The output power can be calculated using the following equation:
 
:$$P_y = P_x \cdot |H_{\rm A}(f = f_x)|^2.$$
 
:$$P_y = P_x \cdot |H_{\rm A}(f = f_x)|^2.$$
*Für&nbsp; $f_x = f_0$&nbsp; ist&nbsp; $P_y = P_x/2 \hspace{0.1cm} \underline{ = 5\hspace{0.1cm} {\rm mW}}$, also ergibt sich am Ausgang nur noch die halbe Leistung.  
+
*For&nbsp; $f_x = f_0$&nbsp; &rArr; &nbsp; $P_y = P_x/2 \hspace{0.1cm} \underline{ = 5\hspace{0.1cm} {\rm mW}}$, so this results in half of the power at the output.  
  
*In logarithmischer Darstellung lautet diese Beziehung:
+
*In logarithmic representation, this relationship is:
 
:$$10 \cdot {\rm lg}\hspace{0.2cm} \frac{P_x(f_0)}{P_y(f_0)} = 3\,{\rm dB}.$$
 
:$$10 \cdot {\rm lg}\hspace{0.2cm} \frac{P_x(f_0)}{P_y(f_0)} = 3\,{\rm dB}.$$
:Deshalb ist für&nbsp; $f_0$&nbsp; auch die Bezeichnung "3dB–Grenzfrequenz" üblich.  
+
:Accordingly, for&nbsp; $f_0$&nbsp; the term&nbsp; "3dB cut-off frequency"&nbsp; is also common.  
  
*Für&nbsp; $f_x = 2f_0$&nbsp; erhält man dagegen einen kleineren Wert:&nbsp; $P_y = P_x/5  \hspace{0.1cm}\underline{=  2\hspace{0.1cm} {\rm mW}}$.  
+
*For&nbsp; $f_x = 2f_0$,&nbsp; on the other hand, a smaller value is obtained:&nbsp; $P_y = P_x/5  \hspace{0.1cm}\underline{=  2\hspace{0.1cm} {\rm mW}}$.  
  
  
  
'''(5)'''&nbsp; Analog zur Teilaufgabe&nbsp; '''(1)'''&nbsp; gilt:  
+
'''(5)'''&nbsp; Analogous to subtask&nbsp; '''(1)'''&nbsp; the following holds:  
 
:$$H_{\rm B}(f) = \frac{Y_{\rm B}(f)}{X_{\rm B}(f)} = \frac{{\rm j}\omega L}{R+{\rm j}\omega L}=\frac{{\rm j2\pi}\cdot f \cdot L/R}{1+{\rm j2\pi}\cdot f \cdot L/R}.$$
 
:$$H_{\rm B}(f) = \frac{Y_{\rm B}(f)}{X_{\rm B}(f)} = \frac{{\rm j}\omega L}{R+{\rm j}\omega L}=\frac{{\rm j2\pi}\cdot f \cdot L/R}{1+{\rm j2\pi}\cdot f \cdot L/R}.$$
*Unter Verwendung der Bezugsfrequenz&nbsp; $f_0 = R/(2πL)$&nbsp; kann hierfür auch geschrieben werden:  
+
*Using the reference frequency&nbsp; $f_0 = R/(2πL)$&nbsp; this can also be written as in the following:  
 
:$$H_{\rm B}(f) = \frac{{\rm j}\cdot f/f_0}{1+{\rm j}\cdot f/f_0}\hspace{0.5cm}\Rightarrow \hspace{0.5cm}|H_{\rm B}(f)| = \frac{|f/f_0|}{\sqrt{1+ (f/f_0)^2}}.$$
 
:$$H_{\rm B}(f) = \frac{{\rm j}\cdot f/f_0}{1+{\rm j}\cdot f/f_0}\hspace{0.5cm}\Rightarrow \hspace{0.5cm}|H_{\rm B}(f)| = \frac{|f/f_0|}{\sqrt{1+ (f/f_0)^2}}.$$
*Daraus erhält man die Zahlenwerte:  
+
*Hence, these numerical values are obtained:  
 
:$$|H_{\rm B}(f = 0)| \hspace{0.15cm}\underline{= 0}, \hspace{0.5cm} |H_{\rm B}( f_0)| \hspace{0.15cm}\underline{=0.707}, \hspace{0.5cm}|H_{\rm B}(2f_0)| \hspace{0.15cm}\underline{= 0.894},
 
:$$|H_{\rm B}(f = 0)| \hspace{0.15cm}\underline{= 0}, \hspace{0.5cm} |H_{\rm B}( f_0)| \hspace{0.15cm}\underline{=0.707}, \hspace{0.5cm}|H_{\rm B}(2f_0)| \hspace{0.15cm}\underline{= 0.894},
 
\hspace{0.5cm}|H_{\rm B}(f \rightarrow \infty)|\hspace{0.15cm}\underline{ = 1}.$$
 
\hspace{0.5cm}|H_{\rm B}(f \rightarrow \infty)|\hspace{0.15cm}\underline{ = 1}.$$
*Der Vierpol $\rm B$ ist demzufolge ein Hochpass.  
+
*Therefore, the circuit&nbsp; $\rm B$&nbsp; is a high-pass filter.  
  
  
  
'''(6)'''&nbsp; Aus obiger Definition der Bezugsfrequenz folgt:
+
'''(6)'''&nbsp; According to the above definition of the reference frequency it follows that:
 
:$$L = \frac{R}{2\pi \cdot f_0} = \frac{{\rm 50\hspace{0.05cm}
 
:$$L = \frac{R}{2\pi \cdot f_0} = \frac{{\rm 50\hspace{0.05cm}
 
\Omega}}{2\pi \cdot{\rm 5000 \hspace{0.05cm} Hz}}= {\rm 1.59 \cdot
 
\Omega}}{2\pi \cdot{\rm 5000 \hspace{0.05cm} Hz}}= {\rm 1.59 \cdot

Latest revision as of 15:11, 9 July 2021

Considered two-port networks

A filter with the frequency response

$$H_{\rm LP}(f) = \frac{1}{1+ {\rm j}\cdot f/f_0}$$

is described as a low-pass filter of first order.  Out of it, a first order high-pass filter can be designed according to the following rule:

$$H_{\rm HP}(f) = 1- H_{\rm LP}(f) .$$

In both cases  $f_0$  represents the so-called  $\text{3 dB}$–cutoff frequency.

The figure shows two two-port networks  $\rm A$  and  $\rm B$.  The task is to clarify which of the two networks has a low-pass characteristic and which has a high-pass characteristic.

The components of circuit  $\rm A$  are given as follows:

$$R = 50 \,\, {\rm \Omega}, \hspace{0.2cm} C = 637 \,\, {\rm nF} .$$

The inductivity  $L$  of circuit  $\rm B$  is to be computed in subtask  (6) .




Please note:

  • The task belongs to the chapter  System Description in Frequency Domain.
  • For the subtask  (4)  cosine-shaped input signals are assumed. The frequency  $f_x$  is variable, the power is  $P_x = 10\,{\rm mW}.$


Questions

1

Compute the frequency response  $H_{\rm A}(f)$  of the two-port network  $\rm A$  and check the following statements.

Two-port network  $\rm A$  is a low-pass filter.
Two-port network  $\rm A$  is a high-pass filter.

2

Compute the reference frequency  $f_0$  using the components  $R$  and  $C$.

$f_0 \ = \ $

 $\text{kHz}$

3

Compute the amplitude response  $|H_{\rm A}(f)|$. What are the numerical values for  $f = f_0$  and  $f = 2f_0$?

$|H_{\rm A}(f = f_0)|\ = \ $

$|H_{\rm A}(f = 2f_0)|\ = \ $

4

What is the power  $P_y$  of the output signal  $y(t)$ if a cosine signal of frequency  $f_x = 5\,{\rm kHz}$  or  $f_x = 10\,{\rm kHz}$  is applied to the input?

$P_y(f_x = 5 \ \rm kHz)\ = \ $

 $\text{mW}$
$P_y(f_x = 10 \ \rm kHz)\ = \ $

 $\text{mW}$

5

Compute the amplitude response  $|H_{\rm B}(f)|$  of the two-port network  $\rm B$  with the components  $R$  and  $L$  using the reference frequency  $f_0 = R/(2πL)$.
What are the values for  $f = 0$,  $f = f_0$, $f = 2f_0$  and for  $f → ∞$?

$|H_{\rm B}(f = 0)|\ = \ $

$|H_{\rm B}(f = f_0)|\ = \ $

$|H_{\rm B}(f = 2f_0)|\ = \ $

$|H_{\rm B}(f → ∞)|\ = \ $

6

What inductivity  $L$  results in the reference frequency  $f_0 = 5 \,\text{kHz}$?

$L\ = \ $

 $\text{mH}$


Sample solution

(1)  Approach 1 is correct:

  • The complex resistance of the capacitance  $C$  is equal to  $1/({\rm j}ωC)$, where  $ω = 2πf$  represents the so-called angular frequency.
  • The frequency response can be calculated according to the voltage divider principle:
$$H_{\rm A}(f) = \frac{Y_{\rm A}(f)}{X_{\rm A}(f)} = \frac{1/({\rm j}\omega C)}{R+1/({\rm j}\omega C)}=\frac{1}{1+{\rm j \cdot 2\pi}\cdot f \cdot R\cdot C}.$$
  • Because of  $H_{\rm A}(f = 0) = 1$  this cannot be a high-pass filter; rather it is a low-pass filter.
  • At low frequencies, the reactance of the capacitance is very large and  $y_{\rm A}(t) ≈ x_{\rm A}(t)$ applies.
  • In contrast to this, at very high frequencies, the capacitor acts like a short circuit and   $y_{\rm A}(t) ≈ 0$  holds.



(2)  By comparing the coefficients between  $H_{\rm LP}(f)$  given in the statement of the task and  $H_{\rm A}(f)$  according to subtask  (1)  the following is obtained:

$$f_0 = \frac{1}{2\pi \cdot R \cdot C} = \frac{1}{2\pi \cdot{\rm 50\hspace{0.05cm} \Omega}\cdot {\rm 637 \cdot 10^{-9}\hspace{0.05cm} s/\Omega}}\hspace{0.15cm}\underline{\approx 5 \, {\rm kHz}}.$$


(3)  The amplitude response is:

$$|H_{\rm A}(f)| = \frac{1}{\sqrt{1+ (f/f_0)^2}}.$$
  • For  $f = f_0$  the numerical value  $1/\sqrt{2}\hspace{0.1cm} \underline{≈ 0.707}$  is obtained, and
  • for  $f = 2f_0$  approximately the value  $1/\sqrt{5}\hspace{0.1cm} \underline{≈ 0.447}$.


(4)  The output power can be calculated using the following equation:

$$P_y = P_x \cdot |H_{\rm A}(f = f_x)|^2.$$
  • For  $f_x = f_0$  ⇒   $P_y = P_x/2 \hspace{0.1cm} \underline{ = 5\hspace{0.1cm} {\rm mW}}$, so this results in half of the power at the output.
  • In logarithmic representation, this relationship is:
$$10 \cdot {\rm lg}\hspace{0.2cm} \frac{P_x(f_0)}{P_y(f_0)} = 3\,{\rm dB}.$$
Accordingly, for  $f_0$  the term  "3dB cut-off frequency"  is also common.
  • For  $f_x = 2f_0$,  on the other hand, a smaller value is obtained:  $P_y = P_x/5 \hspace{0.1cm}\underline{= 2\hspace{0.1cm} {\rm mW}}$.


(5)  Analogous to subtask  (1)  the following holds:

$$H_{\rm B}(f) = \frac{Y_{\rm B}(f)}{X_{\rm B}(f)} = \frac{{\rm j}\omega L}{R+{\rm j}\omega L}=\frac{{\rm j2\pi}\cdot f \cdot L/R}{1+{\rm j2\pi}\cdot f \cdot L/R}.$$
  • Using the reference frequency  $f_0 = R/(2πL)$  this can also be written as in the following:
$$H_{\rm B}(f) = \frac{{\rm j}\cdot f/f_0}{1+{\rm j}\cdot f/f_0}\hspace{0.5cm}\Rightarrow \hspace{0.5cm}|H_{\rm B}(f)| = \frac{|f/f_0|}{\sqrt{1+ (f/f_0)^2}}.$$
  • Hence, these numerical values are obtained:
$$|H_{\rm B}(f = 0)| \hspace{0.15cm}\underline{= 0}, \hspace{0.5cm} |H_{\rm B}( f_0)| \hspace{0.15cm}\underline{=0.707}, \hspace{0.5cm}|H_{\rm B}(2f_0)| \hspace{0.15cm}\underline{= 0.894}, \hspace{0.5cm}|H_{\rm B}(f \rightarrow \infty)|\hspace{0.15cm}\underline{ = 1}.$$
  • Therefore, the circuit  $\rm B$  is a high-pass filter.


(6)  According to the above definition of the reference frequency it follows that:

$$L = \frac{R}{2\pi \cdot f_0} = \frac{{\rm 50\hspace{0.05cm} \Omega}}{2\pi \cdot{\rm 5000 \hspace{0.05cm} Hz}}= {\rm 1.59 \cdot 10^{-3}\hspace{0.05cm} \Omega s}\hspace{0.15cm}\underline{= {\rm 1.59 \hspace{0.1cm} mH}} .$$