Difference between revisions of "Aufgaben:Exercise 3.4: Attenuation and Phase Response"

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{{quiz-Header|Buchseite=Lineare zeitinvariante Systeme/Laplace–Transformation und p–Übertragungsfunktion
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{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function
 
}}
 
}}
  
[[File:P_ID1768__LZI_A_3_4.png|right|Pol–Nullstellen–Diagramm und einige Hilfsgrößen]]
+
[[File:P_ID1768__LZI_A_3_4.png|right|frame|Pole-zero diagram and definition of some auxiliary quantities]]
Wir gehen vom skizzierten Pol–Nullstellen–Diagramm aus, also von den Werten
+
We assume the sketched pole–zero diagram, i.e. the values
$$K = 5, \hspace{0,2cm}Z = 1, \hspace{0,2cm}N = 2, $$
+
:$$K = 5, \hspace{0,2cm}Z = 1, \hspace{0,2cm}N = 2, $$
$$ p_{\rm  o}= 1,\hspace{0,2cm}p_{\rm  x1}= -3 + 3{\rm j},\hspace{0,2cm}p_{\rm  x2}= -3 - 3{\rm j}\hspace{0.05cm} .$$
+
:$$ p_{\rm  o}= 1,\hspace{0,2cm}p_{\rm  x1}= -3 + 3{\rm j},\hspace{0,2cm}p_{\rm  x2}= -3 - 3{\rm j}\hspace{0.05cm} .$$
  
Damit lautet die $p$–Übertragungsfunktion:
+
Thus,  the  $p$–transfer function is:
$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }}
+
:$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }}
 
  {(p - p_{\rm x 1})(p - p_{\rm x 2})}
 
  {(p - p_{\rm x 1})(p - p_{\rm x 2})}
 
  \hspace{0.05cm} .$$
 
  \hspace{0.05cm} .$$
  
Mit der Substitution $p = {\rm j} \cdot 2 \pi f$ lässt sich die herkömmliche Übertragungsfunktion angeben, die auch als Frequenzgang bezeichnet wird:
+
Considering the substitution  $p = {\rm j} \cdot 2 \pi f$,  the conventional transfer function can be specified,  which is also called  "frequency response":
$$H(f) =  H_{\rm L}(p)\Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it
+
:$$H(f) =  H_{\rm L}(p)\Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it
 
  f}} =  {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}
 
  f}} =  {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}
 
  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
  
Aus dieser Gleichung erkennt man auch den Zusammenhang zwischen
+
From this equation it can also be seen the relationship between
* der  Übertragungsfunktion $H(f)$,  
+
*the transfer function  $H(f)$,  
*der Dämpfungsfunktion $a(f)$ und
+
*the attenuation function  $a(f)$  and
*der Phasenfunktion $b(f)$.
+
*the phase function  $b(f)$.  
+
 
  
Für eine durch den Punkt $p = {\rm j} \cdot 2 \pi f$ indirekt vorgegebene Frequenz $f$ kann man die entsprechenden Dämpfungs– und Phasenwerte wie folgt ermitteln:
+
The attenuation and phase values can be determined as follows for a frequency  $f$  indirectly specified by the point  $p = {\rm j} \cdot 2 \pi f$ :
$$a(f)\hspace{0.15cm}{\rm in}\hspace{0.15cm}{\rm Np} \hspace{0.25cm} =  \hspace{0.2cm} -{\rm ln} \hspace{0.1cm} K
+
:$$a(f)\hspace{0.15cm}{\rm in}\hspace{0.15cm}{\rm Np} \hspace{0.25cm} =  \hspace{0.2cm} -{\rm ln} \hspace{0.1cm} K
 
  + {\rm ln} \hspace{0.1cm} |R_{\rm x1}|+{\rm ln} \hspace{0.1cm} |R_{\rm x1}|- {\rm ln} \hspace{0.1cm} |R_{{\rm o} }|\hspace{0.05cm}
 
  + {\rm ln} \hspace{0.1cm} |R_{\rm x1}|+{\rm ln} \hspace{0.1cm} |R_{\rm x1}|- {\rm ln} \hspace{0.1cm} |R_{{\rm o} }|\hspace{0.05cm}
 
  ,$$
 
  ,$$
$$ b(f)\hspace{0.15cm}{\rm in}\hspace{0.15cm}{\rm rad} \hspace{0.25cm} =  \hspace{0.2cm}
+
:$$ b(f)\hspace{0.15cm}{\rm in}\hspace{0.15cm}{\rm rad} \hspace{0.25cm} =  \hspace{0.2cm}
 
  \phi_{\rm x1}+ \phi_{\rm x2}-\phi_{\rm o}
 
  \phi_{\rm x1}+ \phi_{\rm x2}-\phi_{\rm o}
 
  \hspace{0.05cm} .$$
 
  \hspace{0.05cm} .$$
  
Die entsprechenden Beträge $|R_{\rm o}|$,  $|R_{\rm x1}|$ und$|R_{\rm x1}|$ können Sie ebenso wie die Winkel $\phi_{\rm o}$, $\phi_{\rm x1}$ und $\phi_{\rm x2}$ der Grafik entnehmen.
+
The corresponding magnitudes  $|R_{\rm o}|$,   $|R_{\rm x1}|$  and  $|R_{\rm x2}|$ as well as the angles  $\phi_{\rm o}$,  $\phi_{\rm x1}$  and  $\phi_{\rm x2}$  can be taken from the graph .
 +
 
 +
 
 +
 
 +
 
 +
 
 +
Please note:
 +
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function|Laplace Transform and p-Transfer Function]].
 +
 +
 
  
''Hinweise:''
 
*Die Aufgabe gehört zum Kapitel  [[Lineare_zeitinvariante_Systeme/Laplace–Transformation_und_p–Übertragungsfunktion|Laplace–Transformation und p–Übertragungsfunktion]].
 
*Sollte die Eingabe des Zahlenwertes „0” erforderlich sein, so geben Sie bitte „0.” ein.
 
  
  
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Berechnen Sie $H(f)$. Wie groß ist dessen Betrag bei sehr großen Frequenzen?
+
{Compute &nbsp;$H(f)$.&nbsp; What is its magnitude at very large frequencies?
 
|type="{}"}
 
|type="{}"}
$|H(f &#8594; &#8734;)| \ = $ { 0. }
+
$|H(f &#8594; &#8734;)| \ = \ $ { 0. }
  
  
{Berechnen Sie den Betragsfrequenzgang und den Dämpfungswert für $f &#8594 0$.
+
{Compute the magnitude frequency response and the attenuation value for &nbsp;$f &#8594 0$.
 
|type="{}"}
 
|type="{}"}
$|H(f = 0)| \ =$ { 0.278 3% }
+
$|H(f = 0)| \ = \ $ { 0.278 3% }
$a(f = 0) \ =$ { 1.281 3% } $\ \rm Np$
+
$a(f = 0) \ = \ $ { 1.281 3% } $\ \rm Np$
  
  
{Berechnen Sie gemäß der beschriebenen Vorgehensweise den Dämpfungswert bei $f 4/(2 \pi)$ in Neper (Np) und Dezibel (dB).
+
{Compute the attenuation value at &nbsp;$f =4/(2 \pi)$&nbsp; in neper&nbsp; $\rm(Np)$&nbsp; and decibel&nbsp; $\rm(dB)$&nbsp; according to the described approach.
 
|type="{}"}
 
|type="{}"}
$a(f = 2/ \pi)\ =$ { 0.155 3% } $\ \rm Np$
+
$a(f = 2/ \pi)\ = \ $ { 0.155 3% } $\ \rm Np$
$a(f = 2/ \pi)\ =$ { 1.346 3% }$\ \rm dB$
+
$a(f = 2/ \pi)\ = \ $ { 1.346 3% }$\ \rm dB$
  
  
{Berechnen Sie gemäß der beschriebenen Vorgehensweise den Phasenwert bei der Frequenz $f 4/(2 \pi)$.
+
{Compute the phase value at frequency &nbsp;$f = 4/(2 \pi)$ according to the described approach.
 
|type="{}"}
 
|type="{}"}
$b(f = 2/ \pi)\ =$ { -19.3--18.3 } $\ \rm Grad$
+
$b(f = 2/ \pi)\ = \ $ { -19.3--18.3 } $\ \rm Grad$
  
  
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</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Die <i>p</i>&ndash;Übertragungsfunktion lautet:
+
'''(1)'''&nbsp; The&nbsp; $p$&ndash;transfer function is:
 
:$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }}
 
:$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }}
 
  {(p - p_{\rm x 1})(p - p_{\rm x 2})}
 
  {(p - p_{\rm x 1})(p - p_{\rm x 2})}
 
  \hspace{0.05cm} .$$
 
  \hspace{0.05cm} .$$
  
Zur herkömmlichen Übertragungsfunktion (Frequenzgang) kommt man mit der Substitution $p = {\rm j} \cdot 2 \pi f$:
+
*The conventional transfer function&nbsp; (the frequency response)&nbsp; is obtained via the substitution &nbsp;$p = {\rm j} \cdot 2 \pi f$:
$$H(f)= K \cdot \frac {{\rm j \hspace{0.05cm}2\pi \it
+
:$$H(f)= K \cdot \frac {{\rm j \hspace{0.05cm}2\pi \it
 
  f} - p_{\rm o }}
 
  f} - p_{\rm o }}
 
  {({\rm j \hspace{0.05cm}2\pi \it
 
  {({\rm j \hspace{0.05cm}2\pi \it
Line 83: Line 89:
 
  \hspace{0.05cm} .$$
 
  \hspace{0.05cm} .$$
  
Im Grenzfall $f &#8594; \infty$ ergibt sich für den Betrag, die Dämpfung und die Phase:
+
*In the limiting case &nbsp;$f &#8594; \infty$,&nbsp; the following is obtained for the magnitude,&nbsp; attenuation and phase:
$$\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty}  H(f)= \frac{K}{{\rm j \hspace{0.05cm}2\pi \it
+
:$$\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty}  H(f)= \frac{K}{{\rm j \hspace{0.05cm}2\pi \it
  f}}\hspace{0.15cm}\Rightarrow \hspace{0.05cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty}  |H(f)|\hspace{0.15cm}\underline {= 0} \hspace{0.05cm}
+
  f}}\hspace{0.15cm}\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty}  |H(f)|\hspace{0.15cm}\underline {= 0} \hspace{0.05cm}
\Rightarrow \hspace{0.05cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty}  a(f)= \infty,\hspace{0.1cm}
+
\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty}  a(f)= \infty,\hspace{0.1cm}
 
  \lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty}  b(f)\underline {=
 
  \lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty}  b(f)\underline {=
 
  {\pi}/{2}\hspace{0.1cm}(+90^\circ)}
 
  {\pi}/{2}\hspace{0.1cm}(+90^\circ)}
 
  \hspace{0.01cm}.$$
 
  \hspace{0.01cm}.$$
  
'''(2)'''&nbsp; Aus der allgemeinen Gleichung in Teilaufgabe (1) erhält man mit dem Grenzübergang $f &#8594 0$:
+
 
$$|H(f=0)|=  -\frac {K \cdot p_{\rm o }}
+
 
 +
'''(2)'''&nbsp; From the general equation in subtask&nbsp; '''(1)'''&nbsp; the following is obtained with the limit process &nbsp;$f &#8594 0$:
 +
[[File:P_ID1769__LZI_A_3_4_d_neu.png|right|frame|Magnitude&nbsp; (&rArr; Betrag)&nbsp; $|H(f)|$,&nbsp; <br>attenuation&nbsp; (&rArr; Dämpfung)&nbsp; $a(f)$&nbsp; and phase&nbsp; $b(f)$]]
 +
:$$|H(f=0)|=  -\frac {K \cdot p_{\rm o }}
 
  {p_{\rm x 1}\cdot p_{\rm x 2}}
 
  {p_{\rm x 1}\cdot p_{\rm x 2}}
 
  =  \frac {5 \cdot 1}{
 
  =  \frac {5 \cdot 1}{
Line 98: Line 107:
 
  0.278}
 
  0.278}
 
  \hspace{0.05cm} ,$$
 
  \hspace{0.05cm} ,$$
$$a(f=0)=- {\rm ln} \hspace{0.1cm}\hspace{0.15cm}\underline { |H(f=0)|= 1.281\,{\rm Np }}
+
:$$a(f=0)=- {\rm ln} \hspace{0.1cm}\hspace{0.15cm}\underline { |H(f=0)|= 1.281\,{\rm Np }}
 
  \hspace{0.05cm} .$$
 
  \hspace{0.05cm} .$$
[[File:P_ID1769__LZI_A_3_4_d_neu.png|right||H(f)|, a(f) und b(f)]]
 
Der Bildschirmabzug des Flash&ndash;Moduls &bdquo;Kausale Systeme&rdquo; fasst die Ergebnisse dieser Aufgabe zusammen:
 
:*Mittlere Achse (blau): &nbsp; Betrag $|H(f)|$,
 
:*Linke Achse (rot): &nbsp; Dämpfung $a(f)$,
 
:*Rechte Achse (grün): &nbsp; Phase $b(f)$.<br><br>
 
:*Schwarzer Punkt: &nbsp; Werte für $2\pi f = 4.$
 
<br><br><br><br>
 
  
 +
The screen capture of the Flash&ndash;module&nbsp; "Causal Systems"&nbsp; summarizes the results of this exercise:
 +
:*middle axis (blue): &nbsp; magnitude $|H(f)|$, &nbsp; &rArr; &nbsp; here labeled with&nbsp; $|Y(f)|$,
 +
:*left axis (red): &nbsp; attenuation&nbsp;  $a(f)$,
 +
:*right axis (green): &nbsp; phase&nbsp; $b(f)$.
 +
:*black point: &nbsp; values for&nbsp; $2\pi f = 4.$
 +
 +
 +
[[File:P_ID2843__LZI_A_3_4.png|right|frame|Pole-zero diagram and some auxiliary quantities]]
 +
'''(3)'''&nbsp; According to the detailed description in the &nbsp;[[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Graphical_determination_of_attenuation_and_phase|theory part]],&nbsp; the following holds for the attenuation function:
  
'''(3)'''&nbsp; Entsprechend der detaillierten Beschreibung im [[Lineare_zeitinvariante_Systeme/Laplace–Transformation_und_p–Übertragungsfunktion#Grafische_Ermittlung_von_D.C3.A4mpfung_und_Phase|Theorieteil]] gilt für die Dämpfungsfunktion:
+
:$$a(f)=  -{\rm ln} \hspace{0.1cm} K
[[File:P_ID2843__LZI_A_3_4.png|right|Pol–Nullstellen–Diagramm und einige Hilfsgrößen]]
 
$$a(f)=  -{\rm ln} \hspace{0.1cm} K
 
 
  + {\rm ln} \hspace{0.1cm} |R_{\rm x1}|+{\rm ln} \hspace{0.1cm} |R_{\rm x2}|- {\rm ln} \hspace{0.1cm} |R_{{\rm o} }|\hspace{0.05cm} .$$
 
  + {\rm ln} \hspace{0.1cm} |R_{\rm x1}|+{\rm ln} \hspace{0.1cm} |R_{\rm x2}|- {\rm ln} \hspace{0.1cm} |R_{{\rm o} }|\hspace{0.05cm} .$$
  
Zu berücksichtigen ist weiterhin die Zusatzeinheit &bdquo;Neper&rdquo; (Np).
+
*Furthermore,&nbsp; the additional unit&nbsp; "neper"&nbsp; $\rm (Np)$&nbsp; must be taken into account.
 
+
*The attenuation at &nbsp;$f = 2/\pi$&nbsp; is searched-for.&nbsp; For this, we set &nbsp;$p = {\rm j} \cdot 2 \pi f = 4$&nbsp; and determine the following distances:
Gesucht ist die Dämpfung bei $f = 2/\pi$. Dazu setzen wir $p = {\rm j} \cdot 2 \pi f = 4$ und ermitteln folgende Abstände:
+
:$$R_{\rm o} = 1 - 4 \cdot {\rm j}, \hspace{0.2cm}|R_{\rm o}| \hspace{0.25cm} =  \hspace{0.2cm} \sqrt{1^2 + 4^2}= 4.123, \hspace{1.15cm}
$$R_{\rm o} = 1 - 4 \cdot {\rm j}, \hspace{0.2cm}|R_{\rm o}| \hspace{0.25cm} =  \hspace{0.2cm} \sqrt{1^2 + 4^2}= 4.123, \hspace{1cm}
 
 
  {\rm ln} \hspace{0.1cm}|R_{\rm o}|
 
  {\rm ln} \hspace{0.1cm}|R_{\rm o}|
 
  \hspace{0.25cm} =  \hspace{0.2cm}1.417\,{\rm Np }\hspace{0.05cm},$$
 
  \hspace{0.25cm} =  \hspace{0.2cm}1.417\,{\rm Np }\hspace{0.05cm},$$
$$R_{\rm x1} = -3 - 1 \cdot {\rm
+
:$$R_{\rm x1} = -3 - 1 \cdot {\rm
j}, \hspace{0.2cm}|R_{\rm x1}| \hspace{0.25cm} =  \hspace{0.2cm} \sqrt{3^2 + 1^2}= 3.162,\hspace{1cm}
+
j}, \hspace{0.2cm}|R_{\rm x1}| \hspace{0.25cm} =  \hspace{0.2cm} \sqrt{3^2 + 1^2}= 3.162,\hspace{0.5cm}
 
  {\rm ln} \hspace{0.1cm}|R_{\rm x1}|
 
  {\rm ln} \hspace{0.1cm}|R_{\rm x1}|
 
   \hspace{0.25cm} =  \hspace{0.2cm}1.151\,{\rm Np }\hspace{0.05cm},$$
 
   \hspace{0.25cm} =  \hspace{0.2cm}1.151\,{\rm Np }\hspace{0.05cm},$$
$$ R_{\rm x2} = -3 - 7 \cdot{\rm j},
+
:$$ R_{\rm x2} = -3 - 7 \cdot{\rm j},
\hspace{0.2cm}|R_{\rm x2}| \hspace{0.25cm} =  \hspace{0.2cm} \sqrt{3^2 + 7^2}= 7.616,\hspace{1cm}
+
\hspace{0.2cm}|R_{\rm x2}| \hspace{0.25cm} =  \hspace{0.2cm} \sqrt{3^2 + 7^2}= 7.616,\hspace{0.5cm}
 
  {\rm ln} \hspace{0.1cm}|R_{\rm x2}|
 
  {\rm ln} \hspace{0.1cm}|R_{\rm x2}|
 
   \hspace{0.25cm} =  \hspace{0.2cm}2.030\,{\rm Np }\hspace{0.05cm}.$$
 
   \hspace{0.25cm} =  \hspace{0.2cm}2.030\,{\rm Np }\hspace{0.05cm}.$$
Line 131: Line 139:
 
  -{\rm ln} \hspace{0.1cm} 5
 
  -{\rm ln} \hspace{0.1cm} 5
 
  + 1.151+ 2.030-
 
  + 1.151+ 2.030-
  1.417=0.155\,{\rm Np }
+
  1.417\hspace{0.15cm}\underline{=0.155\,{\rm Np }}
 
  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
  
Das entspricht $0.155\  {\rm Np} \cdot  8.686 \  {\rm dB/Np}\underline{= 1.346 \ {\rm dB}}$.
+
This is equivalent to &nbsp; $0.155\  {\rm Np} \cdot  8.686 \  {\rm dB/Np} \hspace{0.15cm} \underline{= 1.346 \ {\rm dB}}$.
 +
 
  
'''(4)'''&nbsp; Nach der detaillierten Beschreibung im Theorieteil gilt wegen $K > 0$ für die Phasenfunktion:
+
'''(4)'''&nbsp; The following holds for the phase function according to the description in the theory section due to &nbsp;$K > 0$&nbsp;:
$$b(f ={2}/\pi) = \phi_{\rm x1} + \phi_{\rm x2}-\phi_{\rm o}\hspace{0.05cm},$$
+
:$$b(f ={2}/\pi) = \phi_{\rm x1} + \phi_{\rm x2}-\phi_{\rm o}\hspace{0.05cm},$$
$$\phi_{\rm x1} ={\rm arctan}\hspace{0.15cm}(1/3) =
+
:$$\phi_{\rm x1} ={\rm arctan}\hspace{0.15cm}(1/3) =
  18.4^\circ\hspace{0.05cm}, \hspace{0.2cm}\phi_{\rm x2} = {\rm arctan}\hspace{0.15cm}(7/3)  =
+
  18.4^\circ\hspace{0.05cm}, \hspace{0.5cm}\phi_{\rm x2} = {\rm arctan}\hspace{0.15cm}(7/3)  =
  66.8^\circ\hspace{0.05cm},$$
+
  66.8^\circ\hspace{0.05cm},\hspace{0.5cm} \phi_{\rm o} = {\rm arctan}\hspace{0.15cm}(-1/4) =
$$ \phi_{\rm o} = {\rm arctan}\hspace{0.15cm}(-1/4) =
+
  180^\circ - 76^\circ = 104^\circ $$
  180^\circ - 76^\circ = 104^\circ$$
+
:$$ \Rightarrow \hspace{0.3cm}b(f ={2}/\pi) =
$$\Rightarrow \hspace{0.3cm}b(f ={2}/\pi) =
 
 
  18.4^\circ + 66.8^\circ - 104^\circ  \hspace{0.15cm} \underline{= -18.8^\circ}
 
  18.4^\circ + 66.8^\circ - 104^\circ  \hspace{0.15cm} \underline{= -18.8^\circ}
 
  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
Line 150: Line 158:
  
  
[[Category:Aufgaben zu Lineare zeitinvariante Systeme|^3.2 Laplace–Transformation und p–Übertragungsfunktion^]]
+
[[Category:Linear and Time-Invariant Systems: Exercises|^3.2 Laplace Transform and p-Transfer Function^]]

Latest revision as of 16:15, 14 October 2021

Pole-zero diagram and definition of some auxiliary quantities

We assume the sketched pole–zero diagram, i.e. the values

$$K = 5, \hspace{0,2cm}Z = 1, \hspace{0,2cm}N = 2, $$
$$ p_{\rm o}= 1,\hspace{0,2cm}p_{\rm x1}= -3 + 3{\rm j},\hspace{0,2cm}p_{\rm x2}= -3 - 3{\rm j}\hspace{0.05cm} .$$

Thus,  the  $p$–transfer function is:

$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }} {(p - p_{\rm x 1})(p - p_{\rm x 2})} \hspace{0.05cm} .$$

Considering the substitution  $p = {\rm j} \cdot 2 \pi f$,  the conventional transfer function can be specified,  which is also called  "frequency response":

$$H(f) = H_{\rm L}(p)\Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it f}} = {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)} \hspace{0.05cm}.$$

From this equation it can also be seen the relationship between

  • the transfer function  $H(f)$,
  • the attenuation function  $a(f)$  and
  • the phase function  $b(f)$.


The attenuation and phase values can be determined as follows for a frequency  $f$  indirectly specified by the point  $p = {\rm j} \cdot 2 \pi f$ :

$$a(f)\hspace{0.15cm}{\rm in}\hspace{0.15cm}{\rm Np} \hspace{0.25cm} = \hspace{0.2cm} -{\rm ln} \hspace{0.1cm} K + {\rm ln} \hspace{0.1cm} |R_{\rm x1}|+{\rm ln} \hspace{0.1cm} |R_{\rm x1}|- {\rm ln} \hspace{0.1cm} |R_{{\rm o} }|\hspace{0.05cm} ,$$
$$ b(f)\hspace{0.15cm}{\rm in}\hspace{0.15cm}{\rm rad} \hspace{0.25cm} = \hspace{0.2cm} \phi_{\rm x1}+ \phi_{\rm x2}-\phi_{\rm o} \hspace{0.05cm} .$$

The corresponding magnitudes  $|R_{\rm o}|$,  $|R_{\rm x1}|$  and  $|R_{\rm x2}|$ as well as the angles  $\phi_{\rm o}$,  $\phi_{\rm x1}$  and  $\phi_{\rm x2}$  can be taken from the graph .



Please note:



Questions

1

Compute  $H(f)$.  What is its magnitude at very large frequencies?

$|H(f → ∞)| \ = \ $

2

Compute the magnitude frequency response and the attenuation value for  $f → 0$.

$|H(f = 0)| \ = \ $

$a(f = 0) \ = \ $

$\ \rm Np$

3

Compute the attenuation value at  $f =4/(2 \pi)$  in neper  $\rm(Np)$  and decibel  $\rm(dB)$  according to the described approach.

$a(f = 2/ \pi)\ = \ $

$\ \rm Np$
$a(f = 2/ \pi)\ = \ $

$\ \rm dB$

4

Compute the phase value at frequency  $f = 4/(2 \pi)$ according to the described approach.

$b(f = 2/ \pi)\ = \ $

$\ \rm Grad$


Solution

(1)  The  $p$–transfer function is:

$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }} {(p - p_{\rm x 1})(p - p_{\rm x 2})} \hspace{0.05cm} .$$
  • The conventional transfer function  (the frequency response)  is obtained via the substitution  $p = {\rm j} \cdot 2 \pi f$:
$$H(f)= K \cdot \frac {{\rm j \hspace{0.05cm}2\pi \it f} - p_{\rm o }} {({\rm j \hspace{0.05cm}2\pi \it f} - p_{\rm x 1})({\rm j \hspace{0.05cm}2\pi \it f} - p_{\rm x 2})} = {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)} \hspace{0.05cm} .$$
  • In the limiting case  $f → \infty$,  the following is obtained for the magnitude,  attenuation and phase:
$$\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} H(f)= \frac{K}{{\rm j \hspace{0.05cm}2\pi \it f}}\hspace{0.15cm}\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} |H(f)|\hspace{0.15cm}\underline {= 0} \hspace{0.05cm} \Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} a(f)= \infty,\hspace{0.1cm} \lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} b(f)\underline {= {\pi}/{2}\hspace{0.1cm}(+90^\circ)} \hspace{0.01cm}.$$


(2)  From the general equation in subtask  (1)  the following is obtained with the limit process  $f → 0$:

Magnitude  (⇒ Betrag)  $|H(f)|$, 
attenuation  (⇒ Dämpfung)  $a(f)$  and phase  $b(f)$
$$|H(f=0)|= -\frac {K \cdot p_{\rm o }} {p_{\rm x 1}\cdot p_{\rm x 2}} = \frac {5 \cdot 1}{ (-3 + 3{\rm j})\cdot (-3 + 3{\rm j})}= \frac {5 }{18}\hspace{0.15cm}\underline {\approx 0.278} \hspace{0.05cm} ,$$
$$a(f=0)=- {\rm ln} \hspace{0.1cm}\hspace{0.15cm}\underline { |H(f=0)|= 1.281\,{\rm Np }} \hspace{0.05cm} .$$

The screen capture of the Flash–module  "Causal Systems"  summarizes the results of this exercise:

  • middle axis (blue):   magnitude $|H(f)|$,   ⇒   here labeled with  $|Y(f)|$,
  • left axis (red):   attenuation  $a(f)$,
  • right axis (green):   phase  $b(f)$.
  • black point:   values for  $2\pi f = 4.$


Pole-zero diagram and some auxiliary quantities

(3)  According to the detailed description in the  theory part,  the following holds for the attenuation function:

$$a(f)= -{\rm ln} \hspace{0.1cm} K + {\rm ln} \hspace{0.1cm} |R_{\rm x1}|+{\rm ln} \hspace{0.1cm} |R_{\rm x2}|- {\rm ln} \hspace{0.1cm} |R_{{\rm o} }|\hspace{0.05cm} .$$
  • Furthermore,  the additional unit  "neper"  $\rm (Np)$  must be taken into account.
  • The attenuation at  $f = 2/\pi$  is searched-for.  For this, we set  $p = {\rm j} \cdot 2 \pi f = 4$  and determine the following distances:
$$R_{\rm o} = 1 - 4 \cdot {\rm j}, \hspace{0.2cm}|R_{\rm o}| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{1^2 + 4^2}= 4.123, \hspace{1.15cm} {\rm ln} \hspace{0.1cm}|R_{\rm o}| \hspace{0.25cm} = \hspace{0.2cm}1.417\,{\rm Np }\hspace{0.05cm},$$
$$R_{\rm x1} = -3 - 1 \cdot {\rm j}, \hspace{0.2cm}|R_{\rm x1}| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{3^2 + 1^2}= 3.162,\hspace{0.5cm} {\rm ln} \hspace{0.1cm}|R_{\rm x1}| \hspace{0.25cm} = \hspace{0.2cm}1.151\,{\rm Np }\hspace{0.05cm},$$
$$ R_{\rm x2} = -3 - 7 \cdot{\rm j}, \hspace{0.2cm}|R_{\rm x2}| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{3^2 + 7^2}= 7.616,\hspace{0.5cm} {\rm ln} \hspace{0.1cm}|R_{\rm x2}| \hspace{0.25cm} = \hspace{0.2cm}2.030\,{\rm Np }\hspace{0.05cm}.$$

$$\Rightarrow \hspace{0.3cm}a(f = \frac{4}{2\pi})= -{\rm ln} \hspace{0.1cm} 5 + 1.151+ 2.030- 1.417\hspace{0.15cm}\underline{=0.155\,{\rm Np }} \hspace{0.05cm}.$$

This is equivalent to   $0.155\ {\rm Np} \cdot 8.686 \ {\rm dB/Np} \hspace{0.15cm} \underline{= 1.346 \ {\rm dB}}$.


(4)  The following holds for the phase function according to the description in the theory section due to  $K > 0$ :

$$b(f ={2}/\pi) = \phi_{\rm x1} + \phi_{\rm x2}-\phi_{\rm o}\hspace{0.05cm},$$
$$\phi_{\rm x1} ={\rm arctan}\hspace{0.15cm}(1/3) = 18.4^\circ\hspace{0.05cm}, \hspace{0.5cm}\phi_{\rm x2} = {\rm arctan}\hspace{0.15cm}(7/3) = 66.8^\circ\hspace{0.05cm},\hspace{0.5cm} \phi_{\rm o} = {\rm arctan}\hspace{0.15cm}(-1/4) = 180^\circ - 76^\circ = 104^\circ $$
$$ \Rightarrow \hspace{0.3cm}b(f ={2}/\pi) = 18.4^\circ + 66.8^\circ - 104^\circ \hspace{0.15cm} \underline{= -18.8^\circ} \hspace{0.05cm}.$$