Difference between revisions of "Aufgaben:Exercise 3.4Z: Various All-Pass Filters"

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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_ID1766__LZI_Z_3_4.png|right|Allpass in zwei verschiedenen Varianten]]
+
[[File:EN_LZI_Z_3_4.png|right|frame|All-pass filter in two different variants<br> &nbsp; &nbsp; $Z_{\rm I}$:&nbsp; Internal resistance of the source <br> &nbsp; &nbsp; $Z_{\rm A}$:&nbsp; Terminating resistor]]
Wir gehen zunächst von einem Vierpol mit der folgenden Übertragungsfunktion aus:
+
We first assume a two-port network with the following transfer function:
$$H_{\rm L}(p)=  \frac {1-{p}/{A}} {1+{p}/{A}}.$$
+
:$$H_{\rm L}(p)=  \frac {1-{p}/{A}} {1+{p}/{A}}.$$
Aus dieser soll der herkömmliche Fourier&ndash;Frequenzgang
+
From this is to be determined the conventional Fourier frequency response
$$H(f) =  {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}$$
+
:$$H(f) =  {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)},$$
ermittelt werden, der sich durch Dämpfungsfunktion $a(f)$ und Phasenfunktion $b(f)$ darstellen lässt.
+
which is representable 
 +
*by the attenuation function &nbsp;$a(f)$&nbsp;
 +
*and the phase function &nbsp;$b(f)$.
  
Die obere Grafik zeigt eine so genannte Allpass&ndash;Schaltung, wobei der komplexe Widerstand $Z_1$ eine Induktivität und $Z_2$  eine Kapazität bezeichnet:
 
$$Z_1 = p \cdot L\hspace{0.05cm},\hspace{0.2cm}Z_2 = \frac{1}{p \cdot C}\hspace{0.05cm}.$$
 
Bei reflexionsfreier Anpassung am Eingang und Ausgang mit
 
$$Z_{\rm I}=Z_{\rm A} = \sqrt{Z_1 \cdot Z_2} = \sqrt{{L}/{C}}$$
 
gilt für die <i>p</i>&ndash;Übertragungsfunktion der Schaltung <b>A</b> (siehe obere Grafik):
 
$$H_{\rm L}(p)=  \frac {Y_{\rm L}(p)} {X_{\rm L}(p)}=  \frac {Z_2-Z_1} {Z_1+2 \cdot \sqrt{Z_1 \cdot Z_2}+Z_2}\hspace{0.05cm}.$$
 
  
Die Schaltung <b>B</b> ist durch die $p$&ndash;Übertragungsfunktion festgelegt. Sie ist dadurch charakterisiert, dass
+
The upper diagram shows a so-called&nbsp; "all-pass circuit"&nbsp; where the complex resistance &nbsp;$Z_1$&nbsp; denotes an inductance and the complex resistance &nbsp;$Z_2$&nbsp; denotes a capacitance:
*alle Pole (in der linken $p$&ndash;Halbebene)  
+
:$$Z_1 = p \cdot L\hspace{0.05cm},\hspace{0.2cm}Z_2 = \frac{1}{p \cdot C}\hspace{0.05cm}.$$
*spiegelbildlich zu den Nullstellen (in der rechten Halbebene) liegen.
+
For reflection-free adaptation at the input and output with
 +
:$$Z_{\rm I}=Z_{\rm A} = \sqrt{Z_1 \cdot Z_2} = \sqrt{{L}/{C}},$$
 +
the following holds for the&nbsp; $p$&ndash;transfer function of circuit&nbsp; $\rm A$&nbsp; (see upper diagram):
 +
:$$H_{\rm L}(p)=  \frac {Y_{\rm L}(p)} {X_{\rm L}(p)}=  \frac {Z_2-Z_1} {Z_1+2 \cdot \sqrt{Z_1 \cdot Z_2}+Z_2}\hspace{0.05cm}.$$
  
 +
Circuit &nbsp;$\rm B$&nbsp; is defined by the pole–zero diagram of the &nbsp;$p$&ndash;transfer function.&nbsp; It is characterized by the fact that
 +
*all poles&nbsp; (in the left $p$&ndash;half-plane)
 +
*are located in a mirror-imaged manner with respect to the zeros&nbsp; (in the right half-plane).
  
''Hinweise:''
 
*Die Aufgabe gehört zum Kapitel  [[Lineare_zeitinvariante_Systeme/Laplace–Transformation_und_p–Übertragungsfunktion|Laplace–Transformation und p–Übertragungsfunktion]].
 
*Sollte die Eingabe des Zahlenwertes &bdquo;0&rdquo; erforderlich sein, so geben Sie bitte &bdquo;0.&rdquo; ein.
 
  
  
===Fragebogen===
+
 
 +
 
 +
Please note:
 +
*The exercise belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function|Laplace Transform and p-Transfer Function]].
 +
 +
 
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Geben Sie die Nullstelle $p_{\rm o}$ und den Pol $p_{\rm x}$ von $H_{\rm L}(p)=  (1 -p/A)/(1 +p/A)$ an. Wie groß ist der konstante Faktor $K$?
+
{Specify the zero &nbsp;$p_{\rm o}$&nbsp; and the pole &nbsp;$p_{\rm x}$&nbsp; of &nbsp;$H_{\rm L}(p)=  (1 -p/A)/(1 +p/A)$&nbsp;. &nbsp;What is the constant factor&nbsp; $K$?
 
|type="{}"}
 
|type="{}"}
$K \ =$  { -1.03--0.97 }
+
$K \ = \ $  { -1.03--0.97 }
$p_{\rm o} \ =$  { 1 3% } $\ \cdot A$
+
$p_{\rm o} \ = \ $  { 1 3% } $\ \cdot A$
$p_{\rm x} \ =$  { -1.03--0.97 } $\ \cdot A$
+
$p_{\rm x} \ = \ $  { -1.03--0.97 } $\ \cdot A$
  
  
{Berechnen Sie die Dämpfungsfunktion $a(f)$. Welche Aussagen treffen zu?
+
{Compute the attenuation function &nbsp;$a(f)$.&nbsp; Which statements are true?
 
|type="[]"}
 
|type="[]"}
- Die Dämpfungsfunktion $a(f)$ zeigt Tiefpassverhalten.
+
- The attenuation function &nbsp;$a(f)$&nbsp; exhibits low-pass filter behavior.
+ Die Dämpfungsfunktion $a(f)$ ist konstant.
+
+ The attenuation function &nbsp;$a(f)$&nbsp; is constant.
+ Das obige Ergebnis gilt allgemein für $p_{\rm x}  = - p_{\rm o}$.
+
+ The above result is generally valid for &nbsp;$p_{\rm x}  = - p_{\rm o}$.
  
  
{Berechnen Sie den Phasenverlauf $b(f)$. Welche Phasenwerte ergeben sich für die angegebenen Frequenzen?
+
{Compute the phase response &nbsp;$b(f)$.&nbsp; What are the phase values for the given frequencies?
 
|type="{}"}
 
|type="{}"}
$b(f = A/2 \pi) \ =$  { 90 3% } $\ \rm Grad$
+
$b(f = A/2 \pi) \ = \ $  { 90 3% } $\ \rm deg$
$b(f = A/ \pi)\ =$ { 126.8 3% } $ \rm Grad$
+
$b(f = A/ \pi)\ = \ $ { 126.8 3% } $ \rm deg$
$b(f &#8594; &#8734;) \ =$  { 180 3% } $ \rm Grad$
+
$b(f &#8594; &#8734;) \ = \ $  { 180 3% } $ \rm deg$
  
  
{Berechnen Sie die $p$&ndash;Übertragungsfunktion von Schaltung <b>A</b>. Welche Aussagen lassen sich daraus ableiten?
+
{Compute the &nbsp;$p$&ndash;transfer function of circuit &nbsp;$\rm A$.&nbsp; What statements can be derived from this?
 
|type="[]"}
 
|type="[]"}
+ Die Dämpfung $a(f)$ ist konstant gleich $0 \ \rm (Np)$.
+
+ The attenuation &nbsp;$a(f)$&nbsp; is constantly equal to&nbsp; $0$&nbsp; (Np).
- Die Phase $b(f)$ steigt linear mit der Frequenz $f$ an.
+
- The phase &nbsp;$b(f)$&nbsp; increases linearly with frequency &nbsp;$f$&nbsp;.
- $b(f)$ ist die Hilbert&ndash;Transformierte von $a(f)$.
+
- $b(f)$&nbsp; is the Hilbert transform of &nbsp;$a(f)$.
  
  
{Welche Aussagen können aus dem Pol&ndash;Nullstellen&ndash;Diagramm von Schaltung <b>B</b> abgeleitet werden?
+
{What statements can be derived from the pole&ndash;zero diagram of circuit &nbsp;$\rm B$&nbsp;?
 
|type="[]"}
 
|type="[]"}
+ Die Dämpfung $a(f)$ ist konstant.
+
+ The attenuation &nbsp;$a(f)$&nbsp; is constant.
+ Die Phasenfunktion $b(f)$ hat bei $f = 0$ den Wert $0$.
+
+ The following holds for the phase function: &nbsp;$b(f =0) =0$.
  
  
Line 67: Line 73:
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
:<b>1.</b>&nbsp;&nbsp;Durch Umformung der angegebenen <i>p</i>&ndash;Übertragungsfunktion ergibt sich
+
'''(1)'''&nbsp; Transforming the given &nbsp;$p$&ndash;transfer function yields
 
:$$H_{\rm L}(p)=  \frac {1-{p}/{A}} {1+{p}/{A}}= -1 \cdot \frac {p-A} {p+A}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm} \underline{K =- 1}, \hspace{0,2cm} \hspace{0.15cm} \underline{p_{\rm  o}/A = 1} ,\hspace{0,2cm}
 
:$$H_{\rm L}(p)=  \frac {1-{p}/{A}} {1+{p}/{A}}= -1 \cdot \frac {p-A} {p+A}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm} \underline{K =- 1}, \hspace{0,2cm} \hspace{0.15cm} \underline{p_{\rm  o}/A = 1} ,\hspace{0,2cm}
 
\hspace{0.15cm} \underline{ p_{\rm  x}/A = -1} \hspace{0.05cm} .$$
 
\hspace{0.15cm} \underline{ p_{\rm  x}/A = -1} \hspace{0.05cm} .$$
:<b>2.</b>&nbsp;&nbsp;Setzt man <i>p</i> = j &middot; 2&pi;<i>f</i>, so erhält man:
+
 
 +
 
 +
 
 +
'''(2)'''&nbsp; <u>Statements 2 and 3</u>&nbsp; are correct:
 +
*If &nbsp;$p = {\rm j} \cdot 2 \pi f$ is set, the following is obtained:
 
:$$H(f)=  \frac {1-{\rm j \cdot 2\pi \it
 
:$$H(f)=  \frac {1-{\rm j \cdot 2\pi \it
 
  f}/A} {1+{\rm j \cdot 2\pi \it
 
  f}/A} {1+{\rm j \cdot 2\pi \it
 
  f}/A}\hspace{0.05cm} .$$
 
  f}/A}\hspace{0.05cm} .$$
:Der Betrag eines Quotienten ist gleich dem Quotienten der Beträge:
+
*The magnitude of a quotient is equal to the quotient of the magnitudes:
 
:$$|H(f)|=  \frac {|1-{{\rm j} \cdot 2\pi
 
:$$|H(f)|=  \frac {|1-{{\rm j} \cdot 2\pi
 
  f}/A|} {|1+{\rm j \cdot 2\pi \it
 
  f}/A|} {|1+{\rm j \cdot 2\pi \it
 
  f}/A|}=  \frac {\sqrt{1+(2\pi
 
  f}/A|}=  \frac {\sqrt{1+(2\pi
 
  f/A)^2}} {\sqrt{1+(2\pi
 
  f/A)^2}} {\sqrt{1+(2\pi
  f/A)^2}}= 1$$
+
  f/A)^2}}= 1\hspace{0.3cm}
:$$\Rightarrow \hspace{0.3cm} a(f)= -{\rm ln} \hspace{0.1cm} |H(f)|= 0\hspace{0.2cm}({\rm  Np \hspace{0.2cm}oder \hspace{0.2cm}dB})\hspace{0.05cm} .$$
+
\Rightarrow \hspace{0.3cm} a(f)= -{\rm ln} \hspace{0.1cm} |H(f)|= 0\hspace{0.2cm}({\rm  Np \hspace{0.2cm}or \hspace{0.2cm}dB})\hspace{0.05cm} .$$
:Richtig ist somit die <u>Aussage 2</u>. Auch die <u>Aussage 3</u> ist richtig, wie aus der Theorieseite Grafische Ermittlung der Dämpfung zu ersehen ist.
+
*However, &nbsp;$\text{statement 3}$&nbsp; is also correct as can be seen from the theory page&nbsp; "Graphical determination of attenuation".
 +
 
 +
 
 +
 
  
:<b>3.</b>&nbsp;&nbsp;Die Phasenfunktion <i>b</i>(<i>f</i>) kann wie folgt berechnet werden:
+
'''(3)'''&nbsp; The phase function &nbsp;$b(f)$&nbsp; can be computed as follows:
:$$b(f)= -{\rm arc} \hspace{0.1cm} H(f) \hspace{0.25cm} =  \hspace{0.2cm} {\rm arctan } \hspace{0.1cm}\frac {2\pi
+
:$$b(f)= -{\rm arc} \hspace{0.1cm} H(f) \hspace{0.25cm} =  \hspace{0.2cm} {\rm arctan } \hspace{0.1cm} ({2\pi
  f}{A} - {\rm arctan } \hspace{0.1cm}\frac {-2\pi
+
  f}/{A}) - {\rm arctan } \hspace{0.1cm} ({-2\pi
  f}{A} = 2 \cdot {\rm arctan } \hspace{0.1cm}\frac {2\pi
+
  f}/{A}) = 2 \cdot {\rm arctan } \hspace{0.1cm} ({2\pi
  f}{A}:\\
+
  f}/{A}),$$
b(f= \frac{A}{2\pi})\hspace{0.25cm} = \hspace{0.2cm}2 \cdot {\rm arctan }
+
:$$b(f= {A}/{2\pi})= 2 \cdot {\rm arctan }
  \hspace{0.1cm}(1) = 2 \cdot 45^\circ\hspace{0.15cm} \underline{ = 90^\circ}\hspace{0.05cm}
+
  \hspace{0.1cm}(1) = 2 \cdot 45^\circ \hspace{0.15cm} \underline{ = 90^\circ}\hspace{0.05cm},$$
,\\
+
:$$ b(f= {A}/{\pi})=2 \cdot {\rm arctan }
b(f= \frac{A}{\pi})\hspace{0.25cm} = \hspace{0.2cm}2 \cdot {\rm arctan }
 
 
  \hspace{0.1cm}(2) = 2 \cdot 63.4^\circ \hspace{0.15cm} \underline{= 126.8^\circ}\hspace{0.05cm}
 
  \hspace{0.1cm}(2) = 2 \cdot 63.4^\circ \hspace{0.15cm} \underline{= 126.8^\circ}\hspace{0.05cm}
  ,\\
+
  ,$$
b(f \rightarrow \infty)\hspace{0.25cm} = \hspace{0.2cm}2 \cdot {\rm arctan }
+
:$$ b(f \rightarrow \infty)=2 \cdot {\rm arctan }
 
  \hspace{0.1cm}(\infty) = 2 \cdot 90^\circ \hspace{0.15cm} \underline{= 180^\circ}\hspace{0.05cm}
 
  \hspace{0.1cm}(\infty) = 2 \cdot 90^\circ \hspace{0.15cm} \underline{= 180^\circ}\hspace{0.05cm}
 
  .$$
 
  .$$
:Zu den gleichen Ergebnissen kommt man nach der Vorgehensweise entsprechend der Seite Grafische Ermittlung der Phase im Theorieteil.
+
*The same results are obtained following the approach according to the page&nbsp; "Graphical determination of phase"&nbsp; in the theory part.
  
:<b>4.</b>&nbsp;&nbsp;Die angegebene <i>p</i>&ndash;Übertragungsfunktion lässt sich wie folgt darstellen:
+
 
:$$H_{\rm L}(p)\hspace{0.25cm} \hspace{0.2cm}  \frac {Z_2-Z_1} {Z_1+2 \cdot \sqrt{Z_1 \cdot Z_2}+Z_2}=\\
+
 
  \hspace{0.25cm} =  \hspace{0.2cm}  \frac {(\sqrt{Z_2}-\sqrt{Z_1})(\sqrt{Z_2}+\sqrt{Z_1})} {(\sqrt{Z_2}+\sqrt{Z_1})^2}=
+
 
 +
'''(4)'''&nbsp; Only <u>statement 1</u>&nbsp; is correct:
 +
*The given &nbsp;$p$&ndash;transfer function can be expressed as follows:
 +
:$$H_{\rm L}(p)=  \frac {Z_2-Z_1} {Z_1+2 \cdot \sqrt{Z_1 \cdot Z_2}+Z_2}=
 +
  \frac {(\sqrt{Z_2}-\sqrt{Z_1})(\sqrt{Z_2}+\sqrt{Z_1})} {(\sqrt{Z_2}+\sqrt{Z_1})^2}=
 
  \frac {\sqrt{Z_2}-\sqrt{Z_1}} {\sqrt{Z_2}+\sqrt{Z_1}}\hspace{0.05cm}.$$
 
  \frac {\sqrt{Z_2}-\sqrt{Z_1}} {\sqrt{Z_2}+\sqrt{Z_1}}\hspace{0.05cm}.$$
:Mit <i>Z</i><sub>1</sub> = <i>p</i> &middot; <i>L</i> und <i>Z</i><sub>2</sub> = 1/(<i>pC</i>) erhält man weiter:
+
*Furthermore,&nbsp; considering &nbsp;$Z_1 = p \cdot L$&nbsp; and &nbsp;$Z_2 = 1/(p \cdot C)$&nbsp; the following is obtained:
 
:$$H_{\rm L}(p)= \frac {\sqrt{{1}/(pC)}-\sqrt{pL}} {\sqrt{{1}/(pC)}+\sqrt{pL}}
 
:$$H_{\rm L}(p)= \frac {\sqrt{{1}/(pC)}-\sqrt{pL}} {\sqrt{{1}/(pC)}+\sqrt{pL}}
 
  = \frac {1- p \cdot \sqrt{LC}} {1+ p \cdot \sqrt{LC}}
 
  = \frac {1- p \cdot \sqrt{LC}} {1+ p \cdot \sqrt{LC}}
  = -1 \cdot \frac {p-\sqrt{{1}/(LC)}} {p+\sqrt{{1}/(LC)}}$$
+
  = -1 \cdot \frac {p-\sqrt{{1}/(LC)}} {p+\sqrt{{1}/(LC)}}\hspace{0.3cm}
:$$\Rightarrow \hspace{0.3cm}A = \sqrt{{1}/(LC)}: \hspace{0.2cm}H_{\rm L}(p)= -1 \cdot \frac {p-A} {p+A}\hspace{0.05cm}.$$
+
\Rightarrow \hspace{0.3cm}A = \sqrt{{1}/(LC)}: \hspace{0.2cm}H_{\rm L}(p)= -1 \cdot \frac {p-A} {p+A}\hspace{0.05cm}.$$
:Es ergibt sich die genau gleiche Übertragungsfunktion, wie in der Teilaufgabe 1) berechnet. Daraus folgt, dass nur die <u>Aussage 1</u> richtig ist:
 
 
 
:<ul class="liste_ohne"><li>Der Dämpfungsverlauf ist <i>a</i>(<i>f</i>) = 0 (Np). Keine Frequenz wird gedämpft oder verstärkt. Man spricht deshalb auch von einem &bdquo;Allpass&rdquo;.
 
  
:<ul class="liste_ohne"><li>Die zweite Aussage ist falsch. Der Phasenverlauf <i>b</i>(<i>f</i>) ist nicht linear, sondern vielmehr wie in der Teilaufgabe 3) berechnet.
+
*The same transfer function as computed in subtask&nbsp; '''(1)'''&nbsp; is obtained.  
  
:<ul class="liste_ohne"><li>Die Hilbert&ndash;Transformierte der Konstanten <i>a</i>(<i>f</i>) = 0 müsste zur Phasenfunktion <i>b</i>(<i>f</i>) = 0 führen, wie in Kapitel 3.1 gezeigt. Das heißt, dass die Aussage 3 falsch ist.
 
  
:<ul class="liste_ohne"><li>Nur bei minimalphasigen Systemen hängen die Dämpfungsfunktion <i>a</i>(<i>f</i>) und Phasenfunktion <i>b</i>(<i>f</i>) über die Hilbert&ndash;Transformation zusammen.
+
From this it follows that only <u>statement 1</u> is correct:
 +
*The attenuation curve is &nbsp;$a(f) = 0\ \rm  (Np)$.&nbsp; No frequency is attenuated or amplified.&nbsp; Therefore, it is called an "all-pass filter".
 +
*The second statement is false.&nbsp; The phase response &nbsp;$b(f)$&nbsp; is not linear but rather curved as computed in subtask&nbsp; '''(3)'''&nbsp;.
 +
*The Hilbert transform of the constant &nbsp;$a(f) = 0$&nbsp; should result in the phase function &nbsp;$b(f) = 0$&nbsp;  as shown in the theory part.&nbsp; That is:&nbsp; statement 3 is false.
 +
*The attenuation function &nbsp;$a(f)$&nbsp; and phase function &nbsp;$b(f)$&nbsp; are related to each other via the Hilbert transformation only for minimum-phase systems.
 +
*However, in such a minimum&ndash;phase system, all poles and zeros lie in the left $p$&ndash;half-plane which is not true here &nbsp; <br>&#8658; &nbsp; '''an all-pass filter is not a minimum&ndash;phase system'''.
  
:<ul class="liste_ohne"><li>Bei einem solchen Minimum&ndash;Phasen&ndash;System liegen aber alle Pole und Nullstellen in der linken <i>p</i>&ndash;Halbebene, was hier nicht zutrifft &nbsp;&#8658;&nbsp; ein Allpass ist kein Minimum&ndash;Phasen&ndash;System.
 
  
:<b>5.</b>&nbsp;&nbsp;<u>Beide Aussagen</u> sind richtig. Wie bereits in der Teilaufgabe 2) festgestellt wurde, ergibt sich dann eine konstante Dämpfung, wenn es zu jedem Pol in der linken <i>p</i>&ndash;Halbebene eine entsprechende Nullstelle in der rechten Halbebene gibt &nbsp;&#8658;&nbsp; die Schaltung <b>B</b> zeigt ebenfalls Allpass&ndash;Charakteristik.
 
  
:Da <i>b</i>(<i>f</i>) stets eine unsymmetrische Funktion ist, gilt <i>b</i>(<i>f</i> = 0) = 0 ganz allgemein, das heißt für jede Spektralfunktion <i>H</i>(<i>f</i>), deren Fourier&ndash;Rücktransformierte reell ist.
+
'''(5)'''&nbsp; <u>Both statements</u>&nbsp; are correct:
 +
*As already determined in subtask&nbsp; '''(2)''': &nbsp; A constant attenuation arises as a result whenever there is a corresponding zero in the right half-plane for each pole in the left $p$&ndash;half-plane &nbsp; &#8658; &nbsp; circuit&nbsp; $\rm B$&nbsp; also exhibits all-pass filter charakteristics.
 +
*Since &nbsp;$b(f)$&nbsp; is always an asymmetric function, &nbsp;$b(f= 0) = 0$&nbsp; holds in general.&nbsp; That is, for any spectral function &nbsp;$H(f)$ whose inverse Fourier transform&nbsp; ("impulse response") &nbsp;is real.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[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 14:54, 15 October 2021

All-pass filter in two different variants
    $Z_{\rm I}$:  Internal resistance of the source
    $Z_{\rm A}$:  Terminating resistor

We first assume a two-port network with the following transfer function:

$$H_{\rm L}(p)= \frac {1-{p}/{A}} {1+{p}/{A}}.$$

From this is to be determined the conventional Fourier frequency response

$$H(f) = {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)},$$

which is representable

  • by the attenuation function  $a(f)$ 
  • and the phase function  $b(f)$.


The upper diagram shows a so-called  "all-pass circuit"  where the complex resistance  $Z_1$  denotes an inductance and the complex resistance  $Z_2$  denotes a capacitance:

$$Z_1 = p \cdot L\hspace{0.05cm},\hspace{0.2cm}Z_2 = \frac{1}{p \cdot C}\hspace{0.05cm}.$$

For reflection-free adaptation at the input and output with

$$Z_{\rm I}=Z_{\rm A} = \sqrt{Z_1 \cdot Z_2} = \sqrt{{L}/{C}},$$

the following holds for the  $p$–transfer function of circuit  $\rm A$  (see upper diagram):

$$H_{\rm L}(p)= \frac {Y_{\rm L}(p)} {X_{\rm L}(p)}= \frac {Z_2-Z_1} {Z_1+2 \cdot \sqrt{Z_1 \cdot Z_2}+Z_2}\hspace{0.05cm}.$$

Circuit  $\rm B$  is defined by the pole–zero diagram of the  $p$–transfer function.  It is characterized by the fact that

  • all poles  (in the left $p$–half-plane)
  • are located in a mirror-imaged manner with respect to the zeros  (in the right half-plane).



Please note:


Questions

1

Specify the zero  $p_{\rm o}$  and the pole  $p_{\rm x}$  of  $H_{\rm L}(p)= (1 -p/A)/(1 +p/A)$ .  What is the constant factor  $K$?

$K \ = \ $

$p_{\rm o} \ = \ $

$\ \cdot A$
$p_{\rm x} \ = \ $

$\ \cdot A$

2

Compute the attenuation function  $a(f)$.  Which statements are true?

The attenuation function  $a(f)$  exhibits low-pass filter behavior.
The attenuation function  $a(f)$  is constant.
The above result is generally valid for  $p_{\rm x} = - p_{\rm o}$.

3

Compute the phase response  $b(f)$.  What are the phase values for the given frequencies?

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

$\ \rm deg$
$b(f = A/ \pi)\ = \ $

$ \rm deg$
$b(f → ∞) \ = \ $

$ \rm deg$

4

Compute the  $p$–transfer function of circuit  $\rm A$.  What statements can be derived from this?

The attenuation  $a(f)$  is constantly equal to  $0$  (Np).
The phase  $b(f)$  increases linearly with frequency  $f$ .
$b(f)$  is the Hilbert transform of  $a(f)$.

5

What statements can be derived from the pole–zero diagram of circuit  $\rm B$ ?

The attenuation  $a(f)$  is constant.
The following holds for the phase function:  $b(f =0) =0$.


Solution

(1)  Transforming the given  $p$–transfer function yields

$$H_{\rm L}(p)= \frac {1-{p}/{A}} {1+{p}/{A}}= -1 \cdot \frac {p-A} {p+A}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm} \underline{K =- 1}, \hspace{0,2cm} \hspace{0.15cm} \underline{p_{\rm o}/A = 1} ,\hspace{0,2cm} \hspace{0.15cm} \underline{ p_{\rm x}/A = -1} \hspace{0.05cm} .$$


(2)  Statements 2 and 3  are correct:

  • If  $p = {\rm j} \cdot 2 \pi f$ is set, the following is obtained:
$$H(f)= \frac {1-{\rm j \cdot 2\pi \it f}/A} {1+{\rm j \cdot 2\pi \it f}/A}\hspace{0.05cm} .$$
  • The magnitude of a quotient is equal to the quotient of the magnitudes:
$$|H(f)|= \frac {|1-{{\rm j} \cdot 2\pi f}/A|} {|1+{\rm j \cdot 2\pi \it f}/A|}= \frac {\sqrt{1+(2\pi f/A)^2}} {\sqrt{1+(2\pi f/A)^2}}= 1\hspace{0.3cm} \Rightarrow \hspace{0.3cm} a(f)= -{\rm ln} \hspace{0.1cm} |H(f)|= 0\hspace{0.2cm}({\rm Np \hspace{0.2cm}or \hspace{0.2cm}dB})\hspace{0.05cm} .$$
  • However,  $\text{statement 3}$  is also correct as can be seen from the theory page  "Graphical determination of attenuation".



(3)  The phase function  $b(f)$  can be computed as follows:

$$b(f)= -{\rm arc} \hspace{0.1cm} H(f) \hspace{0.25cm} = \hspace{0.2cm} {\rm arctan } \hspace{0.1cm} ({2\pi f}/{A}) - {\rm arctan } \hspace{0.1cm} ({-2\pi f}/{A}) = 2 \cdot {\rm arctan } \hspace{0.1cm} ({2\pi f}/{A}),$$
$$b(f= {A}/{2\pi})= 2 \cdot {\rm arctan } \hspace{0.1cm}(1) = 2 \cdot 45^\circ \hspace{0.15cm} \underline{ = 90^\circ}\hspace{0.05cm},$$
$$ b(f= {A}/{\pi})=2 \cdot {\rm arctan } \hspace{0.1cm}(2) = 2 \cdot 63.4^\circ \hspace{0.15cm} \underline{= 126.8^\circ}\hspace{0.05cm} ,$$
$$ b(f \rightarrow \infty)=2 \cdot {\rm arctan } \hspace{0.1cm}(\infty) = 2 \cdot 90^\circ \hspace{0.15cm} \underline{= 180^\circ}\hspace{0.05cm} .$$
  • The same results are obtained following the approach according to the page  "Graphical determination of phase"  in the theory part.



(4)  Only statement 1  is correct:

  • The given  $p$–transfer function can be expressed as follows:
$$H_{\rm L}(p)= \frac {Z_2-Z_1} {Z_1+2 \cdot \sqrt{Z_1 \cdot Z_2}+Z_2}= \frac {(\sqrt{Z_2}-\sqrt{Z_1})(\sqrt{Z_2}+\sqrt{Z_1})} {(\sqrt{Z_2}+\sqrt{Z_1})^2}= \frac {\sqrt{Z_2}-\sqrt{Z_1}} {\sqrt{Z_2}+\sqrt{Z_1}}\hspace{0.05cm}.$$
  • Furthermore,  considering  $Z_1 = p \cdot L$  and  $Z_2 = 1/(p \cdot C)$  the following is obtained:
$$H_{\rm L}(p)= \frac {\sqrt{{1}/(pC)}-\sqrt{pL}} {\sqrt{{1}/(pC)}+\sqrt{pL}} = \frac {1- p \cdot \sqrt{LC}} {1+ p \cdot \sqrt{LC}} = -1 \cdot \frac {p-\sqrt{{1}/(LC)}} {p+\sqrt{{1}/(LC)}}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}A = \sqrt{{1}/(LC)}: \hspace{0.2cm}H_{\rm L}(p)= -1 \cdot \frac {p-A} {p+A}\hspace{0.05cm}.$$
  • The same transfer function as computed in subtask  (1)  is obtained.


From this it follows that only statement 1 is correct:

  • The attenuation curve is  $a(f) = 0\ \rm (Np)$.  No frequency is attenuated or amplified.  Therefore, it is called an "all-pass filter".
  • The second statement is false.  The phase response  $b(f)$  is not linear but rather curved as computed in subtask  (3) .
  • The Hilbert transform of the constant  $a(f) = 0$  should result in the phase function  $b(f) = 0$  as shown in the theory part.  That is:  statement 3 is false.
  • The attenuation function  $a(f)$  and phase function  $b(f)$  are related to each other via the Hilbert transformation only for minimum-phase systems.
  • However, in such a minimum–phase system, all poles and zeros lie in the left $p$–half-plane which is not true here  
    ⇒   an all-pass filter is not a minimum–phase system.


(5)  Both statements  are correct:

  • As already determined in subtask  (2):   A constant attenuation arises as a result whenever there is a corresponding zero in the right half-plane for each pole in the left $p$–half-plane   ⇒   circuit  $\rm B$  also exhibits all-pass filter charakteristics.
  • Since  $b(f)$  is always an asymmetric function,  $b(f= 0) = 0$  holds in general.  That is, for any spectral function  $H(f)$ whose inverse Fourier transform  ("impulse response")  is real.