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

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[[File:EN_LZI_Z_3_4.png|right|frame|All-pass filter in two different variants]]
+
[[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]]
 
We first assume a two-port network with the following transfer function:
 
We first assume a two-port network with the following transfer function:
 
:$$H_{\rm L}(p)=  \frac {1-{p}/{A}} {1+{p}/{A}}.$$
 
:$$H_{\rm L}(p)=  \frac {1-{p}/{A}} {1+{p}/{A}}.$$
From this, the conventional Fourier frequency response
+
From this is to be determined the conventional Fourier frequency response
 
:$$H(f) =  {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)},$$
 
:$$H(f) =  {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)},$$
which is representable by the attenuation function &nbsp;$a(f)$&nbsp; and the phase function &nbsp;$b(f)$, is to be determined.
+
which is representable
 +
*by the attenuation function &nbsp;$a(f)$&nbsp;  
 +
*and the phase function &nbsp;$b(f)$.
  
  
The upper diagram shows a so-called all-pass circuit where the complex resistance &nbsp;$Z_1$&nbsp; denotes an inductance and the complex resistance &nbsp;$Z_2$&nbsp; denotes a capacitance:
+
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:
 
:$$Z_1 = p \cdot L\hspace{0.05cm},\hspace{0.2cm}Z_2 = \frac{1}{p \cdot C}\hspace{0.05cm}.$$
 
:$$Z_1 = p \cdot L\hspace{0.05cm},\hspace{0.2cm}Z_2 = \frac{1}{p \cdot C}\hspace{0.05cm}.$$
For anechoic adjustment at the input and output with
+
For reflection-free adaptation at the input and output with
 
:$$Z_{\rm I}=Z_{\rm A} = \sqrt{Z_1 \cdot Z_2} = \sqrt{{L}/{C}},$$
 
:$$Z_{\rm I}=Z_{\rm A} = \sqrt{Z_1 \cdot Z_2} = \sqrt{{L}/{C}},$$
the following holds for the&nbsp; $p$&ndash;transfer function of the circuit&nbsp; $\rm A$&nbsp; (see upper diagram):
+
the following holds for the&nbsp; $p$&ndash;transfer function of circuit&nbsp; $\rm A$&nbsp; (see upper diagram):
 
:$$H_{\rm L}(p)=  \frac {Y_{\rm L}(p)} {X_{\rm L}(p)}=  \frac {Z_2-Z_1} {Z_1+2 \cdot \sqrt{Z_1 \cdot Z_2}+Z_2}\hspace{0.05cm}.$$
 
:$$H_{\rm L}(p)=  \frac {Y_{\rm L}(p)} {X_{\rm L}(p)}=  \frac {Z_2-Z_1} {Z_1+2 \cdot \sqrt{Z_1 \cdot Z_2}+Z_2}\hspace{0.05cm}.$$
  
The circuit &nbsp;$\rm B$&nbsp; is defined by the &nbsp;$p$&ndash;transfer function.&nbsp; It is characterized by the fact that  
+
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 (in the left $p$&ndash;half-plane)  
+
*all poles&nbsp; (in the left $p$&ndash;half-plane)  
*are located in a mirror-imaged manner with respect to the zeros (in the right half-plane).
+
*are located in a mirror-imaged manner with respect to the zeros&nbsp; (in the right half-plane).
  
  
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+
Please note:  
 
 
 
 
 
 
''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]].
 
*The exercise belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function|Laplace Transform and p-Transfer Function]].
 
   
 
   
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<quiz display=simple>
 
<quiz display=simple>
{Geben Sie die Nullstelle &nbsp;$p_{\rm o}$&nbsp; und den Pol &nbsp;$p_{\rm x}$&nbsp; von &nbsp;$H_{\rm L}(p)=  (1 -p/A)/(1 +p/A)$&nbsp; an. &nbsp;Wie groß ist der konstante Faktor&nbsp; $K$?
+
{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 }
Line 45: Line 43:
  
  
{Berechnen Sie die Dämpfungsfunktion &nbsp;$a(f)$.&nbsp; Welche Aussagen treffen zu?
+
{Compute the attenuation function &nbsp;$a(f)$.&nbsp; Which statements are true?
 
|type="[]"}
 
|type="[]"}
- Die Dämpfungsfunktion &nbsp;$a(f)$&nbsp; zeigt Tiefpassverhalten.
+
- The attenuation function &nbsp;$a(f)$&nbsp; exhibits low-pass filter behavior.
+ Die Dämpfungsfunktion &nbsp;$a(f)$&nbsp; ist konstant.
+
+ The attenuation function &nbsp;$a(f)$&nbsp; is constant.
+ Das obige Ergebnis gilt allgemein für &nbsp;$p_{\rm x}  = - p_{\rm o}$.
+
+ The above result is generally valid for &nbsp;$p_{\rm x}  = - p_{\rm o}$.
  
  
{Berechnen Sie den Phasenverlauf &nbsp;$b(f)$.&nbsp; 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 &nbsp;$p$&ndash;Übertragungsfunktion von Schaltung &nbsp;$\rm A$.&nbsp; 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 &nbsp;$a(f)$&nbsp; ist konstant gleich&nbsp; $0 \ \rm (Np)$.
+
+ The attenuation &nbsp;$a(f)$&nbsp; is constantly equal to&nbsp; $0$&nbsp; (Np).
- Die Phase &nbsp;$b(f)$&nbsp; steigt linear mit der Frequenz &nbsp;$f$&nbsp; an.
+
- The phase &nbsp;$b(f)$&nbsp; increases linearly with frequency &nbsp;$f$&nbsp;.
- $b(f)$&nbsp; ist die Hilbert&ndash;Transformierte von &nbsp;$a(f)$.
+
- $b(f)$&nbsp; is the Hilbert transform of &nbsp;$a(f)$.
  
  
{Welche Aussagen können aus dem Pol&ndash;Nullstellen&ndash;Diagramm von Schaltung &nbsp;$\rm B$&nbsp; abgeleitet werden?
+
{What statements can be derived from the pole&ndash;zero diagram of circuit &nbsp;$\rm B$&nbsp;?
 
|type="[]"}
 
|type="[]"}
+ Die Dämpfung &nbsp;$a(f)$&nbsp; ist konstant.
+
+ The attenuation &nbsp;$a(f)$&nbsp; is constant.
+ Für die Phasenfunktion gilt  &nbsp;$b(f =0) =0$.
+
+ The following holds for the phase function: &nbsp;$b(f =0) =0$.
  
  
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===Solution===
 
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Durch Umformung der angegebenen &nbsp;$p$&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} .$$
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'''(2)'''&nbsp; Richtig sind die <u>Aussagen 2 und 3</u>:
+
'''(2)'''&nbsp; <u>Statements 2 and 3</u>&nbsp; are correct:
*Setzt man &nbsp;$p = {\rm j} \cdot 2 \pi f$, so erhält man:
+
*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
Line 94: Line 92:
 
  f/A)^2}} {\sqrt{1+(2\pi
 
  f/A)^2}} {\sqrt{1+(2\pi
 
  f/A)^2}}= 1\hspace{0.3cm}  
 
  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} .$$
+
\Rightarrow \hspace{0.3cm} a(f)= -{\rm ln} \hspace{0.1cm} |H(f)|= 0\hspace{0.2cm}({\rm  Np \hspace{0.2cm}or \hspace{0.2cm}dB})\hspace{0.05cm} .$$
*Aber auch die &nbsp;$\text{Aussage 3}$&nbsp; 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".
  
  
  
  
'''(3)'''&nbsp; Die Phasenfunktion &nbsp;$b(f)$&nbsp; 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} ({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} ({-2\pi
 
  f}/{A}) - {\rm arctan } \hspace{0.1cm} ({-2\pi
Line 113: Line 111:
 
  \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 nach 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.
  
  
  
  
'''(4)'''&nbsp; Richtig ist nur die <u>Aussage 1</u>:
+
'''(4)'''&nbsp; Only <u>statement 1</u>&nbsp; is correct:
*Die angegebene &nbsp;$p$&ndash;Übertragungsfunktion lässt sich wie folgt darstellen:
+
*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}=
 
:$$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})} {(\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 &nbsp;$Z_1 = p \cdot L$&nbsp; und &nbsp;$Z_2 = 1/(p \cdot C)$&nbsp; 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}}
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\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 gleiche Übertragungsfunktion, wie in der Teilaufgabe&nbsp; '''(1)'''&nbsp; berechnet.  
+
*The same transfer function as computed in subtask&nbsp; '''(1)'''&nbsp; is obtained.  
  
  
Daraus folgt, dass nur die <u>Aussage 1</u> richtig ist:
+
From this it follows that only <u>statement 1</u> is correct:
*Der Dämpfungsverlauf ist &nbsp;$a(f) = 0\ \rm  (Np)$.&nbsp; Keine Frequenz wird gedämpft oder verstärkt.&nbsp; Man spricht deshalb auch von einem "Allpass".
+
*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".
*Die zweite Aussage ist falsch.&nbsp; Der Phasenverlauf &nbsp;$b(f)$&nbsp; ist nicht linear, sondern vielmehr gekrümmt, wie in der Teilaufgabe&nbsp; '''(3)'''&nbsp; berechnet.
+
*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;.
*Die Hilbert&ndash;Transformierte der Konstanten &nbsp;$a(f) = 0$&nbsp; müsste zur Phasenfunktion &nbsp;$b(f) = 0$&nbsp;  führen, wie im Theorieteil gezeigt.&nbsp; Das heißt:&nbsp; die Aussage 3 ist falsch.
+
*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.
*Nur bei minimalphasigen Systemen hängen die Dämpfungsfunktion &nbsp;$a(f)$&nbsp; und Phasenfunktion &nbsp;$b(f)$&nbsp; über die Hilbert&ndash;Transformation zusammen.
+
*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.
*Bei einem solchen Minimum&ndash;Phasen&ndash;System liegen aber alle Pole und Nullstellen in der linken $p$&ndash;Halbebene, was hier nicht zutrifft &nbsp; &#8658; &nbsp; ein Allpass ist kein Minimum&ndash;Phasen&ndash;System.
+
*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'''.
  
  
  
'''(5)'''&nbsp; <u>Beide Aussagen</u> sind richtig:  
+
'''(5)'''&nbsp; <u>Both statements</u>&nbsp; are correct:  
*Wie bereits in der Teilaufgabe&nbsp; '''(2)'''&nbsp; festgestellt wurde, ergibt sich immer dann eine konstante Dämpfung, wenn es zu jedem Pol in der linken $p$&ndash;Halbebene eine entsprechende Nullstelle in der rechten Halbebene gibt &nbsp; &#8658; &nbsp; die Schaltung&nbsp; $\rm B$&nbsp; zeigt ebenfalls Allpass&ndash;Charakteristik.
+
*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.
*Da &nbsp;$b(f)$&nbsp; stets eine unsymmetrische Funktion ist, gilt &nbsp;$b(f= 0) = 0$&nbsp; ganz allgemein.
+
*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.
* Das heißt für jede Spektralfunktion &nbsp;$H(f)$, deren Fourier&ndash;Rücktransformierte&nbsp; ("Impulsantwort") &nbsp;reell ist.
 
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  

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.