Aufgaben:Exercise 3.4: Attenuation and Phase Response - Revision history

Guenter at 15:15, 14 October 2021

2021-10-14T15:15:47Z

Guenter at 14:54, 14 October 2021

2021-10-14T14:54:21Z

Oezer at 16:44, 12 October 2021

2021-10-12T16:44:01Z

Oezer at 16:42, 12 October 2021

2021-10-12T16:42:58Z

Oezer at 16:41, 12 October 2021

2021-10-12T16:41:11Z

Oezer at 16:36, 12 October 2021

2021-10-12T16:36:31Z

Oezer at 15:51, 12 October 2021

2021-10-12T15:51:29Z

Oezer at 15:49, 12 October 2021

2021-10-12T15:49:23Z

Oezer at 15:42, 12 October 2021

2021-10-12T15:42:05Z

Oezer at 15:13, 12 October 2021

2021-10-12T15:13:29Z

@@ Line 80: / Line 80: @@
   \hspace{0.05cm} .$$
-*The conventional transfer function (the frequency response) is obtained via the substitution &nbsp;$p = {\rm j} \cdot 2 \pi f$:
+*The conventional transfer function&nbsp; (the frequency response)&nbsp; is obtained via the substitution &nbsp;$p = {\rm j} \cdot 2 \pi f$:
 :$$H(f)= K \cdot \frac {{\rm j \hspace{0.05cm}2\pi \it
   f} - p_{\rm o }}
@@ Line 89: / Line 89: @@
   \hspace{0.05cm} .$$
-*In the limiting case &nbsp;$f &#8594; \infty$,&nbsp; the following is obtained for the magnitude, attenuation and phase:
+*In the limiting case &nbsp;$f &#8594; \infty$,&nbsp; the following is obtained for the magnitude,&nbsp; attenuation and phase:
 :$$\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty}   H(f)= \frac{K}{{\rm j \hspace{0.05cm}2\pi \it
   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}
@@ Line 100: / Line 100: @@
 '''(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|$|H(f)|$,&nbsp; $a(f)$&nbsp; and&nbsp; $b(f)$]]
+[[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; $b(f)$]]
 :$$|H(f=0)|=  -\frac {K \cdot p_{\rm o }}
   {p_{\rm x 1}\cdot p_{\rm x 2}}
@@ Line 110: / Line 110: @@
   \hspace{0.05cm} .$$
-The screen capture of the Flash&ndash;module "Causal Systems" summarizes the results of this exercise:
+The screen capture of the Flash&ndash;module&nbsp; "Causal Systems"&nbsp; summarizes the results of this exercise:
-:*middle axis (blue): &nbsp; magnitude $|H(f)|$,
+:*middle axis (blue): &nbsp; magnitude $|H(f)|$, &nbsp; &rArr; &nbsp; here labeled with&nbsp; $|Y(f)|$,
-:*left axis (red): &nbsp; attenuation $a(f)$,
+:*left axis (red): &nbsp; attenuation&nbsp;  $a(f)$,
-:*right axis (green): &nbsp; phase $b(f)$.
+:*right axis (green): &nbsp; phase&nbsp; $b(f)$.
-:*black point: &nbsp; values for $2\pi f = 4.$
+:*black point: &nbsp; values for&nbsp; $2\pi f = 4.$
-<br clear=all>
 [[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:
@@ Line 122: / Line 123: @@
   + {\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} .$$
-*The additional unit "neper" $\rm (Np)$ is further to be considered.
+*Furthermore,&nbsp; the additional unit&nbsp;  "neper"&nbsp; $\rm (Np)$&nbsp; must be taken into account.
+*The attenuation at &nbsp;$f = 2/\pi$&nbsp; is searched-for.&nbsp; For this, we set &nbsp;$p = {\rm j} \cdot 2 \pi f = 4$&nbsp; and determine the following distances:
 :$$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}|
@@ Line 142: / Line 142: @@
   \hspace{0.05cm}.$$
-This is equivalent to&nbsp; $0.155\  {\rm Np} \cdot  8.686 \  {\rm dB/Np} \hspace{0.15cm} \underline{= 1.346 \ {\rm dB}}$.
+This is equivalent to &nbsp; $0.155\  {\rm Np} \cdot  8.686 \  {\rm dB/Np} \hspace{0.15cm} \underline{= 1.346 \ {\rm dB}}$.
-<br clear=all>
 '''(4)'''&nbsp; The following holds for the phase function according to the description in the theory section due to &nbsp;$K > 0$&nbsp;:
 :$$b(f ={2}/\pi) = \phi_{\rm x1} + \phi_{\rm x2}-\phi_{\rm o}\hspace{0.05cm},$$
 :$$\phi_{\rm x1} ={\rm arctan}\hspace{0.15cm}(1/3) =
-.4^\circ\hspace{0.05cm}, \hspace{0.2cm}\phi_{\rm x2} = {\rm arctan}\hspace{0.15cm}(7/3)  =
+.4^\circ\hspace{0.05cm}, \hspace{0.5cm}\phi_{\rm x2} = {\rm arctan}\hspace{0.15cm}(7/3)  =
-.8^\circ\hspace{0.05cm},$$
+.8^\circ\hspace{0.05cm},\hspace{0.5cm} \phi_{\rm o} = {\rm arctan}\hspace{0.15cm}(-1/4) =
-:$$ \phi_{\rm o} = {\rm arctan}\hspace{0.15cm}(-1/4) =
+^\circ - 76^\circ = 104^\circ $$
-^\circ - 76^\circ = 104^\circ \hspace{0.3cm}
+:$$ \Rightarrow \hspace{0.3cm}b(f ={2}/\pi) =
 .4^\circ + 66.8^\circ - 104^\circ  \hspace{0.15cm} \underline{= -18.8^\circ}
   \hspace{0.05cm}.$$

@@ Line 8: / Line 8: @@
 :$$ p_{\rm  o}= 1,\hspace{0,2cm}p_{\rm  x1}= -3 + 3{\rm j},\hspace{0,2cm}p_{\rm  x2}= -3 - 3{\rm j}\hspace{0.05cm} .$$
-Thus, the&nbsp; $p$&ndash;transfer function is:
+Thus,&nbsp; the&nbsp; $p$&ndash;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 &nbsp;$p = {\rm j} \cdot 2 \pi f$,&nbsp; the conventional transfer function can be specified, which is also called frequency response:
+Considering the substitution &nbsp;$p = {\rm j} \cdot 2 \pi f$,&nbsp; the conventional transfer function can be specified,&nbsp; which is also called&nbsp; "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, the relationship between
+From this equation it can also be seen the relationship between
 *the transfer function &nbsp;$H(f)$,
 *the attenuation function &nbsp;$a(f)$&nbsp; and
-*the phase function &nbsp;$b(f)$
+*the phase function &nbsp;$b(f)$.
-can also be seen.
-The attenuation and phase values can be determined as follows for a frequency &nbsp;$f$&nbsp; indirectly specified by the point&nbsp;$p = {\rm j} \cdot 2 \pi f$&nbsp;:
+The attenuation and phase values can be determined as follows for a frequency &nbsp;$f$&nbsp; indirectly specified by the point&nbsp; $p = {\rm j} \cdot 2 \pi f$&nbsp;:
 :$$a(f)\hspace{0.15cm}{\rm in}\hspace{0.15cm}{\rm Np} \hspace{0.25cm} =  \hspace{0.2cm} -{\rm ln} \hspace{0.1cm} K
   + {\rm ln} \hspace{0.1cm} |R_{\rm x1}|+{\rm ln} \hspace{0.1cm} |R_{\rm x1}|- {\rm ln} \hspace{0.1cm} |R_{{\rm o} }|\hspace{0.05cm}
@@ Line 33: / Line 32: @@
   \hspace{0.05cm} .$$
-The corresponding magnitudes &nbsp;$|R_{\rm o}|$,  &nbsp;$|R_{\rm x1}|$ &nbsp;and&nbsp; $|R_{\rm x1}|$ as well as the angles &nbsp;$\phi_{\rm o}$, &nbsp;$\phi_{\rm x1}$ &nbsp;and&nbsp; $\phi_{\rm x2}$&nbsp; can be taken from the graph .
+The corresponding magnitudes &nbsp;$|R_{\rm o}|$,  &nbsp;$|R_{\rm x1}|$ &nbsp;and&nbsp; $|R_{\rm x2}|$ as well as the angles &nbsp;$\phi_{\rm o}$, &nbsp;$\phi_{\rm x1}$ &nbsp;and&nbsp; $\phi_{\rm x2}$&nbsp; can be taken from the graph .
@@ Line 40: / Line 38: @@
-''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]].
@@ Line 61: / Line 59: @@
-{Compute the attenuation value at &nbsp;$f =4/(2 \pi)$&nbsp; in neper&nbsp; $\rm(Np)$&nbsp; and decibel&nbsp; $\rm(dB)$ according to the described approach.
+{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="{}"}
 $a(f = 2/ \pi)\ = \ $ { 0.155 3% } $\ \rm Np$

@@ Line 119: / Line 119: @@
 <br clear=all>
 [[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; 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:
 :$$a(f)=  -{\rm ln} \hspace{0.1cm} K

@@ Line 91: / Line 91: @@
   \hspace{0.05cm} .$$
-*In the limiting case &nbsp;$f &#8594; \infty$&nbsp; the following is obtained for the magnitude, attenuation and phase:
+*In the limiting case &nbsp;$f &#8594; \infty$,&nbsp; the following is obtained for the magnitude, attenuation and phase:
 :$$\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}

@@ Line 146: / Line 146: @@
 This is equivalent to&nbsp; $0.155\  {\rm Np} \cdot  8.686 \  {\rm dB/Np} \hspace{0.15cm} \underline{= 1.346 \ {\rm dB}}$.
 <br clear=all>
-'''(4)'''&nbsp; Nach der Beschreibung im Theorieteil gilt wegen &nbsp;$K > 0$&nbsp; 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;:
 :$$b(f ={2}/\pi) = \phi_{\rm x1} + \phi_{\rm x2}-\phi_{\rm o}\hspace{0.05cm},$$
 :$$\phi_{\rm x1} ={\rm arctan}\hspace{0.15cm}(1/3) =

@@ Line 50: / Line 50: @@
 <quiz display=simple>
-{Berechnen Sie &nbsp;$H(f)$.&nbsp; Wie groß ist dessen Betrag bei sehr großen Frequenzen?
+{Compute &nbsp;$H(f)$.&nbsp; What is its magnitude at very large frequencies?
 |type="{}"}
 $|H(f &#8594; &#8734;)| \ = \ $ { 0. }
-{Berechnen Sie den Betragsfrequenzgang und den Dämpfungswert für &nbsp;$f &#8594 0$.
+{Compute the magnitude frequency response and the attenuation value for &nbsp;$f &#8594 0$.
 |type="{}"}
 $|H(f = 0)| \ = \ $ { 0.278 3% }
@@ Line 61: / Line 61: @@
-{Berechnen Sie gemäß der beschriebenen Vorgehensweise den Dämpfungswert bei &nbsp;$f =4/(2 \pi)$&nbsp; in Neper&nbsp; $\rm(Np)$&nbsp; und Dezibel&nbsp; $\rm(dB)$.
+{Compute the attenuation value at &nbsp;$f =4/(2 \pi)$&nbsp; in neper&nbsp; $\rm(Np)$&nbsp; and decibel&nbsp; $\rm(dB)$ according to the described approach.
 |type="{}"}
 $a(f = 2/ \pi)\ = \ $ { 0.155 3% } $\ \rm Np$
@@ Line 67: / Line 67: @@
-{Berechnen Sie gemäß der beschriebenen Vorgehensweise den Phasenwert bei der Frequenz &nbsp;$f = 4/(2 \pi)$.
+{Compute the phase value at frequency &nbsp;$f = 4/(2 \pi)$ according to the described approach.
 |type="{}"}
 $b(f = 2/ \pi)\ = \ $ { -19.3--18.3 } $\ \rm Grad$
@@ Line 77: / Line 77: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Die&nbsp; $p$&ndash;Übertragungsfunktion lautet:
+'''(1)'''&nbsp; The&nbsp; $p$&ndash;transfer function is:
 :$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }}
   {(p - p_{\rm x 1})(p - p_{\rm x 2})}
   \hspace{0.05cm} .$$
-*Zur herkömmlichen Übertragungsfunktion (zum Frequenzgang) kommt man mit der Substitution &nbsp;$p = {\rm j} \cdot 2 \pi f$:
+*The conventional transfer function (the frequency response) is obtained via the substitution &nbsp;$p = {\rm j} \cdot 2 \pi f$:
 :$$H(f)= K \cdot \frac {{\rm j \hspace{0.05cm}2\pi \it
   f} - p_{\rm o }}
@@ Line 91: / Line 91: @@
   \hspace{0.05cm} .$$
-*Im Grenzfall &nbsp;$f &#8594; \infty$&nbsp; 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, attenuation and phase:
 :$$\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty}   H(f)= \frac{K}{{\rm j \hspace{0.05cm}2\pi \it
   f}}\hspace{0.15cm}\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty}   |H(f)|\hspace{0.15cm}\underline {= 0} \hspace{0.05cm}
@@ Line 101: / Line 101: @@
-'''(2)'''&nbsp; Aus der allgemeinen Gleichung in Teilaufgabe&nbsp; '''(1)'''&nbsp; erhält man mit dem Grenzübergang &nbsp;$f &#8594 0$:
+'''(2)'''&nbsp; From the general equation in subtask&nbsp; '''(1)'''&nbsp; the following is obtained with the limit process &nbsp;$f &#8594 0$:
-[[File:P_ID1769__LZI_A_3_4_d_neu.png|right|frame|$|H(f)|$,&nbsp; $a(f)$&nbsp; und&nbsp; $b(f)$]]
+[[File:P_ID1769__LZI_A_3_4_d_neu.png|right|frame|$|H(f)|$,&nbsp; $a(f)$&nbsp; and&nbsp; $b(f)$]]
 :$$|H(f=0)|=  -\frac {K \cdot p_{\rm o }}
   {p_{\rm x 1}\cdot p_{\rm x 2}}
@@ Line 112: / Line 112: @@
   \hspace{0.05cm} .$$
-Der Bildschirmabzug des Flash&ndash;Moduls "Kausale Systeme" fasst die Ergebnisse dieser Aufgabe zusammen:
+The screen capture of the Flash&ndash;module "Causal Systems" summarizes the results of this exercise:
-:*Mittlere Achse (blau): &nbsp; Betrag $|H(f)|$,
+:*middle axis (blue): &nbsp; magnitude $|H(f)|$,
-:*Linke Achse (rot): &nbsp; Dämpfung $a(f)$,
+:*left axis (red): &nbsp; attenuation $a(f)$,
-:*Rechte Achse (grün): &nbsp; Phase $b(f)$.
+:*right axis (green): &nbsp; phase $b(f)$.
-:*Schwarzer Punkt: &nbsp; Werte für $2\pi f = 4.$
+:*black point: &nbsp; values for $2\pi f = 4.$
 <br clear=all>
-[[File:P_ID2843__LZI_A_3_4.png|right|frame|Pol–Nullstellen–Diagramm und einige Hilfsgrößen]]
+[[File:P_ID2843__LZI_A_3_4.png|right|frame|Pole-zero diagram and some auxiliary quantities]]]
 '''(3)'''&nbsp; Entsprechend der detaillierten Beschreibung im &nbsp;[[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Graphical_determination_of_attenuation_and_phase|Theorieteil]]&nbsp; gilt für die Dämpfungsfunktion: