Aufgaben:Exercise 3.2: Laplace Transform - Revision history

Guenter at 11:44, 13 October 2021

2021-10-13T11:44:17Z

Guenter at 11:38, 13 October 2021

2021-10-13T11:38:08Z

Oezer at 15:13, 10 October 2021

2021-10-10T15:13:28Z

Oezer at 15:07, 10 October 2021

2021-10-10T15:07:20Z

Oezer at 15:06, 10 October 2021

2021-10-10T15:06:58Z

Oezer at 15:03, 10 October 2021

2021-10-10T15:03:59Z

Oezer at 14:20, 10 October 2021

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Oezer at 14:17, 10 October 2021

2021-10-10T14:17:51Z

Oezer at 14:08, 10 October 2021

2021-10-10T14:08:04Z

Oezer at 14:03, 10 October 2021

2021-10-10T14:03:28Z

@@ Line 103: / Line 103: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; <u>Suggested solution 2</u> is correct:
+'''(1)'''&nbsp; <u>Suggested solution 2</u>&nbsp; is correct:
-*According to the Laplace&ndash;definition, the following holds with the given equations:
+*According to the Laplace definition, the following holds with the given equations:
 :$$X_{\rm L}(p) =    \int_{0}^{
 \infty}
@@ Line 117: / Line 117: @@
-'''(2)'''&nbsp; <u>Suggested solution 1</u> is correct:
+'''(2)'''&nbsp; <u>Suggested solution 1</u>&nbsp; is correct:
 *Here, the following holds using the same approach as in the subtask&nbsp; '''(1)''':
 :$$Y_{\rm L}(p) =    \int_{0}^{
@@ Line 127: / Line 127: @@
-'''(3)'''&nbsp; <u>Suggested solution 3</u> is correct:
+'''(3)'''&nbsp; <u>Suggested solution 3</u>&nbsp; is correct:
 *The&nbsp; $p$&ndash;transfer function of the causal&nbsp; $\rm si$&ndash;function is as follows considering the integral given above:
 :$$Z_{\rm L}(p) =    \int_{0}^{
@@ Line 135: / Line 135: @@
   \hspace{0.05cm} .$$
 *Suggestion 1 only applies to the Fourier transform of the non-causal&nbsp; $\rm si$&ndash;function.
-*Since here the argument of the arctangent&ndash;function is dimensional, suggestion 2 cannot be true for this reason alone.
+*Since here the argument of the&nbsp; $\rm arctan$&nbsp; function is dimensional, suggestion 2 cannot be true for this reason alone.
-'''(4)'''&nbsp; <u>Suggested solution 1</u> is correct:
+'''(4)'''&nbsp; <u>Suggested solution 1</u>&nbsp; is correct:
 *The following arises as a result from&nbsp; $z(t) = s(t) \cdot \gamma(t)$&nbsp; with the convolution theorem:
 :$$Z(f) = S(f) \star {\it \Gamma}(f) = {1}/{2}
@@ Line 149: / Line 149: @@
 \hspace{0.05cm}.$$
-*The real part of&nbsp; $Z(f)$&nbsp; thus has the same rectangular shape as&nbsp; $S(f)$, but it is only half as high:
+*The real part of&nbsp; $Z(f)$&nbsp; thus has the same rectangular shape as&nbsp; $S(f)$,&nbsp; but it is only half as high:
 :$${\rm Re}\{ Z(f)\}= \left\{ \begin{array}{c} T/2 \\
    \end{array} \right.
@@ Line 162: / Line 162: @@
-'''(5)'''&nbsp; <u>Suggested solution 2</u> is correct:
+'''(5)'''&nbsp; <u>Suggested solution 2</u>&nbsp; is correct:
-*With the result of the last subtask, it follows for the imaginary part:
+*With the result of the last subtask,&nbsp; it follows for the imaginary part:
 :$${\rm Im}\{ Z(f)\} = S(f) \star \frac{(-1)}{{\rm j} \cdot
 \pi f} \hspace{0.05cm}.$$
-*This convolution integral yields the following for sufficiently large frequencies&nbsp; $f \ge 1/(2T)$&nbsp;:
+*This convolution integral yields the following for sufficiently large frequencies&nbsp; $f \ge 1/(2T)$:
 :$${\rm Im}\{ Z(f)\} = -T \cdot \int_{f- 1/(2T)}^{
 f+ 1/(2T)} {  \frac{1}{2\pi x}}\hspace{0.1cm}{\rm

@@ Line 4: / Line 4: @@
 [[File:P_ID1763__LZI_A_3_2.png|right|frame|Three causal time functions]]
-Causal signals and systems are usually described by means of the Laplace transformation. If&nbsp; $x(t)$&nbsp; is identical to zero for all times&nbsp; $t < 0$,&nbsp; then the Laplace transform is:
+Causal signals and systems are usually described by means of the Laplace transformation.&nbsp; If&nbsp; $x(t)$&nbsp; is identical to zero for all times&nbsp; $t < 0$,&nbsp; then the Laplace transform is:
 :$$X_{\rm L}(p) =    \int_{0}^{
 \infty}
@@ Line 19: / Line 19: @@
 *$\sin(t)/t$&ndash;signal with equivalent zero-crossings at a distance of&nbsp; $T$:
-:$$z(t) =  {\rm si} (\pi \cdot {t}/{T})\hspace{0.2cm}{\rm with}\hspace{0.2cm}{\rm si}(x)= {\rm sin}(x)/x
+:$$z(t) =  {\rm si} (\pi \cdot {t}/{T})=  {\rm sinc} ({t}/{T})\hspace{0.4cm}{\rm with}\hspace{0.4cm}{\rm si}(x)= {\rm sin}(x)/x ={\rm sinc}(x)/\pi \hspace{0.05cm}.$$
-The following equation can<u>not</u> be used to calculate the spectral function since&nbsp; $z(t)$&nbsp; is not energy-limited just as the signals&nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp;:
+The following equation can'''<u>not</u>''' be used to calculate the spectral function since&nbsp; $z(t)$&nbsp; is not energy-limited just as the signals&nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp;:
 :$$Z(f) =  Z_{\rm L}(p)\Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it
   f}} .$$
@@ Line 36: / Line 35: @@
 \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad {\it \Gamma}(f) = {1}/{2}
 \cdot \delta (f) + \frac{1}{{\rm j} \cdot 2\pi f}\hspace{0.05cm}.$$
@@ Line 48: / Line 41: @@
 ''Please note:''
 *The exercise belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function|Laplace Transform and p-Transfer Function]].
 *The following definite integrals are given:
 :$$\int_{0}^{
@@ Line 86: / Line 80: @@
 {Compute the Laplace transform&nbsp; $Z_{\rm L}(p)$&nbsp; of the causal&nbsp; $\rm si$&ndash;function&nbsp;  $z(t)$.&nbsp; What is the correct solution?
 |type="()"}
-- $Z_{\rm L}(p)$ has a rectangular shape.
+- $Z_{\rm L}(p)$&nbsp; has a rectangular shape.
 - $Z_{\rm L}(p) = \arctan (1/p)$.
 + $Z_{\rm L}(p) = T/\pi \cdot \arctan (\pi/(pT))$.
@@ Line 93: / Line 87: @@
 {Compute the real part of the spectrum&nbsp; $Z(f)$.&nbsp; Which statements are true?
 |type="[]"}
-+ ${\rm Re}\big[Z(f)\big]$ has a rectangular shape.
++ ${\rm Re}\big[Z(f)\big]$&nbsp; has a rectangular shape.
-- ${\rm Re}\big[Z(f)\big]$ is proportional to $\ln\; \big|(f \cdot T -0.5)/(f \cdot T +0.5)\big|.$
+- ${\rm Re}\big[Z(f)\big]$&nbsp; is proportional to $\ln\; \big|(f \cdot T -0.5)/(f \cdot T +0.5)\big|.$

@@ Line 164: / Line 164: @@
 { |f|> 1/(2T)\hspace{0.05cm},}
 \end{array}
-\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\underline{{\rm Vorschlag \hspace{0.15cm} 1}}.$$
+\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\underline{{\rm suggestion \hspace{0.15cm} 1}}.$$

@@ Line 28: / Line 28: @@
 Rather, the fact that&nbsp; $z(t) =  s(t) \cdot \gamma(t)$&nbsp; holds is to be considered where&nbsp; $s(t)$&nbsp; denotes the conventional symmetric&nbsp; $\rm si$&ndash;function here:
 :$$s(t) =  {\rm si} (\pi \cdot {t}/{T}) \quad
-\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad S(f)$$.
+\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad S(f).$$
 $S(f)$&nbsp; is a rectangular function symmetric about&nbsp; $f  = 0$&nbsp; with height&nbsp; $T$&nbsp; and width&nbsp; $1/T$.

@@ Line 28: / Line 28: @@
 Rather, the fact that&nbsp; $z(t) =  s(t) \cdot \gamma(t)$&nbsp; holds is to be considered where&nbsp; $s(t)$&nbsp; denotes the conventional symmetric&nbsp; $\rm si$&ndash;function here:
 :$$s(t) =  {\rm si} (\pi \cdot {t}/{T}) \quad
-\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad S(f)$$
+\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad S(f)$$.
 $S(f)$&nbsp; is a rectangular function symmetric about&nbsp; $f  = 0$&nbsp; with height&nbsp; $T$&nbsp; and width&nbsp; $1/T$.

@@ Line 26: / Line 26: @@
   f}} .$$
-Rather, the fact that&nbsp; $z(t) =  s(t) \cdot \gamma(t)$&nbsp; is valid is to be considered where&nbsp; $s(t)$&nbsp; denotes the conventional symmetric&nbsp; $\rm si$&ndash;function here:
+Rather, the fact that&nbsp; $z(t) =  s(t) \cdot \gamma(t)$&nbsp; holds is to be considered where&nbsp; $s(t)$&nbsp; denotes the conventional symmetric&nbsp; $\rm si$&ndash;function here:
 :$$s(t) =  {\rm si} (\pi \cdot {t}/{T}) \quad
 \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad S(f)$$
@@ Line 91: / Line 91: @@
-{Berechnen Sie den Realteil des Spektrums&nbsp; $Z(f)$.&nbsp; Welche Aussagen treffen zu?
+{Compute the real part of the spectrum&nbsp; $Z(f)$.&nbsp; Which statements are true?
 |type="[]"}
-+ ${\rm Re}\big[Z(f)\big]$ hat einen rechteckförmigen Verlauf.
++ ${\rm Re}\big[Z(f)\big]$ has a rectangular shape.
-- ${\rm Re}\big[Z(f)\big]$ ist proportional zu $\ln\; \big|(f \cdot T -0.5)/(f \cdot T +0.5)\big|.$
+- ${\rm Re}\big[Z(f)\big]$ is proportional to $\ln\; \big|(f \cdot T -0.5)/(f \cdot T +0.5)\big|.$
-{Berechnen Sie den Imaginärteil von&nbsp; $Z(f)$.&nbsp; Welche Aussagen treffen zu?
+{Compute the imaginary part of&nbsp; $Z(f)$.&nbsp; Which statements are true?
 |type="[]"}
-- ${\rm Im}\big[Z(f)\big]$&nbsp; hat einen rechteckförmigen Verlauf.
+- ${\rm Im}\big[Z(f)\big]$&nbsp; has a rectangular shape.
-+ ${\rm Im}\big[Z(f)\big]$&nbsp; ist proportional zu&nbsp; $\ln\; \big|(f \cdot T -0.5)/(f \cdot T +0.5)\big|.$
++ ${\rm Im}\big[Z(f)\big]$&nbsp; is proportional to&nbsp; $\ln\; \big|(f \cdot T -0.5)/(f \cdot T +0.5)\big|.$
@@ Line 109: / Line 109: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Richtig ist der <u>Vorschlag 2</u>:
+'''(1)'''&nbsp; <u>Suggested solution 2</u> is correct:
-*Entsprechend der Laplace&ndash;Definition gilt mit den vorgegebenen Gleichungen:
+*According to the Laplace&ndash;definition, the following holds with the given equations:
 :$$X_{\rm L}(p) =    \int_{0}^{
 \infty}
@@ Line 119: / Line 119: @@
   d}t = \frac{p}{p^2 + \omega_0^2}
   \hspace{0.05cm} .$$
-*Der Vorschlag 3 scheidet aus, da&nbsp; $X_{\rm L}(p)$&nbsp; die Einheit "Sekunde" aufweisen muss (Integral über die Zeit), während&nbsp; $p$&nbsp; und&nbsp; $\omega_0$&nbsp; jeweils die Einheit "1/s"  besitzen.
+*Suggestion 3 is ruled out since&nbsp; $X_{\rm L}(p)$&nbsp; must have the unit "second" (integral over time) while&nbsp; $p$&nbsp; and&nbsp; $\omega_0$&nbsp; each have the unit "1/s".
-'''(2)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 1</u>:
+'''(2)'''&nbsp; <u>Suggested solution 1</u> is correct:
-*Hier gilt bei gleicher Vorgehensweise wie in der Teilaufgabe&nbsp; '''(1)''':
+*Here, the following holds using the same approach as in the subtask&nbsp; '''(1)''':
 :$$Y_{\rm L}(p) =    \int_{0}^{
 \infty}
@@ Line 133: / Line 133: @@
-'''(3)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 3</u>:
+'''(3)'''&nbsp; <u>Suggested solution 3</u> is correct:
-*Die&nbsp; $p$&ndash;Übertragungsfunktion der kausalen&nbsp; $\rm si$&ndash;Funktion lautet mit dem vorne angegebenen Integral:
+*The&nbsp; $p$&ndash;transfer function of the causal&nbsp; $\rm si$&ndash;function is as follows considering the integral given above:
 :$$Z_{\rm L}(p) =    \int_{0}^{
 \infty}
@@ Line 140: / Line 140: @@
   d}t = \frac{T}{\pi} \cdot {\rm arctan} \; \frac{\pi}{p\cdot T}
   \hspace{0.05cm} .$$
-*Der Vorschlag 1 gilt nur für die Fouriertransformierte der akausalen&nbsp; $\rm si$&ndash;Funktion.
+*Suggestion 1 only applies to the Fourier transform of the non-causal&nbsp; $\rm si$&ndash;function.
-*Der Vorschlag 2 kann schon allein deshalb nicht stimmen, da hier das Argument der Arcustangens&ndash;Funktion dimensionsbehaftet ist.
+*Since here the argument of the arctangent&ndash;function is dimensional, suggestion 2 cannot be true for this reason alone.
-'''(4)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 1</u>:
+'''(4)'''&nbsp; <u>Suggested solution 1</u> is correct:
-*Aus&nbsp; $z(t) = s(t) \cdot \gamma(t)$&nbsp; folgt mit dem Faltungssatz:
+*The following arises as a result from&nbsp; $z(t) = s(t) \cdot \gamma(t)$&nbsp; with the convolution theorem:
 :$$Z(f) = S(f) \star {\it \Gamma}(f) = {1}/{2}
 \cdot S(f) \star \delta (f) + S(f) \star \frac{1}{{\rm j} \cdot
 \pi f}\hspace{0.05cm}.$$
-*Da&nbsp; $S(f)$&nbsp; reell ist, ergibt sich der Realteil von&nbsp; $Z(f)$&nbsp; als der erste Term dieser Gleichung:
+*Since&nbsp; $S(f)$&nbsp; is real, the real part of&nbsp; $Z(f)$&nbsp; is obtained as the first term of this equation:
 :$${\rm Re}[ Z(f)] =  {1}/{2}
 \cdot S(f) \star \delta (f) = {1}/{2} \cdot S(f)
 \hspace{0.05cm}.$$
-*Der Realteil von&nbsp; $Z(f)$&nbsp; hat somit die gleiche Rechteckform wie&nbsp; $S(f)$, ist aber nur halb so hoch:
+*The real part of&nbsp; $Z(f)$&nbsp; thus has the same rectangular shape as&nbsp; $S(f)$, but it is only half as high:
 :$${\rm Re}\{ Z(f)\}= \left\{ \begin{array}{c} T/2 \\
    \end{array} \right.
@@ Line 168: / Line 168: @@
-'''(5)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 2</u>:
+'''(5)'''&nbsp; <u>Suggested solution 2</u> is correct:
-*Mit dem Ergebnis der letzten Teilaufgabe folgt für den Imaginärteil:
+*With the result of the last subtask, it follows for the imaginary part:
 :$${\rm Im}\{ Z(f)\} = S(f) \star \frac{(-1)}{{\rm j} \cdot
 \pi f} \hspace{0.05cm}.$$
-*Für hinreichend große Frequenzen&nbsp; $f \ge 1/(2T)$&nbsp; liefert dieses Faltungsintegral:
+*This convolution integral yields the following for sufficiently large frequencies&nbsp; $f \ge 1/(2T)$&nbsp;:
 :$${\rm Im}\{ Z(f)\} = -T \cdot \int_{f- 1/(2T)}^{
 f+ 1/(2T)} {  \frac{1}{2\pi x}}\hspace{0.1cm}{\rm
   d}x =  \frac{T}{2\pi } \cdot {\rm ln}\hspace{0.15cm}\left |\frac{f- 1/(2T)}{f+ 1/(2T)}\right |
   \hspace{0.05cm}
-\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\underline{{\rm Vorschlag \hspace{0.15cm} 2}}.$$
+\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\underline{{\rm suggestion \hspace{0.15cm} 2}}.$$

@@ Line 71: / Line 71: @@
 <quiz display=simple>
-{Compute the Laplace transform $X_{\rm L}(p)$&nbsp; of the causal cosine function&nbsp; $x(t)$.&nbsp; What is the correct solution?
+{Compute the Laplace transform&nbsp; $X_{\rm L}(p)$&nbsp; of the causal cosine function&nbsp; $x(t)$.&nbsp; What is the correct solution?
 |type="()"}
 - $X_{\rm L}(p) = \omega_0/(p^2 + \omega_0^2)$.
@@ Line 77: / Line 77: @@
 - $X_{\rm L}(p) = 1/(p^2 + \omega_0^2)$.
-{Berechnen Sie die Laplace&ndash;Transformierte&nbsp; $Y_{\rm L}(p)$&nbsp; der kausalen Sinusfunktion&nbsp;  $y(t)$.&nbsp; Wie lautet die richtige Lösung?
+{Compute the Laplace transform&nbsp; $Y_{\rm L}(p)$&nbsp; of the causal sine function&nbsp; $y(t)$.&nbsp; What is the correct solution?
 |type="()"}
 + $Y_{\rm L}(p) = \omega_0/(p^2 + \omega_0^2)$.
@@ Line 84: / Line 84: @@
-{Berechnen Sie die Laplace&ndash;Transformierte&nbsp; $Z_{\rm L}(p)$&nbsp; der kausalen&nbsp; $\rm si$&ndash;Funktion&nbsp;  $z(t)$.&nbsp; Wie lautet die richtige Lösung?
+{Compute the Laplace transform&nbsp; $Z_{\rm L}(p)$&nbsp; of the causal&nbsp; $\rm si$&ndash;function&nbsp;  $z(t)$.&nbsp; What is the correct solution?
 |type="()"}
-- $Z_{\rm L}(p)$ hat einen rechteckförmigen Verlauf.
+- $Z_{\rm L}(p)$ has a rectangular shape.
 - $Z_{\rm L}(p) = \arctan (1/p)$.
 + $Z_{\rm L}(p) = T/\pi \cdot \arctan (\pi/(pT))$.