Aufgaben:Exercise 1.7Z: Overall Systems Analysis - Revision history

Oezer at 12:48, 8 September 2021

2021-09-08T12:48:32Z

Oezer at 12:28, 8 September 2021

2021-09-08T12:28:32Z

Oezer at 12:27, 8 September 2021

2021-09-08T12:27:59Z

Oezer at 12:27, 8 September 2021

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Oezer at 12:26, 8 September 2021

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Oezer at 12:22, 8 September 2021

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Oezer at 12:20, 8 September 2021

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Oezer at 12:18, 8 September 2021

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Oezer at 11:21, 8 September 2021

2021-09-08T11:21:34Z

Oezer at 11:16, 8 September 2021

2021-09-08T11:16:04Z

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 ''Please note:''
 *The exercise belongs to the chapter &nbsp; [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory|Some Low-Pass Functions in Systems Theory]].
-*Reference is made especially to the page&nbsp;   [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory#Gaussian_low-pass_filter|Gaussian low-pass filter]].
+*In particular, reference is made to the page&nbsp;   [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory#Gaussian_low-pass_filter|Gaussian low-pass filter]].

@@ Line 62: / Line 62: @@
 *For this to be possible here, nonlinearity must not play a role.
 *That is, it must be ensured that&nbsp; $|x(t)|$&nbsp; is not greater than&nbsp; $4 \ \rm V$&nbsp;.
-*In contrast to this, the second statement is not true: &nbsp; The bandwidth of&nbsp; $H_3(f)$&nbsp; does not affect whether the non-linearity can be eliminated or not.
+*In contrast to this, the second statement is not true: &nbsp;The bandwidth of&nbsp; $H_3(f)$&nbsp; does not affect whether the non-linearity can be eliminated or not.

@@ Line 60: / Line 60: @@
 '''(1)'''&nbsp; <u>Answers 1 and 3</u> are correct:
 *The first statement is correct: &nbsp; A frequency response can only be specified for a linear system.
-*For this to be possible here nonlinearity must not play a role.
+*For this to be possible here, nonlinearity must not play a role.
 *That is, it must be ensured that&nbsp; $|x(t)|$&nbsp; is not greater than&nbsp; $4 \ \rm V$&nbsp;.
 *In contrast to this, the second statement is not true: &nbsp; The bandwidth of&nbsp; $H_3(f)$&nbsp; does not affect whether the non-linearity can be eliminated or not.

@@ Line 59: / Line 59: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; <u>Answers 1 and 3</u> are correct:
-*The first statement is correct: &nbsp;Only for a linear system a frequency response can be specified.
+*The first statement is correct: &nbsp; A frequency response can only be specified for a linear system.
 *For this to be possible here nonlinearity must not play a role.
 *That is, it must be ensured that&nbsp; $|x(t)|$&nbsp; is not greater than&nbsp; $4 \ \rm V$&nbsp;.

@@ Line 59: / Line 59: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; <u>Answers 1 and 3</u> are correct:
-*The first statement is correct: &nbsp; Only for a linear system a frequency response can be specified.
+*The first statement is correct: &nbsp;Only for a linear system a frequency response can be specified.
 *For this to be possible here nonlinearity must not play a role.
 *That is, it must be ensured that&nbsp; $|x(t)|$&nbsp; is not greater than&nbsp; $4 \ \rm V$&nbsp;.

@@ Line 16: / Line 16: @@
 :$$w(t) = {5\,\rm V}\cdot {\rm e}^{-\pi\hspace{0.05cm}\cdot \hspace{0.05cm}(t/T)^2}.$$
 What needs to be investigated is the range in which the equivalent impulse duration&nbsp; $T$&nbsp; of this Gaussian pulse can vary such that the overall system can be entirely described by the frequency response
-:$$H_{\rm G}(f) = K \cdot {\rm e}^{-\pi\hspace{0.05cm}\cdot \hspace{0.05cm}(f/\Delta f_{\rm G})^2}.$$ Here, the subscript "G" in frequency response and bandwidth stands for "Gesamtsystem" (German for "overall system").
+:$$H_{\rm G}(f) = K \cdot {\rm e}^{-\pi\hspace{0.05cm}\cdot \hspace{0.05cm}(f/\Delta f_{\rm G})^2}.$$

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory}}
-[[File:EN_LZI_Z_1_7.png|right|frame|System with Gaussian low-passe filters and nonlinear characteristic curve]]
+[[File:EN_LZI_Z_1_7.png|right|frame|System with Gaussian low-passe filters and non-linear characteristic curve]]
-Ein Gesamtsystem&nbsp; $G$&nbsp; mit Eingang &nbsp;$w(t)$&nbsp; und Ausgang &nbsp;$z(t)$&nbsp; besteht aus drei Komponenten:
+An overall system&nbsp; $G$&nbsp; with input&nbsp;$w(t)$&nbsp; and output &nbsp;$z(t)$&nbsp; consists of three components:
-*Die erste Komponente ist ein Gaußtiefpass mit der Impulsantwort
+*The first component is a Gaussian low-pass filter with impulse response
 :$$h_1(t) = \frac{1}{\Delta t_1} \cdot {\rm e}^{-\pi\hspace{0.05cm}\cdot \hspace{0.05cm} (t/\Delta t_1)^2}, \hspace{0.5cm} \Delta
   t_1= {0.3\,\rm ms}.$$
-*Danach folgt eine Nichtlinearität mit der Kennlinie
+*This is then followed by a non-linearity with the characteristic curve
 :$$y(t) = \left\{ \begin{array}{c} +{8\,\rm V} \\ 2 \cdot x(t)  \\  {-8\,\rm V} \\  \end{array} \right.\quad \quad \begin{array}{*{10}c}   {\rm{f\ddot{u}r}}  \\ {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\  \end{array}\begin{array}{*{20}c} {x(t) \ge +{4\,\rm V}},  \\
 {{-4\,\rm V} < x(t) < +{4\,\rm V}},  \\ {x(t)\le {-4\,\rm V}}.  \\ \end{array}$$
-:&rArr; &nbsp; Das Eingangssignal&nbsp; $x(t)$&nbsp; der Nichtlinearität wird um den Faktor&nbsp; $2$&nbsp; verstärkt und – falls nötig – auf den Bereich&nbsp; $±8 \ \rm V$&nbsp; begrenzt.
+:&rArr; &nbsp; The input signal&nbsp; $x(t)$&nbsp; of the non-linearity is amplified by the factor&nbsp; $2$&nbsp; and – if necessary – limited to the range&nbsp; $±8 \ \rm V$&nbsp;.
-*Am Ende der Kette folgt wieder ein Gaußtiefpass, der durch seinen Frequenzgang gegeben ist:
+*At the end of the chain there is again a Gaussian low-pass filter given by its frequency response:
 :$$H_3(f) = {\rm e}^{-\pi\hspace{0.05cm}\cdot \hspace{0.05cm}(f/\Delta f_3)^2}, \hspace{0.5cm} \Delta f_3= {2.5\,\rm kHz}.$$
-Das Eingangssignal &nbsp;$w(t)$&nbsp; des Gesamtsystems sei ein Gaußimpuls mit Amplitude&nbsp; $5 \ \rm V$&nbsp; und variabler (äquivalenter) Dauer&nbsp; $T$:
+Let the input signal&nbsp;$w(t)$&nbsp; of the overall system be a Gaussian pulse with amplitude&nbsp; $5 \ \rm V$&nbsp; and variable (equivalent) duration&nbsp; $T$:
 :$$w(t) = {5\,\rm V}\cdot {\rm e}^{-\pi\hspace{0.05cm}\cdot \hspace{0.05cm}(t/T)^2}.$$
-Zu untersuchen ist, in welchem Bereich die äquivalente Impulsdauer&nbsp; $T$&nbsp; dieses Gaußimpulses variieren kann, damit das Gesamtsystem durch den Frequenzgang
+What needs to be investigated is the range in which the equivalent impulse duration&nbsp; $T$&nbsp; of this Gaussian pulse can vary such that the overall system can be entirely described by the frequency response
-:$$H_{\rm G}(f) = K \cdot {\rm e}^{-\pi\hspace{0.05cm}\cdot \hspace{0.05cm}(f/\Delta f_{\rm G})^2}$$
+:$$H_{\rm G}(f) = K \cdot {\rm e}^{-\pi\hspace{0.05cm}\cdot \hspace{0.05cm}(f/\Delta f_{\rm G})^2}$$. Here, the subscript "G" in frequency response and bandwidth stands for "Gesamtsystem" (German for "overall system").
@@ Line 37: / Line 36: @@
 <quiz display=simple>
-{Welche Bedingungen müssen erfüllt sein, damit das Gesamtsystem durch einen einzigen Frequenzgang beschreibbar ist?
+{What conditions must be satisfied for the overall system to be describable by a single frequency response?
 |type="[]"}
-+ Es besteht ein linearer Zusammenhang zwischen&nbsp; $w(t)$&nbsp; und&nbsp; $z(t)$.
++ There is a linear relationship between&nbsp; $w(t)$&nbsp; and&nbsp; $z(t)$.
-- $H_3(f)$&nbsp; muss schmalbandiger sein als&nbsp; $H_1(f)$.
+- $H_3(f)$&nbsp; must be more narrow-band than&nbsp; $H_1(f)$.
-+ Das Signal&nbsp; $x(t)$&nbsp; darf betragsmäßig nicht größer sein als&nbsp; $4 \ \rm V$.
++ The signal&nbsp; $x(t)$&nbsp; must not be greater in magnitude than&nbsp; $4 \ \rm V$.
-{Berechnen Sie den Maximalwert für die äquivalente Impulsdauer&nbsp; $T$, damit die unter&nbsp; '''(1)'''&nbsp; genannten Bedingungen erfüllbar sind.
+{Compute the maximum value for the equivalent impulse duration&nbsp; $T$ so that the conditions given in&nbsp; '''(1)'''&nbsp; are satisfiable.
 |type="{}"}
 $T_{\rm max} \ = \ $ { 0.4 3% } $\ \rm  ms$
-{Geben Sie die Parameter des Gesamtfrequenzgangs&nbsp; $H_{\rm G}(f)$&nbsp; an.
+{Specify the parameters of the overall frequency response&nbsp; $H_{\rm G}(f)$&nbsp;.
 |type="{}"}
 $K \ = \ $ { 2 3% }
@@ Line 58: / Line 57: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Richtig sind die <u>Antworten1 und 3</u>:
+'''(1)'''&nbsp; <u>Answers 1 and 3</u> are correct:
-*Die erste Aussage ist zutreffend: &nbsp; Nur für ein lineares System kann ein Frequenzgang angegeben werden.
+*The first statement is correct: &nbsp; Only for a linear system a frequency response can be specified.
-*Damit dies hier möglich ist, darf die Nichtlinearität keine Rolle spielen.
+*For this to be possible here nonlinearity must not play a role.
-*Das heißt, es muss sicher gestellt sein, dass&nbsp; $|x(t)|$&nbsp; nicht größer als&nbsp; $4 \ \rm V$&nbsp; ist.
+*That is, it must be ensured that&nbsp; $|x(t)|$&nbsp; is not greater than&nbsp; $4 \ \rm V$&nbsp;.
-*Dagegen ist die zweite Aussage nicht zutreffend: &nbsp;  Die Bandbreite von&nbsp; $H_3(f)$&nbsp; hat keinen Einfluss darauf, ob die Nichtlinearität elimimiert werden kann oder nicht.
+*In contrast to this, the second statement is not true: &nbsp; The bandwidth of&nbsp; $H_3(f)$&nbsp; does not affect whether the non-linearity can be eliminated or not.
-'''(2)'''&nbsp; Der erste Gaußtiefpass wird im Frequenzbereich wie folgt beschrieben:
+'''(2)'''&nbsp;  The first Gaussian low-pass filter is described in the frequency domain as follows:
 :$$X(f)  =  W(f) \cdot H_1(f) = {5\,\rm V}\cdot T \cdot {\rm e}^{-\pi(f \cdot T)^2} \cdot {\rm e}^{-\pi(f/\Delta f_1)^2} =  {5\,\rm V}\cdot T \cdot {\rm e}^{-\pi f^2 (T^2 + \Delta t_1^2)}= {5\,\rm V}\cdot T \cdot {\rm e}^{-\pi(f/\Delta f_x)^2}.$$
-*Hierbei bezeichnet&nbsp; $Δf_x$&nbsp; die äquivalente Bandbreite von&nbsp; $X(f)$.
+*Here, &nbsp; $Δf_x$&nbsp; denotes the equivalent bandwidth of&nbsp; $X(f)$.
-*Der Signalwert bei&nbsp; $t = 0$&nbsp; ist gleich der Spektralfläche  und gleichzeitig der Maximalwert des Signals:
+*The signal value at&nbsp; $t = 0$&nbsp; is equal to the spectral area and at the same time to the maximum value of the signal:
-*Dieser soll nicht größer werden als $4 \ \rm V$:
+*This should not exceed $4 \ \rm V$:
 :$$x_{\rm max} = x(t =0) = {5\,\rm V}\cdot T \cdot \Delta f_x \le {4\,\rm V}.$$
-*Daraus folgt durch Koeffizientenvergleich:
+*From this it follows by comparison of coefficients:
 :$$\frac{1}{T \cdot \Delta f_x} > \frac{5}{4}\hspace{0.1cm}  \Rightarrow  \hspace{0.1cm} \frac{1}{T^2 \cdot \Delta f_x^2} > \frac{25}{16}
 \Rightarrow \hspace{0.3cm}\frac{T^2 + \Delta t_1^2}{T^2} > \frac{25}{16}$$
@@ Line 79: / Line 78: @@
 \Delta t_1^2} \le \frac{16}{9}\hspace{0.3cm}\Rightarrow
 \hspace{0.3cm} T \le \frac{4}{3} \cdot \Delta t_1 \hspace{0.15cm}\underline{= {0.4\,\rm ms}}.$$
-*Die Kontrollrechnung ergibt:
+*The control calculation yields:
 :$$\Delta t_x = \sqrt{T^2 + \Delta t_1^2} = \sqrt{({0.4\,\rm ms})^2 + ({0.3\,\rm ms})^2} = {0.5\,\rm ms} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\Delta f_x = {1}/{\Delta t_x}= {2\,\rm kHz}$$
 :$$\Rightarrow \hspace{0.3cm} x(t=0) = {5\,\rm V}\cdot T \cdot \Delta f_x = {5\,\rm V}\cdot {0.4\,\rm ms} \cdot  {2\,\rm kHz} = {4\,\rm V}.$$
-'''(3)'''&nbsp; Die Gaußtiefpässe erfüllen die Bedingung&nbsp; $H_1(f = 0) = H_3(f = 0) = 1$.
+'''(3)'''&nbsp; The Gaussian low-pass filters satisfy the condition&nbsp; $H_1(f = 0) = H_3(f = 0) = 1$.
-*Unter Berücksichtigung der Verstärkung des zweiten Blocks im linearen Bereich erhält man somit für die Gesamtverstärkung:
+*Taking into account the gain of the second block in the linear domain the following is thus obtained for the total gain:
 :$$\underline{K \ = \ 2}.$$
-*Für die äquivalente Impulsdauer des Gesamtsystems gilt:
+*For the equivalent impulse duration of the overall system it holds that:
 :$$\Delta t_{\rm G} = \sqrt{\Delta t_1^2 + \frac{1}{\Delta f_3^2}} = \sqrt{({0.3\,\rm ms})^2 + \left( \frac{1}{{2.5\,\rm
   kHz}}\right)^2}={0.5\,\rm ms} \; \; \Rightarrow  \; \;  \Delta f_{\rm G} = {1}/{\Delta t_{\rm G}} \hspace{0.15cm}\underline{= {2\,\rm kHz}}.$$