Aufgaben:Exercise 3.6Z: Complex Exponential Function - Revision history

Javier: Text replacement - "bandpass" to "band-pass"

2021-05-03T15:12:49Z

Text replacement - "bandpass" to "band-pass"

Guenter at 14:22, 27 April 2021

2021-04-27T14:22:18Z

Guenter at 14:21, 27 April 2021

2021-04-27T14:21:17Z

Guenter at 14:18, 27 April 2021

2021-04-27T14:18:26Z

Guenter at 14:09, 27 April 2021

2021-04-27T14:09:13Z

Noah at 13:40, 27 March 2021

2021-03-27T13:40:12Z

Noah at 13:39, 27 March 2021

2021-03-27T13:39:51Z

Noah at 13:38, 27 March 2021

2021-03-27T13:38:08Z

Javier: Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

2021-03-23T12:39:18Z

Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

Noah at 18:09, 27 January 2021

2021-01-27T18:09:37Z

@@ Line 4: / Line 4: @@
 [[File:P_ID518__Sig_Z_3_6_neu.png|right|frame|Splitting the complex exponential function  in the spectral domain]]
-In connection with&nbsp; [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals|"bandpass systems"]]&nbsp;, one-sided spectra are often used.&nbsp; In the graphic you see such a one-sided spectral function&nbsp; ${X(f)}$, which results in a complex time signal&nbsp; ${x(t)}$.
+In connection with&nbsp; [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals|"band-pass systems"]]&nbsp;, one-sided spectra are often used.&nbsp; In the graphic you see such a one-sided spectral function&nbsp; ${X(f)}$, which results in a complex time signal&nbsp; ${x(t)}$.
 In the sketch below,&nbsp; ${X(f)}$&nbsp; is split into an even component&nbsp; ${G(f)}$&nbsp; &ndash; with respect to the frequency &ndash; and an odd component&nbsp; ${U(f)}$.

@@ Line 62: / Line 62: @@
 is obtained by applying the shifting theorem twice (in the frequency domain):
 :$$U( f ) = {A}/{2} \cdot \delta ( {f - f_0 } ) - {A}/{2} \cdot \delta ( {f + f_0 } )\ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ \ u( t ) = {A}/{2} \cdot \left( {{\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}2{\rm{\pi }}\hspace{0.05cm}\cdot \hspace{0.05cm}f_0\hspace{0.05cm}\cdot \hspace{0.05cm} t}  - {\rm{e}}^{{\rm{ - j}}\hspace{0.05cm}\cdot \hspace{0.05cm}2{\rm{\pi }}\hspace{0.05cm}\cdot \hspace{0.05cm}f_0 \hspace{0.05cm}\cdot \hspace{0.05cm}t} } \right).$$
-*According to&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation_by_Amplidute_and_Phase|Euler's theorem]]&nbsp;, this can also be written.
+*According to&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation_by_magnitude_and_phase|Euler's theorem]]&nbsp;, this can also be written.
 :$$u( t ) = {\rm{j}} \cdot A \cdot \sin ( {2{\rm{\pi }}f_0 t} ).$$
 :*The <u>real part of this signal is always zero</u>.

@@ Line 71: / Line 71: @@
 '''(3)'''&nbsp;  Because&nbsp; $X(f) = G(f)  + U(f)$&nbsp; also holds:
 :$$x(t) = g(t) + u(t) = A \cdot \cos ( {2{\rm{\pi }}f_0 t} ) + {\rm{j}} \cdot A \cdot \sin( {2{\rm{\pi }}f_0 t} ).$$
-This result can be summarised by&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation_by_Amplidute_and_Phase|Euler's theorem]]&nbsp; as follows:
+This result can be summarised by&nbsp; "Euler's theorem"&nbsp; as follows:
 :$$x(t) = A \cdot {\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}2{\rm{\pi }}\hspace{0.05cm}\cdot \hspace{0.05cm}f_0 \hspace{0.05cm}\cdot \hspace{0.05cm}t} .$$
 The given <u>alternatives 1 and 3</u> are correct:

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID518__Sig_Z_3_6_neu.png|right|frame|Complex exponential function and suitable splitting&nbsp; (representation in the spectral domain)]]
+[[File:P_ID518__Sig_Z_3_6_neu.png|right|frame|Splitting the complex exponential function  in the spectral domain]]
 In connection with&nbsp; [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals|"bandpass systems"]]&nbsp;, one-sided spectra are often used.&nbsp; In the graphic you see such a one-sided spectral function&nbsp; ${X(f)}$, which results in a complex time signal&nbsp; ${x(t)}$.
@@ Line 38: / Line 38: @@
 {Which of the statements are true regarding the signal&nbsp; $x(t)$&nbsp;?
 + The signal is&nbsp; $x(t) = A \cdot {\rm e}^{{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi\hspace{0.05cm}\cdot \hspace{0.05cm} f_0 \hspace{0.05cm}\cdot \hspace{0.05cm}t}$.
 - In the complex plane&nbsp; $x(t)$&nbsp; rotates clockwise.
@@ Line 59: / Line 60: @@
 '''(2)'''&nbsp; Starting from the Fourier correspondence
 :$$A \cdot {\rm \delta} ( f )\ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ \ A$$
-is obtained by applying the shift theorem twice (in the frequency domain):
+is obtained by applying the shifting theorem twice (in the frequency domain):
 :$$U( f ) = {A}/{2} \cdot \delta ( {f - f_0 } ) - {A}/{2} \cdot \delta ( {f + f_0 } )\ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ \ u( t ) = {A}/{2} \cdot \left( {{\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}2{\rm{\pi }}\hspace{0.05cm}\cdot \hspace{0.05cm}f_0\hspace{0.05cm}\cdot \hspace{0.05cm} t}  - {\rm{e}}^{{\rm{ - j}}\hspace{0.05cm}\cdot \hspace{0.05cm}2{\rm{\pi }}\hspace{0.05cm}\cdot \hspace{0.05cm}f_0 \hspace{0.05cm}\cdot \hspace{0.05cm}t} } \right).$$
 *According to&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation_by_Amplidute_and_Phase|Euler's theorem]]&nbsp;, this can also be written.

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID518__Sig_Z_3_6_neu.png|right|frame|Representation in the spectral domain: <br>complex exponential function and suitable splitting]]
+[[File:P_ID518__Sig_Z_3_6_neu.png|right|frame|Complex exponential function and suitable splitting&nbsp; (representation in the spectral domain)]]
-In connection with&nbsp; [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals|bandpass systemes]]&nbsp;, one-sided spectra are often used. In the illustration you can see such a one-sided spectral function&nbsp; ${X(f)}$, which results in a complex time signal&nbsp; ${x(t)}$&nbsp;.
+In connection with&nbsp; [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals|"bandpass systems"]]&nbsp;, one-sided spectra are often used.&nbsp; In the graphic you see such a one-sided spectral function&nbsp; ${X(f)}$, which results in a complex time signal&nbsp; ${x(t)}$.
-In the sketch below,&nbsp; ${X(f)}$&nbsp; is split into an even component&nbsp; ${G(f)}$&nbsp; - with respect to the frequency - and an odd component&nbsp; ${U(f)}$&nbsp;.
@@ Line 16: / Line 17: @@
 ''Hints:''
-*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Transform_Laws|Fourier Transform Laws]].
+*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Transform_Theorems|Fourier Transform Theorems]].
-*All of the laws presented there are illustrated with examples in the learning video&nbsp; [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)|Laws of the Fourier transform]].
+*For the first two sub-tasks use the signal parameters&nbsp; $A = 1\, \text{V}$&nbsp; and&nbsp; $f_0 = 125 \,\text{kHz}$.
-*Solve this task with the help of the&nbsp; [[Signal_Representation/Fourier_Transform_Laws#Mapping_Theorem|Mapping Theorem]]&nbsp; and the&nbsp; [[Signal_Representation/Fourier_Transform_Laws#Verschiebungssatz|Verschiebungssatzes]].
+*The&nbsp; [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem|Shifting Theorem]]&nbsp; and the&nbsp; [[Signal_Representation/Fourier_Transform_Theorems#Assignment_Theorem|Assignment Theorem]]&nbsp; – are illustrated with examples in the (German language) learning video<br> &nbsp; &nbsp; &nbsp;[[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)|Gesetzmäßigkeiten der Fouriertransformation]] &nbsp; &rArr; &nbsp;  "Regularities to the Fourier transform".
@@ Line 26: / Line 25: @@
 <quiz display=simple>
-{What is the time function&nbsp; $g(t)$&nbsp; that fits&nbsp; $G(f)$? How large is &nbsp; $g(t = 1 \, &micro; \text {s})$?
+{What is the time function&nbsp; $g(t)$&nbsp; that fits&nbsp; $G(f)$?&nbsp; How large is &nbsp; $g(t = 1 \, &micro; \text {s})$?
 |type="{}"}
 $\text{Re}\big[g(t = 1 \, &micro; \text {s})\big] \ =  \ $ { 0.707 3% } &nbsp;$\text{V}$
@@ Line 32: / Line 31: @@
-{What is the time function&nbsp; $u(t)$&nbsp; that fits&nbsp; $U(f)$? What is the value of&nbsp; $u(t = 1 \, &micro; \text {s})$?
+{What is the time function&nbsp; $u(t)$&nbsp; that fits&nbsp; $U(f)$?&nbsp; What is the value of&nbsp; $u(t = 1 \, &micro; \text {s})$?
 |type="{}"}
 $\text{Re}\big[u(t = 1 \, &micro; \text {s})\big]\ = \ $ { 0. } &nbsp;$\text{V}$

← Older revision		Revision as of 12:39, 23 March 2021
Line 80:		Line 80:

	__NOEDITSECTION__		__NOEDITSECTION__
−	[[Category:~~Exercises for~~ Signal Representation\|^3.3 Fourier Transform Theorems^]]	+	[[Category:Signal Representation: Exercises\|^3.3 Fourier Transform Theorems^]]

@@ Line 4: / Line 4: @@
 [[File:P_ID518__Sig_Z_3_6_neu.png|right|frame|Darstellung im Spektralbereich: <br>komplexe Exponentialfunktion und geeignete Aufspaltung]]
-In Zusammenhang mit den&nbsp; [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals|Bandpass-Systemen]]&nbsp; wird oft mit einseitigen Spektren gearbeitet. In der Abbildung sehen Sie eine solche einseitige Spektralfunktion&nbsp; ${X(f)}$, die ein komplexes Zeitsignal&nbsp; ${x(t)}$&nbsp; zur Folge hat.
+In connection with&nbsp; [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals|bandpass systemes]]&nbsp;, one-sided spectra are often used. In the illustration you can see such a one-sided spectral function&nbsp; ${X(f)}$, which results in a complex time signal&nbsp; ${x(t)}$&nbsp;.
@@ Line 15: / Line 15: @@
+''Hints:''
-''Hinweise:''
+*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Transform_Laws|Fourier Transform Laws]].
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Signal_Representation/Fourier_Transform_Laws|Gesetzmäßigkeiten der Fouriertransformation]].
+*All of the laws presented there are illustrated with examples in the learning video&nbsp; [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)|Gesetzmäßigkeiten der Fouriertransformation]].
-*Alle dort dargelegten Gesetzmäßigkeiten werden im Lernvideo&nbsp; [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)|Gesetzmäßigkeiten der Fouriertransformation]]  an Beispielen verdeutlicht.
+*Solve this task with the help of the&nbsp; [[Signal_Representation/Fourier_Transform_Laws#Mapping_Theorem|Mapping Theorem]]&nbsp; and the&nbsp; [[Signal_Representation/Fourier_Transform_Laws#Verschiebungssatz|Verschiebungssatzes]].
-*Lösen Sie diese Aufgabe mit Hilfe des&nbsp; [[Signal_Representation/Fourier_Transform_Laws#Zuordnungssatz|Zuordnungssatzes]]&nbsp; und des&nbsp; [[Signal_Representation/Fourier_Transform_Laws#Verschiebungssatz|Verschiebungssatzes]].
+*Use the signal parameters&nbsp; $A = 1\, \text{V}$&nbsp; and&nbsp; $f_0 = 125 \,\text{kHz}$ for the first two sub-tasks.
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Wie lautet die zu&nbsp; $G(f)$&nbsp; passende Zeitfunktion&nbsp; $g(t)$? Wie groß ist&nbsp; $g(t = 1 \, &micro; \text {s})$?
+{What is the time function&nbsp; $g(t)$&nbsp; that fits&nbsp; $G(f)$? How large is &nbsp; $g(t = 1 \, &micro; \text {s})$?
 |type="{}"}
 $\text{Re}\big[g(t = 1 \, &micro; \text {s})\big] \ =  \ $ { 0.707 3% } &nbsp;$\text{V}$
@@ Line 33: / Line 32: @@
-{Wie lautet die zu&nbsp; $U(f)$&nbsp; passende Zeitfunktion&nbsp; $u(t)$? Wie groß ist&nbsp; $u(t = 1 \, &micro; \text {s})$?
+{What is the time function&nbsp; $u(t)$&nbsp; that fits&nbsp; $U(f)$? What is the value of&nbsp; $u(t = 1 \, &micro; \text {s})$?
 |type="{}"}
 $\text{Re}\big[u(t = 1 \, &micro; \text {s})\big]\ = \ $ { 0. } &nbsp;$\text{V}$
@@ Line 39: / Line 38: @@
-{Welche der Aussagen sind bezüglich des Signals&nbsp; $x(t)$&nbsp; zutreffend?
+{Which of the statements are true regarding the signal&nbsp; $x(t)$&nbsp;?
-|type="[]"}
++ The signal is&nbsp; $x(t) = A \cdot {\rm e}^{{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi\hspace{0.05cm}\cdot \hspace{0.05cm} f_0 \hspace{0.05cm}\cdot \hspace{0.05cm}t}$.
-+ Das Signal lautet&nbsp; $x(t) = A \cdot {\rm e}^{{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi\hspace{0.05cm}\cdot \hspace{0.05cm} f_0 \hspace{0.05cm}\cdot \hspace{0.05cm}t}$.
+- In the complex plane&nbsp; $x(t)$&nbsp; rotates clockwise.
-- In der komplexen Ebene dreht&nbsp; $x(t)$&nbsp; im Uhrzeigersinn.
++ In the complex plane&nbsp; $x(t)$&nbsp; rotates counterclockwise.
-+ In der komplexen Ebene dreht&nbsp; $x(t)$&nbsp; entgegen dem Uhrzeigersinn.
+- One microsecond is needed for one rotation.
@@ Line 50: / Line 48: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  $G(f)$&nbsp; ist die Spektralfunktion eines Cosinussignals mit der Periodendauer&nbsp; $T_0 = 1/f_0 = 8 \, &micro;\text {s}$:
+'''(1)'''&nbsp;  $G(f)$&nbsp; is the spectral function of a cosine signal with period&nbsp; $T_0 = 1/f_0 = 8 \, &micro;\text {s}$:
 :$$g( t ) = A \cdot \cos ( {2{\rm{\pi }}f_0 t} ).$$
-Bei&nbsp; $t = 1 \, &micro;\text {s}$&nbsp; ist der Signalwert gleich&nbsp; $A \cdot \cos(\pi /4)$:
+At&nbsp; $t = 1 \, &micro;\text {s}$&nbsp; the signal value is equal to&nbsp; $A \cdot \cos(\pi /4)$:
-*Der Realteil ist&nbsp; $\text{Re}[g(t = 1 \, &micro; \text {s})] = \;\underline{0.707\, \text{V}}$,
+*The real part is&nbsp; $\text{Re}[g(t = 1 \, &micro; \text {s})] = \;\underline{0.707\, \text{V}}$,
-*der Imaginärteil ist&nbsp; $\text{Im}[g(t = 1 \, &micro; \text {s})] = \;\underline{0.}$
+*The imaginary part is&nbsp; $\text{Im}[g(t = 1 \, &micro; \text {s})] = \;\underline{0.}$
-'''(2)'''&nbsp; Ausgehend von der Fourierkorrespondenz
+'''(2)'''&nbsp; Starting from the Fourier correspondence
 :$$A \cdot {\rm \delta} ( f )\ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ \ A$$
-erhält man durch zweimalige Anwendung des Verschiebungssatzes (im Frequenzbereich):
+is obtained by applying the shift theorem twice (in the frequency domain):
 :$$U( f ) = {A}/{2} \cdot \delta ( {f - f_0 } ) - {A}/{2} \cdot \delta ( {f + f_0 } )\ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ \ u( t ) = {A}/{2} \cdot \left( {{\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}2{\rm{\pi }}\hspace{0.05cm}\cdot \hspace{0.05cm}f_0\hspace{0.05cm}\cdot \hspace{0.05cm} t}  - {\rm{e}}^{{\rm{ - j}}\hspace{0.05cm}\cdot \hspace{0.05cm}2{\rm{\pi }}\hspace{0.05cm}\cdot \hspace{0.05cm}f_0 \hspace{0.05cm}\cdot \hspace{0.05cm}t} } \right).$$
-*Nach dem&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Darstellung_nach_Betrag_und_Phase|Satz von Euler]]&nbsp; kann hierfür auch geschrieben werden:
+*According to&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation_by_Amplidute_and_Phase|Euler's theorem]]&nbsp;, this can also be written.
 :$$u( t ) = {\rm{j}} \cdot A \cdot \sin ( {2{\rm{\pi }}f_0 t} ).$$
-:*Der <u>Realteil dieses Signals ist stets Null</u>.
+:*The <u>real part of this signal is always zero.</u>.
-:*Bei&nbsp; $t = 1 \, &micro;\text {s}$&nbsp; gilt für den Imaginärteil:&nbsp; $\text{Im}[g(t = 1 \, &micro; \text {s})] = \;\underline{0.707\, \text{V}}$.
+:*At&nbsp; $t = 1 \, &micro;\text {s}$&nbsp; the following applies to the imaginary part:&nbsp; $\text{Im}[g(t = 1 \, &micro; \text {s})] = \;\underline{0.707\, \text{V}}$.
-'''(3)'''&nbsp;  Wegen&nbsp; $X(f) = G(f)  + U(f)$&nbsp; gilt auch:
+'''(3)'''&nbsp;  Because&nbsp; $X(f) = G(f)  + U(f)$&nbsp; also holds:
 :$$x(t) = g(t) + u(t) = A \cdot \cos ( {2{\rm{\pi }}f_0 t} ) + {\rm{j}} \cdot A \cdot \sin( {2{\rm{\pi }}f_0 t} ).$$
-Dieses Ergebnis kann mit dem&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Darstellung_nach_Betrag_und_Phase|Satz von Euler]]&nbsp;  wie folgt zusammengefasst werden:
+This result can be summarised by&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation_by_Amplidute_and_Phase|Euler's theorem]]&nbsp; as follows:
 :$$x(t) = A \cdot {\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}2{\rm{\pi }}\hspace{0.05cm}\cdot \hspace{0.05cm}f_0 \hspace{0.05cm}\cdot \hspace{0.05cm}t} .$$
-Richtig sind die vorgegebenen <u>Alternativen 1 und 3</u>:
+The given <u>alternatives 1 and 3</u> are correct:
-*Das Signal dreht in der komplexen Ebene in mathematisch positiver Richtung, also entgegen dem Uhrzeigersinn.
+*The signal rotates in the complex plane in a mathematically positive direction, i.e. counterclockwise.
-*Für eine Umdrehung benötigt der „Zeiger” die Periodendauer&nbsp; $T_0 = 1/f_0 = 8 \, &micro;\text {s}$.
+*For one rotation, the "pointer" needs the period&nbsp; $T_0 = 1/f_0 = 8 \, &micro;\text {s}$.
 {{ML-Fuß}}