Aufgaben:Exercise 2.5: Half-Wave Rectification - Revision history

Guenter at 13:14, 15 April 2021

2021-04-15T13:14:26Z

Guenter at 13:09, 15 April 2021

2021-04-15T13:09:38Z

Guenter at 13:06, 15 April 2021

2021-04-15T13:06:47Z

Guenter at 13:01, 15 April 2021

2021-04-15T13:01:39Z

Noah at 09:25, 26 March 2021

2021-03-26T09:25:35Z

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

2021-03-23T12:39:29Z

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

Noah at 18:51, 16 January 2021

2021-01-16T18:51:34Z

Noah at 18:41, 16 January 2021

2021-01-16T18:41:36Z

Noah at 18:38, 16 January 2021

2021-01-16T18:38:21Z

Oezdemir: Oezdemir moved page Aufgaben:Aufgabe 2.5: Half-Wave Rectification to Aufgaben:Exercise 2.5: Half-Wave Rectification

2020-11-02T14:16:20Z

Oezdemir moved page Aufgabe 2.5: Half-Wave Rectification to Exercise 2.5: Half-Wave Rectification

@@ Line 66: / Line 66: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; The shifted signal&nbsp; $v(t)$&nbsp; is also even and all sine coefficients are accordingly zero.
-*Nothing changes in the DC signal coefficient either: &nbsp; $A_0 = 1$.
+*Nothing changes in the DC signal coefficient: &nbsp; $A_0 = 1$.
 *From the signal curves it can be seen that&nbsp; $v(t) = u(t - T/4)$&nbsp; applies:&nbsp;
@@ Line 90: / Line 90: @@
-'''(2)'''&nbsp; BecaUSE OF&nbsp; $w(t) = \pi /2 \cdot \sin(\omega_0 t)$&nbsp;  all Fourier coefficients except&nbsp; $B_1 = \pi /2 \hspace{0.1cm}\underline{=1.571}$&nbsp; are zero.
+'''(2)'''&nbsp; Because of&nbsp; $w(t) = \pi /2 \cdot \sin(\omega_0 t)$&nbsp;  all Fourier coefficients except&nbsp; $B_1 = \pi /2 \hspace{0.1cm}\underline{=1.571}$&nbsp; are zero.

@@ Line 47: / Line 47: @@
 <quiz display=simple>
-{Calculate the Fourier coefficients of the signal&nbsp; $v(t)$. What is the value of the coefficient&nbsp; $A_2$?
+{Calculate the Fourier coefficients of the signal&nbsp; $v(t)$.&nbsp; What is the coefficient&nbsp; $A_2$?
 |type="{}"}
 $v(t)$: &nbsp; $A_2\ = \ $  { -0.67--0.66 }
-{Calculate the Fourier coefficients of the signal&nbsp; $w(t)$. What is the value of the coefficient&nbsp; $B_1$?
+{Calculate the Fourier coefficients of the signal&nbsp; $w(t)$.&nbsp; What is the coefficient&nbsp; $B_1$?
 |type="{}"}
 $w(t)$: &nbsp;$B_1\ = \ $ { 1.571 3% }
-{How can&nbsp; $x(t)$&nbsp; be composed of&nbsp; $v(t)$&nbsp; and&nbsp; $w(t)$&nbsp;? Give the corresponding Fourier coefficients of the signal&nbsp; $x(t)$&nbsp; in particular
+{How can&nbsp; $x(t)$&nbsp; be composed of&nbsp; $v(t)$&nbsp; and&nbsp; $w(t)$?&nbsp; Give the corresponding Fourier coefficients of the signal&nbsp; $x(t)$.&nbsp; In particular:
 |type="{}"}
 $x(t)$: &nbsp;$A_0\ = \ $ { 0.5 3% }

@@ Line 11: / Line 11: @@
 It should be noted:
-*The fundamental angular frequency is denoted by&nbsp; $\omega_1$&nbsp;. But since the period of the signals&nbsp; $u(t)$&nbsp; and&nbsp; $v(t)$&nbsp; is&nbsp; $T/2$&nbsp;, &nbsp; $\omega_1 = 2\pi /(T/2) = 4 \pi /T$.
+*The basic circular frequency is denoted by&nbsp; $\omega_1$&nbsp;.&nbsp; But since the period of the signals&nbsp; $u(t)$&nbsp; and&nbsp; $v(t)$&nbsp; is&nbsp; $T/2$, &nbsp; $\omega_1 = 2\pi /(T/2) = 4 \pi /T$.
-*Because in this task the signals&nbsp; $u(t)$,&nbsp; $w(t)$&nbsp; and&nbsp; $x(t)$&nbsp; are to be related to each other, the signal&nbsp; $u(t)$&nbsp; must also be represented with the period duration&nbsp; $T$&nbsp; of the signal&nbsp; $x(t)$&nbsp;.
+*Because in this task the signals&nbsp; $u(t)$,&nbsp; $w(t)$&nbsp; and&nbsp; $x(t)$&nbsp; are to be related to each other, the signal&nbsp; $u(t)$&nbsp; must also be represented with the period duration&nbsp; $T$&nbsp; of the signal&nbsp; $x(t)$.&nbsp;
 *With&nbsp; $\omega_0 = 2\pi /T = \omega_1/2$&nbsp; the same applies:
@@ Line 37: / Line 37: @@
 ''Hints:''
-*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Series|Fourierreihe]].
+*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Series|Fourier Series]].
 *You can find a compact summary of the topic in the two learning videos
-::[[Zur_Berechnung_der_Fourierkoeffizienten_(Lernvideo)|Zur Berechnung der Fourierkoeffizienten]],
+::[[Zur_Berechnung_der_Fourierkoeffizienten_(Lernvideo)|Zur Berechnung der Fourierkoeffizienten]]&nbsp; &rArr; &nbsp; "To calculate the Fourier coefficients",
-::[[Eigenschaften_der_Fourierreihendarstellung_(Lernvideo)|Eigenschaften der Fourierreihendarstellung]].
+:: [[Eigenschaften_der_Fourierreihendarstellung_(Lernvideo)|Eigenschaften der Fourierreihendarstellung]] &nbsp; &rArr; &nbsp;   "Properties of the Fourier series representation".

@@ Line 5: / Line 5: @@
 [[File:P_ID311__Sig_A_2_5.png|right|frame|Rectified cosine functions]]
-We are looking for the Fourier coefficients of the signal&nbsp; $x(t)$, sketched below, which results from the one-way rectification of the sinusoidal signal&nbsp; $w(t)$&nbsp; with amplitude&nbsp; $\pi /2$&nbsp;.
+We are looking for the Fourier coefficients of the signal&nbsp; $x(t)$, sketched below, which results from the one-way rectification of the sinusoidal signal&nbsp; $w(t)$&nbsp; with amplitude&nbsp; $\pi /2$.
-The Fourier series representation of the signal&nbsp; $u(t)$ sketched above is assumed to be known. This was already determined in&nbsp; [[Aufgaben:Aufgabe_2.4:_Gleichgerichteter_Cosinus|Aufgabe 2.4]]&nbsp;. Taking into account the amplitude&nbsp; $\pi /2$&nbsp; the following applies:
+The Fourier series representation of the signal&nbsp; $u(t)$ sketched above is assumed to be known.&nbsp; This was already determined in&nbsp; [[Aufgaben:Aufgabe_2.4:_Gleichgerichteter_Cosinus|Exercise 2.4]].&nbsp; Taking into account the amplitude&nbsp; $\pi /2$&nbsp; the following applies:
 :$$u(t)=1+\frac{2}{3} \cdot \cos(\omega_1t)-\frac{2}{15}\cdot \cos(2\omega_1t)+\frac{2}{35}\cdot \cos(3\omega_1t)-\dots$$

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID311__Sig_A_2_5.png|right|frame|Gleichgerichtete Cosinusfunktionen]]
+[[File:P_ID311__Sig_A_2_5.png|right|frame|Rectified cosine functions]]
 We are looking for the Fourier coefficients of the signal&nbsp; $x(t)$, sketched below, which results from the one-way rectification of the sinusoidal signal&nbsp; $w(t)$&nbsp; with amplitude&nbsp; $\pi /2$&nbsp;.

← Older revision		Revision as of 12:39, 23 March 2021
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	__NOEDITSECTION__		__NOEDITSECTION__
−	[[Category:~~Exercises for~~ Signal Representation\|^2.4 Fourier Series^]]	+	[[Category:Signal Representation: Exercises\|^2.4 Fourier Series^]]

@@ Line 65: / Line 65: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Auch das verschobene Signal&nbsp; $v(t)$&nbsp; ist gerade und alle Sinuskoeffizienten sind dementsprechend Null.
+'''(1)'''&nbsp; The shifted signal&nbsp; $v(t)$&nbsp; is also even and all sine coefficients are accordingly zero.
-*Am Gleichsignalkoeffizienten ändert sich ebenfalls nichts:&nbsp; $A_0 = 1$.
+*Nothing changes in the DC signal coefficient either: &nbsp; $A_0 = 1$.
-*Aus den Signalverläufen ist zu erkennen, dass&nbsp; $v(t) = u(t - T/4)$&nbsp; gilt:&nbsp;
+*From the signal curves it can be seen that&nbsp; $v(t) = u(t - T/4)$&nbsp; applies:&nbsp;
 :$$v(t)=1+\frac{2}{3}\cdot \cos(2\omega_0(t-\frac{T}{4}))-\frac{2}{15}\cdot \cos(4\omega_0(t-\frac{T}{4}))+\frac{2}{35}\cdot \cos(6\omega_0(t-\frac{T}{4}))-\dots$$
-*Die Cosinusterme können nun mit&nbsp; $\omega_0 \cdot T = 2 \pi$&nbsp; umgeformt werden:&nbsp;
+*The cosine terms can now be transformed with&nbsp; $\omega_0 \cdot T = 2 \pi$&nbsp; :&nbsp;
 :$$\cos(2\omega_0(t-\frac{T}{4}))=\cos(2\omega_0t-\pi)=-\cos(2\omega_0t),$$
@@ Line 80: / Line 80: @@
 :$$\cos(6\omega_0(t-\frac{T}{4}))=\cos(6\omega_0t-3\pi)=-\cos(6\omega_0t).$$
-*Damit erhält man für die Fourierreihe:
+*This gives us for the Fourier series:
 :$$v(t)=1-{2}/{3}\cdot \cos(2\omega_0t)-{2}/{15}\cdot \cos(4\omega_0t)-{2}/{35}\cdot \cos(6\omega_0t)-\dots$$
-:bzw. für die Cosinuskoeffizienten mit geradzahligem&nbsp; $n$:
+:or for the cosine coefficients with even-numbered&nbsp; $n$:
 :$$A_n=\frac{-2}{n^2-1}\hspace{0.5cm}\Rightarrow\hspace{0.5cm}A_2=-\hspace{-0.05cm}2/3 \hspace{0.1cm}\underline{= -\hspace{-0.05cm}0.667}.$$
@@ Line 90: / Line 90: @@
-'''(2)'''&nbsp; Wegen&nbsp; $w(t) = \pi /2 \cdot \sin(\omega_0 t)$&nbsp; sind alle Fourierkoeffizienten außer&nbsp; $B_1 = \pi /2 \hspace{0.1cm}\underline{=1.571}$&nbsp; gleich Null.
+'''(2)'''&nbsp; BecaUSE OF&nbsp; $w(t) = \pi /2 \cdot \sin(\omega_0 t)$&nbsp;  all Fourier coefficients except&nbsp; $B_1 = \pi /2 \hspace{0.1cm}\underline{=1.571}$&nbsp; are zero.
-'''(3)'''&nbsp; Aus der grafischen Darstellung erkennt man den Zusammenhang&nbsp; $x(t)={1}/{2} \cdot \big [v(t)+w(t) \big].$ Das bedeutet:
+'''(3)'''&nbsp; From the graphical representation one can see the relationship&nbsp; $x(t)={1}/{2} \cdot \big [v(t)+w(t) \big].$ This means:
 :$$x(t)=\frac{1}{2}+\frac{\pi}{4}\cdot \sin(\omega_0 t)-\frac{1}{3}\cdot \cos(2\omega_0 t)-\frac{1}{15}\cdot \cos(4\omega_0 t)-\frac{1}{35}\cdot \cos(6\omega_0 t)-\ldots$$
-*Die gesuchten Fourierkoeffizienten sind somit:
+*The Fourier coefficients sought are thus:
 :$$A_0 \hspace{0.1cm}\underline{=0.5},\hspace{1cm}
 B_1 = \pi /4 \hspace{0.1cm}\underline{= 0.785},\hspace{1cm}

@@ Line 5: / Line 5: @@
 [[File:P_ID311__Sig_A_2_5.png|right|frame|Gleichgerichtete Cosinusfunktionen]]
-Gesucht sind die Fourierkoeffizienten des unten skizzierten Signals&nbsp; $x(t)$, das sich durch die Einweggleichrichtung des Sinussignals&nbsp; $w(t)$&nbsp; mit der Amplitude&nbsp; $\pi /2$&nbsp; ergibt.
+We are looking for the Fourier coefficients of the signal&nbsp; $x(t)$, sketched below, which results from the one-way rectification of the sinusoidal signal&nbsp; $w(t)$&nbsp; with amplitude&nbsp; $\pi /2$&nbsp;.
-Als bekannt vorausgesetzt wird die Fourierreihendarstellung des oben skizzierten Signals&nbsp; $u(t)$. Diese wurde bereits in der&nbsp; [[Aufgaben:Aufgabe_2.4:_Gleichgerichteter_Cosinus|Aufgabe 2.4]]&nbsp; ermittelt. Unter Berücksichtigung der Amplitude&nbsp; $\pi /2$&nbsp; gilt hierfür:
+The Fourier series representation of the signal&nbsp; $u(t)$ sketched above is assumed to be known. This was already determined in&nbsp; [[Aufgaben:Aufgabe_2.4:_Gleichgerichteter_Cosinus|Aufgabe 2.4]]&nbsp;. Taking into account the amplitude&nbsp; $\pi /2$&nbsp; the following applies:
 :$$u(t)=1+\frac{2}{3} \cdot \cos(\omega_1t)-\frac{2}{15}\cdot \cos(2\omega_1t)+\frac{2}{35}\cdot \cos(3\omega_1t)-\dots$$
-Anzumerken ist:
+It should be noted:
-*Die Grundkreisfrequenz ist mit&nbsp; $\omega_1$&nbsp; bezeichnet. Da aber die Periodendauer der Signale&nbsp; $u(t)$&nbsp; und&nbsp; $v(t)$&nbsp; jeweils&nbsp; $T/2$&nbsp; beträgt, gilt&nbsp; $\omega_1 = 2\pi /(T/2) = 4 \pi /T$.
+*The fundamental angular frequency is denoted by&nbsp; $\omega_1$&nbsp;. But since the period of the signals&nbsp; $u(t)$&nbsp; and&nbsp; $v(t)$&nbsp; is&nbsp; $T/2$&nbsp;, &nbsp; $\omega_1 = 2\pi /(T/2) = 4 \pi /T$.
-*Weil in dieser Aufgabe die Signale&nbsp; $u(t)$,&nbsp; $w(t)$&nbsp; und&nbsp; $x(t)$&nbsp; zueinander in Bezug gebracht werden sollen, muss auch das Signal&nbsp; $u(t)$&nbsp; mit der Periodendauer&nbsp; $T$&nbsp; des Signals&nbsp; $x(t)$&nbsp; dargestellt werden.
+*Because in this task the signals&nbsp; $u(t)$,&nbsp; $w(t)$&nbsp; and&nbsp; $x(t)$&nbsp; are to be related to each other, the signal&nbsp; $u(t)$&nbsp; must also be represented with the period duration&nbsp; $T$&nbsp; of the signal&nbsp; $x(t)$&nbsp;.
-*Mit&nbsp; $\omega_0 = 2\pi /T = \omega_1/2$&nbsp; gilt somit gleichermaßen:
+*With&nbsp; $\omega_0 = 2\pi /T = \omega_1/2$&nbsp; the same applies:
 :$$u(t)=1+\frac{2}{3} \cdot \cos(2\omega_0t)-\frac{2}{15} \cdot \cos(4\omega_0t)+\frac{2}{35} \cdot \cos(6\omega_0t)-\dots$$
-Für die Fourierkoeffizienten bedeutet dies:
+For the Fourier coefficients this means:
-*Der Gleichkoeffizient ergibt sich zu&nbsp; $A_0 = 1$,
+*The DC coefficient results in&nbsp; $A_0 = 1$,
-*Alle Sinuskoeffizienten sind&nbsp; $B_n = 0$,
+*All sine coefficients are&nbsp; $B_n = 0$,
-*Die Cosinuskoeffizienten mit ungeradzahligem&nbsp; $n = 1, \ 3, \ 5, \dots$ sind alle&nbsp; $0$,
+*The cosine coefficients with odd&nbsp; $n = 1, \ 3, \ 5, \dots$ are all&nbsp; $0$,
-*Die Cosinuskoeffizienten mit geradzahligem&nbsp; $n = 2, \ 4, \ 6, \dots$ sind ungleich&nbsp; $0$&nbsp;:
+*The cosine coefficients with even&nbsp; $n = 2, \ 4, \ 6, \dots$ are not equal to&nbsp; $0$&nbsp;:
 :$$A_n=(-1)^{\hspace{0.01cm}n/2+1}\frac{2}{n^2-1}.$$
-Daraus ergeben sich folgende Zahlenwerte:
+This results in the following numerical values:
 :$$A_1=A_3=A_5=\dots=0,$$
@@ Line 36: / Line 36: @@
-''Hinweise:''
+''Hints:''
-*Die Aufgabe gehört zum Kapitel&nbsp; [[Signal_Representation/Fourier_Series|Fourierreihe]].
+*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Series|Fourierreihe]].
-*Eine kompakte Zusammenfassung der Thematik finden Sie in den beiden Lernvideos
+*You can find a compact summary of the topic in the two learning videos
 ::[[Zur_Berechnung_der_Fourierkoeffizienten_(Lernvideo)|Zur Berechnung der Fourierkoeffizienten]],
 ::[[Eigenschaften_der_Fourierreihendarstellung_(Lernvideo)|Eigenschaften der Fourierreihendarstellung]].
@@ Line 44: / Line 44: @@
-===Fragebogen===
+===Questions===
 <quiz display=simple>

← Older revision	Revision as of 14:16, 2 November 2020
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